Samacheer Kalvi 12th Bio Zoology Solutions Chapter 1 Reproduction in Organisms

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Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 1 Reproduction in Organisms

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Samacheer Kalvi 12th Bio Zoology Reproduction in Organisms Text Book Back Questions and Answers

Question 1.
In which type of parthenogenesis are only males produced?
(a) Arrhenotoky
(b) Thelytoky
(c) Amphitoky
(d) Both a and b
Answer:
(a) Arrhenotoky

Question 2.
Animals giving birth to young ones:
(a) Oviparous
(b) Ovoviviparous
(c) Viviparous
(d) Both a and b
Answer:
(c) Viviparous

Question 3.
The mode of reproduction in bacteria is by ___________
(a) Formation of gametes
(b) Endospore formation
(c) Conjugation
(d) Zoospore formation
Answer:
(c) Conjugation

Question 4.
In which mode of reproduction variations are seen?
(a) Asexual
(b) Parthenogenesis
(c) Sexual
(d) Both a and b
Answer:
(c) Sexual

Question 5.
Assertion and reasoning questions:
In each of the following questions there are two statements. One is assertion (A) and other is reasoning (R). Mark the correct answer as
(a) If both A and R are true and R is correct explanation for A
(b) If both A and R are true but R is not the correct explanation for A
(c) If A is true but R is false
(d) If both A and R are false
I. Assertion: In bee society, all the members are diploid except drones.
Reason: Drones are produced by parthenogenesis.
A B C D
II. Assertion: Off springs produced by asexual reproduction are genetically identical to the parent.
Reason: Asexual reproduction involves only mitosis and no meiosis.
A B C D
III. Assertion: Viviparous animals give better protection to their off springs.
Reason: They lay their eggs in the safe places of the environment.
A B C D
Answer:
(I) A (II) A (III) C

Question 6.
Name an organism where cell division is itself a mode of reproduction.
Answer:
Amoeba

Question 7.
Name the phenomenon where the female gamete directly develops into a new organism with an avian example.
Answer:
Parthenogenesis is the phenomenon where the unfertilized female gamete (egg) develops into a new individual,
e.g. Turkey

Question 8.
What is parthenogenesis? Give two examples from animals.
Answer:
Development of an egg into a complete individual without fertilization is known as parthenogenesis. It was first discovered by Charles Bonnet in 1745.
E.g. Honey bees, Aphis.

Question 9.
Which type of reproduction is effective – Asexual or sexual and why?
Answer:
Sexual reproduction is highly effective than asexual reproduction since the off springs produced are genetically different from parents causing variations. Variation leads to evolution.

Question 10.
The unicellular organisms which reproduce by binary fission are considered immortal. Justify.
Answer:
In unicellular organisms during binary fission, the entire cell (organism) divides completely to form two daughter cells which later on develop into adult and the process goes on repeatedly during each division leading to immortality of cell (organism). Hence unicellular organisms like amoeba are ‘biologically immortal’.

Question 11.
Why is the offspring formed by asexual reproduction referred as a clone?
Answer:
Off springs developed by asexual reproduction are referred as clones since they are genetically & morphologically similar to this parent.

Question 12.
Why are the off springs of oviparous animal at a greater risk as compared to off springs of viviparous organisms?
Answer:
Oviparous animals are egg-layers. The eggs containing embryo are laid out of their body and are highly susceptible to environmental factors (temperature, moisture etc.) and predators. Whereas, in viviparous animals, the embryo develops inside the body of female and comes out as young ones. Hence off springs of oviparous animals are at risk compared to viviparous animal.

Question 13.
Give reasons for the following:

  1. Some organisms like honey bees are called parthenogenetic animals
  2. A male honey bee has 16 chromosomes whereas its female has 32 chromosomes

Answer:

  1. Among honey bees, the queen bee and worker bees develop from fertilized eggs whereas the drones develop from unfertilized eggs. Hence the honey bees are parthenogenetic animals showing incomplete parthenogenesis.
  2. Female honey bees (Queen or worker bees) are diploid having 32 chromosomes since they develop from the fertilized egg. On the other hand, the male honey bees (drones) develop from unfertilized egg possessing only 16 chromosomes (i.e., Haploid)

Question 14.
Differentiate between the following:
(a) Binary fission in amoeba and multiple fission in Plasmodium
(b) Budding in yeast and budding in Hydra
(c) Regeneration in lizard and Planaria
Answer:
(a) Binary fission in amoeba and multiple fission in Plasmodium

  1. Binary fission in amoeba: In binary fission of amoeba, the plane of division is hard to observe. The nucleoli disintegrates. The nucleus divides mitotically forming two nucleus. The cell constricts in middle, so the cytoplasm divides forming two daughter cells.
  2. Multiple fission: In Multiple fission of plasmodium, the oocyte or schizont divides into many similar daughter cells simultaneously. Nucleus undergoes repeated mitosis producing many nuclei without the division of cytoplasm. Later the cytoplasm divides & encircles each nucleus forming many daughter cells oocyte undergoes sporogony forming sporozoites. Schizont undergoes schizogony forming merozoites.

(b) Budding in yeast and budding in Hydra:

  1. Budding in Yeast: Yeast is xxxx cellular organism in which the bud develops as a small protuberance following the nuclear division and finally detached to new individual.
  2. Budding in Hydra: Hydra is a multicellular organism where the bud xxxxxx from the parents body, grows gradually and finally gets detached.

(c) Regeneration in lizard and Planaria:

  1. Regeneration of Lizard: If the tail of the lizard is cut and removed, a new tail will regenerate in damaged part. In lizard only the new tail is regenerated.
  2. Regeneration of Planaria: If a planarian worm get cut then each half regenerates the lost part resulting in two worms.
    In planaria, the cut removed part developed into an entire worm.

Question 15.
How is juvenile phase different from reproductive phase?
Answer:

  1. Juvenile phase: Juvenile phase is the period of growth between the brith of an organism and before its reproductive maturity.
  2. Reproduction phase: Reproductive phase is the period of growth after juvenile phase when an individual attain reproductive maturity and reproduces.

Question 16.
What is the difference between syngamy and fertilization?
Answer:
Syngamy & fertilization both are more similar terms with a difference that syngamy refers to the process of fusion of two gametes forming zygote while fertilization refers to the process of being fertile.

Samacheer Kalvi 12th Bio Zoology Reproduction in Organisms Additional Questions and Answers

1 – Mark Questions

I. Choose the Correct Answer

Question 1.
Transverse binary fission is noticed in ___________
(a) Amoeba
(b) Planaria
(c) Ceratium
(d) Vorticella
Answer:
(b) Planaria

Question 2.
Multiple fission occurring in the oocyte of plasmodium is called ___________
(a) Schizogony
(b) Merogamy
(c) Syngamy
(d) Sporogony
Answer:
(d) Sporogony

Question 3.
Taenia solium requires ___________ as a secondary host to complete its life cycle.
(a) Mosquito
(b) pig
(c) dog
(d) human
Answer:
(b) pig

Question 4.
Which type of parthenogenesis only females are produced?
(a) Arrhenotoky
(b) Amphitoky
(c) Thelytoky
(d) Both (a) and (b)
Answer:
(c) Thelytoky

Question 5.
Which among the following animal is not a continuous breeder?
(a) Hen
(b) Rabbit
(c) Honey bees
(d) Frogs
Answer:
(d) Frogs

Question 6.
Match the following.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 1 Reproduction in Organisms
Answer:
a – iii, b – iv, c – ii, d – i

Question 7.
Identify the incorrect statement regarding parthenogenesis.
(a) Development of sperm without fertilization.
(b) It was first discovered by Charles Bonnet.
(c) Honey bees exhibit incomplete parthenogenesis.
(d) Amphitoky is a type of natural parthenogenesis.
Answer:
(a) Development of sperm without fertilization.

Question 8.
Oblique binary fission is seen in ___________
Answer:
Dinoflagellates

Question 9.
The process by which gravid proglottids of tapeworm gets cut off is called ___________
Answer:
apolysis

Question 10.
The concept of regeneration was first noticed in ___________
Answer:
Hydra

Question 11.
Fusion of small sized, morphologically different gametes is called ___________
Answer:
merogamy

Question 12.
Identify the wrong statement.
(a) Oviparous animals lays eggs.
(b) Viviparous animals give rise to young ones.
(c) Ovoviviparous animals lays eggs and then hatch it to young ones.
(d) Amphibians are oviparous animals.
Answer:
(c) Ovoviviparous animals lays eggs and then hatch it to young ones.

Question 13.
Assertion (A): Organisms show three phases in their life cycle.
Reason (R): Juvenile phase is a degenerative phase.
(a) A is correct R but is incorrect.
(b) Both A and R are correct
(c) R is the correct explanation for A
(d) A is not correct but R is correct
Answer:
(a) A is correct R but is incorrect.

Question 14
Match the statements.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 1 Reproduction in Organisms
Answer:
a – iii, b – iv, c-i, d- ii.

Question 15.
The ploidy of males produced by arrhenotoky parthenogenesis is
Answer:
haploid

Question 16.
Identify the mismatched pair.
(a) Paedogenesis – Liver fluke
(b) Strobilation – Aurelia
(c) Amphitoky – Honey bee
(d) Encystment – Amoeba
Answer:
(c) Amphitoky – Honey bee

Question 17.
Identify the proper sequence.
(a) juvenile phase, senescent phase, vegetative phase
(b) juvenile phase, maturity phase, senescent phase
(c) vegetative phase, maturity phase, juvenile phase
(d) senescent phase, juvenile phase, vegetative phase
Answer:
(b) juvenile phase, maturity phase, senescent phase

Question 18.
Match the following:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 1 Reproduction in Organisms
Answer:
a – iv, b – iii, c – ii, d – i

Question 19.
Which of the following types of asexual reproduction is noticed in Amoeba?
(a) Sporulation
(b) Encystment
(c) Binary fission
(d) All the above
Answer:
(d) All the above

Question 20.
Pick out the organism whose fertilization occurs internally.
(a) reptiles
(b) sponges
(c) pisces
(d) amphibians
Answer:
(a) reptiles

Question 21.
Assertion (A): Asexual reproduction is called blastogenic reproduction.
Reason (R): It is accomplished by mitotic and meiotic divisions.
(a) A and R are correct
(b) A is correct but R is incorrect
(c) Both A and R are incorrect
(d) R is the correct explanation for A
Answer:
(6) A is correct but R is incorrect

Question 22.
Egg laying hen is an example for ___________
(a) Thelytoky
(b) Ovovivipary
(c) Vivipary
(d) Ovipary
Answer:
(d) Ovipary

Question 23.
Assertion (A): Syngamy refers to the fusion of two haploid gametes.
Reason (R): Syngamy leads to zygote formation.
(a) A and R are correct.
(b) A and R are incorrect.
(c) R is not the right explanation for A
(d) A is correct but R is incorrect.
Answer:
(a) A and R are correct.

Question 24.
Human beings are an example for ___________ breeders.
Answer:
continuous

2 – Mark Questions

Question 1.
Why asexual reproduction is called as somatogenic reproduction?
Answer:
Asexual reproduction is usually by amitotic or mitotic division of the somatic (body) cells, – hence is also known as somatogenic or blastogenic reproduction.

Question 2.
Name the four types of fission seen in animals.
Answer:
Binary fission, Multiple fission, Sporulation and Strobilation.

Question 3.
Define fission.
Answer:
Fission is the division of the parent body into two or more identical daughter individuals.

Question 4.
Differentiate between transverse binary fission and longitudinal binary fission.
Answer:
Transverse binary fission:

  1. Plane of division runs along the transverse axis of the organism
  2. Example: Paramecium

Longitudinal binary fission:

  1. Plane of division runs along the longitudinal axis of the organism.
  2. Example: Euglena

Question 5.
Define Plasmotomy with example.
Answer:
Plasmotomy is the division of multinucleated parent into many multinucleate daughter individuals with the division of nuclei. Nuclear division occurs later to maintain normal number of nuclei. Plasmotomy occurs in Opalina and Pelomyxa (Giant Amoeba).

Question 6.
What do you mean by regeneration in living organisms? Mention its types.
Answer:
Regeneration is the regrowth of injured region.
It is of two types:

  1. Morphallaxis
  2. Epimorphosis.

Question 7.
How is the fertilization of amphibians differs from aves based on site?
Answer:
In amphibians the fertilization is external (taking place outside the body of female organism) whereas internal fertilization takes place in aves.

Question 8.
What is Paedogamy?
Answer:
Paedogamy is the sexual union of young individuals produced immediately after the division of the adult parent cell by mitosis.

Question 9.
Write a brief note on conjugation.
Answer:
Conjugation is the temporary union of the two individuals of the same species. During their union both individuals, called the conjugants exchange certain amount of nuclear material (DNA) and then get separated. Conjugation is common among ciliates,
e.g. Paramecium, Vorticella and bacteria (Prokaryotes).

Question 10.
Classify animal breeding based on time.
Answer:
On the basis of time, breeding animals are of two types: seasonal breeders and continuous breeders. Seasonal breeders reproduce at particular period of the year such as frogs, lizards, most birds, deers etc., Continuous breeders continue to breed throughout their sexual maturity
e.g. honey bees, poultry, rabbit etc.

Question 11.
Define Vivipary.
Answer:
Vivipary is a condition in which animals give rise to live young ones after being nourished in the uterus though the placenta.
E.g. human.

Question 12.
List out the four types of binary fission.
Answer:

  1. Simple irregular binary fission
  2. Transverse binary fission
  3. Longitudinal binary fission
  4. Oblique binary fission

Question 13.
Repeated fission is a type of multiple fission. Yes or No? Why?
Answer:
Yes. If multiple fission produces four or many daughter individuals by equal cell division and the young ones do not separate until the process is complete, then this division is called repeated fission
e.g. Vorticella.

Question 14.
Define apolysis.
Answer:
The detachment of gravid proglottids either singly or in groups from the body of tapeworm is called apolysis.

3 – Mark Questions

Question 15.
Compare schizogony with sporogony of plasmodium.
Answer:
Schizogony:

  1. In schizogony, the multiple fission occurs in the schizont.
  2. It results in the formation of merozoites.

Sporogony:

  1. In sporogony, the multiple fission occurs in the oocyte.
  2. It results in the formation of sporozoites.

Question 16.
Write a short note on encystment in amoeba.
Answer:
During unfavorable conditions (increase or decrease in temperature, scarcity of food) Amoeba withdraws its pseudopodia and secretes a three-layered, protective, chitinous cyst wall around it and becomes inactive. This phenomenon is called encystment. When conditions become favourable, the encysted Amoeba divides by multiple fission and produces many minute amoebae called pseudopodiospore or amoebulae.

The cyst wall absorbs water and breaks off liberating the young pseudopodiospores, each with a fine pseudopodia. They feed and grow rapidly to lead an independent life.

Question 17.
How exogenous buds are developed by Hydra?
Answer:
When buds are formed on the outer surface of the parent body, it is known as exogenous budding e.g. Hydra. In Hydra when food is plenty, the ectoderm cells increase and form a small elevation on the body surface. Ectoderm and endoderm are pushed out to form the bud. The bud contains an interior lumen in continuation with parent’s gastrovascular cavity. The bud enlarges, develops a mouth and a circle of tentacles at its free end. When fully grown, the bud constricts at the base and finally separates from the parent body and leads an independent life.

Question 18.
Apolysis favours Taenia solium. How?
Answer:
In the tapeworm, Taenia solium the gravid (ripe) proglottids are the oldest at the posterior end of the strobila. The gravid proglottids are regularly cut off either singly or in groups from the posterior end by a process called apolysis. This is very significant since it helps in transferring the developed embryos from the primary host (man) to find a secondary host (pig).

Question 19.
What is autogamy?
Answer:
In autogamy, the male and female gametes are produced by the same cell or same organism and both the gametes fuse together to form a zygote
e.g. Actinosphaerium and Paramecium.

Question 20.
What is exogamy?
Answer:
In exogamy, the male and female gametes are produced by different parents and they fuse to form a zygote. So it is biparental. e.g. Human – dioecious or unisexual animal.

Question 21.
Give the definition for

  1. Arrhenotoky
  2. Thelytoky
  3. Amphitoky

Answer:

  1. Arrhenotoky: In this type only males are produced by parthenogenesis
    eg: honey bees
  2. Thelytoky: In this type of parthenogenesis only females are produced by parthenogenesis.
    e.g: Solenobia
  3. Amphitoky: In this type parthenogenetic egg may develop into individuals of any sex.
    e.g: Aphis.

Question 22.
What is Incomplete parthenogenesis? Explain with example.
Answer:
Incomplete parthenogenesis is a type of reproduction in which both sexual reproduction and parthenogenesis occurs. Example: In honey bees, the fertilized eggs develop into queen bee and worker bees, whereas the unfertilized eggs develop into drones (male).

Question 23.
Explain briefly on the nature of Ovovivipary.
In Ovoviviparous animals, the embryo develops inside the egg and remains in the mother’s body until they are ready to hatch. This method of reproduction is similar to viviparity but the embryos have no placental connection with the mother and receive their nourishment from the egg yolk. Ovoviviparity is seen in fishes like shark.

Question 24.
Point out any six modes of asexual reproduction seen in animals.
Answer:

  1. fission
  2.  budding
  3. fragmentation
  4. sporulation
  5. regeneration
  6. gemmule formation.

Question 25.
Enumerate the types of syngamy.
Answer:

  1. Autogamy
  2. Exogamy
  3. Hologamy
  4. Paedogamy
  5. Merogamy
  6. Isogamy
  7. Anisogamy
  8. Conjugation

Question 26.
Name the types of animals based on embryonic development with an example for each.
Answer:

  1. Oviparous animals
    e.g. Birds
  2. Viviparous animals
    e.g. Human beings
  3. Ovoviviparous animals
    e.g. Shark

Question 27.
Write a short note on phases of life cycle.
Answer:

  1. Juvenile phase – Period of growth between birth of an individual and reproductive maturity.
  2. Reproductive phase – Period of growth when an organism attain reproductive maturity and produces new off springs.
  3. Senescent plane – Period of growth when the structure and functioning of body starts degenerating.

Question 28.
What is a Paedogenesis?
Answer:
In paedogenetic parthenogenesis (paedogenesis) the larvae produce a new generation of larvae ” by parthenogenesis. It occurs in the sporocysts and Redia larvae of liver fluke. It is also seen in the larvae of some insects,
e.g. Gall fly.

Question 29.
Draw and label a gemmule of sponge.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 1 Reproduction in Organisms

Question 30.
Differentiate between asexual and sexual reproduction.
Answer:
Asexual Reproduction:

  1.  Involves only one parent
  2. Only mitotic cell division takes place.
  3. Off springs are genetically identical to parent.
  4. Gametes are not produced

Sexual Reproduction:

  1. Involves two parents (male & female)
  2. Both mitosis & meiosis takes place.
  3. Off springs genetically differ from the parents.
  4. Gametes are produced.

5 – mark Questions

Question 31.
Describe the regeneration process noticed in living organism.
Answer:
Regeneration is regrowth in the injured region. Regeneration was first studied in Hydra by Abraham Trembley in 1740. Regeneration is of two types, morphallaxis and epimorphosis. In morphallaxis the whole body grows from a small fragment e.g. Hydra and Planaria. When Hydra is accidentally cut into several pieces, each piece can regenerate the lost parts and develop into a whole new individual. The parts usually retain their original polarity, with oral ends, by developing tentacles and aboral ends, by producing basal discs.

Epimorphosis is the replacement of lost body parts. It is of two types, namely reparative and restorative regeneration. In reparative regeneration, only certain damaged tissue can be regenerated, whereas in restorative regeneration severed body parts can develop
e.g. star fish, tail of wall lizard.

Question 32.
Given an account on following terms.

  1. Hologamy
  2. Isogamy
  3. Anisogamy
  4. Merogamy
  5. Paedogamy

Answer:

  1. Hologamy: In Hologamy, the adult individuals do not produce gametes, but they themselves act as gametes and fuse to form new individuals.
    E.g : Trichonympha
  2. Isogamy : Fusion of morphologically & physiologically similar gametes.
    E.g : Monocystis.
  3. Anisogamy : Fusion of morphologically & physiologically dissimilar gametes.
    Eg: Vertebrates.
  4. Merogamy : Fusion of small sized morphologically different gametes (merogametes)
  5. Paedogamy : Fusion of young individuals produced immediately after the mitotic division of adult parent cell.

Higher Order Thinking Skills (HOTs) Question

Question 1.
Under threat or attack, garden lizard loses a part of its tail which trembles and avert the attention of predators, so that the lizard escapes later the tail regrown for lizard. The same phenomenon can also be noticed in organisms like starfish etc. What do you call this phenomenon? Define it.
Answer:
Regeneration is the regrowth in the injured region.

Question 2.
Complete the table.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 1 Reproduction in Organisms
Answer:
A – Only males
B – Only Females
C – Both Males & Females

Question 3.
In Vivipary, how the developing embryoes are nourished?
Answer:
In Vivipary, the embryo develops inside the womb of females body, hence they are nourished by the mother through placenta.

Question 4.
How Charles Bonnet and Abraham Trembley contributed to Biological filed?
Answer:
Charles Bonnet discovered the process of parthenogenesis. Abraham Trembley was the first to study the concept of Regeneration in the Hydra

Question 5.
‘A’ and ‘B’ are the male & female sex cells respectively which look alike and performs similar functions. ‘A’ and ‘B’ fuse to form a new individual ‘D’ Which type of gametic fusion does this represent? Give an example.
Answer:
Isogamy Eg: Monocystis

Question 6.
Complete the flow chart by mentioning the ploidy of cells in boxes.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 1 Reproduction in Organisms

  1. Queen Bee Worker Bee – Diploid (2n)
  2. Drones (Male Bee), Egg, sperm – Haploid (n)

Question 7.
Meiosis is a type of cell division where the chromosomal number is reduced to half the number daughter cells. Which type of cellular division occurs in the drones to produces spermatozoa? Why?
Answer:
The gonadal cells of drones undergo mitosis to form sperms. Because the drones are haploid in nature since they develop from unfertilized eggs. To avoid further reduction in chromosome no. and maintain the chromosomal constancy, (instead of undergoing meiosis), mitosis will take place leading to formation of haploid gametes.

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Samacheer Kalvi 12th Commerce Companies Act 2013 Textbook Exercise Questions and Answers

I. Choose the correct answer

Question 1.
The Company will have to issue the notice of situation of Registered Office to the Registrar of Companies with in _____ days from the date of incorporation.
(a) 14 days
(b) 21 days
(c) 30 Days
(d) 60 Days
Answer:
(c) 30 Days

Question 2.
How does a person who envisages the idea to form a company called?
(a) Director
(b) Company Secretary
(c) Registrar
(d) Promoter
Answer:
(d) Promoter

Question 3.
For which type of capital a company pays the prescribed fees at the time of registration?
(a) Subscribed Capital
(b) Authorised Capital
(c) Paid-up Capital
(d) Issued Capital
Answer:
(b) Authorised Capital

Question 4.
Which of the following types of shares are issued by a company to raise capital from the existing shareholders?
(a) Equity Shares
(b) Right Shares
(c) Preference Shares
(d) Bonus Shares
Answer:
(b) Right Shares

Question 5.
Specify the type of resolution to be passed to choose the location of Registered Office of the company within the town or village or city.
(a) Ordinary
(b) Special
(c) Either Ordinary or Special
(d) Board
Answer:
(d) Board

Question 6.
Who can issue stock?
(a) Public
(b) Private
(c) One Person
(d) Small
Answer:
(a) Public

Question 7.
Specify the document which comes under the Negotiable Instrument Act.
(a) Share Certificate
(b) Share
(c) Share Warrant
(d) Stock
Answer:
(c) Share Warrant

Question 8.
The shares which are offered to the existing shareholder at free of cost is known as _____
(a) Bonus Share
(b) Equity Share
(c) Right Share
(d) Preference Share
Answer:
(a) Bonus Share

Question 9.
The shares which are offered first to the existing shareholder at reduced price is known as _____
(a) Bonus Share
(b) Equity Share
(c) Right Share
(d) Preference Share
Answer:
(c) Right Share

Question 10.
The Companies Act 2013 Prohibits the issue of shares at _____ to the public.
(a) Premium
(b) Par
(c) Discount
(d) Both at par and Premium
Answer:
(c) Discount

II. Very Short Answer Questions

Question 1.
Who is called as Promoters?
Answer:
Promotion stage begins when the idea to form a company comes in the mind of a person. The person who envisage the idea is called a ‘promoter’.

Question 2.
What is Share?
Answer:
The term Share is viewed by a layman as a fraction or portion of total capital of the company which have equal denomination.

Question 3.
What do you mean by Equity Share?
Answer:
The share of a company which do not have any preferential rights with regard to dividend and repayment of share capital at the time of liquidation of a company, is called as equity share or ordinary share.

Question 4.
What do you understand by Preference Share?
Answer:
The term ‘preference shares’ means that part of the share capital the holders of which have a preferential right over payment of dividend (fixed amount or rate) and repayment of share capital in the event of winding up of the company.

Question 5.
What is Sweat Equity Shares?
Answer:
Sweat Equity Shares means issue of shares to employees or directors at a lower price for cash or other than Cash.

Question 6.
What is Bonus Shares?
Answer:
Bonus share means to utilize the company’s reserves and surpluses. Issue of shares to existing shareholders without taking any consideration is known as Bonus Shares.

Question 7.
What is Right Shares?
Answer:
The right shares are primarily issued to the existing equity shareholders through a letter of an issue, on pro rata basis.

Question 8.
What is Private placement?
Answer:
Private placement means offer of securities or invitation to subscribe to securities to a select group of persons through private placement offer letter.

Question 9.
Define Share Warrant.
Answer:
A share warrant is a negotiable instrument, issued by the public limited company only against fully paid up shares. It is also termed as a document of title because the holder of the share warrant is entitled to the number of shares mentioned in it.

Question 10.
What is Debentures?
Answer:
When a company needs funds for extension and development purpose without increasing its share capital, it can borrow from the general public by issuing certificates for a fixed period of time and at a fixed rate of interest. Such a loan certificate is called a debenture.

III. Short Answer Questions

Question 1.
Distinguish between shares and stocks.
Answer:

Basis for Difference Shares Stocks
1. Meaning The term ‘share’ means a fraction or unit of the total capital of the company which have equal denomination. Stock is created from fully paid shares by passing resolution in the general meeting. The Articles of Association of the company must permit this conversion.
2. Denomination All the shares are of equal denomination. The denomination of stocks may. differ.
3. Paid up value Shares can be partly or fully paid up. Stock can only be fully paid up.

Question 2.
What do you understand by Issue of Securities at Premium?
Answer:
When shares are issued at a price above the face or nominal value, they are said to be issued at a premium. For example, a share having the face value of Rs. 10 is issued at Rs. 12. Here, Rs.2 is the premium. The amount of share premium has to be transferred to an account called the ‘ Securities Premium Account’

Question 3.
What is issue of shares at discount? What conditions should be fulfilled?
Answer:
When the shares are issued at a price below the face value they are said to be issued at a discount. For example, a share having the face value of Rs 10 is issued at Rs 8. The companies act 2013, prohibits the issue of shares at discount (Section 53), except sweat Equity share.

Question 4.
State condition stipulated for capital subscription at the time of promotion.
Answer:
For capital subscription, steps to be taken are listed below:

  1. The fulfilling formalities to raise necessary capital.
  2. Adhering to SEBI guidelines in this regard.
  3. Observing guidelines for Disclosure and investor protection issued by SEBI.
  4. Issuing prospectus.
  5. Appointing official banker of the company for receiving application from the investors.

Question 5.
Explain different kinds of preference shares.
Answer:
There are eight types of preference shares:

  1. Cumulative Preference shares: As the word indicates, all dividends are carried forward until specified.
  2. Non-cumulative Preference shares: These are opposite of cumulative.
  3. Redeemable Preference shares: Such preference shares can be claimed after a fixed period or after giving due notice. ,
  4. Non-Redeemable Preference shares: Such shares cannot be redeemed during the lifetime of the company.
  5. Convertible Preference shares: The shares can be converted into equity shares after a time period. .
  6. Non-convertible Preference shares: Non-convertible preference shares cannot be, at any time, converted into equity shares.
  7. Participating Preference shares.
  8. Non-Participating Preference shares.

IV. Long Answer Questions

Question 1.
Write the difference between Debentures and Shares.
Answer:
Debentures:

  1. Debentures mean a loan.
  2. Debenture holder gets fixed rate of Interest.
  3. Debentures generally have a charge on the assets of the company.
  4. Debentures can be issued at a discount.
  5. Debenture holders do not have any voting right.
  6. Interest on debentures is payable even if there are no profits.
  7. Interest paid on debenture is a business expense.

Shares:

  1. Shares are part of the capital of a company.
  2. Shareholders gets dividends with a varying rate.
  3. Shares do not carry any such charge.
  4. Shares cannot be issued at a discount.
  5. Shareholders enjoy voting right.
  6. Dividend is payable only if there is profit.
  7. Dividend is not a business expense.

Question 2.
Brief different stages in Formation of a Company.
Answer:
Section 3 (1) of the Act states that a company may be formed for any lawful purpose by-
(a) seven or more persons, where the company to be formed is to be a public company;
(b) two or more persons, where the company to be formed is to be a private company;
(c) one person, where the company to be formed is to be One Person Company.
The process of formation of company consists of different stages:
1. Promotion: Promotion stage begins when the idea to form a company comes in the mind of a person.

2. Registration: The second stage in the formation of the company is incorporation or – registration. In this stage, the promoter has to fix name of the company, prepare the necessary documents (Memorandum and Articles of Association), fix the registered office, and name of the directors. After this, certificate of incorporation is issued.

3. Capital Subscription: A public limited company having its share capital has to pass through two stages. One of them is capital subscription. The steps for this is:

  1. Formalities for raising capital
  2. Issuing prospectus
  3. Appointing official banker
  4. Pass resolution to make allotment

4. Commencement of Business: As per section 11 of the Act, a company having share capital should file with the Registrar, declaration stating that

  1. Every subscriber has paid the value of shares.
  2. Paid up capital is not less than Rs.5 lakhs for a public limited company and Rs.1 lakh in case of a private limited company.
  3. It has filed the Registrar, regarding the verification of registered office. After fulfilling these details, the Registrar will issue certificate of commencement of business.

Question 3.
What are the various kinds of Debentures?
Answer:
Debenture is a document issued by the company for acknowledging the loan from the public. Debentures are classified into different categories on the basis of:

  1. Convertibility of the Instrument
  2. Security of the Instrument
  3. Redemption ability; and
  4. Registration of Instrument.

1. On the basis of convertibility:
(a) Non-Convertible Debentures: These instruments cannot be converted into equity shares.
(b) Partly Convertible Debentures: Apart of these instruments are converted into equity shares.
(c) Fully Convertible Debentures: These are fully convertible into equity shares.
(d) Optionally Convertible Debentures: The investor can have the option to either convert the debentures at a price decided by the issuer or agreed upon at the time of issue.

2. On the basis of Security:
(a) Secured Debentures: These instruments are secured by a charge on the fixed assets of the issuer company.
(b) Unsecured Debentures: These instruments are unsecured against the assets.

3. On the basis of Redeemability:
(a) Redeemable Debentures: It refers to the debentures which will be redeemed in future.
(b) Irredeemable Debentures: It is a debenture, in which no specific time is specified by the companies to pay back the money.

4. On the basis of Registration:
(a) Registered Debentures: These are issued in the name of a particular person, who is registered by the company.
(b) Bearer Debentures: These are issued to the bearer and are negotiable instruments, and are transferred by mere delivery.

Question 4.
What formalities need to be fulfilled for companies having share capital to commence business?
Answer:
A public limited company having its share capital has to pass through two more stages. One of them is capital subscription, steps to be taken at this stage are listed below:

  1. The fulfilling formalities to raise necessary capital.
  2. Following the SEBI guidelines in this regard.
  3. Issuing prospectus.
  4. Appointing official banker of the company for receiving application from the investors.
  5. Passing resolution for making allotment by director.
  6. Despatch allotment letters to allottees.
  7. Filing allotment return with the Registrar.
  8. Issuing share certificates in exchange for their allotment letter.
  9. Ensuring collection of minimum subscription.

Question 5.
Write the difference between Share Certificate and Share Warrant.
Answer:
Share Certificate:

  1. A share certificate is an instrument in writing for the legal proof of the ownership.
  2. Every company must issue share certificate to its shareholders.
  3. Normally, the holder of the share certificate is to be the member of the company.
  4. The share certificate is issued by the company within three months of the allotment of shares.
  5. No need to authorized in the Articles of Association to issue share certificate.

Share Warrant:

  1. A share warrant is a negotiable instrument, issued against fully paid up shares.
  2. There is no compulsion of the issue of share warrants by the company.
  3. Generally, the holder of the share warrant is not the member of the company.
  4. Shares warrant can be issued only when the shares are fully paid up.
  5. The issue of a share warrant must be authorized in the Articles of Association of the company.

Samacheer Kalvi 12th Commerce Companies Act 2013 Additional Questions and Answers

I. Choose the Correct Answer
Question 1.
Match List I with List II and select the correct answer using the codes given below:

Samacheer Kalvi 12th Commerce Solutions Chapter 26 Companies Act 2013

Codes:

Samacheer Kalvi 12th Commerce Solutions Chapter 26 Companies Act 2013
Answer:
(c) 3,1,4,2

Question 2.
The first two stages of formation of a company are
(i) Issue of prospectus
(ii) Promotion
(iii) Issue of share certificate
(iv) Registration
(a) (i) and (ii)
(b) (ii) and (iv)
(c) (ii) and (iii)
(d) (i) and (iii)
Answer:
(b) (ii) and (iv)

We as a team believe the information prevailing regarding the Tamilnadu State Board Solutions for 26th Commerce Chapter 26 Companies Act 2013 Questions and Answers has been helpful in clearing your doubts to the fullest. For any other help do leave us your suggestions and we will look into them. Stay in touch to get the latest updates on Tamilnadu State Board Solutions for different subjects in the blink of an eye.

Samacheer Kalvi 12th Commerce Solutions Chapter 24 Types of Entrepreneurs

Enhance your subject knowledge with Tamilnadu State Board Solutions for 24th Commerce Chapter 24 Types of Entrepreneurs Questions and Answers and learn all the underlying concepts easily. Make sure to learn the subject from Tamilnadu State Board Solutions Chapter 24 Types of Entrepreneurs Questions and Answers PDF on a day to day basis and score well in your exams. You can Download Samacheer Kalvi 24th Commerce Book Solutions Questions and Answers are given after enormous research by people having high subject knowledge and for better scoring grade. You can rely on them and prepare any topic of Commerce as per your convenience easily.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 24 Types of Entrepreneurs

Students those who are looking for Tamilnadu State Board Solutions Chapter 24 Types of Entrepreneurs Questions and Answers Concepts can find them all in one place from our site Tamilnadu State Board Solutions. Simply click on the links available to prepare the corresponding topics of Samacheer Kalvi 24th Commerce Book Solutions Questions and Answers easily. Clarify all your queries from chapter wise different questions to be familiar with the kind of questions appearing in the exam. Thus, you can increase your score and get higher grade in the final exam.

Samacheer Kalvi 12th Commerce Types of Entrepreneurs Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
Choose the type of entrepreneur that isn’t based on function:
(a) Innovative
(b) Classical
(c) Fabian
(d) Drone
Answer:
(c) Fabian

Question 2.
Choose the type of Entrepreneur that is not based on Motivation:
(a) Pure
(b) Corporate
(c) Spontaneous
(d) Induced
Answer:
(c) Spontaneous

Question 3.
Which of the following is the Activity of a Business Entrepreneur?
(a) Production
(b) Marketing
(c) Operation
(d) All of the above
Answer:
(d) All of the above

Question 4.
Find the odd one out in context of Trading Entrepreneur.
(a) Selling
(b) Commission
(c) Buying
(d) Manufacturing
Answer:
(d) Manufacturing

Question 5.
Corporate Entrepreneur is also called as  _________
(a) Intrapreneur
(b) Promoter
(c) Manager
(d) Shareholder
Answer:
(b) Promoter

Question 6.
Poultry, Flowers, Fruits, etc., are called allied products of _________ entrepreneur.
(a) Corporate
(b) Retail
(c) Trading
(d) Agricultural
Answer:
(d) Agricultural

Question 7.
_________ Entrepreneur Supply Services Unlike.
(a) Hoteliers
(b) Banking
(c) Airlines
(d) Livestock
Answer:
(d) Livestock

Question 8.
Motive of a Pure Entrepreneur is _________
(a) Rendering service
(b) Earning profit
(c) Attaining status
(d) Both b and c
Answer:
(d) Both b and c

Question 9.
Which of these is based on Technology?
(a) Modem
(b) Professional
(c) Corporate
(d) Industrial
Answer:
(c) Corporate

Question 10.
Which of the below is not a characteristic of a Fabian Entrepreneur?
(a) Conservative
(b) Risk averse
(c) Sceptical
(d) Adaptive
Answer:
(d) Adaptive

II. Very Short Answer Questions

Question 1.
What is the other name of business entrepreneur?
Answer:
The term entrepreneur means the. person who takes steps for commencing the business. So he is otherwise called as organiser or proprietor.

Question 2.
Mention the other name for corporate entrepreneur.
Answer:
Corporate entrepreneur is called promoter. He/she takes initiative necessary to start an entity under corporate format.

Question 3.
Who are agricultural entrepreneur?
Answer:
Agricultural entrepreneurs are those entrepreneurs who raise farm products and market them.

Question 4.
State the name of the.following ventures:
a. Started by individuals for profit motive
b. Started by Government
c. Started by individuals and Government together
d. Started as a family business
Answer:
a. Private entrepreneur
b. State entrepreneurship
c. Induced entrepreneur
d. Classical entrepreneur

Question 5.
Give some examples of pure entrepreneurs.
Answer:
Dhirubai Ambani, Jamshedji Tata, T.V. Sundaram Iyengar, Seshadriji, Birla, Narayanamurthi, and Azim Premji are few examples of pure entrepreneurship.

III. Short Answer Questions

Question 1.
Who is a private entrepreneur?
Answer:
Ventures started by individual either singly or collectively at their own risk after mobilising – various, resources in order to earn profit are called private entrepreneurship.

Question 2.
What is political environment?
Answer:
To commence business, various factors are needed. Apart from all other factors, political environment is also essential. It means that the concessions, incentives provided by the government drive them to enter into venture. The government also provides support in the form of loans, subsidies and other taxes.

Question 3.
List down few examples of pure entrepreneurship.
Answer:
Pure entrepreneurs are individuals who are propelled to enter into venture by psychological and economic motives. They nurture desire of starting a particular venture and earning high profit there from and thus attaining a social status. They apply their knowledge, skill and insight in making the venture a great’ success in order to earn maximum profit out of the venture. Dhirubai Ambani, Jamshedji Tata, T.V. Sundaram Iyengar, Seshadriji, Birla, Narayanamurthi, and Azim Premji are few examples of pure entrepreneurship.

Question 4.
How does a professional entrepreneur operate?
Answer:
Professional entrepreneur is one who is having a rich expertise in starting a venture but lack interest in continuing the venture as a manager or as a owner. He/she simply sells out the venture started by him to someone else after its successful take-off.

Question 5.
Explain, about the agricultural entrepreneur.
Answer:
Agricultural entrepreneurs are those entrepreneurs who raise farm products and market them. They use the various inputs like labour, fertilizer, insecticide, water technology, etc., to raise the products and market their products either directly or through co-operative entities or through brokers or through tie up with large retailers.

IV. Long Answer Questions

Question 1.
Explain in detail on classification according to the type of business.
Answer:
Classification of Entrepreneur according to the type of business:

(i) Business Entrepreneur: He is called solo entrepreneur. He is the one who finds out an idea for a new product or service and establish a business enterprise.

(ii) Trading Entrepreneur: Trading entrepreneurs are those who restrict themselves to buying and selling finished goods.

(iii) Industrial Entrepreneur: These are entrepreneurs who manufacture products to cater to

(iv) Corporate Entrepreneur: He is called as promoter. He takes initiative necessary to start an entity under corporate format.

(v) Agricultural Entrepreneur: These entrepreneurs are those who raise farm products and market them.

Question 2.
Discuss the nature of functional entrepreneurs.
Answer:
Nature of functional entrepreneurs:

(i) Innovative Entrepreneur: He is a person who introduces new project. They observe the environment regarding information to the new venture.

(ii) Imitative Entrepreneur: He refers to the person who simply imitates existing knowledge or technology already in advance countries. Redesigning of products suited to the local conditions.

(iii) Fabian Entrepreneur: These are said to be conservatives and sceptical about any changes in their organisation. They are of risk-averse.

(iv) Drone Entrepreneur: Drone entrepreneurs are those who are totally opposed to changes in the environment. They used to operate in the-niche market.

Question 3.
Distinguish between the rural and urban entrepreneur.
Answer:

S. No. Rural Entrepreneur Urban Entrepreneur
1. It refers to the person who starts business in rural areas. It refers to the person who commences business in urban areas.
2. These entrepreneurs start doing business in the villages and small towns. They will do their business in state capital, towns, district headquarters, municipalities, etc.
3. They may be agricultural and trading entrepreneurs. They may be industrial or corporate entrepreneur.
4. The availability of material and labour is easy. So the cost of operation tends to be low. The availability of material and labour may be difficult. So the cost of operation may be high.

Samacheer Kalvi 12th Commerce Types of Entrepreneurs Additional Questions and Answers

I. A. Choose the Correct Answer

Question 1.
The entrepreneurs classified on the basis of type of business are __________
(i) Industrial entrepreneur
(ii) Technical entrepreneur
(iii) Professional entrepreneur
(iv) Business entrepreneur
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(b) (i) and (iv)

B. Fill in the blanks

Question 1.
Entrepreneur who commences his entrepreneurial activity in urban areas is called as __________
Answer:
urban entrepreneur

II. Very Short Answer Questions

Question 1.
Who is a service entrepreneur?
Answer:
The entrepreneurs enter into the business of giving service products to end consumers. Example: Banking, Insurance and Transport services.

Question 2.
Write a short note on retail entrepreneurs.
Answer
Retail entrepreneurs are those who enter into venture of distributing the end product to final consumer. They used to buy the goods from numerous wholesalers.

III. Short Answer Questions

Question 1.
Who are industrial entrepreneurs?
Answer:
Industrial entrepreneurs are those persons who manufacture products to cater the needs of the consumers. They may be small, medium and large entrepreneurs. Industrial entrepreneurs mobilise the resources of various types’.

Question 2.
Who are technical entrepreneurs?
Answer:
Technical entrepreneurs are those craftsmen like welder, fitter, turner, carpenter and goldsmith, photographer, weaver who start small business. They manufacture products/service of high quality. They simply focus on production rather than on marketing.

We as a team believe the information prevailing regarding the Tamilnadu State Board Solutions for 24th Commerce Chapter 24 Types of Entrepreneurs Questions and Answers has been helpful in clearing your doubts to the fullest. For any other help do leave us your suggestions and we will look into them. Stay in touch to get the latest updates on Tamilnadu State Board Solutions for different subjects in the blink of an eye.

Samacheer Kalvi 12th Commerce Solutions Chapter 23 Elements of Entrepreneurship

Enhance your subject knowledge with Tamilnadu State Board Solutions for 23th Commerce Chapter 23 Elements of Entrepreneurship Questions and Answers and learn all the underlying concepts easily. Make sure to learn the subject from Tamilnadu State Board Solutions Chapter 23 Elements of Entrepreneurship Questions and Answers PDF on a day to day basis and score well in your exams. You can Download Samacheer Kalvi 23th Commerce Book Solutions Questions and Answers are given after enormous research by people having high subject knowledge and for better scoring grade. You can rely on them and prepare any topic of Commerce as per your convenience easily.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 23 Elements of Entrepreneurship

Students those who are looking for Tamilnadu State Board Solutions Chapter 23 Elements of Entrepreneurship Questions and Answers Concepts can find them all in one place from our site Tamilnadu State Board Solutions. Simply click on the links available to prepare the corresponding topics of Samacheer Kalvi 23th Commerce Book Solutions Questions and Answers easily. Clarify all your queries from chapter wise different questions to be familiar with the kind of questions appearing in the exam. Thus, you can increase your score and get higher grade in the final exam.

Samacheer Kalvi 12th Commerce Elements of Entrepreneurship Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
Which of the below is a factor of production?
(a) Land
(b) Labour
(c) Entrepreneurship
(d) All of the above
Answer:
(d) All of the above

Question 2.
Entrepreneur is not classified as ________
(a) Risk Bearer
(b) Innovator
(c) Employee
(d) Organizer
Answer:
(b) Innovator

Question 3.
What are the characteristics of an entrepreneur?
(a) Spirit of enterprise
(b) Flexibility
(c) Self Confidence
(d) All of the above
Answer:
(d) All of the above

Question 4.
Which of the below is not classified into managerial functions?
(a) Planning
(b) Marketing
(c) Organizing
(d) Controlling
Answer:
(c) Organizing

Question 5.
Which of the below is a commercial function?
(a) Accounting
(b) Coordination
(c) Discovery of idea
(d) Planning
Answer:
(a) Accounting

II. Very Short Answer Questions

Question 1.
Mention any two features of entrepreneurs.
Answer:
Features of Entrepreneurs:

  1. Spirit of Enterprise: Entrepreneur should be bold enough to encounter risk arising from the venture undertaken.
  2. Self Confidence: Entrepreneur should have a self confidence in order to achieve high goals in the business.

Question 2.
List down the managerial functions of entrepreneurs.
Answer:

  1. Planning
  2. Organising
  3. Directing
  4. Controlling
  5. Coordination

Question 3.
List down the promotional functions of entrepreneurs.
Answer:

  1. Discovery of Idea
  2. Determining the business objectives
  3. Detailed Investigation
  4. Choice of form of enterprise
  5. Fullfillment of the formalities
  6. Preparation of Business Plan
  7. Mobilisation of funds
  8. Procurement of Machines and Materials

Question 4.
Define Intrapreneur.
Answer:
Intrapreneur is one who thinks and acts like an entrepreneur for the firm’s development during the course of employment in an organisation.

Question 5.
List the problems faced by the women entrepreneurs.
Answer:
There is a tremendous growth in the women entrepreneurship in India. But there are certain problems met by women entrepreneurs. They are as follows:-

  1. Problem of Finance: The external sources of funds for the women is limited because they do not generally own properties in their own name. They are depending on their own savings and small loans from friends and relatives.
  2. Lack of Education: Illiterate and semi-literate women entrepreneurs face difficulties in respect of accounts, money matters, marketing and day-to-day operations.
  3. Lack of Network Support: The success of business depends on the support of family members, friends and relatives. But it is reported that the women entrepreneurs get very limited support in times of crisis.
  4. Stiff Competition: They have to face acute competition for their goods from organised sector and from their male counterparts.
  5. Lack of Information: The lack of knowledge or limited knowledge about subsidies, concessions and incentives given by Government will affect the business.

III. Short Answer Questions

Question 1.
Define Entrepreneur.
Answer:
The person who establishes business is termed as entrepreneur. The output of an entrepreneurial process ends up in establishing an enterprise.

Question 2.
Distinguish between Entrepreneur and Manager.
Answer:

Basis of difference Entrepreneur Manager
Motive The very motive of an entrepreneur is to start a venture by setting of an entity. The very motive of manager is to render service in an entity setup for execution of venture.
Status Entrepreneur is owner of the entity. Manager is a salaried employee in the entity set up for carrying on the venture.
Risk Bearing Entrepreneur bears the eventual risk and uncertainty in operating the enterprise. Manager doesn’t bear any risk in the venture, where the venture is unsuccessful he/she simply quits the enterprise.

Question 3.
List down the commercial functions of Entrepreneur and explain them shortly.
Answer:
Commercial Functions of Entrepreneur:

(i) Production or Manufacturing: Under production function, entrepreneur has to take decision relating to selection of factory site, design and layout, type of products to be manufactured, research and development, product design, etc.

(ii) Marketing: Entrepreneur has to carry out following functions pertaining to marketing aspect namely consumer research, product planning and development, standardisation, packaging, pricing, warehousing, distribution, promotion etc.

(iii) Accounting: Entrepreneur has to arrange to prepare trading and profit and loss account in order to know the profit or loss incurred out of operation of the business and prepare balance sheet to know the financial status of business at a particular day.

(iv) Finance: In the sphere of financial function, an entrepreneur has to take decisions like choosing the right type of financing, framing the best dividend policy, acquiring of funds, efficiently managing fixed and current assets, maximising shareholders wealth and investing of funds efficiently and effectively.

(v) Human Resource Management: Entrepreneur has to estimate the manpower needs of the enterprise and accordingly decide the size of manpower required for various slots of organisational structure.

Question 4.
Explain the promotional functions of entrepreneur.
Promotional Functions of Entrepreneur
Answer:
(i) Discovery of Idea: The first and foremost function of entrepreneur is idea generation. A person may conceive his own ideas or develop the ideas contributed by others. Ideas can be generated through several ways like own experience and exposure of entrepreneur, keen observation of environment, education, training, market survey, environmental scanning and so on.

(ii) Determining the business objectives: Entrepreneur has to develop business objectives in the backdrop of nature of business and type of business activity i.e. nature of business, manufacturing or trading, type of business organisation chosen so that he/she can organise the venture in accordance with the objectives determined by him/her.

(iii) Detailed Investigation: Entrepreneur should investigate commercial feasibility of the product proposed to be produced and conduct market study to ascertain the potential demand for the product.

(iv) Choice of form of enterprise: Entrepreneur has to choose the appropriate form of organisation suited to implement the venture. There are various forms of organisation namely sole proprietor, partnership, company and co-operatives etc. which are in existence.

(v) Fulfilment of the formalities: Having chosen the appropriate type of organisation, entrepreneur has to take necessary steps to establish the form of organisation chosen. As regards sole trader, the formalities are barest minimum. In the case of partnership firm, entrepreneur has to arrange for partnership deed and he has to get the deed registered.

(vi) Preparation of Business Plan: Entrepreneur has to prepare a business plan or project report of the venture that he is proposing to take up.

(vii) Mobilisation of funds: Entrepreneur has to take steps to mobilise capital needed to implement the venture. Entrepreneur has to estimate the fixed capital and working capital required for running the project.

(viii) Procurement of Machines and Materials: Entrepreneur has to locate the various sources of supply of machineries, equipments and materials.

Question 5.
Explain the commercial functions of entrepreneur.
Answer:
(i) Production or Manufacturing: Under production function, entrepreneur has to take decision relating to selection of factory site, design and layout, type of products to be manufactured, research and development, product design, etc. The efficient and effective performance of production function depends on the proper production planning and control to a major extent.

(ii) Marketing: Entrepreneur has to carry out following functions pertaining to marketing aspect namely consumer research, product planning and development, standardisation, packaging, pricing, warehousing, distribution, promotion etc. The term marketing mix denotes the combination of four components namely product, price, promotion and physical distribution in the case of physical products and three more components are included in the case of service products namely people, process and physical evidence.

(iii) Accounting: Entrepreneur has to arrange to prepare trading and profit and loss account in order to know the profit or loss incurred out of operation of the business and prepare balance sheet to know the financial status of business at a particular day. Besides, cash flow and fund flow statements are prepared to ensure the adequacy of funds and cash for meeting various working capital needs of the business.

(iv) Finance: In the sphere of financial function, an entrepreneur has to take decisions like choosing the right type of financing, framing the best dividend policy, acquiring of funds, efficiently managing fixed and current assets, maximising shareholders wealth and investing of funds efficiently and effectively.

(v) Human Resource Management: Entrepreneur has to estimate the manpower needs of the enterprise and accordingly decide the size of manpower required for various slots of organisational structure. After determining the required manpower, the entrepreneur has to organise for recruitment procedure

selecting manpower, induction and training, determining compensation structure and incentives, designing motivation programmes, structuring wellbeing measures for employees, putting in place safety mechanism at workplace, performance evaluation and career advancement and structuring social security programmes.

IV. Long Answer Questions

Question 1.
How do you classify entrepreneurs?
Answer:
Entrepreneurs are now broadly classified into three groups namely risk bearer, organiser and
innovator.

1. Entrepreneur as a risk bearer : Entrepreneurs acts as an agent combining all factors of production to produce a product or service in order to sell at uncertain price in future.

2. Entrepreneur as an organiser : Entrepreneur is one who brings together various factors of production and creates an entity to produce product or service and supervise and coordinates several functions in the process.

3. Entrepreneur as an innovator : According to Joseph A. Schumpeter in the year 1934 used innovation as a criterion to define an individual as entrepreneur. According to him, entrepreneur is one who

    • Introduces a brand new product in the market.
    • Institutes new technology to produce a new product.
    • Discovers new course of supply of raw materials.

Question 2.
What are the characteristics of an entrepreneur?
Answer:
Characteristics of an Entrepreneur:

1. Spirit of Enterprise: Entrepreneur should be bold enough to encounter risk arising from Elements of Entrepreneurship 145 the venture undertaken.

2. Self-confidence: Entrepreneur should have self-confidence in order to achieve high goals in the business.

3. Flexibility: Entrepreneur should not doggedly stick to decisions in a rigid fashion.

4. Innovation: Entrepreneur should contribute something new or something unique to meet the changing requirements of customers namely new product, new method of production or distribution, adding new features to the existing product, uncovering a new territory for business, innovating new raw material, etc.

5. Resource Mobilisation: Entrepreneur should have the capability to mobilise both tangible inputs like manpower, money materials, technology, market, method etc., which are scattered over a wide area and certain intangible inputs like motivation, morale and innovativeness.

Question 3.
Distinguish between an Entrepreneur and an Intrapreneur.
Answer:

Samacheer Kalvi 12th Commerce Solutions Chapter 23 Elements of Entrepreneurship

Question 4.
Discuss the problems faced by Women Entrepreneurs.
Answer:
There is a tremendous growth in the women entrepreneurship in India. But there are certain problems met by women entrepreneurs.
They are as follows:-

  1. Problem of Finance: The external sources of funds for the women is limited because they do not generally own properties in their own name. They are depending on their own savings and loan from friends and relatives.
  2. Lack of Education: Illiterate and semi-literate women entrepreneurs face difficulties in respect of accounts, money matters, marketing and day-to-day operations.
  3. Lack of Network Support: The success of business depends on the support of family members, friends and relatives. But it is reported that the women entrepreneurs get very limited support in times of crisis.
  4. Stiff Competition: They have to face acute competition for their goods from organised sector and from their male counterparts.
  5. Lack of Information: The lack of knowledge or limited knowledge about subsidies, concessions and incentives given by Government will affect the business.

Question 5.
Explain in detail the various functions of an entrepreneur.
Answer:
The functions of an entrepreneur is divided into
(a) Promotional functions,
(b) Managerial functions
(c) Commercial functions.

(a) Promotional functions:
(i) Discovery of Idea: The first and foremost function of entrepreneur is idea generation. A person may conceive his own ideas or develop the ideas contributed by others. Ideas can be generated through several ways like own experience and exposure of entrepreneur, keen observation of environment, education, training, market survey, environmental scanning and so on.

(ii) Detailed Investigation: Entrepreneur should investigate commercial feasibility of the product proposed to be produced and conduct market study to ascertain the potential demand for the product.

(iii) Mobilisation of funds: Entrepreneur has to take steps to mobilise capital needed to implement the venture. Entrepreneur has to estimate the fixed capital and working capital required for running the project.

(b) Managerial functions:
(i) Planning: In this function, the entrepreneur has to lay down the objectives, goals, vision, mission, policies, procedures, programmes, budget, schedules etc., for enabling the venture to proceed towards established destinations.

(ii) Directing: In this function, the entrepreneur has to motivate, lead, guide and communicate with subordinates on an ongoing basis in order to accomplish pre-set goals.

(c) Commercial functions:
(i) Production or Manufacturing: Under production function, entrepreneur has to take decision relating to selection of factory site, design and layout, type of products to be manufactured, research and development.

(ii) Accounting: Entrepreneur has to arrange to prepare trading and profit and loss account in order to know the profit or loss incurred out of operation of the business and prepare balance sheet to know the financial status of business.

We as a team believe the information prevailing regarding the Tamilnadu State Board Solutions for 23th Commerce Chapter 23 Elements of Entrepreneurship Questions and Answers has been helpful in clearing your doubts to the fullest. For any other help do leave us your suggestions and we will look into them. Stay in touch to get the latest updates on Tamilnadu State Board Solutions for different subjects in the blink of an eye.

Samacheer Kalvi 12th Commerce Solutions Chapter 22 The Negotiable Instruments Act 1881

Enhance your subject knowledge with Tamilnadu State Board Solutions for 22th Commerce Chapter 22 The Negotiable Instruments Act 1881 Questions and Answers and learn all the underlying concepts easily. Make sure to learn the subject from Tamilnadu State Board Solutions Chapter 22 The Negotiable Instruments Act 1881 Questions and Answers PDF on a day to day basis and score well in your exams. You can Download Samacheer Kalvi 22th Commerce Book Solutions Questions and Answers are given after enormous research by people having high subject knowledge and for better scoring grade. You can rely on them and prepare any topic of Commerce as per your convenience easily.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 22 The Negotiable Instruments Act 1881

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Samacheer Kalvi 12th Commerce The Negotiable Instruments Act 1881 Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
Negotiable Instrument Act was passed in the year ________
(a) 1981
(b) 1881
(c) 1994
(d) 1818
Answer:
(b) 1881

Question 2.
Negotiable Instrument is freely transferable by delivery if it is a ________ instrument.
(a) Order
(b) Bearer
(c) Both a and b
(d) None of the above
Answer:
(b) Bearer

Question 3.
The transferee of a Negotiable Instrument is the one ________
(a) Who transfer the instrument
(b) On whose name it is transferred
(c) Who enchases it
(d) None of the above
Answer:
(b) On whose name it is transferred

Question 4.
Number of parties in a bill of exchange are ________
(a) 2
(b) 6
(c) 3
(d) 4
Answer:
(c) 3

Question 5.
Section 6 of Negotiable Instruments Act 1881 deals with ________
(a) Promissory Note
(b) Bills of exchange
(c) Cheque
(d) None of the above
Answer:
(c) Cheque

Question 6.
________ cannot be a bearer instrument.
(a) Cheque
(b) Promissory Note
(c) Bills of exchange
(d) None of the above
Answer:
(a) Cheque

Question 7.
When crossing restrict further negotiation ________
(a) Not negotiable crossing
(b) General Crossing
(c) A/c payee crossing
(d) Special crossing
Answer:
(a) Not negotiable crossing

Question 8.
Which endorsement relieves the endorser from incurring liability in the event of dishonour?
(a) Restrictive
(b) Facultative
(c) Sans recourse
(d) Conditional
Answer:
(b) Facultative

Question 9.
A cheque will become stale after ________ months of its date.
(a) 3
(b) 4
(c) 5
(d) 1
Answer:
(a) 3

Question 10.
Document of title to the goods exclude ________
(a) Lorry receipt
(b) Railway receipt
(c) Airway bill
(d) Invoice
Answer:
(d) Invoice

II. Very Short Answer Questions

Question 1.
What is meant by Negotiable Instrument?
Answer:
A negotiable instrument is a document which entitles a person to a certain sum of money and which is transferable from one person to another by mere delivery or by endorsement and delivery.

Question 2.
Define Bill of Exchange.
Answer:
According to section 5 of the Negotiable Instruments Act, “a bill of exchange is an instrument in writing containing an unconditional order, signed by the maker, directing a certain person to pay a certain sum of money only to, or to the order of a certain person or to the bearer of the – instrument”.

Question 3.
List three characteristics of a Promissory Note.
Answer:
Characteristics of a Promissory Note:

  1. A promissory note must be in writing.
  2. The promise to pay must be unconditional.
  3. It must be signed by the maker.

Question 4.
What is meant by a cheque?
Answer:
According to section 6 of the Negotiable Instruments Act, 1881 defines a cheque as “a bill of exchange drawn on a specified banker and not expressed to be payable otherwise than on demand”.

Question 5.
Define Endorsement.
Answer:
“When the maker or holder of a negotiable instrument signs the name, otherwise that as such maker for the purpose of negotiation, on the back or face thereof, or on a slip of paper annexed thereto.”

III. Short Answer Questions

Question 1.
Explain the nature of a Negotiable Instrument.
Answer:
A negotiable instrument is transferable from one person to another without any formality, such as affixing stamp, registration, etc. When the instrument is held by holder in due course in the process of negotiation, it is cured of all defects in the instrument with respect to ownership. Though a bill, a promissory note or a cheque represents a debt, the transferee is entitled to sue on the instrument in his own name in case of dishonour, without giving notice to the debtor that he has become its holder.

Question 2.
Distinguish between Negotiability and Assignability.
Answer:

Samacheer Kalvi 12th Commerce Solutions Chapter 22 The Negotiable Instruments Act 1881

Question 3.
What are the characteristics of a bill of exchange?
Answer:
Characteristics of a Bill of Exchange:

  1. A bill of exchange is a document in writing.
  2. The document must contain an order to pay.
  3. The order must be unconditional.
  4. The instrument must be signed by the person who draws it.
  5. The name of the person on whom the bill is drawn must .be specified in the bill itself.

Question 4.
Distinguish between Bill of Exchange and Promissory Note.
Answer:

Samacheer Kalvi 12th Commerce Solutions Chapter 22 The Negotiable Instruments Act 1881

IV. Long Answer Questions

Question 1.
Mention the presumptions of Negotiable Instruments.
Answer:
Presumptions of Negotiable Instrument:

  1. Every negotiable instrument is presumed to have been drawn and accepted for consideration.
  2. Every negotiable instrument bearing, a date is presumed to have been made or drawn on such a date.
  3. It is presumed to have been accepted within a reasonable time after the date and before its maturity.
  4. The transfer of a negotiable instrument is presumed to have been made before maturity.
  5. When a negotiable instrument has been lost, it is presumed to have been duly stamped.
  6. The holder of a negotiable instrument is presumed to be a holder in due course.

Question 2.
Distinguish a cheque and a bill of exchange.
Answer:

Samacheer Kalvi 12th Commerce Solutions Chapter 22 The Negotiable Instruments Act 1881

Question 3.
Discuss in detail the features of a cheque.
Answer:
A cheque is a negotiable instrument drawn on a particular banker.
Features:
(i) Instrument in Writings:
A cheque or a bill or a promissory note must be an instrument in writing. Though the law does not prohibit a cheque being written in pencil, bankers never accept it because of risks involved. Alternation is quite easy but detection is impossible in such cases.

(ii) Unconditional Orders:
The instrument must contain an order to pay money. It is not necessary that the word ‘order’ or its equivalent must be used to make the document a cheque. It does not cease to be a cheque just because the world ‘please’ is used before the word pay. Further the order must be unconditional.

(iii) Drawn on a Specified Banker Only:
The cheque is always drawn on a specified banker. A cheque vitally differs from a bill in this respect as latter can be drawn on any person including a banker. The customer of a banker can draw the cheque only on the particular branch of the bank where he has an account.

(v) A Certain Sum of Money Only:
The order must be for payment of only money. If the banker is asked to deliver securities, the document cannot be called a cheque. Further, the sum of money must be certain.

(v) Payee to be Certain:
The cheque must be made payable to a certain person or to the order of a certain person or to the bearer of the instrument. The word, person includes corporate bodies, local authorities, associations, holders of office of an institution etc.

(vi) Signed by the Drawer:
The cheque is to be signed by the drawer. Further, it should tally with specimen signature furnished to the bank at the time of opening the account.

Question 4.
What are the requisites for a valid endorsement?
Answer:
If an endorsement is to be valid, it must possess the following requisites:

  1. Endorsement is to be made on the face of the instrument or on its back.
  2. When there is no space for making further endorsements a piece of paper can be attached
  3. Endorsement for only a part of the amount of the instrument is invalid.
  4. Endorsement is complete only when delivery of the instrument is made.
  5. Signing in block letters does not constitute regular endorsement.
  6. If the payee is an illiterate person, he can endorse it by affixing his thumb impression on the instrument.

Question 5.
Explain the different kinds of endorsements. .
Answer:
When the person signs on the back of the instrument to transfer his interest, it is known as endorsement. The endorsement are of various types:
(i) Blank or general endorsement:
When the endorser puts his mere signature on the back of an instrument without mentioning the name of the person to whom the endorsement is made, it is called Blank Endorsement

(ii) Endorsement in full or special endorsement:
If the endorser, in addition to his signature, mentions the name of the person to whom it is endorsed, is known as endorsement in full or special endorsement.

(iii) Conditional endorsement:
When the endorser of a negotiable instrument makes his liability dependent upon the happening of an event which may or may not happen, it is called conditional endorsement,

(iv) Restrictive endorsement:
When an endorsement restricts or prohibits further negotiability of the instrument, it is called Restrictive Endorsement.

(v) Partial Endorsement:
Where the endorsement seeks to transfer only a part of the amount payable under the instrument, the endorsement is called Partial Endorsement.

Samacheer Kalvi 12th Commerce The Negotiable Instruments Act 1881 Additional Questions and Answers

I. Choose the Correct Answer

Question 1.
A bill of exchange drawn on a specified banker is
(a) promissory note
(b) cheque
(c) hundi
(d) share
Answer:
(b) cheque

Question 2.
Grace days allowed to a Bill of exchange for calculation of due date is
(a) 4
(b) 10
(c) 3
(d) 5
Answer:
(c) 3

We as a team believe the information prevailing regarding the Tamilnadu State Board Solutions for 22th Commerce Chapter 22 The Negotiable Instruments Act 1881 Questions and Answers has been helpful in clearing your doubts to the fullest. For any other help do leave us your suggestions and we will look into them. Stay in touch to get the latest updates on Tamilnadu State Board Solutions for different subjects in the blink of an eye.

Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life

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Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life

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Samacheer Kalvi 12th Chemistry Chapter 15 Chemistry in Everyday Life Textual Evaluation Solved

Samacheer Kalvi 12th Chemistry Chemistry in Everyday Life Multiple Choice Questions

Question 1.
Which of the following is an nalgesic?
(a) Streptornycin
(b) Chloromycetin
(c) Asprin
(d) Penicillin
Answer:
(c)Asprin

Question 2.
Dettol is the mixture of ……………
(a) Chioroxylenol and bithionol
(b) Chioroxylenol and a – terpineol
(c) phenol and iodine
(d) terpineol and bithionol
Answer:
(b) Chioroxylenol and a – terpineol

Question 3.
Antiseptics and disinfectants either kill or prevent growth of microorganisms. Identify which of the following statement is not true.
(a) dilute solutions of boric acid and hydrogen peroxide are strong antiseptics
(b) Disinfectants harm the living tissues
(c) A 0.2% solution of phenol is an antiseptic while 1% solution acts as a disinfectant
(d) Chlorine and iodine are used as strong disinfectants
Answer:
(a) dilute solutions of boric acid and hydrogen peroxide are strong antiseptics

Question 4.
Saccharin, an artificial sweetener is manufactured from ……………..
(a) cellulose
(b) toluene
(c) cyclohexene
(d) starch
Answer:
(b) toluene

Question 5.
Drugs that bind to the receptor site and inhibit its natural function are called …………….
(a) antagonists
(b) agonists
(c) enzymes
(d) molecular targets
Answer:
(a) antagonists

Question 6.
Aspirin is a/an ……………..
(a) acetylsalicylic acid
(b) benzoyl salicylic acid
(c) chlorobenzoic acid
(d) anthranilic acid
Answer:
(a) acetylsalicylic acid

Question 7.
Which one of the following structures represents nylon 6,6 polymer?
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-1
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-2

Question 8.
Natural rubber has ………..
(a) alternate cis – and trans – configuration
(b) random cis – and trans-configuration
(c) all cis – configuration
(d) all trans – configuration
Answer:
(c) all cis – conflguration

Question 9.
Nylon is an example of …………..
(a) polyamide
(b) polythene
(c) polyester
(d) poly saccharide
Answer:
(a) polyamide

Question 10.
Terylene is an example of …………..
(a) polyamide
(b) polythene
(c) polyester
(d) poly saccharide
Answer:
(c) polyester

Question 11.
Which is the monomer of neoprene in the following?
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-3
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-4

Question 12.
Which one of the following is a bio-degradable polymer?
(a) HDPE
(b) PVC
(c) Nylon 6
(d) PHBV
Answer:
(d) PHBV

Question 13.
Non stick cook wares generally have a coating of a polymer, whose monomer is ………….
(a) ethane
(b) prop – 2 – enenitrile
(c) chioroethene
(d) 1, 1, 2, 2 – tetrafluoroethane
Answer:
(d) 1, 1, 2, 2 – tetrafluoroethane

Question 14.
Assertion: 2 – methyl – I ,3 – butadiene is the monomer of natural rubber
Reason: Natural rubber is formed through aniònic addition polymerisation.
(a) If both assertion and reason are true and reason is the correct explanation of assertion
(b) if both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(c) assertion is true but reason is false

Question 15.
An example of antifertility drug is ………….
(a) novesirol
(b) seldane
(c) salvarsan
(d) Chioramphenicol
Answer:
(a) novestrol

Question 16.
The drug used to induce sleep is …………..
(a) paracetamol
(b) bithional
(c) chioroquine
(d) equanil
Answer:
(d) equanil

Question 17.
Which of the following is a co – polymer?
(a) Orlon
(b) PVC
(c) Teflon
(d) PHBV
Answer:
(d) PHBV

Question 18.
The polymer used in making blankets (artificial wool) is ……………
(a) polystyrene
(b) PAN
(c) polyester
(d) polythene
Answer:
(b) PAN

Question 19.
Regarding cross-linked or network polymers, which of the following statement is incorrect?
(a) Examples are Bakelite and melamine
(b) They are formed from bi and tri-functional monomers
(c) They contain covalent bonds between various linear polymer chains
(d) They contain strong covalent bonds in their polymer chain
Answer:
(d) They contain strong covalent bonds in their polymer chain

Question 20.
A mixture of chioroxylenol and terpinecol acts as ……………
(a) antiseptic
(b) antipyretic
(c) antibiotic
(d) analgesic
Answer:
(a) antiseptic

II. Answer the following questions

Question 1.
Which chemical is responsible for the antiseptic properties of dettol?
Answer:
1. Chloroxylenol and

2. Terpineol are the chemicals responsible for the antiseptic properties of dettol. But among these two, chloroxylenol plays more important role. Chioroxylenol is an antiseptic and disinfectant which is used for skin disinfection and cleaning surgical instruments.

