Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

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Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

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Samacheer Kalvi 11th Bio Botany Plant Growth and Development Text Book Back Questions and Answers

Question 1.
Select the wrong statement from the following:
(a) Formative phase of the cells retain the capability of cell division.
(b) In elongation phase development of central vacuole takes place.
(c) In maturation phase thickening and differentiation takes place.
(d) In maturation phase, the cells grow further.
Answer:
(d) In maturation phase, the cells grow further.

Question 2.
If the diameter of the pulley is 6 inches, length of pointer is 10 inches and distance travelled by pointer is 5 inches. Calculate the actual growth in length of plant:
(a) 3 inches
(b) 6 inches
(c) 12 inches
(d) 30 inches
Answer:
(a) 3 inches

Question 3.
In uni sexual plants, sex can be changed by the application of:
(a) ethanol
(b) cytokinins
(c) ABA
(d) auxin
Answer:
(c) ABA

Question 4.
Select the correctly matched one:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
(a) A – (iii), B – (iv), C – (v), D – (vi), E – (i), F – (ii)
(b) A – (v), C – (ii), D – (iv), E – (vi), F – (iii)
(c) A – (iii), B – (v), C – (vi),D – (i), E – (ii), F – (iv)
(d) A – (ii), B – (iii), C – (v), D – (vi), E – (iv), F – (i)
Answer:
(b) A – (v), C – (ii), D – (iv), E – (vi), F – (iii)

Question 5.
Seed dormancy allows the plants to:
(a) overcome un favorable climatic conditions
(b) develop healthy seeds
(c) reduce viability
(d) prevent deterioration of seeds
Answer:
(a) overcome unfavorable climatic conditions

Question 6.
What are the parameters used to measure growth of plants?
Answer:

  • Increase in length or girth (roots and stems)
  • Increase in fresh or dry weight
  • Increase in area or volume (fruits and leaves)
  • Increase in number of cells produced.

Question 7.
What is plasticity?
Answer:
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity,
eg : Heterophylly in cotton and coriander. In such plants, the leaves of the juvenile plant are different in shape from those in mature plants.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
On the other hand, the difference in shapes of leaves produced in air and those produced in water in buttercup also represent he heterophyllous development due to the environment. This phenomenon of heterophylly is an example of plasticity.

Question 8.
Write the physiological effects of Cytokinins.
Answer:

  1. Cytokinin promotes cell division in the presence of auxin (IAA).
  2. Induces cell enlargement associated with IAA and gibberellins
  3. Cytokinin can break the dormancy of certain light-sensitive seeds like tobacco and induces seed germination.
  4. Cytokinin promotes the growth of lateral bud in the presence of apical bud.
  5. Application of cytokinin delays the process of aging by nutrient mobilization. It is known as Richmond Lang effect.
  6. Cytokinin:
    • increases rate protein synthesis
    • induces the formation of inter-fascicular cambium
    • overcomes apical dominance
    • induces formation of new leaves, chloroplast and lateral shoots.
  7. Plants accumulate solutes very actively with the help of cytokinins.

Question 9.
Describe the mechanism of photoperiodic induction of flowering.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
Photoperiodic stimulus is perceived by the leaves. Floral hormone is synthesised in leaves and translocated to the apical tip to promote flowering. This can be explained by a simple experiment on Cocklebur (Xanthium pensylvanicum), a short day plant. Usually Xanthium will flower under short day conditions. If the plant is defoliated and kept under short day conditions it will not flower.

Flowering will occur even when all the leaves are removed except one leaf. If a cocklebur plant is defoliated and kept under long day conditions, it will not flower. If one of its leaves is exposed to short day condition and rest are in long day condition, flowering will occur.

Question 10.
Give a brief account on Pr grammed Cell Death (PCD).
Answer:
Senescence is controlled by plants own genetic program and death of the plant or plant part consequent to senescence is called Programmed Cell Death. In short senescence of an individual cell is called PCD. The proteolytic enzymes involving PCD in plants are phytases and in animals are caspases. The nutrients and other substrates from senescing cells and tissues are remobilized and reallocated to other parts of the plant that survives.

The protoplasts of developing xylem vessels and tracheids die and disappear at maturity to make them functionally efficient to conduct water for transport. In aquatic plants, aerenchyma is normally formed in different parts of the plant such as roots and stems which encloses large air spaces that are created through PCD. In the development of unisexual flowers, male and female flowers are present in earlier stages, but only one of these two completes its development while other aborts through PCD.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Samacheer Kalvi 11th Bio Botany Plant Growth and Development Additional Questions & Answers

I. Choose T he correct answer (1 Mark)
Question 1.
Open form of the growth occurs in:
(a) leaves and flowers
(b) stem and root
(c) leaves and stem
(d) stem and flowers
Answer:
(b) stem and root

Question 2.
Bamboo is classified under:
(a) monocarpi c annual plants
(b) polycarpic perennials
(c) monocarpic perennials
(d) polycarpic annual plants
Answer:
(c) monocarpic perennials

Question 3.
Primary growth of the plant is due to the activity of:
(a) phloem parenchyma
(b) phloem meristem
(c) vascular cambium
(d) apical meristem
Answer:
(d) apical meristem

Question 4.
One single maize root apical meristem can give rise to more than:
(a) 17,500 hew cells per hour
(b) 18,500 new cells per hour
(c) 19,000 new cells per hour
(d) 500 new cells per hour
Answer:
(a) 17,500 hew cells per hour

Question 5.
Thickening and differentiation of cells take place during:
(a) elongation phase
(b) formative phase
(c) maturation phase
(d) flowering phase
Answer:
(c) maturation phase

Question 6.
When die total growth of a plant is plotted against time, the shape of the curve obtained is:
(a) hyperbolic curve
(b) ‘S’ shaped sigmoid curve
(c) linear curve
(d) none of the above
Answer:
(b) ‘S’ shaped sigmoid curve

Question 7.
The total growth of the plant consists of four phases in the following order.
(a) Log phase, lag phase, decelerating phase and maturation phase
(b) Log phase, lag phase, maturation phase and decelerating phase
(c) Lag phase, log phase, maturation phase and decelerating phase
(d) Lag phase, log phase, decelerating phase and maturation phase
Answer:
(d) Lag phase, log phase, decelerating phase and maturation phase

Question 8.
Internal factors, that influences the growth of the plant is:
(a) nutrition
(b) light
(c) C / N ratio
(d) oxygen
Answer:
(c) C / N ratio

Question 9.
Absence of light may lead to yellowish in color in plants and this is called:
(a) venation
(b) etiolation
(c) estivation
(d) vernation
Answer:
(b) etiolation

Question 10.
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called:
(a) plasticity
(b) differentiation
(c) dedifferentiation
(d) redifferentiation
Answer:
(d) redifferentiation

Question 11.
Indicate a plant growth regulator from the following:
(a) cytocin
(b) cytokinins
(c) acetic acid
(d) methylene
Answer:
(b) cytokinins

Question 12.
Some of the polyamines are known to behave like:
(a) growth inhibitors
(b) plant hormones
(c) flowering inhibitors
(d) fruit ripening agent
Answer:
(b) plant hormones

Question 13.
The activity of synergistic effect involves the activity of:
(a) auxin and gibberellins
(b) auxin and ethylene
(c) ABA and gibberellins
(d) none of the above
Answer:
(a) auxin and gibberellins

Question 14.
Phytohormones are usually produced to in tips of:
(a) root alone
(b) stem alone
(c) leaves alone
(d) root, stem and leaves
Answer:
(d) root, stem and leaves

Question 15.
The term auxin was first coined by:
(a) Charles Darwin
(b) Kogl
(c) F.W. Went
(d) Smith
Answer:
(c) F.W. Went

Question 16.
Indole Acetic Acid (IAA) is a:
(a) growth inhibitor
(b) hetero auxin
(c) root inhibitor
(d) synthetic auxin
Answer:
(b) hetero auxin

Question 17.
Indicate a synthetic auxin.
(a) Indole Acetic Acid
(b) Phenyl Acetic Acid
(c) Indole Butyric Acid
(d) Napthalene Acetic Acid
Answer:
(d) Napthalene Acetic Acid

Question 18.
Auxin has a similar chemical structure of:
(a) Indole acetic acid
(b) Napthalene acetic acid
(c) Phenyl acetic acid
(d) 2,4 – Dichloro phenoxy
Answer:
(a) Indole acetic acid

Question 19.
Auxin stimulates:
(a) transpiration
(b) respiration
(c) flowering
(d) none of the above
Answer:
(b) respiration

Question 20.
The term gibberllin was named by:
(a) Brain
(b) Yabuta
(c) Sumiki
(d) kurosawa
Answer:
(b) Yabuta

Question 21.
Who established the structure of gibberellic acid?
(a) Brain etal
(b) Kurosawa
(c) Cross et al
(d) Yabuta and Sumiki
Answer:
(c) Cross et al

Question 22.
Formation of seedless fruits without fertilization is induced by:
(a) auxin
(b) cytokinin
(c) ethylene
(d) gibberellin
Answer:
(d) gibberellin

Question 23.
Cytokinins inducing cell division was first demonstrated by:
(a) Haberlandt
(b) Charles Darwin
(c) Clarke
(d) Hubert
Answer:
(a) Haberlandt

Question 24.
Zeatin is first isolated from unripe grains of:
(a) paddy
(b) wheat
(c) maize
(d) com
Answer:
(c) maize

Question 25.
Indicate correct statements.
(i) Genes are intracellular factors for growth.
(ii) Temperature has no role in the growth of plant.
(iii) Oxygen has a vital role in the growth of plants.
(iv) CIN ratio of soil does not affect the growth of plant.
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Answer:
(c) (i) and (iii)

Question 26.
Aspartic acid is classified under:
(a) freeauxin
(b) precursor of auxin
(c) chemical structure of auxin
(d) bound auxin
Answer:
(d) bound auxin

Question 27.
The stress phytohormones (Abscisic acid) was first isolated by:
(a) Linn et al
(b) Addicott et al
(c) Edward et al
(d) Stone and Black
Answer:
(b) Addicott et al

Question 28.
The chemical structure of abscisic acid resembles the structure of:
(a) indole Acetic Acid
(b) malanic acid
(c) carotenoid
(d) xanthophyll
Answer:
(c) carotenoid

Question 29.
Pick out the correct statement from the following:
(i) Abscisic acid is found abundantly inside the chloroplast of green cells.
(ii) ABA is a powerful growth promotor.
(iii) ABA is formed from pentose phosphate pathway.
(iv) ABA has anti-auxih and anti-gibberellin property.
(a) (i) and (iv)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(a) (i) and (iv)

Question 30.
Abscisic acid induces male flower formation on female plants of:
(a) potato
(b) Cannabis sativa
(c) Vinca rosea
(d) Delomix regia
Answer:
(b) Cannabis sativa

Question 31.
Pea and barley are classified under:
(a) short day plants
(b) short long day plants
(c) long day plants
(d) long short day plants
Answer:
(c) long day plants

Question 32.
The term ‘photoperiodism’ was coined by:
(a) Miller and Amald
(b) Gamer and Allard
(c) Michael and Edward
(d) Darwin and Lamark
Answer:
(b) Gamer and Allard

Question 33.
Usually Xanthiumpensylvanicum will flower under:
(a) long day condition
(b) short long day condition
(c) photoneutral condition
(d) short day condition
Answer:
(d) short day condition

Question 34.
Phytochrome is a:
(a) reddish xanthophyll pigment
(b) bluish biliprotein pigment
(c) rodopsin pigment
(d) none of the above
Answer:
(b) bluish biliprotein pigment

Question 35.
Who found out the phytochrome in plants?
(a) Butler et al
(b) Michell et al
(c) Boumick et al
(d) Gamers and Allard
Answer:
(a) Butler et al

Question 36.
The term “vernalization” was first used by:
(a) Gamer
(b) Michell
(c) Lysenko
(d) Kawasacki
Answer:
(c) Lysenko

Question 37.
Pick out the wrong statement from the following:
(a) Vernalization increases the cold resistance of plants
(b) It increase the resistance of plants to fungal disease
(c) Vemalizatiqn increase the vegetative period of the plant
(d) It accelerates the plant breeding
Answer:
(c) Vemalizatiqn increase the vegetative period of the plant

Question 38.
In Oxalis, the seed viability ranges from:
(a) 10 to 15 years
(b) a few days
(c) more than 100 years
(d) upto 100 years
Answer:
(b) a few days

Question 39.
In apple and plum, the method of breaking seed dormancy involves the process of:
(a) impaction
(b) Scarification
(c) exposing to red light
(d) Stratification
Answer:
(d) Stratification

Question 40.
The proteolytic enzymes involved in – programmed cell death in plants are:
(a) phytochrome
(b) caspases
(c) phytaspases
(d) protolysis
Answer:
(c) phytaspases

II. Answer the following (2 Marks)

Question 1.
Define closed form of growth in plants.
Answer:
Leaves, flowers and fruits are limited in growth or of determinate or closed form growth.

Question 2.
What is meant by grand period of growth in plants?
Answer:
The total period from initial to the final stage of growth is called the grand period of growth. The total growth is plotted against time and ‘S’ shaped sigmoid curve (Grand period curve) is obtained.

Question 3.
Name the phases of growth in ‘S’ shaped growth curve.
Answer:

  • Lag phase
  • Log phase
  • Decelerating phase
  • Maturation phase

Question 4.
Define arithmetic growth rate in plant organ.
Answer:
If the length of a plant organ is plotted against time, it shows a linear curve and this growth is called arithmetic growth.

Question 5.
Distinguish between absolute growth rate and relative growth, rate.
Answer:
Absolute growth rate:
Increase in total growth of two organs measured and compared per unit time is called absolute growth rate.

Relative growth rate:
The growth of the given system per unit time expressed per unit initial parameter is called relative growth rate.

Question 6.
Define the term etiolation
Answer:
Light has its own contribution in the growth of the plant. Light is important for growth and photosynthesis. Light stimulates healthy growth. Absence of light may lead to yellowish in colour. This is called etiolation.

Question 7.
What is meant by redifferentiation of plant cells?
Answer:
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called redifferentiation, eg: Secondary xylem and Secondary phloem.

Question 9.
Mention any two synthetic auxins.
Answer:

  • 2,4 – Dichloro Phenoxy Acetic Acid (2,4 – D)
  • 2, 4, 5 – Trichloro Phenoxy Acetic Acid (2,4,5 – T)

Question 10.
Explain the synergistic effect of phytochromes.
Answer:
The effect of one or more substance in such a way that both promote each others activity, eg: Activity of auxin and gibberellins or cytokinins.

Question 11.
Name the natural auxins present in plants.
Answer:

  • Indole Acetic Acid (IAA)
  • Indole Propionic Acid (IPA)
  • Indole Butyric Acid (IBA)
  • Phenyl Acetic Acid (PAA)

Question 12.
Mention any two physiological effect of auxins in plant.
Answer:

  • They promote cell elongation in stem and coleoptile.
  • At higher concentrations auxins inhibit the elongation of roots but induce more lateral roots. Promotes growth of root only at extremely low concentrations.

Question 13.
Match the following.

(i) Indole acetic acid (a) bolting
(ii) Napthalene acetic acid (b) anti-auxin
(iii) Gibberellins (c) synthetic auxin
(iv) Abscisic acid (d) Natural auxin

Answer:
(i) – (d) Natural auxin
(ii) – (c) synthetic auxin
(iii) – (a) bolting
(iv) – (b) anti-auxin

Question 14.
Where do you find cytokinin hormone in plants?
Answer:
The distribution of cytokinin in plants is not as wide as those of auxin and gibberellins but found mostly in roots. Cytokinins appear to be translocated through xylem.

Question 15.
What is Richmond Lang effect?
Answer:
Application of cytokinin delays the process of aging by nutrient mobilization. It is known as Richmond Lang effect.

Question 16.
What is meant by non-climacteric fruits?
Answer:
All fruits cannot be ripened by exposure to ethylene. Such fruits are called nonclimacteric fruits and are insensitive to ethylene, eg: Grapes, Watermelon, Orange.

Question 17.
Why do you call Abscisic acid (ABA) as stress hormone?
Answer:
It inhibits the shoot growth and promotes growth of root system. This character protect the plants from water stress. Hence, ABA is called as stress hormone.

Question 18.
What is meant by short day plants?
Answer:
The plants that require a short critical day length for flowering are called short day plants or long night plants, eg: Tobacco, Cocklebur, Soybean, Rice and Chrysanthemum.

Question 19.
Write down the importance of photoperiodism in plants.
Answer:

  • The knowledge of photoperiodism plays an important role in hybridisation experiments.
  • Photoperiodism is an excellent example of physiological pre-conditioning that is using an external factor to induce physiological changes in the plant.

Question 20.
Define the term vernalization.
Answer:
Besides photoperiod certain plants require a low temperature exposure in their earlier stages for flowering. Many species of biennials and perennials are induced to flower by low temperature exposure (0°C to 5°C). This process is called Vernalization.

Question 21.
What is meant by Epigeal germination?
Answer:
During epigeal germination cotyledons are pushed out of the soil. This happens due to the elongation of the hypocotyl. eg: Castor and Bean.

Question 22.
Explain the term seed dormancy.
Answer:
The condition of a seed when it fails to germinate even in suitable environmental condition is called seed dormancy.

Question 23.
Define the term phytogerontology.
Answer:
The branch of botany which deals with ageing, abscission and senescence is called Phytogerontology.

Question 24.
Explain the term programmed cell death.
Answer:
Senescence is controlled by plants own genetic programme and death of the plant or plant part consequent to senescence is called Programmed Cell Death. In short senescence of an individual cell is called PCD.

Question 25.
Define the term “Abscission”.
Answer:
Abscission is a physiological process of shedding of organs like leaves, flowers, fruits and seeds from the parent plant body.

III. Answer the following (3 Marks)

Question 1.
Explain the different phases of growth in plants.
Answer:
There are three phases of growth.
1. Formative phase:
Growth in this phase occurs in. meristematic cells of shoot and root tips. These cells are small in size, have dense protoplasm, large nucleus and small vacuoles. Cells divide continuously by mitotic cell division. Some cells retain capability of cell division while other cells enter the next phase of growth.

2. Elongation Phase:
Newly formed daughter cells are pushed out of the meristematic zone and increases the volume. It requires auxin and food supply, deposition of new cell wall materials (intussusception), addition of protoplasm and development of central vacuole take place.

3. Maturation Phase:
During this stage cells attain mature form and size. Thickening and differentiation takes place. After differentiation, the cells do not grow further.

Question 2.
Draw the ‘S’ shaped growth curve and mark the different phases of growth
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 3.
Mention the internal factors, that affect the growth of plant.
Answer:

  • Genes are intracellular factors for growth.
  • Phytohormones are intracellular factors for growth, eg: auxin, gibberellin, cytokinin.
  • C/N ratio.

Question 4.
What are the characteristic features of phytohormones?
Answer:

  • Usually produced in tips of roots, stems and leaves.
  • Transfer of hormones “from one place to another takes part through conductive systems.
  • They are required in trace quantities.
  • AH hormones are organic in nature.
  • There are no specialized cells or organs for their secretion.
  • They are capable of influencing physiological activities leading to promotion, inhibition and modification of growth.

Question 5.
List out the agricultural applications of auxins.
Answer:

  • It is used to eradicate weeds, eg: 2,4 – D and 2,4,5 – T.
  • Synthetic auxins are used in the formation of seedless fruits (Parthenocarpic fruit).
  • It is used to break the dormancy in seeds.
  • Induce flowering in Pineapple by NAA & 2,4 – D.
  • Increase the number of female flowers and fruits in cucurbits.

Question 6.
Mention any three physiological effects of gibberellins.
Answer:

  • It produces extraordinary elongation of stem caused by cell division and cell elongation.
  • Rosette plants (genetic dwarfism) plants exhibit excessive inter modal growth when they are treated with gibberellins. This sudden elongation of stem followed by flowering is called bolting.
  • Gibberellin breaks dormancy in potato tubers.

Question 7.
What are the uses of ethylene in agriculture?
Answer:

  • Ethylene normally reduces flowering in plants except in Pine apple and Mango.
  • It increases the number of female flowers and decreases the number of male flowers.
  • Ethylene spray in cucumber crop produces female flowers and increases the yield.

Question 8.
What is meant by climacteric fruits?
Answer:
In most of the plants, there is sharp rise in respiration rate near the end of the development of fruit, called climacteric  rise. Such fruits are called climacteric fruits. The ripening on demand can be induced in these fruits by exposing them to normal air containing about 1 ppm of ethylene. A liquid called ethephon is being used in fruit ripening as it continuously releases ethylene, eg: Tomato, Apples, Banana, Mango.

Question 9.
Give the classification of plants based on photoperiodism.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 10.
Explain the term photoperiodic induction.
Answer:
An appropriate photoperiod in 24 hours cycle constitutes one inductive cycle. Plants , may require one or more inductive cycles for flowering. The phenomenon of conversion of leaf primordia into flower primordia under the influence of suitable inductive cycles is called photoperiodic induction.
eg: Xanthium (SDP) – 1 inductive cycle and Plantago (LDP) – 25 inductive cycles.

Question 11.
What are the practical applications of vernalization in plants?
Answer:

  • Vernalization shortens’ the vegetative period and induces the plant to flower earlier.
  •  It increases the cold resistance of the plants.
  • It increases the resistance of plants to fungal disease.
  • Plant breeding can be accelerated.

Question 12.
Write down the internal factors, that affect seed germination.
Answer:
1. Maturity of embryo:
The seeds of some plants, when shed will contain immature embryo. Such seeds germinate only after maturation of embryo.

2. Viability:
Usually seeds remain viable or living only for a particular period. Viability of seeds range from a few days (eg: Oxalis) to more than hundred years. Maximum viability (1000 years) has been recorded in lotus seeds. Seeds germinate only within the period of viability.

3. Dormancy:
Seeds of many plants are dormant at the time of shedding.

Question 13.
Mention the factors causing dormancy of seeds.
Answer:

  • Hard, tough seed coat causes barrier effect as impermeability of water, gas and restriction of the expansion of embryo prevents seed germination.
  • Many species of seeds produce imperfectly developed embryos called rudimentary embryos which promotes dormancy.
  • Lack of specific light requirement leads to seed dormancy.
  • A range of temperatures either higher or lower cause dormancy.
  • The presence of inhibitors like phenolic compounds which inhibits seed germination cause dormancy.

Question 14.
What are the significances of abscission?
Answer:

  • Abscission separates dead parts of the plant, like old leaves and ripe fruits.
  • It helps in dispersal of fruits and continuing the life cycle of the plant.
  • Abscission of leaves in deciduous plants helps in water conservation during summer.
  • In lower plants, shedding of vegetative parts like gemmae or plantlets help in vegetative reproduction.

IV. Answer the following (5 Marks)

Question 1.
Describe the geometric growth rate in plants with suitable diagram.
Answer:
Geometric growth rate:
This growth occurs in many higher plants and plant organs and is measured in size or weight. In plant growth, geometric cell division results if all cells of an organism or tissue are active mitotically. eg: Round three in the given figure, produces 8 cells as 23 – 8 and after round 20 there are 220 = 1,048,576 cells. The large plant or animal parts are produced this way. In fact, it is common in animals but rare in plants except when they are young and small. Exponential growth curve can be expressed as,
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

W1 = W0ert
W1 = Final size (weight, height and number)
W0 = Initial size at the beginning of the period
r = Growth rate
t = Tittle of growth
e = Base of the natural logarithms

Here V is the relative growth rate and also a measure of the ability of the plant to produce new plant material, referred to as efficiency index. Hence, the final size of W1 depends on the initial size W0.

Question 2.
Describe the experiment to measure the increase in length of the stem tip using an arc auxanometer.
Answer:
The increase in the length of the stem tip can easily be measured by an arc auxanometer which consists of a small pulley to the axis of which is attached a long pointer sliding over a graduated arc. A thread one end of which is tied to the stem tip and another end to a weight passes over the pulley tightly.

As soon as the stem tip increases in length, the pulley moves and the pointer slide over the graduated arc (Figure). The reading is taken. The actual increase in the length of the stem is then calculated by knowing the length of the pointer and the radius of the pulley. If the radius of the pulley is 4 inches and the length of pointer 20 inches the actual growth is measured as follows:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 3.
Write an essay on the phytochrome, Gibberelf ns in plants.
Answer:
1. Discovery:
The effect of gibberellins had been known in Japan since early 1800 where certain rice plants were found to suffer from ‘Bakanae’ or foolish seedling disease. This disease was found . by Kurosawa (1926) to be caused by a fungus Gibberella fujikuroi. The active substance was separated from fungus and named as gibberellin by Yabuta (1935). These are more than 100 gibberellins reported from both fungi and higher plants. They are noted as GA1, GA2, GA3 and so on. GA3 is the first discovered gibberellin. In 1938, Yabuta and Sumiki isolated gibberellin in crystalline form. In 1955, Brain et al., gave the name gibberellic acid. In 1961, Cross et al., established its structure.

2. Occurrence:
The major site of gibberellin production in plants is parts like embryo, roots and young leaves near the tip. Immature seeds are rich in gibberellins.

3. Precursors:
The gibberellins are chemically related to terpenoids (natural rubber, carotenoids and steroids) formed by 5 – C precursor, an Isoprenoid unit called Iso Pentenyl Pyrophosphate (IPP) through a number of intermediates. The primary precursor is acetate.

4. Chemical structure:
All gibberellins have gibbane ring structure.

5. Transport in plants:
The transport of gibberellins in plants is non-polar. Gibberellins are translocated through phloem and also occur in xylem due to lateral movement between vascular bundles.

6. Bioassay (Dwarf Pea. assay):
Seeds of dwarf pea are allowed to germinate till the formation of the coleoptile. GA solution is applied to some seedlings. Others are kept under control. Epicotyle length is measured and as such, GA stimulating epicotyle growth can be seen.

7. Physiological Effects:
It produces extraordinary elongation of stem caused by cell division and cell elongation.

  • Rosette plants (genetic dwarfism) plants exhibit excessive intermodal growth when they are treated with gibberellins. This sudden elongation of stem followed by flowering is called bolting.
    Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
  • Gibberellin breaks dormancy in potato tubers.
  • Many biennials usually flower during second year of their growth. For flowering to take place, these plants should be exposed to cold season. Such plants could be made to flower without exposure to cold season in the first year itself, when they are treated with gibberellins.

8. Agricultural role:

  • Formation of seedless fruits without fertilization is induced by gibberellins eg: Seedless tomato, apple and cucumber.
  • It promotes the formation of male flowers in cucurbitaceae. It helps in crop improvement.
    Uniform bolting and increased uniform seed production.
  • Improves number and size of fruits in grapes. It increase yield.
  • Promotes elongation of inter-node in sugarcane without decreasing sugar content.
  • Promotion of flowering in long day plants even under short day conditions. .
  • It stimulates the seed germination

Question 4.
What are their physiological effects of Abscisic acid in – plants and its role in agriculture?
Answer:
Physiological effects:

  • It helps in reducing transpiration rate by closing stomata. It inhibits K+ uptake by guard cells and promotes the leakage of malic acid. It results in closure of stomata;
  • It spoils chlorophylls, proteins and nucleic acids of leaves making them yellow.
  •  Inhibition of cell division and cell elongation.
  •  ABA is a powerful growth inhibitor. It causes 50% inhibition of growth in Oat coleoptile.
  • It induces bud and seed dormancy.
  • If promotes, the abscission of leaves, flowers and fruits by forming abscission layers.
  • ABA plays an important role in plants dtiring water stress -and during drought conditions. It results in loss of turgor and closure of stomata,
  • It has anti-auxin and anti-gibberellin property.
  • Abscisic acid promotes senescence in leaves’ by causing loss of chlorophyll pigment decreasing the rate of photosynthesis and changing the rate of proteins and nucleic acid synthesis

Agricultural Role:

  • In Cannabis sativa, induces male flower formation on female plants.
  • Induction of flowers in short day plants.
  • It promotes sprouting in storage organs like Potato.
  • ABA plays an important role in plants during water stress drought conditions.
  • It inhibits the shoot growth and promotes growth of root system. This character protect the plants from water stress. Hence, ABA is called as stress hormone.

Question 5.
Write an essay on the role of ethylene on plant physiology and agriculture.
Answer:
Almost all plant tissues produce ethylene gas in minute quantities.
1. Discovery:
In 1924, Denny found that ethylene stimulates the ripening of lemons. In 1934, R. Gane found that ripe bananas contain abundant ethylene. In 1935, Cocken et al., identified ethylene as a natural plant hormone.

2. Occurrence:
Maximum synthesis occurs during climacteric ripening of fruits and tissues undergoing senescence. It is formed in almost all plant parts like roots, leaves, flowers, fruits and seeds.

3. Transport in plants:
Ethylene can easily diffuse inside the plant through intercellular spaces.

4. Precursor:
It is a derivative of amino acid methionine, linolenic acid and fumaric acid.

5. Bioassay (Gas Chromatography):
Ethylene can be measured by gas chromatography. This technique helps in the detection of exact amount of ethylene from different plant tissues like lemon and orange.

6. Physiological Effects:

  • Ethylene stimulates respiration and ripening in fruits.
  • It stimulates radial growth in sterft and roof and inhibits linear growth.
  • It breaks the dormancy of buds, seeds and storage organs.
  • It stimulates formation of abscission zone in leaves, flowers and fruits. This makes the leaves to shed prematurely.
  • Inhibition of stem elongation (shortening the intemode).
  • In low concentration, ethylene helps in root initiation.
  • Growth of lateral roots and root hairs. This increases the absorption surface of the plant roots.
  • The growth of fruits is stimulated by ethylene in some plants. It is more marked in climacteric fruits.
  • Ethylene causes epinasty.

7. Agricultural role:

  • Ethylene normally reduces flowering in plants except in Pine apple and Mango.
  • It increases the number of female flowers and decreases the number of male flowers.
  • Ethylene spray in cucumber crop produces female flowers and increases the yield.

Question 6.
Explain the two hypothesis of explaining the mechanism of vernalization.
Answer:
Two main theories to explain the mechanism of vernalization are:

1. Hypothesis of phasic development:
According to Lysenko, development of an annual seed plant consists of two phases. First phase is thermostage, which is vegetative phase requiring low temperature and suitable moisture. Next phase is photo stage which requires high temperature for synthesis of florigen (flowering hormone).

2. Hypothesis of hormonal involvement:
According to Purvis (1961), formation of a substance A from its precursor is converted into B after chilling. The substance B is unstable. At suitable temperature B is converted into stable compound D called Vemalin. Vernalin is converted to F (Florigen). Florigen induces flower formation. At high temperature B is converted to C and devemalization occurs.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 7.
Write an essay on the types of senescence, its physiology and the factors affecting senescence.
Answer:
1. Types of Senescence:
Leopold (1961) has recognised four types of senescence.

(a) Overall senescence:
This kind of senescence occurs in annual plants when entire plant gets affected and dies, eg: Wheat and Soybean. It also occurs in few perennials also, eg: Agave and Bamboo.

(b) Top senescence:
It occurs in aerial parts of plants. It is common in perennials, underground and root system remains viable, eg: Banana and Gladiolus.

(c) Deciduous senescence:
It is common in deciduous plants and occurs only in leaves of plants, bulk of the stem and root system remains alive, eg: Elm and Maple.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

(d) Progressive senescence:
This kind of senescence is gradual. First it occurs in old leaves followed by new leaves f then stem and finally root system. It is common in annuals.

2. Physiology of Senescence:

  • Cells undergo changes in structure.
  • Vacuole of the cell acts as lysosome and secretes hydrolytic enzymes.
  • The starch content is decreased in the cells.
  • Photosynthesis is reduced due to loss of chlorophyll accompanied by synthesis and accumulation of anthocyanin pigments, therefore the leaf becomes red.
  • There is a marked decrease in protein content in the senescing organ.
  • RNA content of the leaf particularly rRNA level is decreased in the cells due to increased activity of the enzyme RNAase.
  • DNA molecules in senescencing leaves degenerate by the increased activity of enzyme DNAase.

3. Factors affecting Senescence:

  •  ABA and ethylene accelerate senescence while auxin and cytokinin retard senescence.
  • Nitrogen deficiency increases senescence whereas nitrogen supply retards, senescence. High temperature senescence but low retards senescence.
  • Senescence is rapid in dark than in light.
  • Water stress leads to accumulation of ABA leading to senescence.

Question 8.
Describe the methods of breaking dormancy of seeds in plants.
Answer:
The dormancy of seeds can be broken by different methods. These are:
1. Scarification:
Mechanical and chemical treatments like cutting or chipping of hard tough seed coat and use of organic solvents to remove waxy or fatty compounds are called as Scarification.

2. imp-action:
in some seeds water and oxygen are unable to penetrate micropyle due to blockage by cork cells. These seeds are shaken vigorously to remove the plug which is called Imp-action.

3. Stratification:
Seeds of rosaceous plants (Apple, Plum, Peach and Cherry) will not germinate until they have been exposed to well aerated, moist condition under low temperature (0°C to 10°C) for weeks to months. Such treatment is called Stratification.

4. Alternating temperatures: Germination of some seeds is strongly promoted by alternating daily temperatures. An alternation of low and high temperature improves the germination of seeds.

5. Light:
The dormancy of photoblastic seeds can be broken by exposing them to red light.

Solution To Activity

Textbook Page No : 164

Question 1.
Demonstration of phases of growth.
Answer:
To demonstrate and study the phases of growth, germinate a few seeds of bean on a circular filter paper soaked with water in a petridish. After two days of growth, select a few seedlings with straight radical of 2 to 3 cm length. Dry the surface of radical with a blotting paper and mark the radical from tip to base with at least 2 mm gap using water proof ink. Replace the seedlings in filter paper and observe further growth.

Observation:
The marked area in the radical will grow and increase in length and hence the marked area of 2mm is found to be grow beyond 2mm size due to the growth in the radical.

Textbook Page No: 169

Question 2.
Measurement of growth by direct method.
Answer:
Step 1: Take ordinary scale.
Step 2: Measure ground stem up to the growing point of the plant.
Step 3: Use Indian ink and mark at regular intervals to measure the length of root, stem, and girth of the trunk.

Observation:
At regular intervals measure the increase in length and girth of the trunk and it can be observed that the length of the root, stem and girth of the trunk increased with the increase in the period of growth.

 

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Samacheer Kalvi 11th Bio Zoology Solutions Chapter 9 Locomotion and Movement

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Tamilnadu Samacheer Kalvi 11th Bio Zoology Solutions Chapter 9 Locomotion and Movement

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Samacheer Kalvi 11th Bio Zoology Locomotion and Movement Text Book Back Questions and Answers

Textbook Evaluation Solved
Question 1.
Muscles are derived from ………………..
(a) Ectoderm
(b) Mesoderm
(c) Endoderm
(d) Neuro ectoderm
Answer:
(b) Mesoderm

Question 2.
Muscles are formed by …………………
(a) Myocytes
(b) Leucocytes
(c) Osteocytes
(d) Lymphocytes
Answer:
(a) Myocytes

Question 3.
The muscles attached to the bones are called …………………
(a) Skeletal muscle
(b) Cardiac muscle
(c) Involuntary muscle
(d) Smooth muscles
Answer:
(a) Skeletal muscle

Question 4.
Skeletal muscles are attached to the bones by ………………….
(a) Tendon
(b) Ligament
(c) Pectin
(d) Fibrin
Answer:
(a) Tendon

Question 5.
The bundle of muscle fibres is called ………………….
(a) Myofibrils
(b) Fascicle
(c) Sarcomere
(d) Sarcoplasm
Answer:
(b) Fascicle

Question 6.
The pigment present in the muscle fibre to store oxygen is ……………………
(a) Myoglobin
(b) Troponin
(c) Myosin
(d) Actin
Answer:
(a) Myoglobin

Question 7.
The functional unit of a muscle fibre is …………………..
(a) Sarcomere
(b) Sarcoplasm
(c) Myosin
(d) Actin
Answer:
(a) Sarcomere

Question 8.
The protein present in the thick filament is …………………
(a) Myosin
(b) Actin
(c) Pectin
(d) Leucin
Answer:
(a) Myosin

Question 9.
The protein present in the thin filament is ……………….
(a) Myosin
(b) Actin
(c) Pectin
(d) Leucin
Answer:
(b) Actin

Question 10.
The region between two successive Z-discs is called a …………………
(a) Sarcomere
(b) Microtubule
(c) Myoglobin
(d) Actin
Answer:
(a) Sarcomere

Question 11.
Each skeletal muscle is covered by …………………
(a) Epimysium
(b) Perimysium
(c) Endomysium
(d) Hypomysium
Answer:
(a) Epimysium

Question 12.
Knee joint is an example of ………………..
(a) Saddle joint
(b) Hinge joint
(c) Pivot joint
(d) Gliding joint
Answer:
(b) Hinge joint

Question 13.
Name of the joint present between the atlas and axis is ……………….
(a) Synovial joint
(b) Pivot joint
(c) Saddle joint
(d) Hinge joint
Answer:
(b) Pivot joint

Question 14.
ATPase enzyme needed for muscle contraction is located in ………………
(a) Actinin
(b) Troponin
(c) Myosin
(d) Actin
Answer:
(c) Myosin

Question 15.
Synovial fluid is found in ……………..
(a) Ventricles of the brain
(b) Spinal cord
(c) Immovable joint
(d) Freely movable joints
Answer:
(d) Freely movable joints

Question 16.
Inflammation of joints due to accumulation of uric acid crystals is called as ……………..
(a) Gout
(b) Myasthenia gravis
(c) Dsteoporosis
(d) Osteomalacia
Answer:
(a) Gout

Question 17.
Acetabulum is located in ……………..
(a) Collar bone
(b) Hip bone
(c) Shoulder bone
(d) Thigh bone
Answer:
(b) Hip bone

Question 18.
Appendicular skeleton is ………………..
(a) Girdles and their limbs
(b) Vertebrae
(c) Skull and vertebral column
(d) Ribs and sternum
Answer:
(a) Girdles and their limbs

Question 19.
The type of movement exhibited by the macrophages are …………………
(a) Flagellar
(b) Ciliary
(c) Muscular
(d) Amoeboid
Answer:
(d) Amoeboid

Question 20.
The pointed portion of the elbow is ………………….
(a) Acromion process
(b) Glenoid cavity
(c) Olecranon process
(d) Symphysis
Answer:
(c) Olecranon process

Question 21.
Name the different types of movement?
Answer:

  1. Amoeboid movement
  2. Ciliary movement
  3. Flagellar movement
  4. Muscular movement

Question 22.
Name the filaments present in the sarcomere?
Answer:
Thick and thin filaments are the two types of filaments present inside the sarcomere.

Question 23.
Name the contractile proteins present in the skeletal muscle?
Answer:
Actin and myosin are contractile proteins present in the skeletal muscle.

Question 24.
When describing a skeletal muscle, what does “striated” mean?
Answer:
Each skeletal muscle fibre has a repeated series of dark and light bands. The dark A-bands and light I-bands give a striated appearance to the muscle.

Question 25.
How does an isotonic contraction take place?
Answer:
In isotonic contraction the length of the muscle changes but the tension remains constant. The force produced is unchanged, e.g., lifting dumbbells and weight lifting.

Question 26.
How does an isometric contraction take place?
Answer:
In isometric contraction the length of the muscle does not change but the tension of the muscle changes. The force produced is changed, e.g., pushing against a wall, holding a heavy bag.

Question 27.
Name the bones of the skull?
Answer:
The skull is composed of two sets of bones – cranial and facial bones. It consists of 22 bones of which 8 are cranial bones and 14 are facial bones.

Question 28.
Which is the only jointless bone in human body?
Answer:
Hyoid bone.

Question 29.
List the three main parts of the axial skeleton?
Answer:
The skull, the vertebral column and the ribcage are the three main parts of the axial skeleton.

Question 30.
How is tetany caused?
Answer:
Tetany is caused when rapid muscle spasms occur in the muscles due to deficiency of parathyroid hormone resulting in reduced calcium levels in the body.

Question 31.
How does rigor mortis happen?
Answer:
After the death of an individual, the membrane of muscle cells become more permeable to calcium ions. This happens due to partial contraction of skeletal muscles. The contracted muscles are unable to relax. This condition is known as rigor mortis.

Question 32.
What are the different types of rib bones that form the rib cage?
Answer:
Thoracic vertebrae ribs and sternum together constitute the ribcage.

Question 33.
What are the bones that make the pelvic girdle?
Answer:
Ilium, ischium and pubis make the pelvic girdle.

Question 34.
List the disorders of the muscular system?
Answer:

  1. Myasthenia gravis
  2. Tetany
  3. Muscle fatigue
  4. Atrophy
  5. Muscle pull
  6. Muscular dystrophy

Question 35.
Explain the sliding- filament theory of muscle contraction?
Answer:
Andrew F.Huxley and Rolf Niedergerke proposed the sliding filament theory to explain muscle contraction. According to this theory, overlapping actin and myosin filaments of fixed length slide past one another in an energy requiring process, resulting in muscle contraction.

Question 36.
What are the benefits of regular exercise?
Answer:

  1. The benefits of regular exercise are:
  2. The muscles used in exercise grow larger and stronger.
  3. The resting heart rate goes down.
  4. More enzymes are synthesized in the muscle fibre.
  5. Ligaments and tendons become stronger.
  6. Joints become more flexible.
  7. Protection from heart attack.
  8. Influences hormonal activity.
  9. Improves cognitive functions.
  10. Prevents obesity.
  11. Promotes confidence, esteem.
  12. Aesthetically better with good physique.
  13. Over all well-being with good quality of life.
  14. Prevents depression, stress and anxiety.

In-Text Questions Solved

Question 1.
Which myofilament has the binding sites for calcium? Name the specific molecule that binds with calcium?
Answer:
Actin filament has the binding sites for calcium. Troponin binds to calcium on thin filaments.

Question 2.
All muscles produce movement, but only skeletal muscle is responsible for locomotion. What is meant by this statement?
Answer:
All the muscles, skeletal, smooth and cardiac, produce movement. Smooth muscles control the activities of internal organs like intestine, stomach, lungs, bladder etc. and their actions are involuntary. Cardiac muscles help in the functioning of heart. But only skeletal muscles are responsible for locomotion.