Question 2.
What are antibiotics?
Answer:
Antibiotics is a chemical substance produced by one microorganism, that selectively inhibits the growth of another micro organism. Example : penicillins and cephalosporins.

Question 3.
Name one substance which can act as both analgesic and antipyretic.
Answer:
Aspirin (acetylsalicylic acid) is a chemical substance which lowers body temperature (to normal) and also reduces body pain. Therefore it acts as both antipyretic and analgesic.

Question 4.
Write a note on synthetic detergents.
Answer:
1. Synthetic detergents are formulated products containing either sodium salts of alkyl hydrogen suiphates or sodium salts of long chain alkyL benzene suiphonic acids.

2. Synthetic detergents are three types. They are

  • anionic detergents – sodium lauryl sulphate.
  • Cationic detergents – n – hexadecyltrimethyl ammonium chloride.
  • Non-ionic detergents – Pentaerythrityl stearate.

3. Synthetic detergents can be used even in hard water, while soaps cannot be used in hard water.

4. The cleansing action of detergents are similar to the cleansing action of soaps.

5. When detergents are dissolved in water its hydrocarbon part attaches itself to grease and oil particles. Whereas its ionic part remains attached to water. Therefore when dirty clothes are agitated in solution of detergents then dirty particles sticks to the hydrocarbon part of detergents and at the same time the water loving ionic part pulls away this dirt from clothes.

Question 5.
How do antiseptics differ from disinfectants?
Answer:
An tisep tics

  1. Antiseptics are chemical substance which prevent the growth of micro organizers and may even kill them but are not harmful to living tissues.
  2. They are generally applied to living tissues such as wounds, cuts bulks and diseased surfaces.
  3. All the antiseptics are disinfectants.
  4. They are not ingested or swallowed.
  5. e.g., Povidone – iodine, Benzalkonium – Chloride

Disinfectants

  1. Disinfectants are chemical substances which kill microorganism or stop their growth but are harmful to human tissues.
  2. Disinfectants are applied to inanimated objects such as floors, drainage system, instruments etc.
  3. All the disinfectants are not antiseptics.
  4. They can be injected or swallowed.
  5. e.g. Alcohol, chlorine compunds.

Question 6.
What are food preservatives?
Answer:
Food preservatives are chemical substances are capable of inhibiting, retarding or arresting the process of fermentation, acidification or other decomposition of food by growth of micro organisms.
Examples:

  1. Acetic acid is used mainly as a preservative for preparation of pickles.
  2. Sodium meta suiphite is used as preservative for fresh vegetables and fruits.
  3. Sodium benzoate is used as preservative for juices.

Question 7.
Why do soaps not work ¡n hard water?
Answer:
Soaps are sodium or potassium salts of long – chain falty acids. Hard water contains calcium and magnesium ions. When soaps are dissolved in hard water, these ions displace sodium or potassium from insoluble calcium or magnesium salts of fatty acids. These insoluble salts separate as scum.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-5
This is the reason why soaps do not work in hard water.

Question 8.
What are drugs? How are they classified?
Answer:
A drug is a substance that is used to modify or explore physiological systems or pathological states for the benefits of the recipient. it is used for the purpose of diagnosis, prevention cure or relief of a disease.

Classification of drugs:
Drugs care classified based on their proportions such as chemical structure, pharmacological effect, target system, site of action etc.

1. Classification based on the chemical structure:
In this classification, drugs with a common chemical skelton are classified into a single group. For example, ampicillin, amoxicillin, methiceillin etc. all have similar structure and are classified into a single group called penicillin.

Similarly we have other group of drugs such as opiates, steroids, catecholamines etc. Compounds having similar chemical structure are expected to have similar chemical properties. However, their biological actions are not always similar.

2. Classification based on the pharmacological effect:

  • In this classification, the drugs are grouped based on their biological effect that they produce on the recipient.
  • For example, the medicines that have the ability to kill the pathogenic bacteria are grouped as antibiotics.
  • This kind of grouping will provide the full range of drugs that can be used for a particular disease.

3. Classification based on the target system:

  • In this classification, the drugs are grouped based on the biological system (or) process that they target in the recipient.
  • This classification is more specific than the pharmacological classification.
  • For example, the antibiotics streptomycin and erythromycin inhibit the protein synthesis (target) in bacteria and are classified in a same group. However their mode of action is different.
  • Streptomycin inhibits the initiation of protein synthesis, while erythromycin prevents the incorporation of new aminoacids to the protein.

4. Classification based on the site of action:

  1. The drug molecule interacts with biomolecules such as enzymes, receptors etc, which are referred as drug targets.
  2. We can classify the drug based on the drug target with which it binds.
  3. This classification is highly specific compared to the others. These compounds often have a common mechanism of action, as the target is the same.

Question 9.
How the tranquilizers work in body?
Answer:

  1. They are neurologically active drugs.
  2. Tranquilizer acts on the central nervous system by blocking the neurotransitter dopamine in the brain.
  3. This drug is used for treatment of stress anxiety, depression, sleep disorders and severe mental diseases like schizophrenia.

 

Question 10.
Write the structural formula of aspirin.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-6

Question 11.
Explain the mechanism of cleansing action of soaps and detergents.
Answer:
Mechanism of cleansing action of soaps and detergents:
1. The cleansing action of both soaps and detergents from their ability to lower the surface tension of water, to emulsify oil or grease and to hold them in a suspension in water.

2. This ability is due to the structure of soaps and detergents.

3. In water a sodium soap dissolves to form soap anions and sodium cations. For example, the following chemical equation shows the ionisation of sodium palmitate.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-7

4. A soap anion consists of a long hydrocarbon chain with a carboxylate group on one end. The hydrocarbon chain, which is hydrophobic, is soluble in oils or grease. The ionic part is the carboxylate group which is hydrophilic, is soluble in water.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-8

5. In water, detergent dissolves to form detergent anions and sodium cations. For example the following chemical equations show the ionisation of sodium alkyl sulphate and sodium alkyl benzene sulphate.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-9

6. The following explains the cleansing action of a soap or detergent on a piece of cloth with a greasy stain.

  1. A soap or detergent anion consists of a hydrophobic part and a hydrophilic part.
  2. Soap or detergent reduces the surface tension of water. Therefore the surface of the cloth is wetted thoroughly.
  3. The hydrophobic parts of the soap or detergents anions are soluble in grease.
  4. The hydrophilic parts of the anions are soluble in water.
  5. Scrubbing or mechanical agitation helps to pull the grease away from the cloth and the grease is broken into smaller droplets.
  6. Repulsion between the droplets causes the droplets to be suspended in water, fonning an emulsion.
  7. Thus the droplets do not coagulate or,redeposit on the cloth. Rinsing washes away the droplets.

Question 12.
Which sweetening agent are used to prepare sweets for a diabetic patient?
Answer:
Artificial sweetening agents such as,

  1. saccharin
  2. Alitame
  3. Aspartame

can be used in preparing sweets for diabetic patients.

Question 13.
What are narcotic and non – narcotic drugs. Give examples.
Answer:
1. Narcotic drug is an addictive drug that reduces pain, induces sleep and may alter mood or behaviour. Example: Morphine and codeine.

2. Non – narcotic drug are chemical substance (medications) used to control pain and inflammation. They are available at drugstores without a prescription or by prescription when given at higher doses. Example: Acetaminophen and paracetamol.

Question 14.
What are anti fertility drugs? Give examples.
Answer:
Artificially drugs are chemical substances which suppress the action of hormones that promote pregnancy. These drugs actually reduce the chances of pregnancy and act as a protection. Antifertility drugs are made up of derivatives of synthetic progesterone or a combination of derivatives of oestrogen and progesterone. Example : Ethynylestradiol, menstranol and norcthynodrel etc.

Question 15.
Write a note on co – polymer.
Answer:

  1. A polymer containing two or more different kinds of monomer units is called a co-polymer.
  2. Co – polymers have properties quite different from the homopolymers.
  3. The structural units of co-polymers are derived from the different monomers may be present in regular, alternation or in random order or strings of several units of one kind may alternate with strings of another.
  4. For example, Buna – S, Buna – N, Nylon – 6,6 etc. Buna – S contains styrene and butadiene monomer units.

Question 16.
What are bio degradable polymers? Give examples.
Answer:

  1. The materials that are readily decomposed by micro organisms in the environment are called biodegradable.
  2. Natural polymers degrade on their own after certain period of time but the synthetic polymers do not.
  3. The biopolymers which disintergrates by themselves in biological systems during a
  4. certain period of time by enzymatic hydrolysis and to some extent by oxidation are called biodegradable polymers.

Examples:

  1. Polyhydroxy butyrate (PHB)
  2. Polyhydroxy butyrate – co – hydroxyl valerate (PHBV)
  3. Polylactic acid (PLA)

Question 17.
How is terylene prepared?
Answer:
The monomers are ethylene glycol and terephthalic acid or dimethýlterephthalate. When these monomers are mixed and heated at 500K in the presence of zinc acetate and antimonytrioxide catalyst, terylene (or dacron) is formed.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-10

Question 18.
Write a note on vulcanization of rubber.
Answer:

  1. Natural rubber is very soft and brisky. it has high water absorption capacity and low tensile strength. Its properties can be improved by a process called vulcanization.
  2. Natural rubber is mixed with 3 – 5% sulphur and heated at 100 – 150°C causes cross linking of the cis – 1, 4 polyisoprene chains through disuiphide ( – S – S – ) bonds.
  3. The physical properties of rubber can be altered by controlling the amount of sulphur that is used for vulcanization. When 3 to 10% sulphur is used the resultant rubber is somewhat harder but flexible.
  4. Following properties of rubber CH3

Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-11

Question 19.
Classify the following as linear, branched or cross linked polymers …………………..

  1. Bakelite
  2. Nylon
  3. polvthene

Answer:

  1. Bakelite – cross linked polymer
  2. Nylon – Linear polymer
  3. Polythene – Linear polymer

Question 20.
Differentiate thermoplastic and thermosetting.
Answer:
Difference between thermoplastic and thermosetting:
Thermoplastic

  1. They soften on heating and harden on cooling, and they can be remoulded.
  2. They consists of linear long çhain polymers and low molecular weights polymers.
  3. All the polymer chains are held together by weak Vanderwaals forces.
  4. They are weak, soft and less brittle.
  5. They are formed by additon polymerisation
  6. They are soluble in organic solvents.
  7. Example: PVC, polythene, polystrene etc.

Thermosetting

  1. They do not soften on heating and they cannot be remoulded.
  2. The consist of three dimensional network structure and high molecular weight polymers.
  3. All the polymer chains are linked by strong covalent.
  4. They are strong, hard and more brittle.
  5. They are formed by condensation polymerisation.
  6. They are insoluble in organic solvents.
  7. Example: Bakelite, melamine etc.

Samacheer Kalvi 12th Chemistry Chemistry in Everyday Life Additional Questions

Samacheer Kalvi 12th Chemistry Chemistry in Everyday Life 1 mark Questions and Answers

I. Choose the best answer.

Question 1.
The substance that is used to modify physiological system for the benefit of the recipient is called …………….
(a) a drug
(b) a dye
(c) a food preservative
(d) soap
Answer:
(a) a drug

Question 2.
Which one interacts with macromolecular targets such as proteins to produce a therapeutic and useful biological response?
(a) Detergent
(b) cleansing agent
(c) medicine
(d) food preservative
Answer:
(c) medicine

Question 3.
The ratio between the maximum tolerated dose of a drug and a minimum curative dose is called …………..
(a) iso electric point
(b) therapeutic index
(c) critical point
(d) iso thermal point
Answer:
(b) therapeutic index

Question 4.
Which one of the following does not belong to penicillin group?
(a) Ampicillin
(b) Amoxicillin
(c) catecholamine
(d) mithicillin
Answer:
(c) catecholamine

Question 5.
Which of the following does belongs to penicillin group drugs?
(a) Mithicillin
(b) opiates
(c) steroids
(d) catecholamine
Answer:
(a) Mithicillin

Question 6.
Which one of the following is an antibiotic?
(a) erythromycin
(b) atenolol
(c) amlodipine
(d) propranolol
Answer:
(a) erythromycin

Question 7.
Which one of the following is not an antibiotic?
(a) amoxicillin
(b) cefixime
(c) amlodipine
(d) ampiciflin
Answer:
(c) amlodipine

Question 8.
Which one of the following is an example tbr antihypertensive drug?
(a) atenolol
(b) amoxicillin
(c) cefixime
(d) tetracycline
Answer:
(a) atenolol

Question 9.
Which of the following does not belongs to antihypertensive drug?
(a) atenolol
(b) amlodipine
(c) propranolol
(d) erythromycin
Answer:
(d) erythromycin

Question 10.
Which one of the following inhibits the initiation of protein synthesis?
(a) streptomycin
(b) erythromycin
(c) atenolol
(d) amlodipine
Answer:
(a) streptomycin

Question 11.
Which one of the following prevents the incorporation of new amino acids to the protein?
(a) atenolol
(b) streptomycin
(c) erythromycin
(d) tetracycline
Answer:
(c) erythromycin

Question 12.
Which one of the following inhibits the bacterial growth?
(a) p – amino benzoic acid
(b) sulphanilamide
(c) folic cid
(d) sodium benzoate
Answer:
(b) sulphanilamide

Question 13.
Which of the following is needed by many bacteria to produce folic acid?
(a) PABA
(b) DHPS
(c) TNB
(d) GTN
Answer:
(a) PABA

Question 14.
Which of the following is called PABA?
Answer:
(a) p – nitro benzanilic acid
(b) p – amino butyric acid
(c) p – amino benzene suiphonic acid
(d) p – amido benzene suiphonyl chloride
Answer:
(c) p – amino benzene suiphonic acid

Question 15.
Which one of the following binds to the receptor site should inhibit its natural function?
(a) antacids
(b) antioxidant
(c) antibiotics
(d) antagonists
Answer:
(d) antagonists

Question 16.
Which of the following is used in the reduced sleepiness?
(a) caffeine
(b) morphine
(c) suiphanilide
(d) p – aminobenzene sulphonic acid
Answer:
(a) caffeine

Question 17.
Which one of the following is used as painkiller?
(a) lodoform
(b) chloropicrin
(c) morphine
(d) coffeine
Answer:
(c) morphine

Question 18.
Which of the following is not an example of antacid?
(a) Histamine
(b) cimetidine
(c) ranitidine
(d) erythromycin
Answer:
(d) erythromycin

Question 19.
Which one of the following is used as an antacid?
(a) magnesium hydroxide
(b) aluminium hydroxide
(c) ranitidine
(d) all the above
Answer:
(d) all the above

Question 20.
Which one of the following is used to treat stress, anxiety, depression, sleep disorder and schizopherenia?
(a) Tranquilizer
(b) antibiotic
(c) analgesic
(d) opioids
Answer:
(a) Tranquilizer

Question 21.
Which one of the following is an example for tranquilizer?
(a) cimetidine
(b) diazepam
(c) histamine
(d) PABA
Answer:
(b) diazepam

Question 22.
Identify the medine that is used to treat stress, anxiety. depression and schizophrenia.
(a) valium
(b) cimetidinc
(c) chiorofom
(d) adenosine
Answer:
(a) valium

Question 23.
Which one of the following is used to reduce fever and prevent platelet coagulation?
(a) antibiotic
(b) antiseptic
(c) antioxidant
(d) antipyretic
Answer:
(d) antipyretic

Question 24.
Which one of the following is an anti inflamatory drug?
(a) morphine
(b) coheinc
(c) aspirin
(d) histidine
Answer:
(c) aspirin

Question 25.
Which one of the following is used to cure headache, muscle strain, arthritis?
(a) acetaminophen
(b) ibuprofen
(c) aspirin
(d) all the above
Answer:
(d) all the above

Question 26.
Which one of the foLlowing is used in the prevention of heart attacks?
(a) aspirin
(b) ibuprofen
(c) paracetamol
(d) morphine
Answer:
(a) aspirin

Question 27.
Which one of the following is an example of an àntipyretic?
(a) acetyl salicylic acid
(b) methyl salicylate
(c) paraldehyde
(d) diethyl ether
Answer:
(a) acetyl salicylic acid

Question 28.
Which one of the following is a non steroidal anti inflammatory drug?
(a) aspirin
(b) morphine
(c) haloperidol
(d) ibuprofen
Answer:
(d) ibuprofen

Question 29.
Which of the following is a major tranquilizer?
(a) diazepam
(b) valium
(c) clozapine
(d) alprazolm
Answer:
(c) clozapine

Question 30.
Which of the following is a minor tranquilizer?
(a) haloperidol
(b) clozapine
(c) morphine
(d) valium
Answer:
(d) valium

Question 31.
Consider the following statements
(i) Tranquilizers act on the central nervous system by blocking the neurotransmitter dopamine in the brain.
(ii) Histamines stimulate the secretion of HCI by activating the receptor in the stomach wall.
(iii) The antibiotic cimetidine inhibits the bacterial growth.
Which of the above statement is/are not correct?
(a) (i) only
(b) (¡) & (ii)
(c) (iii) only
(d) (ii) only
Answer:
(c) (iii) only

Question 32.
Consider the following statements
(i) Acetaminophen reduces fever by causing the hypothalamus to override a prostaglandin
(ii) opioids relieve pain and produce sleep and are addictive
(iii) Aspirin finds useful in the pain of terminal cancer.
Which of the above statement is/arc not correct?
(a) (i) only
(b) (ii) only
(c) (ii) & (iii)
(d) (iii) only
Answer:
(d) (iii) only

Question 33.
Which of the following are addictive and poisonous drug?
(a) ibuprofen
(b) aspirin
(c) morphine
(d) paracetamol
Answer:
(c) morphine

Question 34.
Which of the following are used for post operative pain and pain of terminal cancer?
(a) morphine, codeine
(b) ibuprofen, aspirin
(c) methyl salicylate, salicylic acid
(d) histidine, ranitidine
Answer:
(a) morphine, codeine

Question 35.
Which one of the following is an local anaesthetic?
(a) lidocaine
(b) Propofol
(c) iso flurane
(d) ibuprofen
Answer:
(a) lidocaine

Question 36.
Which one of the following is an example of general anaesthetic?
(a) propofol
(b) isoflurane
(c) ranitidine
(d) omeprazole
Answer:
(b) isoflurane

Question 37.
Identify the intraveneous general anaesthetics?
(a) milk of magnesia
(b) lidocaine
(c) omeprazole
(d) iso fharane
Answer:
(d) iso fharane

Question 38.
Which one of the following is an inhalational general anaesthetic?
(a) procain
(b) iso fiurane
(c) lidocaine
(d) rabeprazole
Answer:
(b) iso fiurane

Question 39.
Which one of the following is an antacid?
(a) omeprazole
(b) rabeprazole
(c) milk of magnesia
(d) all the above
Answer:
(d) all the above

Question 40.
Consider the following statements.
(i) Propofol cause a controlled and reversible loss of consciousness by affecting central nervous system.
(ii) Ibuprofen is used for major surgical procedures.
(iii) Lidocaine is used to relieve burning sensation in the chest / throat area.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) & (ii)
(c) (ii) & (iii)
(d) (i) & (iii)
Answer:
(c) (ii) & (iii)

Question 41.
Which one of the following is not an antacid?
(a) propofol
(b) ranitidine
(c) omeprazole
(d) rabeprazole
Answer:
(a) propofol

Question 42.
Which one of the following is used to provide relief from the allergic effects?
(a) cetrizine
(b) ampicillin
(c) erythromycin
(d) milk of magnesia
Answer:
(a) cetrizine

Question 43.
Which one of the following inhibits bacterial cell wall biosynthesis?
(a) eryLhromycin
(b) azithromycin
(c) penicillin
(d) cetrizine
Answer:
(c) penicillin

Question 44.
Which of the following is used to treat respiratory tract infections, genital, gastrointestinal tract and skin infections?
(a) ampicillin
(b) penicillin
(c) terfenadine
(d) azithromycin
Answer:
(d) azithromycin

Question 45.
Which one of the following is used to treat urinary tract infection and respiratory infections?
(a) doxycycline
(b) karamycin
(c) ciprolloxacin
(d) ibuprofen
Answer:
(c) ciprolloxacin

Question 46.
Which of the following is used in the treatment of cholera, acne vulgaris?
(a) fluoro quinolone
(b) aminoglycosides
(c) tetracycline
(d) macrolides
Answer:
(c) tetracycline

Question 47.
Which one of the following is used to treat infections caused by gram negative bacteria?
(a) kanamycin
(b) gentamycin
(c) neomycin
(d) all the above
Answer:
(d) all the above

Question 48.
Which one of the following inhibits bacterial enzyme DNA gyrase?
(a) doxy cycline
(b) kanamycin
(c) ciprofloxacin
(d) aspirin
Answer:
(c) ciprofloxacin

Question 49.
Which one of the following is an antiseptic?
(a) Hydrogen peroxide
(b) alcohol
(c) menstranol
(d) chlorine compounds
Answer:
(a) Hydrogen peroxide

Question 50.
Which one of the following is used to reduce the risk of infection during surgery?
(a) povidone – iodine
(b) ethynyles tradiol
(c) norethindrone
(d) acetyl salicylic acid
Answer:
(a) povidone – iodine

Question 51.
Consider the following statements
(i) Oestrogen, menstranol are synthetic hormones that suppresses ovulation / fertilisation,
(ii) Norethindrone used in birth control pills.
(iii) Chlorine compounds are used to reduce the risk of infection during surgery.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) & (iii)
(c) (iii) only
(d) (i) & (iii)
Answer:
(c) (iii) only

Question 52.
Which one of the following is used as a preservative for the preparation of pickles and preservation of vegetables?
(a) sodium acetate
(b) acetic acid
(c) sodium carbonate
(d) salicylic acid
Answer:
(b) acetic acid

Question 53.
Which one is used as preservatives for fresh vegetables and fruits?
(a) Palmitic acid
(b) Palm oil
(c) sodium meta suiphite
(d) sulphur dioxide
Answer:
(c) sodium meta suiphite

Question 54.
Which one of the following is used as an emulsifier?
(a) sodium meta suiphite
(b) sucrose ester of palmiticacid
(c) sodium benzoate
(d) sodium bi carbonate
Answer:
(b) sucrose ester of palmiticacid

Question 55.
Which method is used to preserve food?
(a) pasteurisation & irration
(b) chilling and freezing
(c) drying and dehydration
(d) all the above
Answer:
(d) all the above

Question 56.
Which one of the following act as an antioxidant?
(a) Palmîtic acid
(b) butyl hydroxy toluene
(c) sodium benzoate
(d) Ascorbic acid
Answer:
(b) butyl hydroxy toluene

Question 57.
Consider the following statements.
(i) Addition of vitamins and minerals reduces the mall nutrient.
(ii) Flouring agents reduces the aroma of the food.
(iii) Antioxidants produce the formation of potentially toxic oxidation products of lipids.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (ii) & (iii)
(d) (i) & (iii)
Answer:
(c) (ii) & (iii)

Question 58.
Which of the following is not a sugar substituent?
(a) Sorbitol
(b) mannitol
(c) xylitol
(d) cresol
Answer:
(d) cresol

Question 59.
Which of the following is not a sugar substituent?
(a) Butyl hydroxy toluene
(b) Butylated hydroxy anisole
(c) Aspartame
(d) Ascorbic acid
Answer:
(c) Aspartame

Question 60.
Identify the artificial sweeteners.
(a) Saccharin, sucralose
(b) culutaric acid, glycollic acid
(c) BHT, BHA
(d) GTN, TNG
Answer:
(a) Saccharin, sucralose

Question 61.
Glyceryl ester of long chain fatty acids are called …………..
(a) soap
(b) detergent
(c) antiseptic
(d) antibiotic
Answer:
(a) soap

Question 62.
Which one of the following describes the quality of soap?
(a) TFT value
(b) TFM value
(c) PPM value
(d) TFP value
Answer:
(b) TFM value

Question 63.
Sodium salt of long chain allyl benzene sulphomc acids are called ……………….
(a) soap
(b) detergent
(c) disinfectant
(d) antiseptic
Answer:
(b) detergent

Question 64.
Which one of the following is an anionic detergent?
(a) n – hexa decyl tri methyl ammonium chloride
(b) Peifla erythntyl stearate
(c) Sodium lauryl sulphate
(d) 3 – hydroxy – 2, 2 bis (hydroxy methyl) propyl heptonoate
Answer:
(c) Sodium lauryl sulphate

Question 65.
Which of the following is an example of catìonic detergent?
(a) Sodium lauryl sulphate
(b) sodium pahnitate
(c) sodium dodecyl benzene suiphonate
(d) n – hexa decyl timethyl ammonium chloride
Answer:
(d) n – hexa decyl timethyl ammonium chloride

Question 66.
Which one of the following is an example of non-ionic detergent?
(a) sodium lauryl sulphate
(b) n – hexa decyl trimethyl ammonium chloride
(c) Penta erythrityl stearate
(d) N, N, N – trimethyl hexa decan – 1 – aminium chloride
Answer:
(c) Penta erythrityl stearate

Question 67.
Which one of the following is a natural polymer?
(a) cellulose, silk
(b) PVC, Polythene
(c) Buna – N, Buna – S
(d) Bakelite, Nylon 6,6
Answer:
(a) cellulose, silk

Question 68.
Which one of the following is a synthetic rubber?
(a) Neoprene
(b) cellulose
(c) silk
(d) poly isoprene
Answer:
(a) Neoprene

Question 69.
Which one of the following is a semisynthetic polymer?
(a) poly isoprene
(b) viscose rayon
(c) nylon
(d) terylene
Answer:
(b) viscose rayon

Question 70.
Which one of the following is not a cross linked polymer?
(a) poly propylene
(b) bakelite
(c) melamine
(d) urea formaldehyde
Answer:
(a) poly propylene

Question 71.
Identify the thermo setting plastic?
(a) nylon 6, 6
(b) neoprene
(c) melamine
(d) bakelite
Answer:
(c) melamine

Question 72.
Which of the following is a thermoplastic?
(a) bakelite
(b) melamine
(c) urea formaldehyde
(d) polystrene
Answer:
(d) polystrene

Question 73.
Which one of the following is an elastomer?
(a) nylon 6,6
(b) terylene
(c) buna – S
(d) bakelite
Answer:
(c) buna – S

Question 74.
Which one of the following is an example for addition polymer?
(a) polyethylene
(b) PVC
(c) teflon
(d) all the above
Answer:
(d) all the above

Question 75.
Which one of the following is an example of condensation polymer?
(a) poly ethylene
(b) polyester
(c) PVC
(d) teflon
Answer:
(b) polyester

Question 76.
Which one of the following is not an additional polymer?
(a) poly ethylene
(b) PVC
(c) Nylon 66
(d) teflon
Answer:
(c) Nylon 66

Question 77.
Consider the following statements
(i) Nylon-6, 6 are polymerchains form fibres by hydrogen bonding.
(ii) Thermoplastic become hard on heating and soft on cooling and cannot be remoulded.
(iii) Cellulose and silk are synthetic polymers.
Which of the above statement is/are not correct? ,
(a) (i) only
(b) (ii) & (iii)
(c) (iii) only
(d) (i) & (iii)
Answer:
(b) (ii) & (iii)

Question 78.
Which one of the following is used as an free radical initiator in the preparation of polystrene?
(a) hydrogen peroxide
(b) methyl chloride
(c) Benzoyl peroxide
(d) Benzyl peroxide
Answer:
(c) Benzoyl peroxide

Question 79.
Which mechanism is followed in the synthesis of polystrene?
(a) free radical polymerisation
(b) cationic polymensation
(c) Anionic polymerisation
(d) SN1 mechanism

Question 80.
Which one of the polymer is used as insulation for cables, making toys?
(a) HDPE
(b) LDPE
(c) teflon
(d) orlon
Answer:
(b) LDPE

Question 81.
Which one of the following catalyst is used in the preparation of high density polyethylene?
(a) benzoyi peroxide
(b) zeigler natta catalyst
(c) ammonium per sulphate
(d) hydrogen peroxide
Answer:
(b) zeigler natta catalyst

Question 82.
Identify the zeiglar natta catalyst.
(a) TiCI4 + (C2H5)3AI
(b) (C2H5)4Pb + TiCl4
(c) AICl3 + HCI
(d) ZnCI2 + Cone. HCI
Answer:
(a) TiCI4 + (C2H5)3AI

Question 83.
Which of the following is used to make bottles and pipes?
(a) LDPE
(b) Terylene
(c) PVC
(d) HDPE
Answer:
(a) LDPE

Question 84.
Which polymer is used in preparing non-sticking utensils?
(a) orlon
(b) PAN
(c) teflon
(d) HDPE
Answer:
(c) teflon

Question 85.
Which one of the following is used as a substitute of wool for making blankets, sweaters?
(a) orlon
(b) terylene
(c) polyester
(d) nylon
Answer:
(a) orlon

Question 86.
What are the raw materials required for the manufacture of Nylon 6, 6?
(a) caprolactam + hydrazine
(b) adipic acid + hexa methylene diamine
(c) methanal + ammonia
(d) phenol + methanal
Answer:
(b) adipic acid + hexa methylene diamine

Question 87.
Which one of the following is not a condensation polymer?
(a) nylon 6, 6
(b) nylon 6
(c) polyethylene
(d) terylene
Answer:
(c) polyethylene

Question 88.
Which one is used in the manufacture of nylon-6?
(a) adipic acid + hexamethylene diamine
(b) succinic acid + hexamethylene tetramine
(c) ∈-amino carproic acid
(d) adipic acid + hexamethylene tetramine
Answer:
(c) ∈-amino carproic acid

Question 89.
Which one of the following is the other name of nylon 6, 6?
(a) poly urethane
(b) urotropine
(c) poly caprolactum
(d) poly hexametheylene adipamide
Answer:
(d) poly hexametheylene adipamide

Question 90.
Which one of the following is used in the manufacture of tyrecards fabrics?
(a) nylon 6, 6
(b) nylon 6
(c) orlon
(d) dacron
Answer:
(b) nylon 6

Question 91.
What are the raw materials required for the manufacture of terylene?
(a) ethylene glycol + terephthalic acid
(b) phthalic auhydride + phenol
(c) adipic acid + hexamethylene diamine
(d) phenol + methanal
Answer:
(a) ethylene glycol + terephthalic acid

Question 92.
Name the catalyst used in the preparation of terylene?
(a) zeiglar natta catalyst
(b) zincacetate + antimony oxide
(c) benzoyi peroxide
(d) ammonium persuiphate
Answer:
(b) zincacetate + antimony oxide

Question 93.
Which one of the following is used as glass reinforcing material in safety helmets?
(a) nylon
(b) bakelite
(c) terylene
(d) orlon
Answer:
(c) terylene

Question 94.
What are the raw materials required for the manufacture of bakelite?
(a) ethane 1, 2 – diol + benzene 1, 4 – dicarboxylic acid
(b) phenol + methanal
(c) adipic acid + hexamethylene diamine
(d) isoprene + methanal
Answer:
(b) phenol + methanal

Question 95.
Linear polymer of phenol formal dehyde is called
(a) novolac
(b) bakelite
(c) terylene
(d) orlon
Answer:
(a) novolac

Question 96.
Which one of the following is used to prepare combs and pens?
(a) navolac
(b) soft bakelite
(c) hard bakelite
(d) neoprene
Answer:
(a) navolac

Question 97.
Which one of the following thermo setting plastic is used in paints?
(a) melamine
(b) hard bakelite
(c) navolac
(d) soft bakelite
Answer:
(c) navolac

Question 98.
Which one of the following is used for making unbreakable crockery?
(a) phenol formal dehyde
(b) melamine formal dehyde
(c) urea formal dehyde
(d) navolac
Answer:
(b) melamine formal dehyde

Question 99.
What are the raw materials required to prepare Buna – S rubber?
(a) phenol + methanal
(b) melamine + methanal
(c) styrene + butadiene
(d) adipic acid + methanal
Answer:
(c) styrene + butadiene

Question 100.
Which one of the following element is used in vulcanization of rubber?
(a) oxygen
(b) nitrogen
(c) carbon
(d) sulphur
Answer:
(d) sulphur

Question 101.
Which one of the following is a natural rubber?
(a) Buna-S
(b) Buna-N
(c) cis – 1, 4 – poly isoprene
(d) neoprene
Answer:
(c) cis – 1, 4 – poly isoprene

Question 102.
The raw material is used in the manufacture of ieoprene?
(a) isoprene
(b) chloroprene
(c) 1, 3 – buta diene
(d) vinyl chloride
Answer:
(b) chloroprene

Question 103.
Which one of the following rubber is used in the manufacture of chemical container and conveyer belts?
(a) Buna – N
(b) neo prene
(c) Buna – S
(d) poly isoprene
Answer:
(b) neo prene

Question 104.
The raw materials required for the manufacture of Buna – N are …………..
(a) acrylonitrile + Buta – 1, 3 – diene
(b) chloro prene + buta – 1, 3 – diene
(c) terephthalic acid + ethane 1, 2 – diol
(d) phenol + methanal
Answer:
(a) acrylonitrile + Buta – 1, 3 – diene

Question 105.
Which of the following are required to prepare Buna – S?
(a) vinyl cyanide + 1, 3 – butadiene
(b) chioro prene + buta -1, 3 – diene
(c) buta – 1, 3 – diene + styrene
(d) isoprene + styrene
Answer:
(c) buta – 1, 3 – diene + styrene

Question 106.
Which of the following used in medical field such as surgical sutures, 1asma substitute?
(a) PHBV
(b) PLA
(c) PCE
(d) all the above
Answer:
(d) all the above

Question 107.
Which one of the following is not an example of biodegradable plastic?
(a) polyhydroxy butyrate
(b) poly glycollic acid
(c) polythene
(d) poly caprolactone
Answer:
(c) polythene

Question 108.
Which of the following is an example for bio degradable plastic?
(a) polystyrene
(b) poly vinyl chloride
(c) bakelite
(d) polylactic acid
Answer:
(d) polylactic acid

Question 109.
Which one of the following is used in orthopaedic devices and in controlled release of drugs?
(a) PHB
(b) PHBV
(c) PGA
(d) PLA
Answer:
(b) PHBV

Question 110.
Glycine and e-amino caproic acid polymenses to give …………..
(a) glycyl amine
(b) nylon 6, 6
(c) Nylon – 2 Nylon 6
(d) orlon
Answer:
(c) Nylon – 2 Nylon 6

Question 111.
Which one of the following is used in making automobiles and foot wear?
(a) Bun – S
(b) Buna – N
(c) natural rubber
(d) neoprene
Answer:
(a) Bun – S

Question 112.
Which one of the following is used as an insulator and making conveyor belts?
(a) terylene
(b) orlon
(c) neoprene
(d) Buna – N
Answer:
(c) neoprene

Question 113.
Which type of nylon is used in making brushes, synthetic fibres, parachute, ropes and carpets?
(a) nylon – 2
(b) nylon – 6
(c) nylon 6,6
(d) nylon – 2, nylon 6
Answer:
(c) nylon 6,6

Question 114.
Which one is used in making non-breakable cups and laminated sheets?
(a) bakelite
(b) urea formaldehyde
(c) PHBV
(d) teflon
Answer:
(b) urea formaldehyde

Question 115.
Which of the polymer is used in making fibres, safety belts, lyre cords and ropes?
(a) terylene
(b) orlon
(c) Nylon
(d) bakelite
Answer:
(a) terylene

Question 116.
Identify the monomer of nylon – 2.
(a) adipic + Hexamethylene tetramine
(b) caprolactam
(c) vinyl chloride
(d) chioroprene
Answer:
(b) caprolactam

Question 117.
Which of the following is a fibre?
(a) nylon
(b) neoprene
(c) PVC
(d) bakelite
Answer:
(a) nylon

Question 118.
Identify the food preservative which is most commonly used by food producers?
(a) sodium cloride
(b) sodium sulphate
(c) baking soda
(d) benzoic acid
Answer:
(a) sodium cloride

Question 119.
Which of the following act as an antiseptic and disinfectant respectiely?
(a) 0.2% phenol, 1% phenol
(b) 1% phenol, 0.2% phenol
(c) 2% phenol, 20% phenol
(d) 20% phenol, 2% phenol
Answer:
(a) 0.2% phenol, 1% phenol

Question 120.
Identify the narcotic which is used as an analgesic.
(a) phenol
(b) equanil
(c) morphine
(d) cetrizine
Answer:
(c) morphine

Question 121.
What type of drug pencillin is?
(a) anaesthetic
(b) antibiotic
(c) antipyretic
(d) analgesic
Answer:
(b) antibiotic

Question 122.
Ranitidine is used as an …………
(a) antioxidant
(b) antiseptic
(c) antacid
(d) antibiotic
Answer:
(c) antacid

Question 123.
Aspirin is chemically named as ………….
(a) methyl salicylate
(b) ethyl salicylate
(c) o – hydroxy benzoic acid
(d) acetyl salicylic acid
Answer:
(d) acetyl salicylic acid

Question 124.
Which of the following can be used an analgesic without causing addiction and any modification?
(a) morphine
(b) n – acetyl paraminophenol
(c) diazepam
(d) tetra hydro catenol
Answer:
(c) diazepam

Question 125.
Tranquilisers are substances used for the treatment of ……………..
(a) cancer
(b) AIDS
(c) mental diseases
(d) blood infection
Answer:

Question 126.
Which of the following represents a synthetic detergent?
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-12
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-13

Question 127.
Which of the following represents a soap?
(a) C17H35COOk
(b) C17H35COOH
(c) C15H31COOOH
(d) (C17H35COO)2Ca
Answer:
(a) C17H35COOk

Question 128.
Which of the following drug is an analgesic?
(a) iodex
(b) valium
(c) analgin
(d) quinine
Answer:
(c) analgin

Question 129.
An antipyretic is …………
(a) chioro quinine
(b) paracetamol
(c) morphine
(d) ranitidine
Answer:
(b) paracetamol

Question 130.
Streptomycin is effective in the treatment of ……………
(a) tuberculosis
(b) malaria
(c) typhoid
(d) cholera
Answer:
(a) tuberculosis

Question 131.
A drug effective in the treatment of pneumonia, bronchitis etc is ………………….
(a) streptomycin
(b) aspirin
(c) penicillin
(d) paracetamol
Answer:
(c) penicillin

Question 132.
The substances which affect the central nervous system and induce sleep are called ………………..
(a) tranquilizers
(b) analgesics
(c) antioxidants
(d) antipyretic
Answer:
(a) tranquilizers

Question 133.
The correct structure of the drug paracetamol is ……………
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-14
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-15

Quesiton 134.
Which of the following acts as an antioxidant in edible oils?
(a) Vitamin B
(b) Vitamin C
(c) Vitamin D
(d) Vitamin E
Answer:
(d) Vitamin E

Question 135.
Which of the following is an antidiabatic drug?
(a) insulin
(b) inulin
(c) chioroquine
(d) aspirin
Answer:
(a) insulin

Question 136.
Which of the following terms means pain killer?
(a) antibiotics
(b) analgesic
(c) antiseptic
(d) antioxidant
Answer:
(b) analgesic

Question 137.
The artificial sweetener containing chlorine that has the appearance and taste as the sugar and is stable at cooking temperature is …………….
(a) aspartame
(b) saccharin
(c) sucralose
(d) alitame
Answer:
(c) sucralose

Question 138.
The role of phosphate in detergent powder is …………..
(a) control pH level of the detergent water mixture
(b) remove Ca2+ and Mg2+ ions from water that causes hardness of water
(c) provide whiteness to the fabric
(d) more soluble in soft water
Answer:
(b) remove Ca2+ and Mg2+ ions from water that causes hardness of water

Question 139.
Which among the following is not an antibiotic?
(a) erythromycin
(b) oxytocin
(c) penicillin
(d) tetracycline
Answer:
(b) oxytocin

Question 140.
Commonly used antiseptic ‘dettoP is a mixture of …………..
(a) O – chloro phenozylenol + terpeneol
(b) O – cresol + terpenol
(c) phenol + terpeneol
(d) chioroxylenol + terpeneol
Answer:
(d) chioroxylenol + terpeneol

II. Fill in the blanks.

  1. The specific treatment of a disease using medicine is known as ……………..
  2. The drug which interacts with macro molecular targets such as proteins to produce a therapautic and useful biological response is called ………………
  3. ……………..is a substance that is used to modifS’ or explore physiological systems for the benefit of the recepient.
  4. Higher the value of …………….. safer is the drug
  5. The medicines that have ability to kill the pathogenic bacteria are grouped as ……………..
  6. Proteins which act as biological catalysts are called …………….. and those which are important for communication systems are called ……………..
  7. Many bacteria need …………….. in order to produce an important coenzyme, folic acid
  8. When adenosine binds to the adenosine receptors, it induces ……………..
  9. Morphine that used as a pain killer suppress the …………….. that causes pain.
  10. Histames stimulate the secretion of …………….. by activating the receptor in the stomach wall.
  11. …………….. acts on the central nervous system by blocking the neuro transmitter dopamine in the brain.
  12. …………….. reduce the pain without causing impairment of consciousness.
  13. …………….. are drugs that used to reduce fever and prevent platelet coagulation
  14. …………….. relieve pain and produces steeps and they are additive.
  15. …………….. neutralise the acid in the stomach that causes acidity.
  16. …………….. cause a controlled and reversible loss of consciousness by affecting central nervous SyStem.
  17. …………….. anaesthetics are often used for major surgical procedures.
  18. …………….. provide relief from allergic effects.
  19.  …………….. inhibits bacterial cell wall biosynthesis.
  20. …………….. inhibits bacterial enzyme DNA gyrase.
  21. …………….. stop or slow down the growth of microorganisms applied to living tissues.
  22. …………….. stop or slow down the growth of microorganisms used on inanimate objects.
  23. The substances which are not naturally a part of the food and added to improve the quality of food are called ……………..
  24. Flavouring agents added to food enhance the …………….. of the food.
  25. …………….. are substances which retard the oxidative deteriorations of food.
  26. Synthetic compounds which imprint a sweet sensation and possess no or negligible nutritional value are called ……………..
  27. Chemically soap is a …………….. or …………….. salt of higher fatty acids.
  28. …………….. is a sodium salt alkyl hydrogen sulphate or alkyl benzene suiphonic acid.
  29. The quality of soap is described interrns of …………….. and the …………….. quantity in the soap better is its quality
  30. …………….. become soft on heating and hard on cooling and they can be remoulded.
  31. …………….. donot become soft on heating but set to an infusible mass upon heating.
  32. In the manufacture of Teflon. the monomer used is ……………..
  33. …………….. is used as a substitute of wool for making blankets, sweaters
  34. …………….. is a monomer which polymerises to give nylon – 6.
  35. Para hydroxyl methyl phenols poíymerises to give a linear polymer called ……………..
  36. The monomer of natural rubber is ……………..
  37. For the vulcanization of natural rubber …………….. is used and heated to 100° 150°C.
  38. …………….. polymers are used in medical field such as surgical sutures, plasma substitute.
  39. A drug that binds to the receptor site should inhibit its natural function is called ……………..
  40. …………….. reduces fever by causing the hypothalamus to override a prostaglandin-induced increase in temperature.

Answer:

  1. chemotheropy
  2. medicine
  3. Drug
  4. therapeutic index
  5. antibiotics
  6. enzymes, receptors
  7. PABA
  8. sleepiness
  9. neuro transmitters
  10. HCI
  11. Tranquilizers
  12. Analgesics
  13. Antipyretic
  14. Narcotic Analgegics (or) opioids
  15. Antacids
  16. General anaesthetics
  17. Inhalational general
  18. Antihistamines
  19. Antimicrobials
  20. Fluoroquinolones
  21. Antiseptic
  22. Disinfectants
  23. food additives
  24. aroma
  25. Antioxidant
  26. artificial sweatness
  27. sodium, potassium
  28. Detergent
  29. TFM, TFM
  30. Thermoplastic
  31. Thermosetting
  32. tetra fluoroethylene
  33. orIon (or) PAN
  34. Caprolactam
  35. novolac
  36. cis – isoprene (OR) 2 – methyl buta- 1, 3 – diene
  37. sulphur
  38. Biodegradable
  39. antagonists
  40. Non steroidal anti inflammatory drugs (or) NSAIDS

III. Match the following

Question 1.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-16
Answer:
(a) 2 4 1 3

Question 2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-17
Answer:
(a) 2 4 1 3

Question 3.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-18
Answer:
(a) 4 3 1 2

Question 4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-19
Answer:
(a) 3 4 2 1

Question 5.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-20
Answer:
(a) 3 2 4 1

Question 6.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-21
Answer:
(a) 2 3 4 1

Question 7.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-22
Answer:
(b) 3 4 1 2

Question 8.

Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-23
Answer:
(a) 2 4 1 3

Question 9.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-64
Answer:
(a) 3 4 1 2

Question 10.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-65
Answer:
(a) 2 4 1 3

Question 11.

Answer:
(a) 2 4 1 3

Question 12.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-24
Answer:
(a) 3 4 1 2

Question 13.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-25
Answer:
(a) 4 1 2 3

Question 14.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-26
Answer:
(a) 2 4 1 3

Question 15.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-27
Answer:
(b) 2 3 4 1

Question 16.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-28
Answer:
(a) 2 3 4 1

IV. Assertion and reasons.

Question 1.
Assertion(A): Higher the value of therapeutic index, safer the drug.
Reason (R): Therapeutic index is defined as the ratio between the maximum tolerated dose of a drug and the minimum curative dose.
(a) Both A and R are correct and R Is the correct explanation of A.
(b) Both A and R are correct but R dOes not explains A.
(c) A is correct but R is wrong.
(d) A is.wrong but R is correct.
Answer:
(a) Both A and R are correct and R Is the correct explanation of A.

Question 2.
Assertion(A): In all living systems, the biochemical reactions are catalysed by enzymes. This principle is applied to kill many pathogens.
Reason (R): The enzyme actions are highly essential for normal functioning of the system.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

Question 3.
Assertion(A): The drugs acts as an inhibitor to the enzyme catalyst.
Reason (R): A drug molecule that has a similar geometry (shape) as the substrate is administered, it can also bind to the enzyme and inhibit its activity.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

Question 4.
Assertion(A): Aspirin is an antipyretic and useful in the prevention of heart attacks.
Reason (R): Aspirin reduces fever and also prevent platelet coagulation.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 5.
Assertion(A): Opioids produces coma and even death.
Reason (R): Opioids releive pain and produce sleep and drugs are addictive and also poisonous in nature.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

Question 6.
Assertion(A): Milk of magnesia and aluminium hydroxide are usually used as antacids.
Reason (R): Mg(OH)2 and Al(OH)3 are weak bases and they neutralise the acid in the stomach that causes acidity.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

Question 7.
Assertion(A): Procaine and Lidocaine are local anaesthetics and cause loss of sensation in the area in which it is applied without losing consciousness.
Reason (R): They block pain perception that is transmitted via peripheral nerve fibres to the brain.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

Question 8.
Assertion(A): Antioxidant such as butyl hydroxy toluene (BHT) and butylated hydroxy anisole (BHA) are added as good additives.
Reason (R): Antioxidants retard the oxidative deterioration of food which contain fat and oils is easily oxidised and turn rancid.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 9.
Assrtion(A): Saccharin, sucralose are artificial sweeteners.
Reason (R): Synthetic compounds which imprint a sweet sensation and possess no or negligible nutritional value are called artificial sweeteners.
(a) Both A and R are correct and R does not explains A.
(b) Both A and R are correct and R explains A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(b) Both A and R are correct and R explains A.

Question 10.
Assertlon(A): Sulphur dioxide and suiphites are also used as food additive.
Reason (R): They act as antimicrobial agents, antioxidant and enzyme inhibitors.
(a) Both A and R are correct but R does not explains A.
(b) Both A and R are correct and R explains A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(b) Both A and R are correct and R explains A.

Question 11.
Asscrtion(A): During soap preparation, common salt is added to the reaction mixture.
Reason (R): Common salt decreases the solubility of soap and it helps to precipitate out from the aqueous solution.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

Question 12.
Assertion(A): Higher the TFM quantity in the soap, better is its quality.
Reason (R): The quality of the soap is described in terms of total fatty matter (TFM value). Grade I soap should have 76% minimum TFM value.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

Question 13.
Assertion(A): Natural rubber becomes strong and elastic when heated with sulphur.
Reason (R): Natural rubber is mixed with 3 – 5% sulphur and heated at 100 – 150°C causes cross linking of the cis – 1, 4 – polyisoprene chains through disulphide – s – s bonds.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

Question 14.
Assertion(A): Artificial sweeteners are added to the food to control the intake of calories.
Reason (R): Most of the artificial sweeteners are inert and do not metabolise in the body,
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

Question 15.
Assertion(A): Penicillin (G) is an antihistamine.
Reason (R): Penicillin G is effective against gram positive as well as gram negative bacteria.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(d) A is wrong but R is correct.

Question 16.
Assertlon(A): Enzymes have active sites that hold substrate molecule for a chemical reaction.
Reason (R): Drugs compete with natural substate by attaching covalently to the active site of enzyme.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(c) A is correct but R is wrong.

Question 17.
Assertion(A): Transparent soaps are made by dissolving soaps in ethanol.
Reason (R): Ethanol made things invisible.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R Is wrong.
(d) A is wrong but R is correct.
Answer:
(c) A is correct but R Is wrong.

Question 18.
Assertion(A): Sodium chloride is added to precipitate soap after saponification.
Reason (R): Hydrolysis of esters of long chain fatty acids by alkali produces soap in colloidal form.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

Question 19.
Assertlon(A): Aspirin has antipyretic properties.
Reason (R): Aspirin gives relief from pain.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

Question 20.
Assertlon(A): Bithional is added to soap as an antiseptic.
Reason (R): Bithional is a suipha drug and destroy bacteria.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R explains A.

VI. Find out the correct pair.

Question 1.
Ampicillin, amoxicillin, methiceillin, cetrizine, cephalosporin.
Answer:
Cetrizine. It is an antihistamine whereas others belongs to penicillin group.

Question 2.
Aluminium hydroxide, magnesium hydroxide, erythromycin, cimetidine, ranitidine.
Answer:
Erythromycin. It is antimicrobial whereas others are antacids.

Question 3.
Halo peridol, clozapine, aiprazolam, aspirin, diazepam.
Answer:
Aspirin. It is an analgesic and antipyretic whereas others are anaesthetics.

Question 4.
Acetamino phenol, ibuprofen, aspirin, morphine.
Answer:
Morphine. It is an opioids (narcotic analgesic) whereas others are non narcotic analgesics.

Question 5.
Morphine, heroin, hydrocodone, codeine, ibuprofen.
Answer:
Ibuprofen. It is a non narcotic analgesic whereas others are narcotic analgesics.

Question 6.
Procaine, lidocaine, cemitidine, propofol, iso flurane.
Answer:
Ccmitidine. It is an antacid whereas others are anaesthetics.

Question 7.
Omeprazole, rabeprazole, iso flurane, ranitidine, cemitidine.
Answer:
Isoflurane. It is an anaesthetic whereas others are antacids.

Question 8.
Cetrizine, levocetrizine, trefenadine, ampicillin, desloratatide.
Answer:
Ampicillin. It is antimicrobial whereas others are antihistamines.

Quesiton 9.
Penicillin, ampicillin, cephalosorins, hydrogen peroxide, carbapenems.
Answer:
Hydrogen peroxide. it is an antiseptic where as other are antimicrobials.

Question 10.
Hydrogen peroxide, povidone – iodine, chlorine compounds, benzalkonium chloride.
Answer:
Chlorine compounds. It is a disinfectant whereas others arc antiseptic.

Question 11.
Ethynylestradiol, menstranol, hydrogen peroxide, norethindrone, norethynodrel.
Answer:
Hydrogen peroxide. It is an antiseptic and a disinfectant whereas otheres are antifertility drugs.

Question 12.
Sodium benzoate, salt of sorbic acid, acetic acid, sodium bi carbonate, sodium meta suiphite.
Answer:
Sodium bi carbonate. It is a baking soda whereas others are food preservatives.

Question 13.
BIIT, BHA, SO2, Vitamin E, sorbitol.
Answer:
Sorbitol. It is a sugar substituent where as others are antioxidants.

Question 14.
Saccharin, butyl hydroxy toluene, aspartane, sucralose, alitaine.
Answer:
Butyl hydroxy toluene, It is an antioxidant where as others are artificial sweetening agents.

Question 15.
Cellulose, polyester, silk.
Answer:
Polyester. It is a synthetic polymer whereas others are natural polymer.

Question 16.
PVC, polythene, LDPE, cellulose, HDPE, bakelite
Answer:
Cellulose, It is a natural polymer whereas others are synthetic polymers.

Question 17.
Polythcne, PVC, Bakelite, polystrene.
Answer:
Bakelite. It is thermosetting plastic whereas others are thermoplastic.

Question 18.
Nylon 66, polyethylene, PVC, teflon.
Answer:
Nylon 66. It is a condensation polymer whereas others are addition polymers.

Question 19.
Neoprene, bakelite, Buna – S, Buna – N.
Answer:
Bakelite. It is a thermosetting plastic whereas others are synthetic rubber.

Question 20.
Nylon 66, Nylon 6, terylene, teflon, bakelite, melamine.
Answer:
Teflon. It is an additional polymer whereas others are condensation polymers.

Samacheer Kalvi 12th Chemistry Chemistry in Everyday Life 2 mark Questions and Answers

Question 1.
Define the term

  1. medicine
  2. chemotherapy

Answer:
1. Medicine:
The drug which interacts with macromolecular targets such as proteins to produce a therapeutic and useful biological response is called medicine.

2. Chemotherapy:
The specific treatment of a disease using medicine is known as chemotherapy.

Question 2.
Define the term therapeutic index.
Answer:
1. Therapeutic index is defined as the ratio between the maximum tolerated dose of a drug (above which it becomes toxîc) and the minimum curative dose (below which the drug is ineffective).

2. Higher the value of therapeutic index, safer is the drug.

Question 3.
Write about the classification of drugs based on the target system.
Answer:
1. In this classification, the drugs are grouped based on the biological system that they target in the recepient. For example, the antibiotics streptomycin and erthyromycin inhibit the protein synthesis in bacteria and are classified in the same group.

2. However their mode of action is different. Streptomycin inhibits the initiation of protein synthesis, while erythromycin prevents the incorporation of new amino acids to the protein.

Question 4.
Explain about the classification of drug based on the site of action.
1. The drug molecule interacts with biomolecules such as enzymes, receptors which are referred as drug targets. The drug is classified based on the drug target with which it binds.

2. This classification is highly specilic compared to others. These compounds often have a common mechanism of action, as the target is the same.

Question 5.
What are

  1. antagonists
  2. agonists.

Answer:

  1. The drugs which binds to the receptor site and inhibit its natural function are called antagonists.
  2. There are drugs which mimic the natural messenger by switching on the receptor. Those type of drugs are called agonists.

Question 6.
What is the difference between an agonist and antagonist?
Answer:

  1. Agonist and antagonist act in opposite directions. Agonist is a substance which combines with cell receptor to produce some reaction that is typical for that substance.
  2. On the other hand antagonist is the chemical which opposes or reduces the natural function.

Question 7.
Explain the action of agonist and antagonist with proper example.
Answer:
When adenosine binds to the adenosine receptors, it induces sleepiness. So adenosine is an agonist. On the other hand, the antogonist drug coffeine binds to the adenosine receptor and makes it inactive. This results in the reduced sleepiness (wakefulness).

Question 8.
Why ranitine is a better antacid than magnesium hydroxide?
Answer:
To treat acidity, weak base such as magnesium hydroxide is used. But this weak base make the stomach alkaline and trigger the production of much acid. This treatment only relieves the symptoms and does not control the cause. But ranitine stimulate the secretion of HCI by activating the receptor in the stomach wall which binds the receptor and inactivate them. So ranitine is a better antacid than magnesium hydroxide.

Question 9.
What is meant by non-steroidal anti inflammatory drugs? Give example.
Answer:
Non – steroidal anti inflammatory drugs reduces fever by causing the hypothalamus to override a prostaglandin – induced increase in temperature. eg., ibuprofen.

Question 10.
What are narcotic analgesics? Give examples.
Answer:

  1. Narcotic analgesics are opioids that relieve pain and produce sleep. These drugs are addictive. In poisonous dose, these produces coma and ultimately death. eg, morphine, codeine,
  2. These drugs are used for short term or long term relief of severe pain. Mainly used for post operative pain. Pain of terminal cancer.

Question 11.
What are general anaesthetics? Give example.
Answer:

  1. General anaesthetics are drugs cause a controlled and reversible loss of consciousness by affecting central nervous system. e.g., propofol, iso flurane.
  2. They are often used for major surgical procedures.

Question 12.
What are local anaesthetics? Give example. Mention its uses.
Answer:

  1. Local anaesthetics cause loss of sensation in the area in which it is applied without losing consciousness. They block pain perception that is transmitted via peripheral nerve fibre to the brain. e.g., procaine, lidocaine
  2. They are often used during minor surgical procedures.

Question 13.
Draw the structure of propofol? Mention its use.
Answer:
Propofol structure:
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-29

Question 14.
What are antihistamines? Give example and mention its use.
Answer:

  1. Antihistamines block histamine release from histamine – 1 receptors.
  2. eg., cetirizine, terfenadine, levocetirizine.
  3. It is used to provide relief from the allergic effects.

Question 15.
What are antimicrobials? Mention its function and its uses.
Answer:

  1. Antimicrobials inhibits bacterial cell wall biosynthesis.
  2. e.g., penicillin, ampicillin.
  3. It is used to treat skin infections, dental infections, ear infections, respiratory tract infections. Pneumonia, urinary tract infections and gonorrhoea.

Question 16.
Draw the structure of penicillin? Give its use.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-30
Penicillin is used to treat all type of infections pneumonia, urinary tract infections.

Question 17.
Draw the structure of ampicillin
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-31

Question 18.
Write a note about macrolids.
Answer:

  1. Macrolids targets bacterial ribosomes and prevent protein production. e.g., erythromycin, azithromycin
  2. It is used to treat respiratory tract infections, genital, gastro intestinal tract and skin infections.

Question 19.
What are fluoroquinolones? Give its function and uses.
Answer:

  1. Fluoro quinolones inhibits bacterial enzyme DNA gyrase.
  2. e.g., clinafloxacin, ciprofloxacin
  3. It is used to treat urinary tract infections, skin infections and respiratory infections, pulmonary infections in cystic fibrosis.

Question 20.
What are tetracydines? Mention its function and uses.
Answer:

  1. Tetracyclines inhibit the bacterial protein synthesis via interaction with the 30 S subunit of the bacterial ribosome. eg., doxycycline, minocycline.
  2. It is used in the treatment of peptic ulcer disease, infections of the respiratory tract, cholera.

Question 21.
What are aminoglycosides? Give its function and uses.
Answer:

  1. Aminoglycosides bind to the 30 S subunit of the bacterial ribosome, thus stopping bacteria from making proteins.
  2. It is used to treat infections caused by gram negative bacteria.

Question 22.
What are food additives? Give example.
Answer:

  1. The substances which are not naturally a part of the food and added to improve the quality of food are called food additives.
  2. e.g., Aroma compounds, antioxidants, preservatives, stabilizers, food colours, buffering substances are food additives.

Question 23.
Explain about antioxidants.
Answer:

  1. Antioxidants are substances which retard the oxidative deteriotations of food. Food containing fats and oils is easily oxidised and turn rancid, .
  2. To prevent the oxidation of fats and oils, chemical BHT (butyl hydroxy toluene), BHA (butylated hydroxy anisole) are added as antioxidants.
  3. These materials readily undergo oxidation by reacting with free radicals generated by the oxidation of oils there by stop the chain reaction of oxidation of food.
  4. Sulphur dioxide, suiphites are also used as antioxidant and also act as antimicrobial agents and enzyme inhibitors.

Question 24.
What are sugarsubstituents? Give example.
Answer:
The compounds that are used like sugars for sweetening, but are metabolised without the influence of insulin are called sugar substituents. e.g., sorbitol, xylitol, mannitol.

Question 25.
What are artificial sweetening agents? Give example.
Answer:
Synthetic compounds which imprint a sweet sensation and possess no or negligible nutritional value are called artificial sweeteners. e.g., saccharin, aspartame, sucralose, alitame.

Question 26.
Define TFM value.
Answer:

  1. The quality of a soap is described in terms of total fatty matter (TFM value). It is defined as the total amount of fatty matter that can be separated from a sample after spliting with mineral acids.
  2. Higher the TFM value in the soap, better is its quality.
  3. As per BIS standards, Grade I soaps should have 76% minimum TFM value.

Quesiton 27.
Write a note about natural rubber and give Its structure.
Answer:
1. Rubber is a naturally occuring polymer. It is obtained from the latex that excludes from cuts in the bark of rubber tree.

2. The monomer unit of natural rubber is cis – iso prene (2 – methyl buta – 1,3 – diene). Thousands of isoprene units are linearly linked together in natural rubber. NaturaL rubber is not so strong (or) elastic. The properties of natural rubber can be modified by the process called vulcanization.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-32

Question 28.
How is neoprene prepared? Give its use.
Answer:
1. The free radical polymerisation of the monomer 2 – chloro buta 1,3 – diene (chioroprene) gives neoprene.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-33

2. It is used in the manufacture of chemical container, conveyer belts.

Question 29.
How Is Buna – N prepared? Give its use.
Answer:
1. Buna – N is prepared by the polymerisation of acrylonitile and buta – 1, 3 – diene
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-34

2. It is used in the manufacture of hoses and tank linings.

Question 30
How would you prepare Buna – S? Give its use.
Answer:
Buna – S is prepared by the polymerisation of buta – 1, 3 – diene and styrene in the ratio of 3 : 1 in the presence of sodium.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-35

Uses:
It is used in making pneumatic tires in shoe heels and soles, and in gaskets.

Question 31.
How will you prepare PHBV? Give its use?
Answer:
1. The biodegrable polymer PHBV (Poly hydroxy butyrate-co hydroxyl valerate) is prepared by the polymerisation of monomers 3 – hydroxy butanoic acid and 3 – hydroxy pentanoic acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-36

2. It is used in orthopacdic devices añd in controlled release of drugs.

Question 32.
How would you prepare Nylon – 2 – Nylon – 6 polymers?
Answer:
Nylon – 2 – Nylon 6 is a co polymer which contains polyamide linkages. It is obtained by the condensation polymerisation of monomers glycine and E-amino caproic acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-37

Question 33.
What are natural and synthetic polimers? Give two examples of each type.
Answer:
Natural polymers:
Polymers which are found in nature, i.e., in animal and plants are called natural polymers. For example, proteins, starch, cellulose etc.

Synthetic polymers:
Man – made polymers are called synthetic polymers. For example plastics, synthetic fibres.

Question 34.
Distinguish between the terms homopolymer and copolymer and give an example of each.
Answer:
Homopolyers:
Polymers whose repeating structural units are dervied from only one type of monomer units are called homopolymers. For example, polythene, PVC, PAN etc.

Copolymers:
Polymers whose repeating units are derived from two or more types of monomer molecules are called co-polymers. For example, l3una – S, Buna – N, Nylon 6, 6 etc.

Question 35.
How can you differentiate between addition and condensation polymerisation?
Answer:
Addition polymerisation:
In this type of polymerisation, a large number of molecules of same or different monomers simply add to the other unit, leading to the formation of macromolecule. Addition polymerisation generally occurs among molecules containing double and triple bonds.

Condensation polymerisation:
In this type of polymerisation two or more bifunctional molecules undergo a series of independent condensation reactions usually with the elimination of simple molecules like water, alcohol, ammonia etc.

Question 36.
What are the monomeric repeating units of Nylon – 6 and Nylon 6, 6?
Answer:
Nylon 6 – Caprolactam. Nylon 6,6 – Adipic acid and Hexamethylenediamine

Question 37.
Write the names and structure of the monomers of the following polymers:

  1. Buna – S
  2. Buna – N
  3. Dacron
  4. Neoporene

Answer:
1. Butadiene, CH2 = CH – CH = CH2 ; Styrene, C6H5 – CH = CH2

2. Butadiene, CH2 = CH – CH = CH2 ; Acrylonitrile, Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-38

3. Terephthalic acid,Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-39 ; Ethylene glycol (Ethane – 1, 2 – diol)Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-40

4. Chloroprene,Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-41 ; 2 – Chloro – 1, 3 – butadine is the monomer of neoprene.

Question 38.
How is dacron obtained from ethylene glycol and terephthalic acid?
Answer:
Dacron is obtained by condensation polymerisation of ethylene glycol and terephthalic acid with the elimination of water molecules:
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-42

Question 39.
What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester.
Answer:
Polymers which disintegrate by themselves over a period of time due to environmental degradation by bacteria, etc. are called biodegradable polymers. Example: PHBV (poly hydroxy butrate – co – β hydroxyvalcrate)

Question 40.
What is the difference between elastomers and fibres? Give one example of each.
Answer:
Elastomers

  1. These are rubber like solids with elastic properties.
  2. These are held by the weak inter – molecular forces.
  3. Example: Buna-S and Buna-N.

Fibres

  1. These are th thread forming solids which possess high tensile strength and high modulus.
  2. These are held together by strong intermolecular forces like hydrogen bonding.
  3. Example: Nylon 6, 6 and polyesters (terylene)

Question 41.
What are thermoplastics and thermosetting polymers? Give one example of each.
Answer:
Thermoplastics:
Thermoplastics are linear polymers which can be repeatedly softened on heating and hardened on cooling and hence can be used again and again without any change in chemical composition and mechanical strength.

Thermosetting polymers:
Thermosetting polymers, are permanently setting polymers. On heating in a mould, they get hardened and set and cannot be softened again. This hardening on heating is due to cross linking between different polymeric chains which give rise to a three dimensional network solid. Example – Bakelite.

Question 42.
Differentiate between addition and condensation polymers based on the mode of polyrnerisation. Give one example of each type. 1
Answer:
Addition polymers

  1. They are formed by adding monomers to a growing polymer chaìn without loss of any molecule.
  2. They are formed from unsaturated compounds.
  3. Example: Polyethene, polypropene.

Condensation polymers

  1. They are formed by combining monomers together with the loss of small molecules like H2O, NH3, CO2 etc.
  2. Monomers have di or polyfunctional groups.
  3. Example: Nylon – 6, 6, Nylon – 6, Terylene.

Question 43.
Distinguish between ‘chain growth potymerisatlon and step growth polymerisation’ and give one example of each.
Answer:
Chain growth polymerisation

  1. Only one repeating unit is added at a time.
  2. Reaction is fast and polymer is formed at once. Example – polythene.

Step growth polymerisatlon

  1. Any two species present can react.
  2. Polymer is formed in gradual steps. Examply – Nylon-6, 6.

Question 44.
How are biopolymers more beneficial than synthetic polymers?
Answer:
Durability of synthetic polymers is advantageous, however it presents a serious waste disposable problem. In renewal of the disposable problem, biodegradable polymers are useful to us. Biopolymers arc safe in use. They disintegrate by themselves in biological system during a certain period of time by enzymatic hydrolysis and to some extent by oxidation and hence, are biodegradable. As a result, they do not cause any pollution.

Question 45.
Give the method of preparation of polyacrylonitrile?
Answer:
The addition polymerisation of acrylonitrile in the presence of a peroxide catalyst leads to the formation of polycrylonitrile. It is used as a substitute for wool in making fibres such as orlon or acrilan.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-43

Samacheer Kalvi 12th Chemistry Chemistry in Everyday Life 3 mark Questions and Answers

Question 1.
Draw the structures of

  1. Suiphanilamide
  2. p – nltro benzoic acid

Answer:
1. Suiphanilamide
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-44

2. p – nitro benzoic acid:
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-45

Question 2.
Draw the structure of

  1. Adenosine (Agonist)
  2. Caffeine (Antagonist)

Answer:
1. Adenosinc (Agonist)
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-46

2. Caffeine (Antagonist)
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-47

Question 3.
Explain about

  1. Analgesics
  2. Antlinflammatory drugs
  3. Antipyretlcs

Answer:
1. Analgesics:
They alleviate pain by reducing local inflammatory response. They reduce the pain without causing impairment of consciousness.
Example: Paracetamol (Crocin).