Locomotion is effected by both bones and muscles attached to the bones and takes place by the coordination between muscular contraction and relaxation and skeletal system. Moreover, the functioning of those skeletal muscles is voluntary.

Question 3.
The pelvic girdle is a heavy, strong girdle. How does its structure reflect its function?
Answer:
The pelvic girdle is a heavy structure specialized for weight bearing. This feature enables it to connect the trunk and the legs, support and balance the trunk, and contain and supports the intestines, the urinary bladder, and the internal sex organs.

Question 4.
An exhausted student was attending a lecture. After 30 minutes or so, he lost interest and he let go with a tremendous yawn. To his great distress he couldn’t close his mouth -his lower jaw was locked open. What do you think would have caused it?
Answer:
When he opened his mouth very wide, the mandibular condyle might have slid forward to the point that the joint might have got dislocated.

Samacheer Kalvi 11th Bio Zoology Locomotion and Movement Additional Questions & Answers

I. Choose The Correct Answer

Question 1.
Which of the following is not related to skeletal muscle?
(a) It is attached to the bone
(b) It is striated
(c) It is an involuntary muscle
(d) It brings about movement of the organ
Answer:
(c) It is an involuntary muscle

Question 2.
The skeletal system is derived from
(a) Ectoderm
(b) Endoderm
(c) Mesoderm
(d) Mesoglea
Answer:
(c) Mesoderm

Question 3.
The cytoplasm of the muscle fibre is ……………….
(a) Sarcolemma
(b) Sarcoplasm
(c) Ectoplasm
(d) Endoplasm
Answer:
(b) Sarcoplasm

Question 4.
The thick filament of muscle fibre is made up of ……………….
(a) Actin
(b) Myosin
(c) Tropomyosin
(d) Troponin
Answer:
(b) Myosin

Question 5.
The cranial bones are ………………..
(a) 22
(b) 14
(c) 8
(d) 3
Answer:
(c) 8

Question 6.
Where is the hyoid bone present?
(a) Cranium
(b) Appendicular skeleton
(c) Pectoral girdle
(d) Base of the buccal cavity
Answer:
(d) Base of the buccal cavity

Question 7.
The number of vertebrates is ………………..
(a) 8
(b) 12
(c) 5
(d) 33
Answer:
(d) 33

Question 8.
How many thoracic vertebrates are there?
(a) 7
(b) 12
(c) 5
(d) 4
Answer:
(b) 12

Question 9.
Rib cage protects ……………….
(a) Brain
(b) Kidney
(c) Lungs, heart, liver
(d) Heart
Answer:
(c) Lungs, heart, liver

Question 10.
The fore arm bones are the ………………
(a) Tibia and fibula
(b) Radius and ulna
(c) Carpals and metacarpals
(d) Tarsal and metatarsals
Answer:
(b) Radius and ulna

Question 11.
The joint between carpal and metacarpals is ………………..
(a) Pivot joint
(b) Ball and socket joint
(c) Saddle joint
(d) Hinge joint
Answer:
(c) Saddle joint

Question 12.
Which of the following allows movement in only one direction?
(a) Pivot joint
(b) Ball and socket joint
(c) Saddle joint
(d) Hinge joint
Answer:
(d) Hinge joint

Question 13.
Which of the following disorders is related to endocrine gland?
(a) Myasthenia gravis
(b) Tetany
(c) Atrophy
(d) Muscular dystrophy
Answer:
(b) Tetany

Question 14.
Which of the following arthritis is related to protein metabolism?
(a) Osteoarthritis
(b) Rheumatoid arthritis
(c) Gouty arthritis
(d) Osteoporosis
Answer:
(c) Gouty arthritis

Question 15.
Which of the following is deficiency disorder?
(a) Osteoarthritis
(b) Rheumatoid arthritis
(c) Gouty arthritis
(d) Osteoporosis
Answer:
(d) Osteoporosis

II. Fill in the blanks

Question 1.
The sperm cells show …………………… movement.
Answer:
Flagellar

Question 2.
Skeletal muscle is attached to the bone by a bundle of collagen fibres known as ……………………
Answer:
Tendon

Question 3.
The cytoplasm of the muscle fibre is called the ……………………
Answer:
Sarcoplasm

Question 4.
…………………… is a red coloured respiratory pigment of the muscle fibre.
Answer:
Myoglobin

Question 5.
…………………… are the granules of stored glycogen.
Answer:
Glycosomes

Question 6.
…………………… is the functional unit of the skeletal muscle.
Answer:
Sarcomere

Question 7.
The thick filaments are composed of the protein ……………………
Answer:
Myosin

Question 8.
The monomer of the myosin molecule is ……………………
Answer:
Meromyosin

Question 9.
The study of muscle is called ……………………
Answer:
Myology

Question 10.
The junction between the motor neuron and the sarcolemma of the muscle fibre is called the ……………………
Answer:
Neuromuscular junction or motor end plate

Question 11.
When nerve impulse reaches a neuromuscular junction, …………………… is released.
Answer:
Acetylcholine

Question 12.
In …………………… contraction of the length of the muscle changes but the tension remains constant.
Answer:
Isotonic

Question 13.
In …………………… contraction of the length of the muscle does not change but the tension of the muscle changes.
Answer:
Isometric

Question 14.
The oxidative fibres are called as …………………… fibres.
Answer:
Red muscle

Question 15.
Glycolytic fibres or white muscle fibres lack ……………………
Answer:
Myoglobin

Question 16.
The skeletal system is derived from the ……………………
Answer:
Mesoderm

Question 17.
The RBCs and WBCs are produced in the ……………………
Answer:
Bone marrow

Question 18.
The large hole in the temporal bone is the ……………………
Answer:
External auditory meatus

Question 19.
The lower jaw bone is called ……………………
Answer:
Mandible

Question 20.
…………………… is the only bone without any joint.
Answer:
Hyoid bone

Question 21.
The upper jaw is formed of the ……………………
Answer:
Maxilla

Question 22.
…………………… is the large opening found at the posterior base of the skull.
Answer:
Foramen magnum

Question 23.
Through foramen magnum the medulla oblongata continues as the ……………………
Answer:
Spinal cord

Question 24.
The first vertebra is called as the ……………………
Answer:
Atlas

Question 25.
The second vertebra is called as the ……………………
Answer:
Axis

Question 26.
The cranium protects the ……………………
Answer:
Brain

Question 27.
…………………… is the flat bone on the mid ventral line of the thorax.
Answer:
Sternum

Question 28.
The first seven pairs of ribs are called ……………………
Answer:
True ribs/vertebro-sternal

Question 29.
The 11th and 12th pairs of ribs are called as …………………… ribs.
Answer:
Floating /vertebral

Question 30.
The 8th, 9th and 10th pairs of ribs are called …………………… ribs.
Answer:
Flase /vertebro-chondral

Question 31.
Rib cage plays a role in ……………………
Answer:
Breathing

Question 32.
…………………… is a depression in the pectoral girdle on which the head of the humerus form the shoulder joint.
Answer:
Glenoid cavity

Question 33.
…………………… is situated at the upper end of the ulna which forms the pointed portion of the elbow.
Answer:
Olecranon process

Question 34.
…………………… is the largest, longest and strongest bone in the body.
Answer:
Femur

Question 35.
The head of femur articulates with the …………………… of the pelvis to form the hip joint.
Answer:
Acetabulum

Question 36.
The bone forming cells are called ……………………
Answer:
Osteoblasts

Question 37.
The bone destroying cells are called the ……………………
Answer:
Osteoclasts

Question 38.
…………………… cells give rise to the osteoblasts.
Answer:
Osteogenic

Question 39.
The internal bone surfaces are covered with a delicate connective tissue membrane called the ……………………
Answer:
Endosteum

Question 40.
Between the epiphysis and diaphysis …………………… is present.
Answer:
Epiphyseal plate/growth plate

Question 41.
The …………………… are points of contact between the bones.
Answer:
Joints

Question 42.
Sutures of the flat skull bones are …………………… joints.
Answer:
Fibrous

Question 43.
…………………… are the freely movable joints.
Answer:
Synovial joints/Diarthroses joints

Question 44.
Between atlas and axis …………………… joint is present.
Answer:
Pivot

Question 45.
Between the carpals …………………… joint is present.
Answer:
Plane/gliding

Question 46.
Between the carpal and metacarpal …………………… joint is present.
Answer:
Addle

Question 47.
Between humerus and pectoral girdle …………………… joint is seen.
Answer:
Ball and socket

Question 48.
…………………… is an autoimmune disorder affecting the action of acetylcholine.
Answer:
Myasthenia gravis

Question 49.
The removal of parathyroid gland results in ……………………
Answer:
Tetany

Question 50.
A traumatic pulling of the fibres produces a tear known as ……………………
Answer:
Sprain

Question 51.
…………………… is an inflammatory or degenerative disease that damages the joints.
Answer:
Arthritis

Question 52.
The wearing away of the bone ends of the knees and other movable joints is called ……………………
Answer:
Osteoarthritis

Question 53.
The inflammation of the synovial membranes is known as ……………………
Answer:
Rheumatoid arthritis

Question 54.
Inflammation of joints due to accumulation of uric acid crystals is known as ……………………
Answer:
Gout

Question 55.
The deficiency of vitamin D and hormonal imbalance cause ……………………
Answer:
Osteoporosis

Question 56.
The deficiency of vitamin D causes …………………… in children.
Answer:
Rickets

Question 57.
The deficiency of vitamin D causes …………………… in adults.
Answer:
Osteomalacia

Question 58.
…………………… increase the breathing and heart rate.
Answer:
Endurance or aerobic activities

Question 59.
…………………… protects us from heart attack.
Answer:
Exercise

III. Short Answer Questions

Question 1.
What is amoeboid movement?
Answer:
The movement of cells by streaming movements of the cytoplasm forming pseudo-podia is known as amoeboid movement, e.g., macrophages.

Question 2.
What is ciliary movement?
Answer:
The movement caused by the cilia is ciliary movement, e.g., ciliated epithelial cells of respiratory passage and genital tracts.

Question 3.
What is flagellar movement?
Answer:
The movement due to the lashing of flagella is known as flagellar movement, e.g., sperm cells.

Question 4.
What is muscular movement?
Answer:
The movement of hands, legs, jaws, tongue are caused by the contraction and relaxation of the muscle which is known as muscular movement.

Question 5.
What is fascicle?
Answer:
Each muscle is made up of bundles of muscle fibres called fascicle.

Question 6.
What are myofibrils?
Answer:
Each muscle fibre contains hundreds to thousands of rod-like structures called myofibrils that run parallel to its length.

Question 7.
What is epimysium?
Answer:
The connective tissue covering the whole muscle is the epimysium,

Question 8.
What is perimysium?
Answer:
The connective tissue covering around each fascicle is the perimysium.

Question 9.
What is endomysium?
Answer:
The connective tissue surrounding the muscle fibre is called the endomysium.

Question 10.
Distinguish between voluntary muscle and involuntary muscle.
Answer:

Voluntary muscle

Involuntary muscle

1. The muscle whose functions are under the control of our will is voluntary muscle. 1. The muscle whose functions are not under the control of our will is the involuntary muscle.
2. It is striated and and hence called striated or striped muscle. 2. It has no stripes and hence it is called smooth muscles or unstriped or non-striated muscles.

Question 11.
What is tendon?
Answer:
Skeletal muscle is attached to the bone by a bundle of collagen fibres known as tendon.

Question 12.
What is myoglobin?
Answer:
Myoglobin is a red-coloured respiratory pigment of the muscle fibre.

Question 13.
What is sarcoplasm?
Answer:
The cytoplasm of the muscle fibre is called the sarcoplasm.

Question 14.
What are Glycosomes?
Answer:
Glycosomes are the granules of stored glycogen that provide glucose during the period of muscle fibre activity.

Question 15.
What is sarcomere?
Answer:
The functional unit of the skeletal muscle is known as sarcomere.

Question 16.
Distinguish between thick filaments and thin filaments?
Answer:

Thick filaments

Thin filaments

 1. The thick filaments are composed of the protein myosin. 1. The thin filaments are composed of protein actin.

Question 17.
What is meromyosin?
Answer:
The monomer of myosin molecule is called meromyosin.

Question 18.
Name the proteins which regulates the contraction of muscles?
Answer:
Actin, myosin, tropomyosin and troponin.

Question 19.
What is myology?
Answer:
The study of muscle is called myology.

Question 20.
What are oxidative fibres?
Answer:
The muscle fibres that contain numerous mitochondria and have a high capacity for oxidative phosphorylation are classified as oxidative fibres. They are also called red muscle fibres.

Question 21.
What are glycolytic fibres?
Answer:
The muscle fibres which contain a few mitochondria but possess a high concentration of glycolytic enzymes and large stores of glycogen are called glycolytic fibres. These lack myoglobin and hence these are pale. These muscle fibres are termed as white muscle fibres.

Question 22.
What is hydrostatic skeleton?
Answer:
The skeleton found in soft-bodied invertebrates is called hydrostatic skeleton. It is a fluid filled cavity encircled by muscles, e.g., earthworm.

Question 23.
Distinguish between exoskeleton and endoskeleton.
Answer:

Exoskeleton

Endoskeleton

1. The rigid hard case present outside the body of animals is exoskeleton. 1. The rigid support structure found inside the body of vertebrates is endoskeleton.
2. e.g., Cockroach 2. e.g., Human being

Question 24.
What is a metaphysis?
Answer:
The bone region where the diaphysis and epiphyses meet is called the metaphysis.

Question 25.
What is periosteum?
Answer:
The external surface of the bone is covered by a double layered membrane called periosteum. What is endosteum?

Question 26.
What is endosteum?
Answer:
Internal bone surfaces are covered with a delicate connective tissue membrane called the endosteum.

Question 27.
What is sternum?
Answer:
Sternum is a flat bone on the mid ventral line of the thorax. It provides space for the attachment of the thoracic ribs and abdominal muscles.

IV. Give Long Answers

Question 1.
Explain the structure of a skeletal muscle fibre?
Answer:

  1. Each muscle fibre is thin and elongated.
  2. It has multiple oval nuclei beneath sarcolemma.
  3. The cytoplasm of the muscle fibre is called as sarcoplasm.
  4. It contains glycosomes, the stored glycogen granules, myoglobin, respiratory pigment and sarcoplasmic reticulum.
  5. Actin and myosin are muscle proteins present in the muscle fibre.
  6. Each myofibril has a repeated series of dark and light bands called A-bands and I-bands.
  7. Each dark band has a lighter region in its middle called the H-zone.
  8. Each H-zone is bisected vertically by a dark line called the M-line.
  9. Each light I-band has a darker mid line area called the Z-disc.
  10. Sarcomere is the functional unit of the skeletal muscle. It is a region of a myofibril between two successive Z-discs.
  11. Sarcomere has thick and thin filaments. The thick filaments extend the entire length of the A-band, the thin filaments extend across the I-band and partly into the A-band.
  12. The invagination of the sarcolemma forms transverse tubules (T-tubules) and they penetrate into the junction between the A and I-bands.

 

Question 2.
Write the schematic presentation of muscle contraction?
Answer:
Samacheer Kalvi 11th Bio Zoology Solutions Chapter 9 Locomotion and Movement img 1

Question 3.
Explain the structure of contractile proteins?
Answer:
The thick filaments of the muscle has the protein mysoin. Each myosin molecule is made up of a monomer called meromyosin. It has a globular head, a short arm and a tail. The short arm constitutes the heavy meromyosin (HMM). The tail portion forms the light meromyosin (LMM). The head bears an actin-binding site and an ATP- binding site. It also contains ATPase enzyme that split ATP to generate energy for the contraction of muscle.

The thin filaments are composed of two intertwined actin molecules. It has polypeptide subunits called globular actin or G-actin and filamentous form or F-actin.

Each thin filament is made of two F-actins helically wound to each other. Each F-actin is a polymer of monomeric G-actins. Tropomyosin and troponin are other proteins which help in regulating the contraction of muscles along with actin and myosin.

Question 4.
Explain the types of skeletal muscle fibres?
Answer:
Skeletal muscle fibres are classified into three types. They are:

  1. Slow – oxidative fibres
  2. Fast – oxidative fibres
  3. Fast – glycolytic fibres

1. Slow – oxidative fibres have low rates of myosin ATP hydrolysis but have the ability to make large amounts of ATP. These fibres are used for prolonged, regular activity such as long distance swimming. Long – distance runners have a high proportion of these fibres in their leg muscles.

2. Fast – oxidative fibres have high myosin ATPase activity and can make large amounts of ATP. They are particularly suited for rapid actions.

3. Fast – glycolytic fibres have myosin ATPase activity but cannot make as much ATP as oxidative fibres, because their source of ATP is glycolysis. These fibres are best suited for rapid, intense actions, such as short sprint at maximum speed.

Question 5.
Write the functions of skeletal system?
Answer:
Functions of skeletal system

  1. Support -It forms a rigid framework and supports the weight of the body against gravity.
  2. Shape – It provides and maintains the shape of the body.
  3. Protection – It protects the delicate internal organs of the body.
  4. Acts as reservoir – It stores minerals such as calcium and phosphate. Fat (triglyceride) is stored in yellow bone marrow and represents a source of stored energy for the body.
  5. Locomotion – It acts as lever along with the muscles attached to it.
  6. Strength – It can withstand heavy weight and absorbs mechanical shock.
  7. Asa haemopoietic tissue – Red and white blood cells are produced in the bone marrow of the ribs, spongy bones of vertebrae and extremities of long bones.

 

Question 6.
Explain the bones that form the skull?
Answer:
The skull is composed of two sets of bones – cranial and facial bones. It consists of 22 bones of which 8 are cranial bones and 14 are facial, bones. The cranial bones form the hard protective outer covering of the brain and called the brain box. The capacity of the cranium is 1500 cm3.

These bones are joined by sutures which are immovable. They are a paired parietal, paired temporal and individual bones such as the frontal, sphenoid, occipital and ethmoid. The large hole in the temporal bone is the external auditory meatus. In the facial bones maxilla, zygomatic, palatine, lacrimal, nasal are paired bones whereas mandible or lower jaw and vomer are unpaired bones. They form the front part of the skull.

A single U-shaped hyoid bone is present at the base of the buccal cavity. It is the only one bone without any joint. Each middle ear contains three tiny bones- malleus, incus and stapes collectively are called ear ossicles. The upper jaw is formed of the maxilla and the lower jaw is formed of the mandible.

The upper jaw is fused with the cranium and is immovable. The lower jaw is connected to the cranium by muscles and is movable. The most prominent openings in the skull are the orbits and the nasal cavity. Foramen magnum is a large opening found at the posterior base of the skull. Through this opening the medulla oblongata of the brain descends down as the spinal cord.

Question 7.
Write a short note on the vertebral column?
Answer:
Vertebral column is also called the back bone. It consists of 33 serially arranged vertebrae which are interconnected by cartilage known as intervertebral disc. The vertebral column extends from the base of the skull to the pelvis and forms the main frame work of the trunk. The vertebral column has five major regions.

They are, the cervical, thoracic, lumbar, sacrum (5 sacral vertebrae found in the infant which are fused to form one bone in the adult) and coccyx (4 coccygeal vertebrae found in the infant which are fused to form one bone in the adult).

Each vertebra has a central hollow portion, the neural canal, through which the spinal cord passes. The first vertebra is called as the atlas and the second vertebra is called as the axis. Atlas is articulated with the occipital condyles. The vertebral column protects the spinal cord, supports the head and serves as the point of attachment for the ribs and musculature of the back.

Question 8.
Write a short note on Rib cage?
Answer:
There are 12 pairs of ribs. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end, hence called bicephalic.

The first seven pairs of ribs are called ‘true ribs or vertebro-stemal ribs. Dorsally they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilages.

The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but joined with the cartilaginous (hyaline cartilage) part of the seventh rib. These are called ‘false ribs’ or vertebro-chondral ribs.

The last 11th and 12th pairs of ribs are not connected ventrally. Therefore, they are called as ‘floating ribs’ or vertebral ribs. Thoracic vertebrae, ribs and sternum together form the ribcage. Rib cage protects the lungs, heart, liver and also plays a role in breathing.

Question 9.
Write a note on Pectoral girdle?
Answer:
The upper limbs are attached to the pectoral girdles. These are very light and allow the upper limbs a degree of mobility not seen anywhere else in the body. The girdle is formed of two halves. Each half of the pectoral girdle consists of a clavicle or collar bone and a scapula.

The scapula is a large, thin, triangular bone situated in the dorsal surface of the ribcage between the second and seventh ribs. It has a slightly elevated ridge called the spine which projects as a flat, expanded process called the acromion. The clavicle articulates with this process.

Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint. Each clavicle is a long slender bone with two curvatures which lies horizontally and connects axial skeleton with appendicular skeleton.

Question 10.
Write a note on the bones of the upper limb?
Answer:
The upper limb consists of 30 separate bones and is specialized for1 mobility. The skeleton of the arm, the region between the shoulder and elbow is the humerus. The head of humerus articulates with the glenoid cavity of the scapula and forms the shoulder joint. The distal end of humerus articulates with the two forearm bones the radius and ulna. The forearm is the region between the elbow and the wrist.

Olecranon process is situated at the upper end of the ulna which forms the pointed portion of the elbow. The hand consists of carpals, metacarpals and phalanges. Carpals, the wrist bones, 8 in number are arranged in two rows of four each.

The anterior surface of the wrist has tunnel-like appearance, due to the arrangement of carpals with the ligaments. This tunnel is termed as carpal tunnel. Metacarpals, the pajm bones are 5 in number and phalanges the digits bones are 14 in number.

Question 11.
Explain the structure of Pelvic Girdle?
Answer:
The pelvic girdle is a heavy structure specialized for weight bearing. It is composed of two hip bones called coxal bones that secure the lower limbs to the axial skeleton. Together, with the sacrum and coccyx, the hip bones form the basin-like bony pelvis.

Each coxal bone consists of three fused bones, ilium, ischium and pubis. At the point of fusion of ilium, ischium, and pubis a deep hemispherical socket called the acetabulum is present on the lateral surface of the pelvis.

It receives the head of the femur or thigh bone at the hip joint and helps in the articulation of the femur. Ventrally the two halves of the pelvic girdle meet and form the pubic symphysis containing fibrous cartilage..

The ilium is the superior flaring portion of the hip bone. Each ilium forms a secure joint with the sacrum posteriorly. The ischium is a curved bar of bone.

The V-shaped pubic bones articulate anteriorly at the pubic symphysis. The pelvis of male is deep and narrow with larger heavier bones and the female is shallow, wide and flexible in nature, and this helps during pregnancy which is influenced by female hormones.

Question 12.
Write a note on the bones of lower limb?
Answer:
The lower limb consists of 30 bones which carries the entire weight of the erect body and is subjected to exceptional forces when we jump or run. The bones of the lower limbs are thicker and stronger than the upper limbs.

The three segments of each.lower limb are the thigh, the leg or the shank and the foot. The femur is the single bone of the thigh. It is the largest, longest and strongest bone in the body.

The head of femur articulates with the acetabulum of the pelvis to form the hip joint. Two parallel bones, the tibia and fibula, form the skeleton of the shank.

A thick, triangular patella forms the knee cap, which protects the knee joint anteriorly and improves the leverage of thigh muscles acting across the knee. The foot includes the bones of ankle, the tarsus, the metatarsus and the phalanges or toe bones.

The foot supports our body weight and acts as a lever to propel the body forward, while walking and running. The tarsus is made up of seven bones called tarsals. The metatarsus consists of five bones called metatarsals. The arrangement of the metatarsals is parallel to each other. There are 14 phalanges in the toes which are smaller than those of the fingers.

Question 13.
Explain the structure of a typical long bone?
Answer:
A typical long bone has a diaphysis, epiphyses and membranes. A tubular diaphysis or shaft, forms the long axis of the bone. It is constructed of a thick collar of compact bone that surrounds a central medullary cavity or marrow cavity. The epiphyses are the bone ends.

Compact bone forms the exterior of epiphyses and their interior contains spongy bone with red marrow. The region where the diaphysis and epiphyses meet is called the metaphysis. The external surface of the entire bone except the joint surface is covered by a double-layered membrane called the periosteum.

The outer fibrous layer is dense irregular connective tissue. The inner osteogenic layer consists of osteoblasts (bone- forming cells) which secrete bone matrix elements and osteoclasts (bone-destroying cells). In addition, there are primitive stem cells, osteogenic cells, that give rise to the osteoblasts.

The periosteum is richly supplied with nerve fibres, lymphatic vessels and blood vessels. Internal bone surfaces are covered with a delicate connective tissue membrane called the endosteum. The endosteum covers the trabeculae of spongy bone and lines the canals that pass through the compact bone. It also contains both osteoblasts and osteoclasts. Between the epiphysis and diaphysis epiphyseal plate or growth plate is present.

Question 14.
What are joints?
Answer:
The joints are points of contact between bones.

Question 15.
Explain the types of joints?
Answer:
(I) Fibrous joints or Synarthroses: They are immovable fixed joints in which no movement between the bones is possible. Sutures of the flat skull bones are fibrous joints.

(II) Cartilaginous joints or Amphiarthroses: They are slightly movable joints in which the joint surfaces are separated by a cartilage and slight movement is only possible, e.g., Joints of adjacent vertebrae of the vertebral column.

(III) Synovial joints or Diarthroses joints: They are freely movable joints, the articulating bones are separated by a cavity which is filled with synovial fluid e.g., Pivot joint – between atlas and axis plane/gliding joint – between the carpals

  • Saddle j oint – between the carpal and metacarpal
  • Ball and socket joint – between humerus and pectoral girdle
  • Hinge joint – knee joint
  • Condyloid or angular or ellipsoid-joint between radius and carpal.

Question 16.
Write a short note on myasthenia gravis?
Answer:
Myasthenia gravis: An autoimmune disorder affecting the action of acetylcholine at neuromuscular junction leadihg to fatigue, weakening and paralysis of skeletal muscles. Acetylcholine receptors on the sarcolemma are blocked by antibodies leading to weakness of muscles. When the disease progresses, it can make chewing, swallowing, talking and even breathing difficult.

Question 17.
Explain muscle fatigue?
Answer:
Muscle fatigue is the inability of a muscle to contract after repeated muscle contractions. This is due to lack of ATP and accumulation of lactic acid by anaerobic breakdown of glucose.

Question 18.
Explain muscle atrophy?
Answer:
A decline or cessation of muscular activity results in the condition called atrophy which results in the reduction in the size of the muscle and makes the muscle to become weak, which occurs with lack of usage as in chronic bedridden patients.

Question 19.
Write a short note on muscle pull?
Answer:
Muscle pull is actually a muscle tear. A traumatic pulling of the fibres produces a tear known as sprain. This can occur due to sudden stretching of muscle beyond the point of elasticity. Back pain is a common problem caused by muscle pull due to improper posture with static sitting for long hours.

Question 20.
Write a short note on muscular dystrophy?
Answer:
The group of diseases collectively called the muscular dystrophy are associated with the progressive degeneration of skeletal muscle fibres, weakening the muscles and leading to death from lung or heart failure. The most common form of muscular dystrophy is called ‘ Duchene Muscular Dystrophy (DMD).

Question 21.
Explain the disorders of skeletal system?
Answer:
Arthritis and osteoporosis are the major disorders of skeletal system.
1. Arthritis: Arthritis is an inflammatory or degenerative disease that damages the joints. There are several types of arthritis.

(I) Osteoarthritis: The bone ends of the knees and other freely movable joints wear away as a person ages. The joints of knees, hip, fingers and vertebral column are affected.

(II) Rheumatoid arthritis: The synovial membranes become inflamed and there is an accumulation of fluid in the joints. The joints swell and become extremely painful. It can begin at any age but symptoms usually emerge before the age of fifty.

(III) Gouty arthritis or gout: Inflammation of joints due to accumulation of uric acid crystals or inability to excrete it. It gets deposited in synovial joints.

2. Osteoporosis: It occurs due to deficiency of vitamin D and hormonal imbalance. The bone becomes soft and fragile. It causes rickets in children and osteomalacia in adult females. It can be minimized with adequate calcium intake, vitamin D intake and regular physical. activities.

Question 22.
Explain the basic categories of exercise and physical activity?
Answer:
Exercise and physical activity fall into four basic categories. Endurance, strength, balance and flexibility. Endurance or aerobic activities increase the breathing and heart rate. They keep the circulatory system healthy and improve overall fitness.

Strength exercises make the muscles stronger. They help to stay independent and cany out everyday activities such as climbing stairs and carrying bags.

Balance exercises help to prevent falls which is a common problem in older adults. Many strengthening exercises also improves balance.

Flexibility exercises help to stretch body muscles for more freedom of joint movements.

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Samacheer Kalvi 11th Economics Rural Economy Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
Which is considered as the basic unit for rural areas?
(a) Panchayat
(b) Village
(c) Town
(d) Municipality
Answer:
(b) Village

Question 2.
Which feature is identified with rural areas?
(a) Low population density
(b) High population density
(c) Low natural resources
(d) Low human resources
Answer:
(a) Low population density

Question 3.
Identify the feature of rural economy.
(a) Dependence on agriculture
(b) High population density
(c) Low level of population
(d) Low level of inequality
Answer:
(a) Dependence on agriculture

Question 4.
What percentage of the total population live in rural area, as per 2011 censes?
(a) 40
(b) 50
(c) 68.84
(d) 70
Answer:
(c) 68.84

Question 5.
How do you term people employed in excess over and above the requirements?
(a) Unemployment
(b) Underemployment or Disguised Unemployment
(c) Full employment
(d) Self-employment
Answer:
(b) Underemployment or Disguised Unemployment

Question 6.
What is the term used to denote the coexistence of two different features in an economy?
(a) Technology
(b) Dependency
(c) Dualism
(d) Inequality
Answer:
(c) Dualism

Question 7.
The process of improving the rural areas, rural people and rural living is defined as
(a) Rural economy
(b) Rural economics
(c) Rural employment
(d) Rural development
Answer:
(d) Rural development

Question 8.
Identify the agriculture related problem of rural economy.
(a) Poor communication
(b) Small size of landholding
(c) Rural poverty
(d) Poor banking network
Answer:
(b) Small size of landholding

Question 9.
The recommended nutritional intake per person in rural areas.
(a) 2100 calories
(b) 2100 calories
(c) 2300 calories
(d) 2400 calories
Answer:
(d) 2400 calories

Question 10.
Indicate the cause for rural poverty.
(a) Lack of non-farm employment
(b) High employment
(c) Low inflation rate
(d) High investment.
Answer:
(a) Lack of non-farm employment

Question 11.
What is the other name for concealed unemployment?
(a) Open
(b) Disguised
(c) Seasonal
(d) Rural
Answer:
(b) Disguised

Question 12.
How do you term the employment occurring only on a particular season?
(a) Open
(b) Disguised
(c) Seasonal
(d) Rural
Answer:
(c) Seasonal

Question 13.
Identify an example for rural industries?
(a) Sugar factory
(b) Mat making industry
(c) Cement industry
(d) Paper industry
Answer:
(b) Mat making industry

Question 14.
How much share of rural families in India is in debt?
(a) Half
(b) One fourth
(c) Two third
(d) Three fourth
Answer:
(d) Three fourth

Question 15.
Identify the cause for rural indebtedness in India.
(a) Poverty
(b) High population
(c) High productivity
(d) Full employment
Answer:
(a) Poverty

Question 16.
In which year, Regional Rural Banks came into existence?
(a) 1965
(b) 1970
(c) 1975
(d) 1980
Answer:
(c) 1975

Question 17.
Identify the year of launch of MUDRA Bank?
(a) 1995
(b) 2000
(c) 2010
(d) 2015
Answer:
(d) 2015

Question 18.
Identify the year in which National Rural Health Mission was launched.
(a) 2000
(b) 2005
(c) 2010
(d) 2015
Answer:
(b) 2005

Question 19.
Identify the advantages of rural roads.
(a) Rural marketing
(b) Rural employment
(c) Rural development
(d) All the above
Answer:
(d) All the above

Question 20.
“An Indian farmer is born in debt, lives in debt, dies in debt and bequeaths debt”- who said this?
(a) Adam Smith
(b) Gandhi
(c) Amartya Sen
(d) Sir Malcolm Darling
Answer:
(d) Sir Malcolm Darling

Part – B
Answer the following questions in one or two sentences

Question 21.
Define Rural Economy.
Answer:
Rural economy refers to villages. Rural economics deals with the application of economic principles in understanding and developing rural areas.

Question 22.
What do you mean by Rural Development?
Answer:
According to the world bank, “Rural development is a strategy designed to improve the economic and social life of a specific group of people rural poor”

Question 23.
Rural Poverty – Define.
Answer:
Rural poverty refers to the existence of poverty in rural areas. Poverty in India is the situation in which an individual fails to earn sufficient income to buy the basic minimum of subsistence.

Question 24.
Define Open Unemployment.
Answer:
In open unemployment, unemployed persons are identified as they remain without work. This type of unemployment is found among agricultural labourers, rural artisans and literate persons.

Question 25.
What is meant by Disguised Unemployment?
Answer:
Disguised unemployment is a situation where many are employed below their productive capacity and even if they are withdrawn from work the output will not diminish.

Question 26.
Define Cottage Industry.
Answer:
Cottage industries are generally associated with agriculture and provide both part time and full time jobs in rural areas.

Question 27.
What do you mean by Micro Finance?
Answer:
Micro finance is a financial service that offers loans, savings and insurance to entrepreneurs and small business owners.

Question 28.
State any two causes of housing problem in rural areas.
Answer:

  1. Rapid adaptation of nuclear families.
  2. Problems in provision of proper water supply, good sanitation, proper disposal of sewage.

Question 29.
Define Rural Electrification.
Answer:
Rural electrification refers to providing electrical power to rural areas.

Question 30.
State any two factors hindering Rural Electrification in India.
Answer:

  1. Lack of funds
  2. Inter – state disputes
  3. Uneven terrain

Part – C
Answer the following questions in one Paragraph

Question 31.
State the importance of Rural Development.
Answer:
Importance of Rural Development:

  1. India cannot be developed by retaining rural as backward
  2. As the rural economy supports the urban sector, the backwardness of the rural sector would be a major impediment to the overall progress of the economy.
  3. Improvements in education, health and sanitation in villages can help avoid many urban problems.
  4. To provide gainful employment and improve food production.
  5. The evils of brain drain and rural urban migration can be reduced.
  6. For the better utilization of resources.
  7. To minimize the gap between rural and urban areas.

Question 32.
Explain the causes for Rural Backwardness.
Answer:

  1. Skewed distribution of land.
  2. Lack of non-farm employment.
  3. Lack of public sector investment.
  4. Inflation.
  5. Low productivity.
  6. Unequal benefit of growth.
  7. Low rate of economic growth.
  8. More emphasis on large industries.
  9. Social evils.

Question 33.
Enumerate the remedial measures to Rural Poverty.
Answer:
Creation of employment opportunities would support elimination of poverty. The poverty eradication schemes implemented in India are

  1. 20 point programme.
  2. Integrated Rural Development Programme (IRDP)
  3. Training Rural Youths for Self Employment (TRYSEM)
  4. Food for Work Programme (FWP)
  5. National Rural Employment Programme (NREP)
  6. RLEGP, JRY and MGN REGS

Question 34.
What are the remedial measures for Rural Unemployment?
Answer:

  1. Subsidiary occupation : To reduce the seasonal unemployment rural people should be encouraged to adopt subsidiary occupation.
  2. Rural works programme : Rural works programme should be planned during off season to provide gainful employment.
  3. Irrigation facilities : Irrigation facilities should be expanded to enable the farmers to adopt multiple cropping.
  4. Rural industrialization : New industries should be set up in rural areas.
  5. Technical education : Employment oriented courses should be introduced to literate the youth to start their own units.

Question 35.
Write a note on Regional Rural Banks.
Answer:
Regional rural banks came into existence in 1975. RRBs are recommended with a view to develop rural economy by providing credit and other facilities to the small and marginal farmers, agricultural labourers, artisans and small entrepreneurs.

RRBs are set up by the joint efforts of the centre and state governments and commercial banks. At present there are 64 RRBs in India. RRBs confine their lending’s only to the weaker sections and their lending rates are at par with the prevailing rate of cooperative societies.

Question 36.
Mention the features of SHGs.
Answer:

  1. SHG is an economically homogeneous group formed through a process of self selection based upon the affinity of its members.
  2. Most SHGs are women’s groups with membership ranging between 10 and 20.
  3. SHGs have well-defined rules and by laws, hold regular, meetings and maintain records and savings and credit discipline.
  4. SHGs are self-managed institutions characterized by participatory and collective decision making.

Question 37.
List out the objectives of MUDRA Bank.
Answer:

  1. Regulate the lender and the borrower of micro finance and bring stability to the micro finance system.
  2. Extend finance and credit support to micro finance institutions.
  3. Register all MFIs and introduce a system of performance rating and accreditation for the past time.
  4. Offer a credit guarantee scheme for providing guarantees to loans being offered to micro businesses.
  5. Introduce appropriate technologies to assist in the process of efficient lending, borrowing and monitoring of distributed capital.

Part – D
Answer the following questions in about a page

Question 38.
‘The features of Rural Economy are peculiar’- Argue.
Answer:
The Rural economy has very peculiar features which are :

  1. Village is an institution : The village is a primary institution and it satisfies almost all the needs of the rural community.
  2. Dependence on agriculture : Agriculture and allied activities are the main occupation in rural areas.
  3. Life of rural people: Life styles in villages are very simple. Public services like education, housing,health and sanitation, transport and communication, banking, roads and markets are limited and unavailable.
  4. Population density : Density of population is very low and houses are scattered in the entire villages.
  5. Employment: There exists unemployment, seasonal unemployment and underemployment in rural areas.
  6. Poverty : According to the 2011 – 12 estimates about 22 crores of people in rural areas live below the poverty line.
  7. Indebtedness : Sir Malcolm Darling stated that an Indian farmer is born in debt, lives in debt, dies in debt and bequeaths debt.
  8. Rural income : As the rural economy is not sufficiently vibrant to provide them with jobs or self employment opportunities the income of the rural people is constrained.
  9. Dependency : Rural households are dependent on social grants and remittances from family members working in urban areas and cities.
  10. Dualism : Dualism means the co-existence of two extremely different features which is the common characteristics in rural area.
  11. Inequality : The distribution of income, wealth and assets are highly skewed among rural people.
  12. Migration : Rural people are forced to migrate to urban areas to seek employment which gives rise to the formation of cities. Enmity and lack of basic amenities in rural areas also push the people to migrate. This is called double poisoning by Schumacher.

Question 39.
Discuss the problems of Rural Economy.
Answer:
The problems of rural economy are.
1. People related problems : The problem consists of illiteracy, lack of technical know how, low level of confidence, dependence on sentiments and beliefs etc.

2. Agriculture related problems : This include lack of awareness, knowledge, skill and attitude, unavailability of inputs, poor marketing facility, insufficient extension of staff and services, small size of land holding, absence of infrastructure, primitive technology, reduced public investment and absence of role for farmers in fixing the prices for their own products.

3. Infrastructural related problems : Poor infrastructure facilities like water, electricity, transport, educational institutions communication, health, employment are found in rural areas.

4. Economics related problems : Inability to adopt high cost technology, high cost of inputs, under privileged rural industries, low income, indebtedness and existence of inequality in land holdings and assets.

5. Leadership related problems : Leadership among the hands of inactive and incompetent people, self-interest of leaders, biased political will, less bargaining power and negotiation skills and dominance of political leaders.

6. Administrative problems : Political interference, lack of motivation and interest, low wages in villages, improper utilization of budget and absence of monitoring and implementation of rural development programme.

Question 40.
Analyse the causes for Rural Indebtedness.
Answer:

1. Poverty of farmers : The vicious circle of poverty fore es the farmers to borrow for consumption, cultivation and celebrations. Thus, poverty’ debt and high rates of interest hold the farmer in the grip of money lenders.

2. Failure of monsoon : Frequent failure of monsoon is a curse to the farmers and they have to suffer due to the failure of nature. Therefore, farmers find it difficult to identify good years to repay their debts.

3. Litigation : Due to land disputes litigation in the court compels them to borrow heavily. Being uneducated and ignorant they are caught in the litigation process and dry away their savings and resources.

4. Money lenders and high rate of interest: The rate of interest charged by the local money lenders is very high and the compounding of interest leads to perpetuate indebtedness of the farmer.

Samacheer Kalvi 11th Economics Rural Economy Additional Questions and Answers

Part-A
Choose the best options

Question 1.
According to the 2011 population census, there are ___________ villages in India.
(a) 6,40,867
(b) 3,40,867
(c) 4,40,867
(d) 5,40,867
Answer:
(a) 6,40,867

Question 2.
Of 121 crore total population ___________ percent lives in rural areas.
(a) 64.88
(b) 68.84
(c) 68.48
(d) 88.64
Answer:
(b) 68.84

Question 3.
___________ are geographical areas located outside towns and cities.
(a) Townships
(b) Corporations
(c) Rural areas
(d) None
Answer:
(c) Rural areas

Question 4.
Total percentage of poverty in 2009-2010
(a) 33.80
(b) 38.30
(c) 33.08
(d) 80.33
Answer:
(a) 33.80

Question 5.
In 2009, the total number of micro finance institutions in India was around ___________
(a) 140
(b) 150
(c) 160
(d) 170
Answer:
(b) 150

Question 6.
SHG Bank Linked Programme (SBLP) was started in ___________
(a) 1991
(b) 1990
(c) 1992
(d) 1993
Answer:
(c) 1992

Question 7.
Micro Units Development and Refinance Agency is ___________
(a) IRDB
(b) MUDRA
(c) BUNDAN
(d) ICICI
Answer:
(b) MUDRA

Question 8.
MUDRA Bank was started in the year ___________
(a) 2011
(b) 2012
(c) 2015
(d) 2013
Answer:
(c) 2015

Question 9.
___________ was the first professor of economics at Madras University
(a) Gilbert Slater
(b) Samuelson
(c) Marshall
(d) Manmohan Singh
Answer:
(a) Gilbert Slater

Question 10.
National Rural Health Mission was launched on ___________
(a) 2001
(b) 1991
(c) 2005
(d) 2000
Answer:
(c) 2005

Question 11.
___________ are setup with a view to develop rural economy by providing credit and other facilities to small and marginal farmers.
(a) MUDRA Bank
(b) RRB
(c) SHG
(d) NABARD
Answer:
(b) RRB

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
(a) 1 2 3 4
(b) 2 1 3 4
(c) 4 3 2 1
(d) 3 4 1 2
Answer:
(b) 2 1 3 4

Question 2.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
(a) 4 3 2 1
(b) 2 1 4 3
(c) 3 4 1 2
(d) 1 2 3 4
Answer:
(c) 3 4 1 2

Choose the correct option

Question 3.
National rural health mission was established in
(a) March 12, 2005
(b) June 12, 2005
(c) April 12, 2005
(d) June 13, 2005
Answer:
(c) April 12, 2005

Question 4.
Gross national happiness index is
(a) GDP
(b) HDF
(c) WEI
(d) GNHI
Answer:
(d) GNHI

Question 5.
Poverty line is calculated on the basis of
(a) Based on inflation
(b) Based on poverty
(c) Based on income or consumption
(d) Based on unemployment
Answer:
(c) Based on income or consumption

Fill in the blanks with suitable option given below

Question 6.
Micro credit is also known as
(a) Micro finance
(b) Small finance
(c) Large finance
(d) None
Answer:
(a) Micro finance

Question 7.
The co-existence of two extremely different features is called as
(a) Unity
(b) Dualism
(c) Trism
(d) None
Answer:
(b) Dualism

Question 8.
The first economics professor of madras university was
(a) A.K. Sen
(b) Charan Singh
(c) Manmohan Singh
(d) Gilbert Slater
Answer:
(d) Gilbert Slater

Choose the incorrect statement

Question 9.
(a) Physical quality of life index is PQLI
(b) The existence of poverty in rural areas is called rural poverty
(c) SSI s are also known as micro, small and medium enterprises (MSMEs)
(d) Petter diamond was the first economics professor of madras university
Answer:
(d) Petter diamond was the first economics professor of madras university

Question 10.
(a) MUDRA is micro units development and Refinance Agency Bank
(b) NRHM is National Rural Health Mission.
(c) The length of roads in India as per 2018 is 24 lakh kms.
(d) PURA is provision of urban facilities for rural areas.
Answer:
(c) The length of roads in India as per 2018 is 24 lakh kms.

Choose the incorrect pair

Question 11.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
Answer:
(c) NHRM (iii) Rural health

Question 12.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
Answer:
(d) All Indian Debt and investment Survey (iv) AIDS

Question 13.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
Answer:
(d) National Food Security (iv) 2013

Question 14.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
Answer:
(a) Adarsh Gram Sadak Yojana (i) 2010

Choose the correct statement

Question 15.
(a) Unemployment is a situation in which a person who is unwilling to work employed,
(b) According to 2011 12 estimates 42 crore village population list below the poverty line.
(c) In 2015, 60 crore people lived in villages.
(d) Sir Malcolm Darling stated that ‘An Indian farmer is bom in debt, lives in debt, dies in debt and bequeaths debt’
Answer:
(d) Sir Malcolm Darling stated that ‘An Indian farmer is bom in debt, lives in debt, dies in debt and bequeaths debt’

Question 16.
(a) Number of people living per sq.km is called population scarcity.
(b) There are 5,40,867 villages in India.
(c) The concept of ‘Double poisoning’ was coined by Schumacher.
(d) In 2009-10, the total percentage of poverty in India is 54.10
Answer:
(c) The concept of ‘Double poisoning’ was coined by Schumacher.

Pick the odd one out

Question 17.
(a) Petter Diamond
(b) A.K.Sen
(c) Dale Mortensen
(d) Christopher Pissarides
Answer:
(b) A.K.Sen

Question 18.
(a) NCC
(b) IRDP
(c) TRYSEM
(d) MGNREGS
Answer:
(a) NCC

Analyse the reason for the following

Question 19.
Assertion (A) : Creation of employment opportunities eliminate poverty’in rural areas.
Reason (R) : Rural unemployment and rural poverty are interrelated.
(a) (A) is true; but (R) is false
(b) Both (A) and (R) are false
(c) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(d) Both (A) and (R) are true (R) is the correct explanation of (A)
Answer:
(d) Both (A) and (R) are true (R) is the correct explanation of (A)

Question 20.
Assertion (A) : Agricultural operations are seasonal in nature.
Reason (R) : To reduce the seasonal unemployment rural people should be encouraged to adopt subsidiary occupations.
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) (A) is false but (R) is true.
Answer:
(b) Both (A) and (R) are true, (R) is not the correct explanation of (A)

Part – B
Answer the following questions in one or two sentences

Question 1.
Name some of the economic indicators?
Answer:
Human Development Index (HDI),Women Empowerment Index (WEI), Gender Disparity Index (GDI), Physical Quality of Life Index (PQLI) and Gross National Happiness Index (GNHI).

Question 2.
How can you measure poverty by nutritional index?
Answer:
On the basis of recommended nutritional intake, persons consuming less than 2400 calories per day is rural areas are considered as rural poor.

Question 3.
What is unemployment?
Answer:
Unemployment is a situation in which a person actively searching for employment but unable to find work at the prevailing wage rate.

Question 4.
Write a note on seasonal unemployment ?
Answer:
In seasonal unemployment, employment occurs only in a particular season and workers remain unemployed in the remaining period of a year.

Question 5.
Write a note on NRHM.
Answer:
The National Rural Health Mission (NRHM) was launched on 12th April 2005, to provide accessible, affordable and quality health care to the vulnerable groups in rural areas.

Question 6.
What is population density?
Answer:
Population density is the number of persons living in per square kilometre.

Part – C
Answer the following questions in one Paragraph

Question 1.
What are the types of rural unemployment ?
Answer:

  1. Open unemployment: Unemployed persons are identified as they remain without work.
  2. Concealed unemployment: Many are employed below their productive capacity and even if they are withdrawn from work the output will not diminish. It is also called as disguised unemployment or under employment.
  3. Seasonal unemployment: Employment occurs only on a particular season and the remaining period of a year the rural people are unemployed or partially employed.

Question 2.
Explain about village industries.
Answer:
Village industries are traditional in nature and depend on local raw material. They cater to the needs of local population.
(Eg.) Gur and Khandsari, Cane and Bamboo basket, Shoe making, Pottery and Leather tanning.

Question 3.
Write a note on rural indebtedness.
Answer:
Rural indebtedness refers to the situation of the rural people who are unable to repay the loan accumulated over a period. Existence of the rural indebtedness indicates the weak financial infrastructure of our country in reaching the needy farmers landless people and the agricultural labourers.

Question 4.
Write a note on rural roads.
Answer:
Rural roads constitute the very life line of rural economy. A well-constructed road network in rural area would bring several benefits including the linking of remote villages with urban centres, reduction in cost of transportation of agricultural inputs and promotion of marketing for rural produces.

Part – D
Answer the following questions in about a page

Question 1.
Explain the causes for rural unemployment.
Answer:
1. Absence of skill development and employment generation : Lack of Government initiatives to give required training and then to generate employment opportunities.

2. Seasonal nature of agriculture : Agricultural operations are seasonal in nature, so non-farm employment opportunities must be created to reduce seasonal unemployment.

3. Lack of subsidiary occupation : Rural people are not able to start subsidiary occupations due to shortages of funds for investment and lack of proper marketing arrangements.

4. Mechanization of agriculture : Mechanization of agricultural operations reduces employment opportunities for the farm labour.

5. Capital – Intensive technology : The expanding private industrial sector in urban areas do not create additional employment opportunities due to the application of capital intensive technologies.

6. Defective system of education : The present System of education has also aggravated the rural unemployment problem. Students want to get degrees only, not any skill. Degrees should be awarded only on the basis of skills acquired.

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Samacheer Kalvi 11th Economics Development Experiences in India Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
Which of the following is the way of Privatisation?
(a) Disinvestment
(b) Denationalization
(c) Franchising
(d) All the above
Answer:
(d) All the above

Question 2.
Countries today are to be ______ for their growth.
(a) Dependent
(b) Interdependent
(c) Free trade
(d) Capitalist
Answer:
(b) Interdependent

Question 3.
The Arguments against LPG is ______
(a) Economic growth
(b) More investment
(c) Disparities among people and regions
(d) Modernization
Answer:
(c) Disparities among people and regions

Question 4.
Expansion of FDI ______
(a) Foreign Private Investment
(b) Foreign Portfolio
(c) Foreign Direct Investment
(d) Forex Private Investment
Answer:
(c) Foreign Direct Investment

Question 5.
India is the largest producer of ______ in the world.
(a) fruits
(b) gold
(c) petrol
(d) diesel
Answer:
(a) fruits

Question 6.
Foreign investment includes ______
(a) FDI only
(b) FPI and FFI
(c) FDI and FPI
(d) FDI and FFI
Answer:
(a) FDI only

Question 7.
The Special Economic Zones policy was announced in ______
(a) April 2000
(b) July 1990
(c) April 1980
(d) July 1970
Answer:
(a) April 2000

Question 8.
Agricultural Produce Market Committee is a ______
(a) Advisory body
(b) Statutory body
(c) Both a and b
(d) none of these above
Answer:
(b) Statutory body

Question 9.
Goods and Services Tax is ______
(a) a multi point tax
(b) having cascading effects
(c) like Value Added Tax
(d) a single point tax with no cascading effects.
Answer:
(d) a single point tax with no cascading effects.

Question 10.
The New Foreign Trade Policy was announced in the year ______
(a) 2000
(b) 2002
(c) 2010
(d) 2015
Answer:
(d) 2015

Question 11.
Financial Sector reforms mainly related to ______
(a) Insurance Sector
(b) Banking Sector
(c) Both a and b
(d) Transport Sector
Answer:
(c) Both a and b

Question 12.
The Goods and Services Tax Act came in to effect on ______
(a) 1st July 2017
(b) 1st July 2016
(c) 1st January 2017
(d) 1st January 2016
Answer:
(a) 1st July 2017

Question 13.
The new economic policy is concerned with the following ______
(a) foreign investment
(b) foreign technology
(c) foreign trade
(d) all the above
Answer:
(d) all the above

Question 14.
The recommendation of Narashimham Committee Report was submitted in the year ______
(a) 1990
(b) 1991
(c) 1995
(d) 2000
Answer:
(b) 1991

Question 15.
The farmers have access to credit under Kisan credit card scheme through the following except ______
(a) co-operative banks
(b) RRBs
(c) Public sector banks
(d) private banks
Answer:
(a) co-operative banks

Question 16.
The Raja Chelliah Committee on Trade Policy Reforms suggested the peak rate on import duties at ______
(a) 25%
(b) 50%
(c) 60%
(d) 100%
Answer:
(b) 50%

Question 17.
The first ever SEZ in India was set up at
(a) Mumbai
(b) Chennai
(c) Kandla
(d) Cochin
Answer:
(c) Kandla

Question 18.
‘The Hindu Rate of Growth’ coined by Raj Krishna refers to ______
(a) low rate of economic growth
(b) high proportion of Hindu population
(c) Stable GDP
(d) none
Answer:
(a) low rate of economic growth

Question 19.
The highest rate of tax under GST ______ is (as on July, 2017)
(a) 18%
(b) 24%
(c) 28%
(d) 32%
Answer:
(c) 28%

Question 20.
The transfer of ownership from public sector to private sector is known as ______
(a) Globalization
(b) Liberalization
(c) Privatization
(d) Nationalization
Answer:
(c) Privatization

Part – B
Answer the following questions in one or two sentences

Question 21.
Why was structural reform implemented in Indian Economy?
Answer:
Indian economy introduced structural reforms to face the economic crisis in the form of balance of payments problem in 1991.

Question 22.
State the reasons for implementing LPG.
Answer:
To correct the weaknesses and rigidities in the various sectors of the economy India implemented LPG which is the triple pillars of New Economic Policy.

Question 23.
State the meaning of Privatization.
Answer:
Privatization means transfer of ownership and management of enterprises from public sector to private sector.

Question 24.
Define disinvestment.
Answer:
Disinvestment means selling of government securities of public sector undertakings to other PSUs or private sector or banks.

Question 25.
Write three policy initiative introduced in 1991 – 1992 to correct the fiscal imbalance.
Answer:

  1. Reduction in fertilizer subsidy.
  2. Abolition of subsidy on sugar.
  3. Disinvestment of a part of the government’s equity holdings in select public sector undertakings.
  4. Expenditures on welfare measures were reduced.

Question 26.
State the meaning of Special Economic Zones.
Answer:
SEZs is an area in which business and trade laws are different from rest of the country mainly aiming at increasing trade, investment and job creation.

Question 27.
State the various components of Central government schemes under post – harvest measures.
Answer:

  1.  Mega food parks, Integrated cold chain, value addition preservation infrastructure, modernization of slaughter house.
  2. Scheme for quality-assurance, codex standards, research and development and other promotional activities.

Part – C
Answer the following questions in one Paragraph

Question 28.
How do you justify the merits of Privatisation?
Answer:
Privatisation was necessitated because of the belief that the private sector was not given enough opportunities to earn more money.

Question 29.
What are the measures taken towards Globalization?
Answer:
Globalization refers to the integration of the domestic economy with the rest of the world. Import liberalization through reduction of tariff and non-tariff barriers, opening the doors to foreign direct investment and foreign portfolio investment are some of the measures towards globalization.

Question 30.
Write a note on Foreign investment policy?
Answer:
The major feature of the economic reform opened gate to foreign investment and foreign technology. Foreign investment including FDI and FPI were allowed.

In 1991, the government announced a specified list of industries where in automatic permission was granted for FDI. Foreign Investment Promotion Board (FIPB) has been set up to negotiate with international firms and approve foreign direct investment in select areas.

Question 31.
Give short note on Cold storage.
Answer:
Problems relating to the marketing of fruits and vegetables is related with their perishability. Perishability is responsible for high marketing costs, market gluts, price fluctuations and other similar problems.

In order to overcome this constraint, the Government of India and the ministry of agriculture promulgated “Cold Storage Order 1964” under section 3 of the Essential Commodities Act, 1955.

However, the cold storage facility is still very poor and highly inadequate.

Question 32.
Mention the functions of APMC.
Answer:

  1. To promote public private partnership in the ambit of agricultural markets.
  2. To provide market led extension services to farmer.
  3. To promote agricultural activities.
  4. To bring transparency in pricing system and market transactions.
  5. To ensure payments to the farmer for the sale on the same day.
  6. To display data on arrivals and rates from time to time in the market.

Question 33.
List out the features of new trade policy.
Answer:
The trade policy of 1 April 1992 freed imports of almost all intermediate and capital goods. Only 71 items remained restricted. This would affect the domestic industries. Rationalization of tariff structure and removal of quantitative restrictions.

The Chelliah Committee’s report had suggested drastic reduction in import duties. As a first step towards a gradual reduction in the tariffs, the 1991 -92 budget had reduced the peak rate of import duty.

Question 34.
What is GST? Write its advantages.
Answer:
GST is a comprehensive indirect tax levied on manufacture, sale and consumption of goods as well as services.

Advantages of GST :

  1. Removing cascading tax effect.
  2. Single point tax.
  3. Higher threshold for registration
  4. Composition scheme for small business.
  5. Online simpler procedure under GST.
  6. Defined treatment for e-commerce.
  7. Increased efficiency in logistics

Question 35.
Explain the objectives and characteristics of SEZs.
Answer:
The Special Economic Zones (SEZs) policy was announced in April 2000.
The major objectives of SEZs are :

  1. To enhance foreign investment especially to attract foreign direct investment (FDI) and thereby increasing GDP.
  2. To increase shares in global export.
  3. To generate additional economic activity.
  4. To create employment opportunities.
  5. To develop infrastructure facilities.
  6. To exchange technology in the global market.

Main characteristics of SEZs :
Geographically demarked area with physical security.

  1. Administrated by single body / authority.
  2. Streamlined procedures.
  3. Having separate custom area
  4. Governed by more liberal economic laws.
  5. Greater freedom to the firms located in SEZS.

Question 37.
Describe the salient features of EXIM policy (2015 – 2020)
Answer:
The new EXIM policy has been formulated focusing on increasing in exports scenario, boosting production and supporting the concepts like Make in India and Digital India.

  1. Reduce export obligations by 25% and give boost to domestic manufacturing supporting the “Make in India” concept.
  2. As a step to Digital India concept, online procedure to upload digitally signed document, and mobile app for filing tax, stamp duty has been developed.
  3. Repated submission of physical copies of documents are not required.
  4. Export obligation period for export items related to defence, military store, aerospace and nuclear energy to be 24 months.
  5. EXIM policy 2015 – 2020 is expected to double the share of India in World Trade from present level of 3% by the year 2020.This appears to the too ambitious.

Samacheer Kalvi 11th Economics Development Experiences in India Additional Questions and Answers

Part-A
Choose the best options

Question 1.
The expansion of FDI is
(a) Foreign comprehensive investment ______
(b) Foreign direct investment
(c) Foreign indirect investment
(d) Foreign private investment
Answer:
(b) Foreign direct investment

Question 2.
India ranked ______ position in Asia’s GDP.
(a) 4
(b) 3
(c) 1
(d) 2
Answer:
(b) 3

Question 3.
The monopoly and Restrictive Trade Practices Act was abolished in ______
(a) 1919
(b) 1971
(c) 1991
(d) 1791
Answer:
(c) 1991

Question 4.
Crop Insurance scheme was launched on
(a) 2009
(b) 2014
(c) 2016
(d) 2012
Answer:
(c) 2016

Question 5.
The expansion of KCC ______
(a) Kisan Crop Card
(b) Kisan Cash Card
(c) Kisan Credit Card
(d) None of the above
Answer:
(c) Kisan Credit Card

Question 6.
EPZ means ______
(a) Export Pay Zone
(b) Export Policy Zone
(c) Export Processing Zone
(d) Special Economic Zone
Answer:
(c) Export Processing Zone

Question 7.
Goods and Services Tax came into effect on ______
(a) 30,h June 2017
(b) 1st July 2016
(c) 1st July 2017
(d) 1st April 2017
Answer:
(c) 1st July 2017

Question 8.
Fiscal deficit should not exceed ______ percent of GDP
(a) 2
(b) 4
(c) 3
(d) 5
Answer:
(c) 3

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India
(a) 1 2 3 4
(b) 2 1 4 3
(c) 3 4 2 1
(d) 1 2 4 3
Answer:
(b) 2 1 4 3

Question 2.

(a) 4 3 2 1
(b) 3 2 4 1
(c) 1 2 3 4
(d) 3 1 4 2
Answer:
(d) 3 1 4 2

Choose the correct option

Question 3.
New economic policy was introduced in
(a) 2010
(b) 1991
(c) 1884
(d) 1894
Answer:
(b) 1991

Question 4.
Kisan card was introduced in
(a) 2000
(b) 2015
(c) 1998
(d) 2017
Answer:
(c) 1998

Question 5.
was one of the first in Asia to recognize the effectiveness of the export processing zone
(a) India
(b) China
(c) America
(d) Soviet Union
Answer:
(a) India

Fill in the blanks with suitable option given below

Question 6.
Disinvestment means
(a) Selling of government securities
(b) Selling of public sectors
(c) Selling of private securities
(d) None
Answer:
(a) Selling of government securities

Question 7.
was set up to encourage FDI
(a) Domestic investment promotion board
(b) Foreign investment promotion board
(c) Both
(d) None
Answer:
(b) Foreign investment promotion board

Question 8.
The share of India to Asia’s GDP is
(a) 7.50%
(b) 6.50%
(c) 8.50%
(d) 9.50%
Answer:
(c) 8.50%

Choose the incorrect statement

Question 9.
(a) Prime minister’s crop insurance scheme was launched on 2016
(b) New industrial policy was established on 1991
(c) Kisan credit card was introduced by RBI and NABARD
(d) Goods and Services Tax was introduced in 2015
Answer:
(d) Goods and Services Tax was introduced in 2015

Question 10.
(a) New foreign trade policy was introduced on 2015
(b) First ever export processing zone was set up by China
(c) Cold storage order was introduced on 1964.
(d) India’s GDP crossed 2 trillion dollars in 2015-16
Answer:
(b) First ever export processing zone was set up by China

Choose the incorrect pair

Question 11.
Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India
(a) 1 2 3 4
(b) 2 1 4 3
(c) 3 4 2 1
(d) 1 2 4 3
Answer:
(b) 2 1 4 3

Question 12.
Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India
(a) 4 3 2 1
(b) 3 2 4 1
(c) 1 2 3 4
(d) 3 1 4 2
Answer:
(c) Chelliah committee (iii) GST

Choose the correct pair

Question 13.
Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India
Answer:
(d) Cash reserve ratio (iv) 1991

Question 14.
Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India
Answer:
(a) Dr.Raj Krishna (i) Low rate of economic growth

Pick the odd one out

Question 15.
(a) Liberalization
(b) Industrialization
(c) Privatization
(d) Globalization
Answer:
(b) Industrialization

Question 16.
(a) Crop insurance
(b) Cold storage
(c) Agriculture subsidy reduction
(d) Kisan credit card
Answer:
(c) Agriculture subsidy reduction

Choose the correct statement

Question 17.
(a) China is the largest producer of fruits.
(b) India is the largest producer of vegetables.
(c) Agricultural produce market committee (ADMC) promote agricultural activities.
(d) Tobacco is a plantation crop
Answer:
(c) Agricultural produce market committee (ADMC) promote agricultural activities.

Question 18.
(a) Privatization measures discourages the continuance of monopoly.
(b) Disinvestment process was fully implemented
(c) New industrial policy abolished MRTP act.
(d) Foreign investment includes FDI and FII
Answer:
(c) New industrial policy abolished MRTP act.

Analyse the reason for the following

Question 19.
Assertion (A) : Agriculture in India is highly prone to risks like droughts and floods.
Reason (R) : Agriculture is a seasonal activity.
(a) (A) and (R) are true; (R) is the correct explanation of (A)
(b) (A) and (R) are true. (R) is not the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) (A) is true (R) is false.
Answer:
(a) (A) and (R) are true; (R) is the correct explanation of (A)

Question 20.
Assertion (A) : GST is a comprehensive indirect tax.
Reason (R) : GST is a multi point tax
(a) (A) is true (R) is false.
(b) (A) is false (R) is true
(c) Both (A) and (R) are true.
(d) Both (A) and (R) are false.
Answer:
(a) (A) is true (R) is false.

Part -B
Answer the following questions in one or two sentences

Question 1.
What is industrial delicensing policy or Red Tapism?
Answer:
The new industrial policy of 1991 abolished the procedure of securing licenses (Red Tapism) to start an industry which is called as industrial delicensing policy.

Question 2.
What is liberalization?
Answer:
Liberalization refers to removal of relaxation of governmental restrictions in all stages in industry.

Question 3.
What is Statutory Liquidity Ratio (SLR) ?
Answer:
SLR refers to the amount that the commercial banks require to maintain in the form of cash or gold or government approved securities before providing credit to the customers. Agricultural Produce Market Committee (APMC) is a statutory body constituted by state government in order to trade in agricultural or horticultural or livestock produce.

Question 5.
Write a note on Kisan Credit Card.
Answer:
A Kisan Credit Card (KCC) is a credit delivery mechanism that is aimed at enabling farmers to have quick and timely access to affordable credit. It was launched in 1998 by RBI and NABARD.

Question 6.
What is globalization ?
Answer:
Globalization stand for the consolidation of the various economies of the world.

Question 7.
Explain agrarian crisis after the economic reforms of 1991.
Answer:

  1. High input costs
  2. Cutback in agricultural subsidies
  3. Reduction of import duties
  4. Paucity of credit facilities.

Part – C
Answer the following questions in one Paragraph

Question 1.
What is crop insurance scheme ?
Answer:
To protect the farmers from natural calamities and ensure their credit eligibility for the next season, the Government of India introduced many agricultural schemes throughout the country.

One of the scheme is the Pradhan Mantri Fasal Bima Yojana which was launched on 18 February 2016. It envisages a uniform premium of 2% for Kharif and 1.5% for Rabi and 5% for commercial and horticultural crops.

Question 2.
State the arguments against LPG.
Answer:

  1. Liberalization measures favor the unrestricted entry of foreign companies which prevents the growth of the local manufactures.
  2. Privatization measures favor the continuance of the monopoly power.
  3. Only the developed countries are favored under globalization but the welfare of (he less developed countries will be neglected.

Question 3.
Explain the major changes in India after 1991.
Answer:

  1. Foreign exchange reserves started rising.
  2. There was a rapid industrialization.
  3. The pattern of consumption started improving.
  4. Infrastructure facilities such as express highways, metro rails, flyovers and airports started expanding (but the local people were thrown away).

Part -D
Answer the following questions in about a page

Question 1.
Explain the measures taken in Ranking system under new economic policy 1991.
Answer:

  1. Reduction in statutory liquidity ratio (SLR) and the Cash Reserve Ratio (CRR) were recommended by the Narasimham Committee Report 1991.
  2. Interest rate liberalization.
  3. Greater competition among public, private and foreign banks and elimination of administrative constraints.
  4. Liberalization of bank branch licensing policy.
  5. Banks were given freedom to relocate branches and open specialised branches.
  6. Guidelines for opening new private sector banks.
  7. New accounting norms regarding classification of assets and provisions of bad debt were introduced.

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Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence

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Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence

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Samacheer Kalvi 11th Economics Indian Economy Before and After Independence Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
The arrival of Vasco da Gama in Calicut, India
(a) 1498
(b) 1948
(c) 1689
(d) 1849
Answer:
(a) 1498

Question 2.
In 1614 Sir Thomas Roe was successful in getting permission from
(a) Akbar
(b) Shajakan
(c) Jahangir
(d) Noorjakhan
Answer:
(c) Jahangir

Question 3.
The power for governance of India was transferred from the East India Company (EIC) to the British crown in
(a) 1758
(b) 1858
(c) 1958
(d) 1658
Answer:
(b) 1858

Question 4.
Ryotwari system was initially introduced in _______
(a) Kerala
(b) Bengal
(c) Tamil Nadu
(d) Maharastra
Answer:
(c) Tamil Nadu

Question 5.
First World War started in the year _______
(a) 1914
(b) 1814
(c) 1941
(d) 1841
Answer:
(a) 1914

Question 6.
When did the Government of India declared its first Industrial Policy ?
(a) 1956
(b) 1991
(c) 1948
(d) 2000
Answer:
(c) 1948

Question 7.
The objective of the Industrial Policy 1956 was
(a) Develop heavy industries
(b) Develop agricultural sector only
(c) Develop private sector only
(d) Develop cottage industries only
Answer:
(a) Develop heavy industries

Question 8.
The industry which was de-reserved in 1993 ?
(a) Railways
(b) Mining of copper and zinc
(c) Atomic energy
(d) Atomic minerals
Answer:
(b) Mining of copper and zinc

Question 9.
The father of Green Revolution in India was _______
(a) M.S. Swaminathan
(b) Gandhi
(c) Visweswaraiah
(d) N.R. Viswanathan
Answer:
(a) M.S. Swaminathan

Question 10.
How many commercial banks were nationalised in 1969 ?
(a) 10
(b) 12
(c) 14
(d) 16
Answer:
(c) 14

Question 11.
The main objective of nationalisation of banks was _______
(a) Private social welfare
(b) Social welfare
(c) To earn
(d) Industries monopoly
Answer:
(b) Social welfare

Question 12.
The Planning Commission was setup in the year _______
(a) 1950
(b) 1955
(c) 1960
(d) 1952
Answer:
(a) 1950

Question 13.
In the first five year plan, the top priority was given to _______ Sector.
(a) Service
(b) Industrial
(c) Agriculture
(d) Bank
Answer:
(c) Agriculture

Question 14.
Tenth Five year plan period was _______
(a) 1992-1997
(b) 2002-2007
(c) 2007-2012
(d) 1997-2002
Answer:
(b) 2002-2007

Question 15.
According to HDR (2016), India ranked _______ out of 188 countries.
(a) 130
(b) 131
(c) 135
(d) 145
Answer:
(b) 131

Question 16.
Annual Plans formed in the year _______
(a) 1989-1991
(b) 1990-1992
(c) 2000-2001
(d) 1981-1983
Answer:
(b) 1990-1992

Question 17.
The Oldest large scale industry in India ________
(a) cotton
(b) jute
(c) steel
(d) cement
Answer:
(a) cotton

Question 18.
Human development index (HDI) was developed _________
(a) Jawaharlal Nehru
(b) M.K.Gandhi
(c) Amartiya Sen
(d) Tagore
Answer:
(c) Amartiya Sen

Question 19.
The main theme of the Twelth Five Year Plan __________
(a) faster and more inclusive growth
(b) growth with social Justice
(c) socialistic pattern of society
(d) faster, more inclusive and sustainable growth
Answer:
(d) faster, more inclusive and sustainable growth

Question 20.
The PQLI was developed by _______
(a) Planning Commission
(b) Nehru
(c) Morris
(d) Morris
Answer:
(c) Morris

Part – B
Answer the following questions in one or two sentences

Question 21.
What are the Phases of colonial exploitation of India?
Answer:
The period of merchant capital, the period of industrial capital, the period of finance capital.

Question 22.
Name out the different types of land tenure existed in India before Independence.
Answer:

  1. Zamindari system
  2. Mahalwari system
  3. Ryotwari system

Question 23.
State the features that distinguish a land tenure system from other system.
Answer:
The land tenure system differs as

  1. Who owns the land ?
  2. Who cultivates the land ?
  3. Who is responsible for paying the land revenue to the government ?

Question 24.
List out the weaknesses on Green Revolution.
Answer:

  1. Indian agriculture was still a gamble of the monsoons.
  2. This strategy needed heavy investment.
  3. The income gap has increased
  4. Increased unemployment among agricultural labourers.
  5. Reduced the soil fertility and spoiled human wealth.

Question 25.
What are the objectives of Tenth five year plan ?
Answer:

  1. Double the per capita income in the next 10 years.
  2. Aimed to reduce the poverty ratio to 15% by 2012.
  3. Growth target was 8.0% but it achieved only 7.2%

Question 26.
What is the difference between HDI and PQLI ?
Answer:

  1. Human Development Index : It is constructed based on life expectancy index, Education index and GDP per capita.
  2. Physical quality of life index : The PQLI is a measure to calculate the quality of life.

Question 27.
Mention the indicators which are used to calculate HDI.
Answer:

  1. Life expectancy index.
  2. Educational index.
  3. GDP per capita.

Part – C
Answer the following questions in one Paragraph

Question 28.
Explain about the Period of Merchant Capital.
Answer:

  1. The period of merchant capital was from 1757 to 1813.
  2. The only aim of the East India company was to earn profit by establishing monopoly trade.
  3. During this period, India had been considered as the best hunting ground for capital
  4. The objective of monopoly trade was fulfilled by achieving political control.
  5. The company administration succeeded in generating huge surpluses which were repatriated to England.

Question 29.
The Handicrafts declined in India in British Period. Why?
Answer:

  1. Through discriminatory tariff policy, the British Government purposefully destroyed the handicrafts.
  2. With the disappearance of Nawabs and Kings, there was no one to protect Indian handicrafts.
  3. Indian products could not compete with machine made products.
  4. Introduction of railways in India increased the domestic market for the British goods.

Question 30.
Elucidate the different types of land tenure system in colonial India.
Answer:
The three types of land tenure system are Zamindari, Mahalwari and Ryotwari system.

1. Zamindari system or the land – lords : This system was created in 1793,after the introduction of permanent settlement act. Under this system the landlords were declared as the owners of the land. They were responsible to pay the land revenue. The share of the rent to government is fixed at 10/11th and the balance as zamindar’s remuneration.

2. Mahalwari system or communal farming : The ownership of the land was maintained by the collective body of villagers which served as a unit of management. They distribute the land and collect revenue and pay it to the state.

3. Ryotwari or the owner – cultivator system : The ownership rights of use and control of land were held by the tiller himself.

Question 31.
State the reasons for nationalization of commercial banks.
Answer:

  1. The main objective of the economic planning aimed at social welfare.
  2. Before independence commercial banks were in the private sector.
  3. These commercial banks failed in helping the government to achieve social objectives of planning.
  4. Therefore, the government decided to nationalize 14 major commercial banks on 1969

Question 32.
Write any three objectives of Industrial Policy 1991.
Answer:

  1. Ensuring rapid industrial development in a competitive environment.
  2. Enchancing support to the small scale sector.
  3. Providing more incentives for industrialisation of the backward areas.

Question 33.
Give a note on Twelfth Five Year Plan.
Answer:

  1. Twelfth five year plan (2012 – 17).
  2. Its main theme is “Faster more inclusive and sustainable growth”
  3. Its growth rate target is 8%

Question 34.
What is PQLI ?
Answer:
Morris D Morris developed the physical quality of life index (PQLI). The PQLI is a measure to calculate the quality of life.
PQLI includes three indicators life expectancy, infant mortality rate and literacy rate. A scale of each indicator ranges from the number 1 to 100. Represents the worst performance and 100 is the best performance.
HDI includes income while PQLI do not. PQLI has only the physical aspects of like.

Part – D
Answer the following questions in about a page

Question 35.
Discuss about the Indian economy during British Period.
Answer:
Britain had exploited India over a period of two centuries of its colonial rule. On the basis of the form of colonial exploitation, economic historians have divided the whole period into three phases namely the period of merchant capital, period of industrial capital and the period of finance capital.

1. Period of merchant capital (1757 to 1813):
The only aim of the East India company was to earn profit by establishing monopoly trade. India was considered as the best hunting ground for capital. By attaining political power the objective of monopoly trade was fulfilled. The company administration succeeded in generating huge surpluses which were repatriated to England.

2. Period of industrial capital (1813 to 1858):
India had become a market for British textiles. Indians were exploited by fixing low price for exports and high price for imports. India’s traditional handicrafts were thrown out of gear.

3. Period of financial capital (Late 19th century – 1947) : Finance imperialism began to entrench. Britain decided to make massive investments in various fields by plundering Indian capital.

4. The land tenure system in India : Land tenure refers to the system of land ownership and managements.
The Zamindari system, the Mahalwari system and the Ryotwari system were the systems of land tenure introduced by British.

5. Problems of British rule : Their profit motives led to drain of resources from India. The handicraft industries were collapsed.
The British rule stunted the growth of Indian enterprise. Capital formation in India were retarded.

Question 36.
Explain the role of SSIs in economic development?
Answer:

1. SSIs provide employment:

  • SSIs use labour intensive techniques thus reduce the problem of unemployment to a great extent.
  • They provide employment to people in villages and unorganized sectors.
  • The employment – capital ratio is high for the SSIs.

2. SSIs bring balanced regional development: SSIs remove regional disparities by industrializing rural and backward areas and bring balanced regional development.

3. Help in mobilization of local resources : SSIs help to mobilize and utilize local resources like small savings, entrepreneurial talent etc.,

4. Pave for optimization of capital: SSIs require less capital per unit of output. They function as a stabilizing force by providing output-capital ratio as well as high employment capital ratio.

5. Promote exports : SSIs earn valuable foreign exchange through exports from India.

6. Complement large scale industries : SSIs serve as ancillaries to large scale units.

7. Meet consumer demands : SSIs serves as an anti-inflationary force by providing goods of daily use.

8. Develop entrepreneurship : They promote self-employment and spirit of self-reliance in the society. They help to increase the per capita income. They help in distributing national income in more efficient and equitable manner.

Question 37.
Explain the objectives of nationalization of commercial banks.
Answer:

  1. The main objective of nationalization was to attain social welfare. Sectors such as agriculture, small and village industries were in need of funds for their expansion and further economic development.
  2. It helped to curb private monopolies in order to ensure a smooth supply of credit to socially desirable sections.
  3. To encourage the banking habit among the rural population.
  4. To reduce the regional imbalances where the banking facilities were not available.
  5. After nationalization, new bank branches were opened in both rural and urban areas, and they created credit facilities mainly to the agriculture sector and its allied activities.

Question 38.
Describe the performance of 12th five year plan in India.
Answer:
Economic planning is the process in which the limited natural resources are used skillfully so as to achieve the desired goals.

1. First five year plan (1951 – 56) : Main focus was on the agricultural development it achieved the growth rate of 3.6%.

2. Second five year plan (1956 – 61) : Focus was on the industrial development of the country and achieved the growth rate of 4.1%.

3. Third five year plan (1961 – 66): To make the economy independent and to reach take off.

4. Plan holiday (1966 – 69): Equal priority was given to agriculture, its allied sectors and the industrial sectors during annual plans.

5. Fourth five year plan (1969 – 74): Growth with stability and progressive achievement of self reliance was the goal but the plan failed.

6. Fifth five year plan (1975 – 79): A successful plan prioritized agriculture and then industry and mines.

7. Rolling plan : This plan was started with an annual plan for 1978-79.

8. Sixth five year plan (1980 – 85): Based on investment yojana, its objective was poverty eradication and technological self reliance.

9. Seventh five year plan (1985 – 90): Establishment of the self sufficient economy and opportunities for productive employment. Private sector got the priority over public sector.

10. Annual plans : Two annual plans were formed in 1990 – 91 & 1991 -92.

11. Eighth five year plan (1992 – 97): Priority was given to development of the human resources. New economic policy was introduced.

12. Ninth five year plan (1997 – 02):

  • Aimed to double the percapita income in the next 10 years.
  • Aimed to reduce the poverty ratio to 15% by 2012.

13. Tenth five year plan (2002 – 07): Growth with justice and equity was the focus but the plan failed with growth rate of 5.6%. Aimed to double the percapita income in the next 10 years. Aimed to reduce the poverty ratio to 15% by 2012.

14. Eleventh five year plan (2007 – 12): Main theme was faster and more inclusive growth.

15. Twelfth five year plan (2012 – 17): Main theme was faster more inclusive and sustainable growth. The five year plans played a’very prominent role in the economic development of the country. These plans had guided the government as to how it should utilise scarce resources so that maximum benefits can be gained.

Samacheer Kalvi 11th Economics Indian Economy Before and After Independence Additional Questions and Answers

Part – A
Choose the best options

Question 1.
Lord Cornwallis introduced “Permanent Settlement Act in” _______
(a) 1793
(b) 1794
(c) 1795
(d) 1796
Answer:
(a) 1793

Question 2.
In the year _______ green revolution started in India.
(a) 1950
(b) 1951
(c) 1948
(d) 1960
Answer:
(d) 1960

Question 3.
KGN. Daber established Mumbai’s spinning and weaving co. in _______
(a) 1810
(b) 1854
(c) 1845
(d) 1948
Answer:
(b) 1854

Question 4.
The oil well of Digboi, Assam was dug in.
(a) 1889
(b) 1898
(c) 1988
(d) 1810
Answer:
(a) 1889

Question 5.
The period of twelfth five year plan _______
(a) 2010 – 15
(b) 2011 – 16
(c) 2012 – 17
(d) 2013 – 18
Answer:
(c) 2012 – 17

Question 6.
NITI Aayog replaced the planning commission in _______
(a) 2013
(b) 2015
(c) 2014
(d) 2016
Answer:
(b) 2015

Question 7.
Human development index is _______
(a) HDI
(b) UNDP
(c) PQLI
(d) HID
Answer:
(a) HDI

Question 8.
Physical Quality of Life Index (PQLI) was developed by _______
(a) HDI
(b) PQLI
(c) UNDP
(d) HID
Answer:
(c) UNDP

Question 9.
In _______ system the ownership rights of use and control of land were held by the tiller himself.
(a) Zamindari
(b) Mahalwari
(c) Ryotwari
(d) Green revolution
Answer:
(c) Ryotwari

Question 10.
The period of industrial capital is _______
(a) 1757 – 1813
(b) 1813 – 1858
(c) 1757 – 1858
(d) 19 th century
Answer:
(b) 1813 – 1858

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
(a) 12 3 4
(b) 2 4 3 1
(c) 2 3 1 4
(d) 4 3 2 1
Answer:
(c) 2 3 1 4

Question 2.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
(a) 12 3 4
(b) 2 1 4 3
(c) 4 3 2 1
(d) 2 3 1 4
Answer:
(b) 2 1 4 3

Choose the correct option

Question 3.
The period of merchant capital is _______
(a) 1842 – 1857
(b) 1918 – 1920
(c) 1757 – 181
(d) 1721 – 1838
Answer:
(c) 1757 – 181

Question 4.
How many types of land tenure system were there before independence ?
(a) 2
(b) 3
(c) 1
(d) 5
Answer:
(b) 3

Question 5.
The first Indian modernized cotton cloth mill was established _______
(a) Chennai
(b) Calcutta
(c) Mumbai
(d) Bengaluru
Answer:
(b) Calcutta

Fill in the blanks with suitable option given below

Question 6.
The plan holiday was _______
(a) 1951-53
(b) 1966-69
(c) 2001 -02
(d) 1976-79
Answer:
(b) 1966-69

Question 7.
The planning commission was replaced by _______
(a) NITI Aayog
(b) Planning group
(c) Finance commission
(d) None of the above
Answer:
(a) NITI Aayog

Question 8.
NITI Aayog was formed in _______
(a) 2001
(b) 2014
(c) 2015
(d) 2016
Answer:
(c) 2015

Choose the incorrect statement

Question 9.
(a) Ryotwari system was first established in Tamilnadu.
(b) Zamindari system was established by British East India Company
(c) Mahalwari system was also called as the owner – cultivator system
(d) Zamindari system was created in 1793
Answer:
(c) Mahalwari system was also called as the owner – cultivator system

Question 10.
(a) The investment of small service enterprises should be more than ? 10 lakhs.
(b) The investment of micro service enterprises should not exceed ?10 lakhs.
(c) The investment of medium service enterprises should be more than ?2 crores,
(d) The investment of medium service enterprises should not exceed ?10 crores.
Answer:
(d) The investment of medium service enterprises should not exceed ?10 crores.