2. Anti inflammatory drug:
They are used for short term pain relief and for modest pain like head ache, muscle strain, bruising or arthritis.
Example: Ibuprofen, Aspirin.

3. Antipyretics:
These drugs have many effects such as reducing fever, and preventing platelet coagulation.
Example: Aspirin.

Question 4.
Explain about anaesthetics with their types.
Answer:
1. Local anaesthetics: It causes loss of sensation in the area in which it is applied without losing consciousness. They block pain perception that is transmitted via peripheral nerve fibres to the brain. Example: Procaine, Li do Caine. They are often used during minor surgical procedures.

2. General anaesthetics:
They cause a controlled and reversible loss of consciousness by affecting central nervous system.
Example: Propofol, Isoflurane. They are often used for major surgical procedures.

Question 5.
Draw the structure of

  1. procaine
  2. Lidocaine

Answer:
1. Procaine
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-48

2. Lidocaine
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-49

Question 6.
Explain about antacids?
Answer:

  1. Antacids neutralise the acid in the stomach that causes acidity.
  2. They are used to relieve burning sensation in the chest, throat area caused by acid reflux. Example – Milk of magnesia, alumminium hydroxide, Ranitidine, Cemitidino, Omeprazole, Rabeprazole.

Question 7.
Distinguish between Antiseptic and Disinfectants.
Answer:
1. Antiseptic:
They are the drugs used to stop (or) slow down the growth of micro organism and they are applied to living tissue (body). Example: H2O2.

2. Disinfectant:
They are the drugs used to stop or slow the growth of micro organism and they are applied on inanimate objects (non living surfaces). Example: Chlorine compounds.

Question 8.
What are the advantages of food additives?
Answer:

  1. Uses of preservatives reduce the product spoilage and extend the shelf-life of food.
  2. Addition of vitamins and minerals reduces the mall nutrient.
  3. Flavouring agents enhance the aroma of the food.
  4. Antioxidants prevent the formation of potentially toxic oxidation products of lipids and other food constituents.

Question 9.
Differentiate soap and detergents?
Answer:
Soap

  1. Soaps are sodium or potassium salt of long chain fatty acid.
  2. Soaps are made from animal (or) plant fats and oils.
  3. Soaps have lesser cleansing action.
  4. Soaps are bio degradable.
  5. Soaps are less effective in hard water.
  6. They have a tendency to form a scum in hard water.
  7. Example: Sodium palmitate.

Detergent

  1. Detergent is sodium salt of alkyl hydrogen sulphate or alkyl benzene suiphonic acid.
  2. Detergents are made from petrochemicals.
  3. Detergents have more cleansing action.
  4. Detergents are non – bio degradable.
  5. Detergents are more effective even in hard water.
  6. They do not form scum with hard water.
  7. Example: Sodium lauryl sulphate.

Question 10.
What is LDPE? Give its preparation and uses.
Answer:
1. LDPE is low density poiy ethylene. It is formed by heating ethene at 2000 to 300°C under oxygen as a catalyst. This reaction follows free radical mechanism. The peroxide formed from oxygen acts as a free radical initiator.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-50

2. LDPE is used as insulation for cables, making toys.

Question 11.
What is HDPE? Cive its preparation and use.
Answer:

  1. IIDPE is high density polyethylene. it is prepared by the polymerisation of ethylene at. 373k and 6 to 7 atm. using zeiglar Natta Catalyst)
  2. Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-51
  3. It is used to make bottles, pipes.

Question 12.
What is Orlon? Give its preparation and use.
Answer:
1. Orlon is poiy acrylonitrite (PAN). It is prepared by the addition of polymerisation of vinyl cyanide using a peroxide initiator.

2.Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-52

3. It is used as a substitute of wool for making blankets, sweaters etc.

Question 13.
How will you prepare Nylon 6,6.? Give its use.
Answer:
1. Nylon 6,6 can be prepared by mixing equimolar adipic acid and hexamethylene diaminc. With the elimination of water to form amide bonds.

2. Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-53

3. It is used in textiles, manufacture of cards.

Question 14.
How will you prepare Nylon – 6? Give its use.
1. Capro tactum on heating at 533k in an inert atmosphere with traces of water gives E amino caproic acid which polymerises to give Nylon 6.

2.Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-54

3. It is used in the manufacture of tyre cards, fabrics.

Question 15.
What is bakeite? How is it prepared? Give its uses.
Answer:
1. Bakelite is a thermo setting plastic. It is prepared from the monomers such as phenol and formaldehyde. The condensalion polymerisation take place in the presence of acid or base catalyst.

2. Phenol reacts with methanal to form ortho or para hydroxyl methyl phenols which on further reaction with phenol gives linear polymer called novolac. Novolac on further healing with formaldehyde undergoes cross linkages to form bakelite.

3.Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-55

4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-56

  1. Novolac is used in paints.
  2. Soft bakelites are used in making glue for binding laminated wooden planks and in varnishes.
  3. Hard bakelites are used to prepare combs, pens.

Question 16.
How Is melamlue prepared? Give its use?
Answer:
Melamine and formaldehyde are the monomers. They undergo condensation polymerisation to form melamine formaldehyde resin.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-57
Uses : It is used in making unbreakable crockery.

Question 17.
How is urea formaldehyde prepared?
Answer:
It is formed by condensation polymerisation of the monomers urea and formaldehyde. Uses – it is used in decorative laminates, textiles, wrinkle resistant fabrics, paper and glue wood.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-58

Question 18.
Mention one use of each of the following:

  1. Ranitidine
  2. Paracetamol
  3. Tincture of iodine.

Answer:

  1. Rnitidine is used as an antacid.
  2. Paracetamol is used to bring down the body temperature during high fever.
  3. Tincture of iodine is used as an antiseptic. It is 2-3% solution of iodine in alcohol and water.

Question 19.
Describe the following with suitable examples:

  1. Preservatives
  2. Artificial sweetening agents.

Answer:
1. Preservatives:
Preservatives are the substances which are used to prevent spoilage of food due to microbial growth. Examples – Sodium benzoate, Common salt.

2. Artificial sweetening agents:
These are the chemical substances which are used to create sweet taste in food items in place of sugar.

Question 20.
Give one important use of each of the following:

  1. Bithional
  2. Chioramphenicol
  3. Streptomycin
  4. Paracetamol

Answer:

  1. Bithional is added to soap so as to impart antiseptic properties to the soap.
  2. Chioramphenicol is a broad spectrum antibiotic used in curing typhoid, meningitis.
  3. Streptomycin is used for the treatment of T.B (Tuberculosis).
  4. Paracetamol is an antipyretic used in bringing down temperature in high fever.

Question 21.
What are detergents? How are they classified? Why are detergents preferred over soaps?
Answer:
Detergents are suiphonate or hydrogen sulphate salts of long chain hydrocarbons containing 12-18 carbon atoms.
Types of detergents

  1. Cationic detergents
  2. Anionic detergents
  3. Non-ionic detergents

Advantages of detergents over soaps: Unlike soaps they work well even with hard water. They can work well even in acidic water. They are more effective than soaps.

Question 22.

  1. What class of drug is Ranitidine?
  2. If water contains dissolved Ca2+ ions, out of soaps and synthetic detergents, which will you use for cleaning clothes?
  3. Which of the following is an antisepctic? 0.2% phenol, 1% phenol.

Answer:

  1. It is an antacid.
  2. In this case we use synthetic detergents because it give foam with hard water.
  3. 0.2% solution of phenol acts as antiseptic.

Question 23.
Define the following by giving one example of each:

  1. Antiseptics
  2. Antioxidants
  3. Narcotic analgesics

Answer:

  1. Antiseptics are the chemicals applied to the living tissues either to kill or prevent the growth of micro organisms. Example : dettol.
  2. Antioxidants are the compounds which retards the action of oxygen on food and reduces its rate of decomposition by oxidation. Example: BHA.
  3. Narcotic analgesics are the chemicals used for the relief of pst operative pain. Example – morphine.

Question 24.
In order to wash clothes which cleaning agent what will you prefer and why: soap or synthetic detergents? Give one advantage of soaps and synthetic detergents each.
Answer:
Soaps have straight hydrocarbon chains and are easily degraded by bacteria present in the sewage water and hence, do not cause water pollution. Most of the detergents are non – biodegradable and hence cause water pollution of rivers and waterways. So, one will prefer soap.

Question 25.
Name the action of the following on the human body.

  1. Aspirin
  2. Penicillin
  3. Phenacetin
  4. Morphine
  5. Analgin
  6. Luminal
  7. Seconal
  8. Streptomycin

Answer:

  1. Aspirin is an analgesic which is used for relieving pain. It also prevents heart attack.
  2. Penicillin is an antibiotic used against large number of infections caused by various cocci, gram positive bacteria, etc. It is an effective drug for pneumonia, bronchitis, sore throat.
  3. Phenacetin is an antipyretic drug used to bring down the temperature of body in high fever.
  4. Morphine is an strong analgesic. It is a narcotic drug. It cause addiction. It gives relief from acute pain, induce sleep and unconsciousness in higher doses.
  5. Analgin is an antipyretic and analgesic. It brings down the temperature of body in fever and give relief from pain.
  6. Luminal produces sleep and it is a habit forming drug. It is also called a sedative tranquilliser.
  7. Seconal is an antidepressant (tranquiliser). Sometimes the patients are highly depressed and loses self – confidence. This drug produces feeling of well being and improved efficiency.
  8. Streptomycin is used as an antibiotic. It is used to cure tuberculosis.

Samacheer Kalvi 12th Chemistry Chemistry in Everyday Life 5 mark Questions

Question 1.
Explain free radical polymerisation with example.
Answer:
1. When alkenes are heated with free radical initiator such as benzoyl peroxide, they undergo polymerisation reaction. For example, styrene polymerises to polystrene when it is heated with a peroxide initator. The mechanism involves the following steps.

2. Initiation – Formation of free radical.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-59

3. Propagation step.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-60

4. Chain growth will continue with the successive addition of several thousands of monomer units.

5. Termination:
The above chain reaction can be stopped by stopping the supply of monomer or by coupling of two chains or reaction with an impurity such as oxygen.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-61

Question 2.
How are polymers classified on the basis of forces operating between their molecules? To which of these classes does nylon-6,6 belong?
Answer:

  1. Elastomers: The polymer chains are held together by weak intermolecular forces. Example – Buna – S, Buna – N, Neoprene.
  2. Fibres: They have strong forces of attraction. Example – Polymides, (Nylon 6,6), polyesters.
  3. Thermoplastics: They are long chain molecules capable of repeatedly softening on heating and hardening on cooling. Example – Polythene, polystyrene.
  4. Thermosetting plastics: They do not become soft on heating and cannot be remoulded. Example – Bakelite, Nylon – 6,6, belong to fibres.

Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-62

Common Errors

1. Medicines – Molecular fonnula are not given only structural fotmula are drawn.

Rectifications

1. It is written by counting the C, H, O, N in the compound. For example. Aspirin on Acetyt salicylic acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life-63

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Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds

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Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds

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Samacheer Kalvi 12th Chemistry Chapter 13 Organic Nitrogen Compounds Text Book Evaluation

Samacheer Kalvi 12th Chemistry 13 Organic Nitrogen Compounds Multiple Choice Questions

Question 1.
Which of the following reagent can be used to convert nitrobenzene to aniline?
(a) Sn / HCl
(b) ZnHg / NaOH
(c) LiAIH4
(d) All of these
Answer:
(a) Sn / HCl

Question 2.
The method by which aniline cannot be prepared is ……………
(a) degradation of benzamide with Br2 / NaOH
(b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.
(c) Hydrolysis of phenylcyanide with acidic solution
(d) reduction of nitrobenzene by Sn / HCI
Answer:
(b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.

Question 3.
Which one of the following will not undergo Hofmann bromamide reaction?
(a) CH3CONHCH3
(b) CH3CH2CONH2
(c) CH3CONH2
(d) C6H5CONH2
Answer:
(a) CH3CONHCH3
Only primary amides undergo hoffmann bromamide reaction

Question 4.
Assertion : Acetamide on reaction with KOH and bromine gives acetic acid
Reason : Bromine catalyses hydrolysis of acetamide.
(a) if both assertion and reason are true and reason is the correct explanation of assertion,
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(d) both assertion and reason are false

Question 5.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-1
(a) bromomethane
(b) a – bromo sodium acetate
(c) methanamine
(d) acetamide
Answer:
(c) methanamine
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-2

Question 6.
Which one of the following nitro compounds does not react with nitrous acid?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-3
Answer:
(c) (CH3)3CNO2 – 30 nitroalkane

Question 7.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-4 this reaction is known as ………………
(a) Friedel – crafts reaction
(b) HVZ reaction
(c) Schotten – Baumann reaction
(d) none of these
Answer:
(c) Schotten – Baumann reaction

Question 8.
The product formed by the reaction an aldehyde with a primary amine
(a) carboxylic acid
(b) aromatic acid
(c) schiff ‘s base
(d) ketone
Answer:
(c) schiff ‘s base

Question 9.
Which of the following reaction is not correct.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-5
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-6
P – nitrosation takes places, the product is Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-7

Question 10.
When aniline reacts with acetic anhydride the product formed is …………….
(a) o – aminoacetophenone
(b) m – aminoacetophcnone
(c) p – aminoacetophenone
(d) acetanilide
Answer:
(d) acetanilide
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-8

Question 11.
The order of basic strength for methyl substituted amine solution is ………….
(a) N(CH3)3 > N(CH3)2H > N(CH3)H2 > NH3
(b) N(CH3)H2 > N(CH3)2H > N(CH3)3 > NH3
(c) NH3 > N(CH3)H2 > N(CH3)2H > N(CH3)3
(d) N(CH3)2H > N(CH3)H2 > N(CH3)3 > NH3
Answer:
(d) N(CH3)2H > N(CH3)H2 > N(CH3)3 > NH3

Question 12.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-9
(a) H3PO2 and H2O
(b) H+ / H2O
(c) HgSO4 / H2SO4
(d) Cu2Cl2
Answer:
(a) H3PO2 and H2O

Question 13.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-10
(a) C6H5 – OH
(b) C6H5 – CH2OH
(c) C6 H5 – CHO
(d) C6H5NH2
Answer:
(a) C6H5 – OH
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-11

Question 14.
Nitrobenzene on reaction with at 80 – 100°C forms which one of the following products?
(a) 1, 4 – dinitrobenzene
(b) 2, 4, 6 – tirnitrobenzene
(c) 1, 2 – dinitrobenzene
(d) 1, 3 – dinitrobenzene
Answer:
(d) 1, 3 – dinitrobenzene
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-12

Question 15.
C5H13N reacts with HNO2 to give an optically active compound – The compound is …………..
(a) pentan – 1 – amine
(b) pentan – 2 – amine
(c) N,N – dimethylpropan – 2 – amine
(d) N – methylbutan – 2 – amine
Answer:
(d) N – methylbutan – 2 – amine
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-13

Question 16.
Secondary nitro alkanes react with nitrous acid to form …………..
(a) red solution
(b) blue solution
(c) green solution
(d) yellow solution
Answer:
(b) blue solution

Question 17.
Which of the following amines does not undergo acetylation?
(a) t – butylamine
(b) ethylamine
(c) diethylamine
(d) triethylamine
Answer:
(d) triethyl amine (3°amine)

Question 18.
Which one of the following is most basic?
(a) 2, 4 – dichloroaniline
(b) 2, 4 – dimethyl aniline
(c) 2, 4 – dinitroaniline
(d) 2, 4 – dibromoaniline
Answer:
(b) 2, 4 – dimethyl aniline
CH3 is a +1 group, all other – I group. +1 group increase the electron density on NH2 and hence increases the basic nature.

Question 19.
When
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-14
is reduced with Sn / HCI the pair of compounds formed are ………..
(a) Ethanol, hydrozylamme hydrochloride
(b) Ethanol, ammonium hydroxide
(c) Ethanol, NH2OH
(d) C3H5NH2, H2O
Answer:
(a) Ethanol, hydrozylamine hydrochloride

Question 20.
TUPAC name for the amine
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-15 is ………………
(a) 3 – Bimethylamino – 3 – methyl pentane
(b) 3 (N,N – Triethyl) – 3 – amino pentane
(c) 3 – N, N – trimethyl pentanamine
(d) 3 – (N, N – Dimethyl amino) – 3 – methyl pentane
Answer:
(d) 3 – (N, N – Dimethyl amino) – 3 – methyl pentane

Question 21.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-16
Product ‘P’ in the above reaction is ………………
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-17
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-18

Question 22.
Ammonium salt of bcnzoic acid is heated strongly and the product so formed is reduced and then treated with NaNO2 / HCl at low temperature. The final compound formed is ……………
(a) Benzene diazonium chloride
(b) Benzyl alcohol
(c) Phenol
(d) Nitrosobenzene
Answer:
(b) Benzyl alcohol
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-19

Question 23.
Identify X in the sequence give below.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-20 + Methanoic acid
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-21
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-22

Question 24.
Among the following, the reaction that proceeds through an electrophilic substitution, is ……………..
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-23
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-24

Question 25.
The major product of the following reaction
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-25
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-26
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-27

Samacheer Kalvi 12th Chemistry 13 Organic Nitrogen Compounds Short Answer Questions

Question 1.
Write down the possible isomers of lthe C4H9NO2 give their IUPA names.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-28
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-29

Question 2.
There are two isomers with the formula CH3NO2. How will you distinguish between them?
Answer:
CH3NO2 has two isomers. They are
1. CH3 – NO2 (Nitromethane)
2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-30

Question 3.
What happens when

  1. 2 – Nitropropane boiled with HCI
  2. Nitrobenezen electrolytic reduction in strongly acidic medium.
  3. Oxidation of tert – butylamine with KMnO4
  4. Oxidation of acetoneoxime with triuluoroperoxy acetic acid.

Answer:
1. 2 – Nitropropane boiled with HCI:
2 – nitropropane upon hydrolysis with boiling HCl give a ketone (2 – propanone) and nitrous oxide.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-31

2. Nitrobenezen electrolytic reduction in strongly acidic medium:
Electrolytic reduction of nitrobenzene in weakly acidic medium gives aniline but in strongly acidic medium, it gives p – aminophenol obviously through the acid – cataLysed rearrangement of the initially formed phenyihydroxylamine.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-32

3. Oxidation of tert – butylamine with XMnO4:
In general, primary amines, in which the – NH2 group is attached to a tertiary carbon can be oxidised with KMnO4 to the corresponding nitro compound in excellent yield. Therefore 3° – butylamine oxidised to give 2 – methyl – 2 – nitropropane.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-33

4. Oxidation ofacetoneoxime with trifluoroperoxy acetic acid:
Oxidation ofacetoneoxime with trifluoroperoxy acetic acid gives 2 – nitropropane.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-34

Question 4.
How will you convert nitrobenzene into

  1. 1, 3, 5 – trinitrobenzene
  2. o and p – nitrophenol
  3. m – nitro aniline
  4. azoxybenzene
  5. hydrozabenzene
  6. N – phenylhydroxylamine
  7. aniline

Answer:
1. Conversion of nitrobenzene into 1, 3, 5 – trinitrobenzene:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-35

2. Conversion of nitrobenzene into o and p – nitrophenol:
(a) Method I:
Nitrobenzene heated with solid KOH at 340 K gives a low yield of a mixture of 0 – and P – nitrophenols.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-36

(b) Method II:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-37

3. Conversion of nitrobenzene into m – nitro aniline:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-38

4. Conversion of nitrobenzene into azoxybenzene:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-39

5. Conversion of nitrobenzene into hydrazobenzene:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-40

6. Conversion of nitrobenzene into N – phenylhydrozylamine:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-41

7. Conversin of nitrobenzene into aniline:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-42

Question 5.
Identify compounds A,B and C in the following sequence of reactions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-43
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-44
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-45
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-46
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-47
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-48

Question 6.
Write short flotes on the following

  1. Hoffmann’s bromide reaction
  2. Ammonolysis
  3. Gabriel phthalimide synthesis
  4. Schotten – Baumann reaction
  5. Carbylamine reaction
  6. Mustard oil reaction
  7. Coupling reaction
  8. Diazotisation
  9. Gomberg reaction

Answer:
1. Hoffmann’s bromide reaction:
When Amides are treated with bromine in the presence of aqueous or ethanolic solution of KOH, primary amines with one carbon atom less than the parent amides are obtained.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-49

2. Ammonolysis:
When Alkyl halides (or) benzylhalides are heated with alcoholic ammonia in a sealed tube, mixtures of 1°, 2° and 3° amines and quaternary ammonium salts are obtained.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-50

3. Gabriel phthalimide synthesis:
Gabriel synthesis is used for the preparation of Aliphatic primary amines. Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-51

4. Schotten – Baumann reaction:
Aniline reacts with benzoylchloride (C6H5COCl) in the presence of NaOH to give N – phenyl benzamide. This reaction is known as Schotten Baumann reaction. The acylation and benzoylation are nucleophilic substitutions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-52

5. Carbylamine reaction:
Aliphatic (or) aromatic primary amines react with chloroform and alcoholic KOH to give isocyanides (carbylamines), which has an unpleasant smell. This reaction is known as carbylamines test. This test used to identify the primary amines.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-53

5. Mustard oil reaction:
When primary amines are treated with carbon disuiphide (CS2), N – alkyldithio carbonic acid is formed which on subsequent treatment with HgCI2, give an alkyl isothiocyanate.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-54

6. Coupling reaction:
Benzene diazonium chloride reacts with electron-rich aromatic compounds like phenol, aniline to form brightly coloured azo compounds. Coupling generally occurs at the para position. If para position is occupied then coupling occurs at the ortho position. Coupling tendency is enhanced if an electron-donating group is present at the para – position to Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-56 group. This is an electrophilic substitution.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-55

7. Diazotisation:
Aniline reacts with nitrous acid at low temperature (273 – 278 K) to give benzene diazonium chloride which is stable for a short time and slowly decompose seven at low temperatures. This reaction is known as diazotization.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-57

8. Gomberg reaction
Benzene diazonium chloride reacts with benzene in the presence of sodium hydroxide to give biphenyl. This reaction in known as the Gomberg reaction.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-58

Question 7.
How will you distinguish between primary secondary and tertiary alphatic amines.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-59

Question 8.
Account for the following

  1. Aniline does not undergo Friedel – Crafts reaction
  2. Diazonium salts of aromatic amines are more stable than those of aliphatic amines
  3. pkb of aniline is more than that of methy lamine
  4. Gabriel phthalimide synthesis is preferred for synthesising primary amines.
  5. Ethylamine is soluble in water whereas aniline is not
  6. Amines are more basic than amides
  7. Although amino group is o – and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m – nitroaniline.

Answer:
1. Aniline does not undergo Friedel – Crafts reaction:
Aniline being a Lewis base reacts with Lewis acid AiCl3 to form a salt.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-60
Due to the presence of a positive charge on N – atom in the salt the group Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-61 acts as a strongly deactivating group. As a result, it reduces the electron density in the benzene ring and which inhibits the electrophilic substitution reaction. Therefore aniline does not under go Friedel – Crafts reaction.

2. Diazonium salts of aromatic amines are more stable than those of aliphatic amines:
The diazonium salts of aromatic amines are more stable than those of aliphatic amines due to dispersal of the positive charge on the benzene ring as shown below.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-62

3. pKb of aniline is more than that of methylamine:
In aniline, the lone pair of electrons on the N – atom is delocalized over the benzene ring. As a result electron density on the nitrogen decreases. In contrast in CH3NH2, +I effect of CH3 increases the electron density on the N-atom. Therefore, aniline is a weaker base than methylamine and hence its pK value is more than that of methyl amine.

4. Gabriel phthalimide synthesis is preferred for synthesising primary amines:
Gabriel phthalimide reaction gives pure 10 amine without any contamination of 2° and 3°-amines. Therefore it is preferred for synthesising primary amines.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-63

5. Ethylamine is soluble in water whereas aniline is not:
Ethylamine when added to water forms intermolecular H – bonds with water. And therefore it is soluble in water. But aniline does not form H – bond with water to a very large extent due to the presence of a large hydrophobic – C6H5 group. Hence, aniline is insoluble in water.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-64

6. Amines are more basic than amides:
In simple amines, the lone pair of electrons is on nitrogen and hence available for protonation. In amides on the other hand, the electron pair on nitrogen is delocalised to the carboxyl oxygen through resonance and thus it is not available for protonation. So amines are more basic than amides.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-65

7. Although amino group is o – and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m – nitroaniline:
Nitration is usually carried out with a mixture of conc HNO3 and conc H2SO4. In the presence of these acids, most of aniline gets protonated to form anilinium ion. Therefore, in the presence of acids, the reaction mixture consists of aniline and anilinium ion.

Now – NH2, group in aniline is O, P – directing and activating while the – NH3 group is anilinium ion is meta – directing and deactivating. Whereas nitration of aniline (due to steric hindrance at o – position) mainly gives p-nitroaniline, the nitration of anilinium ion gives m – nitro aniline. In actual practice, approximately a 1 : 1 mixture of P and m – nitroaniline is formed.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-66

Question 9.
Arrange the following

  1. In increasing order of solubility in water, C6H5NH2, (C2H5)2NH , C2H5NH2
  2. In increasing order of basic strength
    • aniline, p – toludine and p – nitroaniline
    • C6H5NH2, C6H5NHCH3, C6H5NH2, p – Cl – C6H4 – NH2
  3. In decreasing order of basic strength in gas phase.
    C2H5NH2, (C2H5)3NH , (C2H5)3N and NH3
  4. In Increasing order of boiling point
    C2H5OH , (CH3)2NH, C2H5NH2
  5. In decreasing order of the pKb values
    C2H5NH2, C6H5NHCH3, (C2H)2NH4 and CH3NH2
  6. Increasing order of basic strength C6H5NH2, C6H5N(CH3)2, (C6H5)2NH and CH3NH2
  7. In decreasing order of basic strength
    Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-67

Answer:
1. Solubility decreases with increase in molecular mass of amines due to increase in the size of a hydrophobic hydrocarbon part and with decrease in the number of H – atoms on the N – atom which undergo H – bonding.

Now among the given compounds C6H5NH2 has the highest molecular mass of 93 followed by (C2H5)2NH with molecular mass of 73 with C2H5NH2 has the lowest molecular mass of 45. Thus the solubility increases in the order in which molecular mass decreases.

2. (a) The electron – donating groups increases the basic strength of amines while the electron – withdrawing groups decrease the basic strength of amines. Therefore p – nitroaniline is the weakest base followed by aniline while p – toluidine, which has methyl group and therefore it is the strongest base. Basic strength increases in the order. P – nitro aniline < aniline < p – toluidine

(b) Chlorine atom has both – I effect and + R effect since – I effect out weights the + R effect, therefore p – chloro aniline is weak base than aniline. Aikyl groups are electron – donating groups. ”

As a result the electron density on the nitrogen atom increases in the ethylamine and thus they can donate lone pair of electrons niore easily. Therefore Ethylamine is more base than aromatic amines.

Due to delocalization of lone pair of electrons of the N – atom over the benzene ring, C6H5NH, and C6H5NHCH3 are far less basic than C2H5NH2. Further due to +1 effect of the CH3 group, C6H5NHCH3 is little more basic than C6H5NH2. Therefore increasing order basic strength is

3. In the gas phase, solvent effects i.e., stabilization of the conjugate acids due to H – bonding, are absent. Therefore, in the gas phase, basic strength mainly depends upon the +1 effect of the alkyl groups. Since the +1 effect increases with the number of ailcyl groups,

therefore the basic strength of the amines decreases as the number of ethyl groups decreases from three in (C2H5)3N to two in (C2H5)2NH to one in C2H5NH2 and zero in NH3. Basic strength in the gas phase decreases in the order is,
(C2H5)3N > (C2H5)2N > C2H5NH2 > NH3

4. Since the electro negativity of O is higher than that of N, therefore, alcohols form stronger H – 0bonds than amines. In other words, the boiling points of alcohols are higher than those of amines of comparable molecular masses. Therefore the boiling point of C2H5OH (46) is higher than those of (CH3)2NH (45) and C2H5NH2 (45).

Further since the extent of H – bonding depends upon the number of H-atoms on the N-atom. Therefore 1° – amines with two H – atoms on the N – atom have higher boiling points than 2° – amines having only one H – atom. Therefore the boiling point of C2H5NH2 is higher than that of(CH3)2NH. Increasing order of boiling point is,
(CH3)2NH < C2H5NH < CH5OH

5. Due to delocalization of lone pair of electrons of the N – atom over the benzene ring, C6H5NHCH3 is far less basic than C2H5NH2, (C6H5)2NH and CH3NH2. Among C2H5NH2 and (C2H5), NH, (C2H5)2NH is more basic than C2H5NH2 due to greater +1 effect of the two C2H5 groups and stabilization of its conjugate acid by H – bonding.

Compare to Ethyl and methyl group, C2H5 – group has more +1 effect than CH3 – group. Therefore methylamine is weak base than ethylamine. Combining all these facts the relative basic strength of these four amines decreases in the order.

(C2H5)2NH > C2H5NH2 > CH3NH2 > C6H5NHCH3. Since a stronger base has a lower pKb value therefore, pKb values decrease in the reverse order. C6H5NHCH3 > CH3NH2 > C2H5NH2 > (C2H5)2NH

6. Due to delocalization of lone pair of electrons of the N-atom over the benzene ring, all aromatic amines are less basic than alkylamines i.e., CH3NH2. Presence of electron – donating groups ( – CH3) on the N – atom increases the basicity of substituted aniline with respect to C6H5NH2.

In (C6H5)2NH, the lone pair of electrons on the N – atom is delocalized over two benzene rings instead of one in C6H5NH2, therefore (C6H5)2NH is much less basic than C6H5NH2. Combining all the three trends together, the basic strength of the four amines increasing in the order.
(C6H5)2NH < C6H5NH2 < C6H5N (CH3)2 < CH3NH,

7. Aliphatic amines are more basic than aromatic amines. Therefore CH3CH2NH2 and CH3NH2 are more basic. Among the ethylamine and methylamine. ethylamine was experienced more +1 effect than methylamine and hence ethylamine is more basic than methylamine.

Nitrogroup has a powerful electron withdrawing group and they have both – R effect as well as – I effect. As a result, all the nitro anilines are weaker bases than aniline. In P – nitroaniline
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-68
both – R effect and – I effect of the NO2 group decrease the basicity. Therefore decreasing order of basic strength is,
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-69
Ethylamine > Methylamine > Aniline > p – nitro aniline

Question 10.
How will you prepare propan – 1 – amine from

  1. butane nitrile
  2. propanamide
  3. 1 – nitropropane

Answer:
1. Preparation of propan -1- amine from butane nitrile.
Butane nitrile treated with acid hydrolysis followed by Hoffmann’s bromamide degradation. gives propan – 1 – amine.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-70

2. Preparation of propan – 1 – amine from propanamide.
When propanamide is treated with LiAIH4 in the presence of water gives propan – 1 – amine.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-71

3. Preparation of propan – 1 – amine from 1 – nitropropane.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-72
Reduction of 1 – Nitropropane using H2 / Ni or Fe / HCl gives propan – 1 – amine.

Question 11.
Identify A,B,C and D
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-73
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-74

Question 12.
How will you convert dlethylamine into

  1. N, N – dlethylacetamide
  2. N – nitrosodiethylamine

Answer:
1. Conversion of diethylamine into N, N – diethylacetamide.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-75
Diethylamine react with acetyichioride in the presence of pyridine to form N, N – diethyl acetamide.

2. Conversion of diethylamine into N – nitrosodiethylamine.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-76

Question 13.
Identify A,B and C
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-77
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-78

Question 14.
Identify A, B, C and D
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-79
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-80

Question 15.
Complete the following reaction
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-81
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-82

Question 16.
Predict A, B, C and D for the follwing reaction.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-83
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-84

Question 17.
A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO2. (B) on heating with liquid ammonia followed by treating with Br2 / KOH gives (C) which on treating with NaNO2 and HCI at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-85

Question 18.
Identify A to E in the following frequncy of reactions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-86
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-87

Samacheer Kalvi 12th Chemistry 13 Organic Nitrogen Compounds Evaluate Yourself

Question 1.
Write all possible isomers for the following compounds.

  1. C2H5 – NO2
  2. C3H7 – NO2

Answer:
1. Possible isomers for C2H5NO2 as following

(a) CH3 – CH2 – NO2 – Nitroethane

(b) CH3 – CH2 – O – N = O – Ethyl nitrite

(c)Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-88

(d) H2N – CH2 – COOH – Glycine (amino acid)

(e) Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-89

(f) Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-90

(g) Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-91

2. Possible isomers for C3H7NO2 as follows.

(a) CH3 – CH2 – CH2 – NO2 – 1 – Nitropropane

(b) CH3 – CH2 – CH2 – O – N = O – propane – 1 – nitrite

(c) Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-92

(d) Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-93

(e) Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-94

(f) H2N – CH2 – CH2 – COOH – Alanine

Question 2.
Find out the product of the following reactions
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-95
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-96

Question 3.
Predict the major product that would be obtained on nitration of the following compounds.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-97
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-98

Question 4.
Draw the structure of the following compounds

  1. Neopentylamine
  2. Tert – butylamine
  3. α – amino propionaldehyde
  4. tribenzylamine
  5. N – ethyl – N – methylhexan – 3 – amine

Answer:
1. Neopentylamine:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-99

2. Tert – butylamine
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-100

3. α – amino propionaldehyde
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-101

4. tribenzylamine
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-102

5. N – ethyl – N – methylhexan – 3 – amine
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-103

Question 5.
Give the correct IUPAC names for the following amines.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-104
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-105

Samacheer Kalvi 12th Chemistry 13 Organic Nitrogen Compounds Additional Question

Samacheer Kalvi 12th Chemistry 13 Organic Nitrogen Compounds 1 Mark Questions and Answers

I. Choose the correct answer.

Quesiton 1.
Which one of the following play an important role in bioregulation and neurotransmission?
(a) Acid derivatives
(b) Carbonyl compounds
(c) Organic derivatives of ammonia
(d) Aromatic hydro carbons
Answer:
(c) Organic derivatives of ammonia

Question 2.
Which of the following is needed to maintain the health of nerves, skin and red blood cells?
(a) Vitamin B12
(b) Vitamin B6
(c) Vitamin B1
(d) Vitamin C
Answer:
(b) Vitamin B6

Question 3.
Which one of the following is needed to maintain the health of nerves and skin?
(a) Pyridoxine
(b) Cobalamine
(c) Dopamine
(d) Histamine
Answer:
(a) Pyridoxine

Question 4.
Which one of the following is act as neurotransmitter?
(a) Pyridoxine
(b) Histamine
(c) Dopamine
(d) Cyano cobalamine
Answer:
(c) Dopamine

Question 5.
Which one of the following dilates blood vessels?
(a) Histamine
(b) Streptomycin
(c) Penicillin
(d) Dopamine
Answer:
(a) Histamine

Question 6.
Which one of the following is an example of primary nitro alkane?
(a) 2 – nitropropane
(b) Ethyl nitrite
(c) Nitro ethane
(d) 2 – methyl – 2 – nitropropane
Answer:
(c) Nitro ethane

Question 7.
2 – methyl – 2 – nitropropane belongs to
(a) 1° nitro alkane
(b) 3° nitro alkane
(c) 2° nitro alkane
(d) nitro arenes
Answer:
(b) 3° nitro alkane

Question 8.
Which of the following is an example for 2° nitro alkane?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-199
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-199

Question 9.
Which one of the following is an example for nitro arenes?
(a) C6H5 – CH2 NO6
(b) C6H5NH2
(c) CH3 – CH2 – O – NO
(d) C6H5NO2
Answer:
(d) C6H5NO2

Question 10.
The IUPAC name of
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-200 is ………………
(a) 1 – nitro butane
(b) 2 – methyl – 1 – nitro propane
(c) Isobutyl nitrate
(d) 1 – Nitro iso butane
Answer:
(b) 2 – methyl – 1 – nitro propane

Question 11.
Which one of the following is the structure of 2, 2 – dimethyl – 1 – nitro propane?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-201
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-202

Question 12.
1 – nitrobutane and 2 – methyl – 1 – mtropropane are belong to ……………..
(a) position isomerism
(b) functional isomerism
(c) Tautomerism
(d) chain isomerism
Answer:
(d) chain isomerism

Question 13.
Which of the following pair shows functional isomerism?
(a) 1 – nitro butane and 2 – nitro butane
(b) 1 – nitro butane and butyl nitrite
(c) 1 – nitro butane and 2 – methyl – 1 – nitropropane
(d) 2 – nitro butane and 2 – methyl – 2 – nitro propane
Answer:
(b) 1 – nitro butane and butyl nitrite

Question 14.
Which of the following pair shows position isomerism?
(a) 1 – nitro butane and butyl nitrite
(b) Nitro methane and methyl nitrite
(c) 1 – nitro butane and 2 – nitro butane
(d) 1 – nitro butane and 2 – methyl – 1 – nitro propane
Answer:
(c) 1 – nitro butane and 2 – nitro butane

Question 15.
Nitro methane and methyl nitrite are the examples of ……………
(a) Position isomerism
(b) chain isomerism
(c) metarnersm
(d) Tautomerism
Answer:
(d) Tautomerism

Question 16.
Consider the following statements.
(i) Nitro form of alkane dissolves in NaOH instantly
(ii) Nitro form of alkane decolourises FeCI3 solution
(iii) Nitro form of alkane are more acidic
Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (ii) only
Answer:
(c) (i) and (iii)

Question 17.
Consider the following statements.
(i) Aci form of nitro alkanes dissolves in NaOH slowly.
(ii) Aci form of nitro alkane gives reddish brown colour with FeCl3
(iii) Aci form of nitro alkane’s electrical conductivity is low.
Which of the above statement(s) is / are correct?
(a) (ii) only
(b) (i) only
(c) (iii) only
(d) (i) and (iii)
Answer:
(a) (ii) only

Question 18.
Which one of the following does not exhibit tautomerism?
(a) 1 – nitro ethane
(b) Nitro methane
(c) methyl nitrite
(d) 2 – methyl – 2 – nitro propane
Answer:
(d) 2 – methyl – 2 – nitro propane

Question 19.
Identify the compound which does not exhibit tautomerism?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-203
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-204

Question 20.
The correct decreasing order of acidity of nitro alkane is …………….
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-205
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-206

Question 21.
Which one of the following mechanism is followed by the reaction of Ethyl bromide with ethanolic solution of potassium nitrite ……………
(a) SN1
(b) SN2
(c) SNi
(d) E1
Answer:
(b) SN2

Question 22.
Which one of the following is formed when Ethyl bromide reacts with ethanolic solution of potassium nitrite?
(a) Nitro methane
(b) 2 – Nitro propane
(c) 1 – nitro propane
(d) nitro ethane
Answer:
(d) nitro ethane

Question 23.
Which method is used to separate the mixture of nitroalkane?
(a) Crystallization
(b) zone refining
(c) fractional distillation
(d) sublimation
Answer:
(c) fractional distillation

Question 24.
Which of the following is formed when ethane is heated with conc. HNO3 at 675 K?
(a) Nitro propane
(b) Nitro ethane
(c) Nitro methane
(d) both (ii) and (iii)
Answer:
(d) both (ii) and (iii)

Question 25.
What is the product formed when α – chioro acetic acid is boiled with aqueous solution of sodium nitrite?
(a) Nitro ethane
(b) Nitromethane
(c) Acetamide
(d) α – chloro acetamide
Answer:
(b) Nitromethane

Question 26.
Which one of the following reagent is used to convert teritary butylamine to tertiary nitro alkane?
(a) Aqueous KMnO4
(b) Cone HNO3
(c) Sn / HCI
(d) alcoholic KOH
Answer:
(a) Aqueous KMnO4

Question 27.
The reagent used in the conversion of acetaldoxime to nitro ethane (1°) is …………
(a) aqueous KMnO4
(b) trifluoro peroxy acetic acid
(c) alcoholic KOH
(d) Cone. HNO3
Answer:
(b) trifluoro peroxy acetic acid

Question 28.
Which of the following is called oil of mirbane?
(a) Nitro methane
(b) Nitro propane
(c) Nitro benzene
(d) Nitro ethane
Answer:
(c) Nitro benzene

Question 29.
On direct nitration of nitro benzene gives ………………
(a) 0 – dinitro benzene
(b) m – dinitro benzene
(c) p – dinitro benzene
(d) 2, 4, 6 – trinitrobenzene
Answer:
(b) m – dinitro benzene

Question 30.
Amino group can be directly converted into nitro group by ……………
(a) Caro’s acid
(b) Fuming mixture of conc. HNO3 + conc. H2SO4
(c) NaNO2 + HCI
(d) Ethanolic KNO2
Answer:
(a) Caro’s acid

Question 31.
Amino group can be directly converted into nitro group by ………….
(a) Caro’s acid
(b) marshall’s acid
(c) Peroxy trifluoro acetic acid
(d) all the above
Answer:
(d) all the above

Question 32.
Consider the following statements.
(i) Nitro alkanes have high points because of their highly polar nature
(ii) Alkyl nitrites have lower boiling points than nitro alkanes.
(iii) Nitro alkanes are readily soluble in water due to intermolecular hydrogen bonding formation.
Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i) and (ii)
Answer:
(c) (iii) only

Question 33.
The reagent used to convert Nitromethane to methyl amine is ……..
(a) Zn/NH4Cl
(b) Sn/HCI
(c) H2SO5
(d) H2S2O8
Answer:
(b) Sn/HCI

Question 34.
The reagent used to convert Nitromethane to N – methyl hydroxylamine is …………
(a) Sn/HCl
(b) Zn/NH4CI
(c) Ni
(d) Pd/BaSO4
Answer:
(b) Zn/NH4CI

Question 35.
Which one of the following is formed when ethyl nitrite is treated with Sn/HCl?
(a) Nitro ethane
(b) Ethylamine
(c) Ethyl alcohol
(d) Ethanamide
Answer:
(c) Ethyl alcohol

Question 36.
The product formed when nitro ehtane is boiled with conc. HCl is ……………
(a) Acetic acid
(b) Ethyl chloride
(c) Ethanoyl chloride
(d) Amino ehtane
Answer:
(a) Acetic acid

Question 37.
Which one of the following is formed when 2- nitro propane is boiled with conc.HCI?
(a) Ethanoic acid
(b) Propanoic acid
(c) Propanoyl chloride
(d) Acetone
Answer:
(d) Acetone

Question 38.
Which of the following does not react with conc. HCI?
(a) Nitro ethane
(b) 2 – methyl – 2 – nitropropane
(c) 2 – nitro propane
(d) Aniline
Answer:
(b) 2 – methyl – 2 – nitropropane

Question 39.
Acid (or) Basic hydrolysis of ethyl nitrite gives ………….
(a) Ethanoic acid
(b) Nitro ethane
(c) Ethanol
(d) Aceto nitrile
Answer:
(c) Ethanol

Question 40.
Which one of the following is formed when nitro methane reacts with chlorine and NaOH?
(a) CH3CI
(b) CH3COCl
(c) CCI3NO2
(d) CHCI2NO2
Answer:
(c) CCI3NO2

Question 41.
Which one of the following reagent is used to convert Nitrobenzene to aniline?
(a) Sn/HCI
(b) Zn/NH4CI
(c) Fe/H2O(g)
(d) Zn/NaOH
Answer:
(a) Sn/HCI

Question 42.
Which one of the following is the best reagent used to convert Nitrobenzene into Nitroso benzene?
(a) Sn/HCI
(b) Zn/NH4CI
(c) Fe/H2O(g)
(d) SnCl2 + KOH
Answer:
(c) Fe/H2O(g)

Question 43.
Identify the reagent used to convert Nitrobenzene into hydrazo benzene?
(a) Zn/NaOH
(b) Zn/NH4CI
(c) Sn/HCI
(d) SnCI2 + KOH
Answer:
(a) Zn/NaOH

Question 44.
Which one of the following is formed when nitrobenzene is treated with Fe/H2O(Steam)?
(a) Aniline
(b) Phenyl hydroxylamine
(c) Nitroso benzene
(d) Azobenzene
Answer:
(c) Nitroso benzene

Question 45.
Which one of the following is formed when nitrobenzene is treated with Zn/NaOH?
(a) Phenyl amine
(b) Phenyl hydroxylamine
(c) Azo benzene
(d) Hydrazo benzene
Answer:
(d) Hydrazo benzene

Question 46.
Which one of the following is formed when nitrobenzene is treated with SnCI2/KOH?
(a) Azo benzene
(b) Azoxy benzene
(c) Hydrazo benzene
(d) Nitroso benzene
Answer:
(a) Azo benzene

Question 47.
Which of the following is formed when nitro benzene undergoes electrolytic reduction?
(a) Aniline
(b) Phenyl hydroxylamine
(c) p – amino phenol
(d) all the above
Answer:
(d) all the above

Question 48.
Which of the following can be used to reduce nitrobenzene to aniline?
(a) LiAIH4
(b) Sn + HCI
(c) Pt/H2
(d) all the above
Answer:
(d) all the above

Question 49.
What will be the product formed when nitrobenzene is treated with conc. HNO3 and conc.H2SO4 at 373K?
(a) 1, 3 – dinitro benzene
(b) 1, 4 – dinitro benzene
(c) 1, 3, 5 – trinitro benzene
(d) all the above
Answer:
(a) 1, 3 – dinitro benzene

Question 50.
Which one of the following is formed when nitrobenzene is treated with cone. HNO3 and H2SO4 473 K?
(a) 1, 2 – din itro benzene
(b) 1, 4 – dintro benzene
(c) 1, 3 – dinitro benzene
(d) 1, 3, 5 – trinitro benzene
Answer:
(d) 1, 3, 5 – trinitro benzene

Question 51.
What will be the product formed when 1, 3, 5 – trinito toluene is treated with acidified Na2Cr2O7 and sodalime?
(a) TNB
(b) TNT
(c) TNG
(d) GTN
Answer:
(a) TNB

Question 52.
What is the IUPAC name of
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-207
(a) Tertiary butyl amine
(b) Trimethyl amine
(c) N, N – dimethyl methanamine
(d) N – methyl ethanamine
Answer:
(c) N, N – dimethyl methanamine

Question 53.
Which one of the following is called (N – ethyl – N – methyl) propanamine?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-208
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-209

Question 54.
The IUPAC name of
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-210 is ……………..
(a) Methyl iso propyl amine
(b) N – methyl propan – 1 – amine
(c) N, N – dimethyl methanamine
(d) propan – 1 – amine
Answer:
(b) N – methyl propan – 1 – amine

Question 55.
What is the IUPAC name of
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-211
(a) Ethyl methyl isopropylamine
(b) N, N – dimethyl methanamine
(c) N, N – diethyl butan – 1 – amine
(d) N – ethyl – N – methyl propan -2 – amine
Answer:
(d) N – ethyl – N – methyl propan -2 – amine

Question 56.
Which one of the following is called Hexane – 1, 6 – diamine?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-212
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-213

Question 57.
Which one of the following is the TUPAC name of CH2 = CH – CH2 – NH2?
(a) Isopropyl amine
(b) Allylamine
(c) 1 – amino propane
(d) prop – 2 – en – 1 – amine
Answer:
(d) prop – 2 – en – 1 – amine

Question 58.
Which one of the following is the structure of phenyl methanamine?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-214
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-215

Question 59.
Consider the following statements.
(i) Nitrogen atom of amines is trivalent and carries a lone pair of electron.
(ii) Nitrogen atom of amines is Sp2 hybridised.
(iii) Amines posses pyramidal geometry.
Which of the above statement(s) is / are correct?
(a) (i) and (ii)
(b) (i) and (iii)
(c) (iii) only
(d) (ii) only
Answer:
(b) (i) and (iii)

Question 60.
What is the C – N – C bond angle of trimethylamine?
(a) 109°. 5′
(b) 107°
(c) 108°
(d) 108°. 31’
Answer:
(c) 108°

Question 61.
Which one of the following is the geometry of amines?
(a) Tetrahedral
(b) Pyramidal
(c) Planar triangle
(d) square planar
Answer:
(b) Pyramidal

Question 62.
Which one of the following is formed when cyano methane reacts with LiAIH4?
(a) Ethanamine
(b) Methane
(c) Methanoic acid
(d) Acetic acid
Answer:
(a) Ethanamine

Question 63.
The reducing agent used in mendius reaction is …………
(a) H2/Ni
(b) LiAIH4
(c) Na/C2H5OH
(d) Sn/HCI
Answer:
(c) Na/C2H5OH

Question 64.
The product formed when methyl isocyanide is reduced by Na(Hg)/C2H5OH is …………
(a) Isopropyl amine
(b) Ethanamine
(c) N – methyl methanamine
(d) N – ethyl methanamine
Answer:
(c) N – methyl methanamine

Question 65.
Which one of the following is formed when acetamide reacts with LiAlH4 and H2O?
(a) Methyl amine
(b) Ethylamine
(c) Ammonium acetate
(d) N – methyl ethanamine
Answer:
(b) Ethylamine

Question 66.
In which reaction acetamide is changed to methylamine by the action of Br2/KOH?
(a) Gapriel phthalimide synthesis
(b) Hoffmann degration reaction
(c) Mendius reaction
(d) Mustard oil reaction
Answer:
(b) Hoffmann degration reaction

Question 67.
Which one of the reaction is used in the synthesis of aliphatic primary amines?
(a) Hoffmann ammonolysis
(b) Rosenmund’s reduction
(c) Carbylamine reaction
(d) Gabriel phthalimide synthesis
Answer:
(d) Gabriel phthalimide synthesis

Question 68.
The conversion of ethanol into all types of amines by the action of ammonia along with Alumina is …………
(a) HVZ reaction
(b) Sabatier – mailhe method
(c) Carbylamine reaction
(d) Mendius reaction
Answer:
(b) Sabatier – mailhe method

Question 69.
Consider the following statements.
(i) Amines have higher boiling point than alcohols.
(ii) Lower aliphatic amines are colourless gases whereas higher amines have fish like small.
(iii) Aniline and arylamines are colourless but when exposed to air they become coloured due to oxidation.
Which of the above statement(s) is / are correct?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) only
(d) (iii) only
Answer:
(b) (ii) and (iii)

Question 70.
The correct order of basic strength in the case of ailcyl substituted amines is …………
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-217
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-216

Question 71.
The correct order of basic strength in the case of substituted ethyl amines is ………….
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-218
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-219

Question 72.
The relative basicity of amine follows the order as ………….
(a) Alkyl amines > Aralkyl amines > Ammonia > N – aralkylamine > Arylamine
(b) Aralkyl amines > Ammonia > Arylamine > Alkyl amine > N – aralkylamine
(c) Arylamine Alkyl amine N – aralkylanilne
(d) N – aralkylamine < Arylamine < Ammonia < Alkyl amine < Aralkyl amine Answer: (a) Alkyl amines > Aralkyl amines > Ammonia > N – aralkylamine > Arylamine
Answer:
(a) Alkyl amines > Aralkyl amines > Ammonia > N – aralkylamine > Arylamine

Question 73.
Identify the name of the reaction in which aniline reacts with Benzoyl chloride to form N – Phenyl benzamide?
(a) Hoffmann degradation reaction
(b) Gabriel phthalimide synthesis
(c) Schotten – Baumann reaction
(d) Mustard oil reaction
Answer:
(c) Schotten – Baumann reaction

Question 74.
Which one of the product is formed when aniline reacts with benzoyl chloride in the presence of NaOH?
(a) N – Phenyl benzamide
(b) N – Phenyl ethanamide
(c) Benzamide
(d) N – Benzyl aniline
Answer:
(b) N – Phenyl ethanamide

Question 75.
Which one of the following is formed as product when ethylamine reacts with nitrous acid?
(a) Ethyl nitrite
(b) Nitro ethane
(c) Ethanol
(d) Ethane nitrile
Answer:
(c) Ethanol

Question 76.
Identify X in the following reaction?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-220
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-221
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-222

Question 77.
The reaction of aniline with nitrous acid at low temperature is known as …………
(a) Carbylamme reaction
(b) mustard oil reaction
(c) Diazotisation
(d) Sand meyer’s reaction
Answer:
(c) Diazotisation

Question 78.
Which one of the product is formed with N-methyl aniline reacts with nitrous acid?
(a) Anilinium chloride
(b) N – nitroso methyl phenylamine
(c) Benzene diazonium chloride
(d) Benzylamine
Answer:
(b) N – nitroso methyl phenylamine

Question 79.
Which one of the réaction is called Libermann’s nitroso set?
(a) N – methyl aniline Conc. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-223 N – nitroso methyl phenylamine.
(b) Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-224 Benzene diazonium chloride.
(c) Aniline + CHCI3 + 3KOH → Phenyl iso cyanide.
(d) Methyl amine + CHCl3 + 3KOH → Methyl iso cyanide.
Answer:
(a) N – methyl aniline Conc. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-223 N – nitroso methyl phenylamine.

Question 80.
The conversion of N – methyl aniline into N – nitrosomethyl phenyl amine is known as …………
(a) Carbylamine reaction
(b) mustard oil reaction
(c) Diazotisation
(d) Libermann’s nitroso test
Answer:
(d) Libermann’s nitroso test

Question 81.
Which one of the following reaction is used to identify primary amines?
(a) Schotten – Baumann reaction
(b) Carbylamine reaction
(c) Sand meyer’s reaction
(d) Gattermann reaction
Answer:
(b) Carbylamine reaction

Question 82.
The reaction between methylamine and CS2 is known as ……………
(a) mustard oil reaction
(b) Carbylamine reaction
(c) Sand meyer’s reaction
(d) Gabriel phthalirnide synthesis
Answer:
(a) mustard oil reaction

Question 83.
Which one of the following is formed when aniline reacts with CS2 followed by hydrolysis by cone. HCI?
(a) Phenyl isocyanide
(b) phenyl cyanide
(c) Phenvi isothio cyanate
(d) Benzene diazonium chloride
Answer:
(c) Phenvi isothio cyanate

Question 84.
Consider the following reaction.
(i) Aniline does not undergo friedel – crafts reaction
(ii) Aromatic amine on treatment with NaNO2 + HCI gives diazonium slats.
(iii) Aniline is more basic than ammonia.
Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (i) and (ii)
(c) (iii) only
(d) (ii) only
Answer:
(c) (iii) only

Question 85.
Which one of the following is formed when aniline reacts with cone. H2SO4?
(a) Zwitter ion
(b) Acetanilide
(c) Suiphanilic acid
(d) p – sulphonic bcnzoic acid
Answer:
(c) Suiphanilic acid

Question 86.
Consider the following statements.
(i) Benzene diazonium chloride aqueous solution are neutral to litmus.
(ii) The stability of arene diazonium salt is due to the dispersal of the positive charge over the ring.
(iii) Benzenediazonium chloride is reddish brown colour liquid.
Which of the above statement(s) is / are correct?
(a) (iii) and (ii)
(b) (i) and (ii)
(c) (i) and (iii)
(d) (iii) only
Answer:
(b) (i) and (ii)

Question 87.
Identify the product formed when Benzene diazonium chloride reacts with phosphinic acid?
(a) Benzene
(b) Chioro benzene
(c) Phenol
(d) cyano benzene
Answer:
(a) Benzene

Question 88.
The conversion of Benzene diazonium chloride into chlorobenzene is known as …………
(a) Gabriel phthalimide synthesis
(b) Carbylamine reaction
(c) Sand meyer reaction
(d) Coupling reaction
Answer:
(c) Sand meyer reaction

Question 89.
Identify X and Y in the following reaction
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-225
(a) C6H5CI + N2
(b) C6H6 + N2
(c) C6H5CI + NH4CI
(d) C6H5CI + H2
Answer:
(a) C6H5CI + N2

Question 90.
Which one of the following is formed when benzene diazonium chloride is boiled with water?
(a) Benzene
(b) Phenol
(c) Chiorobenzene
(d) Aniline
Answer:
(b) Phenol

Question 91.
Complete the following reaction.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-226
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-1.1
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-106

Question 92.
What is the name of the reaction in which benzene diazonium chloride react with benzene to give Biphenyl?
(a) Sandmeyer’s reaction
(b) Gomberg reaction
(c) Gattermann reaction
(d) Baltz – schiemann reaction
Answer:
(b) Gomberg reaction

Question 93.
Which one of the following reagent reacts with ben.zene diazonium chloride to give biphenyl?
(a) Chioro benzene
(b) Bromobenzene
(c) Benzene
(d) Acetic acid
Answer:
(c) Benzene

Question 94.
Which one of the following is formed when Benzene dia.zonium chloride reacts with Aniline?
(a) p – hydroxy azo benzene
(b) 2 – phenyl azo – 4 – methyl
(c) Biphenyl
(d) p – amino azo benzene
Answer:
(d) p – amino azo benzene

Question 95.
Which one of the following should react with Benzene diazonium chloride to get orange dye?
(a) Aniline
(b) phenol
(c) 0 – cresol
(d) P – cresol
Answer:
(b) phenol

Question 96.
Aniline + Benzene diazonium chloride → X. Identify X.
(a) orange dye
(b) yellow dye
(c) malachite green dye
(d) madder dye
Answer:
(b) yellow dye

Question 97.
Which one of the following is the IUPAC name of CH3 – CH2 – CH2CN?
(a) Propiono nitrite
(b) Butane cyanide
(c) Isobutyro nitnie
(d) Butane nitrile
Answer:
(d) Butane nitrile

Question 98.
Which one of the following is formed when methyl magnesium bromide reacts with cyanogen chloride?
(a) methane nitrile
(b) ethane nitrite
(c) Acetamide
(d) Nitro ethane
Answer:
(b) ethane nitrite

Question 99.
The reagent used in the conversion of CH3CONH2 into CH3CN is …………
(a) Br2/KOH
(b) conc. HNO3
(c) Sn/HCI
(d) P2O5
Answer:
(d) P2O5

Question 100.
The product of complete hydrolysis of Ethane nitrite is …………
(a) Ethane
(b) Ethyl nitrate
(c) Nitro ethane
(d) Acetic acid
Answer:
(d) Acetic acid

Question 101.
What is the name of the reaction that take place between Ethane nitrile and Ethyl propionate?
(a) Coupling reaction
(b) Levine and hauser acetylation
(c) Diazotisation
(d) Acetic acid
Answer:
(b) Levine and hauser acetylation

Question 102.
Which one of the following is formed when methyl iso cyanide ùndergoes acid hydrolysis?
(a) Dimethyl amine + H2O
(b) Acetic acid + Formic acid
(c) Methyl amine + Formic acid
(d) Methyl cyanide + Ammonia
Answer:
(c) Methyl amine + Formic acid

Question 103.
The product formed when methyl isocyanide is heated to 250°C is …………
(a) Methyl amine
(b) Methyl cyanide
(c) Ethyl Cyanide
(d) Amino ethane
Answer:
(b) Methyl cyanide

Question 104.
Which one of the following is used a fuel for cars?
(a) CH3NO2
(b) CH3NH2
(c) CH3N
(d) CH3NC
Answer:
(a) CH3NO2

Question 105.
Chloropicrin is used as ……….
(a) antiseptic
(b) analgesic
(c) insecticide
(d) fertilizer
Answer:
(c) insecticide

Question 106.
Which one of the following is used as a fuel additive and precursor to explosive?
(a) Nitroglycerine
(b) Nitro methane
(c) Nitro benzene
(d) Nitro ethane
Answer:
(d) Nitro ethane

Question 107.
Which one of the following is known as sweet spirit of nitre?
(a) 10 % solution of methyl nitrite
(b) 4% solution of ethyl nitrite
(c) 10% solution of ethyl nitrite
(d) 40% solution of methanal
Answer:
(b) 4% solution of ethyl nitrite

Question 108.
Which one of the following is used as diuretic?
(a) Nitro methane
(b) Nitrobenzene
(c) ethyl nitrite
(d) Oil of mirbane
Answer:
(c) ethyl nitrite

Question 109.
Which of the following is used to produce lubircating oils in motors and machinery?
(a) Nitro benzene
(b) m – dinitro benzene
(c) 1, 3, 5 – trinitro benzene
(d) Nitro glycerine
Answer:
(a) Nitro benzene

Question 110.
Which of the following is used in the manufacture of aniline, synthetic rubber, dyes and explosives like TNT, TNB?
(a) Nitro ethane
(b) Aminobenzene
(c) Nitro benzene
(d) Benzene diazonium chloride
Answer:
(c) Nitro benzene

Question 111.
Which of the following is used in textile industries and also as a solvent in perfume industries?
(a) Alkyl cyanide
(b) Alkyl iso cyanide
(c) Alkyl iso thio cyanate
(d) Alkyl amine
Answer:
(a) Alkyl cyanide

Question 112.
Which one of the following is used as an anticancer agent used to stomach and colon cancer?
(a) Vitamin C
(b) Cobalamine
(c) mitomycin C
(d) Streptomycin
Answer:
(c) mitomycin C

Question 113.
In the Hoffmann bromamide degradation reaction, the number of moles of KOH and Br2 used per mole of amine produced are …………..
(a) four moles of KOH and two moles of Br2
(b) two moles of KOH and two moles of Br2
(c) four moles of KOH and one moles of Br2
(d) one moles of KOH and one moles of Br2
Answer:
(c) four moles of KOH and one moles of Br2

Question 114.
What is the product obtained in the following reaction?

Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-229
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-228

Question 115.
The reagent with which the following reaction is best accomplished is ………….
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-227
(a) H3PO3
(b) H3PO4
(c) H3PO2
(d) NaHSO3
Answer:
(c) H3PO2

Question 116.
The amine “A” when treated with nitrous acid gives yellow oily substance. The amine “A” is …………
(a) Triethylamine
(b) Trimethylamine
(c) aniline
(d) Ethyl methyl amine
Answer:
(d) Ethyl methyl amine

Question 117.
Which one of the following amide will not undergo Hoffmann bromamide reaction?
(a) CH3CONH2
(b) CH3CONHCH3
(c) C6H5CONH2
(d) CH3CH2CONH2
Answer:
(b) CH3CONHCH3

Question 118.
Replacement of diazonium group by fluorine is known as …………..
(a) Gattennann reaction
(b) Sandmeyer reaction
(c) Baltz – Schiemann reaction
(d) Comberg reaction
Answer:
(c) Baltz – Schiemann reaction

Question 119.
Considering the basic strength of amines in aqueous solution, which are has the smallest pK value?
(a) CH3NH2
(b) (CH3)3N
(c) C6H5NH2
(d) (CH3)3NH
Answer:
(d) (CH3)3NH

Question 120.
Which one of the following is the strongest base in aqueous solution?
(a) Trimethyl amine
(c) Dimethyl amine
(d) methyl amine
(b) Aniline
Answer:
(c) Dimethyl amine

Question 121.
Diethyl amine when treated with nitrous acid yields
(a) Diethyl ammonium nitrite
(b) Ethyl alcohol
(c) N – nitroso diethyl amine
(d) Triethyl ammonium nitrate
Answer:
(c) N – nitroso diethyl amine

Question 122.
Which one of the following on reduction with Lithium aluminium hydride yields a secondary amine?
(a) Methyl iso cyanide
(b) Acetamide
(c) Methyl cyanide
(d) Nitro ethane
Answer:
(a) Methyl iso cyanide

Question 123.
The action of nitrous acid on ethylamine gives
(a) Ethane
(b) ammonia
(c) Ethyl alcohol
(d) Ethyl nitrite
Answer:
(c) Ethyl alcohol

Question 124.
Indicate which nitrogen compound amongst the following would undergo Hoffmann reaction with Br2 and strong KOH to furnish primary amine kNH2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-232
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-231

Question 125.
The correct order of basicity of the following compounds is …………..
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-230
(a) (ii) > (i) > (iii) > (iv)
(b) (i) > (iii) > (ii) > (iv)
(c) (iii) > (i) > (ii) > (iv)
(d) (i) > (ii) > (iii) > (iv)
Answer:
(b) (i) > (iii) > (ii) > (iv)

Question 126.
Which of the following would be most reactive towards nitration?
(a) Benzene
(b) nitrobenzene
(c) Toluene
(d) Chiorobenzene
Answer:
(b) nitrobenzene

Question 127.
Aniline reacts with acetaldehyde to form.
(a) Schiff’s base
(b) carbylamine
(c) Imine
(d) acetaldoxime
Answer:
(a) Schiff’s base

Question 128.
Which of the following is the strongest base?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-235
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-234

Question 129.
A primary amine is formed from an amide by the treatment of bromine and alkali. The primary amine has ……………
(a) 1 Carbon atom less than amide
(b) 1 carbon atom more than amide
(c) 1 hydrogen atom less than amide
(d) 1 Hydrogen atom more than amide
Answer:
(a) 1 Carbon atom less than amide

Question 130.
Liebermann’s nitroso reaction is used for testing ………….
(a) 1° amine
(b) 2° amine
(c) 3° amine
(d) all the above
Answer:
(b) 2° amine

Question 131.
A nauseating smell in the carbylamine test for primary amines is due to the formation of …………
(a) iso cyanide
(b) chloroform
(c) cyanide
(d) iso thiocyanate
Answer:
(a) iso cyanide

Question 132.
A positive carbylamine test is given by …………
(a) N, N – dimethyl aniline
(b) 2, 4 – dimethyl aniline
(c) N – methyl – 0 – methyl aniline
(d) p – methyl benzylamine
Answer:
(b) 2, 4 – dimethyl aniline

Question 133.
When primary amine is heated with CS2 in the presence of excess of mercuric chloride, it gives isothiocyanate. This reaction is called ……………..
(a) Hoffmann bromamide reaction
(b) Carbylamine reaction
(c) Perkin’s reaction
(d) Hoffmann mustard oil reaction
Answer:
(d) Hoffmann mustard oil reaction

Question 134.
Diazo – coupling reaction is useful to prepare some
(a) Dyes
(b) proteins
(c) pesticides
(d) plastics
Answer:
(a) Dyes

Question 135.
Carbylamine test is used in the detection of …………
(a) aliphatic 2° amine
(b) Aromatic 1° amine
(c) Aliphatic l amine
(d) both aliphatic and aromatic 1° amine
Answer:
(d) both aliphatic and aromatic 1° amine

Question 136.
Which of the following amine will not react with nitrous acid to give nitrogen?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-233
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-236

Question 137.
Which of the following compound is expected to be more basic?
(a) Aniline
(b) Methylamine
(c) Ethylamine
(d) Hydroxylamine
Answer:
(c) Ethylamine

Question 138.
Nitro group in Nitro benzene is a ………….
(a) ortho directing group
(b) Meta directing group
(c) Para directing group
(d) ortho and para directing group
Answer:
(b) Meta directing group

Question 139.
Which of the following amines would undergo diazotisation?
(a) CH3NH2
(b) C2H5NH2
(c) C6H5NH2
(d) (CH3)2NH
Answer:
(c) C6H5NH2

Question 140.
Primary amines can be distinguished from secondary and tertiary amines by reacting with.
(a) CHCI3 and alkali
(b) CH3I
(c) CHCl3 alone
(d) Zn dust
Answer:
(a) CHCI3 and alkali

Question 141.
A solution of methylamine …………..
(a) turns blue litmus red
(b) turns red litmus blue
(c) does not effect red or blue litmus
(d) Bleaches litmus
Answer:
(b) turns red litmus blue

Question 142.
Oil of mirbane is ………..
(a) Aniline
(b) Nitro ethane
(c) p – amino azo benzene
(d) Nitro benzene
Answer:
(d) Nitro benzene

Question 143.
Identify the product Z in the series of the reaction ………….
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-237
(a) CH3CHO
(b) CH3CONH2
(c) CH3COOH
(d) CH3CH2NHOH
Answer:
(c) CH3COOH
Solution:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-239

Question 144.
Primary and secondary amines are distinguished by ……….
(a) Br2 / KOH
(b) HCIO4
(c) NH3
(d) HNO2
Answer:
(d) HNO2

Question 145.
Aniline on treatment with excess bromine water gives
(a) Anilinium bromide
(b) O – bromo aniline
(c) 2, 4, 6 – tribromo aniline
(d) p – bromo aniline
Answer:
(c) 2, 4, 6 – tribromo aniline

Question 146.
Which of the following is not used as an explosive?
(a) Trinitro toluene
(b) Trinitro benzene
(c) Trinitro glycerine
(d) Nitro benzene
Answer:
(d) Nitro benzene

Question 147.
Which of the following has a pyramidal structure?
(a) Trimethyl amine
(b) Water
(c) Acetylene
(d) Methane
Answer:
(a) Trimethyl amine

Question 148.
Which one of the following reacts with benzaldehyde to give schiff’s base.
(a) Acidified K2Cr2O7
(b) formaldehyde
(c) Aniline
(d) Potassium cyanide
Answer:
(c) Aniline

Question 149.
Ethyl amine can be prepared by the action of bromine and caustic potash on
(a) Acetamide
(b) propionamide
(c) Formamide
(d) Methyl Eyanide
Answer:
(b) propionamide

Question 150.
Which of the following reaction will not give primary amine?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-240
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-238

Question 151.
Which of the following compound is the strongest base?
(a) Ammonia
(b) Aniline
(c) Methyl amine
(d) N – methyl aniline
Answer:
(c) Methyl amine

Question 152.
Azo dyes are prepared from ………….
(a) Phenol
(b) Aniline
(c) Benzaldehyde
(d) Both (i) and (ii)
Answer:
(d) Both (i) and (ii)

II. Fill in the blanks.

  1. ………….. is an organic compound needed to maintain the health of nerves, skin and red blood cells.
  2. Plants synthesis and to protect them from being eaten away by insect ………….. and ………….. other animals.
  3. ………….. compounds are the important constituents of explosives, drugs, dyes, fuels, polymers, synthetic rubbers.
  4. Dopamine act as …………..
  5. ………….. dilates blood vessels.
  6. Tertiary nitro alkanes donot exhibit ………….. due to the absence of a-H atom.
  7. Aci form of nitro alkanes gives ………….. colour with ferric chloride.
  8. Aci form of nitro alkanes are otherwise called ………….. or …………..
  9. Laboratory preparation of Nitro ethane from ethyl bromide follows ………….. mechanism
  10. Except ………….. other alkanes gives a mixture of nitro alkanes due to C – C cleavage by nitration of alkanes.
  11. Oxidation of acetaldoxime with ………….. gives 1 – nitro ethane.
  12. ………….. is suspected to cause genetic damage and be harmful to the nervous system.
  13. Nitro benzene on reduction with SnCl2 + KOH gives …………..
  14. Nitrobenzene on alkaline medium reduction gives …………..
  15. Amines posses ………….. geometry.
  16. The nitrogen atom in amine is ………….. hybridised.
  17. Gabriel phthalimide synthesis is used for the preparation of …………..
  18. Ammonolysis of hydroxyl compounds is called ………….. reaction.
  19. Aniline when exposed to air becomes coloured due to …………..
  20. Alkyl amines are stronger base than …………..
  21. Acylation and benzoylation of Aniline aer ………….. reactions.
  22. Libermann’s nitroso test is used to detect …………..
  23. ………….. test is used to identify primary amine.
  24. Direct nitration of aniline gives O and P – nitro aniline along with ………….. due to oxidation.
  25. The conversion of benzene diazonium chloride to benzene by H3PO2 proceeds through ………….. mechanism.
  26. Benzene diazonium chloride when boiled with water gives …………..
  27. The conversion of Benzene diazonium chloride is Biphenyl is called ………….. reaction.
  28. Coupling reaction generally occurs at ………….. position of Benzene ring.
  29. The condensation reaction of esters with nitrites containing a – hydrogen is known as …………..
  30. Chloropicrin is used as an …………..
  31. 4% solution of ethyl nitrite in alcohol is known as …………..
  32. Sweet spirit of nitre is used as …………..
  33. ………….. is used to produce lubricating oils in motors and machinery.
  34. ………….. an anti cancer agent used to treat stomach and colon cancer.
  35. Mitomycin C contains an ………….. ring.
  36. ………….. is used as percursor to explosive.
  37. An organic nitrogen compound ………….. is used as an insecticide.
  38.  ………….. is known as sweet spirit of nitre.
  39.  Chloropicnn ………….. is used as an insecticide.