Choose the incorrect pair

Question 11.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
Answer:
(c) Vijaya nagar (iii) England

Question 12.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
Answer:
(c) Serampur (iii) Iron and steel

Chooose the correct pair

Question 13.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
Answer:
(a) Industrial policy (i) 1948

Question 14.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
Answer:
(d) White revolution (iv) Milk

Analyse the reason for the following

Question 15.
Assertion (A) : Before the advent of the British, Indian economy was self sufficient.
Reason (R) : Before the advent of the British, India lived in village.
(a) Both (A) and (R) are true
(b) (A) is true (R) is false
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true.
Answer:
(a) Both (A) and (R) are true

Choose the incorrect statement

Question 16.
(a) The industrial policy resolutions in India ushered a socialistic economy
(b) The high yielding varieties programme was called as pilot project of green revolution
(c) The aim of seventh five year plan is ‘Garibi Hatao’
(d) Finance commission replaced the planning commission.
Answer:
(b) The high yielding varieties programme was called as pilot project of green revolution

Question 17.
(a) Switzerland ranks first in HDI.
(b) PQLI includes income whereas HDI excludes income
(c) The process of transforming private assets to government ownership is called privatization
(d) The third plan was called as ‘Gadgil plan’.
Answer:
(d) The third plan was called as ‘Gadgil plan’.

Pick the odd one out

Question 18.
(a) The period of revenue capital
(b) The period of industrial capital
(c) The period of merchant capital
(d) The period of financial capital
Answer:
(c) The period of merchant capital

Question 19.
(a) Investment system
(b) Zamindari system
(c) Mahalwari system
(d) Ryotwari system
Answer:
(a) Investment system

Question 20.
(a) Black revolution
(b) Red revolution
(c) White revolution
(d) Violet revolution
Answer:
(d) Violet revolution

Part – B
Answer the following questions in one or two sentences

Question 1.
Name the classification of industries in India?
Answer:

  1. Public sector
  2. Public-cum-private sector
  3. Controlled private sector
  4. Private and co-operative sectors.

Question 2.
What is the resolution of industrial policy 1948 ?
Answer:
The first industrial policy was declared on 6th April 1948. Its main importance was introducing the system of mixed economy in India.

Question 3.
What is the resolution of industrial policy 1956 ?
Answer:
The policy was shaped by the mahalanobis model of growth with emphasis on heavy industries which would lead to higher growth path.

Question 4.
Write a note on micro manufacturing enterprises.
Answer:
In micro enterprises, the investment in plant and machinery does not exceed Rs. 25 lakhs.

Question 5.
Write a note on small manufacturing enterprises
Answer:
The investment in plant and machinery is more than twenty five lakh rupees but does not exceed Rs. 5 crores.

Question 6.
Write a note on medium manufacturing enterprises.
Answer:
The investment in plant and machinery is more than Rs. 5 crores but not exceeding Rs. 10 crores.

Question 7.
Name the classification of public sector banks.
Answer:

  1. Nationalised Banks.
  2. State Bank and its associates

Part – C
Answer the following questions in one Paragraph

Question 1.
Describe the problems of British rule in Indian economy ?
Answer:

  1. The British rule stunted the growth of Indian enterprise. It retarded the capital formation in India.
  2. The drain of wealth financed capital development in Britain.
  3. Indian agricultural sector became stagnant and deteriorated.
  4. Indian handicraft industries were collapsed.
  5. The system of capitalist firms with profit motives led to drain of resources from India.

Question 2.
Mention the achievements of green revolution.
Answer:

  1. The production of major cereals wheat and rice were boosted. India became food surplus exporting food grains to European countries.
  2. It was confined to high yielding variety cereals.
  3. Production of commercial crops were increased.
  4. Per hectare productivity of all crops had increased due to better seeds.
  5. The revolution had positive effect on development of industries manufacturing agricultural tools.
  6. It brought prosperity to rural people.
  7. Demand for labour increased.
  8. Financial resources were provided by banks and co-operative societies.

Question 3.
Examine the requirements of second green revolution.
Answer:

  1. Introduction of genetically modified (GM) seeds which double the per average production.
  2. Contribution of private sector to market the usage of GM foods.
  3. Government can play a key role in expediting irrigation schemes and managing water resources.
  4. Linking of rivers to transfer surplus water to deficient areas.

Question 4.
Write a note on iron and steel industry.
Answer:

  1. First steel industry at Kulti, near Jharia, West Bengal – Bengal iron works company in 1870.
  2. First large scale steel plant TISCO at Jamshedpur in 1907.
  3. Steel authority of India Ltd. (SAIL) was established in 1974 and was made responsible for the development of the steel industry.
  4. India is the eighth largest steel producing country in the world.

Question 5.
Write a note on petrol and natural gas.
Answer:

  1. First successful oil well was dug in India in 1889 at Digboi, Assam.
  2. At present a number of regions with oil reserves have been identified and oil is being extracted in these regions.
  3. For exploration purpose, Oil and Natural Gas Commission (ONGC) was established in 1956 at Dehradun, Uttarakhand.

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Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

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Samacheer Kalvi 11th Economics Indian Economy Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
The main gold mine region in Karnataka is _________
(a) Kolar
(b) Ramgiri
(c) Anantpur
(d) Cochin
Answer:
(a) Kolar

Question 2.
Economic growth of a country is measured by national income indicated by _________
(a) GNP
(b) GDP
(c) NNP
(d) Per capital income
Answer:
(b) GDP

Question 3.
Which one of the following is a developed nations ?
(a) Mexico
(b) Ghana
(c) France
(d) SriLanka
Answer:
(c) France

Question 4.
The position of Indian Economy among the other strongest economies in the world is _______
(a) Fourth
(b) Seventh
(c) Fifth
(d) Tenth
Answer:
(b) Seventh

Question 5.
Mixed economy means _________
(a) Private sectors and banks
(b) Co-existence of Public and Private sectors
(c) Public sectors and banks
(d) Public sectors only
Answer:
(b) Co-existence of Public and Private sectors

Question 6.
The weakness of Indian Economy is _________
(a) Economic disparities
(b) Mixed economy
(c) Urbanisation
(d) Adequate employment opportunities
Answer:
(a) Economic disparities

Question 7.
A scientific study of the characteristics of population is _________
(a) Topography
(b) Demography
(c) Geography
(d) Philosophy
Answer:
(b) Demography

Question 8.
The year 1961 is known as _________
(a) Year of small divide
(b) Year of Population Explosion
(c) Year of Urbanisation
(d) Year of Great Divide
Answer:
(b) Year of Population Explosion

Question 9.
In which year the population of India crossed one billion mark ?
(a) 2000
(b) 2001
(c) 2005
(d) 1991
Answer:
(b) 2001

Question 10.
The number of deaths per thousand population is called as ________
(a) Crude Death Rate
(b) Crude Birth Rate
(c) Crude Infant Rate
(d) Maternal Mortality Rate
Answer:
(a) Crude Death Rate

Question 11.
The number of births per thousand population is called as _________
(a) Crude death rate
(b) Mortality rate
(c) Morbidity rate
(d) Crude Birth Rate
Answer:
(d) Crude Birth Rate

Question 12.
Density of population = _________
(a) Land area / Total Population
(b) Land area / Employment
(c) Total Population / Land area of the region
(d) Total Population
Answer:
(c) Total Population / Land area of the region

Question 13.
Who introduced the National Development Council in India?
(a) Ambedkar
(b) Jawaharlal Nehru
(c) Radhakrishanan
(d) V.K.R.V. Rao
Answer:
(b) Jawaharlal Nehru

Question 14.
Who among the following propagated Gandhian Ecomomic thinkings.
(a) Jawaharlar Nehru
(b) VKRV Rao
(c) JC Kumarappa
(d) A.K.Sen
Answer:
(c) JC Kumarappa

Question 15.
The advocate of democratic socialism was
(a) Jawaharlal Nehru
(b) P.C. Mahalanobis
(c) Dr. Rajendra Prasad
(d) Indira Gandhi
Answer:
(a) Jawaharlal Nehru

Question 16.
Ambedkar the problem studied by in the context of Indian Economy is ________
(a) Small land holdings and their remedies
(b) Problem of Indian Currency
(c) Economics of socialism
(d) All of them
Answer:
(b) Problem of Indian Currency

Question 17.
Gandhian Economics is based on the Principle _________
(a) Socialistic idea
(b) Ethical foundation
(c) Gopala Krishna Gokhale
(d) Dadabhai Naoroji
Answer:
(b) Ethical foundation

Question 18.
V.K.R.V Rao was a student of _________
(a) J.M. Keynes
(b) Colin Clark
(c) Adam smith
(d) Alfred Marshal
Answer:
(a) J.M. Keynes

Question 19.
Amartya Kumara Sen received the Nobel prize in Economics in the year.
(a) 1998
(b) 2000
(c) 2008
(d) 2010
Answer:
(a) 1998

Question 20.
Thiruvalluvar economic ideas mainly dealt with _________
(a) Wealth
(b) Poverty is the curse in the society
(c) Agriculture
(d) All of them
Answer:
(d) All of them

Part – B
Answer the following questions in one or two sentences

Question 21.
Write the meaning of Economic Growth.
Answer:
A country’s economic growth is usually indicated by Gross Domestic Product (GDP). The GDP is the total monetary value of the goods and services produced by that country over a specific period of time, usually one year.

Question 22.
State any two features of developed economy.
Answer:

  1. High National Income
  2. High per capital Income

Question 23.
Write the short note on natural resources
Answer:
Natural resources are stock or reserve that can be drawn from nature. The major natural resources are land, forest, water, mineral and energy.

Question 24.
Point out any one feature of Indian Economy
Answer:
Indian economy is a mixed economy. In India both the private and public sectors coexist.

Question 25.
Give the meaning of non-renewable energy
Answer:
The sources of energy which cannot be renewed or re-used are called non-renewable energy.
(Eg.) Coal, Oil, Gas etc.

Question 26.
Give a short note on Sen’s ‘Choice of Technique’.
Answer:
In a labour surplus economy, generation of employment cannot be increased at the initial stage by the adaptation of capital intensive technique whatsen called as choice of technique.

Question 27.
List out the reasons for low per capital income as given by V.K.R.V. Rao.
Answer:

  1. Uneconomic holdings.
  2. Low levels of water availability for crops.
  3. Excess population pressure on agriculture due to the absence of a large industrial sector.
  4. Absence of capital.
  5. Absence of autonomy in currency policy.

Part – C
Answer the following questions in one Paragraph

Question 28.
Define Economic Development.
Answer:
The level economic development is indicated not just by GDP, but by an increase in citizen’s quality of life or well-being. The quality of life is being assessed by several indices such as Human Development Index (HDI), Physical Quality of Life Index (PQLI) and Gross National Happiness Index (GNHI)

Question 29.
State Ambedkar’s Economic ideas on agricultural economics.
Answer:
Dr. B.R.Ambedkar was a versatile personality. In 1918, he published a paper “Small holding in India and their remedies” citing Adam Smith’s “Wealth of Nations”, he made a fine distinction between “Consolidation of holdings” and “Enlargement of holdings”.

Question 30.
Write on short note on village sarvodhaya.
Answer:
Village sarvodhaya is the concept of Gandhiji on the development of villages. To Gandhi India lives in villages. He was interested in developing the villages as self – sufficient units.

According to him, “Real India was to be found in villages and not in towns or cities”.
So he suggested the development of self-suffficient, self dependent villages.

Question 31.
Write the strategy of Jawahar lal Nehru in India’s planning.
Answer:

  1. Jawahar lal Nehru was responsible for the introduction of planning in our country.
  2. To Nehru, the plan was essentially an integrated approach for development.
  3. He said the essence of planning is to find the best way to utilize all resources of manpower, of money and so on.
  4. Planning for Nehru was essentially linked up with industrialization and eventual self reliance for the country’s economic growth on a self-accelerating growth.
  5. Nehru carried through this basic strategy of planned development.

Question 32.
Write the V.K.R.V.Rao’s contribution on multiplier concept.
Answer:

  1. Rao’s examination of the “interrelation between investment, income and multiplier in an under developed economy (1952)” was his major contribution to macroeconomic theory.
  2. In it he asserts that keynesian multiplier principle remains inoperative in underdeveloped countries.
  3. V.K.R.V. Rao was the best equipped of all keynes’ pupils.

Question 33.
Write a short note on Welfare Economics given by Amartya Sen.
Answer:

  1. Amartya Sen was awarded the nobel prize for his contributions to welfare economics.
  2. Sen’s major point has been that the distribution of income / consumption among the persons below the poverty line is to be taken into account.
  3. Sen has focused on the poor, viewing them not as objects of pity requiring charitable hand-outs, but as disempowered folk needing empowerment in all aspects.

Question 34.
Explain Social infrastructure.
Answer:

  1. Social infrastructure refers to those structures which are improving the quality of man power and contribute indirectly towards the growth of an economy.
  2. These structures are outside the system of production and distribution.
  3. The development of these social structures help in increasing the efficiency and productivity of man power.
    (Eg.) Schools, Colleges, Hospitals

Part – D
Answer the following questions in about a page

Question 35.
Explain strong features Indian economy
Answer:

  1. India has a mixed economy : In India both private and public sectors coexist.
  2. Agriculture plays the key role : Around 60% of the people in India depend upon agriculture for their livelihood.
  3. An emerging market: India has a high potential for prospective growth which attracts FDI and FII.
  4. Emerging economy : As a result of rapid economic growth Indian economy has a place among the G20 countries.
  5. Fast growing economy : India has emerged as the world’s fastest growing economy in 2016-17 with 7.1% GDP next to China.
  6. Fast growing service sector : The service sector, contributes a lion’s share of the GDP in India.
  7. Large domestic consumption : Due to large domestic consumption the standard of living has considerably improved and life style has changed.
  8. paid growth of urban areas : Improved connectivity in transport and communication, education and health have speeded up the pace of urbanization.
  9. Stable macro economy : The current year’s economic survey represents the Indian economy to be a heaven of macroeconomic stability, resilence and optimism.
  10. Demographic dividend : India is a pride owner of the maximum percentage of youth. This has invited foreign investments to the country and outsourcing opportunities.

Question 36.
Write the importance of mineral resources in India.
Answer:

  1. Iron-ore : Hematite iron is mainly found in Chattisgarh, Jharkhand, Odisha, Goa and Karnataka. Magnetite is found in Western Coast of Karnataka. Some deposits of iron ore are also found in Kerala, Tamil Nadu and Andhra Pradesh.
  2. Coal and Lignite : India ranks third in the coal production. The main centres are West-Bengal, Bihar, Madhya Pradesh, Maharashtra, Odisha and Andhra Pradesh. Bulk production comes from Bengal-Jharkhand coal fields. Lignite is found in Neyveli in Tamil Nadu.
  3. Bauxite : Bauxite is a main source of aluminium. Major reserves are found in Odisha and Andhra Pradesh.
  4. Mica : India stand first in sheet mica production. Andhra Pradesh, Jharkhand, Bihar and Rajasthan are major reserves.
  5. Crude Oil : Oil is being explored in India at many places of Assam and Gujarat.
  6. Gold : India possesses only a limited gold reserve.
  7. Diamond : The total reserves of diamond is estimated at around 4582 carats.

Question 37.
Bring out Jawharlal Nehru’s contribution to the idea of economic development.
Answer:
Jawahar Lal Nehru was one of the Chief builders of modem India. He was a great patriot, thinker and statesman. His ideas of economic development are :

1. Democracy : He was a firm believer in democracy. He believed in free speech, civil liberty, adult franchise and the rule of law and parliamentary democracy.

2. Secularism : Secularism is another signal contribution of Nehru to India. There are so many religions in India but there is no domination by religious majority.

3. Planning : Nehru was responsible for the introduction of planning in our country. The plan was essentially an integrated approach for development. Planning for Nehru was essentially linked up with industrialization and eventual self-reliance for the country’s economic growth on a self-accelerating growth.

4. Advancement of Science : Nehru made a great contribution to the advancement of Science, research, technology and industrial development. In his period, many IITs and research institutions were established. He always insisted on scientific temper.

5. Democratic socialism : Nehru put the country on the road towards a socialistic pattern of society. Nehru’s socialism is democratic socialism.

Question 38.
Write a brief note on the Gandhian economic ideas.
Answer:
Gandhian economics is based on ethical foundations.
Salient features of Gandhian economic thought:

  1. Village republics : Gandhi was interested in developing the villages as self-sufficient units.
  2. On machinery : Gandhi described machinery as ‘Great Sin’.
  3. Industrialism : Gandhi considered industrialism as a curse on mankind.
  4. Decentralization : He advocated a decentralized economy.
  5. Village sarvodaya : Gandhi suggested the development of self-sufficient, self-dependent villages.
  6. Bread labour: Gandhi realized the dignity of human labour. Bread labour or body labour was the expression that Gandhi used to mean manual labour.
  7. The doctrine of trusteeship : Trusteeship provides a means of transforming the present capitalist society into an egalitarian one.
  8. On the food problem : Gandhi was against any sort of food controls. Food controls only create artificial scarcity.
  9. On population : Gandhi was in favour of birth control through brahmacharya or self-control.
  10. On prohibition : Gandhi advocated cent per cent prohibition.

Samacheer Kalvi 11th Economics Indian Economy Additional Questions and Answers

Part -A
Choose the best options

Question 1.
Third world country __________
(a) India
(b) China
(c) Bangladesh
(d) Srilanka
Answer:
(c) Bangladesh

Question 2.
__________ is the year of great divide.
(a) 1921
(b) 1918
(c) 1943
(d) 1969
Answer:
(a) 1921

Question 3.
__________ are BIMARU States.
(a) Tamil Nadu, Kerala, Karnataka, Madhya Pradesh
(b) Bihar, Madhya Pradesh, Rajasthan, Uttarpradesh
(c) West Bengal, Himalchalpradesh, Punjab, Delhi
(d) Andhra Pradesh, Odisa, Gujarat, Chattisgarh
Answer:
(b) Bihar, Madhya Pradesh, Rajasthan, Uttarpradesh

Question 4.
In India, Lignite is found in.
(a) New Delhi
(b) Kolkatta
(c) Ranji
(d) Neyveli
Answer:
(d) Neyveli

Question 5.
Indian educational system comprises of __________ stages.
(a) 6
(b) 12
(c) 3
(d) 4
Answer:
(a) 6

Question 6.
__________ is considered as the best pupil of J.M. Keynes
(a) A.K. Sen
(b) J.C.Kumarappa
(c) V.K.R.V. Rao
(d) Manmohan Singh
Answer:
(c) V.K.R.V. Rao

Question 7.
Indian Railways introduced first wifi facility in __________
(a) Chennai
(b) Calcutta
(c) Bengaluru
(d) Hyderabad
Answer:
(c) Bengaluru

Question 8.
National Harbour board was set up in __________
(a) 1947
(b) 1950
(c) 1948
(d) 1951
Answer:
(b) 1950

Question 9.
A study on poverty and famines and concept of entitlements and capability development was done by __________
(a) Adamsmith
(b) A.K.Sen
(c) Marshall
(d) V.K.R.V. Rao
Answer:
(b) A.K.Sen

Question 10.
Percapital income is __________
(a) National income / output
(b) National income / population
(c) National income / percapital output
(d) National income / male population
Answer:
(b) National income / population

Question 11.
‘India lives in villages’ said by __________
(a) V.K.R.V. Rao
(b) J.C. Kumarappa
(c) Mahatma Gandhi
(d) Jawaharlal Nehru
Answer:
(c) Mahatma Gandhi

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
(a) 1 2 3 4
(b) 3 4 2 1
(c) 1 3 4 2
(d) 2 4 3 1
Answer:
(b) 3 4 2 1

Question 2.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
(a) 4 1 2 3
(b) 3 4 2 1
(c) 2 4 3 1
(d) 1 2 4 3
Answer:
(a) 4 1 2 3

Question 3.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
(a) 4 1 2 3
(b) 1 2 3 4
(c) 4 3 2 1
(d) 2 4 1 3
Answer:
(d) 2 4 1 3

Choose the odd one out

Question 4.
(a) Poverty and famine
(b) Choice of technique
(c) Economics of caste
(d) Concept of capability
Answer:
(c) Economics of caste

Question 5.
(a) Budget by assignment
(b) Budget by assigned revenue
(c) Budget by shared revenues
(d) Budget by fiscal revenues
Answer:
(d) Budget by fiscal revenues

Question 6.
(a) Net Domestic Product (NDP)
(b) Human Development Index (HDI)
(c) Physical Quality of Life Index (PQLI)
(d) Gross National Happiness Index (GNHI)
Answer:
(a) Net Domestic Product (NDP)

Choose the correct statement

Question 7.
(a) Indian economy is the seventh largest in the world
(b) Indian economy is a mixed economy
(c) India ranks third in terms of purchasing power parity
(d) The human capital of India is old
Answer:
(d) The human capital of India is old

Question 8.
(a) The annual addition of population in India equals the total population of Australia.
(b) There exists a huge economic disparity in the Indian economy
(c) Kerala has the highest birth rate
(d) Every 6th person in the world is an Indian
Answer:
(c) Kerala has the highest birth rate

Choose the incorrect statement

Question 9.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
Answer:
(b) Demographic transition – (ii) 2001

Question 10.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
Answer:
(c) 1971 – (iii) 1040

Choose the correct statement

Question 11.
(a) Nehru always insisted on scientific temper
(b) Gandhi advocated a centralized economy
(c) Valluvar has recommended an unbalanced budget
(d) V.K.R.V. Rao’s economic thought is coined as Gandhian Economics.
Answer:
(a) Nehru always insisted on scientific temper

Question 12.
(a) Until 1979, Education in India was in state list
(b) Education in India follows the 10+3 pattern
(c) Sex ratio refers to the number of females per 1000 males.
(d) India ranks fourth in coal production of the world.
Answer:
(c) Sex ratio refers to the number of females per 1000 males.

Analyse the reason for the following 

Question 13.
Assertion (A) : There exist a huge economic disparity in the Indian economy.
Reason (R) : The proportion of income and assets owned by top 10% of Indians goes on increasing.
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is the not correct explanation of (A)
(c) (A) is true and (R) are false.
(d) (A) is false, but (R) is true.
Answer:
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)

Question 14.
Assertion (A) : In Kerala the adult sex ratio is 1084 as on 2011.
Reason (R) : Kerala provides better status to women as compared to other states.
(a) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) None of the above
Answer:
(b) Both (A) and (R) are true, (R) is the correct explanation of (A)

Fill in the blanks with suitable option given below

Question 15.
In India, Lignite is found in _____
(a) New Delhi
(b) kolkata
(c) Ranji
(d) Neyveli
Answer:
(d) Neyveli

Question 16.
Indian educational system comprises of _____ stages
(a) 12
(b) 3
(c) 6
(d) 4
Answer:
(c) 6

Question 17.
Indian railways introduced first wifi facility in _____
(a) Chennai
(b) Bengaluru
(c) Calcutta
(d) Hyderabad
Answer:
(b) Bengaluru

Choose the best option 

Question 18.
_____ is considered as the best pupil of J.M. Keynes
(a) A.K. Sen
(b) J.C.Kumarappa
(c) V.K.R.V.Rao
(d) Nehru
Answer:
(c) V.K.R.V.Rao

Question 19.
Third world country is _____
(a) India
(b) China
(c) Srilanka
(d) Bangladesh
Answer:
(d) Bangladesh

Question 20.
Per capital income is _____
(a) National income / output
(b) National income / population
(c) National income / per capital output
(d) National income / male population
Answer:
(b) National income / population

Part – B
Answer the following questions in one or two sentences 

Question 1.
What is Gross National Happiness Index?
Answer:
GNHI is an indicator of progress, which measures sustainable development, environmental conservation, promotion of culture and good governance.

Question 2.
State any two weakness of Indian Economy.
Answer:

  1. Large population.
  2. Inequality and poverty.

Question 3.
What is sex ratio ?
Answer:
Sex ratio refers to the number of females per 1,000 males.

Question 4.
Explain economic infrastructure.
Answer:
Economic infrastructure is the support system which helps in facilitating production

Question 5.
What is Gandhian Economics ?
Answer:
J.C. Kumarappa developed economic theories based on Gandhism. He developed a school of economic thought coined as Gandhian Economics.

Question 6.
Which are “BIMARU” states ?
Answer:

  1. Bihar
  2. Madhya Pradesh
  3. Rajasthan
  4. Uttarpradesh

Question 7.
What is year of small divide ?
Answer:
In 1951, population growth rate has come down from 1.33% to 1.25%. Hence it is known as year of small divide.

Part – C
Answer the following questions in one Paragraph

Question 1.
What are the aspects of demographic trends in India ?
Answer:

  1. Size of population
  2. Rate of growth.
  3. Birth and death rates
  4. Density of population
  5. Sex – ratio
  6. Life – expectancy at birth
  7. Literacy ratio.

Question 2.
Explain Thiruvalluvar’s ideas on agriculture.
Answer:
According to Thiruvalluvar, agriculture is the most fundamental economic activity. They are the axle-pin of the world, for on their prosperity revolves prosperity of other sectors of other sectors of the economy. He says, ploughmen alone is the freemen of the soil. He believes that agriculture is superior to all other occupation.

Question 3.
Write short note on Birth rate and Death rate.
Answer:
Short Note On Birth Rate And Death Rate:

  1. Birth Rate : It refers to the number of births per thousand of population.
  2. Death Rate : It refers to the number of deaths per thousand of population.

Kerala has the lowest birth rate and Uttarpradesh has the highest birth rate. West Bengal has the lowest and Orissa has the highest death rates.

Part – D
Answer the following questions in about a page

Question 1.
Explain about health in India.
Answer:
Health in India is a state government responsibility. The central council of health and welfare formulates the various health care projects and health department reform policies. The administration of health industry is the responsibility of the ministry of health and welfare.

Health care in India has many forms. But, all medical systems are under one ministry AYUSH.
Health status is better in Kerala as compared to other states. India’s health status is poor compared to Sri Lanka.

Question 2.
Explain Thiruvalluvar’s economic ideas.
Answer:
The economic ideas of Thiruvalluvar are found in his immortal work, Thirukkural in its second part porutpal.

  1. Factors of production : Thiruvalluvar has made many passing references about the factors of production, land, labour, capital, organisation.
  2. Agriculture : Agriculture is the most fundamental economic activity. The prosperity of other sectors depends on the prosperity of agriculture. Valluvar believes that agriculture is superior to all other occupation.
  3. Public finance : He elaborately explained public finance under the headings public revenue, financial administration and public expenditure.
  4. Public expenditure : Valluvar has recommended a balanced budget. He advocates
    1. Defence
    2.  Public works
    3. Social Services
  5. External assistance : Valluvar was against seeking external assistance. He advocated a self-sufficient economy.
  6. Poverty and begging : According to him ‘poverty is the root cause of all other evils which would lead to ever-lasting sufferings.
  7. Wealth: Valluvar has regarded wealth as only a means and not an end.
  8. Welfare state : The important elements of a welfare state are
    1. Perfect health of the people without disease
    2. Abundant wealth
    3. Good crop
    4. Prosperity and happiness
    5. Full security for the people.

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Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis

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Samacheer Kalvi 11th Economics Distribution Analysis Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
In Economics, distribution of income is among the
(a) factors of production
(b) individual
(c) firms
(d) traders
Answer:
(a) factors of production

Question 2.
Theory of distribution is popularly known as ………………………..
(a) Theory of product – pricing
(b) Theory of factor – pricing
(c) Theory of wages
(d) Theory of Interest
Answer:
(b) Theory of factor-pricing

Question 3.
Rent is the reward for the use of
(a) capital
(b) labour
(c) land
(d) organization
Answer:
(c) land

Question 4.
The concept of ‘Quasi – Rent’ is associated with ………………………
(a) Ricardo
(b) Keynes
(c) Walker
(d) Marshall
Answer:
(d) Marshall

Question 5.
The Classical Theory of Rent was propounded by
(a) Ricardo
(b) Keynes
(c) Marshall
(d) Walker
Answer:
(a) Ricardo

Question 6.
‘Original and indestructible powers of the soil’ is the term used by ……………………….
(a) J.S.Mill
(b) Walker
(c) Clark
(d) Ricardo
Answer:
(d) Ricardo

Question 7.
The reward for labour is
(a) rent
(b) wage
(c) profit
(d) interest
Answer:
(b) wage

Question 8.
Money wages are also known as ……………………….
(a) Real wages
(b) Nominal wages
(c) Original wages
(d) Transfer wages
Answer:
(b) Nominal wages

Question 9.
Residual Claimant Theory is propounded by
(a) Keynes
(b) Walker
(c) Hawley
(d) Knight
Answer:
(b) Walker

Question 10.
The reward is given for the use of capital ………………………
(a) Rent
(b) Wage
(c) Interest
(d) Profit
Answer:
(c) Interest

Question 11.
Keynesian Theory of interest is popularly known as
(a) Abstinence Theory
(b) Liquidity Preference Theory
(c) Loanable Funds Theory
(d) Agio Theory
Answer:
(b) Liquidity Preference Theory

Question 12.
According to the Loanable Funds Theory, supply of loanable funds is equal to …………………………
(a) S + BC + DH + DI
(b) I + DS + DH + BM
(c) S+ DS + BM+ DI
(d) S + BM + DH + DS
Answer:
(a) S + BC + DH + DI

Question 13.
The concept of meeting unexpected expenditure according to Keynes is
(a) Transaction motive
(b) Precautionary motive
(c) Speculative motive
(d) Personal motive
Answer:
(b) Precautionary motive

Question 14.
The distribution of income or wealth of a country among the individuals are ………………………….
(a) Functional distribution
(b) Personal distribution
(c) Goods distribution
(d) Services distribution
Answer:
(b) Personal distribution

Question 15.
Profit is the reward for
(a) land
(b) organization
(c) capital
(d) labour
Answer:
(b) organization

Question 16.
Innovation Theory of profit was given by …………………………..
(a) Hawley
(b) Schumpeter
(c) Keynes
(d) Knight
Answer:
(b) Schumpeter

Question 17.
Quasi-rent arises in
(a) Man-made appliances
(b) Homemade items
(c) Imported items
(d) None of these
Answer:
(a) Man-made appliances

Question 18.
“Wages as a sum of money are paid under contract by an employer to a worker for service rendered”- Who said this?
(a) Benham
(b) Marshall
(c) Walker
(d) J.S.Mill
Answer:
(a) Benham

Question 19.
Abstinence Theory of Interest was propounded by
(a) Alfred Marshall
(b) N.W Senior
(c) Bohm-Bawerk
(d) Knut Wicksell
Answer:
(b) N.W Senior

Question 20.
Loanable Funds Theory of Interest is called as …………………………
(a) Classical Theory
(b) Modem Theory
(c) Traditional Theory
(d) Neo-Classical Theory
Answer:
(d) Neo-Classical Theory

Part – B
Answer the following questions in one or two sentences

Question 21.
What is meant by distribution?
Answer:
Distribution means division of income among the four factors of production as rent, wage, interest and profit.

Question 22.
Mention the types of distribution?
Answer:
Personal Distribution: Personal Distribution is the distribution of national income among the individuals.
Functional Distribution: Functional Distribution means the distribution of income among the four factors of production namely land, labour, capital and organization for their services in production process.

Question 23.
Define ‘Rent’.
Answer:
Rent is that part of payment made by a tenant to his landlords for the use of land only.

Question 24.
Distinguish between real and money wages.
Answer:
Real wages :
Wages paid in terms of goods and services. It refers to the purchasing power of money wages.

Money wages :
Nominal wages are referred to as wages paid in terms of money.

Question 25.
What do you mean by interest?
Answer:

  1. Interest is the reward paid by the borrower to the lender for the use of capital.
  2. Interest is the price paid for the use of capital in any market.
  3. Generally speaking, interest is a payment made by a borrower to the lender for the money borrowed.

Question 26.
What is profit?
Answer:
Profit is the amount left with the entrepreneur after he has made payments for all the other factors of production.

Question 27.
State the meaning of liquidity preference?
Answer:

  1. Liquidity preference means the preference of the people to hold wealth in the form of liquid cash rather than in other non – liquid assets like bonds, securities, bills of exchange, land, building, gold etc.
  2. “ Liquidity Preference is the preference to have an amount of cash rather than of claims against others”.

Part – C
Answer the following questions in One Paragraph

Question 28.
What are the motives of demand for money?
Answer:
1. Transaction motive :
It relates to the desire of the people to hold cash for the current transactions.
Mt=f (y)

2. Precautionary motive :
It relates to the desire of the people to hold cash to meet unexpected or unforeseen expenditures.
Mp = f (y)

3. Speculative motive :
It relates to the desire of the people to hold cash in order to take advantage of market movements regarding future changes in the price of bonds and securities in the capital market.
Ms = f (i)

Question 29.
List out the kinds of wages:
Answer:
Wages are divided into four types.

  1. Nominal Wages or Money Wages: Nominal wages are referred to the wages paid in terms of money.
  2. Real Wages: Real wages are the wages paid in terms of goods and services. Hence, real wages are the purchasing power of money wages.
  3. Piece Wages: Wages that are paid on the basis of quantum of work done.
  4. Time Wages: Wages that are paid on the basis of the amount of time that the worker works.

Question 30.
Distinguish between rent and quasi-rent
Answer:
Rent:

  1. Rent accrues to land
  2. The supply of land is fixed forever
  3. It enters into price

Quasi – rent:

  1. Quasi-rent accrues to man-made appliances
  2. The supply of man made appliances is fixed for a short period only
  3. It does not enter into price.

Question 31.
Briefly explain the Subsistence Theory of Wages?
Answer:
The subsistence theory of wages:

  1. Subsistence theory of wages is one of the oldest theories of wages.
  2. According to this theory, wage must be equal to the subsistence level of the labourer and his family.
  3. Subsistence means the minimum amount of food, clothing and shelter which workers and their family require for existence.
  4. If workers are paid higher wages than the subsistence level, the workers would be better off and they will have large families.
  5. Hence, the population would increase.
  6. When the population increases, the supply of labourer would increase and therefore, wages will come down.
  7. If wages are lower than the subsistence level, there would be a reduction in population and thereby the supply of labour falls and wages increase to the subsistence level.

Question 32.
State the Dynamic Theory of Profit.
Answer:

  • J.B. Clark propounded this theory in 1900. To him, profit is the difference between price and cost of production of the commodity.
  • Profit is the reward for dynamic changes in society. He points out that, profit cannot arise in a static society. In static society everything remains stationary.
  • The following changes take place in a dynamic society.
    1. Population is increasing.
    2. Volume of capital is increasing.
    3. Methods of production are improving.
    4. Forms of industrial organization are changing.
    5. The wants of consumer are multiplying.

Question 33.
Describe briefly the Innovation Theory of Profit?
Answer:
Innovation Theory of Profit:

  1. The innovation theory of profit was propounded by Joesph. A.Schumpeter.
  2. Schumpeter says an entrepreneur is not only an undertaker of a business but also an innovator in the process of production.
  3. Profit is the reward for “innovation”.
  4. According to Schumpeter, an innovation may consist of the following:
    • Introduction of a new product.
    • Introduction of a new method of production.
    • Opening up of a new market.
    • Discovery of new raw materials
    • Reorganization of an industry / firm.

When any one of these innovations is introduced by an entrepreneur, it leads to reduction in the cost of production and thereby brings profit to an entrepreneur. To obtain profit continuously, the innovator needs to innovate continuously. The real innovators do so. Imitative entrepreneurs cannot innovate.

Question 34.
Write a note on the Risk-bearing Theory of Profit.
Answer:

  • Risk – the bearing theory was propounded by F.B.Hawley in 1907.
  • Profit is the reward for “risk-taking” in business.
  • Every business involves some risks. So risk-taking is an essential function of the entrepreneur and is the basis of profit.
  • Higher the risks, the greater are the profit.
  • Profit induces entrepreneurs to undertake risks.

Part – D
Answer the following questions in about a page

Question 35.
Explain the Marginal Productivity Theory of Distribution.
Answer:
Marginal productivity theory of distribution was developed by Clark, Wickseed and Walras. This theory explains how the prices of various factors of production are determined. This theory is also known as “General theory of distribution” or “National dividend theory of distribution”.
Assumptions:

  1. All the factors of production are homogenous and can be substituted for each other.
  2. There is perfect competition in both factor and product market.
  3. There is perfect mobility and full employment of factors of production.
  4. This theory is applicable only in the long-run.
  5. The entrepreneurs aim at profit maximization.
  6. There is no government intervention and no technological change.

Explanation of the theory :
Each factor is rewarded according to its marginal productivity.

Marginal product:
The marginal product of a factor of production means the addition made to the total product by employment of an additional unit of that factor.

Marginal physical product (MPP) :
The MPP of a factor is the increment in the total product obtained by the employment of an additional unit of that factor.

Value of marginal product (VMP):
VMP = MPP x price

Statement of the theory:

  1. The price of a factor of production depends upon its productivity.
  2. The price of a factor is determined by and will be equal to marginal revenue product of that factor.
  3. Under certain conditions, the price of a factor will be equal to both average and marginal products of that factor.

Marginal productivity under perfect competition:
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 1
X- axis represents factor units and Y-axis represents the factor price and revenue product.
MRP – Marginal Revenue Product Curve
ARP – Average Revenue Product Curve
AFC – Average Factor cost Curve
MFC – Marginal Factor cost Curve
AFC – is horizontal under perfect competition and MFC coincides with it.
When there is perfect competition in the factor market, the firm is in equilibrium only when MFC = MRP
At the point Q by employing ON units of factors and paying OP price (NQ) where MFC = MRP
At Q, MRP = ARP
Price of the factor (NQ) = Marginal revenue
Product (NQ) = Average revenue product (NQ)
There is no exploitation of factors under perfect competition.
Beyond the point Q the price paid to the factor is more than marginal revenue product and average revenue product, so employer do not employ the factors.

Question 36.
Illustrate the Ricardian Theory of Rent.
Answer:
The classical theory of rent is called “Ricardian theory of rent”.
Definition :
Rent is that portion of the produce of the earth which is paid to the landlord for the use of the original and indestructible powers of the soil.

Assumptions :

  1. Land differs in fertility.
  2. The law of diminishing returns operates in agriculture.
  3. Rent depends upon fertility and location of land.
  4. Theory assumes perfect competition and long period.
  5. There is existence of marginal land or no-rent land.
  6. Land has certain “Original and indestructible powers”.
  7. Land, is used for cultivation only.
  8. Most fertile lands are cultivated first.

Statement of the theory with illustration :
Assume that some people settle in newly discovered island. People will first cultivate the most fertile ‘A’ grade land. They produce 40 bags of paddy.

Suppose after some time if another group of people settle down in the same island. They cultivate ‘B’ grade land which produce 30 bags of paddy. Suppose yet another group of people settle down there they cultivate ‘C’ grade land. It produce 20 bags of paddy.

This surplus of ‘A’ grade land is now raised to 20 bags (40 – 20) and it is the ‘Economic Rent’ of ‘A’ grade land. The surplus of ‘B’ grade land is 10 bags (30 – 20). In ‘C’ grade land cost of production is equal to the price of its produce and it does not yield any rent (20 – 20). Hence ‘C’ grade land is called ‘no-rent land’ or marginal land.

The land which yields rent is called “intra-marginal land”.
Rent indicates the differential advantage of the superior land over the marginal land.
Ricardian Theory’ of Rent :

Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 2
Diagrammatic illustration :
X – axis represents various grades of land and Y axis yield per acre (in bags). OA, AB and BC are the ‘A’, ‘B’, ‘C’ grade lands respectively.

The ‘C’ grade land is the no-rent land. A and B grade lands are “intra-marginal lands. The economic rent yielded by ‘A’ and ‘B’ grade lands is equal to the shaded area of their respective rectangles.

Criticisms :

  1. The order of cultivation from most fertile to least fertile lands is historically wrong.
  2. This theory assumes that rent does not enter into price. But in reality, rent enters into price.

Question 37.
Elucidate the Loanable Funds Theory of Interest.
Answer:

  1. The loanable funds theory, also known as the “Neoclassical theory”. This theory was developed by Swedish economists like Wicksell, Bertilohlin, Viner, Gunnarmyrdal and others.
  2. Interest is the price paid for the use of loanable funds.
  3. The rate of interest is determined by the equilibrium between demand for and supply of loanable funds in the credit market.

Demand for loanable funds :

  1. Demand for Investment (I)
  2. Demand for Consumption (C)
  3. Demand for Hoarding (H)

Supply of loanable funds :
1. Savings (S) :
Savings may be of two types, namely.

  1. Savings planned by individuals are “ex-ante savings”. (Eg.) LIC premium
  2. Unplanned savings are called “ex-post savings”

2. Bank credit:
Commercial banks create credit and supply loan able funds to the investors.

3. Dishoarding (DH) :
Dishoarding means bringing out the hoarded money into use and thus it constitutes a source of supply of loanable funds.

4. Disinvestment (DI):
Disinvestment is the opposite of investment. It means not providing sufficient funds for depreciation of equipment.

Equilibrium :
The rate of interest is determined by the equilibrium between the total demand for and the total supply of loanable funds.
Supply of loanable funds = S + BC + DH + DI
Demand for loanable funds = I + C + H
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 3

Explanation :
X-axis represents the demand for and supply of loanable funds, Y axis represents the rate of interest. The LS curve represents the total supply curve ofloanable funds. The LD curve represents the total demand for loanable funds. The LD and LS curves, intersect each other at the point “E” the equilibrium point. At this point OR rate of interest and OM is the amount ofloanable funds.

Criticisms :

  1. Many factors have been included in this theory’. Still there are many more factors like
    • Asymmetric information
    • Moral Hazard.
  2. It is very difficult to combine real factors with monetary factors.

Question 38.
Explain the Keynesian Theory of Interest.
Answer:
According to Keynes, interest is the reward for parting with liquidity’ for a specified period of time.
Meaning of liquidity preference:
Liquidity preference means the preference of the people to hold wealth in the form of liquid cash rather than in other non-liquid assets.
Motives of demand for money:
The three motives of liquidity preference are.

  1. The transaction motive Mt = f (y)
  2. The precautionary motive Mp = f (y)
  3. The speculative motive Ms = f (i)

Determination of Rate of Interest:
According to Keynes, the rate of interest is determined by the demand for and the supply of money. The demand for money is liquidity preference. The supply of money is determined by the policies of the government and the central bank.

Equilibrium between demand and supply of money:
The equilibrium between liquidity preference and demand for money determine the rate of interest. In short-run, the supply of money is assumed to be constant.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 4

LP is the liquidity preference curve. M2 shows the supply curve of money to satisfy speculative motive. Both curves intersect at the point E which is the equilibrium. Here, rate of interest is I.

If liquidity preference increases from LP to L1 P1 the supply of money remains constant, the rate of interest increase from OI to OI1
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 5

Suppose LP remains constant. If the supply of money is OM2, the interest is OI2 and if the supply of money is reduced from OM2 to OM2, the interest would increase from OM2 to OM4. If the supply of money is increased from OM2 to OM4 the interest would decrease from OI2 to Ol4

Criticisms :
1. This theory does not explain the existence of different interest rates prevailing in the market at the same time.
2. It explains interest rate only in the short – run.

Samacheer Kalvi 11th Economics Distribution Analysis Additional Questions and Answers

Part -A
Choose the best options

Question 1.
Marginal Productivity theory is the …………………….. theory of distribution.
(a) Average
(b) Marginal
(c) Liquidity preference theory
(d) General
Answer:
(d) General

Question 2.
“Interest is the price paid for the use of capital in any market” is said by_______
(a) Adam Smith
(b) Torrance
(c) Walker
(d) Marshall
Answer:
(d) Marshall

Question 3.
The standard of living of workers in a country depends upon the ……………………… wages.
(a) Real
(b) Effective
(c) Direct
(d) Elastic
Answer:
(a) Real

Question 4.
The author of agio-theory of interest.
(a) N.W. Senior
(b) Bohm Bawerk
(c) Walker
(d) Marshall
Answer:
(b) Bohm Bawerk

Question 5.
Who propounded the risk bearing theory of profit?
(a) Bohm – Bawerk
(b) F.B Hawley
(c) Marshall
(d) Walker
Answer:
(b) F.B Hawley

Question 6.
Dynamic theory of profit was propounded by_______
(a) Marshall
(b) J.B.Clark
(c) J.M. Keynes
(d) Walker
Answer:
(b) J.B.Clark

Question 7.
What is meant by MPS?
(a) Marginal production supply
(b) Marginal production sale
(c) Marginal production service
(d) Marginal propensity to save
Answer:
(d) Marginal propensity to save

Question 8.
Time preference theory was given by_______
(a) John Ray
(b) Bohm Bawerk
(c) F.B. Hawley
(d) Irving Fisher
Answer:
(d) Irving Fisher

Question 9.
The demand for labour is ………………………
(a) Effective demand
(b) Direct demand
(c) Derived demand
(d) Elastic demand
Answer:
(c) Derived demand

Question 10.
_______ is the author of modern theory of rent
(a) Joan Robinson
(b) Boulding
(c) Both
(d) None
Answer:
(c) Both

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 6
(a) 3 4 2 1
(b) 4 3 1 2
(c) 1 2 3 4
(d) 2 3 4 1
Answer:
(c) 1 2 3 4

Question 2.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 7
(a) 2 1 4 3
(b) 1 4 2 3
(c) 1 2 3 4
(d) 4 1 3 2
Answer:
(a) 2 1 4 3

Choose the incorrect pair

Question 3.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 8
Answer:
(d) Total demand for money (iv) Mt + Mp

Question 4.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 9
Answer:
(b) Wage is the residual portion (ii) J.S. Mill

Choose the correct statement

Question 5.
(a) The supply of land is fixed forever
(b) Quasi-rent accrues to land
(c) The supply of man-made appliances are fixed
(d) Quasi-rent enters price
Answer:
(a) The supply of land is fixed forever

Question 6.
(a) Rent is the reward for labour
(b) Profit is the reward for labour
(c) Wages are the reward for labour
(d) Interest is the reward for organisation
Answer:
(b) Profit is the reward for labour

Choose the odd one out

Question 7.
(a) Bank credit
(b) Hoarding
(c) Dishoarding
(d) Disinvestment
Answer:
(b) Hoarding

Question 8.
(a) Monopoly profit
(b) Wind fall profit
(c) Functional reward
(d) Reward for labourer
Answer:
(d) Reward for labourer

Question 9.
(a) Nominal wages
(b) Real wages
(c) Direct wages
(d) Piece wages
Answer:
(c) Direct wages

Question 10.
(a) Abstinence theory
(b) Loanable funds theory
(c) Liquidity preference theory
(d) Risk bearing theory
Answer:
(b) Loanable funds theory

Analyse the reason for the following

Question 11.
Assertion (A) : Demand for loanable funds depends on investment, consumption and hoarding.
Reason (R) : Supply of loanable funds depends on hoarding and investment.
(a) (A) and (R) are true, (R) is not the correct explanation of (A)
(b) (A) and (R) are true, (R) is the correct explanation of (A)
(c) (A) is true (R) are false.
(d) (A) is false (R) is true.
Answer:
(c) (A) is true (R) are false.

Question 12.
Assertion (A) : The rate of interest is determined by the demand for money and the supply of money.
Reason (R) : The demand for money is liquidity preference.
(a) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(b) Both (A) and (R) are correct, (R) is the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) (A) is false and (R) is true.
Answer:
(a) Both (A) and (R) are true, (R) is not the correct explanation of (A)

Question 13.
Assertion (A) : Only superior land get rent
Reason (R) : Rent arose on account of differences in the fertility of land.
(a) Both (A) and (R) are true
(b) Both (A) and (R) are false correct explanation of (A)
(c) (A) is true and (R) are false.
(d) (A) is false and (R) is true.
Answer:
(a) Both (A) and (R) are true

Choose the incorrect pair

Question 14.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 10
Answer:
(d) Wage fund theory (iv) J.S. Mill

Question 15.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 11
Answer:
(b) Wages (ii) Labour

Fill in the blanks with suitable option given below

Question 16.
F.A. Walker wrote a book ______ in 1875.
(a) Political economy
(b) Social economy
(c) Principles of economics
(d) Wealth of nations
Answer:
(a) Political economy

Question 17.
The author of Agio theory of interest______
(a) N.W. Seniors
(b) Bohm-Bawerk
(c) Walker
(d) Marshall
Answer:
(b) Bohm-Bawerk

Question 18.
______ is the produced means of production.
(a) Land
(b) Labour
(c) Capital
(d) Organisation
Answer:
(c) Capital

Choose the best option

Question 19.
The theory of factor prices is popularly known as the theory of
(a) Distribution
(b) Exchange
(c) Wages
(d) Profit
Answer:
(a) Distribution

Question 20.
______ is the author of the modern theory of rent
(a) Joan Robinson
(b) Boulding
(c) Both
(d) None
Answer:
(c) Both

Part – B
Answer the following questions in one or two sentences

Question 1.
What is the theory of distribution?
Answer:
The theory of factor prices is popularly known as the theory of distribution.

Question 2.
Define VMP?
Answer:

  1. VMP – means Value of Marginal Product.
  2. The Value of Marginal Product is obtained by multiplying the Marginal Physical Product of the factor by the price of product. Symbolically, VMP = MPP x Price.

Question 3.
Define : Marginal productivity theory of distribution.
Answer:
The theory states that price or the reward for any factor of production is equal to the marginal productivity of that factor.

Question 4.
What is marginal product ? Define marginal physical product ?
Answer:
The addition made to the total product by employment of an additional unit of that factor.

Question 5.
Define modern theory of rent.
Answer:
Rent is the difference between the actual earnings of a factor of production and its transfer earning.
Rent = Actual earning – Transfer earning

Question 6.
What is “Wage Fund“.
Answer:
According to Mill, “Every employer will keep a given amount of capital for payment to the workers”. It is known as “Wage Fund”.

Question 7.
Define Quasi Rent.
Answer:
“Quasi rent is the income derived from machines and other appliances made by man
QR = Total revenue – Total variable cost

Part – C
Answer the following questions in one Paragraph

Question 1.
Explain the types of distribution.
Answer:
1. Personal distribution :
It is the distribution of national income among individuals.

2. Functional distribution :
It means the distribution of income among the four factors of production for their services in the production process.

Question 2.
Explain the standard of living theory of wages.
Answer:
This theory of wages was developed by Torrance is an improved version of the subsistence theory of wage. According to this theory, the wage is equal to the standard of living of the workers. If the standard of living is high, wages will be high and vice versa. The standard of a living wage means the amount necessary to maintain the labourer in the standard of life to which he is accustomed.

Question 3.
Briefly explain the Abstinence theory or waiting theory?
Answer:

  1. This theory was propounded by N.W.Senior.
  2. According to Senior, capital is the result of saving.
  3. Saving involves “ abstinence ” or “ Sacrifice
  4. It is possible to save only if one abstains from present consumption.
  5. Interest is the reward or compensation paid to the saver (capitalist) for his “ abstinence ” or “ sacrifice”.
  6. Marshall accepted the Abstinence theory of interest.
  7. According to him, interest is a reward for waiting.
  8. Saving involves waiting.
  9. Therefore, interest is the reward paid to the saver for his “ waiting ”.

Question 4.
State the theories of interest.
Answer:

  1. Abstinence theory or waiting theory.
  2. Agio theory or the psychological theory.
  3. Loanable funds theory or the neoclassical theory.
  4. Liquidity preference theory or the monetary theory.

Question 5.
What are the four sources of loanable funds.
Answer:
The supply of loanable funds depends upon the following four sources.

  1. Savings (S)
  2. Bank Credit (BC)
  3. Dishoarding (DH)
  4. Disinvestment (DI)

Question 6.
What are the kinds of profit.
Answer:
1. Monopoly profit:
Profit earned by the firm because of its monopoly control.

2. Windfall profit:
Some times, profit arises due to changes in price level. Profit is due to unforeseen factors.

3. Profit as functional reward :
Just like rent, wage and interest, profit is earned by the entrepreneur for his entrepreneurial function.

Question 7.
State the residual claimant theory of wage.
Answer:
This theory was propounded by the American economist F.A. Walkar in 1875, in his book political economy. According to this theory, wage is the residual portion after paying the remuneration of all the other three factors namely land capital and organization.

Criticisms :

  1. This theory does not explain the role of trade unions can secure higher wage for workers.
  2. Demand side of labour in the determination of wages needs to be considered.

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Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

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Samacheer Kalvi 11th Chemistry Basic Concepts of Organic Reactions Textual Evaluation Solved

I. Choose The Best Answer:

Question 1.
For the Following reactions –
(A) CH3 CH2 CH2 Br + KOH → CH3 – CH + KBr + H2O
(B) (CH3 )3 CBr + KOH → (CH3 )3 COH + KBr
(C) Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Which of the following statement is correct’?
(a) (A) is elimination, (B) and (C) are substitution
(b) (A) is substitution, (B) and (C) are elimination
(c) (A) and (B) are elimination and (C) is addition reaction
(d) (A) is elimination, B is substitution and (C) is addition reaction.
Answer:
(d) (A) is elimination, B is substitution and (C) is addition reaction.

Question 2.
What is the hybridisation state of benzyl carbonium ion?
(a) sp2
(b) spd2
(c) sp3
(d) sp2d
Answer:
(a) sp2

Question 3.
Decreasing order of nucleophilicity is –
(a) OH> RNH2 > OCH3> RNH2
(b) NH2> OH> OCH3> RNH2
(c) NH,> CH3O >OH> RNH2
(d) CH3O> NH2> OH> RNH2
Answer:
(b) NH2> OH> OCH3> RNH2+

Question 4.
Which of the tòllowing species is not eLectrophilic in nature?
(a) Cl+
(b) BH3
(c) H3O+
(d) +NO2
Answer:
(c) H3O+

Question 5.
Homolytic fission of covalent bond leads to the formation of –
(a) electrophile
(b) nucicophile
(c) Carbocation
(d) free radical
Answer:
(d) free radical

Question 6.
Hyper Conjugation is also known as –
(a) no bond resonance
(b) Baker – nathan effect
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)

Question 7.
Which of the group has highest +I effect?
(a) CH3
(b) CH3CH2
(c) (CH3)2 -CH-
(d) (CH3)3-C-
Answer:
(d) (CH3)3-C-

Question 8.
Which of the following species does not exert a resonance effect?
(a) C6H5OH
(b) C6H5Cl
(c) C6H5NH2
(d) C6H5NH3
Answer:
(d) C6H5NH3

Question 9.
-I effect is shown by –
(a) -Cl
(b) -Br
(c) both (a) and (b)
(d) -CH3
Answer:
(c) both (a) and (b)

Question 10.
Which of the following carbocation will be most stable?
(a) \({ PH }_{ 3 }\overset { + }{ C- } \)
(b) CH3–\(\overset { + }{ C{ H }_{ 2 } } \)
(c) (CH3)2–\(\overset { + }{ C{ H } } \)
(d) CH2 = CH-CH3
Answer:
(a) \({ PH }_{ 3 }\overset { + }{ C- } \)

Question 11.
Assertion : Tertiary Carbocations are generally formed more easily than primary Carbocations.
Reason : Hyper conjugation as well as inductive effect due to additional alkyl group stabilize tertiary carbonium ions.
(a) both assertion and reason arc true and reason is the correct explanation of assertion.
(b) 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) both assertion and reason are true and reason is the correct explanation of assertion.

Question 12.
Heterolytic fission of C-C bond results in the formation of –
(a) free radical
(b) Carbanion
(c) Carbocation
(d) Carbanion and Carbocation
Answer:
(e) Carbocation

Question 13.
Which of the following represent a set of nucleophiles?
(a) BF3, H2O, NH2-
(b) AlCl3, BF3, NH3
(c) CN, RCH2, ROH
(d) H+, RNH+2, CCl2
Answer:
(c) CN, RCH2, ROH

Question 14.
Which of the following species does not acts as a nucleophile?
(a) ROH
(b) ROR
(c) PCl3
(d) BF3
Answer:
(d) BF3

Question 15.
The geometrical shape of carbocation is
(a) Linear
(b) tetrahedral
(c) Planar
(d) Pyramidal
Answer:
(c) Planar

II. Write brief answer to the following questions.

Question 16.
Write short notes on:
(a) Resonance
(b) Hyper conjugation
Answer:
(a) Resonance:
1. Certain organic compounds can be represented by more than one structure and they differ only in the position of bonding and lone pair of electrons. Such structures are called resonance structure and this phenomenon is called as resonance. This phenomenon is also called as mesomerism or mesomeric effect.

2. For example, the structure of aromatic compounds such as benzene and conjugated system like 1,3 butadiene cannot be represent by a single structure and their observed properties can be explained on the base of a resonance hybrid.

3. Resonance structure of benzene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
(I) and (II) are called as resonance hybridš of benzcnc.

4. For 1,3 butadiene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

(I), (II) and (III) are called as resonance hybrids of 1,3 butadiene.

(b) Hyper conjugation:
1. The denationalization of electrons of σ bond is called as hyper conjugation. it is a special stabilizing effect that results due to the interaction of electrons of a σ bond with the adjacent empty non-bonding P-orbitais resulting in an extended molecular orbital.

2. Hyper conjugation is a permanent effect.

3. For example, in propane. the σ- electrons of C-H bond of methyl group can be delocalised into the it- orbital of doubly bonded carbon as represented below.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 17.
What are electrophiles and nucleophiles? Give suitable examples for each.
Answer:
Electrophiles:
Electrophiles are reagents that are attracted towards negative charge or electron rich center. They are either positively charged ions or electron deficient neutral molecules.
Example:
CO2, AlCl3, BF3, FeCl3, NO+, NO+2,, etc.

Nucleophiles:
Nucleophiles are reagents that has high affinity for electropositive centers. They possess an atom has a unshared pair of electrons. They are usually negatively charged ions or electron rich neutral molecules.
Example.
NH3, R-NH2, R-SH, H2O, R-OH, CN. OH etc.

Question 18.
Show the heterolysis of covalent bond by using curved arrow notation and complete the following equations. Identify the nucleophile In each case.
1. CH3-Br+KOH →
2. CH3-OCH3+HI →
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
In this case, \(\overset { \ominus }{ Br } \) and \(\overset { \ominus }{ OH } \) are nucleophiles.

Another mechanism:
Step – I
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Step-II
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

In this case, OCH3 and -I are nucleophiles
Another mechanism:
Step – I
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Step – II

Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 19.
Explain inductive effect with suitable example.
Inductive effect:
Answer:
1. it is defined as the change ¡n the polarization of a covalent bond due to the presence of adjacent bonded atoms or groups in the molecule. It is denoted as I-effect.

2. Atoms or groups which lose electron towards a carbon atom are said to have a +I effect.
Example:
CH3-,(CH3)2 CH-,(CH3)2 C- etc.

3. Atoms or groups which draw electrons away from a carbon atom are said to have a -I effect.
Example:
-NO2. -I, -Br, -OH, C6H5 etc.

4. For example, consider ethane and ethyl chloride. The C-C bond in ethane is non polar while the C-C bond in ethyl chloride is polar. We know that chlorine is more electronegative than carbon and hence it attracts the shared pair of electrons between C-Cl in ethyl chloride towards it self.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
This develops a slight negative charge on chlorine and a slight positive charge on carbon to which chlorine is attached. To compensate it, the C1 draws the shared pair of electron between itself and C2. This polarization effect is called inductive effect.

Question 20.
Explain electromeric effect.
Electromeric effect:
Answer:
1. The electromeric effect refers to the polarity produced in a multiple bonded the compound when it is attacked by a reagent when a double or a triple bond is exposed to an attack by an electrophile the two π electrons which from the π bond are completely transferred to one atom or the other.

2. When a nucleophile approaches the carbonyl compound, the π-electrons between C and O is instantaneously shifted to the more electronegative oxygen. This make the carbon electron deficient and thus facilitating the formation of a new bond between the incoming nucleophile and the carbonyl carbon atom.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

3. When an electrophile such as H approaches an alkene molecule their electrons are instantaneously shifted to the electrophiie and a new bond is formed between carbon and hydrogen. This makes the other carbon electron deficient and hence it acquires a positive charge.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

4. This effect denotes as E-effect.

Question 21.
Give examples for the following types of organic reactions
1. – elimination
2. Electrophilic substitution.
Answer:
1. β-eIimination:
Elimination reactions involve the cleavage of a a bond and tórmation of air bond. A nucleophilic pair of electrons heads into a new it bond as a leaving group departs. This process is called 3-elimination because the bond Í to the nucleophilic pair of electrons breaks.
Example:
(a) n-propyl bromide on reaction with alcoholic KOH give propene. in this reaction hydrogen and Br are eliminated.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

(b) Acid-catalysed dehydration of alcohols
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

2. Electrophilic substitution:
Substitution reactions when are brought about by electrophiles are called electrophilic substitution reaction.
Example:
(a) Nitration of benzene
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

(b) Bromination of benzene:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Samacheer Kalvi 11th Chemistry Basic Concepts of Organic ReactionsAdditional Questions Solved

I. Choose The Correct Answer.

Question 1.
Statement-I: All the organic molecules contain covalent bonds.
Statement-II: Organic molecules are formed by the mutual sharing of electrons between atoms.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but Statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.

Question 2.
Statement-I : Homolytic cleavage is symmetrical one.
Statement-II : A single covalent bond breaks and each of the bonded atoms retains one electron.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statement-I and II are correct and statement-Il is correct explanation of statement-I.

Question 3.
Which one of the following is an example for free radical initiators?
(a) Benzoyl peroxide
(b) Benzyl alcohol
(c) Benzyl acetate
(d) Benzaldehyde.
Answer:
(a) Benzoyl peroxide.

Question 4.
Which one of the following ¡s correct order of stability of alkyl free radicals?
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 5.
Statement-I: Ileterolytic cleavage is unsymmetrical one.
Statement-II: A covalent bond breaks and one of the bonded atom retains the bond pair of electrons.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct
Answer:
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.

Question 6.
Which of the following is correct order of the stability of carbocations?
(a) +CH3 > +CH2 CH3> +CH(CH3)2 > +C(CH3)3
(b) +CH2CH3 > +CH3 >+ CH(CH3)2 > +C(CH3)3
(c) +C(CH3)3 > +CH(CH3)2 >+CH2CH3CH3
(d) +CH(CH3)2 > +CH3 > +CH2CH3 > +C(CH3)3
Answer:
(c) +C(CH3)3 > +CH(CH3)2 >+CH2CH3CH3

Question 7.
Which one of the following is correct order of the stability of carbanions?
(a) C(CH3)3> CH(CH3)2> CH2-CH3> CH3
(b) CH3> CH2-CH3> CH(CH3)2> C(CH3)3
(c) CH(CH3), > CH3> CH2-CH3 > C(CH3)3
(d) CH2-CH3 > CH(CH3)2> CH3> C(CH3)3
Answer:
(b) CH3> CH2-CH3> CH(CH3)2> C(CH3)3

Question 8.
Which one of the following is not electrophile’?
(a) NH3
(b) AlCl3
(c) FeCl3
(d) R-X
Answer:
(a) NH3

Question 9.
Which one of the following is not nucleophile
(a) H2O
(b) NH3
(c) R-OH
(d) FeCl3
Answer:
(d) FeCl3

Question 10.
Which one of the following is positively charged electrophiles?
(a) CO2
(b) AlCl3
(c) BF3
(d) RX
Answer:
(d) RX

Question 11.
Which one of the following is nucleophile?
(a) BF3
(b) AlCl3
(c) CO2
(d) R-SH
Answer:
(d) R-SH

Question 12.
Which one of the following species has tendency to show-I effect?
(a) -CH3
(b) -CH2-CH3
(c) -CH(CH3)2
(d) -C6H5
Answer:
(d) -C6H5

Question 13.
Which one of the following species has tendency to show +I effect?
(a) -NH2
(b) -Cl
(c) -C6H5
(d) -CH3
Answer:
(d) -CH3

Question 14.
Which one of the following has strongest acidic character?
(a) HCOOH
(b) CH3COOH
(c) CH2ClCOOH
(d) CCl3COOH
Answer:
(d) CCl3COOH

Question 15.
Which one of the following has least acidic character?
(a) HCOOH
(b) CH3COOH
(c) CH2ClCOOH
(d) CCl3COOH
Answer:
(b) CH3COOH

Question 16.
Pick out the correct order of acid strength.
(a) CH3-CH2-COOH > CH3COOH > CH2CICOOH
(b) CH3COOH > CH3-CH2-COOH >CH2CICOOH
(c) CH2ClCOOH > CH3COOH > CH3-CH2-COOH
(d) CH2ClCOOH > CH3-CH2-COOH > CH3COOH
Answer:
(c) CH2ClCOOH > CH3COOH > CH3—CH2-COOH

Question 17.
Statement-I: Fluoro acetic acid is stronger acid than acetic acid
Statement-II: Fluorine has high electronegativity and it is facilitate to dissociate the OH bond easily.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-Il is correct
Answer:
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.

Question 18.
Which one of the following is an example for negative mesomeric effect?
(a) -SH
(b) -SR
(c) -NH2
(d) -NO2
Answer:
(d) -NO2

Question 19.
Which one of the following electrophile used for nitration of benzene?
(a) \({ Br }^{ \oplus }\)
(b) \({ { NO }_{ 2 }^{ \ominus } }\)
(c) -NH2
(d) \({ NO }^{ \ominus }\)
Answer:
(b) \({ { NO }_{ 2 }^{ \ominus } }\)

Question 20.
Identify the one which does not come under the organic addition reaction
(a) Hydration
(b) Dehydration
(c) Halogenation
(d) Hydro halogenation
Answer:
(b) Dehydration

Question 21.
Primary alcohols undergo which type of reaction to form alkenes?
(a) Elimination
(b) Oxidation
(c) Reduction
(d) Hydrolysis
Answer:
(a) Elimination

Question 22.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions Identify the type of reaction
(a) Addition reaction
(b) Elimination reaction
(c) Reduction reaction
(d) Oxidation reaction
Answer:
(d) Oxidation reaction

II. Match the following.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Chapter 11 Fundamentals of Organic Chemistry

III. Fill in the blanks.
Question 1.
The slowest step in the mechanism determines ………..
Answer:
Rate of reaction

Question 2.
Homolytic cleavage occurs under the conditions of ………….
Answer:
High temperature.

Question 3.
During the cleavage of Azobisisobutyronitrile produces species ……….
Answer:
Free radical

Question 4.
The cleavage of C-Br bond in tert-butyl bromide leads to formation of ………..
Answer:
Carbocation

Question 5.
The cleavage of C-H bond in aldehydes Leads to formation of ………….
Answer:
Carbanion

Question 6.
Electron displacement occurring in saturated compounds along a carbon chain is termed as …………
Answer:
Inductive effect

Question 7.
The addition of H to alkene is an example of effect ……….
Answer:
+E

Question 8.
Example for positive mesomeric effect is ………..
Answer:
-OH

Question 9.
Acidity of phenol was explained by …………
Answer:
R-effect

Question 10.
Hydrolysis of alkyl halide is an example for ……….
Answer:
Nucleophilic substitution

Question 11.
4-hydroxy phenol reacts with acidified potassium dichromate to gives
Answer:
p-Benzoquinone

Question 12.
Enzyme present in apple is
Answer:
Poly phenol oxidase

Question 13.
Benzene reacts with H2 in the presence of Pt to give
Answer:
Cyclohexane

Question 14.
Alcohol on refluxing with K2Cr2O7 gives……..
Answer:
Carboxylic acid

Question 15.
Carbonyl compounds especially ketones undergo reduction to form ………..
Answer:
Secondary alcohols

Question 16.
Ethane undergo thrombolytic cleavage to form ………..
Answer:
Methyl free radical

IV. Choose the odd one out.
Question 1.
(a) NH3
(b) H2O
(c) CN
(d) RSH
Answer:
(c) CN It is a negatively charged nucleophile whereas others are neutral nucleophiles.

Question 2.
(a) CO2
(b) RX
(c) MCl3
(d) FeCl3
Answer:
(b) RX. It is a positive charged electrophiles whereas others are neutral clectrophiles.

Question 3.
(a) -F
(b) – Cl
(c) – COOH
(d) CH3O
Answer:
(d) CH3O. It is a group carrying negative charge one electron donating (or) +I group whereas others are -I group.

Question 4.
(a) – C(CH3)3
(b) – COOK
(c) – CH(CH3)2
(d) – COO
Answer:
(b) – COOH. It has -I effect whereas other have +I effect.

Question 5.
(a) -OH
(b) -NH2
(c) -SH
(d) -COOH
Answer:
(d) – COOH. It has negative mesomeric effect whereas others have positive mesomeric effect.

V. Choose the correct pair.
Question 1.
(a) Homolytic clevage : carbocation
(b) Homolytic clevage : carbanion
(c) Homolytic clevage : free radicals
(d) Heterolytic clevage : free radicals
Answer:
(c) Homolytic clevage : free radicals

Question 2.
(a) + C(CH3)3 > +CH(CH3)2> +CH2-CH3 > +CH3 : Relative stability of carbocations.
(b) – C(CH3)3 > CH(CH3)2 > CH2-CH3 > CH3 : Relative stability of carbanion.
(c) OH,RO, RCOO : Neutral nucleophile
(d) AlCl3, BF3, FeCl3 : Positively charged nucleophile
Answer:
(a) + C(CH3)3 > +CH(CH3)2> +CH2-CH3 > +CH3 : Relative stability of carbocations.

Question 3.
(a) CH3 – CH2 – CH2Br + Alcoholic KOH : Substitution reaction
(b) CH3 – CH2 – CH2Br + Alcoholic KOH : Elimination reaction
(c) CH3CHO + Acidic dichromate : Reduction reaction
(d) Benzene + Pt + H2 : Oxidation reaction
Answer:
(b) CH3 – CH2 – CH2Br + Alcoholic KOH : Elimination reaction

VI. Choose the incorrect pair.
Question 1.
(a) – OH,- SH, -NH2 : Positive mesometic effect
(b) – NO2 >CO, – COOH : Negative mesomeric effect
(c) – F, – Cl, – NO2 : Electron withdrawing group
(d) – C(CH3)3, – CH(CH3)3 – CH2 – CH3 : Electron withdrawing group
Answer:
(c) – F, – Cl, – NO2 : Electron withdrawing group

Question 2.
(a) NH3 andAmines : Neutral nucleophile
(b) OH and RCOO : Negative nucleophile
(c) RX and H3O3 : Positive electrophile
(d) MCl3, BF3 : Negative electrophile
Answer:
(d) AlCl3, BF3 : Negative electrophile

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

VII. Assertion and Reason.
Question 1.
Assertion (A) : Neutral molecule SnCl4 can act as an electrophile.
Reason (R) : it has vaccant ‘d’ orbitais which can accommadate the electrons from others.
(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) : The C-C bond in ethane is non-polar while the C-C bond in ethyl chloride is polar.
Reason (R) : Chlorine is more electronegative than carbon and hence it attracts the shared pair of electron between C-C in ethyl chloride and it develops a negative charge on Cl and positive charge on Carbon.
(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) is 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) is the correct explanation of (A).

Question 3.
Assertion (A): Phenol is more acidic than aliphatic alcohols.
Reason (R): The phenoxide ion is more stabilized than phenol by resonance effect and hence resonance favours ionization of phenol to form H and shows acidity.
(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): The cut apple turns brown.
Reason (R): Cut apple exposes its cells to atmospheric oxygen and the oxidizes the phenolic compound present in it. Due to this enzymatic browning, the cut apple turns brown.
(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) is 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) is the correct explanation of (A).

Samacheer Kalvi 11th Chemistry Basic Concepts of Organic Reactions 2 Marks Question and Answers

II. Answer briefly.

Question 1.
What are organic reactions?
Answer:
Substrate is an organic molecule reacts with reagent, which may be an organic, inorganic or any agent like heat, photon etc, that brings about the chemical change to form a product, this is known as organic reactions.

Question 2.
What is mechanism of the reaction?
Answer:
In the organic reactions, that series of simple steps which collectively represent the chemical change, from substrate to product, this is called as the mechanism of the reaction. The slowest step in the mechanism determines the overall rate of the reaction.

Question 3.
Mention the types of fission of a covalent bond?
Answer:
There are three types of fission of a covalent bond. They are,

  • Homolytic cleavage
  • Heterolytic cleavage

Question 4.
Explain the homolytic fission of a covalent bond?
Answer:

  • Homolytic cleavage is the process in which a covalent bond breaks symmetrically in such way that each of the bonded atoms retains one electron.
  • This type of cleavage occurs under high temperature or in the presence of UV light.
  • in a compound containing non-polar covalent bond formed between atoms of similar electronegativity, in such molecules the cleavage of bonds results into free radicals.
  • For example, ethane undergo homolytic fission to produce, two methyl free radicals.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 5.
What are free radical initiators?
Answer:
The types of reagents that promote homolytic cleavage in substrate are called as free radical initiators. They are short lived and are highly reactive.

Question 6.
Mention any two examples for free radical initiators?
Answer:
Two examples for free radical initiators

  • Azobisisobutyronitnie (AIBN)
  • Benzoyl peroxide

Question 7.
Explain the heterolytic fission of a covalent bond?
Answer:

  • Heterolytic cleavage is the process in which a covalent bond breaks unsymmetrically such that one of the bonded atom retains the bond pair of electron.
  • It results in the formation of a cation and an anion of the two bonded atoms the most electronegative.
  • For example in tert-butyibromide, the C-Br bond is polar as bromine is more electronegative than carbon. Hence the C-Br undergoes heterolytic cleavage to form a terl-butyl carbocation and bromide anion.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 8.
What are carbocations?
Answer:
1. Let us consider the heterolytic fission of the bond C-X present in organic molecule. If the atom X has greater electronegativity than the carbon atom, the former takes away the bonding electron pair and becomes negatively charged while the carbon will lose its electron and thus acquire a positive charge.
2. Such a cationic species carrying a positive charge on carbon are known as carbocation or carhonitim ion.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 9.
What are carbanions?
Answer:
Let us consider the heterolytic fission of the bond C-X present in an organic molecule. if the carbon atom has greater electronegativity than the atom X, the former takes away the bonding electron pair and acquires a negative charge. The resulting carbon anion is known as carbanion.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 10.
Differentiate the carbocation and carbanion
Answer:
Carbocation:

  • In a carbocation carbon bearing positive charge.
  • Carbon bearing positive charge has sp2 hybridization.
  • It has a planar structure.
  • Example:
    CH3+ (CH3)2\({ C }^{ \ominus }\)H, (CH3)3C+ etc.

Carbanion:

  • In a carbanion carbon bearing negative charge.
  • Carbon bearing negative charge has sp3 hybridization.
  • It has a pyramidal structure.
  • Example:
    CH3–\({ C }^{ \ominus }\)H2, CH3-CHO, (CH3)2\({ C }^{ \ominus }\)H etc.

Question 11.
Identify which of the following are electrophiles and nucleophiles?

  1. NH3
  2. AlCl3
  3. R-SH
  4. R-X
  5. R-O-R
  6. BF3

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 12.
How will you distinguish between electrophiles and nucleophiles?
Answer:
Electrophiles:

  • They are electron deficient.
  • They are cations.
  • They are lewis acids.
  • Accept an electron pair.
  • Attack on electron rich sites.

Nucleophites:

  • They are electron rich.
  • They are anions.
  • They are lewis bases.
  • Donate an electron pair.
  • Attack on electron deficient sites.

Question 13.
What are all the sources for human body that produces free radicals?
Answer:
Sources for human body produces free radicals,

  • Human body is exposed to X-rays.
  • Cigarette smoke.
  • Industrial chemicals.
  • Air pollutants.

Question 14.
In what way free radical affect the human body?
Answer:

  • Free radicals can disrupt cell membranes.
  • Increase the risk of many forms of cancer.
  • Damage the interior lining of blood vessels.
  • leads to a high risk of heart disease and stroke.

Question 15.
How to reduce the effect of free radicals?
Answer:

  • Use vitamins and minerals to counter the effects of free radicals.
  • Fruits contains antioxidants which decrease the effects of free radicals.

Question 16.
Identify which of the following shows +I and -I effect?

  1. -NO2
  2. -SO2H
  3. -I
  4. -OH
  5. CH2O-
  6. CH2

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 17.
Why chloro acetic acid is stronger acid than acetic acid?
Chioro acetic acid:
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Chioro acetic acid has Cl-group and it has high electronegativity and shows -I effect. Therefore Cl-atom to faciLitate the dissociation of O-H bond very fastly, whereas in the case of acetic acid, has CH3 group and it shows +I effect, theretòre dissociation of O-H bond will be more difficult. Thus chloro acetic acid is stronger acid than acetic acid.

Question 18.
Explain the positive and negative electromeric effects?
Answer:
1. When an electrophile such as W approaches an alkene molecule, the π electrons are instantaneously shifted to the electrophile and a new bond is formed between carbon and hydrogen. This makes the other carbon electron deficient and hence it acquires a positive charge.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
2. When the it electron is transferred towards the attacking reagent, it is called positive electromeric (+E) effect.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
3. Whèn the t electron is transferred away from the attacking reagent it is called negative electromenc (-E) effect.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
For example : The attack of C\({ N }^{ \ominus }\) on a carbonyl carbon.
Samacheer Kalvi 11th Chemistry Chapter 11 Fundamentals of Organic Chemistry

Question 19.
Write a short notes on positive mesomeric effect?
Answer:

  • Positive mesomeric effect occurs, when the electrons move away from substituent attached
    to the conjugated system.
  • It occurs, if the electron releasing substituents are attached to the conjugated system.
  • The attached group has a tendency to release electrons through resonance, these electron
    releasing groups are usually denoted as +R or +M groups
  • Examples: -OH, -SH, -OR, -SR, -NH2 etc.

Question 20.
Write a short notes on negative mesomeric effect?
Answer:

  • Negative mesomeric effect occurs, when the electrons move towards the substituent attached to the conjugated system.
  • It occurs if the electron withdrawing substituents are attached to the conjugated system.
    The attached group has a tendency to withdraw electrons through resonance, these electron withdrawing groups are usually denoted as -R or -M groups.
  • Examples:
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 21.
What are addition reactions? Give an example.
Answer:
All organic compounds having double or triple bond adopt addition reactions in which two substances unite to form a single compound. During the addition reaction the hybridization of the substrate changes as only one bond breaks and two new bonds are formed.
Example:

Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 22.
What are elimination reaction? Give an example.
Answer:
In these reactions, two atoms or groups are removed from a molecule without being substituted by other atoms or groups. A new C-C double bond is formed between the carbon atoms to which the eliminated atoms or groups are previously attached. It is always accompanied with change in hybridization.
Example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 23.
What are organic oxidation reactions? Give an example.
Answer:
Most of the oxidation reaction of organic compounds involves gain of oxygen or loss of hydrogen.
Example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 24.
What are organic reduction reactions? Give an example.
Answer:
Most of the reduction reaction of organic compounds involves gain of hydrogen or loss of oxygen.
Example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 25.
Why cut apple turns a brown colour?
Answer:

  • Apples contains an enzyme called polyphenol oxidase (PPO) also known as tyrosinase.
  • Cutting an apple exposes its cells to the atmospheric oxygen and oxidizes the phenolic compounds present in apples. This is called the “enzymatic browning” that turns a cut apple brown.
  • In addition to apples, enzymatic browning is also evident in bananas, pears, avocados and even potatoes.

Question 26.
What are functional group inter conversions?
Answer:
Organic synthesis involves functional group inter conversions. A particular functional group can be converted into other functional group by reacting it with suitable reagents. For example, the carboxylic acid group (-COOH) present in organic acids can be transformed into -COCl by treating the acid with SOCl2 reagent..

Question 27.
How will you convert alcohol into aldehyde?
Answer:
When primary alcohol reacts with acidified potassium dichromate to gives aldehyde.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 28.
What happen when nitrile undergoes acid hydrolysis?
Answer:
When alkyl nitrile undergoes acid hydrolysis to give amide, which on further hydrolysis to give carboxylic acid.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 29.
Complete the following reactions and identify the products?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 30.
Predict the product for the following reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 31.
How will you convert benzene into cyclohexane?
Answer:
Benzene undergo catalytic hydrogenation in presence of platinum to give cyclohexane.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 32.
Complete the reaction and name the product
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 33.
Identify the product and mention the type of organic reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
This is one of the elimination reaction.

Question 34.
Complete the following reaction and identify the product.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 35.
What are substitution reaction?
Answer:
In this reaction an atom or a group of atoms attached to a carbon atom is replaced by a new atom or a group of atoms.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Here, -Br is replaced by -OH group.

Question 36.
How will substitution reactions are classified?
Answer:
Substitution reactions can be classified as,

  • Nucleophilic substitution reaction.
  • Electrophilic substitution reaction.
  • Free radical substitution reaction.

Question 37.
Draw the resonance structures for the following compounds?
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 38.
Identify the reagents shown in box in the following equations as nucleophiles or electrophiles.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
In (a) \(\overset { \ominus }{ OH } \) is nucleophile
In (b) \(\overset { \ominus }{ CN } \) is nucleophile
In (c) CH3\(\overset { \ominus }{ CO } \) is electrophile

Question 39.
Classify the following reactions in one of the reaction type of organic reaction.
Answer:
(a) CH3CH2Br + \({ SH }^{ \ominus }\) → CH3CH2SH + \({ Br }^{ \ominus }\)
(b) (CH3)2C = CH2 + HCI (CH3)2CCl – CH3
(c) CH3CH2Br + \({ OH }^{ \ominus }\) → CH2 = CH2 + H2O + \({ Br }^{ \ominus }\)
(a) Nucleophilic substitution reaction.
(b) Electrophilic addition reaction.
(c) Elimination reaction.

Question 40.
Which electron displacement effect explain the following correct orders of acidity of the carboxylic acids?
Answer:
CH3CH2COOH > (CH3)2 CHCOOH > (CH3)3C COOH
As the number of alkyl group increases, +I effect increases and it strengthen the O-H bond. i.e., O-H bond dissociation is very difficult.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Therefore +I effect explain the correct orders of acidity of the carboxylic acids.

Question 41.
Which of the two, NO2CH2CH2\({O}^{ \ominus }\) or CH3CH2\({O}^{ \ominus }\) is expected to be more stable and why?
Answer:
v
(I) is more stable than (II) because NO2 group has -I effect and hence it tends to disperse the -ve charge on the 0-atom. In contrast., CH3CH, has +I effect. It therefore, tends to intensify the -ve charge and hence destabilizes it.

Samacheer Kalvi 11th Chemistry Basic Concepts of Organic Reactions 5 Marks Questions and Answers

Question 1.
Explain electron movement in organic reactions.
Answer:
All organic reactions can be understood by following the electron movements.

  • Lone pair becomes a bonding pair.
  • Bonding pair becomes a lone pair.
  • A bond breaks and becomes another bond.

The electron movement depends on the nature of the substrate, reagent and the prevailing conditions.
Type – 1.
A lone Pair to a bonding pair
v
Type – 2.
A bonding pair to a lone Pair
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Type – 3.
A bonding pair to an another bonding pair
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 2.
How does inductive effect influence the reactivity and acidity of carboxylic acids?
Answer:
(a) Reactivity:

  • When a highly electronegative atom such as halogen is attached to a carbon then it makes the C—X bond polar.
  • In such cases the -I effect of halogen facilitates the attack of an incoming nucleophile at the polarized carbon and hence increases the reactivity.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
  • If a -I group is attacher neared to a carbonyl carbon, it decreases the availability of electron density on the carbonyl carbon and hence increases the rate of the nucleophilic addition reaction.

(b) Acidity of carboxylic acid:

  • When a halogen atom is attached to the carbon which is neared to the carboxylic acid group, its -I effect withdraws the bonded electrons towards itself and makes the ionization of H+ easy.
  • The acidity of various chioro acetic acid is in the following order. Cl3C-COOH > Cl2CHCOOH > ClCH2COOH The strength of the acid increases with increase in the effect of the group attached to the carboxyl group.
  • Similarly the following order of acidity in the carboxylic acids is due to the +1 effect of alkyl group. (CH3)3CCOOH < (CH3)2CHCOOH <CH3COOH

Question 3.
Explain the acidic nature of phenol.
Answer:

  • Resonance is useful in explaining certain properties such as acidic of phenol.
  • The phenoxide ion is more stabilized than phenol by resonance effect.
  • The resonance favours ionisation of phenol to form H+ and shows acidity.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
    phenoxide ion resonance structures.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
  • The above structures shows that there is a charge separation in the resonance structure of phenol which needs energy whereas there is no such hybrid structures in the case of phenoxide ion. This increases stability accounts for the acidic character of phenol.