Answer:

  1. Pyridoxine, vitamin B6
  2. alkaloids, biologically active amines
  3. Nitrogen
  4. Neurotransmitter
  5. Histamine
  6. tautomerism
  7. Reddish brown
  8. Pseudo acids (or) Nitronic acids
  9. SN2
  10. Methane
  11. trifluoroperoxy acetic acid
  12. Nitro ethane
  13. Azobenzene
  14. Hydrazobenzene
  15. Pyramidal
  16. SP3
  17. Aliphatic primary amines
  18. Sabatier – mailhe
  19. Oxidation
  20. Ammonia
  21. Nucleophilic substitution (or) Schotten Baumann
  22. Secondary amine
  23. Carbylamine reaction (or) Hoffmann mustard oil reaction
  24. dark coloured tars
  25. Free radical chain
  26. Phenol
  27. Gomberg
  28. Para
  29. Levine and hauser
  30. insecticide
  31. Sweet spirit of nitre
  32. diuretic
  33. Nitrobenzene
  34. Mitomycin C
  35. Aziridine
  36. Nitro ethane
  37. chloropicrm
  38. 4% solution of ethyl nifrite in alcohol
  39. CCI3NO2

III. Match the Column I and II using the code given below the columns.

Question 1.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-243
Answer:
(a) 3 4 2 1

Question 2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-241
Answer:
(b) 3 4 1 2

Question 3.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-244
Answer:
(c) 3 4 1 2

Question 4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-242
Answer:
(a) 4 3 1 2

Question 5.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-245
Answer:
(b) 4 1 2 3

Question 6.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-246
Answer:
(a) 3 4 2 1

Question 7.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-247
Answer:
(a) 3 1 4 2

Question 8.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-250
Answer:
(a) 2 1 4 3

Question 9.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-248
Answer:
(a) 3 4 1 2

Question 10.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-249
Answer:
(a) 4 1 2 3

IV. Assertion and reasons.

Question 1.
Assertion(A): Tertiary nitro alkanes do not exhibit tautomerism.
Reason (R): Tertiary nitro alkanes do not have of α – H atom.
(a) Both A and R are correct and R is explains A
(b) Both A and R are not correct
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is explains A

Question 2.
Assertion(A): Primary and secondary nitroalkanes show an equilibrium mixture of two tautomers namely nitro and aci form.
Reason (R): Both primary and secondary nitroalkanes are having a-H atoms.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct but R is not the correct explanation of ofA
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 3.
Assertion(A): Nitro alkanes dissolve in NaOH solution to form a salt.
Reason (R): The α – H atom of 1° and 2° nitroalkanes show acidic character because of the electron withdrawing effect of NO2 group.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct but R is not the correct explanation of of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 4.
Assertion(A): 2 – nitro propane is more acidic than nitro mehtane.
Reason (R): When the number of alkyl group attached to a carbon increases, acidity decreases. due to +1 effect of alkyl groups.
(a) Both A and R are correct but R is not the correct explanation of of A
(b) Both A and R are correct and R is the correct explanation of A
(c) A is correct but R is wrong.
(d) A is wrong but R is correct
Answer:
(d) A is wrong but R is correct

Question 5.
Assertion(A): Nitrobenzene cannot be prepared from bromo benzene by action of ethanolic solution of potassium nitrite.
Reason (R): The bromine directly attached to the benzene ring cannot be cleaved easily.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct but R is not the correct explanation of of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 6.
Assertion(A): Nitrobenzene undergoes friedel – craft reaction.
Reason (R): Nitrobenzene have strong deactivating – NO2 group.
(a) Both A and R arc correct but R is not the correct explanation of A
(b) Both A and R are correct and R is the correct explanation of A
(c) A is wrong but R ¡s correct
(d) A is correct but R is wrong
Answer:
(c) A is wrong but R ¡s correct

Question 7.
Assertion(A): Amines posses pyramidal geometry.
Reason (R): Nitrogen atoms of amines is trivalent and has four sp3 hybridised orbital. Three sp3 orbitais overlap with orbitais of hydrogen and four sp3 orbitais contain a lone pair of electrons.
(a) Both A and R are correct and R ¡s the correct explanation of A
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong.
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R ¡s the correct explanation of A

Question 8.
Assertion(A): The C – N – C bond angle of trimethyl amine is 108°.
Reason (R): The bond angle of C – N – C is due to the repulsion between the bulky methyl groups.
(a) Both A and R are wrong
(b) A is correct and R is wrong
(c) Both A and R are correct and R is the correct explanation of A
(d) Both A and R are correct but R is not the correct explanation of A
Answer:
(c) Both A and R are correct and R is the correct explanation of A

Question 9.
Assertion(A): Aniline cannot be prepared by Gabriel phthalimide synthesis.
Reason (R): Arylhalides do not undergo nucleophilic substitution with the anion formed by phthalimide
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 10.
Assertion(A): Amines have lower boiling point than alcohols.
Reason (R): Nitrogen has lower electronegative value than oxygen and hence the N – H bond is less polar than – OH bond.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 11.
Assertion(A): Tertiary methyl amine is less soluble in water than methyl amine.
Reason (R): Solubility decreases due to the increase in size of the hydrophobic alkyl group.
(a) Both A and R are wrong
(b) A is correct but R is wrong.
(c) Both A and R are correct and R is the correct explanation of A
(d) Both A and R are correct but R is not the correct explanation of A
Answer:
(c) Both A and R are correct and R is the correct explanation of A

Question 12.
Assertion(A): Aniline reacts with acids to form salts and also reacts with electrophiles.
Reason (R): The lone pair of electrons on nitrogen atom in amines makes them basic as well as nucleophilic.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 13.
Assertion(A): Alkyl amines are stronger base than Ammonia.
Reason (R): When a +I gorup like alkyl group is attached to nitrogen increases the electron density on the nitrogen which makes the electron pair readily available for protonation.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is wrong but R is correct
(d) A is correct but R is wrong
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 14.
Assertion(A): 2° amines are more basic.
Reason (R): Due to + I effect, steric effect and hydration effect cause 2° amines are more basic.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 15.
Assertion(A): Aromatic amines are less basic than ammonia.
Reason (R): The lone pair of electrons on nitrogen atom in aniline (aromatic amine) gets delocalised over the benzene ring and less available for protonation.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong.
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 16.
Assertion(A): Electrophilic substituion in aniline take place at ortho and para position.
Reason (R): The – NH2 group is a strong activating group and lone pair of electrons on the nitrogen atom is in conjugation with benzene ring that increases electron density at ortho and para position.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong.
(d) Both A and R are wrong
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 17.
Assertion(A): Acylation of amines gives a mono substituted product whereas alkylation of amines gives polysubstituted product.
Reason (R): Acyl group sterically hindered the approach to further acyl group.
(a) Both A and R are wrong
(b) A is correct but R is wrong
(c) Both A and R are correct and R is the correct explanation of A
(d) Both A and R are correct but R is not the correct explanation of A
Answer:
(c) Both A and R are correct and R is the correct explanation of A

Question 18.
Assertion(A): Acetanilide is less basic than anime.
Reason (R): Acetylation of aniline results in the decrease of electron density on nitrogen.
(a) Both A and R are correct and R explains A.
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R explains A.

Question 19.
Assertion(A): Aromatic 1° amines can be prepared by Gabriel phthalimide synthesis.
Reason (R): Aryl halides undergo nucleophilic substitution with the anion formed by phthalimide.
(a) Both A and R are correct and R is explains A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(b) Both A and R are wrong

Question 20.
Assertion(A): Aniline does not undergo Friedel – Crafts reaction.
Reason (R): Aniline donates its lone pair of electrons to the Lewis acid AiCl3 to form an adduct which inhibits further electrophilic substitution reaction.
(a) Both A and R are correct and R is explains A.
(b) Both A and R are wrong
(c) A is correct but R is wrong.
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is explains A.

V. Find the odd one out and give the reasons.

Question 1.
Pyridoxine, Dopamine, Histamine, Aspirin.
Answer:
Aspirin:
It is acid derivative and used as medicine whereas other are organic nitrogen compounds used in medicine.

Question 2.
Trinitro glycerine, Glyceryl triacetate, Trinitro benzene, Trinitro toluene
Answer:
Glyceryl triacetate:
It is a an ester of glycerol whereas others are organic nitrogen compounds used in making explosives.

Question 3.
N – methyl methanamine, N – methyl ethanamine, N – phenyl benzamide, N,N – dimethyl methanamine
Answer:
N,N – dimethyl methanamine: It ¡s an example of tertiary amine whereas others are secondary amine.

Question 4.
Propan – 2 – amine, N – ethyl – N – methyl, propan – 2 – amine, N,N – dimethyl methanamine, N, N – diethyl butan – 1 – amine
Answer:
Propan – 2 – amine:
It is a secondary amine whereas others are tertiary amines.

Question 5.
P – hydroxy azo benzene, Hydrazo benzene, P – amino azo benzene, 2 – Phenyl azo – methyl phenol
Answer:
Hydrazo benzene: It is not a dye whereas others are dyes.

Question 6.
Methyl ISo cyanide, Methyl cyanide, Acetic anhydride, Ethyl amine Nitro ethane
Answer:
Acetic anhydride:
It is an acid derivative whereas others are organic nitrogen compounds.

Samacheer Kalvi 12th Chemistry 13 Organic Nitrogen Compounds 2 Mark Questions and Answers

Question 1.
What is called nitro compound? Give one example.
Answer:
1. Ntiro compounds are considered as the derivatives of hydrocarbons. 1f one of the hydrogen atom of hydrocarbon is replaced by the – NO2 group, the resultant organic compound is called a ntiro compound.

2. E.g., CH3 – CH2 – NO2. Nitro ethane

Question 2.
Define Tautomerism. Give example. Why tertiary nitro alkanes do not ethibit tautomerism?
Answer:
1. Tautomerism is an isomerism in which the isomers change into one another with great ease of shifting of proton so that they exist together in equilibrium.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-251

2. Tertiary nitro alkanes do not exhibit tautomerism due to absence of a-H atom.

Question 3.
Differentiate between nitro form and acid form of tautomerism of nitro methane.
Answer:
Nitro form

  1. Less acidic
  2. Dissolves in NaOH slowly.
  3. Decolourises FeCl3 solution.
  4. Electrical conductivity is low.

Aci form

  1. More acidic and also called pseudo acids (or) Nitrolic acids.
  2. Dissolves in NaOH instantly.
  3. With FeCI3 gives reddish brown colour.
  4. Electrical conductivity is high.

Question 4.
Compare the acid strength of the following compounds

  1. Nitro methane
  2. Nitro ethane
  3. 2 – nitro propane.

Answer:
1. The α – H atom of 10 and 20 nitroalkanes show acidic character because of the electron withdrawing effect of NO2 group.

2. Nitroalkanes dissolve in NaOH solution to form a salt.

3. When the number of alkyl group attached to a carbon increases, acidity decreases. due to +1 effect of alkyl groups.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-253

Question 5.
Nitro benzene cannot be prepared from Bromo benzene by direct nitration. Give reason.
Answer:
Nitro benzene cannot be prepared from Bromo benzene because the bromine directly attached to the benzene ring cannot be cleaved easily.

Question 6.
How would you convert Acetaldoxime into Nitroethane?
Answer:
Oxidation of acetaldoxime and with trifluoro peroxy acetic acid gives nitroethane.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-252

Question 7.
How is nitrobenzene from benzene?
Answer:
When benzene is heated with a nitrating mixture (Con.HNO3 + Con.H2SO4), at 330 K, electrophilic substitution takes place to form nitro benzene.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-256

Question 8.
How will you prepare p – dinitrobenzene from p – nitroanlline?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-255

Question 9.
How is amino group can be directly converted into nitro group? Explain with an example.
Answer:
Amino group can be directly converted into nitro group using caro’s acid (H2SO5) (or) persuiphuric acid (H2S2O8) (or) peroxytrifluro acetic acid (F3C.COOOH) as oxidising agent.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-257

Question 10.
Explain the action of tin and hydrochloric acid with ethyl nitrite.
Answer:
Ethyl nitrite on reduction with Sn / HCI gives ethanol.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-254

Question 11.
Explain about the acid (or) basic hydrolysis of ethyl nitrite.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-262

Question 12.
What is Chioropicrin? How is ¡t prepared? Give its uses.
Answer:
CCI3NO2 is Chioropicrin. It is prepared from nitro methane with Cl2 in the presence of NaOH. The a – H atom of nitroalkanes are successively replaced by halogen atoms. It is used as an insecticide.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-259

Question 13.
Explain – Nef carbonly synthesis.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-260

Question 14.
What happens when nitrobenzene Is treated with Ni (or) Pt (or) LiAIH4?
Answer:
Nitrobenzene undergoes reduction with Ni (or) Pt (or) LiAlH4 to give aniline.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-261

Question 15.
Write a note about structure of Amines.
Answer:
1. Nitrogen atom of amines is trivalent and carries a lone pair of electron and sp3 hybridised, out of the four sp3 hybridised orbitais of nitrogen, three sp3 orbitais overlap with orbitais of hydrogen (or) alkyl groups of carbon, the fourth sp3 orbital contains a lone pair of electron. Hence, amines posses pyramidal geometry.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-258

2. Due to presence of lone pair of electron C – N – H (or) C – N – Cbond angle is less than the normal tetrahedral bond angle 109.5v. For example, the C – N – C bond angle of trimethylamine is 108°. This is due to the repulsion between the bulky methyl groups.

Question 16.
How would you convert Nitroethane to ethanamine?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-267

Question 17.
Explain the action of Pt (or) Sn/HCI with nitrobenzene.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-263

Question 18.
Explain mendlus reaction.
Answer:
Reduction of alkyl or aryl cyanides with Na / C2H5OH is used as a reducing agent is called mendius reaction.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-266

Question 19.
Explain the action of sodium amalgum and ethanol with Methyl isocyanide.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-265

Question 20.
What happens when sodium azide is treated with methyl bromide?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-264

Quesion 21.
How would you convert chiorobenzene to aniline?
Answer:
When chiorobenzene is heated with alcoholic ammonia, aniline is obtained.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-269

Question 22.
Explain Sabatier – MalIhe method.
Answer:
when vapour of an alcohol and ammonia are passed over alumina, W2O5 (or) silica at 400°C, all types of amines are formed. This method is called Sabatier – Mailhe method.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-268

Question 23.
Convert phenol into aniline.
Answer:
Phenol reacts with ammonia at 300°C in the presence of anhydrous ZnCl2 to give aniline.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-270

Question 24.
Compare the boiling points of 10, 2° and 3° amInes.
Answer:
1. The boiling point of various amines follows the order,
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-271

2. Due to the polar nature of primary and secondary amines, can form intermolecular hydrogen bonds using their lone pair of electron on nitrogen atom. However, tertiary amines do not form intermolecular hydrogen bond and they have lower boiling point than 1° and 2° amines.

Question 25.
Aniline is basic in nature. Justify this statement.
Answer:
The lone pair of electrons on nitrogen atom in aniline makes it base. Aniline reacts with mineral acids to form salt.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-272

Question 26.
Alkyl amines are stronger bases than ammonia. Justify this statement.
Answer:
When a + I group like and alkyl group is attached to the nitrogen increase the electron density on nitrogen which makes the electron pair readily available for protonation. Hence alkyl amines are stronger than ammonia.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-274

Question 27.
Explain the action of acretyl chloride with ethyl amine?
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-277

Question 28.
What happens when ethylamine reacts with nitrous acid?
Answer:
Ethylamine reacts with nitrous acid to give ethyl diazonium chloride, which is unstable and it is converted to ethanol by liberating N2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-276

Question 29.
Explain the action of nitrous acid with N – methyl aniline.
Answer:
N – methyl aniline react with nitrous acid to give N – nitroso amine as yellow oily liquid which is insoluble in water.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-273

Question 30.
Explain the action of nitrous acid with trimethyl amine.
Answer:
Aliphatic tertiary amine reacts with nitrous acid to form trialkyl ammonium nitrite salt, which is soluble in water.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-275

Question 31.
What happens when nitrous acid is treated iith N, N – dimethyl aniline?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-278

Question 32.
Explain Hoffmann mustard oil reaction. (or) Explain the action of CS2 with aniline.
Answer:
When aniline is treated with CS2, or heated together, S – diphenylthio urea is formed, which on boiling with strong HCI , phenyl isothiocyanate (phenyl mustard oil), is formed.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-281
The above reaction is known as Hoffmann mustard oil reaction.

Question 33.
Explain the action of Br2water with aniline.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-280

Question 34.
How would you prepare p – bromo aniline from aniline?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-279

Question 35.
How would you prepare p – nitro from aniline?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-282

Question 36.
Explain the action of hypophosphrous acid with Benzene diazonium chloride. (or) Explain the action of ethanol with benzene diazonium chloride.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-283

Question 37.
Explain Gattermann reaction.
Answer:
Gattermann reaction.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-284

Question 38.
How would you get iodo benzene form benzene diazonium chloride.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-286

Question 39.
Explain Baltz – schiemann reaction.
Answer:
When benzene diazonium chloride is treated with fluoroboric acid, benezene diazonium tetra fluoroborate is precipitated which on heating decomposes to give fluorobenzene.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-285

Question 40.
Convert Benzene diazonium chloride into phenol.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-287

Question 41.
StartIng form Benzene diazonium chloride, how will you get Nitrobenzene?
Answer:
When benzene diazonium chloride is treated with fluoroboric acid we get diazonium fluoroborate which on treated with sodium nitrite solution in the presence of copper, nitrobenzene is obtained.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-289

Question 42.
Convert benzene diazonium chloride to benzoic acid
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-291

Question 43.
Explain the action of SnCl2 and HCl with benzene diazonium chloride.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-1.0

Question 44.
Starting from benzene diazonium chloride, how would you get bright organge azo dye?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-288

Question 45.
Write the structural formula and TUPAC name of the following compounds.

  1. Isobutyl nitrite
  2. Benzo nitrile

Answer:
1. Isobutyl nitrite
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-294
2 – methyll propane nitrile

2. Benzo nitrile C6H5CN Benzene carbo nitrile

Question 46.
Draw the structural formula of

  1. 3 – cyano butanoic acid
  2. 2 – bromo – 3 – chloro – 3 – methyl pentane nitrile.

Answer:
1. 3 – cyano butanoic acid
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-296

2. 2 – bromo – 3 – chloro
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-293
– 3 – methyl pentane nitrile

Question 47.
How will you get propane nitrile from ethyl bromide?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-295

Question 48.
Starting from methyl magnesium bromide, how would you obtain ethane nitrile?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-292

Question 49.
Explain thrope nitrite condensation.
Answer:
Self condensation of two molecules of alkyl nitrile in the presence of sodium to form iminonitrile is called Thrope nitnie condensation.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-298

Question 50.
Explain Levine and hauser acetylation.
Answer:
The nitrites containing α – hydrogen also undergo condensation with esters in the presence of sodamide in ether to form ketoriitriles. This reaction is known as Levine and hauser acetylation.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-297

Question 51.
How would you prepare the following compounds by carbylamines reaction.

  1. Methyl isocyanide
  2. Phenyl isocyanide

Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-299

2.  Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-302

Question 52.
Complete the following reactions.
 Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-301
Answer:
 Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-303

Question 53.
How is methyl isocyanide changed to methyl cyanide?
Answer:
 Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-300

Question 54.
What are the uses of nitrobenzene?
Answer:

  1. Nitro benzene is used to produce lubricating oils in motors and machinery.
  2. It is used in the manufacture of dyes, drugs, pesticides, synthelic rubber, aniline and explosives like TNT, TNB.

Question 55.
Primary amines have higher boiling points than tertiary amines why?
Answer:
Due to the presence of two H – atom on N – atoms of primary amines they undergo extensive intermolecular H – bonding while tertiary amines due to the absence of a H – atom on the N – atom do not undergo H – bonding. As a result primary amines have higher boiling points than tertiary amines.

Question 56.
How is m – nitroaniline obtained from nitrobenzene?
Answer:
 Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-304

Question 57.
How is aniline obtained from benzoic acid?
Answer:
 Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-305

Question 58.
How will you convert benzene into aniline?
Answer:
 Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-308

Question 59.
How will you distinguish between.
 Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-306
Answer:
1. By reacting with NaNO2 and HCl at the temperature of around 0 to 5°C, Aniline will from diazonium salt. CH3NH2 will form methanol and bubbles of N2 gas will come out of the solution.

2. By Hinsberg’s reagent, C6H5SO2CI(CH3)3N will not react. (CH3)2NH will form a product which is insoluble in alkali.

Question 60.
Account for any two of the following.

  1. Amines are basic substances while amides are neutral.
  2. Aromatic amines are weaker bases than aliphatic amines.

Answer:
1. In amines alkyl group is an electron releasing group which increases the electron density on nitrogen thus making them basic whereas in amides
 Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-307
group is electron withdrawing, therefore they are neutral.

2. It is because aryl group is an electron withdrawing group which decreases electron density on nitrogen atom, making them less basic whereas alkyl group is electron releasing which makes alkylamines more basic.

Question 61.

  1. Assign a reason for the following statements – Alkylamines are stronger bases than arylamines.
  2. How would you convert methylamine into ethylamine?

Answer:
1. It is because in arylamines the – NH2 group is attached directly to the benzene ring. It results in the unshared electron pair on nitrogen atom to be in conjugation with the benzene ring and thus making it less available for protonation whereas alkyl group are electron releasing group.

2.  Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-311

Question 62.

  1. How will you convert an alkyl halide to a primary amine whose molecule has one carbon atom more than the used alkyl halide molecule?
  2. Why are amines more basic than the comparable alcohols.

Answer:
1.
 Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-312

2. Due less electronegativity of oxygen atom as compared to nitrogen, amines are more basic than alcohols.

Question 63.
Aniline gets coloured on standing in air for a long time. Why?
Answer:
Due to strong electron – donating effect (+ R effect) of NH2 group, the electron density on the benzene ring increases. As a result, aniline is easily oxidised on standing in air for a long time to form coloured products.

Question 64.
CH3CONH2 is a weaker base than CH3CH2NH2. Why?
Answer:
Due to resonance, the lone pair of electrons on the nitrogen atom in CH3CONH2 is delocalised over the keto group. There is no such effect ip CH3CH2NH2. Due to reduction in electron density on N – atom of CH3CONH2, it is a weaker base than CH3CH2NH2.

Question 65.
Write chemical equation for the following conversions:

  1. CH3CH2 – Cl into CH3CH2CH2 – NH2
  2. C6H5 – CH2 – Cl into C6H5CH2CH2NH2

Answer:
1.  Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-309

2.  Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-310

Samacheer Kalvi 12th Chemistry 13 Organic Nitrogen Compounds 3 Marks Questions and Answers

Question 1.
Write about the classification of organic nitro compounds.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-107

2. Draw the structural formula of the following compounds.

  1. 2 – methyl – 1 – nltropropane
  2. 2, 2 – dimethyl – 1 – nitropropane
  3. Nitrobenzene

Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-108 2 – methyl – 1 – nitropropane
2. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-109 2, 2 – dimethyl – 1- nitropropane
3. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-110 Nitrobenzene

Question 3.
Draw the structural formula of the following compounds.

  1. 2 – nitro – 1 – methyl benzene
  2. 1, 3, 5 – trinitrobenzene
  3. 2 – phenyl – 1 – nitroethane

Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-111    2 – nitro – 1 – methyl benzene
2. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-112 1, 3, 5 – trinitrobenzene
3. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-1132 – phenyl – 1 – nitroethane

Question 4.
Write the possible isomers for the formula C4H9NO2.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-114

Question 5.
How would you prepare nitro ethane from the following compounds?

  1. CH3 – CH2Br
  2. CH3 – CH3

Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-115
2. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-116

Question 6.
Mention any two methods of preparation of nitro methane.
Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-117
2. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-118

Question 7.
Explain the

  1. acid medium reduction
  2. neutral medium reduction of Nitromethane.

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-119

Question 8.
What happens when hydrochloric acid is treated with

  1. Nitro ethane
  2. 2 – nitropropane
  3. 2 – methyl – 2 – nitro propane

(or)
How would you distinguish 1°, 2°, 3° nitro compounds?
Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-120
2. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-121
3. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-122

Question 9.
Explain the following reactions using nitro benzene.

  1. Chlorination
  2. Nitration
  3. Suiphonation

Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-123
2. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-124
3. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-125

Question 10.
Give the structural formula and IUPAC name of the following compounds.

  1. Isopropvl amine
  2. Allyl amine
  3. Hexamethyl diamine

Answer:
1. Isopropvl amineSamacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-126  Propan – 2 – amine
2. Allyl amine Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-127  Prop – 2 – en – 1 – amine
3. Hexamethyl diamine Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-128 Hexane 1, 6 – diamine

Question 11.
Draw the structural formula and write the IUPAC name of

  1. Methyl isopropyl amine
  2. Diethyl butyl amine
  3. Ethyl methyl ispropylamine

Answer:
1. Methyl isopropyl amine:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-129
N – methyl propan – 1 – amine

2. Diethyl butyl amine:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-130
N, N – diethyl butan – 1 – amine

3. Ethyl methyl ispropylamine:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-131
N – ethyl – N – methyl propan – 2 – amine

Question 12.
Draw the structural formula and write the IUPAC name of …………..

  1. N, N – dimethyl aniline
  2. Benzyl amine
  3. N – methyl benzylamine

Answer:
1. N, N – dimethyl aniline
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-132

2. Benzyl amine
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-133

3. N – methyl benzylamine
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-134

Question 13.
Explain the alkylation reaction of methylamine with equation.
Answer:
Alkylation:
Primary amines reacts with alkyl halides to give successively 2° and 3° amines and quaternary ammonium salts.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-135

Question 14.
Explain about the suiphonation reaction of aniline.
Answer:
Aniline reacts with Conc. H2SO4 to form anilinium hydrogen sulphate which on heating with H2SO4 at 453 – 473K gives p – aminobenzene suiphonic acid, commonly known as suiphanilic acid, as the major product.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-136

Question 15.
Explain Sandmeyer reaction with example.
Answer:
On mixing freshly prepared solution of benzene diazonium chloride with cuprous halides, aryl halides are formed. This reaction is called Sandmeyer reaction.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-137

Question 16.
Write the structural formula and IUPAC name of the following compounds.

  1. Methyl cyanide
  2. Propionitrile
  3. Butyro nitrile

Answer:

  1. CH3CN – Ethane nitrile
  2. CH3 – CH2CN – Propane nitrile
  3. CH3 – CH2 – CH2CN – Butane nitnie.

Question 17.
How would you produce Ethane nitrite form the following compounds?

  1. Acetamide
  2. Ammonium acetate
  3. Acetaldoxime

Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-138
2. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-139
3. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-140

Question 18.
Explain the action of following reagent with ethane nitrile.

  1. Dilute mineral acid
  2. Ni / H2

Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-141
2. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-142

Question 19.
What happens when methyl isocyanide reacts with the following reagents?

  1. Mineral acid
  2. Na + C2H5OH

Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-143
2. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-144

Question 20.
Explain the addition reactions of alkyl Isocyanide with

  1. halogen
  2. sulphur
  3. ozone

Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-145
2. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-146
3. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-147

Question 21.
What are the uses of aliphatic nitro compounds.
Answer:

  1. Nitromethane is used as a fuel for cars.
  2. Chloropicrrn (CCI3NO2) is used as an insecticide
  3. Nitroethane is used as a fuel additive and precursor to explosive and they are good solvents for polymers, cellulose ester, synthetic rubber and dyes etc,

Question 22.
Explain about the structure and uses of Mitomycin.
Answer:

  1. Mitomycin C, and anticancer agent used to treat stomach and colon cancer, contains an aziridine ring.
  2. The aziridine functional group participates in the drug’s degradation by DNA, resulting in the death of cancerous cells.
  3. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-148

Question 23
Complete the following reactions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-149
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-150

Question 24.
Convert Methanamine into Ethanamine.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-151

Question 25.
Convert Ethanamic into Methanamine.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-152

Question 26.
How would you obtain Benzoic acid from aniline?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-153

Question 27.
Complete the following reactions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-154
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-155

Question 28.
Write the reactions of

  1. aromatic and
  2. aliphatic primary amines with nitrous acid.

Answer:
Aromatic primary amines react with HNO2 at 273 – 278 K to form aromatic diazonium salts.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-156
Aliphatic primary amines also react with HNO2 at 273 – 278 K to form aliphatic diazonium salts. But these are unstable even at low temperature and thus decomposes readily to form a mixture of compounds consisting of alkyl chlorides, alkenes and alcohols, out of which alcohols generally predominates.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-157

Question 29
Given plausible explanation for each of the following.

  1. Why are amines less acidic than alcohols of comparable molecular masses?
  2. Why do primary amines have higher boiling point than tertiary amines?
  3. Why are aliphatic amines stronger bases than aromatic amines?

Answer:
1. It is because C2H5O° is more stable than C2H5NH° because oxygen is more electronegative than nitrogen.

2. Due to the presence of two H – atoms on N – atom of primary amines, they undergo extensive intennolecular H – bonding while tertiary amines due to the absence of a H – atoms on the N – atom do not undergo H – bonding. As a result, primary amines have higher boiling point than tertiary amines of comparable molecular masses.

3. It is because there is electron withdrawing, C6H5 group in aromatic amines which makes them less basic than aliphatic amines in which alkyl group is electron releasing.

Question 30.
Account for the following.

  1. Primary amines (R – NH2)have higher boiling point than tertiary amines (R3N).
  2. Aniline does not undergo Friedel – Crafts reaction.
  3. (CH3)2NH Is more basic than (CH3)3N in an aqueous solution.

Answer:
1. Due to maximum intermolecular hydrogen bonding in primary amines (due to presence of more number of H – atoms), primary amines have higher boiling point in comparison to tertiary amines.

2. Aniline does not undergo Friedel-Crafts reaction due to acid-base reaction. Aniline and a Lewis Acid / Protic Acid, which is used in Friedel-crafts reaction.

3. In (CH3)3N there is maximum steric hindrance and least solvation but in (CH3)2NH the solvation is more group; di-methyl amine is still a stronger base than trimethyl amine.

Question 31.
Write the structures of A,B and C in the following reactions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-158
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-159

Question 32.
Predict, giving reasons the order of basicity of the following compounds.

  1. gaseous phase
  2. In aqueous solution (CH3)3N, (CH3)2NH, CH3NH2, NH3

Answer:
1. In gaseous phase, basic character of amines increases with the increase in the number of electron the releasing alkyl groups due to + I effect so the trend of basic character is 3° > 2° > 1° > NH3. Therefore, (CH3)3N > (CH3)3NH > CH3NH2 > NH3.

2. In aqueous phase, solvation of ammonium cation occurs by warer molecules, greater the size of ion, lesser will be the solvation and lesser will be the stability of ion. So on combining + I effect and solvation effect, in aqueous phase trend changes to 2° > 1° > 3°, (CH3)2NH > CH3NH2 > (CH3)3N > NH3.

Question 33.
Identify A and B in the following reactions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-160
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-161

Question 34.
What happens when. (Write reactions only)

  1. Nitroethane is treated with LiAlH4.
  2. Diazonium chloride reacts with phenol in basic medium.

Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-162
2. Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-163

Question 35.
How would you achieve the following conversions.
Answer:

(i) Nitrobenzene to aniline

(ii) An alkyl halide to a quarternary ammonium salt.

(iii) Aniline to benzonitrile.

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-164

Question 36.
What happens when (write reactions only).

(i) Nitropropane is treated with LiAlH4.
(ii) Ethyl lsocyañide undergoes hydrolysis.
(iii) Benzene diazonium chloride reacts with phenol i basic medium.

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-165

Question 37.
Identify A and B in the following reactions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-166
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-167

Question 38.
Identify A and B in the following reactions
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-168
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-169
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-170

Question 39.
A compound ‘x’ having molecular formula C3H7NO reacts with Br2 in the presence of KOH to give another compound ‘y’. The compound ‘y’ reacts with HNO2 to form ethanol. and N2 gas. Identify the compounds x andy and write the reactions involved.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-171

Question 40.
An organic compound ‘A’ having molecular formula C3H5N on hydrolysis gave a nother compound ‘B’ The compound ‘B’ on treatment with HNO2 gave ethyl alcohol. ‘B’ on warming with CHCI3 and alcoholic caustic potash gave an offensive smelling substance ‘C’. Identify ‘A’, ‘B’ and ‘C’.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-172

Samacheer Kalvi 12th Chemistry 13 Organic Nitrogen Compounds 5 Mark Questions and Answers

Question 1.
Explain the various reduction reactions of nitrobenzene.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-173
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-174

Question 2.
What happens when nitrous acid react with
(i) Ethyl amine
(ii) Aniline
(iii) N – methyl aniline
(iv) Trimethyl amine
(v) N, N – dimethyl aniline
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-175

Question 3.
Starting from Benzene diazonium chloride, how would you prepare
(i) Benzene
(ii) Phenol
(iii) Nitro benzene
(iv) Benzolc acid
(v) Fluorobenzene
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-176

Question 4.
Starting from Benzene diazonium chloride, how would you prepare
(i) Biphenyl
(ii) Phenyl hydrazine
(iii) p – hydroxy azo benzene
(iv) p – amino azo benzene
(v) Chioro benzene
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-177

Question 5.
Convert aniline into the following compounds.
(i) N – phenyl benzamide
(ii) Phenyl isothiocyanate
(iii) 2, 4, 6 – tribromo aniline
(iv) Sulphanilinic acid
(v) Phenyl isocyanide
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-178

Question 6.
An organic compound (A) of molecular formula C6H7N on reaction with sodium nitrite and hydrochloric acid at 0°C gives (B) of formula C6H5N2Cl. (B) on treatment with cuprous cyanide give (C) of formula C7H5N. (C) on reaction with sodium and ethanol gives (D) of formula C7H9N. (D) on reaction with nitrous acid gives (E) of molecular formula C7H8O. Identify A, B, C, D and E and explain the reactions involved.
Answer:
1. (A) is identified as Aniline from the molecular formula.
2. Aniline on reaction with sodium nitrite and hydrochloric acid at 0°C, diazotisation take place and the product (B) formed is benzene diazonium chloride.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-179

3. Benzene diazonium chloride on treatment with CuCN produces (C) as cyano benzene.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-180

4. Cyano benzene on reaction with Na and C2H5OH undergoes reduction reaction to give (D) as benzvlamine.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-181

5. Benzylamine on treatment with nitrous acid gives benzyl alcohol as (E).
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-182

Question 7.
Complete the following reactions and identify the A, B and C in these reaction.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-183
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-184

Question 8.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compound A, B and C.
Answer:
Step – 1: To find out the structure of compounds ‘B’ and ‘C’.
1. Since compound ‘C’ with molecular formula C6H7N is formed from compound ‘B’ on treatment with Br2 + KOH (i.e., Hoffmatm bromamide reaction). Therefore, compound ‘B’ must be an amide and ‘C’ must be an amine. The only amine having the molecular formula C6H7N is C6H5NH2 (i.e., aniline or benzenamine).

2. Since ‘C’ is aniline, therefore, the amide from which it is formed must be benzamide (C6H5CONH2). Thus, compound ‘B’ is benzamide. The chemical equation showing the conversion of ‘B’ to ‘C’ is
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-185

Step – 2:
To find out the structure of compound ‘A’. Since compound ‘B’ is formed from compound ‘A’ by treatment with aqueous ammonia and heating. Therefore, compound ‘A’ must be benzoic acid or benzenecarboxylic acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-186

Question 9.
An aromatic compound ‘A’ of molecular formula C7H7ON undergoes a series of reactions as shown below. Write the structures of A,B,C, D and E in the following reactions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-187
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-188

Question 10.
Write reactions and conditions required for the following conversions.
(i) Aniline to benzene
(ii) Methylamine to methylcyanide
(iii) Propanenitrile to ethylamine
(iv) m – Bromoaniline to m – bromophenol
(v) Nitrobenzene to 2, 4, 6 – tribromoaniline.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-189

Question 11.
A compound ‘A’ of molecular formula C3H7O2N on reaction with Fe and cone. HCI gives a compound ‘B’ of molecular formula C3H9N. Compound ‘B’ on treatment with NaNO2 and HCI gives another compound ‘C’ of molecular formula C3H8O. The compound ‘C’ has molecular formula, C3H8O. The compound ‘C’ gives effervescence with Na. On oxidation with CrO3, the compound ‘C’ gives a saturated aldehyde containing three carbon atoms. Deduce the structures of A, B and C and write the equations for the reaction involved.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-190

Question 12.
Identify compounds A, B and C in the following reactions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-191
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-192

Question 13.
A aromatic hydrocarbon (A) of molecular formula C6H6 reacts with Conc.HNO3 and Conc. H2SO4 gives (B) of formula C6H5O2N. (B) on reaction with Sn/HCI gives (C) of formula C6H7N which answers carbylamine reaction. (C) on treatment with chloroform and alkali gives (D) of formula C7H5N. Identify A, B, C, D and explain the reactions involved.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-193

Question 14.
Convert the following.
(i) Nitro benzene → Benzene
(ii) Benzene → Benzoic acid

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-194

Common Errors

  1. TUPAC name of nitro compounds.
  2. If the formula ends as – NO2, the students always think of nitro only.
  3. Reducing agents may be different for different medium.
  4. Basic character of amines is always difficult to remember.
  5. Cyanide and isocyanide formula may get confusing.

Rectifications

1. Both common name and IUPAC names are same for aliphatic and aromatic nitro compounds.

2. – NO2 may be
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-195

3.

  • Acid medium reducing agent: Sn / HCI
  • Neutral medium reducing agent: Zn + NH4CI
  • Basic medium reducing agent: Zn / NaOH
  • Catalytic reducing agent: Ni, Pt, LiAlH4

4. 2° amine > 3° amine > 1° amine > NH3
Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-196
Alkyl group is connected to Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-197 C – Cyanide. If alkyl group connected to Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds-198 Isocyandie.

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Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules

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Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules

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Samacheer Kalvi 12th Chemistry Chapter 14 Biomolecules Textual Evaluation Solved

Samacheer Kalvi 12th Chemistry Biomolecules Multiple Choice Questions

Question 1.
Which one of the following rotates the plane polarized light towards left?
(a) D(+) Glucose
(b) L(+) Glucose
(c) D(-) Fructose
(d) D(+) Galactose
Answer:
(c) D(-) Fructose

Question 2.
The correct corresponding order of names of four aldoses with configuration given below Respectively is ………………..
(a) L – Erythrose, L – Threose, L – Erythrose, D – Threose
(b) D – Threose, D – Erythrose, L – Thrcose, L – Erythrose
(c) L – Etythrose, L – Threose, D – Erythrose, D – Threose
(d) D – Erythrose, D – Threose, L – Erythrose, L – Threose
Answer:
(d) D – Erythrose, D – Threose, L – Erythrose, L – Threose

Question 3.
Which one given below is a non-reducing sugar?
(a) Glucose
(b) Sucrose
(c) maltose
(d) Lactose
Answer:
(b) Sucrose

Question 4.
Glucose(HCN) Product (hydrolysis) Product (HI + Heat) A, the compound A is ………………..
(a) Heptanoic acid
(b) 2 – lodohexane
(c) Heptane
(d) Heptanol
Answer:
(a) Heptanoic acid

Question 5.
Assertion: A solution of sucrose in water is dextrorotatory. But on hydrolysis in the presence of little hydrochloric acid, it becomes levorotatory.
Reason: Sucrose hydrolysis gives unequal amounts of glucose and fructose. As a result of this change in sign of rotation is observed.
(a) If both accretion and reason are true and reason is the correct explanation of assertion
(b) If both assertion and reason are true but reason is not the correct explanation of assertion
(c) If assertion is true but reason is false
(d) if both assertion and reason are false
Answer:
(a) If both accretion and reason are true and reason is the correct explanation of assertion

Qustion 6.
The central dogma of molecular genetics states that the genetic information flows from ……………..
(a) Amino acids Protein DNA
(b) DNA Carbohydrates Proteins
(c) DNA RNA Prnteins
(d) DNA RNA Carbohydrates
Answer:
(c) DNA RNA Proteins

Question 7.
In a protein, various amino acids liked together by …………..
(a) Peptide bond
(b) Dative bond
(c) α – Glycosidic bond
(d) β – Glycosidic bond
Answer:
(a) Peptide bond

Question 8.
Among the following the achiral amino acid is …………..
(a) 2 – ethylalanine
(b) 2 – methyiglycine
(c) 2 – hydroxymethyiscrine
(d) Tryptophan
Answer:
(c) 2 – hydroxymethylserine

Question 9.
The correct statement regarding RNA and DNA respectively is ………………
(a) the sugar component in RNA is an arabinose and the sugar component in DNA is ribose
(b) the sugar component in RNA is 2’ – deoxyribose and the sugar component in DNA is arabinose
(c) the sugar component in RNA is an arabinose and the sugar component in DNA is 2’ – deoxyribose
(d) the sugar component in RNA is ribose and the sugar component in DNA is 2’-deoxyribose
Answer:
(d) the sugar component in RNA is ribose and the sugar component in DNA is 2’ – deoxyribose

Question 10.
In aqueous solution of amino acids mostly exists in ……………….
(a) NH2 – CH(R) – COOH
(b) NH2 – CH(R) – COO
(c) H3N – CH(R) – COOH
(d) H3N+ – CH(R) – COO
Answer:
(d) H3N+ – CH(R) – COO

Question 11.
Which one of the following is not produced by body?
(a) DNA
(b) Enzymes
(c) Hormones
(d) Vitamins
Answer:
(d) Vitamins

Question 12.
The number of sp2 and sp3 hybridised carbon in fructose are respectively ……………..
(a) 1 and 4
(b) 4 and 2
(c) 5 and 1
(d) 1 and 5
Answer:
(d) 1 and 5

Question 13.
Vitamin B2 is also known as …………….
(a) Riboflavin
(b) Thiamine
(c) Nicotinamide
(d) Pyridoxine
Answer:
(a) Riboflavin

Question 14.
The pyrimidine bases present in DNA are …………..
(a) Cytosine and Adenine
(b) Cytosine and Guanine
(c) Cytosine and Thiamine
(d) Cytosine and Uracil
Answer:
(c) Cytosine and Thiamine

Question 15.
Among the following L – serine is …………..
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-1
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-2

Question 16.
The secondary structure of a protein refers to ………………
(a) fixed configuration of the polypeptide backbone
(b) hydrophobic interaction
(c) sequence of a-amino acids
(d) α – helical backbone
Answer:
(d) α – helical backbone

Question 17.
Which of the following vitamins is water soluble?
(a) Vitamin E
(b) Vitamin K
(c) Vitamin A
(d) Vitamin B
Answer:
(b) Vitamin K

Question 18.
Complete hydrolysis of cellulose gives ……………
(a) L – Glucose
(b) D – Fructose
(c) D – Ribose
(d) D – Glucose
Answer:
(d) D – Glucose

Question 19.
Which of the following statement is correct?
(a) Ovalbumin is a simple food reserve in egg-white
(b) Blood proteins thrombin and fibrinogen are involved in blood clotting
(c) Denaturation makes protein more active
(d) Insulin maintains the sugar level of in the human body
Answer:
(c) Denaturation makes protein more active

Question 20.
Glucose is an aldose. Which one of the following reactions is not expected with glucose?
(a) It does not form oxime
(b) It does not react with Grignard reagent
(c) It does not form osazones
(d) It does not reduce tollens reagent
Answer:
(b) It does not react with Grignard reagent

Question 21.
If one strand of the DNA has the sequence ‘ATGCTTGA’, then the sequence of complementary strand would be ……………..
(a) TACGAACT
(b) TCCGAACT
(c) TACGTACT
(d) TACGRAGT
Answer:
(a) TACGAACT

Question 22.
Insulin, a hormone chemically is ………….
(a) Fat
(b) Steroid
(c) Protein
(d) Carbohydrates
Answer:
(c) Protein

Question 23.
a – D (+) Glucose and ß – D (+) glucose are ………………..
(a) Epimers
(b) Anomers
(c) Enantiomers
(d) Conformational isomers
Answer:
(b) Anomers

Question 24.
Which of the following are epimers?
(a) D(+) – Glucose and D(+) – Galactose
(b) D(+) – Glucose and D(+) – Mannose
(c) Neither (a) nor (b)
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 25.
Which of the following amino acids are achiral?
(a) Alanine
(b) Leucine
(c) Proline
(d) Glycine
Answer:
(a) Alanine

II. Answer the following questions.

Question 1.
What type of linkages hold together monomers of DNA?
Answer:

  1. Phospho diester linkages hold together monomers of DNA
  2. Phosphoric acid forms phospho diester bond between neucleotides
  3. The sugar – phosphate linkage forms the backbone of each strand of DNA.

Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-3

Question 2.
Give the differences between primary and secondary structure of proteins.
Answer:
Primary structure of proteins

  1. Linear sequence of amino acids
  2. Composed of peptide bonds formed between amino acids.
  3. Formed during translation.
  4. Involved in post – translational modifications.

Secondary structure of proteins

  1. Folding of the peptide chain into an a-helix and 13-sheet.
  2. Encompasses hydrogen bonds
  3. Forms collagen, elastin action, myosin, and keratin – like fibres.
  4. Involved in forming structures such as cartilages, ligaments, skins etc.

Question 3.
Name the Vitamins whose deficiency cause

  1. rickets
  2. scurvy

Answer:

  1. Vitamin – D deficiency causes rickets disease.
  2. Vitamin – C deficiency causes scurvy disease.

Question 4.
Write the Zwitter ion structure of alanine.
Zwitter ion structure of alanine
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-4

Question 5.
Give any three difference between DNA and RNA.
Answer:
DNA

  1. It is mainly present in nucleus, mitochondria and chloroplast .
  2. It contains deoxyribose sugar
  3. Base pair A = T. G = C
  4. Double stranded molecules
  5. Its life time is high
  6. it is stable and not hydrolysed easily by alkalis
  7. It can replicate itself

RNA

  1. It is mainly present in cytoplasm, nucleolus and ribosomes
  2. It contains ribose sugar
  3. Base pair A = U. C = G
  4. Single stranded molecules
  5. It is Short lived
  6. It is unstable and hydrolyzed easily by alkalis
  7. It cannot replicate itself. It is formed from DNA.

Question 6.
Write a short note on peptide bond.
Answer:
1. The amino acids are linked covalently by peptide bonds.

2. The carbonyl group of the first amino acid react with the amino group of the second amino acid to give an amide linkage (- CONH) between these aminoacids. This amide linkage is called peptide bond.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-5

3. The resulting compound is called a dipeptide. Because, two amino acids are inovived for getting one peptide bond.

4. If large number of amino acids combined through peptide bond, the resulting giant molecule is called a protein.

5. The amino end of the peptide is known as N-terminal, while the carboxy end is called C – terminal.

Question 7.
Give two difference between Hormones and vitamins.
Answer:
hormones

  1. Hormones are produced in the endocrine or ductless glands.
  2. Hormones are not stored in the body. These are produced as and when required.
  3. They are effective in low concentration. Their excess or deficiency may cause hormonal disorders.
  4. Hormones influence the genes to produce specific enzymes required during metabolism.
  5. Example: Insulin

Vitamins

  1. Vitamins (except Vitarnin – D) are not produced in the body. Vitamin must be supplied in the diet.
  2. Vitamins are stored in the body upto certain extent.
  3. They are needed in small quantity. Excess vitamins are excreted. Their deficiency causes
  4. malfunctioning called deficiency diseases or avitaminosis.
  5. They act as co-enzymes and help enzymes to perform their function.
  6. Example: Vitamin A, B, C, D, E and K

Question 8.
Write a note on den atu ration of proteins.
Answer:
Denaturation of proteins.
1. In general, protein has a unique three – dimensional structure formed by interactions such as disuiphide bond, hydrogen bond, hydrophobic and electrostatic interactions.

2. These interactions can be disturbed when the protein is exposed to a higher temperature in certain chemicals such urea, alternation of pH, ionic strength etc. It leads to the loss of the three – dimensional structure.

3. The process of a protein losing its higher order structure without losing the primary structure, It is called denaturation of protein. When a protein denatures, its biological function is also lost.

4. Since the primary structure is intact, this process can be reversed in certain proteins. This can happen spontaneously upon restoring the original conditions or with the help of special enzymes called cheperons.

5. Example: Coagulation of egg white by action of heat.

Question 9.
What are reducing and non – reducing sugars?
Answer:
1. Reducing sugars:
Those carbonhydrates which contain free aldehyde or ketonic group and reduces Fehling’s solution and Tollen’s reagent are called reducing sugars. All monosacehaides whether aldose or ketone are reducing sugars.

2. Non – reducing sugars:
Cabohydrates which do not reduce Tollen’s reagent and Fehling’s solution are called non – reducing sugars. Example Sucrose. They do not have free aldehyde group.

Question 10.
Why carbohydrates are generally optically active?
Answer:
Carbohydrates are generally optically active because they have one or more chiral carbon atoms in their molecules. For example, Glucose has four chiral carbons and therefore it is optically active.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-6

Question 11.
Classify the following into monosaccharides, oligosaccharides and polysaccharldes.

  1. Starch
  2. fructose
  3. sucrose
  4. lactose
  5. maltose

Answer:

  1. Starch – Polysaccharides
  2. Furctose – Mono saccharides
  3. Sucrose – Oligo saccharide
  4. Lactose – Oligo saccharide
  5. Maltose – Oligo saccharide

Surcose, Lactose and Maltose are typical disaccharides

Question 12.
How are vitamins classified?
Answer:
Vitamins are classified into two groups based on their solubility in water and in fat. They are,

  1. Water – soluble vitamins
  2. Oil or fat – soluble vitamins

Water – soluble vitamins – Vitamins which dissolve in water are called water soluble vitamins. Examples – Vitamins of B group and Vitamin C Oil or fat – soluble vitamins: Vitamins which dissolve in oils or fat are called oil or fat – soluble vitamins. Examples – Vitamin A, D, E and K

Question 13.
What are hormones? Give examples.
Answer:
Hormone is an organic substance that is sëcreted by one tissue into the blood stream and induces a physiological response in other tissues. It is an inter cellular signaling molecule. Virtually every process is a complex organism is regulated by one or more hormones. Example, insulin, epinephrine, estrogen, androgen etc.

Question 14.
Write the structure of all possible dipeptides which can be obtained from glycine and atanine.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-7
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-8
Therefore two dipeptides structures are possible from glycine and alanine. They are

  1. glycyl alanine and
  2. Alanyl glycine

Question 15.
Define enzymes.
Answer:
There are many biochemical reactions that occur in our living cells. Digestion of food and harvesting the energy from them, and synthesis of necessary molecules required for various – cellular functions are examples for such reactions. All these reactions are catalysed by special proteins called enzymes.
(or)
Enzymes are bio catalysts produced by the living cells which catalyse many biochemical reactions in animal and plant bodies. They are more specific in their action.

Question 16.
Write the structure of – D (+) glucopyranose.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-9

Question 17.
What are different types of RNA which are found in cell?
Answer:
RNA molecules are classified according to their structure and function into three major types.

  1. Ribosomal RNA (r – RNA)
  2. Messenger RNA (m – RNA)
  3. Transfer RNA (t – RNA)

r-RNA:
rRNA is mainly found in cytoplasm and in ribosomes, which contain 60% RNA and 40% protein. Ribosomes are the sites at which protein synthesis takes place.

t – RNA:
tRNA molecules have lowest molecular weight of all nucleic acids. They consist of 73 – 94 nucleotides in a single chain. The function of tRNA is to carry amino acids to the sites of protein synthesis on ribosomes.

m – RNA:
mRNA is present in small quantity and very short lived. They are single stranded, and their synthesis takes place on DNA. The synthesis of mRNA from DNA strand is called transcription. mRNA carries genetic information from DNA to the ribosomes for protein synthesis.

Question 18.
Write a note on formation of – helix.
AnsweR:
1. In the α – helix sub – structure, the aminoacids are arranged in a right handed helical (spiral) structure and are stabilised by the hydrogen bond between the carbonyl oxygen one aminoacid (nth residue) with amino hydrogen of the fifth residue (n + 4th residue)

2. The side chains of the residues protrude outside of the helix. Each turn of an α – helix contains about 3.6 residues and is about 5.4 A long.

3. The amino acid proline produces a line in the helical structure and often called as a helical breaker due to its rigid cyclic structure.

4. Many fibrous proteins such as ct-Keratin in hair, nails, wool, skin and myosin in muscles have α – helix structure. Stretching property of human hair is due to the helical structure of α – keratin in hair.

5. Structure of a-helix.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-10

Question 19.
What are the functions of lipids in living organism?
Answer:

  1. Lipids are the integral component of cell membrane. They are necessary of structural integrity of the cell.
  2. The main function of triglycerides in animals is as an energy reserve. They yield more energy than carbohydrates and proteins.
  3. They act as protective coating in aquatic organisms.
  4. Lipids of connective tissue give protection to internal organs.
  5. Lipids help in the absorption and transport of fat soluble vitamins.
  6. They are essential for activation of enzymes such as lipases.
  7. Lipids act as emulsifier in fat metabolism.

Question 20.
Is the following sugar, D – sugar or L – sugar?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-11

Samacheer Kalvi 12th Chemistry Biomolecules Additional Questions

Samacheer Kalvi 12th Chemistry Biomolecules 1 mark Questions and Answers

I. Choose the best answer.

Question 1.
Which of the following is the most abundant organic compounds in every living organism?
(a) Fats
(b) Proteins
(c) Carbohydrates
(d) Hormones
Answer:
(c) Carbohydrates

Question 2.
What is the general chemical name of carbohydrates?
(a) Poly hydroxy aldehyde or ketones
(b) Poly hydroxy esters
(c) Poly amino acids
(d) Poly carboxylic esters.
Answer:
(a) Poly hydroxy aldehyde or ketones

Question 3.
Which process is utilized in the synthesis of carbohydrates in green plants?
(a) Oxidation
(b) Redox reaction
(c) Photosynthesis
(d) Reduction
Answer:
(c) Photosynthesis

Question 4.
Which of the following compounds are optically active?
(a) Glycine
(b) Carbohydrates
(c) Ethanol
(d) Meso tartaric acid
Answer:
(b) Carbohydrates

Question 5.
Which of the following is optically inactive?
(a) 2 – butanol
(b) Glyceraldehyde
(c) Glucose
(d) Meso tartaric acid
Answer:
(d) Meso tartaric acid

Question 6.
How many isomers are possible for glucose that have 4 asymmetic carbon atoms?
(a) 8 isomers
(b) 16 isomers
(c) 2 isomers
(d) 4 isomers
Answer:
(b) 16 isomers

Question 7.
How many asymmetric carbon atoms are in glucose?
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
(a) 4

Question 8.
Which of the following rotates the plane polarised light in clockwise direction?
(a) L(-) Glucose
(b) D (glucose)
(c) L – fructose
(d) L – Glyceraldehyde
Answer:
(b) D (glucose)

Question 9.
Which one of the following is levorotatory?
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-12
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-13

Question 10.
Which one of the following is not a monosaccharide?
(a) Fructose
(b) Ribose
(c) Erythrose
(d) Maltose
Answer:
(d) Maltose

Question 11.
Which one of the following is a monossachande?
(a) Glucose
(b) Maltose
(c) Surcose
(d) Cellulose
Answer:
(a) Glucose

Question 12.
The number of carbon atoms present in Erythrose is
(a) 6
(b) 4
(c) 3
(d) 5
Answer:
(b) 4

Question 13.
What is the amount of glucose present in human blood?
(a) 150 mg/dI
(b) 50 mg/dl
(c) 100 mg/dl
(d) 1000 mg/dl
Answer:
(c) 100 mg/dl

Question 14.
Which one of the following is called blood sugar?
(a) Erythrose
(b) Ribose
(c) Ribulose
(d) Glucose
Answer:

Question 15.
Acid hydrolysis of starch at high temperature and pressure produces
(a) fructose
(b) glucose
(c) both fructose and glucose
(d) maltose
Answer:
(b) glucose

Question 16.
The other name of glucose is …………
(a) dextrose
(b) blood sugar
(c) aldohexose
(d) all the above
Answer:
(d) all the above

Question 17.
Which one is formed as major product when glucose is on reduction with concentrated HI and red P at 3 73K?
(a) 2 – iodohexane
(b) 3 – iodohexane
(c) n – hexane
(d) 4 – iodohexane
Answer:
(c) n – hexane

Question 18.
Which one of the product is formed when glucose reacts with bromine water?
(a) n – hexane
(b) Gluconic acid
(c) Saccharic acid
(d) Hexanoic acid
Answer:
(c) Saccharic acid

Question 19.
Which one of the following is formed when glucose react with Conc. HNO3?
(a) Gluconic acid
(b) Glutaric acid
(c) Saccharic acid
(d) Hexanoic acid
Answer:
(c) Saccharic acid

Question 20.
Which one of the following will reduce Tollen’s reagent and Fehling’s solution?
(a) Glucose
(b) Fructose
(c) Sucrose
(d) Maltose
Answer:
(a) Glucose

Question 21.
Which of the following form pentacetate with acetic anhydide?
(a) Glucose
(b) Fructose
(c) Lactose
(d) Both a & b
Answer:
(d) Both a & b

Question 22.
Which one of the reagent does not react with glucose?
(a) Acetic anhydride
(b) Tollen’s reagent
(c) Sodium bi suiphite
(d) Bromine water
Answer:
(c) Sodium bi suiphite

Question 23.
The specific rotation of pure α and β (D) glucose are respectively.
(a) 18.7°, 112°
(b) 112°, 18.7°
(c) 90°, 90°
(d) 120°, 20°
Answer:
(b) 112°, 18.7°

Question 24.
Sugar differing in configuration at an asymmetric centre is known as ……………….
(a) epimers
(b) isomers
(c) anomers
(d) monomers
Answer:
(a) epimers

Question 25.
Which enzyme is utilised in the conversion of galactose to glucose?
(a) Maltose
(b) Epimerase
(c) Invertase
(d) Zymase
Answer:
(b) Epimerase

Question 26.
The other name of fructose is ……………
(a) Ketohexose
(b) fruit sugar
(c) levulose
(d) all the above
Answer:
(d) all the above

Question 27.
Hydrolysis of mutin in acidic medium gives ……………
(a) glucose
(b) fructose
(c) both a & b
(d) maltose
Answer:
(b) fructose

Question 28.
Invert sugar is a mixture of equal amount of ……………..
(a) lactose + maltose
(b) diastose + galactose
(c) glucose + fructose
(d) starch + cellulose
Answer:
(c) glucose + fructose

Question 29.
Which enzyme is used in the conversion of sucrose into glucose and fructose?
(a) Zymase
(b) Invertase
(c) Diastase
(d) Maltase
Answer:
(b) Invertase

Question 30.
Which one of the following is the sweetest of all known sugars?
(a) Lactose
(b) Glucose
(c) Fructose
(d) Sucrose
Answer:
(c) Fructose

Question 31.
Which is the product formed when fructose undergoes partial reduction with sodium amalgam and water?
(a) Sorbital + mannitol
(b) D – mannose + D – galactose
(c) Gluconic acid + saccharic acid
(d) Aldehyde + ketone
Answer:
(a) Sorbital + mannitol

Question 32.
Which one of the following reagent is used to convert fructose into sorbitol and mannitol?
(a) LiAlH4
(b) Hl / Red P
(c) Na / Hg
(d) Conc. HNO3
Answer:
(c) Na / Hg

Question 33.
Fructose on oxidation with concentrated nitric acid gives
(a) glyceric acid + oxalic acid
(b) glycoffic acid + tartaric acid
(c) tartronic acid + mesoxalic acid
(d) acetic acid + hexanoic acid
Answer:
(b) glycoffic acid + tartaric acid

Question 34.
How many asymmetric carbon atoms are present in fructose?
(a) 4
(b) 3
(c) 2
(d) 6
Answer:
(b) 3

Question 35.
Two monosaccharides are linked by to form a disaccharide.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-14
Answer:
(a) glycosidic linkage

Question 36.
Which of the following is not an example of disaccharide?
(a) Sucrose
(b) Lactose
(c) Maltose
(d) Cellulose
Answer:
(d) Cellulose

Question 37.
The enzyme that catalyses the hydrolysis of sucrose to glucose and fructose is …………….
(a) zymase
(b) invertase
(c) diastase
(d) maltase
Answer:
(b) invertase

Question 38.
Which one of the following contains a mixture of glucose, fructose and sucrose?
(a) Sugarcane
(b) Bectroot
(c) Honey
(d) Mango
Answer:
(c) Honey

Question 39.
Consider the following statements
(i) in sucrose, C1 of a – D – glucose is joined to C2 of D – fructose.
(ii) Two monosaccharides are linked by glycosidic linkage.
(iii) In sucrose, C2 of a – D – glucose is joined to C1 of D – fructose.
Which of the above statement is / are correct?
(a) (iii) only
(b) (i) & (ii)
(c) (ii) only
(d) (i) & (iii)
Answer:
(b) (i) & (ii)

Question 40.
Which one of the following is an example of non – reducing sugar?
(a) Glucose
(b) Dextrose
(c) Lactose
(d) Sucrose
Answer:
(d) Sucrose

Question 41.
Which one of the following on hydrolysis give galactose and glucose?
(a) Maltose
(b) Cellulose
(c) Lactose
(d) Sucrose
Answer:
(c) Lactose

Question 42.
Which one of the following is called milk sugar?
(a) Lactose
(b) Glucose
(c) Sucrose
(d) Raffinose
Answer:
(a) Lactose

Question 43.
Which of the following is the major source of maltose?
(a) Honey
(b) Apple
(c) Sprouting barley
(d) Grapes
Answer:
(c) Sprouting barley

Question 44.
Which one is produced during digestion of starch by the enzyme α – amylase?
(a) Maltose
(b) Glucose
(c) Fructose
(d) Lactose
Answer:
(a) Maltose

Question 45.
Consider the following statements
(I) Maltose consists two molecules of α – D glucose units linked by an α – 1, 4 glycosidic bond.
(ii) Maltose act as non-reducing sugar.
(iii) Maltose is produced during digestion of cellulose by the enzyme a-amylase.
Which of the above statement is / are not correct?
(a) (i) only
(b) (i) & (iii)
(c) (iii) only
(d) (ii) & (iii)
Answer:
(d) (ii) & (iii)

Question 46.
Which one of the following is hetero polysaccharide?
(a) Starch
(b) Heparin
(c) Cellulose
(d) Glycogen
Answer:
(b) Heparin

Question 47.
Which one of the following is a homopolysaccharide?
(a) Hyaluronic acid
(b) Heparin
(c) Both (a) & (b)
(d) Starch
Answer:
(d) Starch

Question 48.
Consider the following statements
(I) Starch contains 80% of amylase and about 20% amylopectin.
(ii) Polysaccharides are called sugars.
(iii) Lactose act as a reducing agent.
Which of the above statement is / are not correct?
(a) (i) & (ii)
(b) (iii)
(c) (ii) & (iii)
(d) (ii) only
Answer:
(a) (i) & (ii)

Question 49.
Which one of the following gives blue colour with amylose and purple colour with amylopectin?
(a) Tollen’s reagent
(b) Fehling’s solution
(c) Iodine solution
(d) Bromic water
Answer:
(c) Iodine solution

Question 50.
Which colour is formed when amylose is treated with iodine solution?
(a) Purple
(b) Red
(c) Blue
(d) Violet
Answer:
(c) Blue

Question 51.
Which colour is formed when amylopectin is treated with iodine solution?
(a) Purple
(b) Blue
(c) Green
(d) Red
Answer:
(a) Purple

Question 52.
Which one of the following is the major constituent of plant cell walls?
(a) Starch
(b) Cellulose
(c) Glycogen
(d) Amylose
Answer:
(b) Cellulose

Question 53.
Consider the following statements
(i) Cellulose is a straight chain polysaccharide.
(ii) The glucose molecules in cellulose arc linked by β(1, 4) glycosidic bond.
(iii) Cotton is almost pure starch.
Which of the above statement is / are correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i) & (ii)
Answer:
(d) (i) & (ii)

Question 54.
Which of the following enzyme can hydrolyse the cellulose?
(a) Invertase
(b) Glycosidase
(c) Zymase
(d) Diastase
Answer:
(b) Glycosidase

Question 55.
Which one of the following is called gun cotton?
(a) Nitrated ester of cellulose
(b) Cellulose acetate
(c) Glyceryl trinitrate
(d) Trinitrotoluene
Answer:
(a) Nitrated ester of cellulose

Question 56.
Which one of the following is called animal starch?
(a) Cellulose
(b) Glycogen
(c) Lactose
(d) Fat
Answer:
(b) Glycogen

Question 57.
Consider the following statements
(i) The excessive glucose in the body is stored in the form of arnylose and amylopectin.
(ii) Glycogeri is present in the liver and muscle of animals.
(iii) Protein is stored in the body as glycogen and in plant as starch.
Which of the above statement is / are not correct?
(a) (¡) & (iii)
(b) (ii) & (iii)
(c) (i) only
(d) (iii) only
Answer:
(d) (iii) only

Question 58.
Which one of the following is stored in the body as glycogen and in plant as starch?
(a) Protein
(b) Vitamin
(c) Fat
(d) Carbohydrates
Answer:
(d) Carbohydrates

Question 59.
Which one of the following act as shock absorber and lubricant?
(a) Glycosamino glycans
(b) Glycogen
(c) Cellulose nitrate
(d) Rayon explosive
Answer:

Question 60.
Which biomolecule is the most abundant in all living organisms?
(a) Carbohydrates
(b) Vitamins
(c) Hormones
(d) Proteins
Answer:
(d) Proteins

Question 61.
Which of the following is mainly present in proteins?
(a) 3 – keto acid
(b) 1 – amino acid
(c) α, β – ketol
(d) amide and acids
Answer:
(b) 1 – amino acid

Question 62.
Which of the amino acid is optically inactive?
(a) Alanine
(b) Valine
(c) Glycine
(d) Proline
Answer:
(c) Glycine

Question 63.
Proteins are generally …………..
(a) poly amides
(b) polyesters
(c) polymer
(d) poly peptide
Answer:
(d) poly peptide

Question 64.
Which one of the following is an example for fibrous protein?
(a) Myoglobin
(b) Insulin
(c) Keratin
(d) Enzymes
Answer:
(c) Keratin

Question 65.
Which one of the following is an example for globular protein?
(a) Kerating
(b) Myoglobin
(c) Collagen
(d) Etastin
Answer:
(b) Myoglobin