Question 4.
How does hyper conjugation effect explain the stability of alkenes?
Answer:

  • The relative stability of various classes of carhonium ions may be explained by the number of no bond resonance structures that can be written for them.
  • Such structures are arrived by shifting the bonding electrons from an adjacent C-H bond to the electron deficient carbon.
  • in this way, the positive charge originally on carbon is dispersed to the hydrogen. This manner of electron release by assuming no bond character in the adjacent C-H bond is called hyper conjugation or Baker- Nathan effect.
  • The greater the hyper conjugation, the greater will be the stability of the compound. The increasing order of stability can be shown as:
    Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
  • Alkyl group increases in the C=C double bond carbon, hyper conjugation increases and stability of that organic compound also increases.

Question 5.
Explain the types of addition reactions?
Answer:
Addition reactions are classified into three types. They are,

  1. Electrophi lic addition reaction
  2. Nucleoph lije addition reaction
  3. Free radical addition reaction

1. Electrophilic addition reaction:
An electrophilic addition reaction can be described as an addition reaction in which a reactant with multiple bonds as in a double or triple bond undergoes has it air bond broken and two new a bond are formed.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
2. Nucleophilic addition reaction:
A nucleophilic addition reaction is an addition reaction where a chemical compound with an electron deficient or electrophilic double or triple bond, a it bond, reacts with a nucleophilic which is an electron rich reactant with the disappearance of the double bond and creation of two new single or a bonds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

3. Free radical addition reaction:
it is an addition reaction in organic chemistry involving free radicals. The addition may occur between a radical and a non radical or between two radicals.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 6.
Explain the types of substitution reaction?
Answer:
Substitution reactions are classified into three types. They are,

  1. Nucleophilic substitution reaction.
  2. Electrophilic substitution reaction.
  3. Free radical substitution reaction.

1. Nucleophilic substitution reaction:
A nucleophilic substitution reaction in organic chemistry is a type of reaction where a nucleophilic gets attached to the positive charged atoms or molecules of the other substance. A good example of a nucleophilic substitution reaction is the hydrolysis of alkyl bromide, under the basic conditions where in the nucleophile is nothing but the base H. whereas the leaving group is the Br°. The reaction for the following is as give below.
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

2. Electrophilic substitution reaction:
The electrophilic substitution reaction involves the electrophiles. The electrophilic reactions occur mostly with the aromatic compounds. This types of substitution reaction arc basically defined as those chemical reactions where the electrophile replaces the functional group in a compound but not the hydrogen atom, sometimes hydrogen atom can be also replaced by electrophiles.
For example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

3. Free radical substitution reaction:
Free radical substitution reaction involving free radicals arc a reactive intermediate.
A – X + Y → A – Y + X
CH4 + Cl → CH + HCl
CH3 + Cl → CH3Cl (methyl chloride)

Question 7.
For the following bond cleavages use curved-arrows to show the electron flow and classify each as homolytic or heterolytic fission. Identify reactive intermediate produced as free radical,, carbocation and carbanion?
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Question 8.
An organic compound (A) has a molecular formula C2H6O it is one of the primary alcohol. A reacts with acidified potassium dichromate to give B. B on further undergoes to oxidation reaction to give C. C on reacts with SOCl3 to give D which ¡s chlorinated product. Identify A,B,C and D, explain with equation.
Answer:
1. C2H6O is CH3-CH2-OH which is a primary alcohol (A)
2. CH3-CH2-OH (A) reacts with H+/ K2Cr2O7 to give acetaldehyde (B)
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
3. CH3CHO (B) which on further oxidation to give acetic acid (C).
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
4. Acetic acid reacts with SOCl2 to give acetyl chloride (D).
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 9.
An organic compound (A) of a molecular formula C2H4 which is a simple alkene. A reacts with dil H2S04 to give B. A again reacts with Cl2 to give C. Identify AB and C and write the equations.
Answer:
1. C2H2 is CH2=CH2 is a simple alkene. A is ethylene.
2. Ethylene (A) reacts with dii H2SO4 to give ethanol (B)
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
3. Ethylene (A) reacts with Cl, to give 1.2 dichloro ethane (C)
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Question 10.
Complete the reactions and identify the products.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

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Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Multiple Choice Questions

Question 1.
Select the molecule which has only one ir bond.
(a) CH3-CH=CH-CH3
(b) CH3-CH=CH-CHO
(c) CH3-CH=CH-COOH
(d) All of these
Answer:
(a) CH3-CH=CH-CH3

Question 2.
In the hydrocarbon Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistrythe state of hybridisation of carbon 1,2,3,4 and 7 are in the following sequence.
(a) sp, sp, sp3, sp2, sp3
(b) sp2, sp, sp3, sp2, sp3
(c) sp, sp, sp2, sp, sp3
(d) none of these
Answer:
(a) sp, sp, sp3, sp2, sp3

Question 3.
The general formula for alkadiene is ……….
(a) CnH2n
(b) CnH2n-1
(c) CnH2n-2
(d) CnHn-2
Answer:
(c) CnH2n-2

Question 4.
Structure of the compound whose IUPAC name is 5, 6 – dimethylhept-2-ene is ……….
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 5.
The IUPAC name of the compound is …………
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(a) 2,3 – Diemethyiheptane
(b) 3 – Methyl – 4 – ethyloctane
(c) 5 – ethyl – 6 – methyloctanc
(d) 4 – Ethyl -3 – methyloctane.
Answer:
(d) 4 – Ethyl – 3 – methyloctane.

Question 6.
Which one of the following names does not fit a real name?
(a) 3 – Methyl – 3 – hexanone
(b) 4- Methyl – 3 – hexanone
(c) 3 – Methyl – 3 – hexanol
(d) 2 – Methyl cyclo hexanone.
Answer:
(a) 3 – Methyl – 3 – hexanone

Question 7.
The TUPAC name of the compound CH3— CII = CH -C CH is ……………
(a) Pent – 4 – yn – 2 – ene
(b) Pent -3-en – 1- yne
(c) pent – 2 – en – 4 – yne
(d) Pent – 1 – yn – 3 – ene
Answer:
(b) Pent -3-en – 1- yne

Question 8.
IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is ……….
(a) 3, 4, 4 – Trimethylheptane
(b) 2 – Ethyl – 3, 3 – dimethyl heptane
(c) 3, 4, 4 – Trimethyloctane
(d) 2 Butyl – 2 -methyl – 3 – ethyl-butane.
Answer:
(c) 3, 4, 4 – Trimethyloctane

Question 9.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is ………
(a) 2 – Hydroxypropionic acid
(b) 2, 4. 4 – Trimethylpent -3-ene
(c) Propan – 2 – ol 1 – oie acid
(d) 2, 2, 4 – Trimethylpent -2-ene
Answer:
(b) 2 – Hydroxy Propanoic acid

Question 10.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is ………
(a) 3 – Ethyl – 2- hexene
(b) 3 – Propyl – 3. hexene
(c) 4 – Ethyl – 4 – hexene
(d) 3 – Propyl -2-hexenc
Answer:
(a) 3 – Ethyl – 2- hexene

Question 11.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is ……..
(a) 2 – Hydroxypropionic acid
(b) 2 – Hydroxy Propanoic acid
(c) Propan 2 – ol – 1 – oic acid
(d) 1 – Carboxyethanol.
Answer:
(b) 2 – Hydroxy Propanoic acid

Question 12.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is ………
(a) 2 – Bromo – 3 – methylbutanoic acid
(b) 2 – methyl – 3 – bromobutanoic acid
(c) 3 – Bromo – 2 – methylbutanoic acid
(d) 3 – Bromo – 2. 3 – dimethyl propanoic acid.
Answer:
(c) 3 – Bromo – 2 – methylbutanoic acid

Question 13.
The structure of isobutyl group in an organic compound is ………
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 14.
The number of stereoisomers of 1, 2-dihydroxycyclopentane is ……..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 15.
Which of the following is optically active?
(a) 3 – Chioropentane
(b) 2 – Chioropropane
(c) Meso – tat-tat-ic acid
(d) Glucose
Answer:
(d) Glucose

Question 16.
The isomer of ethanol is ……….
(a) acetaldehyde
(b) dimethylether
(c) acetone
(d) methyl carbinol
Answer:
(b) dimethylether

Question 17.
How many cyclic and acyclic isomers are possible for the molecular formula C3H6O?
(a) 4
(b) 5
(c) 9
(d) 10
Answer:
(c) 9

Question 18.
Which one of the following shows functional isomerism?
(a) ethylene
(b) Propane
(c) ethanol
(d) CH2Cl2
Answer:
(c) ethanol

Question 19.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry are ………
(a) resonating structure
(b) taulomers
(c) optical isomers
(d) conformers
Answer:
(b) tautomers

Question 20.
Nitrogen detection in an organic compound is earned out by Lassaigne’s test. The blue colour formed is due to the formation of ………….
(a) Fe3[Fe(CN)6]2
(b) Fe4[Fe(CN)6]3
(c) Fe4[Fe(CN)6]2
(d) Fe3[Fe(CN)6]3
Answer:
(b) Fe4[Fe(CN)6]3

Question 21.
Lassaigne’s test for the detection of nitrogen fails in ………..
(a) H2N -CO – NH. NH2. HCl
(b) NH2 – NH2. HCl
(c) C6H5 – NH – NH2. HCl
(d) C6H5CONH2
Answer:
(c) C6H5 – NH – NH2. HCl

Question 22.
Connect pair of compounds which give blue colouration/precipitate and white precipitate respectively, when their Lassaigne’s test is separately done.
(a) NH2NH2HCl and CICH2 – CHO
(b) NH2CS NH2 and CH3 – CH2Cl
(c) NH2CH2 COOH and NH2CONH2
(d) C6H5NH2 and ClCH2 – CHO
Answer:
(d) C6H5NH2 and ClCH2 – CHO

Question 23.
Sodium nitropruside reacts with suiphide ion to give a purple colour due to the formation of ………..
(a) [Fe(CN)5N0]3-
(b) [Fe(NO)5CN]+
(c) [Fe(CN)5NOS]4-
(d) [Fe(CN)5NOS]3-
Answer:
(c) [Fe(CN)5NOS]4-

Question 24.
An organic compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the compound will be close to ……….
(a) 46%
(b) 34%
(c) 3.4%
(d) 4.6%
Answer:
(b) 34%

Question 25.
A sample of 0.5g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 mL of 0.5M H2SO4. The remaining acid after neutralisation by ammonia consumed 80 mL of 0.5 M NaOH, The percentage of nitrogen in the organic compound is ……….
(a) 14%
(b) 28%
(c) 42%
(d) 56%
Answer:
(b)28%

Question 26.
In an organm compound, phosphorus is estimated as ……….
(a) Mg2P2O7
(b) Mg3(PO4)2
(c) H3PO4
(d) P2O5
Answer:
(a) Mg2P2O7

Question 27.
Ortho and para-nitro phenol can be separated by ………….
(a) azeotropic distillation
(b) destructive distillation
(c) steam distillation
(d) cannot be separated
Answer:
(c) steam distillation

Question 28.
The purity of an organic compound is determined by …………
(a) Chromatography
(b) Crystallisation
(c) melting or boiling point
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

Question 29.
A liquid which decomposes at its boiling point can be purified by …………
(a) distillation at atmospheric pressure
(b) distillation under reduced pressure
(c) fractional distillation
(d) steam distillation
Answer:
(b) distillation under reduced pressure

Question 30.
Assertion: Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is 3-carbethoxy -2- butenoicacid.
Reason: The principal functional group gets lowest number followed by double bond (or) triple bond.
(a) both the assertion and reason are true and the reason is the correct explanation of assertion.
(b) both assertion and reason are true and the reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d both the assertion and reason are false.
Answer:
(a) both the assertion and reason are true and the reason is the correct explanation of assertion.

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Short Answer Questions

Question 31.
Give the general characteristics of organic compounds.
Answer:

  • All organic compounds are covalent compounds of carbon and are insoluble in water and soluble in organic solvents.
  • They are inflammable (except CCl4).
  • They possess low boiling and melting points due to their covalent nature.
  • They are characterised by functional groups.
  • They exhibit isomerism.

Question 32.
Describe the classification of organic compounds based on their structure.
classification of organic compounds based on the structure
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 33.
Write a note on homologous series.
Answer:

  • A series of organic compounds each containing a characteristic functional group and the successive members differ from each other in molecular formula by a CH2 group is called homologous series.
  • e.g., Alkanes, Methane (CH4) ethane (C2H6), Propane (C3H8) etc.
  • Compounds of the homologous series are represented by a general formula. e.g., Alkanes: C2H2nAlkene: CnH2n
  • They can be prepared by general methods.
  • They show regular gradation in physical properties but have almost similar chemical properties.

Question 34.
What is meant by a functional group? Identify the functional group in the following compounds.
(a) acetaldehyde
(b) oxalic acid
(c) dimethyl ether
(d) methylamine
Answer:
1. A functional group ¡s an atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 35.
Give the general formula for the following classes of organic compounds
(a) Aliphatic monohydric alcohol
(b) Aliphatic ketones
(c) Aliphatic amines.
Answer:
(a) Aliphatic monohydric alcohol – CnH2n+1 + OH
(b) Aliphatic ketones – CnH2nO
(c) Aliphatic amines – C2H2n+1NH2

Question 36.
Write the molecular formula of the first six members of homologous series of nitro – alkanes.
Nitroalkanes:
Answer:

  • CH2NO2 Nitromethane
  • CH2-CH2NO2 Nitroethane
  • CH3-CH2-CH2NO2 1- nitropropane
  • CH3-CH2-CH2-CH2-NO2 1- nitrobutane
  • CH3-CH2-CH2-CH2-CH2-NO2 1 – nitropentane
  • CH2-CH2-CH2-CH2-CH2-CH2-CH2-NO2 1- nitrohexane

Question 37.
Write the molecular and possible structural formula of the first four members of homologous series of carhoxylic acids.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 38.
Give the IUPAC names of the following compounds.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(i) 2,3,5-tnmethylhexane
(ii) 2-bromo-3-methylbutane
(iii) methoxymethane
(iv) 2-hydroxybutanal
(v) buta-1,3-diene
(vi) 4-chioropent-2-yne
(vii) 1 -bromobut-2-ene
(viii) 5-oxohexanoic acid
(ix) 3-ethyl-4-ethenylheptane
(x) 2, 4, 4-trimethylpent-2-ene
(xi) 2- methyl-I -phenyipropan- I -amine
(xii) 2,2- dimethyl-4-oxopentanenitrile
(xiii) 2-ethoxypropane
(xiv) I -fluoro-4-methyl-2-nitrobenzene
(xv) 3-bromo-2-methylpentanal

Question 39.
Give the structure for the following compound.
Answer:
(i) 3 – ethyl – 2 methyl -1 – pentene
(ii) 1, 3, 5 – Tnmethyl cyclohex – 1 – ene
(iii) tertiary butyl iodide
(iv) 3 – Chlorobutanal
(y) 3 – Chlorobutanol .
(vi) 2 – Chloro – 2 – methyl propane
(vii) 2, 2-dimethyl- 1 – chioropropane
(viii) 3 – methylbut – 1 – ene
(ix) Butan – 2, 2 – diol
(x) Octane – 1 ,3 – diene
(xi) 1 ,5 – Dimethylcyclohexane
(xii) 2 – Chlorobut – 3 – ene
(xiii) 2 – methylbutan – 3 – ol
(xiv) acetaldehyde

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 40.
Describe the reactions involved in the detection of nitrogen in an organic compound by Lassaigne method.
Answer:
Detection of Nitrogen:
The following reactions are involved in the detection of nitrogen with formation of prussian blue precipitate conforming the presence of nitrogen in an organic compound.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 41.
Give the principle involved in the estimation of halogen in an organic compound by Carius method.
Estimation of halogens:
Answer:
carius method:
1. A known mass of the organic compound is heated with fuming HNO3 and AgNO3.
2. C, H and S gets oxidised to CO2, H2O and SO2 and halogen combines with AgNO3 to form a precipitate of silver halide
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
3. The precipitate AgX is filtered, washed, dried and weighed.
4. From the mass of AgX and the mass of organic compound taken, the percentage of halogens are calculated.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 42.
Give a brief description of the principles of:
1. Fractional distillation
2. Column Chromatography
Answer:
1. Fractional distillation:
This method is used to purify and separate liquids present in the mixture having their boiling point close to each other. The process of separation of the components in liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. This method is applied in distillation of petroleum. coal tar and crude oil.

2. Column chromatography:
(a) The principle behind chromatography is selective distribution of mixture of organic substances between two phases a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. when the stationary phase is a solid, the moving phase is a liquid or gas.

(b) If the stationary phase is solid, the basis is adsorption, and when it is a liquid, the basis is partition.

(c) Chromatography is defined as a technique for the separation of a mixture brought about by differential movement of the individual component through porous medium under the influence of moving solvent.

(d) In column chromatography, the above principle is carried out in a long glass column.

Question 43.
Explain paper chromatography.
Answer:
Paper chromatography:
1. It is an example of partition chromatography. A strip of paper acts as an adsorbent. This method involves continues differential partioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatographic paper is used. This paper act as a stationary phase.

2. A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which acts as the mobile phase. The solvent rises up and flows over the spot. The paper selectivity retains different components according to their different partition in the two phases where a chromatograrn is developed.

3. The spots of the separated coloured components are visible at different heights from the position of initial spots on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent.

Question 44.
Explain various types of constitutional isomerism (structural isomerism) in organic compounds.
Answer:
Constitutional isomers:
These isomers having the same molecular formula but differ in their bonding sequence. It is classified into 6 types:
1. Chain (or) nuclear (or) skeletal isomerism:
The phenomenon in which the isomers have similar molecular formula but differ in the nature of carbon skeleton (i.e., straight (or)
branched)
e.g., C5H12:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. Position isomerism:
If different compounds belonging to same homologous series with the same molecular formula and carbon skeleton but differ in the position of substituent or functional group or an unsaturated linkage are said to exhibit position isomerism.
e.g., C5H10:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

3. Functional isomerism:
Different compounds having same molecular formula but different functional groups are said to exhibit functional isomerism.
e.g., C3H6O:
(i) CH3-CH2-CH2-CH=CH2 propanal (Aldehyde group)
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

4. Metamerism:
This isomerism anses due to the unequal distribution of carbon atoms on either side of the functional group or different alkyl groups attached to either side of the same functional group and having same molecular formula.
e.g., C4H10O:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

5. Tautomerism:
It is an isomerism in which a single compound exists in two readily inter convertible structures that differ markedly in the relative position of atleast one atomic nucleus generally hydrogen.
e.g., C2H4O:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

6. Ring chain isomerism:
It is an isomerism in which compounds having same molecular formula but differ in terms of bonding of carbon atom to form open chain and cyclic structures.
e.g., C2H6:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 45.
Describe optical isomerism with suitable example.
Answer:
Optical isomerism:
1. Compounds having same physical and chemical property but differ only in the rotation of plane of polarised light are known as optical isomers and the phenomenon is known as optical isomerism.

2. Glucose have the ability to rotate the plane of plane polarised light and it is said to be an optically active compound and this property of any compound is called optical activity.

3. The optical isomer which rotates the plane of plane polarised light to the right or in clockwise direction is said to be dextrorotatory and is denoted by the sign (+).

4. The optical isomer which rotates the plane of plane polarised light to the left or in anti- clockwise direction is said to be laveo rotatory and is denoted by the sign (-).

5. Dextrorotatory compounds are represented as ‘d’ (or) by (+) sign and leave rotatory compounds are represented as l (or) by (-) sign.

6. The optical isomers which are non-superimposible mirror images of each other are called enantiomers.

Question 46.
Briefly explain geometrical isomerism in alkenes by considering 2- butene as an example.
Answer:
2-butene: Geometrical isomerism : CH3  -CH = CH – CH3

1. Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.

2. In 2-butene, the carbon-carbon double bond is sp2 hybridised. The carbon-carbon double bond consists of a a bond and a it bond. The presence of it bond lock the molecule in one position. Hence, rotation around C = C bond is not possible.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
4. These two compounds are termed as geometrical isomers and are termed as cis and trans form.

5. The cis-isomer is the one in which two similar groups arc on the same side otthe double bond. The trans-isomer is that in which two similar groups are on the opposite side of the double bond. Hence, this type of isomerism is called cis-Irans isomerism.

Question 47.
0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water calculate the percentage of carbon and hydrogen in it.
Answer:
Weight of organic compound = 0.30 g
Weight of carbon-dioxide = 0.88 g
Weight of water = 0.54 g
Percentage of hydrogen:
18 g of water contains 2 g of hydrogen
0.54 g of water contain = \(\frac {2}{18}\) × 0.54 g of hydrogen
% of hydrogen = \(\frac {2}{18}\) × \(\frac {0.54}{0.30}\) × 100 = \(\frac {2}{18}\) × \(\frac {54}{0.3}\)
% of H = 0.111 × 180 = 19.888 ≈ 20%

Percentage of carbon:
44 g of CO2 contains 12 g of carbon
0.88 g of CO2 contains = \(\frac {12}{44}\) × 0.88 g of carbon
% of carbon = \(\frac {12}{44}\) × \(\frac {0.88}{0.30}\) × 100 = \(\frac {12}{44}\) × \(\frac {88}{0.3}\) = \(\frac {24}{0.3}\)
% of carbon = 80 % .

Question 48.
The ammonia evolved form 0.20 g of an organic compound by kjeldahl method neutralised 15m1 of N/20 Sulphuric acid solution. Calculate the percentage of Nitrogen.
Answer:
Weight of organic compound = 0.20 g
Volume of sulphuric acid taken = 15 ml
Strength of sulphuric acid taken = \(\frac {N}{20}\) = 0.05 N
Percentage of nitrogen = \(\frac{14 \times \mathrm{NV}}{1000 \times 10} \times 100\)
= \(\frac{14 \times 0.05 \times 15}{1000 \times 0.20} \times 100\)
= \(\frac {1050}{200}\) = 5.25
% of nitrogen = 5.25%

Question 49.
0.32 g of an organic compound. after heating with fuming nitric acid and barium nitrate crystals is a scaled tube gave 0.466 g of barium sulphate. Determine the percentage of sulphur in the compound.
Answer:
Weight of organic compound = 0.32 g
Weight of BaSO4 formed = 0.466 g
233 g of BaSO4 contains = 32 g of sulphate
0.466 g of l3aSO4 contain = \(\frac {32}{233}\) x \(\frac {0.466}{2.32}\) x 100
= \(\frac {32}{233}\) x \(\frac {46.6}{0.32}\) = 19.999 g of sulphur
% of sulphur = 20 %

Question 50.
024 g of an organic compound gave 0.287 g of silver chloride in the carius method. Calculate the percentage of chlorine in the compound.
Answer:
Weight of organic compound = 0.24 g
Weight of silver chloride = 0.287 g
143.5 g of AgCl contains = 35.5 g of Cl
0.287 g of AgCl contains = \(\frac {35.5}{143.5}\) x 0.287 g of Cl
% of chlorine = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.24}\) x 100 = 29.58 %

Question 51.
In the estimation of nitrogen present in an organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at 15°C and 760 mm Hg pressure. Calculate the percentage of nitrogen ¡n the compound.
Answer:
Weight of organic compound = 0.35 g
Volume of moist nitrogen (V1) = 20.7 ml = 20.7 x 10-3 L
Temperature = T1 = 15°C = 273 + 15°C = 288K
Pressure of moist nitrogen P1 = 760 mmHg
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
V0 = 19.62 x 10-3L
Percentage of nitrogen = \(\frac {28}{22.4}\) x \(\frac{\mathrm{V}_{0}}{\mathrm{W}}\) x 100
= \(\frac {28}{22.4}\) x \(\frac{19.62 \times 10^{-3}}{0.35}\) x 100
= \(\frac {28}{22.4}\) x = \(\frac {19.62}{0.35}\) x 10-1
= 56.05 x 10-3 x 100 = 7.007%
Percentage of nitrogen = 7.007%

In Text Questions – Evaluate Yourself

Question 1.
Give two examples for each of the following type of organic compounds.
1. non-benzonoid aromatic
2. aromatic heterocyclic
3.  alicycic
4. aliphatic open chain.
Answer:
1. Non benzenoid aromatic compounds
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. Aromatic heterocyclic compounds
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

3. Alicyclic compounds
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

4. Aliphatic open chain compounds

  • CH3-CH2-CH2-CH2-CH3 n-pentane
  • CH3-CH2-CH2OH 1-propanol

Question 2.
Write structural formula for the following compounds
1. Cyclohexa-1, 4-diene
2. Ethynykyclohexane
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 3.
Write structural formula for the following compounds
1. m – dinitrobenzene
2. p-dichlorobenzene
3. 1, 3, S- Trimethytbeuzene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 4.
Write all the possible isomers of molecular formula C4H10O and identify the isomerisms found in them.
Answer:
C4H10O isomers:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 5.
0.2346 g of an organic compound containing C, H & O, o comhution giweb 0.2754 g of H2O and 0.4488 g CO2. Calculate the % composition of C, H & O in the organic compound.
Answer:
Weight of organic substance (w) = 0.2346 g
Weight of water (x) = 0.2754 g
Weight of CO2 (y) = 0.4488 g
Percentage of carbon = \(\frac {12}{44}\)  x  \(\frac {y}{w}\) x 100
= \(\frac {12}{44}\) – \(\frac {0.4488}{0.2346}\) x 100 = 52.17%
Percentage of hydrogen = \(\frac {2}{18}\) x \(\frac {y}{w}\) x 100
= \(\frac {2}{18}\) x \(\frac {0.2754}{0.2346}\) x 100 = 13.04%
Percentage of oxygen = [100- (52.17 +13.04)] = 100 – 65.21 = 34.79%

Question 6.
0.16 g of an organic compound was heated in a carlus tube and H2SO4 acid formed was precipitated with BaCl4. The mass of BaSO4 was 0.35 g. Find the percentage of sulphur.
Answer:
Weight of organic substance (w) = 0.16 g
Weight of Barium sulphate (x) = 0.35 g
Percentage of Sulphur = \(\frac {32}{233}\) x \(\frac {x}{w}\) x 100
= \(\frac {32}{233}\) x \(\frac {0.35}{0.16}\) x 100 = 30.04%

Question 7.
0.185 g of an organic compound when treated with Conc. HNO3 and silver nitrate gave 0.320 g of silver bromide. Calculate the % of bromine in the compound.
Answer:
Weight of organic substance (w) 0.185 g ;
Weight of silver bromide (x) = 0.320 g
Percentage of bromine = \(\frac {80}{188}\) x \(\frac {x}{w}\) x 100 = \(\frac {80}{188}\) x \(\frac {0.32}{0.185}\) x 100 = 73.6%

Question 8.
0.40 g of an iodo-substituted organic compound gave 0.235 g of Agi by carius method. Calculate the percentage of iodine in the compound. (Ag = 108, I = 127).
Answer:
Weight of organic substance (w) = 0.40 g
Weight of silver iodide (x) = 0.235 g
127 x 127 0.235
Percentage of iodine = \(\frac {127}{235}\) x \(\frac {x}{w}\) x 100
= \(\frac {x}{w}\) x \(\frac {0.235}{0.40}\) x 100 = 31.75%

Question 9.
0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg2P2O7 by the analysis. Calculate the percentage of P in the compound.
Answer:
Weight of organic compound = 0.33g ;
Weight of Mg,P,07 = 0.397g
222 g of Mg2P2O7 contains 62 g of phosphorous.
0.397 g of Mg2P2O7 will contain \(\frac {62}{222}\) x 0.397 g of P.
0.33 g of organic compound contains \(\frac {62}{222}\) x 0.397 g of P
100 g of organic compound will contain \(\frac {62}{222}\) x \(\frac {0.397}{0.33}\) x 100
= \(\frac {2,461.4}{73.26}\) = 33.59 %
Percentage of phosphorous = 33.59 %

Question 10.
0.3 g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralise 30 mL of 0.1N H2SO4. Calculate the percentage of nitrogen in the compound.
Answer:
Weight of organic compound (w) = 0.3 g
Strength of sulphuric acid used (N) = 0.1 N
Volume of sulphuric acid used (V) = 30 mL
30 ml of 0.1 N sulphuric acid 30 ml of 0.1 N ammonia
Percentage of nitrogen = \(\left(\frac{14 \times \mathrm{NV}}{1000 \times w}\right)\) x 100
= \(\left(\frac{14 \times 0.1 \times 30}{1000 \times 0.3}\right)\) x 100 = 14%

Example Problems

Question 1.
Classify the following compounds based on the structure
1. CH≡C-CH2-C≡CH
2. CH3-CH2-CH2-CH2-CH3
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. Unsaturated open chain compound
  2. Saturated open chain compound
  3. Aromatic benzenoid compound
  4. Alicyclic compound

Question 2.
0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.
Answer:
Weight of organic compound = 0.26 g
Weight of water = 0.039 g
Weight of CO2 = 0.245 g
Percentage of hydrogen:
18 g of water contains 2 g of hydrogen
0.039 g of water contains = \(\frac {2}{18}\) x \(\frac {0.039}{0.26}\) of H
% of hydrogen = \(\frac {0.039}{0.26}\) x \(\frac {2}{18}\) x 100 = 1.66%
Percentage of carbon:
44 g of CO2 contains 12 g of C
0.245 g of CO2 contains = \(\frac {12}{44}\) x \(\frac {0.245}{0.26}\) g of C
% of Carbon = \(\frac {12}{44}\) x \(\frac {0.245}{0.26}\) x 100 = 25.69%

Question 3.
In an estimation of sulphur by Carius method, 0.2 175 g of the substance gave 0.5825 g of BaSO4, calculate the percentage composition of S ¡n the compound.
Answer:
Weight of organic compound = 0.2 175 g
Weight of BaSO4 = 0.5825 g
233 g of BaSO4 contains = 32 g of S
0.5825 g of BaSO4 contains = \(\frac {32}{233}\) x \(\frac {0.5825}{0.2175}\) g of S
Percentage of S = \(\frac {32}{233}\) x \(\frac {0.5825}{0.2175}\) x 100 = 36.78%

Question 4.
0.284 g of an organic substance gave 0.287 g AgCl in Carius method for the estimation of halogen. Find the percentage of Cl in the compound.
Answer:
Weight of the organic substance = 0.284 g
Weight of AgCl = 0.287 g
143.5 g of AgCl contains 35.5 g of chlorine
0.287 g of AgCl Contains = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.284}\)
% of chlorine = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.284}\) x 100 = 24.98%

Question 5.
0.24 g of organic compound containing phosphorous gave 0.66 g of Mg2P2O7 by the usual analysis. Calculate the percentage of phosphorous ¡n the compound
Answer:
Weight of an organic compound = 0.24 g
Weight of Mg2P2O7 = 0.66 g
222 g of Mg2P2O7 contains = 62 g of P
o. 66 g contains = \(\frac {62}{222}\) x 0.66 g of P
Percentage of P = \(\frac {62}{222}\) x \(\frac {0.66}{0.24}\) x 100 = 76.80%

Question 6.
0.1688 g when analysed by the Dumas method yield 31.7 mL of moist nitrogen measured at 14°C and 758mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14°C =12 mm of Hg).
Answer:
Weight of Organic compound = 0.168 g
Volume of moist nitrogen (V1) = 31.7 ml. = 31.7 x 10-3 L
Temperature (T1) = 14°C = 14 + 273 = 287 K
Pressure of Moist nitrogen (P) = 758 mm Hg
Aqueous tension at 14°C = 12 mm of Hg
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{0} V_{0}}{T_{0}}\)
V0 = \(\frac{746 \times 31.7 \times 10^{-3}}{287} \times \frac{273}{760}\)
V0 = 29.58 x 10-3 L
Percentage of nitrogen:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
= 21.90 %

Question 7.
0.6 g of an organic compound was Kjeldhalised and NH3 evolved was absorbed into 50 mL of semi-normal solution of H2SO4. The residual acid solution ws diluted with distilled water and the volume made up to 150 mL. 20 mL of this diluted solution required 35 mL of \(\frac {N}{2}\) NaOH solution for complete neutralisation. Calculate the % of N in the compound.
Answer:
Weight of Organic compound = 0.6 g
Volume of sulphuric acid taken = 50 mL
Strength of sulphuric acid taken = 0.5 N
20 ml of diluted solution of unreacted sulphuric acid was neutralised by 35 mL of 0.05 N Sodium hydroxide
Strength of the diluted sulphuric acid = \(\frac {35 x 0.05}{20}\) = 0.0875 N
Volume of the sulphuric acid remaining after reaction with ammonia = V1 mL
Strength of H2SO4 = 0.5 N
Volume of the diluted H2SO4 = 150 mL
Strength of the diluted sulphuric acid = 0.0875 N
V1 = \(\frac {150 x 0.087}{0.5}\) = 26.25 mL
Volume of H2SO2consumed by ammonia = 50 – 26.25 = 23.75 mL
23.75 mL of 0.5 N H2SO4 = 23.75 mL of 0.5N NH3
The amount of Nitrogen present in the 0.6 g of organic compound
= \(\frac{14 \mathrm{g}}{1000 \mathrm{mL} \times 1 \mathrm{N}}\) x 23.75 x 0.5 N = 0.166 g
Percentage of Nitrogen \(\frac {0.166}{0.6}\) x 100 = 27.66%

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Additional Questions Solved

I. Choose the correct answer.
Question 1.
Statement 1. The tendency of an atom to form a chain of bonds with the atoms of the same element is called catenation.
Statement 2. The high strength of C-C bond is responsible for its catenation property.
(a) Statement 1 & 2 are correct and statement 2 is the correct explanation of statement 1.
(b) Statement 1 & 2 arc correct but statement 215 not the correct explanation of statement 1.
(c) Statement 1 is correct but statement 2 is wrong.
(d) Statement 1 is wrong but statement 2 is correct.
Answer:
(a) Statement 1 & 2 are correct and statement 2 is the correct explanation of statement 1.

Question 2.
Which of the following is not an organic compound?
(a) DNA
(b) Lipid
(c) Glycogen
(d) Bronze
Answer:
(d) Bronze
Solution:
It is an alloy and a mixture of metals and all other are organic compounds.

Question 3.
Which of the following is an example of an organic reaction?
(a) Rusting of iron
(b) Combustion of magnesium
(c) Biochemical reactions
(d) All the above
Answer:
(c) Biochemical reactions

Question 4.
Which of the following is an example of heterocylic aromatic compound?
(a) THF
(b) Cyclopropane
(c) Pyridine
(d) Azulene
Answer:
(c) Pyridine

Question 5.
Which of the following is an example of non-benzenoid aromatic compound?
(a) Tolucnc
(b) Phenol
(c) Benzyl alcohol
(cl) azulene
Answer:
(d) azulene

Question 6.
Which of the following pair is an example of aromatic compounds?
(a) Benzene, Toluene
(b) Cyclopropane, Cyclobuane
(c) Pyridine, Pyrrole
(d) Propane, Butane
Answer:
(a) Benzene, Toluene

Question 7.
Which of the following is an example of carbocyclic alicyclic compound?
(a) Pyrrole
(b) Thiophene
(c) Cyclopropane
(d) Phenol
Answer:
(c) Cyclopropane

Question 8.
Which one of the following is the functional group of ketone?
(a)-CHO
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(c) -O-
(d)-OH
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 9.
Which one of the following indicates isothiocyanate functional group?
(a) -NC
(b) -NCS
(c) -SCN
(d) -NCO
Answer:
(b) -NCS

Question 10.
Which of the following represent thiol?
(a) -SH
(b) -OH
(c) -SR
(d) -SCN
Answer:
(a) -SH

Question 11.
Which structure ¡s named as 3-chlorocyclobut-1-ene?
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry-

Question 12.
Which one of the following is called 2-cyclobutyipropanal?
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 13.
Which one of the following is called cyclopentylbenzene’?
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 14.
Which one of the following is commonly called mesitylene?
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 15.
Which one of the following is called benzylchloride?
(a) C6H5CH2Cl
(b) C6H5CHCl2
(c) C6H5CCl3
(d) C6H5Cl
Answer:
(a) C6H5CH2Cl

Question 16.
Which of the following pair are called functional isomers?
(a) methyl propyl ether and diethyl ether
(b) 2-pentanone & 3-pentanone
(c) propanoic acid and methyl acetate
(d) I -butanol and 2-butanol
Answer:
(c) propanoic acid and methyl acetate

Question 17.
Which of the following does not show optical isomerism’?
(a) Glucose
(b) Tartane acid
(c) Lactic acid
(d) Methane
Answer:
(d) Methane

Question 18.
Which metal is used to prepare Lassaigne’s extract?
(a) Copper
(b) Sodium
(c) Aluminium
(d) Iron
Answer:
(b) Sodium

Question 19.
Which colour is formed in the Lassaigne’s test for nitrogen?
(a) Purple
(b) Black
(c) Prussian blue
(d) Violet
Answer:
(c) Prussian blue

Question 20.
Which one of the following is called feme ferrocyanide?
(a) Na4[Fe(CN)6]
(b) Na4[Fe(CN)6]3
(c) Fe4[Fe(CN)6]
(d) Fe4[Fe(CN)6]3
Answer:
(d) Fe4[Fe(CN)6]3

Question 21.
What is the colour formed in Lassaigne’s test of an organic compound containing N and S?
(a) Prussian blue colour
(b) Blood red colour
(c) Purple colour
(d) Canary yellow colour
Answer:
(b) Blood red colour

Question 22.
Which one of the following is the formula of sodium nitroprusside?
(a) Na4[Fe(CN)5N05]
(b) Na4[Fe(CN)5SON]
(c) Na4[Fe(CN)6]
(d) Fe4[Fe(CN)6]3
Answer;
(a) Na4[Fe(CN)5N05]

Question 23.
Identify the colour formed when Lassigne’s extract of sulphur containing organic compound is mixed with sodium nitroprusside solution?
(a) Prussian blue
(b) Black
(c) Green
(d) Purple
Answer:
(d) Purple

Question 24.
Which one of the following solutions are added to Lassaigne’s extract to identify halogens?
(a) Acetic acid + Lead acetate
(b) dil HNO3 + AgNO3
(c) Fe(OH)2 + FeCl3
(d) Na2CO3 + KNO3
Answer:
(b) dil HNO3 + AgNO3

Question 25.
Which one of the following is not identified by Lassaigne’s test?
(a) nitrogen
(b) sulphur
(c) halogens
(d) phosphorous
Answer:
(d) phosphorous

Question 26.
Which one of the following test is used to detect phosphorous in an organic compound?
(a) Silver nitrate test
(b) Copper oxide test
(c) Ammonium molybdate test
(d) Lassaigne’s test
Answer:
(c) Ammonium molybdate test

Question 27.
Identify the colour formed in the test kr phosphorous using ammonium molybciate.
(a) Crimson red colour
(b) Deep violet colour
(c) Prussian blue colour
(d) Canary yellow colour
Answer:
(d) Canary yellow colour

Question 28.
Which of the following will absorb CO2?
(a) Conc. H2SO4
(b) KOH
(c) HCl
(d) Copper
Answer:
(b) KOH

Question 29.
Which of the following is used as moisture absorbent?
(a) Potash
(b) Soda
(c) Conc. H2SO4
(d) Na3CO4
Answer:
(c) Conc. H2SO4

Question 30.
Which method is used to estimate sulphur?
(a) Lassaigne’s test
(b) Oxide test
(c) Carius method
(d) Kjedahl’s method
Answer:
(c) Cari us method

Question 31.
Which method is used to estimate nitrogen?
(a) Dumas method and Kjeldahl’s method
(b) Carius method & Oxide method
(c) Lassaignes test & Copper oxide test
(d) Ammonium molybdate test & Silver nitrate test
Answer:
(a) Dumas method and Kjeldahl’s method

Question 32.
Which of the following is not purified by sublimation method?
(a) Camphor
(b) Benzoic acid
(c) Naphthalene
(d) Nitrobenzene
Answer:
(d) nitrobenzene

Question 33.
Which of the following is used to decolourise the organic compounds?
(a) Chlorine
(b) Bleaching powder
(c) Animal charcoal
(d) Iodine
Answer:
(c) Animal charcoal

Question 34.
Which method is used to extract essential oils from plants and flowers?
(a) Crystallization
(b) Sublimation
(c) Steam distillation
(d) Differential extraction
Answer:
(c) Steam distillation

Question 35.
Which of the following is used as adsorbent?
(a) silica gel and alumina
(b) glass wool and cotton
(c) glass plate and paper
(d) glucose and fructose
Answer:
(a) silica gel and alumina

Question 36.
Which of the following compounds gives prussian blue colour in Lassaigne’s test?
(a) CH4 and CH3OH
(b) CH3NH2 and CH3NO2
(c) CH3Cl and CHCl3
(d) CH3CHO and CH3COCH3
Answer:
(b) CH3NH2 and CH3NO2

Question 37.
Which of the following compounds gives curdy white precipitate in Lassaigne’s test?
(a) CH3Br
(b) C2H5I
(c) CH3Cl
(d) C6H5NO2
Answer:
(c) CH3Cl

Question 38.
Which one of the following is not used as air adsorbent in chromatography?
(a) Alumina
(b) Silica gel
(c) Magnesia
(d) Sucrose
Answer:
(d) Sucrose

Question 39.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry ……..
(a) 2-methyl butanal
(b) butan-2-aldehyde
(c) 2-ethyipropanal
(d) 3-methyl isobutraldehyde
Answer:
(c) 2-ethyipropanal

Question 40.
Which of the following compounds will exhibit cis-trans isomerism?
Answer:
(a) 2-Buiene
(b) 2-Butyne
(c) 1-Butene
(d) 2-Butanol
Answer:
(a) 2-Butene

Question 41.
Which of the following sodium fusion extract of organic compound gives brilliant violet colour with sodium nitroprusside solution?
(a) Urea
(b) Thiourea
(c) Benzoic acid
(d) Aniline
Answer:
(b) Thiourea

Question 42.
Which of the following reagent is used to distinguish between halogens (Cl, Br, I) in an organic compound?
(a) NaOH
(b) FeCl3
(c) H2SO4
(d) NH4OH
Answer:
(d) NH4OH

Question 43.
In which of the following, functional group isomerism is not possible?
(a) Alcohols
(b) Aldehydes
(c) Alkyl halides
(d) Cyanides
Answer:
(c) Alkyl halides

Question 44.
Which one of the following is used as a column in the separation of pigments of chlorophyll by chromatography technique?
(a) Petroleum ether
(b) CaCO3
(c) Activated charcoal
(d) Ethanoic acid
Answer:
(b) CaCO3

Question 45.
Which one of the following compound does not give Prussian blue colour in Lassaigne’s test?
(a) C6H5NH2
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(c) C6H5CONH2
(d) C6H5COCl
Answer:
(d) C6H5COCl

Question 46.
Which one of the following shows geometrical isomerism?
(a) n-Butane
(b) 1-butene
(c) 2-butene
(d) butyne
Answer:
(c) 2-butene

Question 47.
Which one of the following shows functional group isomerism?
(a) Ethene
(b) Acetone
(c) Ethane
(a) Propane
Answer:
(b) Acetone

Question 48.
Which of the following pair gives curdy white precipitate and yellow precipitate respectivety in their Lassaignes test?
(a) C2H5I and C2H5Br
(b) C2H5NO2 and C2H5NH2
(c) C6H5Cl and CH3
(d) CH4 and CH3OH
Answer:
(c) C6H5Cl and CH3

II. Match the following.

Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 6.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 7.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 8.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

III. Fill in the blanks.

Question 1.
The property catenation is possible in ……..
Answer:
Carbon.