Questin 66.
Consider the following statement
(i) The amino acids arc linked electro valently by peptide bonds in proteins.
(ii) Fibrous proteins are linear molecules similar to fibres.
(iii) Globular proteins have a linear shape.
Which of the above statement is / are not correct?
(a) (iii) only
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (ii) only
Answer:
(b) (i) & (iii)

Question 67.
Consider the following statement
Answer:
(I) The relative arrangement of amino acids in the polypeptide chain called the secondary structure of protein.
(ii) α – Helix and β – strands are two most common sub structures formed by proteins.
(iii) α – Helix and β – strands further folds to form the three dimensional arrangement in tertiary structure of proteins.
Which of the above statement is / are correct?
(a) (i) only
(b) (ii) & (iii)
(c) (i) & (iii)
(d) (ii) only
Answer:
(b) (ii) & (iii)

Question 68.
Which of the following act as structural backbones?
(a) Keratine, collagen
(b) Myoglobin, insulin
(c) Glycine, proline
(d) Alanine, cysteine
Answer:
(a) Keratine, collagen

Question 69.
Which protein control the glucose level in the blood?
(a) Kerating, collagen
(b) Insulin, glucagon
(c) Glycine, proline
(d) Alanine, myoglobin
Answer:
(b) Insulin, glucagon

Question 70.
Which one of the following act as catalyst in the interconversion of carbonic acid to water and carbondioxide?
(a) Lactose
(b) Carbonic anhydrase
(c) Glycosidase
(d) Invertase
Answer:
(b) Carbonic anhydrase

Question 71.
Which enzyme catalyses the hydrolysis of sucrose to fructose and glucose?
(a) Lactase
(b) invertase
(c) Sucrase
(d) Zymase
Answer:
(c) Sucrase

Question 72.
Lactase enzyme hydrolyses the lactose into its constituent as …………..
(a) glucose, fructose
(b) glucose, galactose
(c) fructose only
(d) glucose only
Answer:
(b) glucose, galactose

Question 73.
Consider the following statement
(i) Lipids are the principal components of cell membranes including cell walls.
(ii) Enzymes are biocatalysts that catalyse a specific biochemical reaction.
(iii) Carbonic anhydrase enzyme catalyses the hydrolysis of sucrose to fructose and glucose.
Which of the above statement is / are correct?
(a) (i) only
(b) (iii) only
(c) (i) & (iii)
(d) (ii) & (iii)
Answer:
(c) (i) & (iii)

Question 74.
Which one help in the absorption and transport of fat soluble vitamins?
(a) Lipids
(b) Protein
(c) Enzyme
(d) Water
Answer:
(a) Lipids

Question 75.
Which one act as emulsifier in fat metabolism?
(a) Enzymes
(b) Fats
(c) Lipids
(d) Proteins
Answer:
(c) Lipids

Question 76.
Which one of the following is fat soluble vitamin?
(a) Vitamin B1
(b) Vitamin B6
(c) Vitamin C
(d) Vitamin A
Answer:
(d) Vitamin A

Question 77.
Which one of the following is a water soluble vitamin?
(a) Vitamin A
(b) Vitamin D
(c) Vitamin C
(d) Vitamin K
Answer:
(c) Vitamin C

Question 78.
Which one of the following deficient disease of Vitamin A?
(a) Cheilosis
(b) Xerophthalmia
(c) Convulsions
(d) Perncious Anaemia
Answer:
(b) Xerophthalmia

Question 79.
Which Vitamin deficiency leads to cheilosis?
(a) Vitamin B12
(b) Vitamin B6
(c) Vitamin B2
(d) Vitamin B5
Answer:
(c) Vitamin B2

Question 80.
Which one of the following vitamin deficiency leads to Rickets?
(a) Vitamin A
(b) Vitamin B1
(c) Vitamin C
(d) Vitamin D
Answer:
(d) Vitamin D

Question 81.
Which vitamin deficiency leads to Hair loss, muscle pain?
(a) Biotin
(b) Niacin
(c) Riboflavin
(d) Thiamine
Answer:
(a) Biotin

Question 82.
The deficiency of Vitamin B12 leads to the disease ………………..
(a) convulsions
(b) beriberi
(c) Pernicious anaemia
(d) pellagra
Answer:
(c) Pernicious anaemia

Question 83.
Which of the following is the chemical name of vitamin B12?
(a) Folic acid
(b) Cobalamin
(c) Pyridoxime
(d) Riboflavin
Answer:
(b) Cobalamin

Question 84.
Night blindness and kertinisation of skin is the chemical name of Vitamin B12?
(a) vitamin B1
(b) vitamin C
(c) vitamin A
(d) vitamin B12
Answer:
(c) vitamin A

Question 85.
Which vitamin deficiency leads to the disease megaloblastic anaemia?
(a) Vitamin B9
(b) Vitamin B6
(c) Vitamin B12
(d) Vitamin B2
Answer:
(a) Vitamin B9

Question 86.
Which one of the following is rich in liver oil, carrot, mango and papaya?
(a) Vitamin B1
(b) Vitamin C
(c) Vitamin A
(d) vitamin D
Answer:
(c) Vitamin A

Question 87.
Which of the vitamin deficiency leads to photosensitive dermatitis (or) pellagra?
(a) Vitamin B5
(b) Vitamin B6
(c) Vitamin B3
(d) Vitamin D
Answer:
(c) Vitamin B3

Question 88.
Depression, Hair loss, muscle pain are due to the deficiency of vitamin ……………..
(a) A
(b) B12
(c) B2
(d) B7
Answer:
(d) B7

Question 89.
The chemical name of vitamin B9 is …………….
(a) biotin
(b) folic acid
(c) niacin
(d) thiamin
Answer:
(b) folic acid

Question 90.
Which of the following is rich in citrus, fruits, tomato, amia and leafy vegetables?
(a) Vitamin C
(b) Vitamin E
(c) Vitamin A
(d) Vitamin D
Answer:
(a) Vitamin C

Question 91.
Consider the following statement
(i) Vitamin D functions in the adsorption and maintenance of calcium.
(ii) Vitamin E act as an antioxidant.
(iii) Vitamin C functions in blood clotting.
Which of the above statement is / are correct?
(a) (iii) only
(b) (ii) & (iii)
(c) (i) & (ii)
(d) (i) only
Answer:
(c) (i) & (ii)

Question 92.
Which vitamin is rich in cotton seed oil, sunflower oil, wheat germ oil and all vegetable oils?
(a) Vitamin C
(b) Vitamin E
(c) Vitamin A
(d) Vitamin D
Answer:
(b) Vitamin E

Question 93.
Which vitamin deficiency leads to the disease osteomalacia?
(a) Vitamin D
(b) Vitamin A
(c) Vitamin C
(d) Vitamin K
Answer:
(a) Vitamin D

Question 94.
Which one of the following is mainly required for blood clotting?
(a) Vitamin E
(b) Vitamin B12
(c) Vitamin C
(d) Vitamin K
Answer:
(d) Vitamin K

Question 95.
Consider the following statement
(i) Nucleic acid are biopolymers of nucleotides.
(ii) Controlled hydrolysis of DNA and RNA yield 3 components namely a nitrogeneous base, a pentose sugar and sulphate group.
(iii) DNA and RNA are the molecular repositories that carry genetic information in every organism.
Which of the above statement is / are correct?
(a) (i) only
(b) (ii) only
(c) (i) & (iii)
(d) (ii) & (iii)
Answer:
(c) (i) & (iii)

Question 96.
Which one of the following is found in cytoplasm and in nbosomers which contain 60% RNA and 40% protein.
(a) Ribosomal RNA
(b) Messenger RNA
(c) Transfer RNA
(d) DNA
Answer:
(a) Ribosomal RNA

Question 97.
Consider the following statement .
(i) Ribosomes are the sites at which protein synthesis takes place.
(ii) Messenger RNA carried genetic information from DNA to the ribosomes for protein synthesis.
(iii) t RNA consist of 20 – 40 nucleotides in a single chain.
Which of the above statement is / are not correct?
(a) (i) only
(b) (i) & (ii)
(c) (iii) only
(d) (ii) & (iii)
Answer:
(c) (iii) only

Question 98.
What is the name of the process of synthesis of mRNA from DNA strand?
(a) Transpiration
(b) Transcription
(c) Transformation
(d) Trans esterification
Answer:
(b) Transcription

Question 99.
Consider the following statement
(i) DNA mainly present in cytoplasm, nucleolus and ribosomes.
(ii) RNA is stable and not hydrolysed easily by alkalis.
(iii) DNA can replicate itself.
Which of the above statement is / are correct?
(a) (iii) only
(b) (i) only
(c) (i) & (ii)
(d) (ii) & (iii)
Answer:
(c) (i) & (ii)

Questin 100.
Who invented DNA finger printing?
(a) Sir Alec Jeffrey
(b) Rosalind Franklin
(c) Watson and Crick
(d) Maurice Wilkins
Answer:
(a) Sir Alec Jeffrey

Question 101.
Which one of the following can act as energy carriers?
(a) GTN
(b) ATP
(c) FAD
(d) Cyclic AMP
Answer:
(b) ATP

Questin 102.
Adenosine 3’, 5’- cyclic monophosphate a chemical messenger is otherwise called …………….
(a) ATP
(b) cyclic ADP
(c) cyclic AMP
(d) 3’P – ADP
Answer:
(c) cyclic AMP

Question 103.
Consider the following statement
(i) Endocrine glands make hormones.
(ii) Hormones may be classified as either protein (or) steroids
(iii) Hormones are intracellular signalling molecule.
Which of the above statement is / are not correct?
(a) (i) only
(b) (ii) & (iii)
(c) (iii) only
(d) (i) & (iii)
Answer:
(c) (iii) only

Question 104.
Which one of the following is a steroid?
(a) Insulin
(b) Epinephrine
(c) mutin
(d) Estrogen
Answer:
(d) Estrogen

Question 105.
Which one of the following is a protein hormone?
(a) Insulin
(b) Androgen
(c) Cortisol
(d) Estrogen
Answer:
(a) Insulin

Question 106.
The nucleic acid base having two possible binding sites is …………….
(a) thymine
(b) cytosine
(c) guanine
(d) adenine
Answer:
(c) guanine

Question 107.
DNA multiplication is called ………….
(a) transcription
(b) transformation
(c) transduction
(d) replication
Answer:
(d) replication

Question 108.
Insulin is a protein which plays the role of ……………
(a) an antibody
(b) a hormone
(c) an enzyme
(d) a transporting agent
Answer:
(b) a hormone

Question 109.
Which metal is present in Vitamin B12?
(a) Ca (II)
(b) Zn (II)
(c) Fe (II)
(d) Co (III)
Answer:
(d) Co (III)

Question 110.
The helical structure of protein is stablized by ………………
(a) oxygen bonds
(b) peptide bonds
(c) dipeptide bonds
(d) hydrogen bonds
Answer:
(d) hydrogen bonds

Question 111.
The cell membranes are mainly composed of ………….
(a) carbohydrates
(b) proteins
(c) phospholipids
(d) fats
Answer:
(c) phospholipids

Question 112.
Which one of the following is a polysaccharide?
(a) Nylon
(b) Amylose
(c) Ribose
(d) Polyethene
Answer:
(b) Amylose

Question 113.
Ribose ¡san example of …………….
(a) keto hexose
(b) aldohexose
(c) aldo pentose
(d) disaccharide
Answer:
(c) aldo pentose

Question 114.
Sucrose molecule is made up of …………..
(a) a gluco pyranose and fructo pyranose
(b) a glyco pyranose and fructo furanose
(c) a gluco furanose and fructo pyranose
(d) a gluco furanose and fructo furanose
Answer:
(b) a glyco pyranose and fructo furanose

Question 115.
A nucleotide consists of ……………….
(a) base and sugar
(b) base and phosphate
(c) sugar and phosphate
(d) base, sugar and phosphate
Answer:
(d) base, sugar and phosphate

Question 116.
Which of the following is responsible for heredity character?
(a) DNA
(b) RNA
(c) Proteins
(d) Hormones
Answer:
(a) DNA

Question 117.
The base adenine present in …………….
(a) DNA only
(b) RNA only
(c) Both DNA & RNA
(d) Protein
Answer:
(c) Both DNA & RNA

Question 118.
The protein which maintains the blood sugar level in the human body is …………..
(a) haemoglobin
(b) oxytocin
(c) insulin
(d) ptyalin
Answer:
(c) insulin

Question 119.
Ascorbic acid is a …………
(a) vitamin
(b) enzyme
(c) protein
(d) hormone
Answer:
(a) vitamin

Question 120.
Which of the following is not a constitutent of RNA?
(a) Ribose
(b) Phosphate
(c) Adenine
(d) Pyridine
Answer:
(d) Pyridine

Question 121.
Which one is fouñd in ATP ribonucleotide?
(a) Guanine
(b) Uracil
(c) Adenine
(d) Inulin
Answer:
(c) Adenine

Question 122.
Which substance is not present in nucleic acid?
(a) Cytosine
(b) Adenine
(c) Thymine
(d) Guanidine
Answer:
(d) Guanidine

Question 123.
In nucleic acid, the correct sequence is ……………..
(a) base – phosphate sugar
(b) phosphate – base – sugar
(c) sugar – base – phosphate
(d) base – sugar – phosphate
Answer:
(d) base – sugar – phosphate

Question 124.
The double helical structure of DNA was proposed by ……………….
(a) Watson and Crick
(b) Meicher
(c) Emil Fischer
(d) Khorana
Answer:
(a) Watson and Crick

Question 125.
Which substance is not present in nucleic acid?
(a) Cytosine
(b) Adenine
(c) Thymine
(d) Guanidine
Answer:
(d) Guanidine

Question 126.
In DNA, the complementary bases are ……………….
(a) Uracil and adenine; cytosine and guanine
(b) Adenine and thymine; guanine and cytosine
(c) Adenine and guanine; thymine and cytosine
(d) adenine and guanine; thymine and uracil
Answer:
(b) Adenine and thymine; guanine and cytosine

Question 127.
The structure of DNA is …………….
(a) linear
(b) single helix
(c) double helix
(d) triple helix
Answer:
(c) double helix

Question 128.
A gene is a segment of molecule of ……………
(a) DNA
(b) m – RNA
(c) t – RNA
(d) protein
Answer:
(a) DNA

Question 129.
The deficiency of vitamin C causes …………….
(a) scurvy
(b) rickets
(c) pyrrohea
(d) pellagra
Answer:
(a) scurvy

Question 130.
Which sugar is present in DNA?
(a) Deoxyribose
(b) Ribose
(c) D – fructose
(d) D – glucose
Answer:
(a) Deoxyribose

Question 131.
The base present ¡n DNA but not in RNA is …………..
(a) guanne
(b) adenine
(c) uracil
(d) thymine
Answer:
(d) thymine

Question 132.
Mutation of DNA occurs due to changes in the sequence of one of the following.
(a) Bases
(b) Ribose units
(c) Phosphate units
(d) Sugar units
Answer:
(a) Bases

Question 133.
Blood calcium level can be increased by the administration of ……………….
(a) glucogon
(b) calcitonin
(c) thyroxine
(d) paratharmone
Answer:
(d) paratharmone

Question 134.
The first hormone chemically synthesised in the laboratory is …………….
(a) cortisone
(b) insulin
(c) adrenaline
(d) eastrone.
Answer:
(b) insulin

Question 135.
RNA is different from DNA because RNA contains ……………
(a) Ribose sugar and tymine
(b) Ribose sugar and uracil
(c) Doxyribose sugar and thymine
(d) Deoxy ribose sugar and uracil
Answer:
(b) Ribose sugar and uracil

Question 136.
The hormone that helps in the conversion of glucose to glycogen is ……………..
(a) adrenaline
(b) insulin
(c) cortisone
(d) bile acid
Answer:
(b) insulin

Question 137.
Enery is stored in our body in the form of ………….
(a) ATP
(b) ADP
(c) Fats
(d) carbohydrates
Answer:
(a) ATP

Question 138.
Nucleic acid is a polymer of …………….
(a) Nucleosides
(b) a – aminoacids
(c) nucleotides
(d) glucose
Answer:
(c) nucleotides

Question 139.
Which one of the following is named as peptides?
(a) Esters
(b) Salts
(c) Amides
(d) Ketones
Answer:
(c) Amides

Question 140.
Irreversible precipitation of proteins is called ……………
(a) denaturation
(b) hydrolysis
(c) transformation
(d) trans esterification
Answer:
(a) denaturation

Question 141.
Which of the following is not an essential amino acid?
(a) Valine
(b) Lysinc
(c) Histidine
(d) Glycine
Answer:
(d) Glycine

Question 142.
Proteins are hydrolysed by enzymes into …………….
(a) dicarboxylic acid
(b) hydroxy acids
(c) amino acids
(d) aromatic acids
Answer:
(c) amino acids

Question 143.
Which one of the protein transports oxygen in the blood stream?
(a) Myoglobin
(b) Insulin
(c) Albumin
(d) Haemoglobin
Answer:
(d) Haemoglobin

Question 144.
Enzymes in the living systems …………..
(a) provide energy
(b) provide immunity
(c) catalyse biological process
(d) transport oxygen
Answer:
(c) catalyse biological process

Question 145.
Which compound can exist in a dipolar state?
(a) C6H5 CH2 CH (N = CH2) COOH
(b) (CH3)2 CH – CH (NH2) COOH
(c) C6H5CONH CH2 COOH
(d) HOOC – CH2 – CH2 – CO – COOH
Answer:
(b) (CH3)2 CH – CH (NH2) COOH

Question 146.
Haemoglobin is …………….
(a) an enzyme
(b) a globular protein
(c) a vitamin
(d) carbohydrate
Answer:
(b) a globular protein

Question 147.
The number of essential amino acid in man is …………….
(a) 8
(b) 10
(c) 20
(d) 18
Answer:
(b) 10

Question 148.
Which one of the biomolecule is insoluble in water?
(a) Keratin
(b) Haemoglobin
(c) Ribonuclease
(d) Aclenine
Answer:
(a) Keratin

Question 149.
Which of the following is used in our body as a fuel for muscles and nerves and to build and repair body tissues?
(a Cane sugar
(b) Fructose
(c) Proteins
(d) Glucose
Answer:
(c) Proteins

Question 150.
The bond that determines the secondary structure of proteins is …………..
(a) coordinate bond
(b) covalent bond
(c) hydrogen bond
(d) peptide bond
Answer:
(c) hydrogen bond

Question 151.
Which of the following monosaccharide is a pentose?
(a) Galactose
(b) Glucose
(c) Fructose
(d) Arabinose
Answer:
(d) Arabinose

Question 152.
Which of the following is a carbohydrate?
(a) Leucine
(b) Albumin
(c) Inulin
(d) Maltase
Answer:
(c) Inulin

Question 153.
Glucose gives silver mirror with Tollens reagent. it shows the presence of ……………..
(a) an acidic group
(b) an alcoholic group
(c) a ketonic group
(d) an aldehydic group
Answer:
(d) an aldehydic group

Question 154.
The compound which does not contain an asymmetric carbon atom is …………….
(a) glyceraldehyde
(b) glycine
(c) glucose
(d) fructose
Answer:
(b) glycine

Question 155.
Which one of the following compounds is found abundantly in nature?
(a) Fructose
(b) Starch
(c) Glucose
(d) Cellulose
Answer:
(d) Cellulose

Question 156.
Blood sugar is the same as ……………
(a) glucose
(b) galactose
(c) glycogen
(d) fructose
Answer:
(a) glucose

Question 157.
Which of the following is an aldohexose?
(a) Sucrose
(b) Cellulose
(c) Glucose
(d) Raffinose
Answer:
(c) Glucose

Question 158.
Glucose and mannose are …………..
(a) epimers
(b) anomers
(c) keto hexoses
(d) disaccharides
Answer:
(a) epimers

Question 159.
Which of the following is the sweetest sugar?
(a) Glucose
(b) Fructose
(c) Lactose
(d) Sucrose
Answer:
(b) Fructose

Question 160.
In fructose, the possible optical isomers are …………..
(a) 12
(b) 16
(c) 8
(d) 4
Answer:
(c) 8

Question 161.
Which one of the following is not used to convert glucose into gluconic acid?
(a) Br2 water
(b) Cone. HNO3
(c) Tollen’s reagent
(d) Fehling’s solution.
Answer:
(b) Cone. HNO3

II. Fill in the blanks.

  1. Chemically, carbohydrates are defined as …………. or …………. with a general formula
  2. …………. are synthesised by green leaves during photo synthesis.
  3. Almost all …………. are optically active as they have one or more chiral carbon.
  4. …………. are carbohydrates that cannot be hydrolysed further and are also called simple sugars.
  5. Erythrose is an example of ………….
  6. Glyceraldehyde is an example of …………. in monosaccharides.
  7. Glucose in human blood about …………. and it also known as ………….
  8. Ribulose is an example of ………….
  9. Glucose solution is optically active and it rotates the plane polarised light in …………. direction and so it called
  10. Glucose when oxidised with con.HNO3 gives ………….
  11. The reaction of glucose with Tollents reagent or Fehlmg’s solution confirms the presence of …………. group in glucose.
  12. The exact special arrangement of – OH groups in glucose was given by ………….
  13. The two isomers differ only in the configuration of C1 carbon are called ………….
  14. The cyclic structure of glucose with 5 carbon and one oxygen atom is called ………….
  15. The slow interconversion of α – D glucose and β – D glucose via open chain form under equilibrium is called ………….
  16. Sugars differing in configuration at an asymmetric centre is known as ………….
  17. …………. is present abundantly in fruts and hence it is also called fruit sugar.
  18. The solution having equal amount of glucose and fructose is termed as ………….
  19. Partial reduction of fructose with sodium amalgam and water produces …………. and …………. which are at second carbon.
  20. The reaction sodium amalgam and water with fructose confirms the presence of ………….
  21. The cyclic form of fructose is called ………….
  22. Disaccharides have general formula ………….
  23. In disaccharides, two monosaccharides are linked by …………. called ………….
  24. …………. is primary mixture of glucose, fructose and sucrose.
  25. Sucrose is also called …………. sugar.
  26. …………. is produced during digestion of starch by the enzyme a-amylase
  27. Starch contains about 20% …………. and about 80% of .
  28. Starch is used for …………. in plants.
  29. Cotton is almost pure ………….
  30. …………. is the storage polysaccharides of animals.
  31. …………. act as shock absorber and lubricant
  32. Proteins are polymers of ………….
  33. Orinithine and citrulline are called ………….
  34. At a specific pH value the net charge of an amino acid in neutral is called ………….
  35. Except …………. all other amino acids are optically active.
  36. In proteins, the amino acids are linked covalently by ………….
  37. The process of a protein, losing its higher order structure without losing the primary structure is called ………….
  38. Proteins such as …………. , …………. act as structural back bones.
  39. …………. and …………. controls the glucose level in the blood.
  40. …………. are biocatalysts that catalyse a specific biochemical reaction.
  41. …………. act as protective coating in aquatic organisms.
  42. Lipids act as …………. in fat metabolism.
  43. …………. help in the absorption and transport of fat soluble vitamins.
  44. Vitamin A, D, E and K …………. are vitamins.
  45. The chemical name of Vitamin A is ………….
  46. …………. deficiency leads to the disease cheilosis.
  47. …………. deficiency leads to the disease pellagra.
  48. …………. is a part of coenzyme A in carbohydrate protein and fat metabolism.
  49. …………. is rich in mushroom, avocada, egg yolk, sunflower oil.
  50. …………. deficiency leads to pernicious Anaemia.
  51. All citrus fruits and amla are rich in vitamin ………….
  52. …………. functions in blood clotting.
  53. Nucleic acids are bio polymers of ………….
  54. Both DNA and RNA have two major purine bases …………. and ………….
  55. The recurring deoxyribonucleotie units of DNA contains …………. and the ribonucleotide units of RNA contain
  56. The molecule with the phosphate group is called a ………….
  57.  The specific association of the two chains of the double helix in DNA is known as ………….
  58. t RNA molecule consists of …………. nucleotides in a single chain
  59. The synthesis of mRNA from DNA strand is called ………….
  60. …………. carries genetic information from DNA to the ribosomes for protein synthesis.
  61. …………. was first invented by Sir Alec Jeifry.

Answer:

  1. polyhydroxy aldehydes, ketoses, C(H2O)
  2. Carbohydrates
  3. carbohydrates
  4. Monosaccharides
  5. monosaccharide
  6. aldotriose
  7. 100 mg/dI, blood sugar
  8. ketopentose
  9. clockwise, dextrose
  10. glucaric acid (or) saccharic acid
  11. aldehyde
  12. Emil Fischer
  13. anomers
  14. pyranose form
  15. mutarotation
  16. epimers
  17. fructose
  18. invert sugar
  19. sorbitol, mannitol, epimers
  20. keto group
  21. furanose form
  22. Cn(H2O)n-1
  23. oxide, linkage, glycosidic linkage
  24. honey.
  25. invert (or) non reducing
  26. maltose
  27. amylose, amylopectin
  28. energy storage
  29. cellulose
  30. glycogen
  31. hyaluronate (or) glycosaminoglycans
  32. α – amino acids
  33. non – protein amino acids
  34.  iso electric point
  35. glycine
  36. peptide bonds
  37. denaturation
  38. keratin, collagen
  39. Insulin, glucagon
  40. Enzymes
  41. Lipids
  42. emulsifier
  43. Lipids
  44. fact – soluble
  45. retinol
  46. Vitamin B2
  47. Vitamin B3
  48. Vitamin B5
  49. Vitamin B5
  50. Vitamin B12
  51. C
  52. Vitamin K
  53. nucleotides
  54. adenine, guanine
  55.  2’ – doxy – D -ribose, D-ribose
  56.  nucleoside
  57. complementary base pairing
  58. 73 -94
  59. transcription
  60. mRNA
  61. DNA finger printing.

III. Match the column I and Column II using the code given below the column.

Question 1.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-15
Answer:
(a) 4 3 2 1

Question 2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-16
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-17
Answer:
(b) 2 1 4 3

Question 3.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-18
Answer:
(a) 3 1 4 2

Question 4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-19
Answer:
(d) 2 3 4 1

Question 5.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-20
Answer:
(a) 3 4 2 1

Question 6.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-21
Answer:
(d) 2 3 4 1

Question 7.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-22
Answer:
(a) 3 4 1 2

Question 8.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-23
Answer:
(a) 3 1 4 2

Question 9.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-24
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-25
Answer:
(a) 3 4 1 2

Question 10.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-26
Answer:
(a) 3 4 1 2

Question 11.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-27
Answer:
(a) 2 4 1 3

Question 12.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-28
Answer:
(a) 2 3 4 1

IV. Assertion and reasons.

Question 1.
Assertlon(A): Almost all carbohydrates are optically active.
Reason (R): All carbohydrates have one or more chiral carbon atoms.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 2.
Assertion(A): Glucose is called blood sugar.
Reason (R): Human blood contains about 100 mg/di of glucose hence it is called blood sugar.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R ¡s the correct explanation of A.

Question 3.
Assertlon(A): Glucose is called aldohexose as well as dextrose.
Reason (R): Glucose contain an aldehyde group and it rotates the plane polarised light in the clockwise direction.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 4.
Assertion(A): Glucose contains an aldehyde group and it occupies one end of the carbon chain.
Reason (R): When glucose is oxidised by bromine water, it gets oxidised to gluconic acid confirms the position of aldehyde group.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 5.
Assertion(A): Glucose contains one primary alcohol group at the end of the carbon chain.
Reason (R): When glucose is oxidised by strong oxidising agent conc.HNO3 it gives glucaric acid proves the presence of- CH2OH group at one end of carbon chain in glucose.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 6.
Assertion(A): Glucose and mannose are epimers.
Reason (R): Sugars differing in configuration at an asymmetric centre are called epimers.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 7.
Assertion(A): Fructose is called levulose and keto hexose
Reason (R): Fructose contains a ketone group and fructose rotates the plane polarised light in anic1ockwise direction.
(a) Both A and R are wrong
(b) Both A and R are correct and R is the correct explanation of A.
(c) A is wrong but R is correct.
(d) A is correct hut R is wrong.
Answer:
(b) Both A and R are correct and R is the correct explanation of A.

Question 8.
Assertion(A): Sucrose is called invert sugar.
Reason (R): During hydrolysis of sucrose, the optical rotation of the reaction mixture changes from dextro to levo.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 9.
Assertion(A): Sucrose is a non reducing sugar.
Reason (R): in sucrose. C1 of α – D glucose and C2 of D-fructose are joined together by glycosidic bond. Both the carbonyl carbons are involved in glycosidic bonding.
(a) Both A and R are correct but R is not the correct explanation of A.
(b) Both A and R are correct and R the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(b) Both A and R are correct and R the correct explanation of A.

Question 10.
Assertion(A): A disaccharide lactose act as reducing sugar.
Reason (R): In lactose, β – D galactose and β – D glucose are linked by β – 1, 4 – glycosidic bond in which aldehyde group is not involved.
(a) A is correct but R is wrong.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) Both A and R are correct and R is the correct explanation of A.
(d) A is wrong but R is correct.
Answer:
(c) Both A and R are correct and R is the correct explanation of A.

Question 11.
Assertion(A): Lactose is referred to as milk sugar.
Reason (R): It is extracted from sprouted barley.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are worng.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(c) A is correct but R is wrong.

Question 12.
Assertion(A): Maltose, a disaccharide acts as a reducing sugar.
Reason (R): Maltose consists of two molecules of α – D glucose with linked by α, 1, 4 – glycosidic bond and one glucose has the carbonyl group.
(a) Both A and R are correct and R ¡s the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 13.
Assertion(A): Except glycine all other amino acids are optically active.
Reason (R): Glycine does not contain chira! carbon atom whereas in all other amino acids have chiral carbon atom.
(a) Both A and R are wrong.
(b) A is correct but R is wrong
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explantion of A.
Answer:
(c) Both A and R are correct and R is the correct explanation of A.

Question 14.
Assertion(A): Enzymes have active sites and substrates, reactive sites on their surfaces respectively.
Reason (R): Active and reactive sites push the enzyme and substrate molecules away from each other.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 15.
Assertion(A): Enzymes are defined as biological proteins.
Reason (R): Chemically all enzymes are globular proteins.
(a) Both A and R are correct and R ¡s the correct explanation of A.
(b) Both A and R are wrong.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R ¡s the correct explanation of A.

Question 16.
Assertion(A): DNA and RNA molecules are found in the molecules of the cell.
Reason (R): On heating, enzymes do not lose their specific activity.
(a) Both A and R are correct and R explains A.
(b) Both A and R are wrong.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(b) Both A and R are wrong.

Question 17.
Assertion(A): Vitamin D can be stored in our body.
Reason (R): Vitamin D is a fat soluble vitamin.
(a) A is correct but R not explains A.
(b) Both A and R are correct and R explains A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(b) Both A and R are correct and R explains A.

Question 18.
Assertion(A): Glycine must be taken through diet.
Reason (R): It is an essential amino acid.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explain A.
(c) A is wrong but R is correct.
(d) Both A and R are wrong.
Answer:
(d) Both A and R are wrong.

Question 19.
Assertion(A): In proteins, amino acids are linked through peptide bonds.
Reason (R): Peptide bonds are glycosidic (or) oxygen bridges.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explains A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(c) A is correct but R is wrong.

Question 20.
Assertion(A): Monosaccharides are held by glycocidic bonds.
Reason (R): Monosaccharides are macro molecules.
(a) Both A and R are correct and R explains A.
(b) Both A and R are correct but R does not explains A.
(c) A is wrong but R is correct.
(d) Both A and R are wrong.
Answer:
(d) Both A and R are wrong.

V. Find the odd one out and give the reasons.

Question 1.
Glucose, fructose, galactose, mannose, sucrose.
Answer:
Sucrose, It is a disaccharide whereas other are monosaccharides.

Question 2.
Glucose, aldo hexose, dextrose, blood sugar, fruit sugar.
Answer:
Fruit sugar, It is the name of fructose, All others are indicating glucose only.

Question 3.
Fructose, fruit sugar, milk sugar, levulose, ketohexose.
Answer:
Milk sugar, It is the name of lactose, All others are indicating fructose only.

Question 4.
Mannose, sucrose, lactose, maltose, diastose.
Answer:
Mannose, It is a monosaccharide whereas otheres are disaccharide.

Question 5.
Keratm, glucose, mannose, starch, cellulose.
Answer:
Keratin, It is a protein whereas others are carbohydrates.

Question 6.
Keram, collagen, glycine, alanine, inulin, instilin.
Answer:
Inulin, It is a carbohydrates whereas others arc proteins.

Question 7.
Glycine, alanine, histidine, cultamine, proline, serine.
Answer:
Histidine. it is an essential amino acid whereas others are non essentîal amino acids.

Question 8.
Valine, phenyl alanine, histidine, lysine, alanine.
Answer:
Alanine, It is non essential amino acid whereas others are essential amino acids.

Question 9.
Invertase, maltase, zymase, maltose, lactase.
Answer:
Maltose, It is a carbohydrates whereas others are enzymes.

Question 10.
Vitamin A, Vitamin D, Vitamin C, Vitamin E, Vitamin K.
Answer:
Vitamin C is water soluble vitamins whereas others are fat soluble vitamins.

Samacheer Kalvi 12th Chemistry Biomolecules 2 Mark Questions and Answers

VI. Answer the following.

Question 1.
Define carbohydrates. Give example.
Answer:

  1. Carbohydrates are defined as polyhydroxy aldehydes (or) ketoses with a general formula Cn(H2O)n.
  2. They are considered as hydrates of carbon containing hydrogen and oxygen in the same ratio as in water.
  3. Example: Glucose, sucrose, cellulose.

Question 2.
Draw the structure of

  1. D – Glucose
  2. D – Fructose.

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-29

Question 3.
Draw the structure of sucrose
Answer:
Sucrose C12H22O11 is a disaccharide
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-30

Question 4.
Explain photosynthesis.
Answer:
Carbohydrates are synthesised by green leaves during photosynthesis, a complex process in which sunlight provides the energy to convert carbon dioxide and water into glucose and oxygen. Glucose is then converted into other carbohydrates and is consumed by animals.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-31

Question 5.
Draw and explain the structure of glyceraldehyde.
Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-32

2. D – Glyceraldehyde and L – Glyceraldehyde are two enantiomers.

3. D (or) (+) is dextre rotatory and L (or) (-) is levo rotatory.

Question 6.
What is meant by dextro and levo rotatory?
Answer:
1. Fischer has devise a projection formula to reláte the two enantiomeric forms. Base on this, carbohydrates are named as D or L.

2. + and – sign indicates the dextro rotatory and levo rotatory respectively.

3. Dextre rotatory compound rotate the plane of plane polarised light in clockwise direction while the levo rotatory compounds rotate in anticlockwise direction.

4. Dextro rotatory compounds are represented as D (+) or L (+) and the levo rotatiory compounds as D (-) or L (-).
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-33

Question 7.
Give examples for the following.

  1. Aldotriose
  2. Ketotriose
  3. Aldotetrose
  4. Ketotetrose

Answer:

  1. Aldotriose – Glyceraldehyde
  2. Ketotriose – Dihydroxy acetone
  3. Aldotetrose – Erythrose
  4. Ketotetrose – Erythrulose

Question 8.
Give examples for the following.

  1. Aldo pentose
  2. Keto pentose
  3. Aldo hexose
  4. Ketohexose

Answer:

  1. Aldo pentose – Ribose
  2. Keto pentose – Ribulose
  3. Aldo hexose – Glucose
  4. Keto hexose – Fructose

Question 9.
Explain the action of Conc.HNO3 with fructose with equation.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-34

Question 10.
Write a note about glycogen?
Answer:
1. Glycogcn is the storage polysaccharide of animals. It is present in the liver and muscles of animals. Glycogen is called as animal starch.

2. Glycogen on hydrolysis gives glucose molecules. Structurally glycogen resembles amylo pectin with more branching. In glycogen the branching occurs every 8 – 14 glucose units opposed to 24 – 30 units in amylopectin. The excessive glucose in the body is stored in the form of glycogen.

Question 11.
What are amino acid? Give its structure.
Answer:
1. Amino acids are compounds which contain an amino group and a carboxylic group. Protein molecules are made up of α, β, γ – amino acids.

2. The protein molecules are mde up of a-amino acids which can be represented by the following general formula.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-35

3. There are 20 amino acids commonly found in protein molecules.

Question 12.
Define iso electric point.
Answer:
At a specific pH value, the net charge of an amino acid in neutral is called iso electric point. At a pH value above the iso electric point the amino acid will be negatively charged and positively charged at pH values below the iso electric point.

Question 13.
Whatis zwitter ion? Give its structure.
Answer:
1. At aqueous solution, the proton from cabroxyl group can be transferred to the amino group of an amino acid leaving these groups with opposite charges.

2. Despite having both positive and negative charges, this molecule is neutral and has amphotenc behaviour. These ions are called zwitter ions.

3. Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-36

Question 14.
What are hormones? Mention their functions. Name some hormones.
Answer:
1. Hormones are organic compounds (eg. peptide or a steroid) that is secreted by endocrine glands. It is an inter cellular signalling molecule and induces a physiological response. Hormones maintain blood pressure, blood volume and electrolyte balance, embryogenesis, hunger, eating behaviour and digestion.

2. The major endogrine glands are the pituitary, pineal, thymus. thyroid, adrenal glands and pancreas. In addition men produces hormones in their testes and women produces hormones in their ovary.

Question 15.
What are the expected products of hyrolysis of lactose?
Answer:
Since lactose is a disaccharide, therefore on hydrolysis it gives two molecules of monosaccharides, i.e., one molecule each of D – (+) – glucose and D – (+) – galactose
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-37

Question 16.
Glucose or sucrose are soluble in water by cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
Answer:
Glucose and sucrose molecules contain many – OH groups and hence are capable of forming H – bond with water molecules. On the other hand, cyclohexane and benzene are hydrocarbon and hence they cannot form H-bond with water. Therefore, glucose and sucrose are soluble in water whereas cyclohexane and benzene are insoluble in water.

Question 17.
How do you explain the absence of aldehyde group in the pentaacetate of D – glucose?
Answer:
Since pentaacetate of D – glucose are not oxidised either by Tollen’s reagent or Fehling’s solution. This indicates the absence of aldehyde group.

Question 18.
Why cannot Vitamin C be stored in our body?
Answer:
Vitamin C is water soluble, therefore it is readily excreted in urine and hence cannot be stored in the body.

Question 19.
What do you understand by the term glycosidic linkage?
Answer:
In disaccharides or polysaccharides, monosaccharides are joined together by an oxide linkage formed by loss of a water molecule. Such a linkage between two monosaccaride units through oxygen atom is called glycosidic linkage.

Question 20.
What are essential and non-essential amino acids? Give two examples of each type.
Answer:
Essential amino acids:
Amino acids which cannot be synthesised in the body and must be obtained through regular diet are known as esential amino acids. Foe example: Valine, Leucine.

Non – essential amino acids:
The amino acids which can be synthesised in the body by itself are known as non – essential amino acids. For example: glycine, alanine.

Question 21.
What is the effect of denaturaion on the structure of proteins?
Answer:
During the denaturation process, 2° and 3° structure of proteins are destroyed by 1° structure remains intact. For example: curdling of milk.

Question 22.
How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Answer:
Vitamins are classified into two groups depending upon their solubility in water or fat.
1. Water soluble vitamins:
These include vitamins of ‘B’ group (except B12) and vitamin ‘C’.

2. Fat soluble vitamin:
These include vitamins A, D, E and K. They are stored in liver and adipose tissues. Vitamin ‘K’ is responsible for the coagulation of blood.

Question 23.
What are nucleic acids? Mention their two important functions.
Answer:
Nucleic acids are polymers of nucleotides containing a pentose sugar, heterocyclic base and a phosphate group. They help in synthesis of proteins. They are also responsible for the transfer of genetic characters from one generation to the next generation.

Question 24.
What is the difference between a nucleoside and a nucleotide?
Answer:
Nucleoside is formed by the condensation of a purine or pyrimidine base with pentose sugar at position 1. When nucleoside is linked to phosphoric acid at 5 position of sugar moiety, we get a nucleotide. Hence a nucleotide has three units – Phosphate group, pentose sugar and a base, whereas nucleoside has two units – pentose sugar and a base.

Question 25.
Write two main functions of carbohydrates in plants.
Answer:
Main functions of carbohydrates in plants:

  1. Carbohydrates are used as storage molecules as starch in plants.
  2. Cell wall of bacteria and plants is made up of cellulose.

Question 26.
Name two components of starch. How do they differ from each other structurally?
Answer:
The two components of starch are:

  1. Amylose
  2. Amylopectin

Amylose is a straight chain polymer of α – D – (+) glucose, while amylopectin is a branched chain polymer of α – D – glucose.

Question 27.
Name the four bases present in DNA. Which one of these is not present in RNA?
Answer:
DNA contains four bases, viz; adenine (A), guanine (G), cytosine (C), thymine (T). RNA also contain four bases, first three bases are same as in DNA but the fourth one is uracil(U).

Question 28.
Name two fat soluble vitamins, their sources and the diseases caused due to their deficiency in diet.
Answer:

Vitamins

  1. Vitamin A
  2. Vitamin D

Sources

  1. Fish, liver oil, carrot
  2. Sunlight, milk, egg yolk

Deficiency Diseases

  1. Night blindness
  2. Rickets and osteomalacia

Question 29.
Name two water soluble vitamins, their sources and the diseases caused due to their deficiency in diet.
Answer:

Vitamins

  1. Vitamin B<sub>1</sub>
  2. Vitamin C

Sources

  1. Yeast. milk, vegetables.
  2. Citrus fruits, amia and green leafy vegetables.

Deficiency Diseases

  1. Ben – ben
  2. Scurvy (bleeding gums)

Samacheer Kalvi 12th Chemistry Biomolecules 3 mark Question and Answers

VII. Answer the following questions.

Question 1.
Explain the methods of preparation of glucose.
Answer:
1. When sucrose is boiled with diI-H2SO4 in alcoholic solution, hydrolysis take place and glucose and fructose are formed.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-38

2. Glucose is produced commercially by the hydrolysis of starch with dilute HCL at high temperature and pressure.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-39

Question 2.
What happens when glucose reacts with

  1. Br2 / H2O
  2. Conc.HNO3

Answer:
1.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-40

2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-41

Question 3.
How will you prove the presence of aldehyde group in glucose?
Answer:
Glucose is oxidised to gluconic acid with ammoniacal silver nitrate (Tollen’s reagent) and alkaline copper sulphate (Fehling’s solution). Tollen’s reagent is reduced to metallic silver and Fehling’s solution to cuprous oxide (red precipitate). These reactions confirm the presence of an aldehye group in glucose.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-42
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-43

Question 4.
Define

  1. Epimers
  2. Epimerisation.

Answer:
1. Sugar differing at an asymmetric centre is known as epimers.

2. The process by which one epimer is converted into other is called epimerisation and it requires the enzyme epimersase.

3. Galactose is converted to glucose by this manner in our body.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-44

Question 5.
Explain the methods of preparation of fructose with equations.
Answer:
1. Fructose is obtained from sucrose by heating with dilute H2SO4 (or) with the enzyme invertase.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-45

2. Fructose is prepared commercially by the hydrolysis of Inulin (a polysaccharide) in acidic medium
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-46

Question 6.
What happens when fructose is treated with sodium amalgam and water?
Answer:
When fructose is treated with sodium amalgam and water, partial reduction take place and the products formed are epimers of sorbitol and mannitol. New asymmetric carbon is formed at C – 2. This reaction confirms the presence of keto group in fructose.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-47

Question 7.
Explain about the cyclic structure of fructose?
Answer:
Fructose forms a five membered ring similar to furan. Hence it is called furanose form. When fructose is a component of a saccharide as in sucrose, it usually occurs in furanose form.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-48

Question 8.
Explain about the structure, nature and properties of sucrose.
Answer:
1. Sucrose commonly known as table sugar is the most abundant disaccharide. It is obtained mainly from juice of sugar cane and sugar beets. Insects such as honey bees have the enzyme mvetase that catalyses the hydrolysis of sucrose into glucose and fructose mixture.

2. Honey is primarily a mixture of glucose, fructose and sucrose. On hydrolysis sucrose yields equal amount of glucose and fructose units.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-79

3. Sucrose (+66.6°) and glucose (52.5°) are dextrorotatory compounds while fructose is levo rotatory (- 92.4°)

4. During hydrolysis of sucrose the optical rotation of the reaction mixture changes from dextro to levo. Hence sucrose is also as invert sugar.

5. Structure:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-49
In sucrose C1 of α – D glucose is joined to C2 of D – Fructose. The glycosidic bond thus formed is called α, 1, 2 – glycosidic bond. Since both the carbonyl carbons are involved in the glycosidic bonding, sucrose is a non – reducing sugar.

Question 9.
Prove that sucrose is

  1. invert sugar
  2. non – reducing sugar.

Answer:
1. Sucrose is an invert sugar: Sucrose (+66.6°) and glucose (+52.5°) are dextro rotatory compounds while fructose is levo rotatory (- 92.4°). During hydrolysis of sucrose, the optical rotation of the reaction mixture changes from dextro to levo. Hence sucrose is also called invert sugar.

2. In sucrose, C1 of α – D glucose is joined to C2 of α – D fructose. The glycosidic bond thus formed is called α – 1, 2 – glycosidic bond. Since both the carbonyl carbons (reducing groups) are involved in the glycosidic bonding, sucrose is a non – reducing sugar.

Question 10.
Write a note about lactose
Answer:
1. Lactose is a disaccharide found in milk of mammals and hence it is referred to as milk sugar.

2. On hydrolysis, it yeilds galactose and glucose. The β – D galactose and β – D glucose are linked by β – 1, 4 – glycosidic bond. the aldehyde carbon is not involved in the glycosidic linkage. Hence it retains its reducing property and is called a reducing sugar.

3. Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-50

Question 11.
Lactose act as reducing sugar. Justify this statement.
Answer:

  1. Lactose is a disaccharide and contains one galactose unit and one glucose unit.
  2. In lactose, the β – D galactose and β – D glucose are linked by β – 1, 4 – glycosidic bond.
  3. The aldehyde carbon is not involved in the glycosidic bond hence it retains its reducing property and is called a reducing sugar.

Question 12.
Write about maltose with its structure.
Answer:
1. Maltose is extracted from malt and it is called malt sugar. Malt from sprouting barely is the major source of maltose. Maltose is produced during digestion of starch by the enzyme α – analyse.

2. Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-51

3. Maltose consists of two molecules of α – D glucose units linked by an α – 1, 4 – glycocidic bond between anomeric carbon ofone unit and C – 4 of the other unit. Since one of the glucose has the carbonyl group intact it act as a reducing sugar.

Question 13.
Sucrose and maltose are disaccharides but sucrose Is a non reducing sugar while maltose Is a reducing sugar. Give reason.
Answer:
1. Sucrose is a disaccharide that composed of α – D glucose and β – D fructose. In sucrose Cl of α – D glucose is joined to C2 of D – fructose. The glycosidic bond thus formed is called α – 1, 2- glycosidic bond. Since both the carbonyl carbons (reducing carbons) are involved in the glycosidic bonding, sucrose is a non-reducing sugar.

2. But maltose contains two molecules of x – D glucose units that are linked by an α – 1, 4 – glycosidic bond. Anomeric carbon of one unit and C – 4 of other unit are connected together. Smce one of the glucose bas the carbonyl group intact it acts as reducing sugar.

Question 14.
Give brief account of nature and structure of cellulose.
Answer:
1. Cellulose is the major constituent of plant cell walls. Cotton is almost pure cellulose. On hydrolysis, cellulose yields D – glucose molecules

2. Cellulose is a straight chain polysaccharide. The glucose molecules are linked by β (1, 4 -) glycoside bond.

3. Structure of cellulose.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-52

Question 15.
What are the uses of cellulose?
Answer:

  1. Cellulose is used extensively in manufacturing paper, cellulose fibres and rayon explosive.
  2. Gun cotton – nitrated ester of cellulose an explosive is prepared from cellulose.
  3. Cellulose act as food for animals

Question 16.
Human cannot use cellulose as food – Why?
Answer:
Human cannot use cellulose as food because our digestive systems do not contain the necessary enzymes such as glycosidases (or) cellulases that can hydrolyse the cellulose. But animals contain cellulose enzyme in their digestive system and they can digest cellulose. So cellulose can used as food for animals but not for human.

Question 17.
What are the major classification of proteins? Give example?
Answer:
1. Proteins are classified based on their structure into two major types. They are fibrous protein and glubular protein.

2. Fibrous proteins are linear molecules similar to fibres. They are generally insoluble in water and are held together by disulphide bridges and weak inter molecular hydrogen bonds. These proteins often used as structural proteins. Example, Keratin, Collagen.

3. Globular proteins have an overall spherical shape. The poly peptide chain is folded into a spherical shape. These proteins are usually soluble in water and have many functions including catalysis.

Question 18.
Explain the mechanism of enzyme action?
Answer:
1. Enzymes are bio catalysts that catalyse a specific bio chemical reaction. They generally activate the reaction by reducing the activation energy by stabilising the transition state.

2. In a typical reaction, enzyme E binds with the substrate molecule leversity to produce an enzyme – sybstate complex. During this stage the substrate is converted into product and

3. Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-53

Question 19.
Explain about the nature, classification and properties of lipids (or) write a note about lipids.
Answer:
1. Lipids are organic molecules that are soluble in organic solvents such as chloroform and methanol and are insoluble in water. Lipids means fat. They are the principal component of cell membrane including cell walls.

2. Lipids act as energy source for living systems. Fat provide 2 – 3 fold higher energy compared to carbohydrates or proteins.

3. Based on their structure, lipids can be classified as simple lipids, compound lipids and derived lipids.

4. simple lipids can be further classified into fats, which are esters of long chain fatty acids fats, with glycerol (triglycerides) and waxes which are esters of fatty acid with long chain monohydric alcohols (Bees wax)

5. Compound lipids are the esters of simple fatty acid with glycerol which contain additional group. Based on the groups attached, they are further classified into phospholipids, glycolipids and lipproteins. Phospho lipids contain a phospho ester linkage while the glycolipids contain a sugar molecule attached. The iipo proteins are complexes of lipid with proteins.

Question 20.
Write the chemical name, sources and deficient disease of the following.

  1. Vitamin D
  2. Vitamin E
  3. Vitamin K

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-54

Question 21.
What are the biological functions of nucleic acids?
Answer:

  1. Energy carriers (ATP)
  2.  Components of enzyme cofactors. Example – Co enzyme A, NAD, FAD
  3. Chemical messengers. Example – Cyclic AMP, CAMP

Question 22.
What happens when D – glucose is treated with the following reagents?

  1. HI
  2. Bromine water
  3. HNO3

Answer:
1. Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-55

2.Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-56

3. Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-57

Question 23.
Define the following as related to proteins.

  1. Peptide linkage
  2. Primary structure
  3. Denaturation

Answer:
1. Peptide linkage:
Amino acids are bifunctional molecules with NH2 group at its one end and COOH at the other. Therefore, the COOH of one molecule and NH2 of another molecule can interact with elimination of a H,O molecule to form an amide like linkage called peptide bond or peptide linkage.

2. Primary structure:
The sequence in which amino acids are linked together in a polypeptide chain forms the primary structure.

3. Denaturation:
The process by which secondary and tertiary structure of proteins get disturbed by change of pH or temperature, so that they are not able to perform their functions, is called denaturation of proteins.

Question 24.
Difference between Globular and fibrous proteins.
Answer:
Globular proteins

  1. They form α – helix structure
  2. They are soluble in water.
  3. They are cross lined condensation polymers of acidic and basic amino acids.

Fibrous Proteins

  1. They have – plcated structure.
  2. They are insoluble in water.
  3. They are linear condensation polymeric proteins.

Question 25.
Explain what is meant by

  1. a peptide linkage
  2. a glycosidic linkage.

Answer:
1. Peptide linkage:
Polymers of x – amino acids are connected to each other by peptide bond or peptide linkage. Chemically peptide linkage is an amide formed between – COOH group and NH2 group.

2. Glycosidic linkage:
The two monosaccharide units are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two mono saecharide units through oxygen atom is called glycosidic linkage.

Question 26.
What are essential and non-essential amino acids? Give one example of each type.
Answer:
1. Essential amino acids:
Amino acids which are not synthesised by the human body are called essential amino acids. Example – Valine, Leucinc.

2. Non – essential amino acids:
Amino acids which are synthesised by human body are called non – essential amino acids. Example – Glycine, Aspartic acid, etc.

Question 27.
Mention the type of linkage responsible for the formation of the following.

  1. Primary structure of proteins
  2. Cross linking of polypeptide chains.
  3. α – helix formation
  4. β – sheet structure.

Answer:
Type of Structure

  1. Primary structure of proteins
  2. Cross linking of polypeptide chains
  3. α – helix
  4. β – sheet structure

Type of linkages

  1. Peptide bond or peptide linkage
  2. Polypeptide linkage
  3. Hydrogen bond
  4. Intermolecular hydrogen bond

Question 28.
Name the chemical components which constitute nucleotides. Write any two functions of nucleotides in a cell.
Answer:
Nucleotides are made up of a heterocyclic base containing nitrogen, a five carbon atom – moeity and a phosphate group.
Example:

  1. AMP (adenosine monophosphate)
  2. ADP (adenosine diphosphate) and
  3. ATP (adenosine triphospate)

Functions:

  1. Act as energy carriers
  2. They synthesise proteins

Question 28.
Name the main disease caused due to lack of vita min and its source in each of the following. A, B6 and E.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-58

Question 29.
Define the following and give one example of each.

  1. Isoelectric point
  2. Mutarotation
  3. Enzymes

Answer:
1. Isoelectric point:
It is the pH at which +ve and – ve charges on zwitter ion are equal. Example amino acid exists as zwitter ion at pH = 5.5 to 6.3.

2. Mutarotation:
It is a spontaneous change in optical rotation when an optically active substance is dissolved in water. Example α – glucose, when dissolved in water, then its optical rotation changes from 1110 to 52.5°.

3. Enzymes:
Enzymes are biocatalysts which speeds up the reactions in biosystems. They are highly specific and selective in their action. Chemically all enzymes are proteins.

Question 30.
What is denaturation and renaturation of proteins? Give reason: Amylose present in the saliva becomes inactive in the stomach.
Answer:
1. The process of disruption of 2° and 3° structure of proteins without changing its primary structure is called denaturation.

2. In stomach the pH is acidic, therefore amylose becomes inactive. That is why digestion of carbohydrates does not take place in stomach.

Question 31.
Define the following terms.

  1. Nucleotide
  2. Anomers
  3. Essential amino acids

Answer:
1. Nucleotide:
It is the monomer unit of DNA which is formed by a nitrogenous base, deoxyribose sugar and phosphoric acid.

2. Anomers:
Anomers are cyclic monosaccharides which are differing from each other in the configuration of C – 1 if it is an aldose or in the configuration at C – 2 if it is a ketose.

3. Essential amino acids:
The amino acids cannot be synthesised by the body and are essential for the body.

Question 32.
Which one of the following is disaccharide.

  1. Starch, Maltose, Fructose, Glucose.
  2. Write the name of vitamin whose deficiency causes bone deformities in children.

Answer:

  1. Maltose
  2. Vitamin D

Quesiton 33.
Write the major classes in which the carbohydrates are divided depending upon whether thee undergo hydrolysis and if so, the number of products formed.
Answer:
On the basis of hydrolysis, carbohydrates are divided into three major classes:
1. Monosaccharides. These cannot be hydrolysed into simpler molecules. These are further classified as aldoses, and ketoses.

2. Oligosaccharides. These carbohydrates on hydrolysis give 2-10 units of mono-saccharides. For example – Sucrose.

3. Polysaccharides. These are high molecular mass carbohydrates which give many molecules of monosaccharides on hydrolysis. Form example: Cellulose, starch.

Question 34.

  1. What changes occur in the nature of egg proteins on boiling?
  2. Name the type of bonding which stabilises the a-helix structure in proteins.

Answer:
1. On boiling, protein of egg gets denatured. Thus, due to coagulation water get absorbed.

2. Hydrogen bonding between
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-59

Question 35.
Answer the following questions briefly.

  1. What are reducing sugars?
  2. What is meant by denatu ration of a protein?
  3. How is oxygen replenished in our atmosphere?

Answer:
1. Reducing sugar:
All those carbohydrates which reduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars. All monosaccharides whether aldose or ketose are reducing sugars.

2. Denaturation of a protein:
When 2° and 3° structure of a protein is destroyed due to the physical changes like temperature, change in pH, it is called denaturation of a protein.
Example: Coagulation of egg white on boiling.

3. We take oxygen from atmosphere and release CO2. Plants take up CO2 and H2O from the atmosphere to prepare their food in the presence of sunlight and release O2 thus O2 is replenished in atmosphere.

Samacheer Kalvi 12th Chemistry Biomolecules 5 mark Questions Answers

VIII.Answer the following questions.

Question 1.
How would you prove the structure of glucose? (OR) Elucidate the structure of glucose.
Answer:
1. Elemental analysis and molecular weight determination show that the molecular formula of glucose is C6H12O6.

2. On reduction with Conc.HI and red P at 373K, glucose gives a mixture of n – hexane and 2 – iodohexane indicating that the six carbon atoms are bonded linearly.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-60

3. Glucose reacts with hydroxylamine to form oxime and with HCN to form cyanohydrin. These reactions indicate the presence of carbonyl group in glucose.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-61

4. Glucose gets oxidised to gluconic acid with bromine water with bromine water suggests that the carbonyl group is an aldchyde group and it occupies one end of the carbon chain. When glucose is oxidised by conc.HNO3, glucaric acid is formed and it suggest that the other end Id occupied by a primary alcoholi.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-62

5. Glucose is oxidised to gluconic acid with ammoniacal silver nitrate (Toiles reagent) and alkaline copper sulphate (Fehling’s solution). These reactions further confirm the presence of an aldehyde group.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-63

6. Glucose forms penta acetate with acetic anhydride suggesting the presence of five alcohol groups.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-64

7. Glucose is a stable compound and does not undergo dehydration easily. It indicates that not more than one hydroxyl group is bonded to a single carbon atom. Thus the five hydroxyl groups are attached to five different carbon atoms and the sixth carbon is an aldehyde group.

8. The exact special arrangement of – OH groups was given by Emil Fischer as follows.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-65

9. D (+) glucose has D configuration and it is dextro rotatoly.

Question 2.
Explain about the cyclic structure of Glucose.
Answer:
1. Fischer identified that the open chain Penta hydroxyl aldehyde structure of glucose that he proposed did not completely explain its chemical behaviour.

2. Unlike simple aldehyde, glucose did not form crystalline hisuiphite compound with sodium bisuiphite. Glucose does not give Schifis test and pent.a acetate derivative of glucose was not oxidised by Tollen’s reagent. This behaviour could not be explained by open chain structure.

3.Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-66

4. In order to explain these it was proposed that one of the hydroxyl group reacts with aldehyde group to form a cyclic structure (hemiacetal form). This also results in the conversation of the achiral aldehyde carbon into a chiral one leading the possibility of two isomers. These two isomers differ only in the configuration of C1 carbon. These isomers are called anomers.

5. The two anomeric forms of glucose are called – and β – forms. This cyclic structure of glucose is similar to pyran, a cyclic compound with 5 carbon and one oxygen atom, and hence is called pyranose form.

6. The specific rotation of pure α – and 3 – (D) glucose are 112° & 18.7° respectively. However, when pure form any one of these sugars dissolved in water, slow interconversion of – D glucose and β – D glucose via open chain form until equilibrium is established giving constant specific rotation + 53°. This phenomenon is called mutarotation.

Question 3.
Explain about the structure of Fructose. (OR) Elucidate the structure of Fructose.
Answer:
1. Elemental analysis and molecular weight determination of fructose show that it has the molecular formula C6H12O6.

2. Fructose on reduction with Hl and red phosphoms gives a mixture of n – hexane (major product) and 2 – iodohexane (minor product). This reaction indicates that the six carbon atoms in fructose are in a straight chain.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-67

3. Fructose reacts with NH2OH and HCN. It shows the presence of carbonyl groups in the molecules of fructose.

4. Fructose reacts with acetic anhydride in the presence of pyridine to form penta acetate. This reaction indicates the presence of’ five hydroxyl groups in a fructose molecule.

5. Fructose is not oxidized by bromine water. This rules out the possibility of presence of an aldehyde (-CHO) group.

6. Partial reduction of fructose with sodium amalgam and water produces mixtures of sorbitol and mannitol which are epimers at second carbon. New asymmetric carbon is formed at C – 2. This confirms the presence of keto group.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-68

7. On oxidation with nitric acid, it gives glycolic acid and tartane acids which contain smaller number of carbon atoms than in fructose. This shows that a keto group is present in C – 2. It also shows the presence of 10 alcoholic groups at C- 1 and C- 6. From the above reaction the structure of fructose is
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-69

This shows that a keto group is present in C – 2. It also shows the lpresence of 10 alcoholic groups at C – 1 and C – 6. From the above reaction the structure of fructose is
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-70

Question 4.
Describe about the structure, nature and properties of starch.
Answer:

  1. Starch is used for energy storage in plants. Potatoes, corn, wheat and rice are the rich sources of starch.
  2. It is a polymer of glucose in which glucose molecules are lined by α (1,4) glycosidic bonds.
  3. Starch can be separated into two fractions namely, water soluble amylose and water insoluble amylo pectin. Starch contain about 20% of amylase and about 80% amylopectin.
  4. Amylose is composed of unbranched chains upto 4000 α – D glucose molecules joined by α (1,4) glycosidic bonds.
  5. At branch points, new chains of 24 to 30 glucose molecules are linked by α (1,6) glycosidic bonds with iodine solution amylose gives blue colour while amylo pectin gives a purple colour.

Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-72

Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-73

Question 5.
Explain about the structure of proteins.
Answer:
1. Proteins are polymers of amino acids. Their three dimensional structure depends mainly on the sequence of amino acids. The protein structure can be described at four hierarchal levels called primary, secondary, tertiary and quaternary structures.

Primary structure of proteins:
Proteins are polypeptide chains made up of amino acids connected through peptide bonds. The relative arrangement of the amino acids in the polypeptide chain is called the primary structure of the protein.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-74

Secondary structure of proteins:
The amino acids in the polypeptide chain forms highly, regular shapes through the hydrogen bond between the carbonyl oxygen and the
neighbouring amine hydrogen of the main chain, α – Helix and β – strands or sheets are two most common substructures formed by proteins.

α – Helix:
In the α – helix sub – structure, the amino acids are arranged in a righthanded helical (spiral) structure and are stabilised by the hydrogen bond. The side chains of the residues protrude outside of the helix. Each turn of an α – helix contains about 3.6 residues and is about 5.4 A long.

β – Strands:
β – Strands are extended peptide chain rather than coiled. The hydrogen bond occur between main chain carbonyl group one such strand and the amino group of the adjacent strand resulting in the formation of a sheet like structure. This arrangement is called β – sheets.

Tertiary structure:
The secondary structure elements (α – helix & β – sheets) further folds to form a three dimensional arrangement. This tertiary structure of proteins are stabilised by the interactions between the side chains of the amino acids. These interactions include the disulphide bridges between cysteine residues, electrostatic, hydrophobic, hydrogen bonds and van der Waals interactions.

Quaternary Structure:
The oxygen transporting protein, haemoglobin contains four polypeptide chains while DNA polymerase enzyme that make copies of DNA, has ten polypeptide chains. In these proteins the individual polypeptide chains interacts with each other to form the multimeric structure which known as quaternary structure. The interactions that stabilises the tertiary structures also stabilises the quaternary structures.

Question 6.
What are the biological importance of proteins?
Answer:
Proteins are the functional units of living things play vital role in all biological processes

  1. All biochemical reactions occur in the living systems are catalysed by the catalytic proteins called enzymes.
  2. Proteins such as keratin, collagen acts as structural back bones.
  3. Proteins are used for transporting molecules (Haemogiobin), organelles (Kinesins) in the cell and control the movement of molecules in and out of the cells (Transporters).
  4. Antibodies help the body to fight various diseases.
  5. Proteins are used as messengers to coordinate many functions. Insulin & glucagon controls the glucose level in the blood.
  6. Proteins act as receptors that detect presence of certain signal molecules and activate the proper response.
  7. Proteins are also used to store metals such as iron (Femtin) etc.

Question 7.
Write the chemical name, source and deficient disease of the following

  1. Vitamin A
  2. Vitamin B1
  3. Vitamin B2
  4. Vitamin B3
  5. Vitamin B5

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-75

Question 8.
Write the chemical name, source and deficient disease of the following

  1. Vitamin B6
  2. Vitamin B7
  3. Vitamin B9
  4. Vitamin B12
  5. Vitamin C

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-76

Question 9.
Explain about the composition and structure of nucleic acids.
Answer:
1. Nucleic acids are biopolymers of nucleotides. Controlled hydrolysis of DNA and RNA yields three components namely a nitrogeneous base, a pentose sugar and phosphate group.

2. Nitrogen base.
(a) These are nitrogen containing organic compounds which are derivatives of two parent compounds, pyrimidine and purine.
(b) Both DNA and RNA have two major purine bases, adenine (A) and guanine (G). In both DNA and RNA, one of the pyrimidines is cytosinc (C), but the second pyrimidine is thymine (T) ¡n DNA and uracil (U) in RNA.

3. Pentose Sugar.
Nucleic acids have two types ofpentoses. The recurring deoxyribonucleotide units of DNA contain 2’ – dcoxy – D – ribose and the ribonucleotide units of RNA contain D – ribose. In nucleotides, both types of pentoses are in their – furanose form.

4. Phosphate group.
Phosphoric acid forms phosphor diester bond between nucleotides. Based on the number of phosphate group present in the nucleotides, they are classified mono nucleotide, dinucleotide and trinucleotide.

5. The molecule without the phosphate group is called a nucleoside. A nucleotide is derived from a nucleoside by the addition of a molecule of phosphoric acid.

6. Sugar + Base → Nucleoside
Nucleoside + Phosphate → Nucleotidc
Nucleotide → Polynucleotide (Nucleic Acid)

Question 10.
Describe about the double strand helix structure of DNA.
Answer:
1. Watson and Crick postulated a 3 – dimensional model of RNA structure which consisted of two antiparallel helical DNA chains wound around the same axis to form a right handed double helix.

2. Th e hydrophilic backbones of alternating deoxyribose and phosphate groups are on the outside of the double helix, facing the surrounding water. Th e purine and pyrimidine bases of both strands are stacked inside the double helix,

with their hydrophobic and ring structures very close together and perpendicular to the long axis, thereby reducing the repulsions between the charged phosphate groups. The offset pairing of the two strands creates a major groove and minor groove on the surface of the duplex.

3. The model revealed that there are 10.5 pairs (36A°) per turn of the helix and 3.4A° between the stacked bases. They also found that each base is hydrogen bonded to a base in opposite strand to form a planar base pair.

4. Two hydrogen bonds are formed between adenine and thymine and three hydrogen bonds are formed between guanine and cytosine. Other pairing tends to destablize the double helical structure. This specific association of the two chains of the double helix is known as complementary base pairing.

5. The DNA double or duplex is held together by two forces.

  • Hydrogen bonding between complementary base pairs.
  • Base – stacking interactions.

The complementary between the DNA strands is attributable to the hydrogen bonding between base pairs but the base stacking interactions are largely non-specific, make the major contribution to the stability of the double helix.
Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-77

Question 11.
Explain about the types of RNA molecules.
Answer:
1. Ribonucleic acids are similar to DNA. Cells contain upto eight times high quantity of RNA than DNA. RNA is found in large amount in the cytoplasm and a lesser amount in the nucleus.

2. RNA molecules are classified according to their structure and function into three major types.

  1. Ribosomal RNA (r – RNA)
  2. Messenger RNA (m – RNA)
  3. Transfer RNA (t – RNA)

3. r – RNA:
r – RNA is mainly found in cytoplasm and in ribosomes, which contain 60% RNA and 40% protein. Ribosomes are the sites at which protein synthesis takes place.

4. t – RNA:
t – RNA molecules have lowest molecular weight of all nucleic acids. They consist of 73 – 94 nucleotides in a single chain. The function of t – RNA is to carry amino acids to the sites of protein synthesis on ribosomes.

5. m – RNA:
m – RNA is present in small quantity and very short lived. They are single stranded and their synthesis take place on DNA. The synthesis m-RNA from DNA strand is called transcription. m – RNA carries genetic information from DNA to the ribosomes for protein synthesis.

Question 12.
Explain about DNA finger printing process.
Answer:
1. DNA finger printing is one of the most accurate methods for placing an individual at the scene of a crime has been a finger print.

2. The DNA finger print is unique for every person and can be extracted from traces of sample from blood, saliva, hair etc. By using this method, we can detect the individual specific variation in human DNA.

3. In this method, the extracted DNA is cut at specific points of varying lengths in the formation of DNA fragments of varying lengths which were analysed by technique called gel electrophoresis. This method separates the fragments based on their size. The gel containing the DNA fragments are then transferred to a nylon sheet using a technique called blotting. Then the fragments will undergo autoradiography in which they were exposed to DNA probes.

4. A piece of X-ray film was then exposed to the fragments, and a dark mark was produced at any point where a radioactive probe had become attached. The resultant pattern of marks could then be compared with other samples.

5. DNA finger printing is based on slight sequence differences between individuals. These methods are providing decisive in court cases world wide.

Common Errors

  1. Carbohydrates & Enzyme names are different.
  2. Peptide bond and amide bond are different.
  3. Vitamins: Solubility based classification
  4. Main source of vitamin

Rectifications

1. Carbohydrates name should end as – ose.
Eg. Diatose, maltose
Enzyme name should end as – ase.
Eg. Diastase, maltase

2. Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules-78

3. Vitamin A, D, E, K – fat soluble. Vitamin C, B complex are water soluble.

4. Milk:
Vitamin A, B1, B2, B6, B7, D (6 Vitamins)
Green Vegetables:
B1, B3, B7, K (4 vitamins)
Liver:
B1, B2, B3, B7

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Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics

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Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Textual Evaluation Solved

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Multiple Choice Questions

Question 1.
Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is
(a) 14.4\(\frac { Z }{ V }\) Å
(b) 14.4\(\frac { V }{ Z }\) Å
(c) 1.44\(\frac { Z }{ V }\) Å
(d) 14.4\(\frac { V }{ Z }\) Å
Answer:
(c) 1.44\(\frac { Z }{ V }\) Å

Question 2.
In a hydrogen atom, the electron revolving in the fourth orbit, has angular momentum equal to
(a) h
(b) \(\frac { h }{ π }\)
(c) \(\frac { 4h }{ π }\)
(d) \(\frac { 2h }{ π }\)
Answer:
(d) \(\frac { 2h }{ π }\)
Hint:
Angular momentum of an electron is an integral multiple of \(\frac { h }{ 2π }\)
According to Bohr atom model,
Angular momentum of an electron mvr = \(\frac { nh }{ 2π }\)
n = 4th orbit = \(\frac { 4h }{ 2π }\)
mvr = \(\frac { 2h }{ π }\)

Question 3.
Atomic number of H-like atom with ionization potential 122.4 V for n = 1 is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3
Hint:
The ionisation energy of a hydrogen atom is, IE = \(\frac {{ 13.6z }^{2}}{{ n }^{2}}\)
∴ z2 = \(\frac{I E \times n^{2}}{13.6}\) = \(\frac{122.4 \times(1)^{2}}{13.6}\) = 9

Question 4.
The ratio between the first three orbits of hydrogen atom is
(a) 1 : 2 : 3
(b) 2 : 4 : 6
(c) 1 : 4 : 9
(d) 1 : 3 : 5
Answer:
(c) 1 : 4 : 9
Hint:
En = \(\frac {{ -13.6×z }^{2}}{{ n }^{2}}\) eV / atom
n = 1; E1 = 13.6 eV / atom
n = 2; E2 = 3.4 eV / atom
n = 3; E3 = 151 eV / atom
The ratio of theree orbits
E1 : E2 : E3 = 13.6 : 3.4 : 1.51
= 1 : 4 : 9

Question 5.
The charge of cathode rays is
(a) positive
(b) negative
(c) neutral
(d) not defined
Answer:
(b) negative

Question 6.
In J.J. Thomson e/m experiment, a beam of electron is replaced by that of muons (particle with same charge as that of electrons but mass 208 times that of electrons). No deflection condition is achieved only if
(a) B is increased by 208 times
(b) B is decreased by 208 times
(c) B is increased by 14.4 times
(d) B is decreased by 14.4 times
Answer:
(c) B is increased by 14.4 times
Hint:
In the condition of no deflection \(\frac { e }{ m }\) = \(\frac {{ E }^{2}}{{ 2vB }^{2}}\)
If m is increased by 208 times then B should be increased \(\sqrt { 208 } \) = 14.4 times

Question 7.
The ratio of the wavelengths for the transition from n =2 to n = 1 in Li++, He+ and H is
(a) 1 : 2 : 3
(b) 1 : 4 : 9
(c) 3 : 2 : 1
(d) 4 : 9 : 36
Answer:
(d) 4 : 9 : 36
Hint:
According to Rydberg formula, the wavelength
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-1

Question 8.
The electric potential between a proton and an electron is given by V = V0 In \(\left( \frac { r }{ { r }_{ 0 } } \right) \), where r0 is a constant. Assume that Bohr atom model is applicable to potential, then variation of radius of nth orbit rn with the principal quantum number n is
(a) rn ∝\(\frac { 1 }{ n }\)
(b) rn ∝ n
(c) rn ∝ \(\frac { 1 }{{ n }^{2}}\)
(d) rn ∝ n2
Answer:
(b) rn ∝ n
Hint:
Electric potential between proton and electron in nth orbit is given as,
V = V0 In \(\left( \frac { { r }_{ n } }{ { r }_{ 0 } } \right) \)
Thus the coulomb force |Fc| = e \(\left( \frac { { dv } }{ dr } \right) \) = e \(\left( \frac { { V }_{ 0 } }{ { r }_{ n } } \right) \)
This coulomb force is balance by the centripetal force
\(\frac {{ mv }^{2}}{{r}_{n}}\) = e \(\left( \frac { { V }_{ 0 } }{ { r }_{ n } } \right) \left( \frac { { dv } }{ dr } \right) \) ⇒ V = \(\sqrt { \frac { e{ V }_{ 0 } }{ m } } \)
Now from
mvrn = \(\frac { nh }{2π}\)
rn ∝ n

Question 9.
If the nuclear radius of 27 Al is 3.6 fermi, the approximate unclear radius of64 Cu is
(a) 2.4
(b) 1.2
(c) 4.8
(d) 3.6
Answer:
(c) 4.8
Hint:
\(\frac {{ R }_{Al}}{{ R }_{Cu}}\) = \(\frac{(27)^{1 / 3}}{(64)^{1 / 3}}\) = \(\frac { 3 }{ 4}\)
Rcu = \(\frac { 4 }{ 3}\) RAl = \(\frac { 4 }{ 3}\) x 3.6 fermi
Rcu = 4.8 fermi

Question 10.
The nucleus is approximately spherical in shape. Then the surface area of nucleus having mass number A varies as
(a) A2/3
(b) A4/3
(c) A1/3
(d) A5/3
Answer:
(a) A2/3
Hint:
Volume of nucleus is proportional to mass number 4
\(\frac { 4 }{ 3 }\) πR3 ∝ A = R0 A1/3
So, πR2 = RR0 A2/3 ⇒ 4πR2 ∝ A2/3
Surface area is proportional to (mass number)2/3

Question 11.
The mass of a \(_{ 3 }^{ 7 }{ Li }\) nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of \(_{ 3 }^{ 7 }{ Li }\) nucleus is nearly
(a) 46 MeV
(b) 5.6 MeV
(c) 3.9 MeV
(d) 23 MeV
Answer:
(b) 5.6 MeV
Hint:
If w = 1 u, C = 3 x 108 ms-1 then, E = 931 MeV
1 u = 931 Mev
Binding energy = 0. 042 x 931
= 39. 10 MeV
∴ B.E 39.10
Binding energy per nucleon = \(\frac { B.E }{ A }\) = \(\frac { 39.10 }{ 7 }\) = 5.58 = 5.6 MeV

Question 12.
denotes the mass of the proton and Mn denotes mass of a neutron. A given nucleus of binding energy B, contains Z protons and N neutrons. The mass M (N, Z) of the nucleus is given by(where c is the speed of light)
(a) M (N, Z) = NMn + ZMp – Bc2
(b) M (N, Z) = NMn + ZMp + Bc2
(c) M (N, Z) = NMn + ZMp – B / c2
(d) M (N, Z) = NMn + ZMp + B / c2
Answer:
(c) M (N, Z) = NMn + ZMp – B / c2
Hint:
Binding energy, B = [ZMp + NMn – M (N, Z)] C2
M(N,Z) = ZMp + NMn – \(\frac { B }{{ C }^{ 2 }}\)

Question 13.
A radioactive nucleus (initial mass number A and atomic number Z) emits 2α and 2 positrons. The ratio of number of neutrons to that of proton in the final nucleus will be
(a) \(\frac { A-Z-4 }{ Z-2 }\)
(b) \(\frac { A-Z-2 }{ Z 6 }\)
(c) \(\frac { A-Z-4 }{ Z-6 }\)
(d) \(\frac { A-Z-12 }{ Z-4 }\)
Answer:
(b) \(\frac { A-Z-2 }{ Z 6 }\)

Question 14.
The half-life period of a radioactive element A is same as the mean life time of another radioactive element B. Initially both have the same number of atoms. Then
(a) A and B have the same decay rate initially
(b) A and B decay at the same rate always
(c) B will decay at faster rate than A
(d) A will decay at faster rate than B.
Answer:
(c) B will decay at faster rate than A
Hint:
(t1/2)A = (tmean )B
\(\frac { 0.6931 }{{ λ }_{A}}\) = \(\frac { 1 }{{ λ }_{B}}\)
λA = 0.6931 λB
λA < λB

Question 15
A system consists of N0 nucleus at t = 0. The number of nuclei remaining after half of a half-life (that is, at time t =\(\frac { 1 }{ 2 }\) T\(\frac { 1 }{ 2 }\))
(a) \(\frac {{ N }_{0}}{ 2 }\)
(b) \(\frac {{ N }_{0}}{ √2 }\)
(c) \(\frac {{ N }_{0}}{ 4 }\)
(d) \(\frac {{ N }_{0}}{ 8 }\)
Answer:
(b) \(\frac {{ N }_{0}}{ √2 }\)
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-2

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Short Answer Questions

Question 1.
What are cathode rays?
Answer:
A cathode ray is a stream of electrons that are seen in vaccum tubes. It is called a “cathode ray” because the electrons are being emitted from the negative charged element in the vaccum tube called the cathode.

Question 2.
Write the properties of cathode rays.
Answer:

  • Cathode rays possess energy and momentum and travel in a straight line with high speed of the order of 107 m s-1
  • It can be deflected by application of electric and magnetic fields.
  • When the cathode rays are allowed” to fall on matter, they produce heat.
  • They affect the photographic plates and also produce fluorescence when they fall on certain crystals and minerals.
  • When the cathode rays fall on a material of high atomic weight, x-rays are produced.
  • Cathode rays ionize the gas through which they pass.
  • The speed of cathode rays is up to( \(\frac { 1 }{ 10 }\))th

Question 3.
Give the results of Rutherford alpha scattering experiment.
Answer:

  • Most of the alpha particles are undeflected through the gold foil and went straight.
  • Some of the alpha particles are deflected through a small angle.
  • A few alpha particles (one in thousand) are deflected through the angle more than 90°.
  •  Very few alpha particles returned back (back scattered) -that is, deflected back by 180°.

Question 4.
Write down the postulates of Bohr atom model.
Answer:
Postulates of Bohr atom model:

  1. The electron in an atom moves around nucleus in circular orbits under the influence of Coulomb electrostatic force of attraction. This Coulomb force gives necessary centripetal force for the electron to undergo circular motion.
  2. Electrons in an atom revolve around the nucleus only in certain discrete orbits called stationary orbits where it does not radiate electromagnetic energy. Only those discrete orbits allowed are stable orbits.

Question 5.
What is meant by excitation energy?
Answer:
The energy required to excite an electron from lower energy state to any higher energy state is known as excitation energy.

Question 6.
Define the ionization energy and ionization potential.
Answer:
The ionization energy and ionization potential are:

  1. Ionization energy: The minimum energy required to remove an electron from an atom in the ground state is known as binding energy or ionization energy.
  2. Ionization potential: Ionization potential is defined as ionization energy per unit charge.

Question 7.
Write down the draw backs of Bohr atom model.
Answer:
Limitations of Bohr atom model:
The following are the drawbacks of Bohr atom model:

  1. Bohr atom model is valid only for hydrogen atom or hydrogen like-atoms but not for complex atoms.
  2. When the spectral lines are closely examined, individual lines of hydrogen spectrum is accompanied by a number of faint lines. These are often called fine structure. This is not explained by Bohr atom model.
  3. Bohr atom model fails to explain the intensity variations in the spectral lines.
  4. The distribution of electrons in atoms is not completely explained by Bohr atom model.

Question 8.
What is distance of closest approach?
Answer:
The minimum distance between the centre of the nucleus and the alpha particle just before it gets reflected back through 180° is defined as the distance of closest approach r0 (also known as contact distance).

Question 9.
Define impact parameter.
Answer:
The impact parameter is defined as the perpendicular distance between the centre of the gold nucleus and the direction of velocity vector of alpha particle when it is at a large distance.

Question 10.
Write a general notation of nucleus of element X. What each term denotes?
Answer:
The nucleus of any element, we use the following general notation \(_{ Z }^{ A }X\)
where X is the chemical symbol of the element, A is the mass number and Z is the atomic number.

Question 11.
What is isotope? Give an example.
Answer:
Isotopes are atoms of the same element having same atomic number Z, but different mass number A. For example, hydrogen has three isotopes and they are represented as \(_{ 1 }^{ 1 }H\) (hydrogen), \(_{ 1 }^{ 2 }H\) (deuterium),and \(_{ 1 }^{ 3 }H\) (tritium).

Question 12.
What is isotone? Give an example.
Answer:
Isotones are the atoms of different elements having same number of neutrons. \(_{ 5 }^{ 12 }B\) and \(_{ 6 }^{ 13 }B\) are examples of isotones which 7 neutrons.

Question 13.
What is isobar? Give an example.
Answer:
1. Isobar: Isobars are the atoms of different elements having the same mass number A, but different atomic number Z.
2. For example \(_{ 16 }^{ 40 }S\), \(_{ 17 }^{ 40 }Cl\), \(_{ 18 }^{ 40 }Ar\),\(_{ 19 }^{ 40 }K \) and \(_{ 20 }^{ 40 }Ca\) are isobars having same mass number 40 and different atomic number.

Question 14.
Define atomic mass unit u.
Answer:
One atomic mass unit (u) is defined as the 1/12th of the mass of the isotope of carbon \(_{ 6 }^{ 12 }C\).

Question 15.
Show that nuclear density is almost constant for nuclei with Z > 10.
Answer:
Nuclear density,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-3
The expression shows that the nuclear density is independent of the mass number A. In other words, all the nuclei (Z > 10) have the same density and it is an important characteristics of the nuclei.

Question 16.
What is mass defect?
Answer:
The mass of any nucleus is always less than the sum of the mass of its individual constituents. The difference in mass Am is called mass defect.
∆m = (Zmp + Nmn) – M.

Question 17.
What is binding energy of a nucleus? Give its expression.
Answer:
when Z protons and N neutrons combine to form a nucleus, mass equal to mass defect disappears and the corresponding energy is released. This is called the binding energy of the nucleus (BE) and is equal to (∆m) c2.
BE = (Zmp + Nmn – M ) c2

Question 18.
Calculate the energy equivalent of 1 atomic mass unit.
Answer:
We take, m = 1 amu = 1.66 x 10-27 kg
c = 3 x 108ms-1
Then, E = mc2 = 1.66 x 10-27 x (3 x 108)2 J
\(\frac{1.66 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{2}}{1.6 \times 10^{-19}} e \mathrm{V}\)
E ≈ 981 MeV
∴ 1 amu = 931 MeV.

Question 19.
Give the physical meaning of binding energy per nucleon.
Answer:
The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus.

Question 20.
What is meant by radioactivity?
Answer:
The phenomenon of spontaneous emission of highly penetrating radiations such as α, β and γ rays by an element is called radioactivity.

Question 21.
Give the symbolic representation of alpha decay, beta decay and gamma decay.
Answer:
1. Alpha decay:
The alpha decay process symbolically in the following way
\(_{ Z }^{ A }X\) → \(_{ Z-2 }^{ A-4 }Y\) + \(_{ 2 }^{ 4 }He\)

2. Beta decay:
β decay is represented by \(_{ Z }^{ A }X\) → \(_{ Z-1 }^{ A }Y\) +e+ + v

3. Gamma decay:
The gamma decay is given by \(_{ Z }^{ A }{{ X }^{ * }}\) → \(_{ Z }^{ A }X\) + gamma (γ ) rays

Question 22.
In alpha decay, why the unstable nucleus emits \(_{ 2 }^{ 4 }He\) He nucleus? Why it does not emit four separate nucleons?
Answer:
After all \(_{ 2 }^{ 4 }He\) He consists of two protons and two neutrons. For example, if \(_{ 92 }^{ 238 }U\) nucleus decays into \(_{ 90 }^{ 234 }U\) Th by emitting four separate nucleons (two protons and two neutrons), then the disintegration energy Q for this process turns out to be negative. It implies that the total mass of products is greater than that of parent (\(_{ 92 }^{ 238 }U\)) nucleus. This kind of process cannot occur in nature because it would violate conservation of energy. In any decay process, the conservation of energy, conservation of linear momentum and conservation of angular momentum must be obeyed.

Question 23.
What is mean life of nucleus? Give the expression.
Answer:
The mean life time of the nucleus is the ratio of sum or integration of life times of all nuclei to the total number nuclei present initially.
The expression for mean life time, τ = \(\frac { 1 }{ λ }\).

Question 24.
What is half-life of nucleus? Give the expression.
Answer:
The half-life T1/2 is defined as the time required for the number of atoms initially present to reduce to one half of the initial amount.
T1/2 = \(\frac { In 2 }{ λ }\) = \(\frac { 0.6931 }{ λ }\).

Question 25.
What is meant by activity or decay rate? Give its unit.
Answer:
The activity (R) or decay rate is defined as the number of nuclei decayed per second and it is denoted as R = \(\left| \frac { dN }{ dt } \right| \)
The SI unit of activity R is Becquerel.

Question 26.
Define curie.
Answer:
One curie was defined as number of decays per second in 1 g of radium and it is equal to 3.7 x 1010 decays/s.

Question 27.
What are the constituent particles of neutron and proton?
Answer:
Protons and neutrons are Baryon which are made up of three Quarks. According to quark model, proton is made up of two up quarks and one down quark and neutron is made up of one up quark and two down quarks.

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Long Answer Questions

Question 1.
Explain the J.J. Thomson experiment to determine the specific charge of electron.
Answer:
In 1887, J. J. Thomson made remarkable improvement in the scope of study of gases in discharge tubes. In the presence of electric and magnetic fields, the cathode rays are deflected. By the variation of electric and magnetic fields, mass normalized charge or the specific charge (charge per unit mass) of the cathode rays is measured.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-4
A highly evacuated discharge tube is used and cathode rays (electron beam) produced at cathode are attracted towards anode disc A. Anode disc is made with pin hole in order to allow only a narrow beam of cathode rays. These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage.

Further, this gas discharge tube is kept in between pole pieces of magnet such that both electric and magnetic fields are perpendicular to each other. When the cathode rays strike the screen, they produce scintillation and hence bright spot is observed. This is achieved by coating the screen with zinc sulphide.

(i) Determination of velocity of cathode rays:
For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-5
This means that the magnitude of electric force is balanced by the magnitude of force due to magnetic field. Let e be the charge of the cathode rays, then
eE = eBv
⇒ v = \(\frac { E }{ B }\) ….. (1)

(ii) Determination of specific charge:
Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode. Let V be the potential difference between anode and cathode, then the potential energy is eV. Then from law of conservation of energy,
eV = \(\frac { 1 }{ 2 }\) mv2 ⇒ \(\frac { e }{ m }\) = \(\frac {{ v }^{ 2 }}{ 2V }\)
Substituting the value of velocity from equation (1), we get
\(\frac { e }{ m }\) = \(\frac { 1 }{ 2V }\) = \(\frac {{ E }^{ 2 }}{{ B }^{ 2 }}\) …… (2)
Substituting the values of E, B and V, the specific charge can be determined as
\(\frac { e }{ m }\) = 1.7 x 1011 C kg-1

(iii) Deflection of charge only due to uniform electric field:
When the magnetic field is turned off, the deflection is only due to electric field. The deflection in vertical direction is due to the electric force.
Fe = eE ….. (3)
Let m be the mass of the electron and by applying Newton’s second law of motion, acceleration of the electron is
ae = \(\frac { 1 }{ m }\) Fe …. (4)
Substituting equation (4) in equation (3),
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-6
ae = \(\frac { 1 }{ m }\) eE = \(\frac { e }{ m }\) E
Lety be the deviation produced from original position on the screen. Let the initial upward velocity of cathode ray be u = 0 before entering the parallel electric plates. Let t be the time taken by the cathode rays to travel in electric field. Let t be the length of one of the plates, then the time taken is
t = \(\frac { 1 }{ v }\) ….. (5)
Hence, the deflection y’ of cathode rays is (note : u = 0 and ae = \(\frac { e }{ m }\) E)
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-7
Therefore, the deflection y on the screen is
y ∝ y’ ⇒ y = Cy’
where C is proportionality constant which depends on the geometry of the discharge tube and substituting y’ value in equation (6), we get
y = C\(\frac { 1 }{ 2 }\) \(\frac { e }{ m }\) \(\frac{l^{2} B^{2}}{E}\) …… (7)
Rerranging equation (7) as
\(\frac { e }{ m }\) = \(\frac{2 y E}{C l^{2} B^{2}}\) ……. (8)
Substituting the values on RHS, the value of specific charge is calculated as
\(\frac { e }{ m }\) = 1.7 x 1011 Ckg-1

(iv) Deflection of charge only due to uniform magnetic field:
Suppose that the electric field is switched off and only the magnetic field is switched on. Now the deflection occurs only due to magnetic field. The force experienced by the electron in uniform magnetic field applied perpendicular to its path is
Fm = evB (in magnitude)
Since this force provides the centripetal force, the electron beam undergoes a semicircular path. Therefore, we can equate Fm to centripetal force
\(\frac {{ mv }^{2}}{ R }\)
Fm = evB = m \(\frac {{ v }^{2}}{ R }\)
where v is the velocity of electron beam at the point where it enters the magnetic field and R is the radius of the circular path traversed by the electron beam.
eB = m \(\frac { v }{ R }\) ⇒ \(\frac { e }{ m }\) = \(\frac { v }{ BR }\) …… (9)
Further, substituting equation (1) in equation (9), we get
\(\frac { e }{ m }\) = \(\frac{E}{B^{2} R}\) ……. (10)
By knowing the values of electric field, magnetic field and the radius of circular path, the value of specific charge\(\left( \frac { e }{ m } \right) \) can be calculated.

Question 2.
Discuss the Millikan’s oil drop experiment to determine the charge of an electron.
Answer:
Millikan’s oil drop experiment is another important experiment in modem physics which is used to determine one of the fundamental constants of nature known as charge of an electron. By adjusting electric field suitably, the motion of oil drop inside the chamber can be controlled – that is, it can be made to move up or down or even kept balanced in the field of view for sufficiently long time.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-8
1. The apparatus consists of two horizontal circular metal plates A and B each with diameter around 20 cm and are separated by a small distance 1.5 cm. These two parallel plates are enclosed in a chamber with glass walls. Further, plates A and B are given a high potential difference around 10 kV such that electric field acts vertically downward.

2. A small hole is made at the centre of the upper plate A and atomizer is kept exactly above the hole to spray the liquid. When a fine droplet of highly viscous liquid (like glycerine) is sprayed using atomizer, it falls freely downward through the hole of the top plate only under the influence of gravity.

3. Few oil drops in the chamber can acquire electric charge (negative charge) because of friction with air or passage of x-rays in between the parallel plates. Further the chamber is illuminated by light which is passed horizontally and oil drops can be seen clearly using microscope placed perpendicular to the light beam. These drops can move either upwards or downward.

4. Let m be the mass of the oil drop and q be its charge. Then the forces acting on the droplet are
(a) gravitational force Fg = mg
(b) electric force Fe = qE
(c) buoyant force Fb

(a) Determination of radius of the droplet: When the electric field is switched off, the oil drop accelerates downwards. Due to the presence of air drag forces, the oil drops easily attain its terminal velocity and moves with constant velocity. This velocity can be carefully measured by nothing down the time taken by the oil drop to fall through a predetermined distance. The free body diagram of the oil drop), we note that viscous force and buoyant force balance the gravitational force.
Let the gravitational force acting on the oil drop (downward) be Fg = mg.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-9
Let us assume that oil drop to be spherical in shape. Let ρ be the density of the oil drop, and r be the radius of the oil drop, then the mass of the oil drop can be expressed in terms of its density as
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-10
The gravitational force can be written in terms of density as
Fg = mg ⇒ Fg = ρ \(\left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) \)g
Let σ be the density of the air, the upthrust force experienced by the oil drop due to displaced air is
Fb = σ \(\left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) \)g
Once the oil drop attains a terminal velocity υ, the net downward force acting on the oil drop is equal to the viscous force acting opposite to the direction of motion of the oil drop. From Stokes law, the viscous force on the oil drop is
Fr = 6πr vη
From the free body diagram as shown in Figure (a), the force balancing equation is Fg = Fb + Fv
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-11
Thus, equation (1) gives the radius of the oil drop.

(b) Determination of electric charge: When the electric field is switched on, charged oil drops experience an upward electric force (qE). Among many drops, one particular drop can be chosen in the field of view of microscope and strength of the electric field is adjusted to make that particular drop to be stationary. Under these circumstances, there will be no viscous force acting on the oil drop. Then, from the free body diagram, the net force acting on the oil droplet is
Fe = Fb + Fg
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-12
Substituting equation (1) in equation (2), we get
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-13
Millikan repeated this experiment several times and computed the charges on oil drops. He found that the charge of any oil drop can be written as integral multiple of a basic value, -1.6 x 10-19C, which is nothing but the charge of an electron.

Question 3.
Derive the energy expression for hydrogen atom using Bohr atom model.
Answer:
The energy of an electron in the nth orbit
Since the electrostatic force is a conservative force, the potential energy for the nth orbit is
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-14
The kinetic energy for the nth orbit is
KEn = \(\frac { 1 }{ 2 }\) \({ mv }_{ n }^{ 2 }\) \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\) \(\frac{Z^{2}}{n^{2}}\)This implies that Un = -2 KEn. Total energy in the nth orbit is
En = kEn + Un = KEn – 2KEn = – KEn
En = \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\) \(\frac{Z^{2}}{n^{2}}\)
For hydrogen (Z = 1),
En = \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\) \(\frac { 1 }{{ n }^{ 2 }}\) joule ….. (1)
where n stands for principal quantum number. The negative sign in equation (1) indicates that the electron is bound to the nucleus.

Substituting the values of mass and charge of an electron (m and e), permittivity’ of free space ε0and Planck’s constant h and expressing in terms of eV, we get
En = -13.6 \(\frac { 1 }{{ n }^{ 2 }}\) eV
For the first orbit (ground state), the total energy of electron is E1 = – 13.6 eV. For the second orbit (first excited state), the total energy of electron is E2 = -3.4 eV. For the third orbit (second excited state), the total energy of electron is E3 = -1.51 eV and so on.

Notice that the energy of the first excited state is greater than the ground state, second excited state is greater than the first excited state and so on. Thus, the orbit which is closest to the nucleus (r1) has lowest energy (minimum energy compared with other orbits). So, it is often called ground state energy (lowest energy state). The ground state energy of hydrogen (-13.6 eV ) is used as a unit of energy called Rydberg (1 Rydberg = -13.6 eV). The negative value of this energy is because of the way the zero of the potential energy is defined. When the electron is taken away to an infinite distance (very far distance) from nucleus, both the potential energy and kinetic energy terms vanish and hence the total energy also vanishes.

Question 4.
Discuss the spectral series of hydrogen atom.
Answer:
The spectral lines of hydrogen are grouped in separate series. In each series, the distance of separation between the consecutive wavelengths decreases from higher wavelength to the lower wavelength, and also wavelength in each series approach a limiting value known as the series limit. These series are named as Lyman series, Balmer series, Paschen series, Brackett series, Pfund series, etc. The wavelengths of these spectral lines perfectly agree with the equation derived from Bohr atom model.
\(\frac { 1 }{ λ }\) R \(\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \) … (1)
where \(\bar { v } \) is known as wave number which is inverse of wavelength, R is known as Rydberg constant whose value is 1.09737 x 107 m-1 and m and n are positive integers such that m > n. The various spectral series are discussed below:

(a) Lyman series:
Put n = 1 and m = 2, 3, 4 …..in equation (1). The wave number or wavelength of spectral lines of Lyman series which lies in ultra-violet region is
\(\bar { v } \) \(\frac { 1 }{ λ }\) R \(\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \)

(b) Balmer series:
Put n = 2 and m = 3, 4, 5 …. in equation (1). The wave number or wavelength of spectral lines of Balmer series which lies in visible region is
\(\bar { v } \) \(\frac { 1 }{ λ }\) R \(\left(\frac{2}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \)

(c) Paschen series:
Put n = 3 and m = 4, 5, 6…. in equation (1). The wave number or wavelength of spectral lines of Paschen series which lies in infra-red region (near IR) is
\(\bar { v } \) \(\frac { 1 }{ λ }\) R \(\left(\frac{3}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \)

(d) Brackett series:
Put n = 4 and m = 5, 6, 7 ….in equation (1). The wave number or wavelength of spectral lines of Brackett series which lies in infra-red region (middle IR) is
\(\bar { v } \) \(\frac { 1 }{ λ }\) R \(\left(\frac{4}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \)

(e) Pfund series:
Put n = 5 and m = 6, 7, 8 …. in equation (1). The wave number or wavelength of spectral lines of Pfund series which lies in infra-red region (far IR) is
\(\bar { v } \) \(\frac { 1 }{ λ }\) R \(\left(\frac{5}{n^{2}}-\frac{1}{m^{2}}\right)\) = \(\bar { v } \)

Question 5.
Explain the variation of average binding energy with the mass number by graph and discuss its features.
Answer:
We can find the average binding energy per nucleon \(\overline { BE } \). It is given by
\(\overline { BE } \) = \(\frac{\left[Z m_{H}+N m_{n}-M_{\mathrm{A}}\right] c^{2}}{\mathrm{A}}\)
The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus. \(\overline { BE } \) is plotted against A of all known nuclei.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-15
Important inferences from of the average binding energy curve:

(i) The value of \(\overline { BE } \) rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A = 56 (iron) and then it slowly decreases.

(ii) The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number between A = 40 and 120. These elements are comparatively more stable and not radioactive.

(iii) For higher mass numbers, the curve reduces slowly and BE for uranium is about 7.6 MeV. They are unstable and radioactive.
If two light nuclei with A < 28 combine with a nucleus with A < 56, the binding energy per nucleon is more for final nucleus than initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb.

(iv) If a nucleus of heavy element is split (fission) into two or more nuclei of medium value A, the energy released would again be large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled.

Question 6.
Explain in detail the nuclear force.
Answer:
Nucleus contains protons and neutrons. From electrostatics, we leamt that like charges repel each other. In the nucleus, the protons are separated by a distance of about a few Fermi (1 0-15 m), they must exert on each other a very strong repulsive force. For example, the electrostatic repulsive force between two protons separated by a distance 10-15 m
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-16
The acceleration experienced by a proton due to the force of 230 N is
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-17
This is nearly 1028 times greater than the acceleration due to gravity. So if the protons in the nucleus experience only the electrostatic force, then the nucleus would fly apart in an instant. From this observation, it was concluded that there must be a strong attractive force between protons to overcome the repulsive Coulomb’s force. This attractive force which holds the nucleus together is called strong nuclear force. A few properties of strong nuclear force are

(i) The strong nuclear force is of very short range, acting only up to a distance of a few Fermi. But inside the nucleus, the repulsive Coulomb force or attractive gravitational forces between two protons are much weaker than the strong nuclear force between two protons. Similarly, the gravitational force between two neutrons is also much weaker than strong nuclear force between the neutrons. So nuclear force is the strongest force in nature.

(ii) The strong nuclear force is attractive and acts with an equal strength between proton-proton, proton-neutron, and neutron – neutron.

(iii) Strong nuclear force does not act on the electrons. So it does not alter the chemical properties of the atom.

Question 7.
Discuss the alpha decay process with example.
Answer:
When unstable nuclei decay by emitting an α-particle (\(_{ 2 }^{ 4 }{ He }\) nucleus), it loses two protons and two neutrons. As a result, its atomic number Z decreases by 2, the mass number decreases by 4. We write the alpha decay process symbolically in the following way
\(_{ Z }^{ A }{ X }\) → \(_{ Z-2 }^{ A-4 }{ Y}\) +\(_{ 2 }^{ 4 }{ He }\)
Here X is called the parent nucleus and Y is called the daughter nucleus.

Example:
Decay of Uranium \(_{ 92 }^{ 238 }{ U }\) to thorium \(_{ 92 }^{ 234 }{ Th }\)with the emission of \(_{ 2 }^{ 4 }{ He }\) nucleus (α-particle)
\(_{ 92 }^{ 238 }{ U }\) → \(_{ 92 }^{ 234 }{ Th }\) + \(_{ 2 }^{ 4 }{ He }\)
As already mentioned, the total mass of the daughter nucleus and \(_{ 2 }^{ 4 }{ He }\) nucleus is always less than that of the parent nucleus. The difference in mass Q = (∆mx – my – mα) is released as energy called disintegration energy Q and is given by Q = (∆mx – my – mα) c2

Note that for spontaneous decay (natural radioactivity) Q > 0. In alpha decay process, the disintegration energy is certainly positive (Q > 0). In fact, the disintegration energy Q is also the net kinetic energy gained in the decay process or if the parent nucleus is at rest, Q is the total kinetic energy of daughter nucleus and the 2 He nucleus. Suppose Q < 0, then the decay process cannot occur spontaneously and energy must be supplied to induce the decay.

Question 8.
Discuss the beta decay process with examples.
Answer:
In beta decay, a radioactive nucleus emits either electron or positron. If electron (e) is emitted, it is called β decay and if positron (e+) is emitted, it is called p+ decay. The positron is an anti-particle of an electron whose mass is same as that of electron and charge is opposite to that of electron – that is, +e. Both positron and electron are referred to as beta particles.

1. β decay:
In β decay, the atomic number of the nucleus increases by one but mass number remains the same. This decay is represented by
\(_{ Z }^{ A }{ X }\) → \(_{ Z+12 }^{ A }{ Y}\) + e + \(\bar { v } \) …(1)
It implies that the element X becomes Y by giving out an electron and antineutrino (\(\bar { v } \)). In otherwords, in each β decay, one neutron in the nucleus of X is converted into a proton by emitting an electron (e) and antineutrino. It is given by
n → p + e + \(\bar { v } \)
Where p -proton, \(\bar { v } \) -antineutrino. Example: Carbon (\(_{ 6 }^{ 14 }{ C }\)) is converted into nitrogen (\(_{ 7 }^{ 14 }{ N }\)) through β- decay.
\(_{ 6 }^{ 14 }{ C }\) → \(_{ 7 }^{ 14 }{ N }\) + e + \(\bar { v } \)

2. β+ decay:
In p+ decay, the atomic number is decreased by one and the mass number remains the same. This decay is represented by
\(_{ Z }^{ A }{ X }\) → \(_{ Z-12 }^{ A }{ Y}\) + e+ + v
It implies that the element X becomes Y by giving out an positron and neutrino (v). In otherwords, for each β+ decay, a proton in the nucleus of X is converted into a neutron by emitting a positron (e+) and a neutrino. It is given by
p → n + e+ + v

However a single proton (not inside any nucleus) cannot have β+ decay due to energy conservation, because neutron mass is larger than proton mass. But a single neutron (not inside any nucleus) can have β decay.
Example: Sodium (\(_{ 11 }^{ 23 }{ Na }\)) is converted into neon (\(_{ 10 }^{ 22 }{ Ne }\)) decay.
\(_{ 11 }^{ 23 }{ Na }\) → \(_{ 10 }^{ 22 }{ Ne }\) + e+ + v

Question 9.
Discuss the gamma decay process with example.
Answer:
In a and p decay, the daughter nucleus is in the excited state most of the time. The typical life time of excited state is approximately 10-11 s. So this excited state nucleus immediately returns to the ground state or lower energy state by emitting highly energetic photons called 7 rays. In fact, when the atom is in the excited state, it returns to the ground state by emitting photons of energy in the order of few eV. But when the excited state nucleus returns to its ground state, it emits a highly energetic photon (γ rays) of energy in the order of MeV. The gamma decay is given by
\(_{ Z }^{ A }{ { X }^{ * } }\) → \(_{ Z }^{ A }{ X}\) + gamma (γ) rays
Here the asterisk (*) means excited state nucleus. In gamma decay, there is no change in the mass number or atomic number of the nucleus.
Boron (\(_{ 5 }^{ 12 }{ B }\)) has two beta decay modes:

(i) it undergoes beta decay directly into ground state carbon by emitting an electron of maximum of energy 13.4 MeV.

(ii) it undergoes beta decay to an excited state of carbon (\(_{ 6 }^{ 12 }{{ C}^{ * }}\)) by emitting an electron of maximum energy 9.0 MeV followed by gamma decay to ground state by emitting a photon of energy 4.4 MeV.
It is represented by
\(_{ 5 }^{ 12 }{ B }\) → \(_{ 6 }^{ 12 }{ C }\) + e+ + \(\bar { v } \)
\(_{ 6 }^{ 12 }{{ C }^{ * }}\) → \(_{ 6 }^{ 12 }{ C }\) + γ

Question 10.
Obtain the law of radioactivity.
Answer:
Law of radioactive decay:
At any instant t, the number of decays per unit time, called rate of decay \(\left( \frac { dN }{ dt } \right) \) is proportional to the number of nuclei (N) at the same instant.
\(\frac { dN }{ dt } \) ∝ N
By introducing a proportionality constant, the relation can be written as
\(\frac { dN }{ dt } \) = -λN …… (1)
Here proportionality constant λ is called decay constant which is different for different radioactive sample and the negative sign in the equation implies that the N is decreasing with time. By rewriting the equation (1), we get
dN = -λNdt …… (2)
Here dN represents the number of nuclei decaying in the time interval dt. Let us assume that at time t =0 s, the number of nuclei present in the radioactive sample is N0. By integrating the equation (2), we can calculate the number of undecayed nuclei N at any time t. From equation (2), we get
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-18
Taking exponentials on both sides, we get
N = N0 e-λt ….. (4)
[Note: eInx = ey ⇒ x = ey]
Equation (4) is called the law of radioactive decay. Here N denotes the number of undecayed nuclei present at any time t and N0 denotes the number of nuclei at initial time t = 0. Note that the number of atoms is decreasing exponentially over the time. This implies that the time taken for all the radioactive nuclei to decay will be infinite. Equation (4) is plotted.
We can also define another useful quantity called activity (R) or decay rate which is the number of nuclei decayed per second and it is denoted as R = \(\left| \frac { dN }{ dt } \right| \).
Note: that activity R is a positive quantity. From equation (4), we get
R = \(\left| \frac { dN }{ dt } \right| \) = λ N0 e-λt ….. (5)
R = R0 e-λt ….. (6)
Where R = λ N0
The equation (6) is also equivalent to radioactive law of decay. Here R0 is the activity of the sample at t = 0 and R is the activity of the sample at any time t. From equation (6), activity also shows exponential decay behavior. The activity R also can be expressed in terms of number of undecayed atoms present at any time t. From equation (6), since N = N0 e-λtwe write
R = λ N …… (7)
Equation (4) implies that the activity at any time t is equal to the product of decay constant and number of undecayed nuclei at the same time t. Since N decreases over time, R also decreases.

Question 11.
Discuss the properties of neutrino and its role in beta decay.
Answer:
Neutrino:
Initially, it was thought that during beta decay, a neutron in the parent nucleus is converted to the daughter nuclei by emitting only electron as given by
\(_{ Z }^{ A }{ X }\) → \(_{ Z+1 }^{ A }{ X}\) Y+e

1. But the kinetic energy of electron coming out of the nucleus did not match with the experimental results. In alpha decay, the alpha particle takes only certain allowed discrete energies whereas in beta decay, it was found that the beta particle (i.e, electron) have a continuous range of energies.

2. But the conservation of energy and momentum gives specific single values for electron energy and the recoiling nucleus Y. It seems that the conservation of energy, momentum are violated and could not be explained why energy of beta particle have continuous range of values. So beta decay remained as a puzzle for several years.

3. After a detailed theoretical and experimental study, in 1931 W. Pauli proposed a third particle which must be present in beta decay to carry away missing energy and momentum. Fermi later named this particle the neutrino (little neutral one) since it has no charge, have very little mass.

4. For many years, the neutrino (symbol v , Greek nu) was hypothetical and could not be verified experimentally. Finally, the neutrino was detected experimentally in 1956 by Fredrick Reines and Clyde Cowan. Later Reines received Nobel prize in physics in the year 1995 for his discovery.
The neutrino has the following properties

  • It has zero charge
  • It has an antiparticle called anti-neutrino.
  • Recent experiments showed that the neutrino has very tiny mass.
  • It interacts very weakly with the matter. Therefore, it is very difficult to detect. In fact, in every second, trillions of neutrinos coming from the sun are passing through our body without any interaction.

Question 12.
Explain the idea of carbon dating.
Answer:
Carbon dating:
1. The interesting application of beta decay is radioactive dating or carbon dating. Using this technique, the age of an ancient object can be calculated. All living organisms absorb carbon dioxide (CO2) from air to synthesize organic molecules. In this absorbed CO2, the major part is \(_{ 6 }^{ 12 }{ C }\) and very small fraction (1.3 x 10-12) is radioactive \(_{ 6 }^{ 14 }{ C }\) whose half-life is 5730 years.
Carbon-14 in the atmosphere is always decaying but at the same time, cosmic rays from outer space are continuously bombarding the atoms in the atmosphere which produces \(_{ 6 }^{ 14 }{ C }\). So the continuous production and decay of \(_{ 6 }^{ 14 }{ C }\) in the atmosphere keep the ratio of
\(_{ 6 }^{ 14 }{ C }\) to \(_{ 6 }^{ 12 }{ C }\) always constant.

2. Since our human body, tree or any living organism continuously absorb CO2 from the atmosphere, the ratio of \(_{ 6 }^{ 14 }{ C }\) to \(_{ 6 }^{ 12 }{ C }\) in the living organism is also nearly constant. But when the organism dies, it stops absorbing C2.

3. Since \(_{ 6 }^{ 14 }{ C }\) starts to decay, the ratio of \(_{ 6 }^{ 14 }{ C }\) to \(_{ 6 }^{ 12 }{ C }\) in a dead organism or specimen decreases over the years. Suppose the ratio of \(_{ 6 }^{ 14 }{ C }\) to \(_{ 6 }^{ 14 }{ C }\) in the ancient tree pieces excavated is known, then the age of the tree pieces can be calculated.

Question 13.
Discuss the process of nuclear fission and its properties.
Answer:
1. When uranium nucleus is bombarded with a neutron, it breaks up into two smaller nuclei of comparable masses with the release of energy.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-19

2. The process of breaking up of the nucleus of a heavier atom into two smaller nuclei with the release of a large amount of energy is called nuclear fission.

3. The fission is accompanied by the release of neutrons. The energy that is released in the nuclear fission is of many orders of magnitude greater than the energy released in chemical reactions.

4. Uranium undergoes fission reaction in 90 different Neutrons ways. The most common fission reactions of \(_{ 92 }^{ 235 }{ U }\) nuclei are shown here.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-20

5. Here Q is energy released during the decay of each uranium nuclei. When the slow neutron is absorbed by the uranium nuclei, the mass number increases by one and goes to an excited state \(_{ 92 }^{ 236 }{{ U}^{ * }}\). But this excited state does not last longer than 10-12s and decay into two daughter nuclei along with 2 or 3 neutrons. From each reaction, on an average, 2.5 neutrons are emitted.

Question 14.
Discuss the process of nuclear fusion and how energy is generated in stars.
Answer:
Nuclear Fusion:
1. When two or more light nuclei (A < 20) combine to form a heavier nucleus, then it is called nuclear fusion.

2. In the nuclear fusion, the mass of the resultant nucleus is less than the sum of the masses of original light nuclei. The mass difference appears as energy. The nuclear fusion never occurs at room temperature unlike nuclear fission. It is because when two light nuclei come closer to combine, it is strongly repelled by the coulomb repulsive force.

3. To overcome this repulsion, the two light nuclei must have enough kinetic energy to move closer to each other such that the nuclear force becomes effective. This can be achieved if the temperature is very much greater than the value 107 K. When the surrounding temperature reaches around 107 K, lighter nuclei start fusing to form heavier nuclei and this resulting reaction is called thermonuclear fusion reaction.

Energy generation in stars:
1. The natural place where nuclear fusion occurs is the core of the stars, since its temperature is of the order of 107 K. In fact, the energy generation in every star is only through thermonuclear fusion. Most of the stars including our Sun fuse hydrogen into helium and some stars even fuse helium into heavier elements.

2. The early stage of a star is in the form of cloud and dust. Due to their own gravitational pull, these clouds fall inward. As a result, its gravitational potential energy is converted to kinetic energy and finally into heat.

3. When the temperature is high enough to initiate the thermonuclear fusion, they start to release enormous energy which tends to stabilize the star and prevents it from further collapse.

4. The sun’s interior temperature is around 1.5 x 107 K. The sun is converting 6 x 1011 kg hydrogen into helium every second and it has enough hydrogen such that these fusion lasts for another 5 billion years.

5. When the hydrogen is burnt out, the sun will enter into new phase called red giant where helium will fuse to become carbon. During this stage, sun will expand greatly in size and all its planets will be engulfed in it.

6. According to Hans Bethe, the sun is powered by proton-proton cycle of fusion reaction. This cycle consists of three steps and the first two steps are as follows:
\(_{ 1 }^{ 1 }{ H }\) + \(_{ 1 }^{ 1 }{ H }\) → \(_{ 1 }^{ 2 }{ H }\) + e+ + v …… (1)
\(_{ 1 }^{ 1 }{ H }\) + \(_{ 1 }^{ 2 }{ H }\) → \(_{ 2 }^{ 3 }{ H }\) + γ …… (2)
A number of reactions are possible in the third step. But the dominant one is
\(_{ 2 }^{ 3 }{ H }\) + \(_{ 12}^{ 3 }{ H }\) → \(_{ 2 }^{ 4}{ H }\) + \(_{ 1 }^{ 1 }{ H }\) + \(_{ 1 }^{ 1 }{ H }\)…… (3)
The overall energy production in the above reactions is about 27 MeV. The radiation energy we received from the sun is due to these fusion reactions.

Question 15.
Describe the working of nuclear reactor with a block diagram.
Answer:
Nuclear reactor:
1. Nuclear reactor is a system in which the nuclear fission takes place in a self-sustained controlled manner and the energy produced is used either for research purpose or for power generation.

2. The main parts of a nuclear reactor are fuel, moderator and control rods. In addition to this, there is a cooling system which is connected with power generation set up.

Fuel:
1. The fuel is fissionable material, usually uranium or plutonium. Naturally occurring uranium contains only 0.7% of \(_{ 92 }^{ 235 }{ U }\) and 99.3% are only If \(_{ 92 }^{ 238 }{ U }\). So the \(_{ 92 }^{ 238 }{ U }\) must be enriched such that it contains at least 2 to 4% of \(_{ 92 }^{ 235 }{ U }\).

2. In addition to this, a neutron source is required to initiate the chain reaction for the first time. A mixture of beryllium with plutonium or polonium is used as the neutron source. During fission of \(_{ 92 }^{ 235 }{ U }\), only fast neutrons are emitted but the probability of initiating fission by it in another nucleus is very low. Therefore, slow neutrons are preferred for sustained nuclear reactions.

Moderators:
1. The moderator is a material used to convert fast neutrons into slow neutrons. Usually the moderators are chosen in such a way that it must be very light nucleus having mass comparable to that of neutrons. Hence, these light nuclei undergo collision with fast neutrons and the speed of the neutron is reduced

2. Most of the reactors use water, heavy water (D2O) and graphite as moderators. The blocks of uranium stacked together with blocks of graphite (the moderator) to form a large pile.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-21

Control rods:
1. The control rods are used to adjust the reaction rate. During each fission, on an average 2.5 neutrons are emitted and in order to have the controlled chain reactions, only one neutron is allowed to cause another fission and the remaining neutrons are absorbed by the control rods.

2. Usually cadmium or boron acts as control rod material and these rods are inserted into the uranium blocks. Depending on the insertion depth of control rod into the uranium, the average number of neutrons produced per fission is set to be equal to one or greater than one.

3. If the average number of neutrons produced per fission is equal to one, then reactor is said to be in critical state. In fact, all the nuclear reactors are maintained in critical state by suitable adjustment of control rods. If it is greater than one, then reactor is said to be in super-critical and it may explode sooner or may cause massive destruction.

Shielding:
1. For a protection against harmful radiations, the nuclear reactor is surrounded by a concrete wall of thickness of about 2 to 2.5 m.
Cooling system:

2. The cooling system removes the heat generated in the reactor core. Ordinary water, heavy water and liquid sodium are used as coolant since they have very high specific heat capacity and have large boiling point under high pressure.

3. This coolant passes through the fuel block and carries away the heat to the steam generator through heat exchanger. The steam runs the turbines which produces electricity in power reactors.

Question 16.
Explain in detail the four fundamental forces.
Answer:
Fundamental forces of nature:

  1. It is known that there exists gravitational force between two masses and it is universal in nature. Our planets are bound to the sun through gravitational force of the sun.
  2. The force between two charges there exists electromagnetic force and it plays major role in most of our day-today events.
  3. The force between two nucleons, there exists a strong nuclear force and this force is responsible for stability of the nucleus.
  4. In addition to these three forces, there exists another fundamental force of nature called the weak force. This weak force is even shorter in range than nuclear force. This force plays an important role in beta decay and energy production of stars.
  5. During the fusion of hydrogen into helium in sun, neutrinos and enormous radiations are produced through weak force.
  6. Gravitational, electromagnetic, strong and weak forces are called fundamental forces of nature.

Question 17.
Briefly explain the elementary particles of nature.
Answer:
Elementary particles:
1. An atom has a nucleus surrounded by electrons and nuclei is made up of protons and neutrons. Till 1960s, it was thought that protons, neutrons and electrons are fundamental building blocks of matter.

2. In 1964, physicist Murray Gellman and George Zweig theoretically proposed that protons and neutrons are not fundamental particles; in fact they are made up of quarks. These quarks are now considered elementary particles of nature. Electrons are fundamental or elementary particles because they are not made up of anything.
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-22
In the year 1968, the quarks were discovered experimentally by Stanford Linear Accelerator Center (SLAC), USA. There are six quarks namely, up, down, charm, strange, top and bottom and their antiparticles. All these quarks have fractional charges.
For example, charge of up quark is +\(\frac { 2 }{ 3 }\)e and that of down quark is –\(\frac { 1 }{ 3 }\)e.

3. According to quark model, proton is made up of two up quarks and one down quark and neutron is made up of one up quark and two down quarks.

4. The study of elementary particles is called particle physics.

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Exercises

Question 1.
Consider two hydrogen atoms HA and HB in ground state. Assume that hydrogen atom HA is at rest and hydrogen atom HB is moving with a speed and make head-on collide on the stationary hydrogen atom HA. After the strike, both of them move together. What is minimum value of the kinetic energy of the moving hydrogen atom HB, such that any one of the hydrogen atoms reaches one of the excitation state.
Solution:
Collision between hydrogen HA and hydrogen HB atom will be inelastic if a part of kinetic energy is used to excite atom.
If u1 and u2 are speed of HA and HB atom after collision, then
mu = mu1 + mu2 …… (1)
\(\frac { 1 }{ 2 }\) mu2 = \(\frac { 1 }{ 2 }\) \({ mu }_{ 1 }^{ 2 }\) + \(\frac { 1 }{ 2 }\) \({ mu }_{ 2 }^{ 2 }\) + ∆ E …… (2)
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-23
The minimum K.E of the moving hydrogen atom HB is 20.4 eV.

Question 2.
In the Bohr atom model, the frequency of transitions is given by the following expression υ = Rc \(\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\), Where n < m,
Consider the following transitions:
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-24
Show that the frequency of these transitions obey sum rule (which is known as Ritz combination principle)
Solution:
In the Bohr atom model, the frequency of transition
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-25
IIIrd transition, m = 3 and n = 1
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-26
According to Ritz combination principle, the frequency transition of single step is the sum of frequency transition in two steps
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-27

Question 3.
(a) A hydrogen atom is excited by radiation of wavelength 97.5 nm. Find the principal quantum number of the excited state.
(b) Show that the total number of lines in emission spectrum is \(\frac { n(n-1) }{ 2 }\) and compute the total number of possible lines in emission spectrum.
Solution:
(a) Wavelength, λ = 97.5 nm = 97.5 x 10-9 m
Principle quantum number n = ?
According to Bohr atom model,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-28

(b) A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level
So total number of lines in emission spectrum is \(\frac { n(n-1) }{ 2 }\)
= \(\frac { (4(4-1) }{ 2 }\) = \(\frac { 4×3) }{ 2 }\) = 6
So the total number of possible lines in emission spectrum is 6.

Question 4.
Calculate the radius of the earth if the density of the earth is equal to the density of the nucleus. [mass of earth 5.97 x 1024 kg].
Solution:
The density of the nucleus of an atom
ρN = 2.3 x 1017 kg m-3
ρN = ρE = 2.3 x 1017 kg m-3
Mass of the earth ME = 5.97 x 1024 kg
Density of the earth,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-29
r3 = 0.62155 x 107 m3
r3 = 183.85 m
r ≈ 180 m.

Question 5.
Calculate the mass defect and the binding energy per nucleon of the \(_{ 47 }^{ 108 }{ Ag }\) nucleus, [atomic mass of Ag = 107.905949]
Solution:
Mass of proton, mp = 1.007825 amu
Mass of neutron, mn = 1.008665 amu
Mass defect, ∆m = Zmp + Z mN – MN
= 47 x 1.007825 + 61 x 1.008665 – 107.905949
= 108.89634- 107.905949
∆m = 0.990391 u
Binding energy per nucleon of the \(_{ 47 }^{ 108 }{ Ag }\) nucleus
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-30

Question 6.
Half lives of two radioactive elements A and B are 20 minutes and 40 minutes respectively. Initially, the samples have equal number of nuclei. Calculate the ratio of decayed numbers of A and B nuclei after 80 minutes.
Solution:
80 minutes = 4 half lives of A = 2 half live of B
Let the initial number of nuclei in each sample be N.
NN after 80 minutes = \(\frac { N }{{ 2 }^{ 4 }}\)
Number of A nuclides decayed = \(\frac { 15 }{16}\)N
NB after 80 minutes = \(\frac { N }{{ 2 }^{ 4 }}\)
Number of B nuclides decayed = \(\frac { 3 }{4}\)N
Required ratio = \(\frac { 15 }{16}\) x \(\frac { 4 }{3}\) = \(\frac { 5 }{4}\)
NN : NB = 5 : 4.

Question 7.
On your birthday, you measure the activity of the sample 210Bi which has a half-life of 5.01 days. The initial activity that you measure is lμCi . (a) What is the approximate activity of the sample on your next birthday? Calculate (b) the decay constant (c) the mean life (d) initial number of atoms.
Solution:
(a) A year of 365 days is equivalent to 365 d/5.01 d ≈ 73 half-lives. Thus, the activity will be reduced after one year to approximately (1/2)73 (1.000 μCi) ~ 10-22 μCi.

(b) Initial measure R0 = 1.000 μCi
= 10-6 x 3.7 x 1010
= 3.7 x 104 Bq
After 1 year, the measure R = 10-22 μCi.
= 10-22 x 10-6 x 3.7 x 1010
= 3.7 x 10-18 Bq
decay constant,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-31

(c) Mean life
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-32

(d) Initial number of atoms
R0 = λN ; N = \(\frac {{ R }_{ 0 }}{ λ }\)
= \(\frac{3.7 \times 10^{4}}{1.6 \times 10^{-6}}\) ; N = 2.31 x 1010

Question 8.
Calculate the time required for 60% of a sample of radon undergo decay. Given T1/2 of radon = 3.8 days.
Solution:
Here consider Rn – 222 with a half life of 3.823 days.
From decay equation,
Current amount = Initial amount x (2)-n
N = N0 (2)-n
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-33

Question 9.
Assuming that energy released by the fission of a single \(_{ 92 }^{ 235 }{ U }\) nucleus is 200MeV, calculate the number of fissions per second required to produce 1 watt power.
Solution:
The fission of a single \(_{ 92 }^{ 235 }{ U }\) nucleus releases 200 MeV of energy
Energy released in the fission is given by the formula,
E = \(\frac { Pt }{ n }\) ⇒ \(\frac { n }{ t }\) = \(\frac { P }{ E }\)
E = 200 MeV = 200 x 106 x 1.6 x 10-19
E = 3.2 x 10-11 J
\(\frac { n }{ t }\) = \(\frac { P }{ E }\) = \(\frac{1}{3.2 \times 10^{-11}}\) = 0.3125 x 1011 = 3.125 x 1010
\(\frac { n }{ t }\) = 3.125 x 1010

Question 10.
Show that the mass of radium (\(_{ 88 }^{ 226 }{ Ra }\)) with an activity of 1 curie is almost a gram. Given T1/2 = 1600 years.
Solution:
The activity of the sample at any time t
R = λN
Here, λ = \(\frac{0.6931}{\mathrm{T}_{1 / 2}}\)
R = 1 Ci = 3.7 x 1010 dis s-1
T1/2 = 1600 year = 1600 x 3.16 x 107 dis
∴ The amount of radium,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-34
= 26990.62 x 1017
N = 2.7 x 1021 atoms
As 226 g of radium contains 6.023 x 1023 atoms so the amount of required strength.
= \(\frac{226 \times 2.7 \times 10^{21}}{6.023 \times 10^{23}}\)
= 101.311 x 10-2
= 1.013 g ≈ 1 g.

Question 11.
Characol pieces of tree is found from an archeological site. The carbon-14 content of this characol is only 17.5% that of equivalent sample of carbon from a living tree. What is the age of tree?
Solution:
R0 = 100%
R = 17.5%
λ = \(\frac{0.6931}{\mathrm{T}_{1 / 2}}\)
T1/2 = 5730 years
According to radioactive law
R = R0 e-λt
e-λt = \(\frac {{ R }_{ 0 }}{ R }\)
Talking log on both sides
t = \(\frac {1}{ λ }\) in \(\left( \frac { { R }_{ 0 } }{ R } \right) \)
Half life of carbon, T1/2 = 5730 years
t = \(\frac{\mathrm{T}_{1 / 2}}{0.6931}\) In \(\left(\frac{1}{0.175}\right)\)
= \(\frac { 5730 years }{ 0.6931 }\) x 1.74297
= 14409.49 years
t = 1.44 x 104 years.

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Additional Questions

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Multiple Choice Questions

Question 1.
The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 mA. Then the number of electrons striking the target per second is
(a) 2 x 1016
(b) 5 x 1018
(c) 1 x 1017
(d) 4 x 105
Answer:
(a) 2 x 1016
Hint:
n = \(\frac { It }{ e }\) = \(\frac{3.2 \times 10^{-3} \times 1}{1.6 \times 10^{-19}}\) = 2 x 1016.

Question 2.
The allowed energy for the particle for a particular value of n is proportional to
(a) a-2
(b) a-3/2
(c) a-1
(d) a2
Answer:
(a) a-2
Hint:
For the standing wave, a = n \(\frac { λ }{ 2 }\) or λ= \(\frac { 2a }{ n }\)
P = \(\frac {h}{ λ }\) = \(\frac { nh }{ 2a }\) ; E = \(\frac {{ p }^{2}}{ 2m}\) = \(\frac{n^{2} h^{2}}{2 a^{2} m}\) ; E ∝ a-2.

Question 3.
A diatomic molecular has moment of inertia I. By Bohr’s quantization condition its rotational energy in the nth level (n = 0 is not allowed) is
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-35
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-36
Hint:
Angular momentum, L = \(\frac { nh }{ 2π }\)
Rotation K.E = \(\frac {{ L }^{2}}{ 2I}\) = \(\frac{n^{2} h^{2}}{8 \pi^{2} I}\).

Question 4.
The speed of the particle, that can take discrete values is proportional to
(a) n-3/2
(b) n-1
(c) n1/2
(d) n
Answer:
(d) n
Hint:
P = mv = \(\frac { nh }{ 2a }\) ; V ∝ n.

Question 5.
If 13.6 eV energy is required to 10 is the hydrogen atom, then energy required to remove an electron from n = 2 is
(a) 10.2 eV
(b) 0 eV
(c) 3.4 eV
(d) 6.8 eV
Answer:
(c) 3.4 eV
Hint:
En = \(\frac { 13.6 }{ n }^{2}\)eV
∴ ∆E = E – E2 = 0 + \(\frac { 13.6 }{ n }^{2}\) = 3.4 eV.

Question 6.
Which of the following transitions in hydrogen atoms emits photon of highest frequency?
(a) n = 1 to n = 2
(b) n = 2 to n = 6
(c) n = 6 to n = 2
(d) n = 2 to n = 1
Answer:
(d) n = 2 to n = 1
Hint:
The energy difference E2 – E1 is maximum as calculated in the above problem.

Question 7.
The wavelengths involved in the spectrum of deuterium \(_{ 1 }^{ 2 }{ H }\) are slightly different from that of hydrogen spectrum because
(a) sizes of the two nuclei are different
(b) masses of the two nuclei are different
(c) attraction between the electron and the nucleus is different in the two cases
(d) nuclear forces are different in the two cases
Answer:
(b) masses of the two nuclei are different
Hint:
It is because the masses of the two nuclei are different.

Question 8.
Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is
(a) 12.1 eV
(b) 36.3 eV
(b) 36.3 eV
(c) 108.8 eV
Answer:
(c) 108.8 eV
Hint:
En = – 13.6 \(\frac { { Z }^{ 2 } }{ { n }^{ 2 } } \)
∆E = E3 – E2 = 13.6 (3)2 \(\left[ \frac { 1 }{ { 1 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 2 } } \right] \)
= \(\frac { 13.6×9×8 }{ 9 } \) = 108.8 eV.

Question 9.
Minimum energy required to take out the only one electron from ground state of He+ is
(a) 13.6 eV
(b) 54.4 eV
(c) 27.2 eV
(d) 6.8 eV
Answer:
(b) 54.4 eV
Hint:
Ionisation energy, E = 13.6 Z2 eV
Fe He+, Z = 2
∴ E= 13.6 x (2)2 = 13.6 x 4 = 54.4 eV.

Question 10.
Energy of characteristic X-ray is a consequence of
(a) energy of projectile electron
(b) thermal energy of target
(c) transition in target atoms
(d) none of the above
Answer:
(c) transition in target atoms.

Question 11.
How much energy is needed to excite an electron in H-atom from ground state to first excited state?
(a) – 13.6 eV
(b) – 10.2 eV
(c) + 10.2 eV
(d) + 13.6 eV
Answer:
(c) + 10.2 eV
Hint:
E1 = – 13.6 eV,
E2 = – 13.6/222 = – 3.4 eV
Required excitation energy
= E2 – E2 = – 3.4 + 13.6 = + 10.2 eV.

Question 12.
For an electron in the second orbit of hydrogen, what is the moment of momentum as per the Bohr’s model?
(a) 2πh
(b) πh
(c) h / π
(d) 2h / π
Answer:
(c) h / π
Hint:
In second orbit of hydrogen, L = 2 \(\left( \frac { h }{ 2\pi } \right) \) = \(\frac { h }{ π }\).

Question 13.
The total energy of an electron in the first excited state of hydrogen atom is about -3.4 eV. Its kinetic energy in this state is
(a) 3.4 eV
(b) 6.8 eV
(c) – 3.4 eV
(d) – 6.8 eV
Answer:
(a) 3.4 eV
Hint:
K.E = – Total energy = +3.4 eV.

Question 14.
The energy of the ground electronic state of hydrogen atom is 13.6 eV. The energy of the first excited state will be
(a) – 27.2 eV
(b) – 52.4 eV
(c) – 3.4 eV
(d) – 6.8 eV
Answer:
(c) – 3.4 eV
Hint:
For the first excited state, n = 2
∴ E2 = \(\frac {{ E }_{ 1 }}{{ E }_{ 2 }}\) = \(\frac {-13.6 eV}{4}\) = -3.4 eV.

Question 15.
The total energy of electron in the ground state of hydrogen atom is – 13.6 eV. The kinetic energy of an electron in the first excited state is
(a) 6.8 eV
(b) 13.6 eV
(c) 1.7 eV
(d) 3.4 eV
Answer:
(d) 3.4 eV
Hint:
Total energy in the first excited state,
E2 = \(\frac {{ E }_{ 1 }}{{ E }_{ 2 }}\) = \(\frac {{ E }_{ 1 }}{{ 2 }^{ 2 }}\) = \(\frac {-13.6 }{4}\) = -3.4 eV
K.E = -E2 = 3.4 eV.

Question 16.
Bohr’s theory of hydrogen atom did not explain fully
(a) diameter of H-atom
(b) emission spectra
(c) ionisation energy
(d) the fine structure of even hydrogen spectrum
Answer:
(d) the fine structure of even hydrogen spectrum
Hint:
Bohr theory could not explain the five structure of hydrogen spectrum.

Question 17.
In Bohr’s model of an atom, which of the following is an integral multiple of \(\frac { h }{ 2\pi } \) ?
(a) Kinetic energy
(b) Radius of an atom
(c) Potential energy
(d) Angular momentum
Answer:
(d) Angular momentum
Hint:
L = mvr = \(\frac { nh }{ 2\pi } \).

Question 18.
According to Bohr’s theory, relation between n and radius of orbit is:
(a) r ∝ \(\frac { 1 }{ n } \)
(b) r ∝ n
(c) r ∝ n2
(d) r ∝ \(\frac { 1 }{{ n }^{2}} \)
Answer:
(c) r ∝ n2
Hint:
r = \(\frac{n^{2} h^{2}}{4 \pi^{2} m K Z e^{2}}\) i.e., r ∝ n2.

Question 19.
In Bohr’s model of hydrogen atom, the radius of the first electron orbit is 0.53 Å. What will be the radius of the third orbit?
(a) 4.77 Å
(b) 47.7 Å
(c) 9 Å
(d) 0.09 Å
Answer:
(a) 4.77 Å
Hint:
r3 = (3)2 r1 = 9 x 0.53 = 4.77 Å.

Question 20.
In Bohr model of hydrogen atom, which of the following is quantised?
(a) linear velocity of electron
(b) angular velocity of electron
(c) linear momentum of electron
(d) angular momentum of electron
Answer:
(d) angular momentum of electron.

Question 21.
In Bohr’s model, the atomic radius of the first orbit is r0. Then, the radius of the third orbit is
(a) r0/9
(b) r0
(c) 9r0
(d) 3r0
Answer:
(c) 9r0
Hint:
rn = r1 n2, where r1 = r0
∴ v3 = r0 (3)2 9r0

Question 22.
What is ratio of Bohr magneton to the nuclear magneton?
(a) \(\frac {{ m }_{ p }}{{ m }_{ e }}\)
(b) \(\frac{m_{p}^{2}}{m_{e}^{2}}\)
(c) 1
(d) \(\frac {{ m }_{ e }}{{ m }_{ p }}\)
Answer:
(a) \(\frac {{ m }_{ p }}{{ m }_{ e }}\)
Hint:
Bohr magneton, μB = \(\frac {eh}{{ 2m }_{ e }}\)
Nuclear magneton, μN = \(\frac {eh}{{ 2m }_{ p }}\)
∴ \(\frac {{ μ }_{ B }}{{ μ }_{ N }}\) = \(\frac {{ m }_{ p }}{{ m }_{ e }}\).

Question 23.
In terms of Bohr radius a0, the radius of the second Bohr orbit of a hydrogen atom is given by
(a) 4a0
(b) 8a0
(c) √2a0
(d) 2a0
Answer:
(a) 4a0
Hint:
rn = r1 n2
r2 = a0 (2)2 =4a0

Question 24.
If an a-particle collides head on with a nucleus, what is impact parameter?
(a) zero
(b) infinite
(c) 10-10 m
(d) 1010 m
Answer:
(a) zero

Question 25.
One femtometre is equivalent to
(a) 1015 m
(b) 10-15 m
(c) 10-12 m
(d) 1011 m
Answer:
(b) 10-15 m

Question 26.
Wavelength of Kα line of X-ray spectra varies with atomic number as
(a) λ ∝ Z
(b) λ ∝ √Z
(c) λ ∝ \(\frac { 1 }{{ Z }^{2}}\)
(d) λ ∝ \(\frac { 1 }{ √Z }\)
Answer:
(c) λ ∝ \(\frac { 1 }{{ Z }^{2}}\)
Hint:
ccording to moseley’s law, √V = a(Z – b) or V = \(\frac { c }{ λ }\) = a2 (Z – b)2
∴ (c) λ ∝ \(\frac { 1 }{{ Z }^{2}}\).

Question 27.
The shortest wavelength of X-rays, emitted from a X-ray tube, depend upon
(a) current in the tube
(b) voltage applied to the tube
(c) nature of glass material in the tube
(d) atomic number of the target material
Answer:
(b) voltage applied to the tube
Hint:
λmin = \(\frac { 12375 }{V (volt)}\) Å ; λmin ∝ \(\frac { 1 }{ V }\).

Question 28.
During X-ray formation, if voltage is increased
(a) minimum wavelength decreases
(b) minimum wavelength increases
(c) intensity decreases
(d) intensity increases
Answer:
(a) minimum wavelength decreases
Hint:
As λmin ∝ \(\frac { 1 }{ V }\) if voltage is increased, the minimum wavelength of X-rays emitted decreases.