Question 2.
Acetic acid was synthesised by ………
Answer:
Kolbe

Question 3.
Methane was synthesised in laboratory by ………
Answer:
Berthiot
CH3

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry is an example of ………
Answer:
aromatic benzenoid compound

Question 5.
2-butene is an example of compound.
Answer:
unsaturated open chain

Question 6.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis …….
Answer:
2, 2, 5-trimethyl heptane

Question 7.
The IUPAC name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis ……..
Answer:
3, 4-diethyl, 4-methyiheptane

Question 8.
The name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis ……….
Answer:
1-cyclobutyl-2-cyclopropylethene

Question 9.
The name of Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
2-(cyclobut-2-en-1-yl)-propanal

Question 10.
The name ofSamacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis …………
Answer:
2-cyclopentyipropanal

Question 11.
Esopentane and neopentane are the examples for ………
Answer:
chain isomerism

Question 12.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry are called ……….
Answer:
functional isomers

Question 13.
Copper oxide test is used to detect ………..
Answer:
Carbon & Hydrogen

Question 14.
The formula of feme suiphocyanide is ………
Answer:
Fe(CNS)3

Question 15.
In Lassaigne’s test for halogens, if colour of the precipitate is curdy white, the halogen present is ……….
Answer:
chlorine

Question 16.
The formula of ammonium phospho molybate is ……….
Answer:
(NH4)3.PO4. MO3

Question 17.
Silver nitrate test is used to detect the presence of ……….
Answer:
Halogens

Question 18.
During the estimation of carbon and hydrogen, presence of nitrogen can be avoided by using ………
Answer:
a spiral of copper

Question 19.
In Carius method, the sulphur in an organic compound is oxidised to ……….
Answer:
H2SO4

Question 20.
The method used to estimate nitrogen in foods and fertilisers is ………
Answer:
Kjeidahl’s method

Question 21.
The mixture of diethyl ether and ethanol can he purified by ………
Answer:
simple distillation

Question 22.
The method used to purify petroleum. coal-tar and crude oil is ………
Answer:
fractional distillation

Question 23.
The method used in the manufacture of aniline and turpentine is ……….
Answer:
steam distilation

Question 24.
The mixture of exhanol and water are separated by ………..
Answer:
azeotropic distillation

Question 25.
The different coloured constituents of chlorophyll are separated by ………
Answer:
chromatography

Question 26.
The large number of organic compounds is due to of carbon ………..
Answer:
catenation

Question 27.
The IUPAC name of the compound shown below is:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Answer:
1, 1-dichloropropane

Question 28.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryname of this compound is ………..
Answer:
2-chloro-3-ethyl- 1, 4-pentadiene

Question 29.
Carboxylic acids are isomenc with ……….
Answer:
esters

Question 30.
Alcohols are isomeric with ……….
Answer:
ethers

Question 31.
The correct IUPAC name for the following structure is ……….
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
5-hex-i -en-3-ol

Question 32.
The Prussian blue colour confirms the presence of nitrogen in an organic compound is due to the formation of ………
Answer:
Fe4[Fe(CN)6]I

Question 33.
The principle involved in paper chromatography is ……….
Answer:
partition

Question 34.
Steam distillation is used for the extraction of ……..
Answer:
essential oils

Question 35.
In chromatography, if the stationary phase is solid, the basis is ……….
Answer:
adsorption

Question 36.
In chromatography, if the stationary phase is liquid, the basis is ………
Answer:
partition

Question 37.
The isomer of acetaldehyde is ……….
Answer:
acetone

Question 38.
The general formula of alkyne is ……….
Answer:
CnH2n-2

Question 39.
The IUPAC name of (CH3)2CH-CH2-CH(CH3)2-CH(CH3)2 is ……..
Answer:
2, 3, 5-trimethyl hexane

IV. Choose the odd one out.

Question 1.
(a) THF
(b) Pyridine
(c) Phenol
(d) Thiophen
Answer:
(c) Phenol. ft is a homocyclic compound whereas others are heterocyclic compounds.

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 3.
(a) Azulene
(b) Propane
(c) Butane
(d) Ethene
Answer:
(a) Azulene. It is a non benzenoid aromatic homocyclic compound whereas others are aliphatic compounds.

Question 4.
(a) Dyes
(b) Polymers
(c) Cosmetics
(d) Common salt
Answer:
(d) Common salt. It is an inorganic compound whereas others are organic compound.

Question 5.
(a) Renzene
(b) Water
(c) Ether
(d) Chloroform
Answer:
(b) Water. It is a polar solvent whereas others are non-polar solvents.

V Choose the correct pair.

Question 1.
(a) Benzene : Aliphatic compound
(b) Propane : Aromatic compound
(c) Pyridine : Heterocyclic compound
(d) Cyclohexane : Polycyclic compound
(c) Pyridine : Heterocyclic compound

Question 2.
(a)-OH : Ketone
(b)-CHO : Carboxylic acid
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(d)-NO2 : Amine
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 3.
(a) Organic compounds : inflammable
(b) Organic compounds : ionic compound
(c) Organic compounds : high boiling point and high melting point
(d) Organic compounds : soluble in water
Answer:
(a) Organic compounds : inflammable

Question 4.
(a) CnH2n+2 : C2H4
(b) CnH2n : C3H6
(c) CnH2n-2 : C2H6
(d) CnH2n+2 : C3H4
Answer:
(b) CnH2n : C3H6

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 6.
(a) n-pentane and iso pentane : position isomerism
(b) neopentane and n-pentane : chain isomerism
(c) propanal and propanone : position isomerism
(d) propañoic acid and methyl acetate : chain isomerism
Answer:
(b) neopentane and n-pentane : chain isomerism

VI. Choose The incorrect pair.

Question 1.
(a) Dumas method : Estimation of nitrogen
(b) Kjeldahls method : Estimation of nitrogen
(c) Carius method : Estimation of halogens
(d) Dumas method : Estimation of sulphur
Answer:
(d) Dumas method : Estimation of sulphur

Question 2.
(a)-CHO : Aldehyde
(b)-COOH : Carboxylic acid
(c)-NH2 : Nitro group
(d)-O- : Ether
Answer:
(c)-NH2 : Nitro group

Question 3.
(a) Benzene and nitro benzene : Distillation
(b) Coal tar and crude oil : Fractional distillation
(c) Aniline and turpentine : Steam distillation
(d) Naphthalcne and benzoic acid : Crystallization
Answer:
(d) Naphthalene and benzoic acid: Crystallization

Question 4.
(a) BaSO4 : White colour precipitate
(b) Ag2S : Black colour precipitate
(c) Fe(CNS)3 : Prussian blue colour
(d) PbS : Black colour precipitate
Answer:
(c) Fe(CNS)3 : Prussian blue colour

Question 5.
(a) propanal and propropane : Functional isomerism
(b) Nitrite fòrrn and nitro form : Tautomerism
(c) Pent- 1-ene and pcnt-2-ene : Chain isomerism
(d) Propanoic acid and methyl acetate: Functional isomerism
Answer:
(c) Pent-1-ene and pent-2-ene : Chain isomerism

VII. Assertion & Reason.

Question 1.
Assertion (A) : Carbon cannot form ionic bond.
Reason (R) : It is not possible for the carbon to form either C4t or C ions, as it requires large amount of energy.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct hut R is not the correct explanation of A.
(c) A is.correct but R is wrong.
(d) A is wrong but Ris correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 2.
Assertion (A) : Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and propanone (boiling point 56°C).
Reason (R) : Liquids with a dîftèrence of more than 30°C in their boiling points can be separated by simple distillation.
(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but Reason is correct.
Answer:
(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Question 3.
Assertion (A) : Pent-1 -ene and pent-2-ene are position isomers.
Reason (R) : Position isomers ditTer in the position of functional group or substituent.
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.
(b) Both assertion and reason are correct but reason is not the correct explanation of assertion.
(c) Assertion is correct but reason is wrong.
(d) Assertion is wrong but reason is correct.
Answer:
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.

VIII. Choose the correct statement.

Question 1.
(a) All organic compounds are ionic compounds.
(b) All organic compounds have high boiling point and high melting point.
(c) Many of the organic compounds are inflammable.
(d) Organic compounds are mostly soluble in water.
Answer:
(c) Many of the organic compounds are inflammable.

Question 2.
(a) Propane is heterocyclic compound.
(b) Azulene is a non benzenoid and aromatic homocyclic compound.
(c) Pyridine is a homocyc lic compound.
(d) Cyclopropane is an aromatic compound.
Answer:
(b) Azulene is a non henzenoid and aromatic homocyclic compound.

Question 3.
(a) CH≡CH-CH2-C≡CH is a saturated open chain compound.
(b) CH3-CH2-CH2-CH2-CH3 is an aromatic benzenoid compound.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis an aromatic benzenoid compound.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistryis an aromatic benzenoid compound.

Question 4.
(a) Organic compounds are covalent and generally insoluble in water.
(b) Organic compounds are ionic but generally soluble in water.
(c) Organic compounds non-inflammable
(d) Organic compounds do not show catenation.
Answer:
(a) Organic compounds are covalent and generally insoluble in water.

Question 5.
(a) Fe4[Fe(CN)6] is prussian blue precipitate.
(b) Ag2S is a white precipitate.
(c) PbS is a blood red colour precipitate.
(d) BaSO4 is a black colour precipitate.
Answer:
(a) Fe4[Fe(CN)6] is prussian blue precipitate.

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 2 Mark Question and answers

Write brief answer to the following questions:

Question 1.
What is catenation?
Answer:
The tendency of an atom to form a chain of bonds with the atoms of the same element is called catenation. The high strength of C-C bond is responsible for its catenation property.

Question 2.
Almost all compounds of carbon form covalent bonds. Give reason.
Answer:
Carbon (Z = 6) have electronic configuration of is2 2s2 2p2. It is not possible for the carbon to form either C4+ or C4- ions to attain the nearest noble gas configuration as it requires large amount of energy. This implies that carbon cannot form ionic bond. So almost in all compounds of carbon, it form four covalent bonds.

Question 3.
What is meant by functional group? Give two example.
Answer:
A functional group is an atom or a specific combination of bonded atoms that react in a characteristic way. irrespective of organic molecule in which it is present. The reaction of an organic compound takes place at the functional group.
e.g.. Alcohol -OH group
Ether -O- group

Question 4.
Classify the following compounds based on the structure.
(i) CH2=CH-CH=CH2
(ii) CH3-CH2-CH2-CH2-CH2-CH2-CH3
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) Unsaturated open chain compound
(ii) Saturated open chain compound
(iii) Aromatic benzenoid compound
(iv) Alicylic compound

Question 5.
Give one example for each of the following type of organic compounds.

  1. Non-benzeoid
  2. Aromatic heterocclic
  3. Alicyclic
  4. Aliphatic open chain

Answer:
1. Non-benzenoid aromatic compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. Alicyclic compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

3. Aromatic heterocyclic Compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

4. Atiphatic open chain compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 6.
Give two examples for each of the following type of organic compounds.

  1. Aromatic bomocyclic compound
  2. Aromatic heterocyclic compound

Answer:
1. Aromatic homocyclic compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. Aromatic heterocyclic compound
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 7.
Write the functional group of the following compounds

  1. Aldehyde
  2. Ester
  3. Ether
  4. alcohol

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 8.
Write the functional group of –

  1. cyanide
  2. socyanide
  3. cyanate
  4. isocyanate

Answer:

  1. cyanide -CN
  2. isocyanide -NC
  3. cyanate -CNO
  4. isocyanate -NCO

Question 9.
Write the functional group of –

  1. thiocyanate
  2. isothiocyanate
  3. thiols
  4. thioether

Answer:

  1. Thio cyanale -SCN
  2. Isothiocyanate -NCS
  3. thiols -SH
  4. thicethers -S-

Question 10.
Write the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Answer:

  1. 3-methylpentane
  2. 2, 2, 5-trimethylheptane

Question 11.
Write the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. 3-ethyl-2-methyl pentane
  2. 2-methylbutanal

Question 12.
What are the IUPAC names of the following compounds?
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. 2-ethyl-but-4-ene-oic acid
  2. 2, 2-dimethyl-hexanoic acid

Question 13.
Write the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. 2-ethyl-3-propyl pentane-dioic acid
  2. 3-methy-1 hexane

Question 14.
Predict the IUPAC names of the following compounds
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. cyclo butane
  2. cyclo pentane
  3. cyclo hutene
  4. cyclo octane

Question 15.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. Ethyl cyclobutanc
  2. Methyl cyclohexane
  3. Cyclo hexanol

Question 16.
Write the ¡UPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. 2-cyclobutyl propanol
  2. 3-cyclohexyl pentan-2-one

Question 17.
Write the structural formula for the following compounds.
(i) cyclohexa- 1, 3-diene
(ii) methyl cyclopentane
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 18.
Draw the structures of –
(i) o-xylene
(ii) m-xylene
(iii) p-xylene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 19.
Draw the structure of –
(i) Mesitylene
(ii) 1,2, 3-trimethyl benzene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 20.
Write the structure of –
(i) p-dinitrobenzene
(ii) o-dichlorobenzene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 21.
Draw the structure of –
(i) 2-cyclopentyl propanal
(ii) 2-(cyclo-but-cnyl) propanal
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 22.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:

  1. N, N-dimethylbenzene amine
  2. N-ethyl-N-methylpropan-I-amine

Question 23.
Draw the structure of 4-hydroxy-3(2-hydroxy ethyl) hexanoic acid.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 24.
Explain about dash line structure with a suitable example.
Answer:
The line bond structure is obtained by representing the two electron covalent bond by a dash or line (-) in a lewis structure. A single line or dash represents a single covalent bond.
e.g., n- propanal:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 25.
What ¡s meant by condensed structure? Explain with an example.
Answer:
The bond line structure can be further abbreviated by omitting all these dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula.
e.g., 1, 3-butadiene.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 26.
What are bond line structures? Give one example.
Answer:
Condensed structural formula is simplified in which only lines are used. In this type of representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bond are shown in zig zag fashion. The only atoms specially written are oxygen, chlorine, nitrogen etc. Example, Ten. butyl chloride
condensed structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
bond line structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 27.
Define isomerism. Give example.
Answer:
Isomerism represents the existence of two or more compounds with the same molecular formula but different structure and properties. Compounds exhibiting this isomerism are called isomers.
e.g., C2H6O:

  1. CH3-CH2OH Ethanol
  2. CH3-O-CH3 Methoxyrnethane

Question 28.
Write the possible isomers for the formula C5H10 with their name and type of isomerism present in it.
Answer:
C5H10:

  1. CH3-CH2-CH2-CH = CH2 (Pent-1-ene)
  2. CH3-CH2-CH=CH-CH3 (Pent-2-ene)
    The type of isomerism present above is position isomerism.

Question 29.
Write the possible isomers for the formula C5H10O with their name indicating position isomerism.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Question 30.
Draw the functional isomers for the formula C3H6O2 with their names.
Answer:
C3H6O2:

  1. CH3-CH2-COOH (Propanoic acid)
  2. CH3-COOCH2 (Methyl acetate)

Question 31.
What is metamerism? Give an example.
Answer:
Metamerism is one kind of structural isomerism which arises due to the unequal distribution of carbon atoms on either side of the functional group or diffèrent alkyl groups attached to the either side of the same functional group and having the same molecular formula.
e.g., C4H10O:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 32.
What is tautomerism?
Answer:
It is a special type of functional isomerism in which single compound exists in two readily interconvertible structures that differ markedly in the relative position of atleast one atomic hydrogen. The two different structures are known as tautomers.

Question 33.
What ¡s meant by dyad system? Explain with example.
Answer:
In this system, hydrogen atom oscillates between two directly linked polyvalent atoms.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 34.
What is triad system? Give example.
Answer:
1. In this system hydrogen atom oscillates between three polyvalent atoms. It involves 1. 3-migration of hydrogen atom from one polyvalent atom to other with in the molecule.
2. The most important type of triad system is keto-enol tautomerism and the two groups of tautomers are keto form and enol form.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 35.
What is enolisation? What is labile form?
Answer:
Enolisation is a process in which keto form is converted into eno! form. Both tautomeric forms are equally stable. The less stable form is known as labile form.

Question 36.
Give the structures of Nitro-aci tautomerism.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 37.
Explain ring chain isomerism with the formula C4H8.
Answer:
In ring chain isomerism, compounds have the same molecular formula but differ in terms of bonding of carbon atoms to form open chain and cyclic structures.
C4H8:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 38.
Define stereo-isomerism.
Answer:
The isomers which have same bond connectivity but different arrangement of groups or atoms in space are known as stereoisomers. This phenomenon is known as stereoisomerism.

Question 39.
Define geometrical isomerism with an example.
Answer:
Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 40.
Trans isomer is more stable than cis isomer. Justify this statement.
Answer:
Trans isomer is more stable than cis isomer. This is because in the cis isomer, the bulky groups are on the same side of the double bond. The steric repulsion of the groups makes the cis isomers less stable than the trans isomers in which bulky groups are on the opposite side.

Question 41.
Draw the cis, trans isomeric structures of 1, 3-butadiene.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 42.
What are the condition for optical isomerism (or) enantiomerism.
Answer:
1. A carbon atom whose tetravalency in satisfied by four different substituents (atoms (or) groups) is called asymmetric carbon (or) chiral carbon. The optical isomer should have one or more chiral carbon to show optical activity.

2. The molecule possessing chiral carbon atom and is non-superimposable its own mirror image is said to be chiral ntolecule and the property is called chirality or dissymmetry.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 43.
How will you prepare Lassaigne’s extract?
Lassagine’s extract preparation:
Answer:

  • A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is gently heated. When it melts to a shining globule, a pinch of organic compound is added.
  • The tube is heated till reaction ceases arid become red hot. Then it is plunged in 50 ml of distilled water taken in a china dish and the bottom of the tube is broken by striking it against the dish.
  • The contents of the dish is boiled for 10 minutes and then it is filtered. The filtrate is known as Lassaigne’s extract.

Question 44.
What Is the need for purification of organic compounds?
Answer:
In order to study the structure, physical properties, chemical properties and biological properties of organic compounds, they must be in the pure state. So organic compounds must be purified.

Question 45.
Define sublimation. Give two examples.
Answer:
The process of conversion of solid to vapour without melting or heating and on cooling the vapours getting back solids, such phenomenon is known as sublimation.
e.g., Naphthalene, Camphor.

Question 46.
Explain the process of chromatography in chlorophyll.
Answer:
The separation of different coloured constituents of chlorophyll is done by chromatography by M.S. Tswelt. He achieved it by passing a petroleum ether solution of chlorophyll present in leaves through a column of CaCO3 firmly packed into a narrow glass tube. Different components of the pigments got separated and lands to form zones of different colours.

Question 47.
Draw the first six members of the carboxylic acid homologous series.
Answer:

  1. HCOOH
  2. CH3COOH
  3. CH3-CH2-COOH
  4. CH3-CH2-CH2-COOH
  5. CH3-CH2-CH2-CH2COOH
  6. CH3-CH2-CH2-CH2-CH2-COOH

Question 48.
Give the condensed formula and bond line formula of 2, 2,4- trimethylpentane.
Answer:
2, 2, 4 –  irimethylpentane
(CH3)3CCH2CH(CH3)2-Condensed formula
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 49.
Differentiate between the principle of estimation of nitrogen in an organic compound by

  1. Dumas method
  2. Kjeldahl’s method.

Answer:
1. Dumas method:
The organic compound is heated strongly with excess of CuO (Cupic Oxide) in an atmosphere of CO2 where free nitrogen, CO2 and H2O are obtained.

2. Kjeldahl’s method:
A known mass of the organic compound is heated strongly with conc. H2SO4, a little amount of potassium sulphate and a little amount of mercury (as catalyst). As a result of reaction, the nitrogen present in the organic compound is converted to ammonium sulphate.

Question 50.
Explain the principle of paper chromatography.
Answer:
This is the simplest form of chromatography. Here a strip of paper acts as an adsorbent. It is based on the principle which is partly adsorption. The paper is made of cellulose fibres with molecules of water adsorbed on them. This acts as stationary phase. The mobile phase is the mixture of the components to be identified whose solution is prepared in a suitable solvent.

Question 51.
Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Answer:
Organic compound is fused with sodium metal so as to convert organic compounds into NaCN, Na2S, NaX and Na3PO4. Since these are ionic compounds and become more reactive and thus can be easily tested by suitable reagents.

Question 52.
Name a suitable technique of separation of the components from a mixture of calcium, sulphate and camphor.
Answer:
Sublimation. Because camphor can sublime whereas CaSO4 does not.

Question 53.
Explain, why an organic liquid vapourises at a temperature below its boiling point on steam distillation?
Answer:
It is because in steam distillation the sum of vapour pressure of organic compound and steam should be equal to atmospheric pressure.

Question 54.
Will CCl4give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Answer:
No. CCl4 is a completely non-polar covalent compound whereas AgNO3 is ionic in nature. Therefore they are not expected to react and thus a white ppt. of silver chloride will not be formed.

Question 55.
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Answer:
CO2 is acidic in nature and therefore it reacts with the strong base KOH to form K2CO3:
2KOH + CO2 – K2CO2 + H2O

Question 56.
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Answer:
Sulphur sodium extract is acidified with acetic acid because lead acetate is soluble and does not interfere with the test.
Pb(OCOCH3)2 + H2SO4 → PbSO4-+ 2CH3COOH

Question 57.
Why is an organic compound fused with sodium for testing nitrogen, halogens and sulphur?
Answer:
On Ilising with sodium metal the clements present in an organic compound are converted into sodium salts which are water soluble which can be filtered and detected by the respective tests.

Question 58.
Under what conditions can the process of team distillation is used?
Answer:
Steam distillation is used to purify the Liquids which are steam volatile and not miscible with water.

Question 59.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
1, 2-dichloropropane

Question 60.
Write bond-line formulas for: Isopropyl alcohol, 2, 3-dimethvlbutanal, Heptan-4-one.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 3-Mark Questions

Question 1.
Write the functional group of the following comopounds:
(i) carboxylic acid
(ii) Acid anhydride
(iii) Acyichioride
(iv) Amide
(v) imines
(vi) Nitroso compound
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 2.
What are the general molecular formula and functional group of the following compounds?
Answer:
(i) Ilydrazines
(ii) Hydrazo compound
(iii) Imide
(iv) Phenols
(v) Amine
(vi) Nitroalkane
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 3.
Write the tUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) Pentan-2a1
(ii) Pentan-(2-ene-2-propyl)- i -oic acid
(iii) 4-methyl-i -cyanohexane

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(I) Hex-4-ene-2-oI
(ii) 3-ethyl-5-methyl heptane

Question 5.
Draw the structure of
(i) 1-ethyl-2-methyl cyclopentane
(ii) 1-ethyl-2, 3-dimethyl cyclohexane
(iii) 5-ethyl-2-methylcyclohcx- I -ene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 6.
Draw the structures of:
(i) 2-cyclobutyl propane
(ii) 2-cyctopropyl butane
(iii) chiorocyclo but-2-eue
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 7.
Give the IUPAC name of the following compounds:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 2-(cyclo but-2-en-1-yl)-propanal
(ii) 4-(cyclopent-3-en- 1-yl )-3-methylbutanoic acid
(iii) 3-(3-nitro cyclopenryl)-prop-2-enoic acid

Question 8.
Write the IUPAC names of the following compounds
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(I) 2-(2-hydroxypropyl) cyclohexan- 1-01
(ii) CyclopentyÍ benzene
(iii) Cyclohexane carboxyl Ic acid

Question 9.
Draw the structure
(i) 1-(cyclo bytyl)-2 (cylopropyl) ethane
(ii) 2-carbamyl cyclobutane-1-carboxylic acid
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 10.
Draw the structures of:
(i) Bromohenzene
(ii) 1, 2-dichlorobenzene
(iii) 1-chloro-3-methvlbenzene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 11.
Draw the structures of –
(i) Benzvl chloride
(ii) Benzal dichloride
(iii) Benzotrichloride
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 12.
Write the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 3-methylpentane
(ii) 2, 2. 5-trimethylheptane
(iii) 2-methylbutanal

Question 13.
Write the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 3-ethyl-2-methylpentane
(ii) 2-ethyl but-3-enoic acid
(iii) 2-forrnyl-2-methylheptanoic acid

Question 14.
Draw the structures of
(i) 3-methylpentanal
(ii) 5-hydroxy 2,2-dimethyl heptanoic acid
(iii) 2-ethyl-4-propy Ipentane-d ioic acid
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 15.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 3-methylhexane
(ii) 2-methylbutanal
(iii) 2-ethylbut-3-enoic acid

Question 16.
Give the IUPAC name Of –
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 4-methyl/hexanenitrite
(ii) 2-methyl but-3-en-amide
(iii) I-Iex-4-en-2-ol

Question 17.
Draw the structures of –
(i) 3-ethyl-5-methylheptane
(ii) 3-ethyl-2-methylhexane
(iii) 2, 4-dimethylpent-2-ene
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 18.
Draw the structures of:
(i) 3-methylhcpta 1, 3, 5-triene
(ii) pent-1-yne
(iii) 2-methylpropan-2-oI
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 19.
Give the IUPAC name of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) 4-mcthylpcntan-1-ol
(ii) 2, 2-dimethylpropan-1-ol
(iii) Propanoic acid

Question 20.
Draw the structural formula of:
(i) 4-methylpent-3-en-2rone
(ii) pent-I-yne-3-one
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 21.
Write the IUPAC names of the following compounds.
CH3-CH2-CH2-NH-CH3

Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(i) N-methylpropan- 1 -amine
(ii) N-rnethylpropan-2-amine
(iii) N, N-dimethylpropan- 1-amine

Question 22.
Draw the structurai formula of the following compounds.
(i) N-cthyl-N-methylpropan-1-amine
(ii) N, N-dimethyl benzenamine
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 23.
Draw the complete structural formula, condensed structure and bond line structure of
(i) n-propanol
(ii) 1, 3-butadiene.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 24.
Draw the dash line structure, condensed structure and bond line structure of 1, 3-dimethyl cyclopentane.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 25.
What is wedge formula? Explain with suitable example.
Answer:
1. The simplest convention is solid and dashed wedge formula in which 3-D image of a molecule can be perceived from two dimensional picture.

2. In this representation, a tetrahedral molecule with four atoms or groups a, b, e and d bonded to it can be represented by wedge formula as follows.

3. A solid wedge Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
or a heavy line is used to indicate a bond projecting above the plane of the paper and dashed wedge Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
or a dashed line is used to depict the bond below the plane. The bonds lying in the plane of the paper are shown by normal lines.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 26.
Draw the fisher projection formula for tartaric acid.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 27.
Explain the advantage of sawhorse projection formula over the fisher projection formula with an example.
Answer:
1. The fisher projection fonnula inadequately portrays the spatial relationship between ligands attached to the atoms. The sawhorse projection attempts to clarify the relative location of the groups.

2. In sawhorse projection formula, the bond between two carbon atoms ¡s drawn diagonally and slightly elongated. The lower left hand carbon is considered lying towards the front and the upper right hand carbon towards the back.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 28.
Explain about the Newmann projection formula with an example.
Answer:
1. In this method, the molecules are viewed from the front along the carbon-carbon bond axis.

2. The two carbon atom forming the G bond is represented by two circles. One behind the other so that only the front carbon is seen. The front carbon atom is shown by a point where as the carbon lying farther from the eye is represented by the origin of the circle.

3. Therefore the C-H bonds of the front carbon are depicted from the circle while the C H bonds of the back carbon are drawn from the circumferance of the circle with an angle of 120 to each other.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 29.
Write the possible isomers for the formula C5H12 with their names and structures.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(i), (ii) and (iii) are chain isomers.

Question 30.
What are the possible isomers for the formula C4H9Cl? Give their structures and IUPAC names.
Answer:
C4H9Cl:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 31.
Write the metamers for the formula C5H10O with their IUPAC names.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 32.
Explain about the geometrical isomerism possible in oximes.
Answer:
1. Restricted rotation around C = N (oximes) gives rise to geometrical isomerism in oximes. Here syn and anti are used instead of cis and trans respectively.

2. In the syn isomer the H atom of a doubly bonded carbon and OH group of doubly bonded nitrogen lie on the same side of the double bond, while in the anti isomer, they lie on the opposite side of the double bond.

3. for e.g.,
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 33.
What are enantiomers?
Answer:
1. An optically active substance may exist in two or more isomeric forms which have same physical and chemical properties but diflèr in terms of direction of rotation of plane polarised light, such optical isomers which rotate the plane polarised light with equal angle but in opposite directions are known as enantiomers and (he phenomenon is known as enantiomerism.

2. Isomers which are non-super impossible mirror images of each other are called enantiomers.

Question 34.
How would you detect sulphur?
Answer:
1. To a portion of the Lassaigne’s extract, freshly prepared sodium nitroprusside solution is added. If deep violet or purple colour is formed, the presence of sulphur is confirmed.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. To another portion of Lassaignes extract, acetic acid and lead acetate solution are added. If black precipitate is formed, sulphur presence is confirmed.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 35.
Explain about the oxidation test for sulphur.
Answer:
1. Oxidation test:
The organic substances are fused with a mixture of KNO3 and Na2CO3. The sulphur if present is oxidised to sulphate.
Na2CO3 + S + 3(O) Na2SO4 + CO2

2. The fused mass is extracted with water, acidified with HCl and the BaCl2 solution is added to it. A while precipitate indicates the presence of sulphur.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 36.
How would you detect the halogens in an organic compound?
Answer:

  • To a portion of the Lassaigne’s filtrate. dii. HNO3 is added, warmed gently and AgNO3 solution is added.
  • Appearance of curdy white precipitate soluble in ammonia solution indicates the presence of chlorine.
  • Appearance of pale yellow precipitate sparingly soluble in ammonia solution indicates the presence of bromine.
  • Appearance of yellow precipitate insoluble in ammonia solution indicates the presence of iodine.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 37.
Why nitric acid is added in the Lassaigne’s test for halogens?
Answer:
1. if N (or) S is present in the organic compound along with the halogen, we might obtain Na2S and NaCN in the solution which interfere with the detection of the halogen in the AgNO3 test.

2. Therefore we boil the Lassiagne’s extract with HNO3 which decomposes NaCN and Na2 S as follows:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 38.
Explain about the test for phosphorous in an organic compound.
Answer:

  • A solid organic compound is strongly heated with a mixture of Na2CO3 and KNO3. Phosphorous present in the compound is oxidised to sodium phosphate.
  • The residue is extracted with water and boiled with cone. HNO3. A solution of ammonium molyixiate is added to this solution.
  • A canary yellow precipitate shows the presence of phosphorous.

Question 39.
Explain about principle and reactions involved in carius method of estimation of sulphur. Carius method:
Answer:
1. Principle:
A known mass of the organic substance is heated strongly with fuming HNO3. C and H get oxidized to CO2. and H2O while sulphur is oxidised to sulphuric acid as follows.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

2. The resulting solution is treated with excess of BaCl2 solution, H2SO4 present in the solution is converted into BaSO4. From the mass of BaSO4, the percentage of sulphur can be calculated.

Question 40.
Explain about the procedure and calculation behind the carius method of estimation of sulphur.
Answer:
Carius method:
(I) Procedure:
A known mass of the organic compound is taken in a clean carius tube and few mL of fuming HNO3 is added and then the tube is sealed. It is then placed in an iron tube and bcatcd for 5 hours. The tube is allowed to cool and a hole is made to allow gases to escape. The carius tube is broken and the content collected in a beaker. Excess of BaCl, is added to the beaker. BaSO4 furmed is converted to BaSO4 (white ppt.) The precipitate is filtered. washed, dried and weight. From the mass of BaSO4, percentage of S is calculated.

(ii) Calculation:
Mass of organic compound = Wg
Mass of BaSO4 formed = r g
233 g of BaSO4 contains 32 g of sulphur
∴ x g of BaSO4 contain \(\frac {32}{233}\) x x g of sulphur
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 41.
How are naphthalene and camphor purified?
Answer:
1. Naphthalcne, camphor and benzoic acid when heated, pass directly from solid to vapour without melting. On cooling the vapours will give back solid. This phenomenon is known as sublimation. This technique is used to purify naphthalene, camphor from non volatile impurities.

2. Substances to he purified is taken in a beaker. It is covered with a watch glass. The beaker is heated for a while and the resuJting vapours condense on the bottom of the watch glass. Then the watch glass is removed and the crystals are collected.

Question 42.
Explain how nitrobenzene and benzenc arc separated and purified. (or) How will you separate the mixture of diethyl ether and ethanol?
Answer:
Distillation:

  • The process of distillation involves the impure liquid when boiled gives out vapour and the vapour so formed is collected and condensed to give back the pure liquid in the receiver.
  • This method ¡s to purify liquids from non-volatile impurities and used for separating the constituents of a liquid mixture which differ in their boiling points.
  • In this simple distillation process, liquids with large difference in boiling point (about 40K) and do not decompose under ordinary pressure can be purified.
  • e.g., the mixture of C6H5NO2 nitrohenzene (h.p. 484 K) and C6H6 benzene (b.p. 354 K) can be urified and separated. Similarly the mixture of diethyL ether (b.p. 308K) and ethyl alcohol (h.p. 351 K) can he purified and separated.

Question 43.
Explain about differential extraction.
Answer:

  • The process of removing a substance from its aqueous solution by shaking with a suitable organic solvent is termed extraction.
  • When an organic substance present as solution in water can be recovered from the solution by means of a separating funnel.
  • The aqueous solution is taken in a separating funnel with little quantity of ether or chloroform (CHCl3). The organic solvent immiscible with water will form a separate layer and the contents are shaken gently.
  • The solute being more soluble in the organic solvent is transferred to it.
  • The solvent layer is then separated by opening the tap of separating funnel and the substance is recovered.

Question 44.
Explain about the principle involved in chromatography. Give its types.
Answer:

  1. The principle behind chromatography is selective distribution of the mixture of organic substances between two phases-a stationary phase and a moving phase. The stationary phase can be a solid or liquid while the moving phase is a liquid or a gas.
  2. if the stationary phase is solid, the basis is adsorption and when it is a liquid, the basis is partition.
  3. Chromatography is defined as technique for the separation of a mixture brought about by differential movement of the individual compound through porous medium under the influence of moving solvent.
  4. The various methods of chromatography are:
    • Column chromatography (CC)
    • Thin layer chromatography (TLC)
    • Paper chromatography (PC)
    • Gas liquid chromatography (GLC)
    • Ion exchange chromatography

Question 45.
Describe about adsorption chromatography.
Answer:

  • The principle involved is different compounds are adsorbed on an adsorbent to different degree.
  • Silica gel and alumina are the commonly used adsorbent. The components of the mixture move by varying distances over the stationary phase.

Question 46.
What are hybridisation states of each carbon atom ¡n the following compounds?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 47.
Give the IUPAC names of the following compounds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Answer:
(a) Propylbenzene
(b) Methylpentanenitrile
(c) 2, 5-Dimethyllieptanc
(d) 3-Bromo-3-chloroheptane
(e) 3-Chioropropanal
(f) 2, 2-Dichloroethanol

Question 48.
Write the formulas for the first five members of each homologous series beginning with the following compound. CH3COCH3
Answer:
CH3COCH3
CH3COCH2CH2CH3
CH3COCH2CH2CH2CH3
CH3COCH2CH2CH2CH3
CH3CO(CH2)4CH3

Question 49.
Write the formulas for the first five members of each homologous series beginning with the following compound: H-CH=CH2
Answer:
H-CH=CH2
CH3CH=CH2
CH3CH2CH=CH2
CH3CH2CH2CH=CH2
CH3CH2CH2CH2CH=CH2

Question 50.
Identify the functional groups ¡n the following compounds.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 51.
What is the difference between distillation, distillation under reduced pressure and steam distillation?
Answer:
Distillation:
it is used in case of volatiLe liquid mixed with a non-volatile impurities.

Distillation under reduced pressure:
This method is used to purify such liquids which have very high boiling points and which decompose at or below their boiling points.

Steam distillation:
it is used to purify steam volatile liquids associated with water immiscible impurities.

Question 52.
(a) What is Lassaigne’s extract? Will NaCN give a positive Lassaigne’s test for nitrogen?
(b) Which colour will appear in the Lassaigne’s test if the compound contains both nitrogen and sulphur.
(c) Why is Lassaigne’s extract prepared in distilled water? Can we detect oxygen iii a compound by Lassaigne’s test?
Answer:
(a) When organic compounds is fused with sodium metal and then extracted by water it is called Lassaigne’s extract. Yes.
(b) Blood red colour.
(c) Lassaigne’s extract is prepared in distilled water since ta water contains Cl’ ions. No, oxygen cannot be detected by Lassaigne’s test.

Question 53.
0.3780 g of an organic compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine in the compound.
Answer:
Mass of the compound = 0.3780
Mass of silver chloride 0.5740 g
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 54.
In an estimation of sulphur by Carius method, 0.468 of an organic sulphur compound gave 0.668 g of barium sulphate. Find the percentage of sulphur in the compound.
Answer:
Percentage of sulphur
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 5 marks Questions and Answers

Question 1.
Explain about the steps involved in naming an organic compound as per IUPAC nomenclature.
Answer:
The following steps should be followed for naming an organic compound as per IUPAC nomenclature.

  • Choose the longest carbon chain (Root word). Consider all other groups attached to this chain as substituents.
  • Numbering of the longest carbon chain.
  • Naming the substituents (prefixes) or (suffixes).
  • Arrange the substituents in the alphabetical order.
  • Write the name of the compound as below.

Prefix + Root word Primary suffix + r Secondary suffix

Question 2.
How will you detect the presence of carbon and hydrogen in an organic compound?
Answer:
Copper oxide test:
1. The organic substance is mixed with three times its weight of dry copper oxide by grinding. The mixture is placed in a hard glass test rube fitted with a bent delivery tube. The other end of which is dipping into lime water in an another test tube. The mixture is heated strongly.

2. 2CuO + C → CO2 + 2Cu
CuO + 2H → H2O + Cu

3. Thus if carbon is present, it is oxidized to CO2 which turns lime water milky. If hydrogen is also present, it will be oxidized to water and condenses in small droplets on the cooler wall of the test tube and inside the bulb. Water is collected on white anhydrous CuSO4 which turns blue.

4. This confirms the presence of C and H in the compound.

Question 3.
Explain about lassaigne’s test for detection of nitrogen ¡n an organic compound.
Answer:
I step:
Preparation of sodium fusion extract:
A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is heated. When it melts to a shining globule, a pinch of organic compound is added to it. The tube is then heated till the reaction ceases and becomes red hot.

Then the test tube is plunged in about 50 ml of distilled water taken in a china dish and break the bottom of the tube by striking against the dish. The contents of the dish is boiled for about 10 minutes and then filtered. This filtrate is known as lassaigne’s extract (or) sodium fusion extract.

II step :
Test for Nitrogen:
If Nitrogen is present, it gets converted to sodium cyanide which reacts with freshly prepared ferrous sulphate and ferric ion followed by cone. HCl and gives a Prussian blue colour (or) green coloured precipitate, it confirms the presence of nitrogen. HCl is added to dissolve the greenish precipitate of ferrous hydroxide produced by the action of NaOH on FeSO4 which would otherwise mark the Prussian blue precipitate.

Reactions involved:
Na + C + N → NaCN
FeSO4 + 2NaOH → Fe(OH)2 + Na2SO4
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
if both N & S are present. a blood red colour is obtained due to the following reactions
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Question 4.
Explain about the estimation of carbon and hydrogen.
1. Principle:
A known weigIt of organic substance is brunt in excess of oxygen and the carbon and hydrogen present in it are oxidised to CO2 and H2O respectively.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
The weight of carbon dioxide and water thus formed are determined and the amount of carbon and hydrogen in the organic substance are calculated.

2. Description of the apparatus:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
(a) The oxygen supply (b) combustion tube (c) Absorption tube

Oxygen supply:
To remove the moisture from oxygen. it is allowed to bubble through sulphuric acid and then passed through a U-tube containing sodalime to remove CO2. The oxygen gas free from moisture and CO2 enters the combustion tube.

Combustion tube:
A hard glass tube open at both ends used for the combustion. It contains (i) an oxidized copper gauze to prevent the backward diffusion of the products of combustion (ii) a porcelain boat containing a known weight of the organic substance (iii) coarse copper oxide on either side and (iv) an oxidized copper gauze placed towards the end of the combustion tube. The combustion tube is heated by a gas burner.

Absorption apparatus:
The combustion products containing moisture and CO2 are then passed through the absorption apparatus which consists of(i) a weighed U-tube packed with pumice soaked in conc. H2SO4 to absorb water (ii) a set of bulbs containing a Strong solution of KOH to absorb CO2 and finally (iii) a guard tube filled with anhydrous CaCl2 to prevent the entry of moisture from atmosphere.

3. Procedure:
The combustion tube is heated strongly to dry its content. It is then cooled and connected to absorption apparatus. The other end of the combustion tube is open for a while and the boat containing weighed organic substance is introduced. The tube is again heated strongly till all the substance in the boat is burnt away. This takes about 2 hours. Finally a strong current of oxygen is passed. Then the U-tube and potash bulbs are then
detached and increase in weight of each of them is determined.