Question 29.
What would be the radius of second orbit of He+ ions?
(a) 1.058 Å
(b) 3.023 Å
(c) 2.068 Å
(d) 4.458 Å
Answer:
1.058 Å
Hint:
rn = \(\frac {{ n }^{2}}{ Z }\) r1
For He+ ion, n = 2, Z = 2
∴ r2 = \(\frac {4}{ 2 }\) x 0.59 Å = 1.058 Å.

Question 30.
The minimum wavelength of the X-rays produced by electrons accelerated through a potential difference of V volts is directly proportional to
(a) \(\frac { 1 }{ √V }\)
(b) \(\frac { 1 }{ V }\)
(c) √V
(d) V2
Answer:
(b) \(\frac { 1 }{ V }\)
Hint:
\(\frac { hc }{ λ }\) =eV or λ = \(\frac { hc }{ eV }\), i.e., λ ∝ \(\frac { 1 }{ V }\).

Question 31.
Which source is associated with a line emission spectrum?
(a) Electric fire
(b) Neon street sign
(c) Red traffic light
(d) Sun
Answer:
(b) Neon street sign

Question 32.
Which one of the relation is correct between time period and number of orbits while an electron is revolving in a orbit?
(a) T ∝ \(\frac { 1 }{{ n }^{2}}\)
(b) T ∝ n2
(c) T ∝ n3
(d) T ∝ \(\frac { 1 }{{ n }^{2}}\)
Answer:
(c) T ∝ n3
Hint:
In Bohr’s atomic model, T ∝ n3.

Question 33.
The size of atom is proportional to
(a) A
(b) A1/3
(c) A2/3
(d) A-1/3
Answer:
(b) A1/3

Question 34.
If an electron jumps from 1st orbit to 3rd orbit, then it will
(a) not lose energy
(b) not given energy
(c) release energy
(d) absorb energy
Answer:
(d) absorb energy
Hint:
Only by absorbing energy, an electron jumps from first orbit to third orbit.

Question 35.
According to uncertainty principle for an electron, time measurement will become uncertain if following is measured with high certainty
(a) energy
(b) momentum
(c) location
(d) velocity
Answer:
(a) energy
Hint:
According to uncertainty principle, ∆E.∆t ≥ \(\frac { h }{ 2π }\).

Question 36.
According to Rutherford’s atomic model, the electrons inside an atom are
(a) stationary
(b) centralized
(c) non-stationary
(d) none of these
Answer:
(c) non-stationary
Hint:
According to Rutherford model, the electron inside an atom cannot be stationary.

Question 37.
Wavelength of a light emitted from second orbit to first orbit in a hydrogen atom is
(a) 1.215 x 10-7 m
(b) 1.215 x 10-5 m
(c) 1.215 x 10-4 m
(d) 1.215 x 10-3 m
Answer:
(a) 1.215 x 10-7 m
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-37

Question 38.
In terms of Rydberg constant R, the wave number of the first Balmer line is
(a) R
(b) 3R
(c) \(\frac { 5R }{ 36 }\)
(d) \(\frac { 8R }{ 9 }\)
Answer:
(c) \(\frac { 5R }{ 36 }\)
Hint:
For the first Balmer line, \(\bar { v } \) =\(\frac { 1 }{ λ }\) = R\(\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)\) =\(\frac { 5R }{ 36 }\).

Question 39.
The K X-ray emission line of tungsten occurs at λ = 0.021 nm. The energy difference between K and L levels in this atom is about
(a) 0.51 MeV
(b)1.2MeV
(c) 59 keV
(d) 136
Answer:
(c) 59 keV
Hint:
E = \(\frac { hc }{ λ }\) = \(\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{0.021 \times 10^{-9}}\) eV = 589.3 x 102 eV ≈ 59 KeV.

Question 40.
The radius of an electron orbit in a hydrogen atom is of the order of
(a) 10-8 m
(b) 10-9 m
(c) 10-11 m
(d) 10-13 m
Answer:
(c) 10-11 m

Question 41.
Which of the following atoms has the lowest ionisation potential?
(a) \(_{ 7 }^{ 14 }{ N }\)
(b) \(_{ 55 }^{ 133 }{ Cs }\)
(c) \(_{ 18 }^{ 40 }{ Ar }\)
(d) \(_{ 8 }^{ 16 }{ O }\)
Answer:
(b) \(_{ 55 }^{ 133 }{ Cs }\)
Hint:
In \(_{ 55 }^{ 133 }{ Cs }\), the outermost electron is farthest from the nucleus and so minimum energy is required to remove this electron from the atom. Hence \(_{ 55 }^{ 133 }{ Cs }\) has lowest concision potential.

Question 42.
The transition from the state n = 4 to n = 3 in a hydrogen like atom result in ultraviolet radiation. Infrared radiation will be obtained in the transition from
(a) 2 → 1
(b) 3 → 2
(c) 4 → 2
(d) 5 → 4
Answer:
(d) 5 → 4
Hint:
The energy gap between 4th and 3rd states is more than the gap between 5th and 4th states.

Question 43.
The number of waves, contained in unit length of the medium, is called
(a) elastic wave
(b) wave number
(c) wave pulse
(d) electromagnetic wave
Answer:
(b) wave number
Hint:
The number of waves contained in a unit length of the medium is called a wave number.

Question 44.
When hydrogen atom is in its first excited level, its radius is
(a) sarhe
(b) half
(c) twice
(d) four times
Answer:
(d) four times
Hint:
r2 = r1 (2)2 = 4r1

Question 45.
The ground state energy of hydrogen atom is -13.6 eV. What is the potential energy of the electron in this state?
(a) 0 eV
(b) -27.2 eV
(c) 1 eV
(d) 2 eV
Answer:
(b) -27.2 eV
Hint:
PE = 2 x Total energy = 2 x (-13.6) = – 27.2 eV.

Question 46.
For ionising an excited hydrogen atom, the energy required (in eV) will be
(a) a little less than 13.6
(b) 13.6
(c) more than 13.6 eV
(d) 3.4 or less
Answer:
(d) 3.4 or less
Hint:
The energy of the electron is – 3.4 eV in first excited state and the its magnitude is less for higher excited state.

Question 47.
What is the energy of He+ electron in first order?
(a) 40.8 eV
(b) -27.2 eV
(c) -54.4 eV
(d)-13.6eV
Answer:
(c) -54.4 eV
Hint:
For hydrogen like atoms or ions, En = \(\frac{-13.6 Z^{2}}{n^{2}}\) eV
For He+, Z = 2 and n = 1
E1 = \(\frac{-13.6 \times 2^{2}}{12}\) 54.4 eV.

Question 48.
If voltage across on X-ray tube is doubled, then energy of X-ray emitted by
(a) be doubled
(b) be quadrupled
(c) become half
(d) remain the same
Answer:
(d) remain the same
Hint:
The energy of the X-rays depends on the nature of the target material. Thus the energy of the X-rays remain the same.

Question 49.
When hydrogen atom is in its first excited level, its radius is of the Bohr radius.
(a) twice
(b) 4 times
(c) same
(d) half
Answer:
(b) 4 times
Hint:
For first excited level, n = 2
r2 = (2)2 r0 = 4r0

Question 50.
The ionisation energy of hydrogen atom is 13.6 eV, the ionisation energy of a singly ionsed helium atom would be
(a) 13.6 eV
(b) 27.2 eV
(c) 6.8 eV
(d) 54.4 eV
Answer:
(d) 54.4 eV
Hint:
\({ E }_{ 2 }^{ 1 }\) = (2)2 E1 = 4 x 13.6 = 54.4 eV.

Question 51.
When an electron makes transition from n = 4 to n = 2, then emitted line spectrum will be
(a) first line of lyman series
(b) second line of Balmer series
(c) first line of paschen series
(d) second line of paschen series
Answer:
(b) second line of Balmer series
Hint:
The transition from n = 4 to n = 2 emits second line of Balmer series.

Question 52.
Maximum frequency of emission is obtained for the transition
(a) n = 2 to n = 1
(b) n = 6 to n = 2
(c) n = 1 to n = 2
(d) n = 2 to n = 6
Answer:
(a) n = 2 to n = 1
Hint:
The energy difference E2 – E1 is maximum, so photon of maximum frequency is emitted in transition n = 2 to n = 1.

Question 53.
Hydrogen atoms are excited from ground state to the state of principle quantum number 4. Then the number of spectral lines observed will be
(a) 3
(b) 6
(c) 5
(d) 2
Answer:
(b) 6
Hint:
Here n = 4
∴ The number of spectral lines emitted \(\frac { n(n-1) }{ 2 }\) = \(\frac { 4×3 }{ 2 }\) = 6

Question 54.
The radius of hydrogen atom, in the ground state is of the order of
(a) 10-18 cm
(b) 10-7 cm
(c) 10-6 cm
(d) 10-4 cm
Answer:
(a) 10-18 cm
Hint:
Radius of first orbit of H-atom = 0.53 Å ≈ 10-8 cm.

Question 56.
According to Bohr’s theory of the hydrogen atom, the speed vn of the electron in a stationary orbit is related to the principal quantum number n as (c is a constant)
(a) vn = c/n2
(b) vn = c/n
(c) vn = c x n
(d) vn = c x n2
Answer:
(b) vn = c/n
Hint:
Speed of electron in nth orbit, υn= c/n.

Question 57.
Out of the following which one is not possible energy for a photon to be emitted by hydrogen atom according to Bohr’s atomic model?
(a) 13.6 eV
(b) 0.65 eV
(c) 1.9 eV
(d) 11.1 eV
Answer:
(d) 11.1 eV
Hint:
For no two energy levels of hydrogen atom, E2 – E1 = 11.1 eV.

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Short Answer Questions

Question 1.
Write down the drawbacks of Rutherford model.
Answer:
1. Drawbacks of Rutherford model:
Rutherford atom model helps in the calculation of the diameter of the nucleus and also the

2. Size of the atom but has the following limitations:
(a) This model fails to explain the distribution of electrons around the nucleus and also the stability of the atom. According to classical electrodynamics, any accelerated charge emits electromagnetic radiations. Due to emission of radiations, it loses its energy.

Hence, it can no longer sustain the circular motion. The radius of the orbit, therefore, becomes smaller and smaller (undergoes spiral motion) and finally the electron should fall into the nucleus and the atoms should disintegrate. But this does not happen. Hence, Rutherford model could not account for the stability of atoms.

(b) According to this model, emission of radiation must be continuous and must give continuous emission spectrum but experimentally we observe only line (discrete) emission spectrum for atoms.

Question 2.
Define excitation potential.
Answer:
Excitation potential is defined as excitation energy per unit charge.

Question 3.
What is meant by atomic number?
Answer:
The number of protons in the nucleus is called the atomic number and it is denoted by Z.

Question 4.
What is meant by neutron number?
Answer:
The number of neutrons in the nucleus is called neutron number (N).

Question 5.
What is meant by mass number?
Answer:
The total number of neutrons and protons in the nucleus is called the mass number and it is denoted by A. Hence, A = Z + N.

Question 6.
Write down the properties of neutrino.
Answer:
The neutrino has the following properties:

  1. It has zero charge
  2. It has an antiparticle called anti-neutrino.
  3. Recent experiments showed that the neutrino has very tiny mass.
  4. It interacts very weakly with the matter. Therefore, it is very difficult to detect. In fact, in every second, trillions of neutrinos coming from the sun are passing through our body without any interaction.

Samacheer Kalvi 12th Physics Atomic and Nuclear Physics Numerical Problems

Question 1.
What is the distance of closest approach when a 5 MeV proton approaches a gold nucleus.
Solution:
q1 = ze
q2 = e
At the distance r0 of closest approach,
K.E of a Proton = P.E. of proton and the gold nucleus
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-38

Question 2.
Calculate the impact parameter of a 5 MeV particle scattered by 90° when it approaches.
Solution:
KE = 5 MeV = 5 x 106 x 1.6 x 10-19 J
θ = 90°
For gold, Z = 79
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-39

Question 3.
What is the angular momentum of an electron in the third orbit of an atom?
Solution:
Here n = 3; h = 6.6 x 10-34 Js
Angular momentum,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-40

Question 4.
Write down the expression for the radii of orbits of hydrogen atom. Calculate the radius of the smallest orbit.
Solution:
The radius of the nth orbit of a hydrogen atom is given by
r = \(\frac{n^{2} h^{2}}{4 \pi^{2} m K e^{2}}\)
Radius of innermost orbit, called Bohr’s radius, is obtained by putting n = 1. It is denoted by r0
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-41
r0 = 0.53 x 10-10 m = 0.53 A°.

Question 5.
Calculate the frequency of the photon, which can excite the electron to – 3.4 eV from -13.6 eV.
Solution:
Energy of photon, hυ = E2 – E1
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-42
υ = 2.47 x 1015 Hz

Question 6.
The ground state energy of hydrogen atom is -13.6 eV. If an electron makes a transition from an energy level -0.85 eV to -1.51 eV, Calculate the wavelenth of the spectral line emitted. To which series of hydrogen spectrum does this wavelenth belong?
Solution:
Here ∆E = E2 – E1 = -0.85-(-1.51).
= 0.66 eV
∆E = 0.66 x 1.6 x 10-19 J
λ = \(\frac { hc }{ ∆E }\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{0.66 \times 1.6 \times 10^{-19}}\)
= 18.84 x 10-7
λ = 18840 Å
This wavelength belongs to the Pachen series of the hydrogen spectrum.

Question 7.
Express 16 rag mass into equivalant energy in eV.
Solution:
Here m = 16 mg = 16 x 10-16 kg, C = 3 x 108 ms-1
Equivalent energy, E = mc2
= 16 x 10-16 x (3 x 108)2 J
= \(\frac{16 \times 10^{-6} \times\left(3 \times 10^{8}\right)^{2}}{1.6 \times 10^{-19}} \mathrm{eV}\)
E = 9 x 1030 eV.

Question 8.
The nuclear mass of \(_{ 26 }^{ 56 }{ Fe }\) is 55.85 amu. Calculate its nuclear density.
Solution:
Here MFe = 55.85 amu = 55.85 x 1.66 x 10-27 kg
= 9.27 x 10-26 kg
Nuclear Mass = R0 A1/3 = 1.1 x 10-15 x (56)1/3 m
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-43
ρnu = 2.9 x 1017 kg m-3.

Question 9.
Calculate the density of hydrogen nuclear in SI units. Given R0 = 1.1 fermi and mp = 1.007825 amu.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-44
ρ = 2.98 x 1017 kg m-3.

Question 10.
Express one atomic mass unit in energy units, first in Joules and then in MeV. Using this, express the mass defect of \(_{ 8 }^{ 16 }{ O }\) in MeV.
Solution:
We have, m = 1 amu = 1.66 x 10-27 kg, c = 3 x 108 ms-1
E = mc2 = 1.66 x 10-27 x (3 x 108)2
= 14.94 x10-11 J
= \(\frac{1.494 \times 10^{-10}}{1.6 \times 10^{-13}} \mathrm{MeV}\) [ 1 MeV = 1.6 x 10-13]
= 931.5 MeV
The \(_{ 8 }^{ 16 }{ O }\) nucleus contains 8 protons and 8 neutrons
Mass of 8 protons = 8 x 1.00727 = 8. 05816 amu
Mass of 8 neutrons = 8 x 1.00866 = 8. 06928 amu
Total Mass = 16.12744 amu
Mass of \(_{ 8 }^{ 16 }{ O }\) nucleus = 15.99053 amu
Mass defect = 0.13691 amu
∆Eb = 0.13691 x 931.5 Mev
∆Eb = 127.5 Mev

Question 11.
The decay constant, for a given redioactive sample is 0.3465 / day. What percentage of this sample will get decayed in a period of 4 years?
Solution:
Here λ, = 0.3465/day; t = 4 years
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-45
Hence sample left undecayed after a period of 4 years,
Samacheer Kalvi 12th Physics Solutions Chapter 8 Atomic and Nuclear Physics-46

Question 12.
If 200 MeV energy is released in the fission of a single nucleus of \(_{ 92 }^{ 235 }{ U }\), how many fissions must occur to produce a power of 1 kW?
Solution:
Let the number of fissions per second be n.
Then, Energy released per second = n x 200 MeV
= n x 200 x 1.6 x 10-13 J
Energy required per second = Power x Time
= 1kW x 1 s = 1000 J
Energy released = Energy required
n x 200 x 1.6 x 10-13 = 1000
n = 3.125 x 10-13

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Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

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Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

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Samacheer Kalvi 12th Chemistry Chapter 10 Surface Chemistry Textual Evaluation Solved

Samacheer Kalvi 12th Chemistry Surface Chemistry Multiple Choice Questions

Question 1.
For freudlich isotherm a graph of log \(\frac{x}{m}\) is plotted against log P. The slope of the line and its y – axis intercept respectively corresponds to
(a) \(1 / n\), k
(b) log \(1 / n\), k
(c) \(1 / n\), log k
(d) log \(1 / n\), log k
Answer:
(c) \(1 / n\), log k
\(\frac{x}{m}\) = \(\mathrm{k} \cdot \mathrm{p}^{1 / \mathrm{n}}\)
log\((\frac{x}{m})\) = log k + \(\frac { 1 }{ n }\)log p
y = c + mx
m = \(\frac { 1 }{ n }\) and c = log k

Question 2.
Which of the following is incorrect for physisorption?
(a) reversible
(b) increases with increase in temperature
(c) low heat of adsorption
(d) increases with increase in surface area
Answer:
(b) increases with increase in temperature
Physisorption is an exothermic process. Hence increase in temperature decreases the physisorption.

Question 3.
Which one of the following characteristics are associated with adsorption?
(a) ∆G and ∆H are negative but ∆S is positive
(b) ∆G and ∆S are negative but ∆H is positive
(c) ∆G is negative but ∆H and ∆S are positive
(d) ∆G. AH and ∆S all are negative.
Answer:
(d) ∆G, ∆H and ∆S all are negative.
Adsorption leads to decrease in randomness (entropy).i.e. ∆S < 0 for the adsorption to occur, ∆G should be – ve. We know that ∆G = ∆H – T∆S if ∆S is – ve, T∆S is + ve. It means that ∆G will become negative only when ∆H is – ve and ∆H > T∆S

Question 4.
Fog is colloidal solution of ……………..
(a) solid in gas
(b) gas in gas
(c) liquid in gas
(d) gas in liquid
Answer:
(c) liquid in gas
dispersion medium-gas, dispersed phase-liquid

Question 5.
Assertion: Coagulation power of Al3+ is more than Na.
Reason: greater the valency of the flocculating ion added, greater is its power to cause precipitation
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(a) if both assertion and reason are true and reason is the correct explanation of assertion. (Hardy-Sechuize nile)

Question 6.
Statement: To stop bleeding from an injury, ferric chloride can be applied. Which comment about the statement is justified?
(a) It is not true, ferric chloride is a poison.
(b) It is true, Fe3+ ions coagulate blood which is a negatively charged sol
(c) It is not true; ferric chloride is ionic and gets into the blood stream.
(d) It is true, coagulation takes place because of formation of negatively charged sol with Cl.
Answer:
(b) It is true, Fe3+ ions coagulate blood which is a negatively charged sol

Question 7.
Hair cream is …………..
(a) gel
(b) emulsion
(c) solid sol
(d) sol.
Answer:
(b) emulsion
Emulsion dispersed phase, Dispersion medium -liquid

Question 8.
Which one of the following is correctly matched?
(a) Emulsion – Smoke
(b) Gel – butter
(c) foam – Mist
(d) whipped cream – sol
Answer:
(b) Gel – butter

Question 9.
The most effective electrolyte for the coagulation of As2S3 Soils
(a) NaCI
(b) Ba(NO3)2
(c) K3[Fe(CN)6]
(d) AI3+
Answer:
(d) AI3+
As2S3 is a – vely charged colloid. It will be most effectively coagulated by the cation with greater valency. i.e., Al3+.

Question 10.
Which one of the is  not a surfactant?
(a) CH3 – (CH2)15 – N – (CH3)2CH2Br
(b) CH3 – (CH2)15 – NH2
(c) CH3 – (CH2)16 – CH2OSO2 – Na+
(d) OHC – (CH2)14 – CH2 – COONa+
Answer:
(b) CH3 – (CH2)15 – NH2

Question 11.
The phenomenon observed when a beam of light is passed through a colloidal solution is ………….
(a) Cataphoresis
(b) Electrophoresis
(c) Coagulation
(d) Tyndall effect
Answer:
(d) Tyndall effect-scattering of light

Question 12.
In an electrical field, the particles of a colloidal system move towards cathode. The coagulation of the same sol is studied using K2SO4
(i). Na3PO4
(ii). K4[Fe(CN)6]
(iii). and NaCI
(iv). Their coagulating power should be …………..
(a) II > I >IV > III
(b) III > II > I > IV
(c) I > II > III > IV
(d) none of these
Answer:
(b) III > II > I > IV

Question 13.
Collodion is a 4% solution of which one of the following compounds in alcohol – ether mixture?
(a) Nitroglycerine
(b) Cellulose acetate
(c) Glycoldinitrate
(d) Nitrocellulose
Answer:
(a) Nitrocellulose
pyroxylin (nitro cellulose)

Question 14.
Which one of the following is an example for homogeneous catalysis?
(a) manufacture of ammonia by Haber’s process
(b) manufacture of sulphuric acid by contact process
(c) hydrogenation of oil
(a) Hydrolysis of sucrose in presence of all HCI
Answer:
(a) Hydrolysis of sucrose in presence of all HCl
Both reactant and catalyst are in same phase. i.e. (1)

Question 15.
Match the following.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-1
Answer:
(a) (iv) (i) (ii) (iii)

Question 16.
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of AS2S3 are given below
(I) (NaCI) = 52
(II) (BaCl) = 0.69
(III) (MgSO4) = 0.22
The correct order of their coagulating power is ……….
(a) III > II > I
(b) I > II > III
(c) I >III > II
(d) II > III > I
Answer:
(a) III > II > I
coagulating power ± \(\frac{1}{\text { coagulation value }}\)

Question 17.
Adsorption of a gas on solid metal surface is spontaneous and exothermic, then ……………
(a) ∆H increases
(b) ∆S increases
(c) ∆G increases
(d) ∆S decreases
Answer:
(a) ∆S decreases – ∆S is -ve

Question 18.
If x is the amount of adsorbate and m is the amount of adsorbent, which of the following relations is not related to adsorption process?
(a) x/m = f(P) at constant T
(b) x/m = f(T) at constant P
(c) P = f(T) at constant x/m
(d) x/m = PT
Answer:
(d) x/m = mPT

Question 19.
On which of the following properties does the coagulating power of an ion depend?
(a) Both magnitude and sign of the charge on the ion.
(b) Size of the ion alone
(c) the magnitude of the charge on the ion alone
(d) the sign of charge on the ion alone.
Answer:
(a) Both magnitude and sign of the charge on the ion.

Question 20.
Match the following.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-2
Answer:
(d) (iv) (iii) (ii) (i)

Samacheer Kalvi 12th Chemistry Surface Chemistry Short Answer Questions

Question 1.
Give two important characteristics of physisorption.
Answer:
Important characteristics of physisorption:

  1. It is reversible
  2. It has low heat of adsorption
  3. It has weak van der Waals forces of attraction with adsorbent.
  4. It increases with increase in pressure.
  5. It forms multimolecular layer.

Question 2.
Differentiate physisorption and chemisorption.
Answer:
Chemical adsorption orChemisorption or Activated adsorption

  1. It is very slow
  2. It is very specific depends on nature of adsorbent and adsorbate.
  3. chemical adsorption is fast with increase pressure, it can not alter the amount.
  4. When temperature is raised chemisorption first increases and then decreases.
  5. Chemisorption involves transfer of electrons between the adsorbent and adsorbate,
    Heat of adsorption is high i.e., from 40 – 400kJ/mole.
  6. Monolayer of the adsorbate is formed.
  7. Adsorption occurs at fixed sites called active centres. It depends on surface area .
  8. Chemisorption involves the formation of activated complex with appreciable activation energy.
  9. Physical adsorption or van der waals adsorption or Physisorptlon
  10. It is irreversible

Physical adsorption or van der waals adsorption or physisorption.

  1. It is instantaneous
  2. It is non-specific
  3. In Physisorption. when pressure increases the amount of adsorption increases.
  4. Physisorption decreases with increase in temperature.
  5. No transfer of electrons
  6. Heat of adsorption is low in the order of 4OkJ/mole.
  7. Multilayer of the adsorbate is formed on the adsorbent.
  8. It occurs on all sides.
  9. Activation energy is insignificant.
  10. It is reversible.

Question 3.
In case of chemisorption, why adsorption first increases and then decreases with temperature?
Answer:
1. Chemisorption involves a high activation energy so it is also referred to as activated adsorption.

2. It is found in chemisorption that it first increases and then decreases with increase in temperature. When adsorption is plotted, the graph first increaes and then decreases with temperature.

3. The initial increase illustrates the requirement of activation of the surface for adsorption is due to fact that formation of activated complex requires certain energy. But later it decreacs at high temperature is due to desorption as the kinetic energy of the adsorbate increases (exothermic nature)

Question 4.
Which will be adsorbed more readily on the surface of charcoal and why; NH3 or CO2?
Answer:
1. The gaseshaving low critical temperature are adsorbed slowly, while gases with high critical temperature are adborbed readily.

2. Among CO2, and NH3, NH3 will be more readily adsorbed on the surface of the charcoal. This is because the critical temperature of ammonia gas is quite high than the CO2. Hence, it easily combines with the materials than the CO2 whether it is solid, liquid or any gases.

Question 5.
Heat of adsorption is greater for chemisorptions than physisorption. Why?
Answer:
Chemisorption has higher heat of adsorption. because in chemisorption the chemical bonds are much stronger. In adsorbed state the adsorbate is hold on the surface of adsorbent by attractive forces (bond). And chemisorption is irreversible one. Therefore, heat of adsorption is greater for chenil sorptions than physisorption. Chemisorption, heat of adsorption range 40 – 400kJ/mole.

Question 6.
In a coagulation experiment 10 mL of a colloid (X) is mixed with distilled water and 0.1M solution of an electrolyte AB so that the volume is 20 mL. It was found that all solutions containing more than 6.6 mL of AB coagulate with in 5 minutes. What is the flocculation values of AB for sol (X)?
Answer:
A minimum of 6.6mL of AB is required to coagulate the sol. The moles of AB in the sol is
\(\frac{6.6 \times 0.01}{20}\) = 0.033 moles
This means that a minimum of 0.033 moles or 0.0033 x 1000 = 3.3 milli moles are required for coagulating one litre of sol. Flocculation value of AB for X = 3.3

Question 7.
Peptising agent is added to convert precipitate into colloidal solution. Explain with an example.
Answer:
1. Ions either positive or negative of peptizing agent (electrolyte) are adsorbed on the particles of precipitate. They repel and hit each other and break the particles of the precipitate into colloidal size.

2. For example, when we add a small volume of very dilute hydrochloric acid solution peptising agent to a fresh precipitate of a silver chloride, it leads to formation of silver chloride colloidal solution,
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-3

Question 8.
What happens when a colloidal sol of Fe(OH)3 and As2S3 are mixed?
Answer:
On mixing Fe(OH)3 positive sol and As2S3 negative sol, mutual coagulation occurs which causes precipitation. When these sol got mixed with each other, due to Fe3+ and S2- ions neutralisation of charges will happen and precipitate will be formed.
Fe(OH)3 + As2S3 → Fe2S3 + As(OH)3

Question 9.
What is the difference between a sol and a gel?
Answer:
Sol

  1. The liquid state of a colloidal solution is called sol.
  2. The sol does not have a definite structure.
  3. The dispersion medium of the sol may be water.
  4. The sol can be converted to gel by cooling The sol can be easily dehydrated.
  5. The viscosity of the sol is very low.
  6. Sol is categorized into lyophobic and lyophilic sols.
  7. Example: Blood

Gel

  1. The solid or semi solid state of a colloidal solution is called gel.
  2. The gel possesses honey comb like structure.
  3. The dispersion medium of gel will be hydrated colloid particles.
  4. The gel can be converted sol by heating.
  5. The gel cannot be dehydrated.
  6. The viscosity of the gel is very high.
  7. There is no such classification of gel.
  8. Example: Fruit jelly, cooked gelatin jelly.

Question 10.
Why are lyophillic colloidal sols are more stable than lyophoblic colloidal sol?
Answer:
1. A lyophilic colloidal sols are stable due to the charge and the hydration of sol particles.

2. Lyophilic sols are more stable than lyophobilc sols because they are highly hydrated in the solution. And since more is the hydration more will be its stability.

3. Lyophilic sols are stabilized by electrostatic charge and hydration where as lyophobile sols are only stabilized by charge, so they easily gets coagulated and requires a stabilising agent. Hence, lyophilic sols are more stable than the lyophobilc sols.

Question 11.
Addition of Alum purifies water. Why?
Answer:
Purification of drinking water is activated by coagulation of suspended impurities in water using alums containing Al<sup>3+</sup>. That is why we are adding to purify water.

Question 12.
What are the factors which influence the adsorption of a gas on a solid?
Answer:
Factors which influence the adsorption of a gas on a solid is as follows:
1. Nature of the gas:
Easily liquifiable gases such as NH3, HCl etc are adsorbed to a great extent in comparison to gases such as H2, O2 etc. This is because van der Waal’s forces are stronger is easily liquifiable gases.

2. Surface area of the solid:
The greater the surface area of the adsorbant, the greater is the adsorption of gas on the solid surface.

3. Effect of pressure:
Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.

4. Effect of temperature:
Adsorption is an exothermic process. Thus in accordance with Le – Chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Question 13.
What are enzymes? Write a brief note on the mechanism of enzyme catalysis.
Answer:
Enzymes are complex protein molecules with three dimensional structures. They catalyse the chemical reaction in living organism. They are often present in colloidal state and extremely specific in catalytic action.

Each enzyme produced in a particular living cell can catalyse a particular reaction in the cell. Mechanism of enzyme catalysis: Mechanism of enzyme catalysed reaction is known as lock and key mechanism.
1. Enzymes arc highly specific in their action.

2. These specificity is due to the pressure of active sites. The shape of active site of any given enzyme is like cavity such that only a specific substrate can fit into it.

In the same way a key fit into lock. The specific binding needs to the formation of an enzyme substrate complex which accounts for high specificity of enzyme catalysed reactions.

3. Once the proper orientation is attained the substrate molecules reacts to form the product in two steps.

4. Since product molecule do not have any affinity for the enzyme they leave the enzyme surface making room for fresh substrate.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-4

step 1: Formation of enzyme – substrate complex
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-5
Step 2: Dissociation of enzyme – substrate complex to form product
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-6
The rate of the formation of product depends upon concentration of ES.

Question 14.
What do ou mean by activity and selectivity of catalyst?
Answer:
1. Activity of Catalyst:
The activity ofcatalyst is its ability to increase the rate ofa particular reaction, Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants in the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.

2. Selectivity of the catalyst:
The ability of the catalyst to dircct a reaction to yield a particular product is referred to as the selectivity of the catalyst. For example, by using different catalysts, we can get different products for the reaction between H2 and CO.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-7

Question 15.
Describe some feature of catalysis b Zeoliles.
Answer:
1. Zeolites are microporous, crystalline, hydrated aluminosilicates, made of silicon and aluminium tetrahedra.

2. There are about 50 natural zeolites and 1 50 synthetic zeolites. As silicon is tetravalent and aluminium is trivalent, the zeolite matrix carries extra negative charge. To balance the negative charge, there are extra framework cations for example, H+ or Na+ ions.

3. Zeolites earring protons are used as solid acids, catalysis and they are extensively used in the petrochemical industry for cracking heavy hydrocarbon fractions into gasoline, diesel,etc.

4. Zeolites earring Na ions are used as basic catalysis.

5. One of the most important applications of zeolites is their shape selectivity. In zeolites, the active sites namely protons are lying inside their pores. So, reactions occur only inside the pores of zeolites.

Question 16.
Give three uses of emulsions.
Answer:

  1. The cleansing action of soap is due to emulsions.
  2. It is used in the preparation of vanishing cream.
  3. It is used in the preparation of cold liver oil.

Question 17.
Why does bleeding stop by rubbing moist alum
Answer:
Blood is a colloidal sol. When we nib the injured part with moist alum then coagulation of blood takes place. Hence main reason is coagulation, which stops the bleeding. Therefore bleeding stop by rubbing moist alum.

Question 18.
Why is desorption important for a substance to act as good catalyst?
Answer:
Desorption is important for a substance to act as a good catalyst, so that after the reaction, the products found on the surface separate out (desorbed) to create free surface again for other reactant molecules to approach the surface and react. If desorption does not occur then other reactants are left with no space on the catalysts surface for adsorption and reaction will stop.

Question 19.
Comment on the statement: Colloid is not a substance but it is a state of substance.
Answer:
The statement is true. Because the same substance may exist as a colloid under certain conditions and as a crystalloid under certain other conditions. For example. NaCl in water behaves as a crystalloid while in benzene, it behaves as a colloid. Similarly, dilute soap solution behaves

like a crystalloid while concentrated solutions behaves as a colloid. It is the size of the particles which matters. That is the state in which the substance exists. If the size of the particles lies in the range 1 nm to 1oo nm, it is in the colloidal state.

Question 20.
Explain any one method for coagulation
Answer:
The flocculation and setting down of the sol particles is called coagulation. Various method of coagulation are given below:

  1. Addition of electrolytes
  2. Electrophoresis
  3. Mining oppositively charged sols
  4. Boiling.

Addition of electrolytes
A negative ion causes the precipitation of positively charged sol and vice versa. When the valency of ion is high, the precipitation power is increased. For example, the precipitation power of some cations and anions varies in the following order
Al3+ > Ba2+ > Na+, Similarly [Fe(CN)6]-3 > SO4-2 > Cl
The precipitation power of electrolyte is determined by finding the minimum concentration (millimoles / lit) required to cause precipitation of a sol in 2hours. This value is called flocculation value. The smaller the flocculation value greater will be precipitation.

Question 21.
Write a note on electro osmosis
Answer:
Electro osmosis:
A sol is electrically neutral. Hence the medium carries an equal but opposite charge to that of dispersed particles. When sol particles are prevented from moving, under the influence of electric field the medium moves in a direction opposite to that of the sol particles. This movement of dispersion medium under the influence of electric potential is called electro osmosis.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-8

Question 22.
Write a note on catalytic poison
Answer:
Catalytic poison:
Certain substances when added to a catalysed reaction, decreases or completely destroys the activity of catalyst and they are often known as catalytic poisons. For example, In the reaction,
2SO2 + O2 → 2SO3 with a Pt catalyst, the poison is AS2O3.
i.e., AS2O3 destroys the activity of pt. AS2O3 blocks the activity of the catalyst. So, the activity is lost.

Question 23.
Explain intermediate compound formation theory of catalysis with an example.
Answer:
The intermediate compound formation theory:
A catalyst acts by providing a new path with low energy of activation. in homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the catalyst is regenerated.

Consider the reactions:
A + B → AB ……………(1)
A + C → AC (intermediate) ………….(2)
C is the catalyst
AC + B → AB + C …………(3)
Activation energies for the reactions (2) and (3) are lowered compared to that of (1). Hence the formation and decomposition of the intermediate accelerate the rate of the reaction.
Example:
The mechanIsm of Fridel crafts reaction is given below
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-9
The action of catalyst is explained as follows .
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-10

This theory describes,

  1. The specificity of a catalyst.
  2. The increase in the rate of the reaction with increase in the concentration of a catalyst.

Limitations

  1. The intermediate compound theory fails to explain the action of catalytic poison and activators (promoters).
  2. This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Question 24.
What is the difference between homogenous and hetrogenous catalysis?
Answer:
Hornogenous Catalysis:

  1. In a catalysed reaction the reactants, products and catalyst are present in the same phase.
  2. For example.
    Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-11
    Hence NO act as catalyst.
  3. Homogeneous catalysis explained by intermediate compound formation theory.

Heterogeneous Catalysis:

  1. In a reaction, the catalyst is present in a different phase. i.e., catalyst is not present in the same phase as that of reactants and products.
  2. For example.
    Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-12
    Hence Pt(s) act as catalyst.
  3. Hetenogeneous catalysis explained by adsorption theory.

Question 25.
Describe adsorption theory of catalysis.
Answer:
Adsorption theory:
Langmuir explained the action of catalyst in heterogeneous catalysed reactions based on adsorption. The reactant molecules are adsorbed on the catalyst surfaces, so this can also he called as contaçt catalysis.

According to this theory, the reactants arc adsorbed on the catalyst surface to form an activated complex which subsequently decomposes and gives the product. The various steps involved in a heterogeneous catalysed rcacton arc given as follows:

  1. Reactant molecules diffuse from bulk to the catalyst surface.
  2. The reactant molecules are adsorbed on the surface of the catalyst.
  3. The adsorbed reactant molecules are activated and form activated complex which is decomposed to form the products.
  4. The product molecules are desorbed.
  5. The product diffuse away from the surface of the catalyst.

Advantages of adsorption theory:
The adsorption theory explains the following .
1. Increase in the activity of a catalyst by increasing the surface area. Increase in the surface area of metals and metal oxides by reducing the particle size increases the rate of the reaction.

2. The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.

3. A promoter or activator increases the number of active centres on the surfaces.

Samacheer Kalvi 12th Chemistry Surface Chemistry Additional Questions

Samacheer Kalvi 12th Chemistry Surface Chemistry 1 mark Questions and Answers

I. Choose the correct answer and write it. Answers are in bold it.

Question 1.
Which one of the following is used to absorb colourants from sugar?
(a) Silica gel
(b) Magnesia
(c) Charcoal
(d) Alumina
Answer:
(c) Charcoal

Question 2.
Silica gel is usually adsorb ………….
(a) Colourants
(b) Hydrogen
(c) Liquid Helium
(d) Water
Answer:
(d) Water

Question 3.
Which one of the following is called adsorbate?
(a) Charcoal
(b) Silica gel
(c) Ammonia
(d) Magnesia
Answer:
(c) Ammonia

Question 4.
Which of the following can act as adsorbent?
(a) Silica gel
(b) Ammonia
(c) Colourants
(d) Water
Answer:
(a) Silica gel

Question 5.
The surface of separation of two phases where the concentration of adsorbed molecule is high is known as …………..
(a) adsorbate
(b) adsorbent
(c) interface
(d) residual phase
Answer:
(c) interface

Question 6.
Consider the following statement.
(i) High adsorption is the result of high surface area of the adsorbent.
(ii) The process of removing an adsorbed substance is called absorption.
(iii) Adsorbed substance is called an adsorbate.

Which of the above statement is / are not correct?
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (ii) only
(d) (iii) only
Answer:
(c) (ii) only

Question 7.
Which metal cannot act as adsorbent?
(a) Pt
(b) Ag
(c) Pd
(d) Al
Answer:
(d) Al

Question 8.
Consider the following statements.
(i) Adsorption is spontaneous process.
(ii) Adsorption is always accompanied by increase in free energy.
(iii) Adsorption is an endothermic reaction.

Which of the above statement is / are not correct?
(a) (i) only
(b) (ii) & (iii)
(c) (ii) & (i)
(d) (i) only
Answer:
(b) (ii) & (iii)

Question 9.
Absorption and adsorption if simultaneously occurs, it is called ……………
(a) occlusion
(b) sorption
(c) desorption
(d) dissolution
Answer:
(b) sorption

Questioin 10.
The process of sorption of gases on metal surface is called ……………
(a) Desorption
(b) Dissolution
(c) Occlusion
(d) Condensation
Answer:
(c) Occlusion

Question 11.
When gas molecules are held to the surface by the formation of chemical bond the heat energy released is nearly equal to
(a) 40 kJ/mole
(b) 800 kJ/mole
(c) 400 kJ/mole
(d) 4 kJ/mole
Answer:
(c) 400 kJ/mole

Question 12.
Which of the following is physical adsorption?
(a) Adsorption of H2 on nickel
(b) Friedel crafts reaction
(c) Synthesis of SO3 in the presence of NO
(d) Corrosion of iron
Answer:
(a) Adsorption of H2 on nickel

Question 13.
Which one of the following is chemical adsorption?
(a) Adsorption of O2 on tungsten
(b) Adsorption of ethyl alcohol vapours on nickel
(c) Adsorption of N2 on mica
(d) Rusting of iron
Answer:
(d) Rusting of iron

Question 14.
Which of the following occurs at low temperature?
(a) Adsorption of O2 on tun gsen
(b) Adsorption of N2 on mica
(c) Adsorption of ethyl alcohol vapours on nickel
(d) Adsorption of H2 on nickel
Answer:
(b) Adsorption of N2 on mica

Question 15.
Consider the following statements:
(i) Chemical adsorption is an instantaneous process
(ii) Multilayer of the adsorbate is formed on the adsorbent
(iii) Chemisorption involves the formation of activated complex.

Which of the above statement is / are not correct?
(a) (i) & (ii)
(b) (iii) only
(c) (i) only
(d) (ii) only
Answer:
(a) (i) & (ii)

Question 16.
Consider the following statements:
(i) In chemisorption, heat of adsorption is high
(ii) Monolayer of the adsorbate is formed during chemisorption
(iii) Physisorption increases with increase in temperature.

Which of the above statement is / are not correct’?
(a) (i) & (ii)
(b) (iii) only
(c) (ii) only
(d) (i) only
Answer:
(b) (iii) only

Question 17.
The extent of surface adsorption does not depend on
(a) Nature of the adsorbent
(b) Pressure
(c) Temperature
(d) Density
Answer:
(d) Density

Question 18.
Which of the following gases is not a permanent gas?
(a) NH3
(b) H2
(c) N2
(d) O2
Answer:
(a) NH3

Question 19.
Which of the following is liquefiable gas?
(a) SO2
(b) H2
(c) N2
(d) O2
Answer:
(a) SO2

Question 20.
Which one of the following is a permanent gas?
(a) NH3
(b) SO3
(c) N2
(d) CO2
Answer:
(c) N2

Question 21.
Consider the following statements.
(i) When pressure increases, the amount of physisorplion also increases.
(ii) Permanent gases like H2, N2 and O2 cannot be liquefied easily.
(iii) Lesser is the surface area, higher is the amount adsorbed.

Which of the above statement is / are correct?
(a) (i) & (ii)
(b) (iii) only
(c) (ii) only
(d) (i) only
Answer:
(a) (i) & (ii)

Question 22.
Which one of the following is used in blast furnace for drying air?
(a) Activated charcoal
(b) Silica gel
(c) Alumina
(d) Permutit
Answer:
(b) Silica gel

Question 23.
Which is employed in the softening of hardwater to absorb Ca2+ and Mg2+ ions?
(a) Alumina
(b) Silica gel
(c) Permutit
(d) Charcoal
Answer:
(c) Permutit

Question 24.
The formula for permutit is …………
(a) Ca Al2 Si4 O12
(b) CaAl3SiO2. xH2O
(c) Na2 Al2 Si4 O12
(d) Na2 SiO3
Answer:
(c) Na2 Al2 Si4 O12

Question 25.
Which one of the following is used to regenerate permutit in softening of hard water?
(a) Common salt
(b) Baking soda
(c) Washing soda
(d) Quick lime
Answer:
(a) Common salt

Question 26.
Which of the following is used to demineralise water?
(a) Permutit
(b) Common salt
(c) Ion exchange resin
(d) Charcoal

Question 27.
Which of the following is used during world war as gas masks?
(a) Permutit
(b) Silica gel
(c) Fuller’s earth
(d) Charcoal
Answer:
(d) Charcoal

Question 28.
Which of the following is used in petroleum refining and refining of vegetable oils?
(a) Charcoal
(b) Silica gel
(c) Pcrmutit
(d) Nickel
Answer:
(b) Silica gel

Question 29.
The catalyst used in the hydrogenation of oils to obtain vanaspathi is …………
(a) Iron
(b) Molybdenum
(c) Nickel
(d) Copper
Answer:
(c) Nickel

Question 30.
The catalyst and promoter used in Haber’s process are respectively ………..
(a) Mo, Fe
(b) Fe, Mo
(c) Pt, H2S
(d) Pt, V2O5
Answer:
(b) Fe, Mo

Question 31.
Which method is used for identification, detection and estimation of many substances even if they are in micro quantities?
(a) Lassaigne’s test
(b) Canus method
(c) Kjeldhals method
(d) Chromatography
Answer:
(d) Chromatography

Question 32.
Which one of the following is used in the identification of Al3+ ion in Al(OH)3?
(a) Red litmus
(b) Blue litmus
(c) Phenol red
(d) Sodium hydroxide
Answer:
(b) Blue litmus

Question 33.
Which ores are concentrated by froth floatation process?
(a) Oxide ore
(b) Carbonate ore
(c) Sulphate ores
(d) Suiphide ores
Answer:
(d) Suiphide ores

Question 34.
In froth floatation process, the lighter ore particles are wetted by …………..
(a) Olive oil
(b) Pine oil
(c) Soap oil
(d) Neem oil
Answer:
(b) Pine oil

Question 35.
Which one of the following is an example for homogeneous catalysis?
(a) Decomposition of acetaldehyde by 12 catalysts
(b) Hydrolysis of cane sugar with mineral acid
(c) Ester hydrolysis with alkali
(d) All the above
Answer:
(d) All the above

Question 36.
Which one of the following is an example for homogeneous catalysis?
(a) Manufacture of sulphuric acid by contact process
(b) Manufacture of ammonia by Haber’s process
(c) Oxidation of ammonia carried out in the presence of platinum gauze
(d) Hydrolysis of cane sugar with mineral acid
Answer:
(d) Hydrolysis of cane sugar with mineral acid

Question 37.
Which one of the following is an example for heterogeneous catalysis?
(a) Decomposition of acetaldehyde by I2 catalyst
(b) Decomposition of H2O2 in the presence of Pt catalyst
(c) Acid hydrolysis of ester
(d) Hydrolysis of cane sugar with mineral acid
Answer:
(b) Decomposition of H2O2 in the presence of Pt catalyst

Question 38.
Which one of the following is not an example for homogeneous catalysis?
(a) Contact process of manufacture of H2SO4
(b) Haber’s process of manufacture of NH3
(c) Acid hydrolysis of ester
(d) Freidel crafts reaction
Answer:
(c) Acid hydrolysis of ester

Question 39.
Consider the following statements:
(i) A catalyst needed in very small quantity
(ii) A catalyst can initiate a reaction
(iii) Catalyst are highly specific in nature

Which of the above statement is / are not correct?
(a) (i) & (iii)
(b) (ii) & (iii)
(c) (iii) only
(d) (ii) only
Answer:
(d) (ii) only

Question 40.
Consider the following statements.
(i) A solid catalyst will be more effective if it is taken in a finely divided form
(ii) A catalyst cannot initiate a reaction
(iii) For a chemical reaction4 catalyst is needed in very large quantity

Which of the above statement is / are not correct?
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) only
(d) (i) & (iii)
Answer:
(a) (i) & (ii)

Question 41.
The catalyst poison in contact process of manufacture of SO3 is …………
(a) As2O3
(b) H2S
(c) CO
(d) As2S3
Answer:
(a) As2O3

Question 42.
In Haber’s process of manufacture of ammonia, the Fe catalyst is poisoned by the pressure of …………….
(a) Mo
(b) Co
(c) H2S
(d) As2O3
Answer:
(c) H2S

Question 43.
In the reaction 2H2 + O2 → 2H2O acts as a catalytic poison for Pt catalyst.
(a) Co
(b) Mo
(c) As2O3
(d) H2S
Answer:
(a) Co

Question 44.
The negative catalyst in the decomposition of H2O2 is …………..
(a) Ethanol
(b) Acetic acid
(c) Ethanoic acid
(d) Methanol
Answer:
(a) Ethanol

Question 45.
Which one of the following is an example tar an autocalalysis?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-13

Question 46.
In the decomposition of hydrogen peroxide which acts as a negative catalyst?
(a) Dilute acid
(b) Glycerol
(c) a (or) b
(d) Ethanol
Answer:
(c) a (or) b

Question 47.
The energy required for the reactants to reach the activated complex is called …………
(a) threshold energy
(b) activation energy
(c) internal energy
(d) Gibbs free energy
Answer:
(b) activation energy

Question 48.
Which of the following is explained by intermediate compound formation theory?
(a) Mechanism of friedel crafts reaction
(b) Thermal decomposition of KClO3 in the presence of MnO2
(c) Oxidation of HCl by air in the presence of CuCl2
(d) Manufacture of NH3 by Haber’s process
Answer:
(d) Manufacture of NH3 by Haber’s process

Question 49.
Consider the following statements.
(i) Intermediate compound theory describes the specificity of a catalyst.
(ii) Intermediate compound theory explains the action of catalytic poison and activators.
(iii) Intermediate compound theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Which of the above statement is / are not correct?
(a) (ii) only
(b) (i) & (iii)
(c) (iii) only
(d) (i) & (ii)
Answer:
(a) (ii) only

Question 50.
Who explained the action of catalyst in adsorption theory?
(a) Berzellius
(b) Langmuir
(c) Thomas Graham
(d) Dalton
Answer:
(b) Langmuir

Question 51.
Consider the following statements.
(i) The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.
(ii) A promoter decreases the number of active centres on the surfaces.
(iii) Increase in the activity of a catalyst by increasing the surface area.

Which of the above statement is / are correct?
(a) (ii) only
(b) (iii) only
(c) (i) & (iii)
(d) (ii) & (iii)
Answer:
(c) (i) & (iii)

Question 52.
Which of the following catalyse the chemical reaction in living organism?
(a) enzymes
(b) protein
(c) lipids
(d) serum
Answer:
(c) lipids

Question 53.
Which of the following enzyme catalyse the hydrolysis of starch into maltose?
(a) maltase
(b) irivertase
(c) diastase
(d) zymase
Answer:
(c) diastase

Question 54.
Which enzyme catalyses the conversion of glucose into ethanol?
(a) maltase
(b) invertase
(c) diastase
(d) zymase
Answer:
(c) diastase

Question 55.
Which of the following act as catalyst in the oxidation of alcohol into acetic acid?
(a) pepsin
(b) diastase
(c) micoderma
(d) urease
Answer:
(c) micoderma

Question 56.
Which catalyst is used in the hydrolysis of urea?
(a) micoderma
(b) zymase
(c) pepsin
(d) urease
Answer:
(d) urease

Question 57.
Which of the following enzyme is present in soya beans?
(a) urease
(b) zymase
(c) pepsin
(d) lactase
Answer:
(a) urease

Question 58.
Consider the following statements.
(i) Enzymes are complex protein molecules with three dimensional structures.
(ii) Enzymes catalyse the chemical reaction in living organism.
(iii) Enzymes arc not specific in catalytic action.

Which of the above statement is J are correct?
(a) (iii) only
(b) (ii) & (iii)
(c) (i) & (ii)
(d) (i) & (iii)
Answer:
(c) (i) & (ii)

Question 59.
Consider the following statements.
(i) Enzyme catalysed reaction has maximum rate at optimum temperature
(ii) Enzyme catalysis is highly specific in nature
(iii) Catalytic activity of enzyme is decreased by coenzyrnes or activators.

Which of the above statement is / are not correct?
(a) (iii) only
(b) (i) only
(c) (ii) only
(d) (i) & (ii)

Question 60.
The temperature at which enzyme activity is high (or) maximum is called ………….
(a) critical temperature
(b) optimum temperature
(c) low temperature
(d) high temperature
Answer:
(b) optimum temperature

Question 61.
Enzymes can be active in human body at a temperature of ………….
(a) 98°F
(b) 105°F
(c) 37°F
(d) 50°F
Answer:
(a) 98°F

Question 62.
Consider the following statements.
(i) Zeolites are alumino silicates made of silicon and aluminium tetrahedra.
(ii) Zeolites carrying Na ions are used as basic catalyst.
(iii) As silicon is tetravalent and aluminium is trivalent, the zeolite matrix carries extra positive charge.

Which of the above statement is / are correct?
(a) (i) & (ii)
(b) (i), (ii) & (iii)
(c) (iii) only
(d) (ii) only
Answer:
(a) (i) & (ii)

Question 63.
Which one of the following is used in petrochemical industry for cracking heavy hydrocarbon fractions into gasoline, diesel etc.?
(a) permutit
(b) zeolite
(c) pepsin
(d) protein
Answer:
(b) zeolite

Question 64.
Which one of the following is used as a catalyst in the conversion of Lindane to cyclohexane?
(a) Fe°/Pd°
(b) Ni
(c) Zn + HCl
(d) LiAIH4
Answer:
(a) Fe°/Pd°

Question 65.
Which one of the following is used as catalyst in homogeneous and heterogeneous catalysis?
(a) enzymes
(b) zeolite
(c) nano catalyst
(d) coenzyme
Answer:
(c) nano catalyst

Question 66.
Who studied and analysed about colloids?
(a) Berzelius
(b) Thomas Graham
(c) Langmuir
(d) Robert Brown
Answer:
(b) Thomas Graham

Question 67.
Which one of the following is lyophillic colloid?
(a) Protein sol
(b) Gold sol
(c) Silver sol
(d) Platinum sol
Answer:
(a) Protein sol

Question 68.
Which one of the following is lyophobic colloid?
(a) Protein sol
(b) Starch sol
(c) Gel
(d) Gold sol
Answer:
(d) Gold sol

Question 69.
An example of liquid aerosol is ………..
(a) Soda water
(b) Milk
(c) Fog
(d) Inks
Answer:
(c) Fog

Question 70.
Which of the following is an example of Emulsion?
(a) mayonnaise
(b) shaving cream
(c) fumes
(d) paint
Answer:
(a) mayonnaise

Question 71.
The dispersed phase and dispersion medium in smoke, fumes and dust are …………..
(a) gas, solid
(b) solid, gas
(c) gas, liquid
(d) solid, liquid
Answer:
(b) solid, gas

Question 72.
Inks, paints arc considered as …………
(a) liquid in solid
(b) solid in liquid
(c) gas in gas
(d) solid in solid
Answer:
(b) solid in liquid

Question 73.
Which of the following is an example for gel?
(a) Pumice stone
(b) Pearls
(c) Coloured glass
(d) Butter
Answer:
(c) Coloured glass

Question 74.
Which one of the following is an example for solid sol?
(a) Butter
(b) Cheese
(c) Pearls
(d) Pumice stone
Answer:
(c) Pearls

Question 75.
Soda water is an example for ………..
(a) gel
(b) emulsion
(c) foam
(d) sol
Answer:
(c) foam

Question 76.
Colloidal ink and graphite are prepared by …………
(a) colloid mill
(b) Bredig’s arc
(c) ultrasonic homogenizer
(d) peptisation
Answer:
(a) colloid mill

Question 77.
Which method is used to prepare metal sols?
(a) ultrasonic dispersion
(b) mechanical dispersion
(c) Bredigs arc method
(d) peptisation
Answer:
(c) Bredigs arc method

Question 78.
Who prepared non aqueous inflammable liquids like Benzene and ether by Bredig’s arc method?
(a) George Bredig
(b) Sved berg
(c) Thomas Graham
(d) Robert Brown
Answer:
(b) Sved berg

Question 79.
Which method is used to prepare mercury colloid?
(a) peptisation
(b) mechanical dispersion
(c) ultrasonic dispersion
(d) Bredig’s arc method
Answer:
(c) ultrasonic dispersion

Question 80.
Mercury sol is obtained by subjecting it to sound waves of frequency more than ………..
(a) 20 Hz
(b) 20 kHz
(c) 200 kHz
(d) 2000 kHz
Answer:
(b) 20 kHz

Question 81.
The conversion of a precipitate into colloid is called …………..
(a) coagulation
(b) hydrolysis
(c) condensation
(d) peptisation
Answer:
(d) peptisation

Question 82.
Gold sol is prepared by reduction of auric chloride using …………..
(a) water
(b) HCHO
(c) CH3CHO
(d) CH3COOH
Answer:
(b) HCHO

Question 83.
Which method is suitable to prepare I2 sol and Se sol?
(a) Reduction
(b) Hydrolysis
(c) oxidation
(d) peptisation
Answer:
(c) oxidation

Question 84.
Which condensation method is used to prepare sulphur sol?
(a) Hydrolysis
(b) Decomposition
(c) Reduction
(d) Peptisation
Answer:
(b) Decomposition

Question 85.
Arsenic sulphide colloid is prepared by ………..
(a) hydrolysis
(b) reduction
(c) double decomposition
(d) decomposition
Answer:
(c) double decomposition

Question 86.
By which method phosphorous colloid can be prepared’?
(a) Decomposition
(b) Exchange of solvent
(c) Hydrolysis
(d) Reduction
Answer:
(b) Exchange of solvent

Question 87.
Which one of the following is not used to purify colloids?
(a) Dialysis
(b) Peptisation
(c) Electro dialysis
(d) Uhrafilteration
Answer:
(b) Peptisation

Question 88.
The process of conversion of colloidal solution into precipitate is known as …………..
(a) peptisation
(b) dispersion
(c) coagulation
(d) decomposition
Answer:
(c) coagulation

Question 89.
Which one of the following is named collodion?
(a) 4% solution of nitro cellulose in a mixture of alcohol and water
(b) 40% solution of cellulose acetate in acetic acid.
(c) agar-agar along with gel
(d) semipermeable membrane
Answer:
(a) 4% solution of nitro cellulose in a mixture of alcohol and water

Question 90.
Which of the following is the size of the colloidal particle?
(a) 100 μm diameter – 1000 μm diameter
(b) 1 mμ to 1 μm diameter
(c) 1 mμ to 100 μm diameter
(d) 1 μm to 1 μm diameter
Answer:
(b) 1 mμ to 1 μm diameter

Question 91.
Consider the following statements.
(i) Colloidal solutions are quite stable and are not affected by gravity
(ii) Colloids diffuse more readily through membranes
(iii) Colloidal solutions show colligative properties

Which of the above statement is / are correct?
(a) (i) & (iii)
(b) (ii) & (iii)
(c) (ii) only
(d) (iii) only
Answer:
(a) (i) & (iii)

Question 92.
The shape of tungstic acid W3O5 sol is ………….
(a) spherical
(b) disc
(c) plate like
(d) rod like
Answer:
(d) rod like

Question 93.
Which one of the following colloid has spherical shape?
(a) AS2S3
(b) Fe(OH)3
(c) W3O5
(d) dust
Answer:
(a) AS2S3

Question 94.
Tyndall effect is possible in colloid due to ……………
(a) absorption of light
(b) adsorption of light
(c) scattering of light
(d) reflection of light
Answer:
(c) scattering of light

Question 95.
Which one of the following does not show Tyndali effect and Brownian movement?
(a) Milk
(b) common salt solution
(c) smoke
(d) tungstic acid sol
Answer:
(b) common salt solution

Question 96.
The migration of sol particles under the influence of electric field is called ……………
(a) electro osmosis
(b) electro dialysis
(c) electrophoresis
(d) dialysis
Answer:
(c) electrophoresis

Question 97.
Which one of the following is used for detection of pressure of charge on sol particles?
(a) Cataphoresis
(b) Electro dialysis
(c) Dialysis
(d) Ultrafilteration
Answer:
(a) Cataphoresis

Question 98.
Which of the following is positively charged colloid?
(a) haemoglobin
(b) starch
(c) clay
(d) AS2S3
Answer:
(a) haemoglobin

Question 99.
Which one of the following is a positively charged colloid?
(a) Ag
(b) AU
(c) Basic dyes
(d) Clay
Answer:
(c) Basic dyes

Question 100.
Which one of the following is a negatively charged colloid?
(a) Pt
(b) Al(OH)3
(c) Fe (OH)3
(d) Basic dyes
Answer:
(a) Pt

Question 101.
Which one of the following is a negatively charged colloid?
(a) Ferric hydroxide
(b) Clay
(c) Basic dyes
(d) Haemoglobin
Answer:
(b) Clay

Question 102.
The movement of dispersion medium under the influence of electric potential is called ………….
(a) Electrophoresis
(b) Cataphoresis
(c) Electro osmosis
(d) Electro dialysis
Answer:
(c) Electro osmosis

Question 103.
Which one of the following is added to gold sol to protect it?
(a) Gelatine sol
(b) Gum
(c) Starch
(d) Basic dye
Answer:
(a) Gelatine sol

Question 104.
Consider the following statements.
(i) Smaller the gold number, greater the protective power
(ii) Greater the gold number, greater the protective power
(iii) Colloidal sols with opposite charges are mixed, mutual coagulation takes place.

Which of the above statement is / are not correct?
(a) (i) only
(b) (i) & (iii)
(c) (ii) only
(d) (ii) & (iii)
Answer:
(c) (ii) only

Question 105.
Which one of the following can act as emulsifier?
(a) glue
(b) dye
(c) water
(d) starch
Answer:
(a) glue

Question 106.
Which one of the following is not used to identify the types of emulsion?
(a) dye test
(b) viscosity test
(c) conductivity test
(d) Tollen’s test
Answer:
(d) Tollen’s test

Question 107.
By adding which one of the following oil in water emulsion containing potassium soap can be converted into water in oil emulsion?
(a) MCl3
(b) NaCI
(c) KCI
(d) C6H5Cl
Answer:
(a) MCl3

Question 108.
Which of the following colloid is used as a medicine for stomach troubles?
(a) colloidal Au
(b) colloidal Ca
(c) milk of magnesia
(d) silver sol
Answer:
(c) milk of magnesia

Question 109.
Which one of the following is used in the purification of drinking water?
(a) silver sol protected by gelatine
(b) milk of magnesia
(c) Alum containing Al3+
(d) Argyrol
Answer:
(c) Alum containing Al3+

Question 110.
Which of the following is used as tonics?
(a) milk of magnesia
(b) Argyrol
(c) colloidal Au & colloidal Ca
(d) Alum
Answer:
(c) colloidal Au & colloidal Ca

Question 111.
Which one of the following is used in tanning of leather?
(a) chromium salt
(b) colloidal Au
(c) Argyrol
(d) Fe (OH)3
Answer:
(a) chromium salt

Question 112.
Carbon dust in air is solidified by ………..
(a) cottrell’s precipitator
(b) colloidal mill
(c) Bredig’s arc
(d) peptisation
Answer:
(a) cottrell’s precipitator

Question 113.
Which of the following voltage is used in cottrell’s precipitator?
(a) 5000 V
(b) 50,000 V
(c) 1,000V
(d) 10,000V
Answer:
(b) 50,000 V

Question 114.
The blue colour of the sky is due to …………
(a) coagulation
(b) peptisation
(c) Tyndall effect
(d) Brownian movement
Answer:
(c) Tyndall effect

Question 115.
Which one of the following is used to distinguish Natural honey and artificial honey?
(a) Ammoniacal AgNO3
(b) Fehling’s solution
(c) Arsenic suiphide sol
(d) gelatin
Answer:
(a) Ammoniacal AgNO3

Question 116.
Which one of the following is the catalyst poison in Haber’s process?
(a) AS2S3
(b) AS2O3
(c) Co
(d) H2S
Answer:
(d) H2S

Question 117.
Which one of the following is an example for water in oil emulsion?
(a) Milk
(b) Vanishing cream
(c) Butter
(d) Soap
Answer:
(c) Butter

Question 118.
Which of the following is contributed towards the extra stability of lyophillic colloid?
(a) Hydration
(b) Charge
(c) Colour
(d) Tyndall effect
Answer:
(a) Hydration

Question 119.
A catalyst is a substance which
(a) increases the equilibrium concentration of the product
(b) changes the equilibrium constant of the reaction
(c) shortens the time to reach equilibrium
(d) supplies energy to the reaction
Answer:
(c) shortens the time to reach equilibrium

Question 120.
The ability of an ion to bring about coagulation of a given colloid depends upon …………
(a) its size
(b) magnitude of its charge
(c) the sign of its charge
(d) both the magnitude and sign of the charge
Answer:
(d) both the magnitude and sign of the charge

Question 121.
Which one of the following is an incorrect statement for physisorption?
(a) It is a reversible process
(b) It requires less heat of adsorption
(c) It requires activation energy
(d) It take place at low temperature
Answer:
(c) It requires activation energy

Question 122.
Which is not a colloid?
(a) Chlorophyll
(b) Egg
(c) Ruby glass
(d) Milk
Answer:
(a) Chlorophyll

Question 123.
Which of the following electrolytes is most effective in the coagulation of gold sol?
(a) NaNO3
(b) K4[Fe(CN)6]
(c) Na3PO4
(d) MgCl2
Answer:
(b) K4[Fe(CN)6]

Question 124.
Gold number gives ………………..
(a) the amount of gold present in the colloid
(b) the amount of gold required to break the colloid
(c) the amount of gold required to protect the colloid
(d) the measure of protective power of a lyophillic colloid
Answer:
(d) the measure of protective power of a lyophillic colloid

Question 125.
Identify the gas which is readily adsorbed by activated charcoal?
(a) N2
(b) SO2
(c) H2
(d) O2
Answer:
(b) SO2

Question 126.
Starch dispersed in hot water is an example of …………..
(a) emulsion
(b) hydrophobic sol
(c) lyophilic sol
(d) associated colloid
Answer:
(c) lyophilic sol

Question 127.
Which one is an example of gel?
(a) soap
(b) cheese
(c) milk
(d) fog

Question 128.
The random, zig-zag motion of colloidal particles in the dispersion medium is referred to as …………..
(a) Electrophoresis
(b) Brownian movement
(c) Tyndall effect
(d) Electro osmosis
Answer:
(b) Brownian movement

Question 129.
Which of the following electrolytes is least effective in causing flocculation of ferric hydroxide sol?
(a) K4 [Fe(CN)6]
(b) K2CrO4
(c) KBr
(d) K2SO4
Answer:
(c) KBr

Question 130.
Gelatin is mostly used in making icecream in order to …………..
(a) prevent making of colloid
(b) to stabilize the colloid and to prevent the crystallization
(c) to stabilise the mixture
(d) to enrich the aroma
Answer:
(b) to stabilize the colloid and to prevent the crystallization

Question 131.
Which one of the following is not a colloidal solution?
(a) smoke
(b) ink
(c) air
(d) coffee
Answer:
(c) air

Question 132.
Milk can be preserved by adding a few drops of …………
(a) HCOOH
(b) HCHO
(c) CHCOOH
(d) CH3CHO
Answer:
(b) HCHO

Question 133.
Bleeding is stopped by the application of ferric chloride. This is because …………
(a) ferric chloride seal the blood cells
(b) blood starts flowing in ohter direction
(c) blood is coagulated and blood vessel is sealed
(d) blood is peptised
Answer:
(c) blood is coagulated and blood vessel is sealed

Question 134.
Delta at the rivers are formed due to …………
(a) peptisation
(b) coagulation
(c) hydrolysis
(d) precipitation
Answer:
(b) coagulation

Question 135.
Alum purifies muddy water by …………..
(a) dialysis
(b) adsorption
(c) coagulation
(d) forming a true solution
Answer:
(c) coagulation

Question 136.
Reactions of zeolite catalysts depend on ………….
(a) pores
(b) apertures
(c) size of cavily
(d) all of these
Answer:
(d) all of these

Question 137.
chemisorption ……………
(a) increases with increase in temperature
(b) decreases with increase in temperature
(c) remains unaffected by the, change of temperature
(d) first increases and then decreases.
Answer:
(d) first increases and then decreases.

Question 138.
Adsorption is always …………
(a) endothermic
(b) exothermic
(c) iso thermic
(d) either a (or) b
Answer:
(a) endothermic

Question 139.
Which one of the following can be explained by the adsorption theory?
(a) Homogeneous catalysis
(b) Acid-base catalysis
(c) Heterogeneous catalysis
(d) Enzyme catalysis
Answer:
(c) Heterogeneous catalysis

Question 140.
Physical adsorption is inversly proportional to ……………..
(a) volume
(b) concentration
(c) temperature
(d) all of these
Answer:
(c) temperature

Question 141.
Noble gases are adsorbed by ……………
(a) anhydrous CaCl2
(b) Fe(OH)3
(c) Conc. H2SO4
(d) activated charcoal
Answer:
(d) activated charcoal

Question 142.
Animal charcoal is used in decolourising agent in the manufacture of sugar because it is a good …………..
(a) adsorbate
(b) adsorbent
(c) oxidising agent
(d) dehydrating agent
Answer:
(a) adsorbate

Question 143.
Gold number is associated only with …………
(a) lyophobic colloids
(b) lyophilic colloids
(c) both lyophobic and lyophilic colloids
(d) Au in water
Answer:
(b) lyophilic colloids

Question 144.
Which of the following forms a colloidal solution with water?
(a) NaCl
(b) Glucose
(c) Starch
(d) Barium sulphate
Answer:
(c) Starch

Question 145.
Which one of the following is an example for homogeneous catalysis?
(a) Hydrogenation of oil
(b) manufacture of NH3 by Haber’s process
(c) manufacture of sulphuric acid by contact process
(d) hydrolysis of sucrose in the presence of dilute hydrochloric acid
Answer:
(d) hydrolysis of sucrose in the presence of dilute hydrochloric acid

Question 146.
Which of the following does not involve coagulation?
(a) peptisation
(b) formation of delta regions
(c) treatment of drinking water by potash alum
(d) clotting of blood by the use of ferric chloride
Answer:
(a) peptisation

Question 147.
Among the electrolytes Na2SO4, CaCI2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent for Sb2S3 Sol is ……………
(a) Na,SO4
(b) CaCl2
(c) Al2(SO4)3
(d) NH4Cl
Answer:
(c) Al2(SO4)3

Question 148.
Which of the following statement is incorrect regarding physisorption?
(a) It occurs because of Van der Waals forces
(b) more easily liquefiable gases are adsorbed easily
(c) under high pressure, it resuLts into multimolecular layer on adsorbent surface
(d) enthalpy of adsorption is low and positive
Answer:
(d) enthalpy of adsorption is low and positive

Question 149.
Rate of physical adsorption increase with ………..
(a) increase in temperature
(b) decrease in pressure
(c) decrease in temperature
(d) decrease in surface are
Answer:
(c) decrease in temperature

Question 150.
Gold numbers of protective eolloids, A, B, C and D are respectively 0.50, 0.01, 0.10 and 0.005. The correct order of the stability of colloids is ……..
(a) B < D < A < C
(b) D < A < C < B
(c) C < B < D < A
(d) A < C < B < D
Answer:
(d) A < C < B < D

II. Fill in the blanks

  1. The surface of separation of the two phases where the concentration of adsorbed molecule is high is known as …………..
  2. In adsorption, if the concentration of a substance in the interface is high, it is called ……………
  3. The process of removing a adsorbed substance from the surface is called ……………
  4. Adsorption is always accompanied by decrease in ……………
  5. A term …………… is used for sorption of gases on metal surfaces
  6. M.C. Bain introduced a term …………… to represent the simultaneous adsorption and absorption.
  7. In chemical adsorption, gas molecules are held to the surface by formation of chemical bond and nearly …………… is given out as heat of adsorption
  8. In physical adsorption …………… exist between adsorbent and adsorbate
  9. Heat of adsorption is low hence physical adsorption occurs at ……………
  10. …………… involves the formation of activated complex with appreciable activation energy.
  11. Adsorption occurs at fixed sites called ……………
  12. Gases like NH3, SO3 and CO2 are …………… as have greater Van der Waals force of attraction.
  13. gases like H2, N2, O2 have low critical temperature and ………….. slowly
  14. A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorbate at constant temperature is called …………..
  15. Freundlich adsorption isothermal equation is applicable fo adsorption of …………..
  16. During World War I ………….. gas mask was employed
  17. ………….. is used to create high vaccum in vessels.
  18. In blast furnace ………….. is used for drying air
  19. For dehydration and also purification of gases like CO2, N2, CI2, O2 and He ………….. and are employed.
  20. …………..is employed in the softening of hard water to adsorb Ca2+ and Mg2+ ions
  21. In the process of softening of hard water, exhausted permutit is regenerated by adding a solution of …………..
  22. ………….. are used to demineralise water.
  23. ………….. and ………….. are used in petroleum refining and refining of vegetable oil.
  24. ………….. is used to decolourising agent in manufacture of sugar from molasses.
  25. In forth floation process, the suiphide are particles are wetted by …………..
  26. ………….. is defined as a substance which alters the rate of a chemical reaction without itself undergoing chemical change.
  27. The decomposition of acetaldehyde by I2 catalyst is an example of ………….. catalysis.
  28. Manufacture of sulphuric acid by contact process is an example of ………….. catalysis.
  29. Friedel crafts reaction is an example of ………….. catalysis.
  30. The substances increases the activity of a catalyst are called …………..
  31. ………….. or ………….. used as promoter for iron in Haber’s process
  32. ………….. destroys the activity of platinum in contact process
  33. In the reaction CH3COOC2H5 + H2O → CH3COOH + C2H5OH ………….. is act as auto calyst.
  34. The decomposition of H2O2 rate is decreased by …………..
  35. As …………..is lowered in the presence of catalyst, more molecules take part in the reaction and hence the rate of the reaction increases.
  36. The mechanism of friedel crafts reaction is explained by ………….. theory.
  37. The catalyst used for the oxidation of HCl by air is …………..
  38. Thermal decomposition of KClO3 in the presence of ………….. follows ………….. theory.
  39. Intermediate compound formation theory is unable to explain the mechanism of …………..
  40. Hydrogenation of ethylene in the presence of nickel catalyst follows ………….. theory.
  41. ………….. are complex protein molecules and catalyse the chemical reaction in living organism.
  42. The enzyme ………….. hydrolyses starch into maltose.
  43. The enzyme ………….. oxidises alcohol into acetic acid.
  44. Enzyme catalysed reaction has ………….. rate at optimum temperature.
  45. ………….. in hibits the action of bacteria and used for curing pneumonia.
  46. ………….. are microporous, hydrated alumino silicates.
  47. Zeolites carrying ………….. are used as basic catalysts
  48. Like heterogeneous catalyst ………….. can be recovered and recycled.
  49. Sols of gold, silver, platinum and copper are examples of …………..
  50. So is of protein and starch are examples of …………..
  51. A liquid in liquid colloid is called …………..
  52. Pearls, opals and Ruby red glass are belong the colloid named …………..
  53. Rubber forns colloidal solution with …………..
  54. In colloid mill, the two metal plates rotating in opposite directions of nearly ………….. revolution per minute.
  55. Colloidal solutions of ink and graphite are prepared by ………….. method.
  56. A brown colloidal solution of platinum was prepared by ………….. in 1898.
  57. ………….. is added as an stabilising agent for making platinum colloid.
  58. Metal hydroxide is added as an ………….. for making noble metal sols.
  59. Claus obtained ………….. by subjecting ………….. to high frequency ultrasonic vibrations,
  60. In the preparation of AgCI colloid from AgCl precipitate, the peptising or dispersing agent used is …………..
  61. Gold sol is prepared by reduction of auric chloride using …………..
  62. Arsenic sulphide colloid can be prepared by ………….. method.
  63. 12 sol is obtained from HIO3 by ………….. method.
  64. The process of conversion of colloidal solution into precipitate is called …………..
  65. In the dialysis of kidney. recycling of patient’s blood is done through semipermeable tube in an ………….. solution
  66. Collodion is 4% solution of ………….. in a mixture of …………..
  67. The size of colloidal particles ranges from …………..
  68. The shape of blue gold sol (or) Fe(OH)3 sol is …………..
  69. Pollen grains suspended in water showed …………..
  70. The flocculation and setting down of sol particles is called …………..
  71. When the valency of ion is high ………….. power is increased in colloids.
  72. The smaller the ………….. value, greater will be precipitation of colloids.
  73. ………….. is added to gold sol to protect it.
  74. ………….. introduced the term gold number as a measure of protecting power of a colloid.
  75. An oil in watrer emulsion containing potassium soap as emulsifying agent can be converted into water in oild emulsion by adding ………….. or …………..
  76. Synthetic polymers like polystyrene, silicones and PVC, are …………..
  77. ………….. colloid is used as eye lotion.
  78. ………….. protected by gelatin is known as Argyrol.
  79. ………….. salts arc used in tanning of leather.
  80. Natural honey is distinguished artificial honey by adding …………..

Answer:

  1. interface
  2. positive adsorption
  3. desorption
  4. free energy
  5. occlusion
  6. sorption
  7. 400kJ/rnole
  8. Van der Waals forces of attraction
  9. low temperature
  10. Chemisorption
  11. active centres
  12. easily liquefiable
  13. Permanent, adsorbed
  14. Adsorption isotherm
  15. gases on solid surface
  16. charcoal
  17. Activated charcoal
  18. silica gel
  19. alumina, silica
  20. Permutit
  21. common salt
  22. Con exchange resins
  23. Fuller’s earth, silica gel
  24. Animal charcoal
  25. pine oil
  26. catalyst
  27. homogeneous
  28. heterogeneous
  29. heterogeneous
  30. promoters
  31. Mo (or) Al2O3
  32. AS2O3 catalyst poison
  33. CH3COOH
  34. ethanol (or) Glycerol, negative catalyst
  35. Activation energy
  36. intermediate compound formation
  37. CuCI2
  38. MnO2, intermediate compound formation theory
  39. heterogeneous catalysis
  40. adsorption
  41. enzymes
  42. diastase
  43. micoderma aceti
  44. maximum
  45. penicillin
  46. zeolites
  47. Na ions
  48. nano catalyst
  49. lyophobic sols (or) irreversible sols
  50. lyophilic sols (or) reversible sols
  51. emulsion
  52. solid sol
  53. benzene
  54. 7000
  55. mechanical dispersion
  56. George Bredig
  57. alkali hydroxide
  58. stabilising agent
  59. mercury sol, mercury
  60. HCl
  61. formaldehyde
  62. double decomposition
  63. oxidation
  64. coagulation
  65. isotonic saline
  66. nitro cellulose, alcohol and water
  67. 1 mi to llim diameter
  68. disc or plate like
  69. Brownian movement
  70. coagulation
  71. precipitation
  72. flocculation
  73. geLatine sol
  74. Zsigmondy
  75. CaCl2, AlCl3
  76. colloids
  77. Argyrol
  78. Silver sol
  79. chromium
  80. Tollens reagent (or) Ammoniacal AgNO3

III. Match the column I & II using the code given below the coulum.

Question 1.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-14
Answer:
(a) 4 3 1 2

Question 2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-15
Answer:
(c) 3 4 1 2

Question 3.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-16
Answer:
(a) 4 3 1 2

Question 4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-17
Answer:
(a) 2 3 4 1

Question 5.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-19
Answer:
(b) 3 4 2 1

Question 6.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-20
Answer:
(a) 3 1 4 2

Question 7.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-21
Answer:
(a) 4 1 2 3

Question 8.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-23
Answer:
(a) 4 3 2 1

Question 9.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-25
Answer:
(a) 2 4 1 3

Question 10.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-26
Answer:
(a) 4 3 1 2

Question 11.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-27
Answer:
(c) 3 1 4 2

Question 12.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-28
Answer:
(a) 4 3 1 2

Question 13.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-29
Answer:
(d) 3 1 4 2

Question 14.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-30
Answer:
(b) 4 3 2 1

Question 15.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-31
Answer:
(a) 2 4 1 3

Question 16.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-32
Answer:
(b) 3 4 2 1

Question 17.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-33
Answer:
(a) 4 1 2 3

Question 18.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-34
Answer:
(a) 3 1 4 2

IV. Assertion and Reason

Question 1.
Assertion (A) : Absorption is a bulk phenomenon.
Reason (R) : The absorbed molecules are distributed throughout the absorbent.
(a) Both A and R are correct and R is the correct explanation of A.
(b) A is correct but R is wrong.
(c) A is wrong but R is correct.
(d) Both A and R are wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 2.
Assertion (A) : Adsorption is a spontaneous process.
Reason (R) : Adsorption is always accompanied by decrease in free energy. When molecules are adsorbed, there is always a decrease in randomness of the molecules.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) Both A and R are wrong.
(d) A is correct but R is wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 3.
Assertion (A) : Chemical adsorption is an exothermic process.
Reason (R) : In chemical adsorption, gas molecules are held to the surface by formation of chemical bonds. Since strong bond is formed, nearly 400kJ/mole is given out as heat adsorption.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) Both A and R are wrong.
(d) A is correct but R is wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 4.
Assertion (A) : Physical adsorption occurs at low temperature.
Reason (R) : ¡n physical adsorption, weak Van der Waals force of attraction exist. Other weak forces exist in physical adsorption are dipole-dipole interaction and dispersion forces. As these forces are weak, heat of adsorption is low.
(a) Both A and R are correct but R is not correct explanation of A.
(b) Both A and R are correct and R is the correct explanation of A.
(c) Both A and R are wrong.
(d) A is correct but R is wrong.
Answer:
(b) Both A and R are correct and R is the correct explanation of A.

Question 5.
Assertion (A) : Platinised asbestos is a better adsorbent than platinum block.
Reason (R) : Higher the surface area, higher is the amount adsorbed. In platinum coated asbestos the surface area is more and so it act as a better adsorbent.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are wrong.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 6.
Assertion (A) : Gases like SO2, NH3 and CO2 are readily adsorbed.
Reason (R) : SO2, NH3 and CO2 are easily liquefiable as have greater van der WaaI’s forces of attraction and adsorbed readily.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and Rare wrong.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 7.
Assertion (A) : Permanent gases like H2, N2 and O2 cannot be adsorbed readily.
Reason (R) : Permanent gases having low critical temperature and adsorbed slowly.
(a) Both A and R are wrong.
(b) A is correct and R is the correct explanation of A.
(c) A is wrong but R is correct.
(d) A is correct but R is wrong.
Answer:
(b) A is correct and R is the correct explanation of A

Question 8.
Assertion (A) : Chromatography is a very effective method and used for identification, detection and estimation of micro quantities of many substances.
Reason (R) : Chromatography technique is applied for separation and detection of components in a mixture it is mainly based on adsorption of components on the surface of adsorbents.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 9.
Assertion (A) : Ester hydrolysis of acid (or) alkali catalyst is an example of homogeneous catalysis.
Reason (R) : Ester, H2O acid (or) alkali and the products are in liquid form.
(a) Both A and R are correct but R is the correct explanation of A.
(b) Both A and R are wrong.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct,
Answer:
(a) Both A and R are correct but R is the correct explanation of A.

Question 10.
Assertion (A) : The manufacture of sulphuric acid by contact process is an example of heterogeneous catalysis.
Reason (R) : The catalyst Pt (or) VO5, reactants and products are in different phases in contact process.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) Both A and R are wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 11.
Assertion (A) : Acid hydrolysis of ethylacetate by water to produce acetic acid and ethanol is an example of auto catalysis.
Reason (R) : In acid hydrolysis of ester, the product acetic acid act as catalyst and this process is called autocatalysis.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are wrong.
(c) A is correct hut R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 12.
Assertion (A) : Effective and efficient conversion is the special characteristic of enzyme catalysed reactions.
Reason (R) : An enzyme may transform a million molecules of reactants to products in a minute.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 13.
Assertion (A) : lyophillic colloids will not get precipitated easily.
Reason (R) : In lyophillic colloids, definite attractive forces exists between dispersion medium and dispersed phase and they are more stable.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) Both A and R are wrong.
(d) A is correct but R is wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 14.
Assertion (A) : Lyophobic colloids like sols of gold will precipitate readily.
Reason (R) : In lyophobic colloids, no attractive force exists between the dispersed phase and dispersion medium and are less stable.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) Both A and R are correct.
(d) A is correct but R is wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 15.
AssertIon (A) : Iron colloid cannot be prepared by Bredig’s arc method.
Reason (R) : iron cannot react with alkali hydroxide stabilising agent added in water.
(a) Both A and R are correct and R is the correct explanation of A.
(b) A is correct but R is wrong.
(c) A is wrong but R is correct.
(d) Both A and R are wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

V. Find the odd one out.

Question 1.
He, Ne, O2, N2, Pt
Answer:
Pt. It is adsorbent and all others are adsorbates.

Question 2.
SO2, NH3, NaCI, Silica gel
Answer:
Silica gel. It is an adsorbent whereas others are adsorbates.

Question 3.
Silica gel, Pt, Ag, Pd, NH3
Answer:
NH3 It is an adsorbate whereas others are adsorbents.

Question 4.
Coconut charcoal, silica gel, mica, SO2, Animal charcoal
Answer:
SO2 It is an adsorbate whereas others are adsorbents.

Question 5.
Mica, Nickel, Charcoal, Tungsten, Ethyl alcohol vapours
Answer:
Ethyl alcohol vapours. It is an adsorbate whereas others are adsorbents.

Question 6.
Pt, Glycerol, MnO2, Ni, I2
Answer:
Glycerol. It is a negative catalyst whereas others are positive catalyst.

Question 7.
Fe, Anhydrous AlCl3, V2O5, Pt, Ethyl alcohol
Answer:
Ethyl alcohol is a negative catalyst whereas others are positive catalyst.

Question 8.
Decomposion of acetaldehyde by I2 Decomposition of H2O2 by Pt, Ester hydrolysis with acid, Hydrolysis of cane sugar.
Answer:
Decomposition of H2O2 by Pt. It is heterogeneous catalysis whereas others are homogeneous catalysis.

Question 9.
Friedel crafts reactopm, Haner’s process, Hydrolysis of cane sugar, contact process.
Answer:
Hydrolysis of cane sugar. It is homogeneous catalysis whereas others are heterogeneous catalysis.

Question 10.
Pepsin, Zymase, Maltase, Diastase, Maltose, Urease.
Answer:
Maltose. It is a carbohydrate whereas others are enzyme catalysts.

Question 11.
Pt, Ni, Fe°/Pd°, Fe/Mo, Sn/HCI
Answer:
Fe°/Pd°. It is a nano catalyst where as others ordinary catalyst.

Question 12.
Milk, coffee, smoke, common salt solution, dust.
Answer:
Common salt solution. It is a true solution whereas others are colloids.

Question 13.
Soda water, Butter, Starch solution, Cheese, Cream.
Answer:
Starch solution. It is a suspension whereas others are colloids.

Question 14.
Ink, Milk, Cream, Mayonnaise.
Answer:
Ink. It is a sol whereas others are Emulsion.

Question 15.
Pearls, Opals, Coloured glass, Alloys, Pumice stone.
Answer:
Pumice stone. It is a solid foam whereas others are solid sol.

Question 16.
Smoke, Froth, Fumes, Dust, Air pollutants.
Answer:
Froth. It is a foam whereas others are solid aerosol.

Question 17.
Pumice stone, Foam, Milk, Rubber band.
Answer:
Milk. It is an emulsion whereas others are solid foam.

Question 18.
Mechanical dispersion, Bredigts arc method, Peptisation, Double decomposition, Ultrasonic dispersion.
Answer:
Double decomposition. It is a condensation method of preparation of colloids whereas others are dispersion methods of preparation of colloids.

Question 19.
Oxidation, Peptisation, Reduction, Decomposition, Hydrolysis.
Answer:
Peptisation. It is a dispersion method of preparation ofcoiloid whereas others are condensation methods of preparation of colloids.

Question 20.
Dialysis, electrophoresis, Ultrafi iteration. Electrodialysis.
Answer:
Electrophoresis is an electrical property of colloids but others are purification methods of colloids.

Samacheer Kalvi 12th Chemistry Surface Chemistry 2 Marks Questions and Answers

Question 1.
Define

  1. Adsorbent
  2. Adsorbate with an example.

Answer:

  1. Adsorbent is a material on which adsorption takes place. eg., silica gel and metals like Ni, Cu, Pt.
  2. Adsorbate is a substance which is adsorbed on the adsorbent. e.g., Gaseous molecules like He, Ne, O2, N2 and solutions of NaCI (or) KCI.

Question 2.
Define

  1. Interface
  2. Desorption.

Answer:
1. Interface:
The surface of separation of the two phases where the concentration of adsorbed molecule is high is known as interface.

2. Desorption:
The process of removing an adsorbed substance from the substance is called desorption.

Question 3.
What is adsorption? What is meant by positive and negative adsorption?
Answer:
Adsorption is the odhesion of atoms, ions or molecules from a gas, liquid or dissolved solid to a surface. In adsorption, if the concentration of a substance in the interface is high, then it is called positive adsorption. 1f it is less, then it is called negative adsorption.

Question 4.
Define chemical adsorption? Give example.
Answer:
Chemical adsorption is a process in which gas molecules are held to the surface by formation of chemical bonds. e.g., Adsorption of O2 on tungsten, Adsorption of H2 on nickel.

Question 5.
Chemical adsorption is an exothermic process. Justify this statement.
Answer:
In chemical adsorption, gas molecules are held to the surface by the formation of strong bond. Due to this 400kJ/mole is given out as heat of adsorption. Since heat is evolved, it is an exothermic process.

Question 6.
Why physical adsorption take place at low temperature?
Answer:
In physical adsorption, the forces between the adsorbent and adsorbate are very weak, heat of adsorption is low and hence physical adsorption occurs at low temperature.

Question 7.
What are the forces exist in physical adsorption?
Answer:
In physical adsorption, physical forces like van der Waals force of attraction exist between adsorbent and adsorbate.
The other forces that can cause physical adsorption are

  1. dipole-dipole interaction
  2. dispersion forces.

Question 8.
What is physical adsorption? Give example.
Answer:
Physcial adsorption is a process in which adsorbate are attached to adsorbent by means of weak van der Waals forces of attraction. e.g.,

  1. Adsorption of N2 on mica
  2. Adsorption of gases on charcoal.

Question 9.
Finely divided Nickel is a better adsorbent than Nickel crystal.
Answer:
As the adsorption is a surface phenomenon it depends on the surface area of adsorbent. i.e., higher the surface area, higher is the amount adsorbed. Finely divided Nickel has longer surface area and it is a better adsorbent than Nickel crystal.

Question 10.
NH3, CO2 are readily adsorbed where as H2, N2 are slowly adsorbed. Give reason.
Answer:
1. The nature of adsorbate can influence the adsorption. Gases like NH3 CO2 are easily liquefiable as have greater van der Waals forces of attraction and hence readily adsorbed due to high critical temperature.

2. But permanent gases like H2, N2 can not be easily liquefied and having low critical temperature and adsorbed slowly.

Question 11.
What is meant by adsorption isotherm?
Answer:
A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorbate at constant temperature is called adsorption isotherm.

Question 12.
Mention Freundlinch adsorption isothermal equation.
Answer:
Freundlinch adsorption isotherm
\(\frac{x}{m}=k p^{1 / n}\)
where x is the amount of adsorbate or adsorbed on ‘m’ gm of adsorbent at a pressure of p. k and n arc constants.

Question 13.
What are the limitations of Freundlich adsorption isotherm?
Answer:

  1. The Freundlich adsorption isotherm is purely empirical and valid over a limited pressure range.
  2. The values of constants ‘k’ and ‘n’ also found vary with temperature. No theoritical explanations were given.

Question 14.
How is adsorption applied in the decolourisation of sugar?
Answer:
Sugar prepared from molasses is decolourised to remove coloured impurities. Animcal charcoal is applied to adsorb colouring impurities and sugar is purified.

Question 15.
What is chromatography?
Answer:
1. The chromatographic technique is a process of separation of components in a mixture which is mainly based on adsorption of components on the surface of adsorbents.

2. This method is very effective and used for identification, detection and estimation of many substances even if they are contained in micro quantities.

Question 16.
Explain the application of adsorption ¡n qualitative analysis with an example.
Answer:
1. In the identification of Al3+ ion in Al(OH)3 is obtained as blue lake. Due to adsorption of blue litmus solution added to the aluminium ion solution.

2. A red colouration is seen due to the acidic nature of the solution. Ammonium hydroxide is added to it until the bkie colour develops.

3. During this addition, Al(OH)3 precipitate is formed with blue colour due to the adsorption of litmus compound.

Question 17.
Define catalyst. Give example.
Answer:
A catalyst is defined as a substance which alter the rate of chemical reaction without itself undergoing chemical change. The phenomenon which involves the action of a catalyst is called catalysis.
e.g.,Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-35

Question 18.
What is homogeneous catalysis? Give example.
Answer:
Homogeneous catalysis is a process in which the reaction, products and catalyst are present in the same phase. NO
e.g. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-36
In the above reaction, the catalyst NO, the reactants SO2, O2 and product SO3 are present in the gaseous form

Question 19.
What is heterogeneous catalysis? Give example.
Answer:
Heterogeneous catalysis is a process in which the reactants, products and catalyst are present in different phases.
e.g. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-37

Question 20.
What are promoters? Explain with example.
Answer:
1. In a catalysed reaction, the presence of a certain substance increases the activity of a catalyst. Such substance is called a promoter.

2. For example, in Haber’s process of manufacture of ammonia, the activity of the iron catalyst is increased by the presence of molybdenum. Hence molybdenum is called a promoter.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-38

Question 21.
What is meant by catalyst poison?
Answer:
1. The substances when added to a catalysed reaction decreases or completely destroys the activity of a catalyst are often known as catalytic poisons.

2. In the reaction 2SO2 + O2 → 2SO3 with Pt catalyst, the catalyst poison is AS2O3.

Question 22.
Which is the catalyst and catalyst poison In Haber’s process?
Answer:
In the Haber’s process of the manufacture of ammonia, the Fe is catalyst and it is poisoned by the presence of H2S.
3H2 + N2 → 2NH3 Fe – catalyst, H2S – Catalyst poison

23. In the reaction 2H2 + O2 → 2H2O, which ¡s the catalyst and catalyst poison?
Answer:

  1. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-39 Pt – catalyst
  2. CO act as catalytic poison for Pt catalyst.

Question 24.
Explain the relation between activation energy and the rate of the reaction using catalyst.
Answer:
1. For a chemical reaction to occur, the reactants are to be activated to form a activated complex. The energy required for the reactants to reach the activated complex is called the activation energy. The activation energy can be decreased by increasing the reaction temperature.

2. In the presence of catalyst, the reactants are activated at reduced temperatures. i.e., the activation energy is lowered. The catalyst adsorbs the reactants activates them by weakening the bonds and allow them to react to form products.

3. As activation energy is lowered in the presence of a catalyst, more molecules take part in the reaction and hence the rate of the reaction increases.

Question 25.
What are the merits and limitations of the intermediate compound theory?
Answer:
Merits:

  1. The specificity of a catalyst
  2. The mercase in the rate of the reaction with increase in the concentration of catalyst.

Limitations:

  1. This theory fails to explain the action of catalytic poison and promoters.
  2. This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Question 26.
What are active centres?
Answer:

  1. The surfiice of a catalyst is not smooth. It bears steps, cracks and corners. Hence the atoms on such locations of the surface are coordinatively unsaturated.
  2. The have residual force of attraction. Such sites are called active centres. So the surface cames high surface free energy.

Question 27.
Enzyme catalysis are more effective and efficient than ordinary catalysis. Prove this statement.
Answer:
1. Effective and efficient conversion ¡s the special chareteristic of enzyme catalysed reactions. An enzyme may transform a million molecules of reactant to product in a minute.
For e.g. 2H2O2 → 2H2O + O2

2. For this reaction colloidal platinum as a catalyst the activation energy is 11 .7 k.cal/mole. But with the enzyme catalyst the activation energy of this reaction is less than 2k.cal/mole.

Question 28.
Enzyme catalysis is highly specific in nature. Justify this statement.
Answer:
1. Urea can be hydrolysed to urea by the enzyme urease.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-40
The catalyst mease (enzyme) catalyses urea but it does not hydrolyse methyl urea.

Question 29.
Enzyme catalysed reaction has maximum rate at optimum temperature. Prove it.
Answer:
1. At first rate of reaction increases with the increase of temperature, but above a particular temperature, the activity of enzyme is destroyed. The rate may even drop to zero. The temperature at which enzymic activity in high or maximum is called optimum temperature.

2. For e.g., enzymes involved in human body have an optimum temperature 37°C/98°F.

Question 30.
What are the types of colloids based on dispersion medium?
Answer:

1. If the dispersion medium considered as water, then the colloids are referred as hydrosols (or) aquasols.
2. If the dispersion medium is an alcohol, the colloid is termed as alcosol. and if benzene is the dispersion medium, it is called benzosol.

Question 31.
Explain about

  1. Liquid aerosol
  2. solid aerosol with example.

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-41

Question 32.
Explain oxidation method of preparation of colloids with two examples.
Answer:
Oxidation:
1. When hydrolodic acid is treated with iodic acid 12 sol is obtained.
HIO3 + 5HI → 3H2O + 3I2 (sol)

2. When O2 is passed through H2Se, a sol of selenium is obtained.
H2Se + O2 → 2H2O + Se (sol)

Question 33.
Explain the method of preparation of gold sol by reduction method.
Answer:
Gold sol is prepared by the reduction of auric chloride using formaldehyde.
2AuCl3 + 3HCHO + 3H2O → 2Au (sol) + 6HCI + 3HCOOH

Question 34.
How would you prepare ferric hydroxide sol by hydrolysis method?
Answer:
Fe(OH)3 sol can be preparcd by the hydrolysis of FeCl3
FeCI3 + 3H2O → Fe(OH)3 (soL) + 3 HCI

Question 35.
How would you prepare colloid by the exchange of solvent method?
Answer:
1. Colloidal solution of few substances like phosphorous or sulphur is obtained by the solutions in alcohol and pouring them into water.
2. As they are insoluble in water, they form colloidal solution.
P in alcohol + water → Psol.

Question 36.
Why colloids are to be purified? If not what will happen?
Answer:

  1. The colloidal solutions due to their different methods ofpreparation may contain impurities.
  2. If they are not removed, they may destabilise and precipitate the colloidal solution. This is called coagulation.
  3. Hence the impurities mainly electrolytes should be removed to increase the stabilisation of colloid.

Question 37.
How is human kidney dialysis take place?
Answer:

  1. Kidney malfunction results in the building up of electrolyte concentration within the blood to toxic levels.
  2. In the dialysis. recycling of patient’s blood is done through considerable length of semipermeable tube in an isotomic saline solution.

Question 38.
Write a note about Helmoholtz double laver.
Answer:
The surface of a colloidal particle adsorbs one type of ion due to preferential adsorption. This Stem layer
layer attracts the oppositely charged ions in the. medium and hence the boundary separating the Slpping plane two electrical double layers are set up. This is called Helmholtz electrical double layer. As the particles nearby are having similar charges. they cannot come close and condense. Hence this helps to explain the stability of the colloid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-42

Question 39.
What is meant by gold number?
Answer:

  1. Zsigmondy introduced the term “Gold number” as a measure of protective power of a colloid.
  2. Gold number is defined as the number of milligrams of hydrophilic colloid that will just prevent the precipitation of 10 ml of gold sol on the addition of 1 ml of 10% NaCI solution.

Question 40.
Potato starch is less stable than gelatin. Why?
Answer:

  1. The gold number of gelatin is 0.005 – 1 and the gold number of potato starch is 25.
  2. Smaller the gold number, greater the protective power.

Question 41.
Write a note about Cortrell’s precipitator.
Answer:
Carbon dust in air is solidified by Cortrell’s precipitator. In it. a high potential difference of about 50.000V is used. The change on carbon is neutralised and solidified. Thus the air is free from carbon particles.

Question 42.
Explain about

  1. blue colour of the sky
  2. formation of delta.

Answer:

  1. The blue colour of the sky in nature is due to Tyndall effect of air particles.
  2. The electrolytes in sea and river water coagulate the solid particles in river water at their intersection. By this way delta is formed.

Question 43.
Distinguish between the meaning of the terms adsorption and absorption. Give one example each.
Answer:
Adsorption

  1. Substance is concentrated only at the surface and does notpcnctrate from the surface to the bulkofthe adsorbent.
  2. It is a surface phenomenon.
    Example: Silical gel absorbs water vapour on its surface.

Absorption

  1. Substnce is uniformly distributed throughout the bulk of the solid.
  2. It is a bulk phenomenon.
    Example: Anhydrous CaCl, absorbs water vapours in it.

Question 44.
Explain the following term giving a suitable example, emulsification.
Answer:
The process of making emulsion using a mixture of two immiscible or partially miscible liquids is called emulsification. For example. Cod liver oil is an emulsion made up of water in oil.

Question 45.
What is the reason for the stability of colloidal sols?
Answer:
Reason for the stability of colloidal sols are:

  1. Coagulation of the colloidal sol is prevented because of the presence of equal and similar charges on the colloidal particles.
  2. Colloidal particles are covered by a sheath of liquid in which they are extensively solvated because of which they acquire stability.

Question 46.
Dialysis is a method of purification of sols. But prolonged dialysis of the sol makes it unstable. Why?
Answer:
Traces of electrolytes in the sol, impart charge to the dispersed phase particles making it stable. Prolonged dialysis removes all the electrolytes, thus making the sol unstable.

Question 47.
Why the sun looks red at the time of setting? Explain on the basis of colloidal properties.
Answer:
At the time of setting, the sun is at the horizon. The light emitted by the sun has to travel a longer distance. As a result the blue part of the light is scattered away by the dust particles in the atmosphere. Hence the red part is visible.

Question 48.
What are emulsions? What are their different types? Give one example of each type.
Answer:
An emulsion is a colloidal dispersion in which both the dispersed phase and the dispersion medium are liquids and the two liquids involved are otherwise immiscible.
1. Oil in water type, in which oil is a dispersed phase and water is the dispersion medium, For example: milk is as emulsion of liquid fat dispersed in water.

2. Water in oil type. in which water is the dispersed phase and oil is thedispersion medium. For example: cod liver oil is an emulsion of water in oil in which water is the dispersed phase and oil is the dispersion medium.

Question 49.
How does chemical adsorption of a gas on a solid vary with temperature?
Answer:
Chemical adsorption first increases with increase in temperature and then it decreases, if the pressure remains constant.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-43

Question 50.
What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Answer:
Lyophilic Sols:
Colloidal sols directly formed by rnixing substances like gums, gelatin, starch, rubber. etc. with a suitable liquid (The dispersion medium) are lyophilic sols. An important characteristic of these sols is that if the dispersion medium is separated from the dispersed phase

(say by evaporation) the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols. These sols are quite stable and cannot be easily coagulated.

Lyophobic sols:
These colloidal sols can only be prepared by some special methods. These sols are readily precipitated on the addition of small amount of electrolytes, by heating or by shaking and hence are not stable.

Hydrophobic sols are water hating. They are formed by indirect method. These sols are irreversible sols. These sols are readily precipitated by the addition of small amount of electrolytes, by heating or by shaking and hence are not stable.

Samacheer Kalvi 12th Chemistry Surface Chemistry 3 Marks Questions and Answers

Question 1.
What are the characteristics of adsorption?
Answer:
1. Adsorption can occur in all interfacial faces i.e., the adsorption can occur in between gas – solid, liquid – solid, liquid – liquid, solid – solid and gas – liquid.

2. Adsorption is always accompanied by decrease in free energy. When ∆G reaches zero, the equilibrium is attained.

3. Adsorption is a spontaneous process.

4. When molecules are getting adsorbed, there is always decrease in randomness of the molecules.
∆G = ∆H – T∆S where ∆G = change in free energy
∆H change in enthalpy
∆S change in entropy
∆H = ∆G + T∆S

5. Adsorption is exothermic and it is a quick process.

6. If simultaneous adsorption and absorption take place, it is termed as ‘sorption and sorption of gases on metal surface is called occlusion.

Question 2.
Explain graphical representation of chemical adsorption and physical adsorption.
Answer:
1. Adsorption isotherms represents the variation of adsorption at constant temperature.

2. When amount ?f adsorption is plotted versus temperature at constant pressure is called adsorption iso bar
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-44
3. In physical adsorption x/m decreases with increase in T. But in chemical adsorption x/m increases with rise in temperature and then decreases. The increase illustrate the requirement of activation of the surface for adsorption is due to the fact that formation of activated complex require certain energy. The decrease at high temperature is due to desorption, as the kinetic energy of the adsorbate increases.

Question 3.
Write any 3 applications of adsorption.
Answer:

  1. Gas masks: During World War I. charcoal gas mask was employed.
  2. To create high vaccum in vessels, activated charcoal is used. For dehydration and purification of gases like CO2, N2, O2 and He, alumina and silica gel are employed.
  3. In blast furnace, silica gel is used for drying air.

Question 4.
Explain the function of permutit in the softening of hard water.
Answer:
1. Permutit is employed for softening 6 hard water. It adsorbs Ca2 and Mg2 ions in its surface, there is an ion exchange as shown below it occurs on the surface.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-45

2. Exhausted permutit is generated by adding a solution of commonsalt.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-46

Question 5.
Explain about the application of ion exchange resins In adsorption.
Answer:

  1. Ion exchange resins are working only based on the process of adsorption.
  2. Ion exchange resins are used to demmeralise water. This process is carried out by passing water through two columns of cation and anion exchange resins.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-47

Question 6.
What is catalysis? Explain with two examples.
Answer:
Catalysis is a process in which a chemical reaction takes place in a faster rate with the help of catalyst.
Haber’s process:
N2 + 3H2 → 2NH3 In this process. Fe is the catalyst and Mo is the promoter.
The surface of Fe catalyses the reaction.

Vanaspathi preparation:
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-48
Nickel surface catalyses the reaction.

Question 7.
In the following fields, how adsorption ¡s applied?

  1. Medicine
  2. Metallurgy
  3. Mordant & Dyes
  4. indicators

Answer:
1. Medicine:
Drugs cure diseases by adsorption on body tissues.

2. Metallurgy:
Sulphide ores are concentrated by a process called froth floation in which lighter ore particles are adsorbed by pine oil. .

3. Mordants and Dyes:
Most of the dyes are adsorbed on the surface of the fabric. Mordants are the substances used for fixing dyes onto the fabric.

4. In the precipitation titrations, the end point is indicated by an external indicator which changes its colour aller getting absorbed on precipitate. It is used to indicate the end point of filtration.

Question 8.
Differentiate homogeneous and heterogeneous catalysis.
Answer:
Homogeneous catalysis

  1. The reaction in which, the reactants, products and catalyst are present in the same phase, is called homogeneous catalysis.
  2. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-49
  3. Homogeneous catalysis mechanism is ex plained by intermediate compound formation theory.

Heterogeneous catalysis

  1. The reaction in which the reactants, products and catalyst are present in different phases is called heterogeneous calalysis.
  2. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-50
  3. Heterogeneous catalysis mechanism is explained by adsorption theory.

Question 9.
Give three examples for homogeneous catalysis.
Answer:
1. Decomposition of acetaldehyde by I2 catalyst.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-51

2. Hydrolysis of cane sugar with mineral acid
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-52

3. Acid hydrolysis of an ester
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-53

Question 10.
Give three examples for heterogeneous catalysis
Answer:
1. hydrogenation of ethylene in the presence of Nickel
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-54

2. Decomposition of H2O2 with Pt catalyst
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-55

3. Friedel crafts reaction
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-56

Question 11.
What is auto ctalysis? Give two examples
Answer:
The process in which one of the product formed act as a catalyst is termed auto catalysis
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-57
Acetic acid acts as auto catalyst.
Decomposition of Arsenic hydride
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-58
Arsenic act as auto catalyst.

Question 12.
What is negative catalysis? Explain with example.
Answer:

1. In certain reactions, presence of certain substances decreases the rate of the reaction. Such substances are called negative catalyst and the process is called negative catalysis.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-59

2. In oxidation of chloroform, ethanol decreases the rate of the reaction and ethanol act as negative catalyst.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-60

Question 13.
Explain the formation of water with copper catalyst by Intermediate compound formation theory.
Answer:
1. Formation of water due o the reaction of H2 and O2 in the presence of Cu can be given as

2. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-61

3. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-62

Question 14.
Explain the mechanism of oxidation of HCl by air in the presence of CuCl2
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-63

Question 15.
Explain the thermal decomposition of potssium chlorate by intermediate compound formation theory.
Answer:
1. Thermal decomposition of KClO3 in the presence of MnO2 proceeds as follows.

2. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-64

Question 16.
Describe the action of active centres present in the catalyst.
Answer:
1. Active centres increases the rate of the reaction by adsorbing and activating the reactants.

2. Increase in the activity of a catalyst by increasing the surface area. Increase in the surface area of metals and metal oxides by reducing the particle size increases the rate of the reaction.

3. The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.

4. A promoter (or) activator increases the number of active centres on the surface.

Question 17.
Write a note about nano catalyst.
Answer:

  1. Nano materials such as metallic nano particles, metal oxide are used as catalyst in many chemical transformation.
  2. Nano catalysts carry the advantages of both homogeneous and heterogeneous catalysis.
  3. Like homogeneous catalyst the nino catalyst give 100% selective transformations and excellent yield and show extremely high reactivity.
  4. Like the heterogeneous catalyst, nano catalysts can be recovered and recycled.
  5. Nano catalysts are actually soluble heterogeneous catalyst.Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-65

Question 18.
Differentiate lyophillic and lyophobic colloids
Answer:
Lyophiltic colloids

  1. In lyophillic colloids (or) sols definite attractive force (or) affinity exists between dispersion medium and dispersed phase.
  2. e.g., Sols of protein, starch
  3. They are more stable and will not get precipitated readily.
  4. They can be brought back to colloidal solution even after the precipitation by addition of dispersion medium
  5. They arc called reversible sols

Lyophobic colloids

  1. In lyophobic colloid no attractive force exists between dispersed phase and dispersed phase and dispersion medium.
  2. e.g., sols of gold, silver
  3. They are less stable and get precipitated readily
  4. They cannot be produced again by adding the dispersion medium.
  5. They are called irreversible sols.

Question 19.
Explain about dispersion mediu, dispersed phase and example of

  1. foam,
  2. emulsion
  3. sol

Answer:

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-66

Question 20.
Explain about dispersion medium, dispersed phase and example of

  1. solid foam
  2. Gel
  3. Solid sol.

Answer:

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-67

Question 21.
How would you prepare colloids of ink and graphite? (OR)
Answer:
Explain about mechanical ‘dispersion method.
1. Using a colloid mill, the solid is ground to colloidal dimension. Coarse dispersion The colloid mill consists of two metal plates rotating in opposite directions at very high speed of nearly 7000 revolution per minute.

2. The colloidal particles of required colloidal size is obtained by adjusting the distance between the two plates.

3. By this method, colloidal solutions of ink and graphite and prepared.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-68

Question 22.
Explain about Bredic’s arc method (or) Electro dispersion method (or) How would you prepare colloids of noble metals?
Answer:
1. An electrical arc is struck between electrodes dispersed in water surrounded by ice. When a current of I amp! 100V is passed, an arc produced forms vapours of metal which immediately condense to from colloidal solution.

2. By this method, colloidal solution of many metals like copper, silver, gold. platinum can be prepared.

3. Alkali hydroxide is an added an stabilising agent for the colloid solution.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-69

Question 23.
Explain about ultrasonic dispersion. (or) How would you prepare mercury colloid?
Answer:
1. Sound waves of frequency more than 20kHz (audible limit) could cause transformation of coarse suspension to colloidal solution.

2. Claus obtained mercury sol by subjecting mercury to sufficiently high frequency ultrasonic vibrations.

3. The ultrasonic vibrations produced by generator spread the oil and transfer the vibration to the vessel with mercury in water.

Question 24.
Explain the methods of preparation of colloids of

  1. AS2S3
  2. S.

Answer:
1. Double decomposition:
When hydrogen suiphide gas is passed through a solution of arsenic oxide, a yellow coloured arsenic sulphide is obtained as a colloidal solution.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-70

2. Decomposition:
When few drops of an acid is added to a dilute solution of sodium thio sulphate, the insoluble free sulphur produced by the decomposition of sodium thiosuiphate accumulates into clusters which impart various colours blue, yellow and even red to the system depending on their growth within the size of colloidal dimensions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-71

Question 25.
Describe about

  1. Dialysis
  2. Electro dialysis.

Answer:
1. Dialysis:
Thomas Graham separated the electrolyte from a colloid using a semipermeable membrane. In this method, the colloidal solution is taken in a bag made up of semipermeable membrane. It is suspended in a trough of flowing water, the electrolytes diffuse out of the membrane and they are carried away by water. membrane and they are carried away by water.

2. Electro dialysis:
The presence of electric field increases the speed of removal of electrolytes from colloidal solution. The colloidal solution containing an electrolyte as impurity is placed between two dialysing membranes enclosed into two compartments filled with water.

When current is passed, the impurities pass into water compartment and get removed periodically. This process is faster than dialysis, as the rate of diffusion of electrolytes is increased by The application of electricity.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-72

Question 26.
Explain about ultrafiltration.
Answer:
1. The pores of ordinary filter papers permit the passage of colloidal solutions. In ultrafilteration, the membranes are made by using collodion, cellophane or visiking.

2. Vhen a colloidal solution is filtered using such a filter, colloidal particles are separated on the filter and the impurities are removed as washings.

3. This process is quickened by application of pressure. The separation of sol particles from electrolyte by fikration through an ultrafilter is called ultrafiltration.

4. Collodian is 4% solution of nitrocellulose in a mixmre of alcohol and water.

Question 27.
Write a note about shape of colloidal particles.
Answer:
Colloidal particles possess various shapes.
Collidal particle

  1. AS2S3
  2. Fe(OH)3 (blue gold sol)
  3. W3O5 sol(tungstic acid sol)

Shape

  1. Spherical
  2. Disc (or) plate like
  3. Rod like

Question 28.
What is meant by Tyndall effect? (or) Explain about the optical property of colloid.
Answer:
Answer:

  1. Colloids have optical property. When a homogeneous solution is seen in the direction of light, it appears clear but it appears dark in a perpendicular direction.
  2. When light passes through colloidal solution, it is scattered in all direction and it is known as Tyndall effect.
  3. The colloidal particles absorbs a portion of light and remaining portion is scattered from the surface of the colloid. Hence the path of light is made clear.

Question 29.
What is meant by Brownian movement?
Answer:
1. Robert Brown observed that when the pollen grains suspended in water were viewed through an ultra microscope, they showed a random, zigzag, ceaseless motion. This is called Brownian movement of colloidal particles.

2. The colloidal sol particles are continuously bombard with the molecules of dispersion mediuri and hence they follow a zigzag. random, continuous movement.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-73

Question 30.
Mention the uses of Brow nian movement.
Answer:
1. Brownian movement enables us to calculate Avogadro Number.

2. It is used to confirm kinetic theory which considers the ceaseles rapid movement of molecules that increases with increase in temperature.

3. It is used to understand the stability of colloids. As the particles are in continuous rapid movement, they do not come close and hence not get condensed. That is Brownian movement does not allow the particles to be acted on by force of gravity.

Question 31.
What is coagulation? Mention the method used to coagulate a colloid.
Answer:

  1. The flocculation and setting down of the sol particles is called coagulation.
  2. Various methods of coagulation are
  3. addition of an electrolyte
  4. Electrophoresis
  5. mixing oppositely changed sols
  6. boiling

Question 32.
The precipitation power of ions are in the order Al3+ > Ba2+ > Na+.
Similarly [Fe(CN)6]4-SO42- Cl. Give the reason behind this.
Answer:
1. A negative ion causes the precipitation of positively changed sol and vice versa.

2. When the valency of ion is high, the precipitation power increased. So in cations the precipitation power order is Al3+ > Ba2+ > Na and in anions the precipitation power order is [Fe(CN)6]4- > SO42- > Cl.

3. The precipitation power of electrolyte is determined by finding the minimum concentration required to cause precipitation of sol in 2hrs. This value is called flocculation value. The smaller the flocculation value, greater will be precipitation.

Question 33.
Explain how coagulation of colloid is carried out by

  1. Flectrophoresis
  2. By mixing two oppositely changed sols
  3. By boiling.

Answer:
1. Electrophoresis:
In eleetrophoresis, charged particles migrate to the electrodes of opposite sign. It is due to neutralization of the charge of the colloids. The particles are discharged and sothey get precipitated.

2. By mixing two oppositely charged sols: When colloidal sols with opposite charges are mixed, mutual coagulation takes place. It is due to migration of ions from the surface of the particles.

3. By boiling – When colloidal sol is boiled, due to increased collisions, the sol particles combine and settle down.

Question 34.
Explain about protective action of colloid.
Answer:
1. Lyophobic sols are precipitated readily even with small amount of electrolytes. But they are stabilised by the addition of small amount of lyophillic colloid.

2. A small amount of gelatine sol is added to gold sol to protect gold sol.

3. Gold number is a measure of protecting power of a colloid, Gold number is defined as the number of milligrams of hydrophillic colloid that will just prevent the precipitation of 10ml of gold sol on the addition of imi of 10% NaG solution. Smaller the gold number, greater the protective power.

Question 35.
What are emulsions? Give its types. Explain with examples.
Answer:

  1. Emulsions are colloidal solution in which a liquid is dispersed in another liquid.
  2. There are two types of emulsions.
    • Oil in water (0/W)
    • Water in Oil (W/O)
  3. Oil in Water : Oil dispersed in water. e.g., mayonnaise. icecrearn.
    Water in Oil: Water dispersed in Oil. e.g., stiff grease, butter, cold cream.

Question 36.
Write 3 examples for emulsifiers.
Answer:

  1. Most of the lyophillic colloids act as emulsifiers. Example, glue. gelatin.
  2. Long chain compounds with polar groups like soap and suiphonic acid.
  3. Insoluble powders like clay and lamp black also act as emulsifier,

Question 37.
What is meant by inversion of phase? Explain with example.
Answer:

  1. The change of W/O emulsion into O/W emulsion is called inversion of phase.
  2. Example: An oil in water emulsion containing potassium soap as emulsifying agent can be converted into water in oil emulsion by adding CaCl2 (or) AlCl3. This is called inversion of phase.

Question 38.
Write a note about medicinal applications of colloids.
Answer:

  1. Antibodies such as penicillin and streptomycin are produced in colloidal form for suitable injections. They cure pneumonia.
  2. Colloidal gold and colloidal calcium are used as tonics.
  3. Milk of magnesia is used for stomach troubles.
  4. Silver sol protected by gelatine known as Argyrol is used as eye lotion.

Question 39.
How colloids are used in

  1. Tanning of leather
  2. Rubber industry
  3. Sewage disposal.

Answer:
1. Tanning of leather:
Skin and hides are protein containing positively charged particles which are coagulated by adding tannin to give hardened leather for further application. Chromium salts are used for tanning of leather. Chrome tanning can produce soft and polishable leather.

2. Rubber industry:
Latex is the emulsion of natural rubber with negative particles. By heating rubber with sulphur, vulcanized rubbers are produced for tyres, tubes etc.

3. Sewage disposal:
Sewage containing dirt, mud and wastes dispersed in water. Th e passage of electnic current deposits the wastes materials which can be used as a manure.

Question 40.
How would you distinguish natural honey from artificial honey?
Answer:

  1. Natural honey is a colloidal sol. It is distinguished from artificial one by adding ammonia cal AgNO3.
  2. In the case of natural honey, a metallic silver is produced assumes a reddish yellow colour due to the traces of albumin or ethereal oil which acts as a protective colloid,
  3. In case of artificial honey, a dark yellow (or) greenish yellow precipitate is formed.

Question 41.
Give four uses of emulsions.
Answer:

  1. The cleansing action of soap is due to emulsions.
  2. It is used in the preparation of vanishing cream.
  3. It is used in the preparation of cod liver oiL.
  4. It is used in the preparation of butter, cream etc.

Question 42.

  1. Adsorption of a gas on the surface of solid is generally accompanied by a decrease in entropy. Still ¡t is a spontaneous process. Explain.
  2. How does an increase in temperature affect both physical as well as chemical adsorption?

Answer:
1. According to the equation ∆G = ∆H – T∆S For a process to be spontaneous, ∆G should be negative. ∆H of adsorption is always negative. For a gas. ∆S is also negative. Thus, in an adsorption process, which is spontaneous, a. combination of these two factors always makes ∆G negative.

2. On increasing the temperature desorption occurs in case of physical adsorption. Chemical adsorption first increases and then decreases with increase in temperature.

Question 43.

  1. What is the difference between a colloidal solution and an emulsion? Give one example of each.
  2. What are emulsifiers?

Answer:
1. In a colloidal solution, the dispersed phase is a solid and the dispersion medium is liquid. In an emulsion, both the dispersed phase and the dispersion medium are liquid. Example – Colloidal sol — cell fluids, muddy water. Emulsion – milk, cold cream

2. Emulsifiers – The substance which are added to stabilise the emulsions are called emulsifiers. Example – Soaps of various kinds and lyophilic colloids (proteins, gum etc.)

Question 44.
Explain what is observed when.

  1. KCl, an electrolyte, is added to an hydrated ferric hydroxide sol.
  2. An electric current is passed through a colloidal solution.
  3. A beam of light is passed through a colloidal solution.

Answer:

  1. Feme hydroxide Fe(OH)3 is a positively charged sol, so it gets coagulated by the chloride ions released from KCI solution.
  2. When an electric current is passed through a colloidal solution due to charge on the colloidal particles they migrate towards the oppositely charged electrode.
  3. When a beam of light is passed through a colloidal solution the path of light becomes visible.

Question 45.
Write three distinct differences between physical adsorption and chemisorption.
Answer:
Physical adsorption

  1. Forces of attraction between the adsorbent and adsorbate molecules are weak Van der Waals’ forces,
  2. Heat of adsorption is low (20-40 kJ mol1)
  3. It is temporary and reversible.

Chemical adsorption

  1. Forces between the adsorbent and the adsorbate are strong chemical bond.
  2. Heat of adsorption is high (80-24OkJ mol1)
  3. It is permanent and irreversible.

Question 46.
Explain the following observations.
Answer:

  1. Lyophilic colloid is more stable than lyophobic colloid.
  2. Coagulation takes place when sodium chloride solution added to a colloidal solution of ferric hydroxide.
  3. Sky appears blue in colour.

Answer:
1. A lyophilic sol is stable duc to the charge and the hydration of the sol particles. Such a sol can only be coagulatd by removing the water and adding solvents like alcohol, acetone, etc. and then an electrolyte. On the other hand, a lyophobic soils stable due to charge only and hence it can he easily coagulated by adding small amount of an electrolyte.

2. The colloidal particles get precipitated. i.e., ferric hydroxide is precipitated.

3. The atmospheric particles of colloidal range scatter blue component of the white sunlight preferentially. That is why the sky appears blue.

Question 47.

  1. Heat of adsorption is greater for chemisorption than physisorption. Why?
  2. What is colloldion?
  3. Differentiate between peptization and coagulation.

Answer:

  1. Due to the formation of chemical bond between adsorbate and adsorbent.
  2. 4% solution of nitro cellulose in a mixture of alcohol and ether.
  3. Peptisation is the process of converting a precipitate into colloidal sol by adding an electrolyte. Rut coagulation is the settling of colloidal particles.

Question 48.
Give reasons for the following.

  1. Enzyme catalysts are highly specific in their action.
  2. The path of light beconies visible wheti it is passed through As2S3 sol ¡n water.
  3. The enthalpy in case of chemisorption is usually higher than that of physisorption.

Answer:

1. This is because each enzyme has a specific active site on which only a particular substrate can bind. ,

2. This is because of Tyndall effect caused due to the scattering of light by colloidal particles of As2S3 sol.

3. Chemisorption involves the formation ofa chemical bond between adsorbent and adsorbate molecule which involves high energy changes while in physisorption, the molecules of adsorbate and adsorbent arc held by weak van der Waals’ interactions.

Question 49.
What is adsorption? How does adsorption ola gas on a solid surface vary with pressure? Illustrate with the help of an appropriate graph.
Answer:
Adsorption is the phenomenon of attracting and retaining the molecules of a substance on the surface of a liquid or a solid, resulting in higher concentration of the molecules on the surface.

Effect of pressure:
At constant temperature, the adsorption of gas increases with increase of pressure of the gas. At low pressure. it increases rapidly. At an equilibrium pressure. the extent of adsorption (x/m) reaches its maximum value after which adsorption is independent of pressure.
\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{1 / \mathrm{n}}\)
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-74

Question 50.
How do size of particles of adsorbent, pressure of a gas and prevailing temperature influence of extent of adsorption of a gas on a solid?
Answer:
1. Effect of size of the particles of adsorbent: Greater the specific area of the solid available for adsorption of adsorbent, greater would be its adsorbing power. That is why porous or finely divided forms of adsorbents more strongly. However, the size of pores should be large enough to allow the diffusion of gas molecules.

2. Effect of pressure:
Increase in pressure initially increases the adsorption which later attains equilibrium at high pressure.
\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{1 / \mathrm{n}}\) (n>1)

Samacheer Kalvi 12th Chemistry Surface Chemistry 5 Mark Questions and Answers

Question 1.
What is adsorption isotherm? Explain about Freundlich adsorption isotherm.
Answer:
1. Adsorption isotherms represents the variation of adsorption at constant temperature. Adsorption isoterm can be studied quantitatively.

2. A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorbate at constant temperature is called adsorption isotherms.

3. Freundlich adsorption isotherm. According to Freundlich
\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{1 / \mathrm{n}}\)
Where x is the amount of adsorbate (or) adsorbed on ‘m’ gm of adsorbent at a pressure of P. k and n are constants. Value of’n’ is always less than unity.

4. This equation is applicable for adsorption of gases on solid surfaces. The Same equation becomes \(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kc}^{1 / \mathrm{n}}\) when used for adsorption in solutions with ‘c’ as concentration.

5. These equation quantitatively predict the effect of pressure (or concentration) on the adsorption of gases (or adsorbates) at constant temperature.

6. Taking log on both sides of equation
\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{1 / \mathrm{n}}\)
log \(\frac { x }{ m }\) = log k + \(\frac { 1 }{ n }\) log P
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-75

7. Hence the intercept represents the value of log k and the slope \(\frac { b }{ q }\) gives \(\frac { 1 }{ n }\)

8. This equation explains the increase of with increase in pressure. But experimental values shows the deviation at low pressure.

9. Limitations:

  1. This equation is purely empirical and valid over a limited pressure range.
  2. The value of k and n also found vary with temperatures. No theoretical explanations were given.

Question 2.
Define catalyst. What are the characteristics of catalysts?
Answer:
A catalyst is defined as a substance which alters the rate of chemical reaction without itself undergoing chemical change.
Characteristics of catalyst:
1. For chemical reaction, catalyst is needed in very small quantity. Generally, a pinch of catalyst is enough for a reaction in bulk.

2. There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.

3. A catalyst itself cannot initiate a reaction. It means it cannot start a reaction which is not taking place. Rut, lithe reaction is taking place in a slow rate it can increase its rate.

4. A solid catalyst will be more effective if it is taken in a finely divided form.

5. A catalyst can catalyse a particular type of reaction, hence they are said to be specific in nature.

6. In an equilibrium reaction, presence of catalyst reduces the time for attainment of equilibrium and hence it does not affect the position
of equilibrium and the value of equilibrium constant.

7. A catalyst is highly effective at a particular temperature called as optimum temperature.

8. Presence of a catalyst generally does not change the nature of products.

9. For example 2SO2 + O2 → 2SO3
This reaction is slow in the absence of a catalyst, but fast in the presence of Pt catalyst.

Question 3.
What is enzyme catalysis? Give the characteristics of enzyme catalysed reaction?
Answer:
Enzymes are complex protein molecules with three dimensional structures. They catalyse the chemical reaction in living organism. They are often present in colloidal state and extremely specific in catalytic action. This process is called enzyme catalysis.
Special characteristics of enzyme catalysis:
1. Effective and efficient conversion is the special characteristic of enzyme catalysed reactions. An enzyme may transform a million molecules of reactant in a minute.
For e.g., 2H2O2 → 2H2O + O2
For this reaction, activation energy is 1 8k.cal/mole without a catalyst. With colloidal platinum as a catalyst, the activation energy is 11.7 k.cal/mole. But with the enzyme catalyst. the activation energy of this reaction is less than 2 k.cal / mole.

2. Enzyme catalysis is highly specific in nature.
H2N – CO – NH2 + H2O → 2NH3 + H2O
The enzyme urcase which catalyses the reaction of urea does not catalyse the hydrolysis of methyl urea (NH2 – CO – NHCH3)

3. Enzyme catalysed reaction has maximum rate at optimum temperature. At first rate of the reaction increases with increase of temperature, but above a particular temperature, the activity of enzyme is destroyed. The rate may even drop to zero. The temperature at which enzyrnic activity is high (or) maximum is called as optimum temperature. e.g., enzyme involved in human body have an optimum temperature 37°C / 98°F.

4. The rate of enzyme catalysed reactions varies with the pH of the system. The rate is maximum at a pH called optimum pH.

5. Enzymes can be inhibited i.e., poisoned. Activity of an enzyme is decreased and destroyed by a poison. The physiological action of drugs is related to their inhibiting action. e.g.. sulpha drugs, penicillin inhibits the action of bacteria and used for curing diseases like
pneumonia, dysentery, cholera.

6. Catalytic activity of enzymes is increased by coenzymes or activators. A small non protein (vitamin) called a coenzyme promotes the catalytic activity of enzyme.

Question 4.
Explain about phase transfer catalysis.
Answer:
1. Consider the reactant of a reaction is present in one solvent and the other reactant is present in an another solvent. The reaction between them is very slow, if the solvents are immisible.

2. As the solvents form separate phases, the reactants have to migrate across the boundary to react. But migration of reactants across the boundary is not easy. For such situation, a third solvent is added which is miscible with both. So, the phase boundary is eliminated the reactants freely mix and react fast.

3. But for large scale preparation of any product, use of a third solvent is not convenient as it may be expensive. For such problems, phase transfer catalysis provides a simple solution, which avoids the use of solvents.

4. it directs the use of a phase transfer catalyst (a phase transfer reagent) to facilitate transport of a reactant in one solvent to other solvent where the second reactant is present. As the reactants are now brought together, they rapidly react and form the product.

5. Example – Substitution of C1 and CN in the following reaction.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-76
R Cl = 1 – chloro octane
R CN = 1 – cyano octane

6. By direct heating of two phase mixture of organic 1 – chioro Octane with aqueous sodium cyanide for several days, 1 – cyano octane is not obtained. However, if a small amount of quarternary ammonium salt like tetra alkyl ammonium cation which has hydrophobic and hydrophilic ends, transports CN – from the aqueous phase to the organic phase using its hydrophilic end and facilitates the reactions with 1 – chloro octane as shown below
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-77

7. So phase transfer catalyst, speeds up the reaction by transporting one reactant from one phase to another.

Question 5.
Explain about the classification of colloids based on the physical state of dispersed phase and dispersion medium with example.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-78

Question 6.
Describe about condensation methods of preparation of colloids. (OR) Describe chemical methods of preparation of colloids. When the substance for colloidal particle is present as smalL sized particle, molecule or ion, they are brought to the colloidal dimension by condensation methods.

Answer:
1. Oxidation method:
When hydroiodic acid is treated with iodle acid J, sol is obtained.
HIO3 + 5HI → 3H2O + 3I2(sol)

2. Reduction method:
Gold sol is prepared by reduction of aune chloride using formaldehyde.
2 AuCl3 + 3HCHO + 3H2O → 2 Au(sol) + 6HCI + 3HCOOH

3. Hydrolysis:
Ferric chloride is hydrolysed to get ferric hydroxide colloid
FeCl3 + 3H2O → Fe(OH)3(sol) + 3HCI

4. Double decomposition:
When hydrogen suiphide gas is passed through a solution of arsenic oxide, a yellow coloured arsenic sulphide is obtained as a colloidal solution.
As2O3 – 3H2S → As2S3 + 3H2O

5. Decomposition:
When few drops of an acid is added to a dilute solution of sodium thiosulphate, sulphur colloid is produced by the decomposition of sodium thio sulphate.
S2O32- + 2H+ → S(sol) + H2O + SO2

7. Describe about the properties of colloids.
Answer:
1. Colour:
The colour of a sol is not always the same as the colour of the substance in the bulk. For example, bluish tinge is given by diluted milk in reflected light and reddish tinge in transmitted light.

2. Size:
The size of colloidal particles ranges from Imp to 1pm diameter.

3. Colloidal solutions are heterogeneous in nature:
They have two distinct phases. Experiments like dialysis. ultrafiltration show the heterogeneous in nature but they are considered as borderline cases.

4. Filterability:
As the size of pores in ordinary filter paper are large, the colloidal particles easily pass through the ordinary filter papers.

5. Non setting nature:
Colloidal solutions are quite stable i.e., they are not affected by gravtiy.

6. Concentration and density:
When the colloidal solution is dilute, it is stable. When the volume of medium is decreased, coagulation occurs. Density of sol decreases with
decrease in the concentration.

7. Diffusability:
Unlike true solution, colloids diffuse less readily through membranes.

8. Colligative properties:
The colloidal solutions show colligative properties such as elevation of boiling point, depression in freezing point and osmotic pressure. These. properties are used to determine molecular weight of colloidal particles.

9. Shape of colloidal particles:
Colloidal particles have various shapes

Collidal particle

  1. AS2S3
  2. Fe(OH)3 (blue gold sol)
  3. W3O5 sol(tungstic acid sol)

Shape

  1. Spherical
  2. Disc (or) plate like
  3. Rod like

10. Optical property:
The path of the light is visible when it is passes through a colloidal solution due to the scattering of light by colloidal particles. This is known as Tyndall effect.

11. Kinetic property:
When colloidal solution is viewed through an ultra microscope, they showed a random, zigzag ceaseless motion which is called Brownian movement.

12. Electrical property:
Helmholtz double layer. electrophoresis and electro osmosis are electrical properties of colloids.

Question 8.
Explain about Electrophoresis (or) Cataphoresis (or) How would you detect the presence of charges on sol particles? (or) Explain about the method used to detect the presence of charge on sol particles. Electrophoresis (or) cataphoresis:

Answer:
1. When electric potential is applied across two platinum electrodes dipped in a hydrophilic sol, the dispersed particles move toward one or other electrode. This migration of sol particles under the influence of electric field is called electrophoresis.

2. If the sol particles migrate to the cathode, then they possess positive charges and if the sol particles migrate to the anode, they have negative charges. Thus from the direction Afiet of migration of sol particles, we can determine the charge of the sol particles. Hence electrophoresis is used for the detection of presence of change on the sol particles.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-79

3.
Positively charged coiloids

Fe(OH)3, Al(OH)3
Basic dyes, Haemoglobin

Negatively charged colloids

Ag, Au & Pt, AS2S3,
Clay, Starch

Question 9.
What are emulsion? Mention its type with example. What is emulsification? How the types of emulsions are identified?
Answer:
1. Emulsions arc colloidal solution in which a liquid dispersed in another liquid.

2. Two types of emulsions

  • Oil in Water (O/W) e.g., mayonnaise
  • Water in Oil (W/O) e.g., Butter

3. The process of preparation of emulsion by the dispersal of one liquid in another liquid is called emulsification.

4. Identification of types of emulsions. The two types of emulsions can be identified by the following tests

  • Dye test – A small amount of dye soluble in oil is added to emulsion. The emulsion is shaken well. The aqueous emulsion will not take the colour whereas oily emulsion will take up the colour of the dye.
  • Viscosity test – Viscosity of the emulsion is determined by experiments. Oily emulsions will have higher value than aqueous emulsions.
  • Conductivity test – Conductivity of aqueous solutions are always higher than oily emulsions.
  • Spreading test – Oily emulsions spread readily than aqueous emulsion when spread on an oily surface.

Question 10.
What is deeniulsification? Explain about the various techniques of deemulsification.
Answer:
Emulsion can be separated into two separate layers. This process is called Deemulsification. Various deemulsilication techniques are given below.

  1. Distilling of one component
  2. Adding an electrolyte to destroy the charge.
  3. Destroying the emulsifier using chemical methods.
  4. Using solvent extraction to remove one component.
  5. By freezing one of the components.
  6. By applying centrifugal force.
  7. Adding dehydrating agents for water in oil type.
  8. Using ultrasonic waves.
  9. Heating at high pressures.

Question 11.

(a) How can a colloidal solution and a true solution of the same colour be distinguished from each other?
(b) List four applications of adsorption.

Answer:

  1. The path of light becomes visible when passed through a colloidal solution while it is not visible in case of a true solution. This is because of Tyndall effect caused by the scattering of light by colloidal particles.
  2. Applications of adsorption:
    • Activated charcoal is used in gas masks to remove poisonous gases such as CH4, CO, etc.
    • Animal charcoal is used as decoloriser in the manufacture of sugar.
    • Silica is used for removing moisture.
    • The ion exchange resins are used for removing hardness of water.

Question 12.
Illustrate with examples.

  1. Lyophilic and Lyophobic sols
  2. Homogeneous and Heterogeneous catalysis.

Answer:
1. The substances such as proteins, starch, rubber, etc. directly passes into the colloidal state when brought in contact with the solvent. Such colloids are known as lyophillic sols, The substances like metals, their suiphides. hydroxides, etc. do not form colloidal sol readily when mixed with dispersion medium. The colloidal sols can only be prepared by some special methods. Such sols are called lyophobic sols.

2. Homogeneous catalysis:
Here the reactants and catalyst are present in the same phase. For example, lead chamber process for the manufacture of H2SO4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-80

Heterogeneous catalysis:
Here the reactants and catalyst are present in different phases. For example, contact process for manufacture of H2SO4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-81
Common Errors

  1. Students may get confused with the words absorption and adsorption
  2. There may be a confusion between physical adsorption and chemical adsorption
  3. Catalyst poison and negative catalyst are different.
  4. Colloids : There may be a doubt between dispersed phase and dispersion medium.
  5. Electro dialysis, Electrophoresis, Electro osmosis may get confused.

Rectifications

1. In surface chemistry, we should use only adsorption. Adsorption mainly explains surface phenomenon.
Physical adsorption – Weak Vander walls forces, multilayer formation, Heterogenous catalysis.

2. Chemical adsorption:
Bond formation, mono-layer formation, Homogeneous catalysis.
Catalyst poison: decreases the catalytic activity of catalyst. Negative catalyst: reduces the rate of a chemical reaction.

3. Dispersion medium:
Larger in amount. Dispersed phase – Smaller in amount. Colloid is dispersed phase in dispersion

4. medium. For e.g., smoke is colloid in which carbon solid particles dispersed in air dispersion medium.
Electrodialysis – Purification of colloid by electric current.

5. Electrophoresis:
Movement of colloidal particles (dispersed phase) towards the oppositely charged electrode under the influence of electric current. Electro osmosis: Movement of dispersion medium towards the oppositely charged electrode under the influence of electric current.

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