4. Calculation:
Weight of organic substance = W g
Increase in weight of H2O = x g
Increase in weight of CO2 = y g
18g of H2O contains 2 g of hydrogen
∴ x g of H2O contain \(\frac {1}{2}\) × x g of hydrogen.
∴ Percentage of hydrogen = (\(\frac {2}{18}\) x \(\frac {x}{w}\) x 100
44g of CO2 contains 12 g of carbon
∴ y g of CO2 contain \(\frac {12}{44}\) × y g of carbon
∴ Percentage of carbon = (\(\frac {12}{44}\) × \(\frac {12}{44}\) × 100)%

Question 5.
Explain about the estimation of halogens by carius method.
Answer:
Carius method:
A known mass of the substance is taken along with fuming HNO3 and AgNO3 taken in a clean carius tube. The open end of the carius tube is sealed and placed in a iron tube for 5 hours in a range at 530 to 540 K. Then the tube is allowed to cool and a small hole is inade in the tube to allow the gases to excape. The tube is broken and the precipitate is filtered, washed, dried and weighed. From the mass of AgX produced percentage of halogen in the organic compound is calculated.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Calculation:
Weight of the organic compound = Wg
Weight of AgCl = a g
143.5 g AgCl contains 35.5g of Cl
a g of AgCl contain \(\frac {35}{143.5}\) x 9 g of Cl
Wg of organic compound contains \(\frac {35}{143.5}\) × 9 g of Cl
% of chlorine = (\(\frac {35}{143.5}\) × \(\frac {a}{w}\) × 100)%
Weight of silver bromide = b g
188 g of AgBr contains 80 g of Br
b g of AgBr contain = \(\frac {80}{188}\) × b g of Br
% of Bromine = (\(\frac {80}{188}\) × \(\frac {b}{w}\) × 100)
Weight of silver iodide = c g
235 g of AgI contains 127 g of I
c g of AgI contain = \(\frac {80}{188}\) × c g of I
% of Iodine = (\(\frac {80}{188}\) × \(\frac {c}{w}\) × 100)

Question 6.
How will you estimate phosphorous in an organic compound?
Answer:
Carius method:
Procedure:
A known mass of organic compound (wg) containing phosphorus is heated with fuming HNO3 in a sealed tube where C is converted into CO2 and H to H2O. Phosphorous present in the organic compound is oxidised to phosphoric acid which is precipitated as ammonium phospho molybdate by heating with conc. HNO3 and by adding ammonium molybdate.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
The precipitate of ammonium phrospho molybdate is filtered, washed, dried and weighed.

Calculation:
Mass of organic compound = Wg
Mass of ammonium phospho molybdate = x g
Molar mass of ammonium phospho molybdate 1877 g
1877 g of ammonium phospho molybdate contains 31 g of phosphorous
x g of ammonium phospho molybdate contain = \(\frac {31}{1877}\) × x g of phosphorous
% of phosphorous = (\(\frac {31}{1877}\) × \(\frac {x}{w}\) × 100)% of phosphorous
In an alternate method, phosphoric acid is precipitated as magnesium-ammoniumphosphate by adding magnesia mixture. The ppt. is washed dried and ignited to get magnesium pyrophosphate which is washed, dried and weighed.

Weight of magnesium pyrophosphate = y g
Molar mass of magnesium pyrophosphate = 222 g
222 g of magnesium pyrophosphate contains 62 g of P
y g of magnesium pyrophosphate contain = \(\frac {62}{222}\) × y g of P
% of phosphorous = (\(\frac {31}{1877}\) × \(\frac {y}{w}\) × 100)%

Question 7.
Explain Dumas method of estimation of nitrogen.
Answer:
Dumas method:
Principle:
This method is based on the fact that nitrogeneous compound when heated with cupric oxide in an atmosphere of CO2 yields free nitrogen.
Traces of nitrogen are reduced to elemental nitrogen by passing over heated copper spiral.
Description of the apparatus:
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
CO2 Generator:
CO2 needed in this process’is prepared by heating magnetite or sodium bicarbonate contained in a hard glass tube (or) by the action of dil. HCl on marble in a kipps apparatus. The gas is passed through the combustion tube after dried by bubbling through cone. H2SO4.

Combustion tube:
The combustion tube is heated in a furnace is charged with (a) A roll of oxidised copper gauze to prevent the back diffusion of products of combustion and to heat the organic substance mixed with CuO by radiation (b) a weighed amount of organic substance mixed with excess of CuO (c) a layer of CitO packed in about 2/3 length of the tube and kept in position by loose asbestos plug on either side and (d) a reduced copper piral which reduces any oxides of nitrogen formed during combustion of nitrogen.

Schiff’s nitromctc:
The nitrogen gas obtained by the decomposition of the substance in the combustion tube is mixed with considerable excess of CO2. It is estimated by passing nitro meter when CO2 is absorbed by KOH and the nitrogen gas gets collected in the upper part of the graduated tube.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Calculation:
Weight of the substance taken Wg
Volume of nitrogen = V1L
Room temperature = T1K
Atmospheric pressure = p mm Hg
Aqueous tension al room temperature P’ nun of Hg
Pressure of dry nitrogen = P—P’ = P’1 mm Fig
P0, V0 and T0 be the pressure, volume and temperature respectively of dry nitrogen at S.T.P.
Then, \(\frac{P_{0} V_{0}}{T_{0}}=\frac{P_{1} V_{1}}{T_{1}}\)
V0 = \(\frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{0}}{P_{0}}\)
V0 = \(\frac{P_{1} V_{1}}{T_{1}} \times \frac{273 \mathrm{K}}{760 \mathrm{mm} \mathrm{Hg}}\)
22.4 L of N2 at STP weigh 28 g of N2
V0L of N2 at STP weigh \(\frac {28}{22.4}\) x V0
W g of organic compound contain \(\frac {28}{22.4}\) x V0 g of N2
100 g of organic contain \(\frac {28}{22.4}\) x \(\frac{V_{o}}{w}\) x 100 = % of Nitrogen

Question 8.
Explain Kjeldahl’s method.
Answer:
Principle:
This method is based on the fact that an organic compound containing nitrogen is heated with cone. H2SO4. the nitrogen is conerted Lo ammonium sulphate. The resultant liquid is heated with excess of alkali and then liberated ammonia gas is absorbed in excess of standard acid. The amount of ammonia (nitrogen) is determined by finding the amount of acid neutralised by back titration with same standard alkali.

Procedure:
A weighed quantity of the substance 0.3 to 0.5g is placed in a special long necked Kjeldahl flask made of pyrex glass. About 25 ml of cone. H2SO4 together with a little K2SO4 and CuSO4 [catalyst] are added to il. the flask is loosely stoppered by a glass bulb and heated gently in an inclined position.

The heating is continued till the brown colour of the liquid disappears leaving the content clear as before. At this point all the nitrogen is converted to ammonium sulphate. The kjeldahl flask is cooled aiid its contents are diluted with distilled water and carefully transferred into a I litre round bottom flask. An excess NaOH is poured down the side of the flask and it is filled with a kjeldhals trap and a water condenser.

The lower end of the condenser dips in a measured volume of excess of \(\frac {N}{20}\) H2SO4 solution. The liquid in the round bottom flask is heated and liberated ammonia is distilled to sulphuric acid. When no more ammonia passes over (test the distillate with red litmus) the receiver is removed. The excess of acid is then determined by titration with alkali, using phenolphthalein as the indicator.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Calculation:
Weight of the substance = Wg
Volume of H2SO4 required for the complete neutralisation of evolved NH3 = V ml
Strength H2SO4 used to neutralise NH3 = N.
Let the volume and strength of NH3 formed are V1 and N1 respectively.
V1N1 = VN
The amount of nitrogen present in W g of organic compound = \(\frac{14 \times \mathrm{NV}}{1 \times 1000 \times w}}\)
Percentage of nitrogen = \(\frac{14 \times \mathrm{NV}}{1000 \times w} \times 100\) = \(\frac {1.4NV}{w}\) %

Question 9.
Explain the various steps involved in crystallization method.
Answer:
Most solid organic compounds are purified by crystallization method. This process is carried out by the following steps.
1. Selection of solvent:
rganic substances being covalent do not dissolve in water, hence selection of suitable solvent becomes necessary. Hence the powdered organic substance is taken in a test tube and the solvent is added little by little with constant stirring and heating, till the amount added is sufficient to dissolve the organic compound.

If the solid dissolves upon heating and throws out maximum crystals on cooling, then the solvent is suitable. This process is repeated with benzene, ether, acetone and alcohol various solvent. till the most suitable one is sorted out.

2. Preparation of solution:
The organic compound is dissolved in minimum quantity of suitable solvent small amount of animal charcoal can he added to decolonize any colored substance. The solution may be prepared by heating over a wire gauze or water bath.

3. Filtration of hot solution:
The hot solution so obtained is filtered through a fluted filter paper placed in a funnel.

4. Crystallization:
The hot filtrate is then allowed to cool. Most of the impurities are removed on the filter paper. the pure solid substance separate as crystal. If the rate of crystallization slow, it is induced either by scratching the walls of the beaker with a glass rod or by adding a few crystals of pure compounds to the solution.

5. Isolation and drying of crystals:
The crystals are separated from the mother liquor by filtration is done under reduced pressure using a Buchner funnel. Finally the crystals are washed with small amount of pure cold solvent and dried.

Question 10.
Explain about steam distillation (or) How is essential oils are recovered from plants and flowers.
Steam distillation:
This method is applicable for solids and liquids. If the compound to be steam distilled and it should not decompose at any steam temperature should have a fairly high vapour pressure at 273 K, it should be insoluble in water and the impurities present should be non-volatile.

The impure liquid along with little water is taken in a round bottomed flask which is connected to a boiler on one side and water condenser on the other side, the flask is kept in a slanting position so that no droplets of the mixture will enter into the condenser on the brisk boiling and bubbling of steam.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
The mixture in the flask is heated and then a current of steam passed into it. The vapours of the compound mix up with the steam and escape into the condenser. The condensate obtained is a mixture of water and organic compound which can be separated. This method is used to recover essential oils from plants and flowers also in the manufacture of aniline and turpentine oil.

Question 11.
Explain about azeotropic distillation.
Answer:
The mixture of liquids that cannot be separated by fractional distillation can be purified by azeotropic distillation. The mixture are called azeotropes. These azeotropes are constant boiling mixture which distill as a single component at a fixed temperature for example ethanol and water in the ratio of 95.87: 4.13. In this method, the presence of a third component C6H6, CCl4, ether, glycol glycerol which act as dehydrating agent depress the partial pressure of one component of azeotropic mixture and raises the boiling point of that component and thus the other component will distil over. Substance like C6H6, CCl4 have low b.pt. and reduce the partial vapour pressure of alcohol more than that of water while substance like glycerol and glycol have high boiling point and reduce the partial vapour pressure of water more than that of alcohol.

Question 12.
Explain about thin layer chromatography.
Answer:
A sheet of glass is coated with a thin layer of adsorbent (cellulose, silica gel (or) Alumina). This sheet of glass is called chromplate or thin layer chromatography plate. After drying the plate, a drop of the mixture is placed just above one edge and the plate is then placed in a closed jar containing eluant (solvent) .

The eluant is drawn up the adsorbent layer by capillary action. The components of the mixture move up along with the eluent to different distances depending upon their degree of adsorption of each component of the mixture. It is expressed in terms of its retention factor (Rf) value.
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

The spots of coloured compounds are visible on TLC plate due to their original colour. The colourless compounds are viewed under Uy light or in another method using Iodine crystals or by using appropriate reagent.

Question 13.
An organic compound contains 69% carbon and 4.9% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this compound is subjected to complete combustion.
Answer:
Step I.
Calculation of mass of CO2 produced
Mass of compound = 0.20 g
Percentage of carbon = 69 g
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Mass of CO2 formed = \(\frac{69 \times 44 \times(0.20 \mathrm{g})}{12 \times 100}\) = 0.506g

Step II.
Calculation of mass of H2O produced
Mass of compound = 0.20 g
Percentage of hydrogen = 4.8%
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
Mass of H2O formed = \(\frac{4.8 \times 18 \times(0.20 \mathrm{g})}{2 \times 100}\) = 0.0864 g

Question 14.
0.50 g of an organic compound was Kjeldahlished. The ammonia eolved was passed in 50cm3 of IN H2SO2. The residual acid required 60 cm3 of N/2 NaOH solution. Calculate the percentage of nitrogen in the compound.
Answer:
Step 1.
Calculation of volume of unused acid
Volume of NaOli solution required = 60 cm3
Normality of NaOH solution = \(\frac {1}{2}\) N
Normality of H2SO4 solution = \(\frac {1}{N}\)
Volume of unused acid can be calculated by applying normality equation
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
1 x V = \(\frac {1}{2}\) x 60 = 30cm3

Step II.
Calculation of volume of acid used
Volume of acid added = 50 cm3
Volume of unused acid = 30 cm3
Volume of acid used = (50 – 30) = 20 cm3

Step III.
Calculation of percentage of nitrogen
mass of compound = 0.05 g
Volume of acid used = 20 cm3
Normality of acid used = 1 N
Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry
percentage of nitrogen = \(\frac{1.4 \times 20 \times 1}{0.50}=56 \%\)

Question 15.
In a Dumas nitrogen estimation method, 0.30 g of an organic compound gave 50 cm3 of N2 collected at 300 K and 715 mm Hg pressure. Calculate the percentage composition of nitrogen ¡n the compound. (‘apour pressure of water at 300 K is 15 mm Hg)
Answer:
P1 = 715 – 15 = 700mm Hg, P1 = 760 mm Hg
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
T1 = 300 K, T2 = 273K, V1 = 50 cm3, V2 = ?
V2 = \(\frac{700 \times 50 \times 273}{300 \times 760}\) = 41.9 cm3
% of N = \(\frac {28}{22400}\) x 41.9 x \(\frac {100}{W}\) = 17.46%

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Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding

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Samacheer Kalvi 11th Chemistry Chapter 10 Chemical Bonding Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Chemical Bonding Multiple Choice Questions

Question 1.
In which of the following compounds does the central atom obey the octet rule?
(a) XeF4
(b) AICI3
(c) SF6
(d) SO2
Answer:
(d) SCl2
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-1
Hence in (d) SCl2 octet rule is followed

Question 2.
In the molecule OA = C = OB the formal charge on OA, C and OB are respectively.
(a) – 1, 0, +1
(b) +1, 0, – 1
(c) – 2, 0, +2
(d) 0, 0, 0
Answer:
(d) 0, 0, 0
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-2
Formal charge of Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-3
Formal charge of C = 4 – \(\left( 0+\frac { 8 }{ 2 } \right)\) = 4 – 4 = 0

Question 3.
Which of the following is electron deficient?
(a) PH3
(b) (CH3)2
(c) BH3
(d) NH3
Answer:
(c) BH3
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-3 – electron rich,
CH3 – CH3 – Covalent neutral molecule,
BH3 – electron deficient

Question 4.
Which of the following molecule contain no π bond?
(a) SO2
(b) NO2
(c) CO2
(d) H2O
Answer:
(d) H2O
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-5
Water (H2O) contains only σ bonds and no π bonds.

Question 5.
The ratio of number of sigma (σ) and pi (π) bonds in 2- butynal is …………..
(a) 8/3
(b) 5/3
(c) 8/2
(d) 9/2
Answer:
(d) 9/2
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-6
no. of σ bonds = 8 [4C – H; 3C – C; 1C— O]
no.of π bonds = 3 [2C – C; 1C – O]
∴ratio = \(\frac { 8 }{ 3 }\)

Question 6.
Which one of the following is the likely bond angles of sulphur tetrafluoride molecule?
(a) 120°, 80°
(b) 109°. 28
(c) 90°
(d) 89°, 117°
Answer:
(d) 89°, 117°
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-7
Solution:
Normal bond angle in regular trigonal bipyramidal are 90° and 120°. Due to l.p – b.p repulsion, bond angle is reduced to 89°, 117° option (d).

Question 7.
Assertion: Oxygen molecule is paramagnetic.
Reason: It has two unpaired electron in its bonding molecular orbital.
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) 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
Correct statement: Oxygen molecule is paramagnetic
Correct Reason: It has two unpaired electrons in its antibonding molecular orbital.

Question 8.
According to Valence bond theory, a bond between two atoms is formed when ……………….
(a) fully filled atomic orbitals overlap
(b) half filled atomic orbitals overlap
(c) non-bonding atomic orbitals overlap
(d) empty atomic orbitals overlap
Answer:
(b) half filled atomic orbitals overlap

Question 9.
In CIF3, NF3 and BF3 molecules the chlorine, nitrogen and boron atoms are …………………….
(a) sp3 hybridised
(b) sp3, sp3 and sp2 respectively
(c) sp2 hybridised
(d) sp3d, sp3 and sp2 hybridised respectively
Answer:
(d) sp3d, sp3 and sp2 hybridised respectively
Solution:
CIF3 – sp3d hybridisation
NF3 – sp3 hybridisation
BF3 – sp2 hybridisation

Question 10.
When one s and three p orbitais hybridise,
(a) four equivalent orbitais at 900 to each other will be formed
(b) four equivalent orbitais at 1090 28’ to each other will be formed.
(c) four equivalent orbitals, that are lying the same plane will be formed
(d) none of these
Answer:
(b) four equivalent orbitals at 109° 28′ to each other will be formed.

Question 11.
Which of these represents the correct order of their increasing bond order?
(a) C2 < C22- < O2
(b) C22- < C2+ < O2 < O22-
(c) O22- < O2+ < O2 < C22-
(d) O22- < C2+ < O2 < C22-
Answer:
(d) O22- < C2+ < O2 < C22-
Solution:
bond order = (nb – na)
bond order of O22- = \(\frac { 1 }{ 2 }\) (8 – 6) = 1
bond order of C2+ = \(\frac { 1 }{ 2 }\) (5 – 2) = 1.5
bond order of O2 = \(\frac { 1 }{ 2 }\) (8 – 4) = 2
bond order of C22- = \(\frac { 1 }{ 2 }\) (8 – 2) = 3

Question 12.
Hybridisation of central atom in PCl5 involves the mixing of orbitais.
(a) s, px, py, dx2, dx2-y2
(b) s, px, py, pxy, dx2-y2
(c) s, px, py, pz, dx2-y2
(d) px, py, pxy, dx2-y2
Answer:
(c) s, px, py, pz, dx2-y2
Solution:
PCl5 – sp3d hybridisation s, px, py, pz, dx2-y2

Question 13.
The correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is ……………..
(a) H2O2 > O3 > O2
(b) O2 > O3 > H2O
(c) O2 > H2O2 > O3
(d) O3 > O2 > H2O2
Answer:
(b) O2 > O3 > H2O2
Solution:
The bond order for O2, O3 and H2O2 decreases in the order 2 > 1.5 > 1

Question 14.
Which one of the following is diamagnetic?
(a) O2
(b) O22-
(c) O22+
(d) None of these
Answer:
(b) O22-
Solution:
O22- is diamagnetic. Additional two electrons are paired in anti-bonding molecular orbits π*2py and π*2pz

Question 15.
Bond order of a species is 2.5 and the number of electrons are in its bonding molecular orbital is found to be 8. The no. of electrons in its anti-bonding molecular orbital is ………………….
(a) three
(b) four
(c) zero
(d) cannot be calculated form the given information.
Answer:
(a) three
Solution:
Bond order = \(\frac { 1 }{ 2 }\) (nb – na)
2.5 = \(\frac { 1 }{ 2 }\) (8 – na)
⇒ 5 = 8
⇒ na = 8 – 5 = 3

Question 16.
Shape and hybridisation of IF5 are ………….
(a) Trigonal bipyramidal, sp3d2
(b) Trigonal bipyramidal, sp3d
(c) Squai2e pyramidal, sp3d2
(d) Octahedral, sp3d2
Answer:
(c) Square pyramidal, sp3d2
Solution:
IF5 – 5 bond pair + 1 lone pair
∴ hybridisation sp3d2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-10

Question 17.
Pick out the incorrect statement from the following.
(a) sp3 hybrid orbitais are equivalent and are at an angle of 1 09°28’ with each other.
(b) dsp2 hybrid orbitais are equivalent and bond angle between any two of them is 900.
(c) All five sp3d hybrid orbitais arc not equivalent. Out of these five sp3d hybrid orbitais, three are at an angle of 120°, remaining two are perpendicular to the plane containing the other three
(d) none of these
Answer:
(c) All five sp3d hybrid orbitals are not equivalent. Out of these five sp3d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three.

Question 18.
The molecules having same hybridisation, shape and number of lone pairs of electrons are ………………..
(a) SeF4, XeO2F2
(b) SF4, XeF2
(c) XeOF4, TeF4
(d) SeCI4, XeF4
Answer:
(a) SeF4, XeO2F2
Solution:
SeF4, XeO2F2 – sp3d hybridisation
T – shaped, one lone pair on central atom.

Question 19.
In which of the following molecules / ions BF3, NO2, H2O the centrai atom is sp2 hybridised?
(a) NO2 and H2O
(b) NO2 and H2O
(c) BF3 and NO2
(d) BF3 and NH2
Answer:
(c) BF3 and NO2
Solution:
H2O – Central atom sp3 hybridised
NO2 – Central atom sp2 hybridised
BF3 – Central atom sp2 hybridised
NH2  – Central atom sp3 hybridised

Question 20.
Some of the following properties of two species, NO3 and H3O+ are described below. Which one of them is correct?
(a) dissimilar in hybridisation for the central atom with different structure.
(b) isostnictural with same hybridisation for the Central atom.
(c) different hybridisation for the central atom with same structure
(d) none of these
Answer:
(a) dissimilar in hybridisation for the central atom with different structure.
Solution:
NO3 – sp2 hybridisation, planar
H3O+ – sp3 hybridisation, pyramidal

Question 21.
The types of hybridisation on the five carbon atom from right to left in the, 2,3 pentadiene.
(a) sp3, sp2, sp, sp2, sp3
(b) sp3, sp, sp, sp, sp3
(c) sp2, sp, sp2, sp , sp3
(d) sp3, sp3, sp2, sp3, sp3
Answer:
(a) sp3, sp2, sp, sp2, sp3
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-11

Question 22.
XeF2 is isostructural with ………..
(a) SbCl2
(b) BaCl2
(c) TeF2
(d) ICl2
Answer:
(d) ICl2
Solution:
XeF2 is isostructural with ICI2

Question 23.
The percentage of s-character of the hybrid orbitais in methane, ethane, ethene and ethyne are respectively ………………..
(a) 25, 25, 33.3, 50
(b) 50, 50, 33.3, 25
(c) 50, 25, 33.3, 50
(d) 50, 25, 25. 50
Answer:
(a) 25, 25, 33.3, 50
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-12

Question 24.
Of the following molecules, which have shape similar to carbon dioxide?
(a) SnCI2
(b) NO2
(c) C2H2
(d) All of these
Answer:
(c) C2H2
Solution:
CO2 – Linear
C2H2 – Linear

Question 25.
According to VSEPR theory, the repulsion between different parts of electrons obey the order …………………
(a) l.p – l.p > b.p – b.p > l.p – b.p
(b) b.p – b.p > b.p – 1.p > l.p – b.p
(c) l.p – l.p > b.p – l.p > b.p – b.p
(d) b.p – b.p > l.p – l.p > b.p – l.p
Answer:
(c) l.p – l.p > b.p – l.p > b.p – b.p

Question 26.
Shape of CIF3 is ……………………..
(a) Planar triangular
(b) Pyramidal
(c) ‘T’ Shaped
(d) none of these
Answer:
(c) ‘T’ Shaped
Solution:
dF3 – sp3d hybridisation

Question 27.
Non- Zero dipole moment is shown by …………………
(a) CO2
(b) p – dichlorobenzene
(c) carbon tetrachloride
(d) water
Answer:
(d) water
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-13

Question 28.
Which of the following conditions is not correct for resonating structures?
(a) the contributing structure must have the same number of unpaired electrons.
(b) the contributing structures should have similar energies.
(c) the resonance hybrid should have higher energy than any of the contributing structure.
(d) none of these
Answer:
(c) the resonance hybrid should have higher energy than any of the contributing structure.
Solution:
Correct statement is – the resonance hybrid should have lower energy than any of the contributing structure.

Question 29.
Among the following, the compound that contains, ionic, covalent and coordinate linkage is …………………
(a) NH4Cl
(b) NH3
(c) NaCl
(d) none of these
Answer:
(a) NH4Cl

Question 30.
CaO and NaCl have the same crystal structure and approximately the same radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is ………….
(a) U
(b) 2U
(c) U/2
(d) 4U
Answer:
(d) 4U

Samacheer Kalvi 11th Chemistry Chemical Bonding Short Answer Questions.

Question 31.
Define the following

  1. Bond order
  2. Hybridisation
  3. a- bond

Answer:
1. Bond order:
Bond orderSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-14
The number of bonds formed between the two bonded atoms in a molecule is called bond order.

2. Hybridisation:
It is a process of mixing of atomic orbitais of the same atom with. comparable energy to form equal number of new equivalent orbitais with same energy. The resultant orbitais are called hybridised orbitais and they possess maximum symmetry and definite orientation in space so as to minimise the force of repulsion between their electrons.

3. σ – bond:
When two atomic orbitais overlap linearly along the axis, the resultant bond is called sigma (σ) bond.

Question 32.
What is a pi bond?
Answer:
Pi – bond:
When two atomic orbitals overlap sideways, the resultant covalent bond is called a pi (π) bond.

Question 33.
In CH4, NH3 and H2O, the central atom undergoes sp3 hybridisation-yet their bond angles are different, why?
Answer:
1. In CH4, NH3 and H2O the central atom undergoes sp3 hybridisation. But their bond angles are different due to the presence of lone pair of electrons.

2. It can be explained by VSEPR theory. According to this theory, even though the hybridisation is same, the repulsive force between the bond pairs and lone pairs arc not same.

3. Bond pair-Bond pair < Bond pair – Lone pair < Lone pair – Lone pair So due to the varying repulsive force the bond pairs and lone pairs are distorted from regular geometry and organise themselves in such a way that repulsion will be minimum and stability will be maximum.

4. In case of CH4, there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry. i.e., tetrahedral with bond angle 109° 28’.

5. H2O has 2 bond pairs and 2 lone pairs. There is large repulsion between lp – lp. Again repulsion between lp – bp is more than that of 2 bond pairs. So 2 bonds are more restricted to form inverted V shape (or) bent shape molecule with a bond angle of 104° 35’.

6. NH3 has 3 bond pairs and 1 lone pair. There is repulsion between lp – bp. So 3 bonds are more restricted to form pyramidal shape with bond angle equal to 107° 18’.

Question 34.
Explain sp2 hybridisation in BF3
Answer:
1. sp2 hybridisation in boron trifluoride – Boron atom – B. Electronic configuration [H2]2s22p2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-15

2. In boron, the s orbital and two p orbitals in the valence shell hybridises to generate three equivalent sp2 orbitais. These 3 orbitaIs lie in the same xy plane and the angle between any two orbitals is equal to 120°.

3. The 3 sp2 hybridised orbitais of boron now overlap with the 2pz orbitais of fluorine (3 atoms). This overlap takes place along the axis.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-16
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-17

Question 35.
Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-18

  1. Electronic configuration of O atom is is 1s2 2s2 2P4
  2. Electronic configuration of O, molecule is
    Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-19
  3. Bond order = Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-20
  4. Molecule has two unpaired electrons, hence it is paramagnetic.

Question 36.
Draw MO diagram of CO and calculate its bond order.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-21

  1. Electronic configuration of C atom: 1s2 2s2 2p2
    Electronic configuration of O atom: 1s2 2s2 2p4
  2. Electronic configuration of CO molecule is: σ1s2 σ*1s2 σ2s2 σ*22 π2py2 π2pz2 σ2px2
  3. Bond order Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-22
  4. Molecule has no unpaired electron, hence it is diamagnetic.

Question 37.
What do you understand by Linear combination of atomic orbitais in MO theory.
Answer:
Linear combination of atomic orbitais (LCAO):
1. The wave functions for the molecular orbitais can be obtained by solving Schrodinger wave equation for the molecule. Since solving Schrodinger wave equation is too complex, a most common method linear combination of atomic orbitais (LCAO) is used to obtain wave function for molecular orbitals.

2. Atomic orbitais are represented by wave functions ψ. Consider two atomic orbitals represented by the wave functions ψA and ψB with comparable energy that combines to form two molecular orbitals.

3. One is bonding molecular orbitai (ψ bonding) and the other is anti-bonding molecular orbital (ψ anti-bonding).

4. The wave function for molecular orbitais, ψA and ψB can be obtained by the LCAO as shown below:
ψbonding = ψA + ψB
ψanti-bonding = ψA – ψB

5. The formation of bonding molecular orbital can be considered as the result of constructive interference of the atomic orbitais and the formation of anti-bonding molecular orbital can be the result of the destructive interference of the atomic orbitais.

6. The formation of two molecular orbitals from two is orbitals is show below.

Constructive interaction:
The two is orbitals are in phase and have the same signs.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-23

Destructive interaction:
The two is orbitals are out of phase and have opposite signs
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-24
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-25

Question 38.
Discuss the formation of N2 molecule using MO Theory.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-26

  1. Electronic configuration of N atom 1s2 2s2 2p3.
  2. Electronic configuration of N, molecule is: σ1s2 σ*1s2 σ2s2 σ*22 π2py2 π2pz2 σ2px2
  3. Bond order Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-27
  4. Molecule has no unpaired electrons hence, it is diamagnetic.

Question 39.
What is dipole moment?
Answer:
1. The polarity of a covalent bond can be measured in terms of dipole moment which is defined as: µ = q x 2d, where µ is the dipole moment, q is the charge, 2d is the distance between the two charges.

2. The dipole moment is a vector quantity and the direction of the dipole moment points from the negative charge to positive charge.

3. The unit of dipole moment is Coulomb metre (C m). It is usually expressed in Debye unit (D).

4. 1 Debye = 3.336 x 10-30 Cm

Question 40.
Linear form of carbon dioxide molecule has two polar bonds. yet the molecule has zero dipole moment, why?
Answer:

  1. The linear form of carbon dioxide has zero dipole moment, even though it has two polar bonds.
  2. In CO2, there are two polar bonds [C = O], which have dipole moments that are equal in magnitude but have opposite direction.
  3. Hence the net dipole moment of the CO2 is µ = µ1 + µ2 = µ1 +( – µ1) = 0
  4. Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-28
  5. In this case Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-187

Question 41.
Draw the Lewis structures for the following species.

  1. NO3
  2. SO42-
  3. HNO3
  4. O3

Answer:
1. NO3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-29

2. SO42-
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-30

3. HNO3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-31
4. O3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-32

Question 42.
Explain the bond formation in BeCl2 and MgCl2. BeCl2 bond formation:
Answer:
1. Electronic confiuration of Be(Z = 4) is 1s2 2s2 and electronic configuration of Cl (Z = 17) is 1s2 2s2 2p6 3s2 3p5.

2. Beryllium has 2 electrons in its valence shell and chlorine atoms (2) have 7 electrons in their valence shell.

3. By losing two electrons, Beryllium attains the inert gas configuration of Helium and becomes a dipositive cation, Be2+ and each chlorine atom accepts one electron to become (Cl) uninegative anion and attains the stable electronic configuration of Argon.

4. Then Be2+ combine with 2Cl ions to form an ionic crystal in which they are held together by electrostatic attractive forces.

5. During the formation of 1 mole of BeCl2, the amount of energy released is – 468 kJ/mol. This favours the formation of BeCl, and its stabilisation.

MgCI2 bond formation:
1. Electronic configuration of Mg (z = 12) is 1s2 2s2 2p6 3s2.
Electronic configuration of Cl (z = 17) is 1s2 2p6 3p6 3p5

2. Magnesium has 2 electrons in its valence shell and chlorine has 7 electrons in its valence shell.

3. By losing two electrons, magnesium attains the inert gas configuration of Neon and becomes a dipositive cation (Mg2+) and two chlorine atoms accept these electrons to become two uninegative anions [2Cl] by attaining the stable inert gas configuration of Argon.

4. These ions, Mg2+ and 2Cl combine to form an ionic crystal in which they are held together by electrostatic attractive forces.

5. The energy released during the formation of 1 mole of MgCl2 is – 783 kJ/mole. This favours the formation of MgCI2 and its stabilisation.

Question 43.
Which bond is stronger or π? Why?
Answer:
1. Sigma bonds (σ) are stronger than Pi bonds (π). Because, sigma bonds are formed from bonding orbitals directly between the nuclei of the bonding atoms, resulting in greater overlap and a strong sigma bond (axial overlapping).

2. π bonds results from overlap of atomic orbitals that are in contact through two areas of overlap (lateral overlapping). Pi bonds are more diffused bonds than sigma bonds.

Question 44.
Define bond energy.
Answer:
Bond energy:
Bond energy (or) Bond enthalpy is defined as the minimum amount of energy required to break one mole of a bond in molecules in their gaseous state. The unit of bond energy is kJ mol-1

Question 45.
Hydrogen gas is diatomic whereas inert gases are monoatormic – explain on the basis of MO theory.
Answer:
1. Hydrogen gas is diatomic. According to MO theory. which is based on quantum mechanics H2 molecule can be represented in terms of the following diagram called M.O. diagram.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-33
H – H. i.e.. H2 molecule has two atoms which are connected by 1 σ bond. So it is diatomic.

2. But in the case of inert gases. the valence shell is fully filled i.e.. an octet (8 electrons) (or) duplet (2 electrons) in case of Helium, due to which they are in monoatomic state and remain stable. So they do not combine with any atom (neither of same or of different elements). Due to this they do no exist in diatomic state and always exist in mono – atomic state.

Question 46.
What is Polar Covalent bond? Explain with example.
Answer:
1. If a covalent bound is formed between atoms having different electronegativities. the atom with higher electronegativity will have greater tendency to attract the shared pair of electrons towards itself than the other atom. As a result, the cloud of shared electron pair gets distorted and polar covalent bond is formed.

2. Example – HF – Hydrogen fluoride:
The electronegativities of hydrogen and fluorine on Pauling’s scale are 2.1 and 4 respectively. It means that fluorine attract the shared pair of

electrons approximately twice as much as hydrogen which leads to partial negative charge on fluorine atom and partial positive charge on hydrogen atom. Hence, the H – F bond is said to be a polar covalent bond.

Question 47.
Considering x-axis as molecular axis, which out of the following will form a sigma bond.

  1. 1s and 2py
  2. 2px and 2px
  3. 2px and 2pz
  4. 1s and 2Pz

Answer:
Along X-axis as molecular axis, only 2p and 2p can form a sigma bond
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-34

Question 48
Explain resonance with reference to carbonate ion
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-35
1. For the above structure, we can draw two additional lewis structures by moving the lone pairs from the other two oxygen atoms OB and OC. and thus creating three similar structures in which the relative positive of the atoms are same.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-36

2. They only differ in the position of bonding and lone pair of electrons. Such structures are called resonance structures and this phenomenon is called resonance.

3. It is evident from the experimental results that all carbon-oxygen bonds in carbonate ion are equivalent. The actual structure of the molecule is said to be a resonance hybrid, an average of these 3 resonance forms. The following structure gives a qualitative idea about the correct structure of CO32- (carbonate) ion.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-37

Question 49.
Explain the bond formation in ethylene and acetylene.
Answer:
Bonding in Ethylene, C2H4
1. Bonding in ethylene can he explained by hybridisation concept.

2. The valency of carbon is 4. The electronic configuration of carbon is 1s2 2s2 2px1 2py1 2pz0. One electron from 2s orbital is promoted to 2pz. orbital in the excited state to satisfy the valency of carbon.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-38

3. In ethylene both the carbon atoms undergo sp2 hybridisation involving 2s, 2px and spy orbitals resulting in 3 equivalent sp2 hybridised orbitals lying in the XY plane at an angle of 1200 to each other. The unhybridised 2pz orbital lies perpendicular to the xy plane.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-39

4. One of the sp2 hybndised orbitals of each carbon atoms lying along the X – axis linearly overlaps with each other resulting in the formation of C – C sigma bond. The other two sp2 hybridised orbitals of both carbon atom linearly overlap with the four is orbitals of four hydrogen atoms leading to the formation of two C – H sigma bonds on each carbon atom.

5. The unhybridised 2pz orbital of both carbon atoms can overlap only sideways as they are not in the molecular axis. This lateral overlap results in the formation of a pi bond between the two carbon atoms.

Bonding in acetylene (C2H2):
1. The electronic configuration of valence shell of carbon atom in the ground state is [He] 2s2 2px1 2pz0. One electron from 2s orbital is promoted to 2pz orbital in the excited state to satisfy the valency of carbon.

2. In acetylene molecule, both the carbon atoms are in sp hybridised state. The 2s and 2px orbitals resulting in two equivalent sp hybridised orbitals are formed lying in a straight line along the X – axis. The unhybridised 2py, and 2pz orbitais lie perpendicular to the X-axis.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-41
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-42
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-43

3. One of the two sp hybridised orbitals of each carbon atom linearly overlaps with each other resulting in the formation of a C – C sigma bond. The other sp hybridised orbital of both carbon atoms linearly overlap with the two is orbitals of two hydrogen atoms leading to the formation of one C – H sigma bond on each carbon atom.

4. The unhybridised 2py and 2pz orbitals of each carbon atom overlap sideways. This lateral overlap results in the formation of two pi bonds. (py – py) and (pz – pz) between the two carbon atoms.

Question 50.
What type of hybridisations are possible ¡n the following geometeries?

  1. octahedral
  2. tetrahedral
  3. square planar.

Answer:

  1. Octahedral geometry is possible by sp3d2 (or) d2sp3 hybridisation.
  2. Tetrahedral geometry is possible by sp3 hybridisation.
  3. Square planar geometry is possible by dsp2 hybridisation.

Question 51.
Explain VSEPR theory. Applying this theory to predict the shapes of F7 and SF6.
Answer:
VSEPR theory:
1. The shape of the molecules depend on the number of valence shell electron pair around the central atom.

2. There are two types of electron pairs namely, bond pairs and lone pairs.

3. Each pair of valence electrons around the central atom repel each other and hence they are located as far away as possible in three dimensional space to minimise the repulsion between them.

4. The repulsive interaction between the different types of electron pairs is in the following
order:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-44

5. The lone pair of electrons are localised only on the central atom and interact with only one nucleus whereas the bond pairs are shared between two atoms and they interact with two nuclei. Because of this, the lone pairs occupy more space and have greater repulsive power than the bond pairs in a molecule.

IF7:
It is an AB7 type molecule. This molecule has 7 bond pair of electrons and no lone pair of electrons. Due to bond pair-bond pair interaction of electrons, IF7 has pentagonal bipyramidal shape.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-45

SF6:
It is an AB6 type molecule. This molecule has 6 bond pairs of electrons and no lone pair of electrons. Due to bond pair-bond pair interaction of electrons, SF6 has octahedral shape.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-46

Question 52.
CO2 and H2O both are triatomic molecules but their dipole moment values are different. Why?
Answer:
1. Linear form of carbon dioxide has zero dipole moment. In CO2 the dipole moment of two polar bonds are equal in magiitude but have opposite direction. Hence, the net dipole moment of the CO2 molecule is
µ = µ1 + µ2
µ = µ1 + (- µ1) = 0
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-47
In this caseSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-48

2. But in the case of water, net dipole moment is the vector sum µ1 + µ2 as follows:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-49
Dipole moment in water is found to be 1.85 D.

3. CO2 and H2O both are triatomic molecules but their dipole moment values are zero and 1.85 D respectively.

Question 53.
Which one of the following has highest bond order? N2, N2+ or N2 ?
Answer:
N2 (14 electrons)
Bond order = 3, Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-50
N2+ (13 electrons)
Bond order = 2.5, Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-51
N2(15 electrons)

Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-52
So N2 has the highest bond order.

Question 54.
Explain the covalent character in ionic bond.
Answer:
1. Ionic compounds like lithium chloride shows covalent character and it is soluble in organic solvents such as ethanol.

2. The partial covalent character in ionic compounds can be explained on the basis of a phenomenon called polarisation.

3. In an ionic compound, there is an electrostatic attractive force between the cation and anion. The positively charged cation attract the valence electrons of anion while repelling the nucleus.

This cause a distortion in the electron cloud of the anion and its electron density drills towards the cation, which results in some sharing of valence electrons between these ions. Thus, a partial covalent character is developed between them. This phenomenon is called polarisation.

4. Thus due to polarisation, ionic compounds shows covalent character.

Question 55.
Describe Fajan’s rule.
Answer:
1. The ability of a cation to polarise an anion is called its polarising ability and the tendency of an anion to get polarised is called its polarisibility. The extent of polarisation in an ionic compound is given by the Fajans rule.

2. To show greater covalent character, both the cation and anion should have high charge on them. Higher the positive charge on the Cation greater will be the attraction on the electron cloud of the anion.

Similarly higher the magnitude of negative charge on anion, greater is its polansability. For example, Na+ < Mg2+ < Al3+, the covalent character also follows the order – NaCI < MgCl2 < AICI3

3. The smaller cation and larger anion show greater covalent character due to the greater extent of polarisation. e.g., LiCl is more covalent than NaCI.

4. Cation having ns2np6nd10 configuration shows greater polarising power than the cations with ns2np6 configuration. e.g., CuCI is more covalent than NaCl.

Samacheer Kalvi 11th Chemistry Chemical Bonding In Text Questions Evaluate yourself

Question 1.
Draw the lewis structures for

  1. Nitrous acid (HNO2)
  2. Phosphoric acid
  3. Sulphur troxide (SO3)

Answer:
1. Nitrous acid (HNO2)
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-53

2. Phosphoric acid
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-54

3. Sulphur troxide (SO3)
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-55

Question 2.
Calculate the formal charge on each atom of carbonyl chloride (COCl2)
Answer:
Formal charge  Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-188
Carbonyl chloride COCl2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-56
Formal charge on carbon atom = 4 – \(\left[ 0+\frac { 8 }{ 2 } \right]\) = 4 – 4 = 0
Formal charge on chlorine atom = 7 – \(\left[ 6+\frac { 2 }{ 2 } \right]\) = 7 – 7 = 0
Formal charge on oxygen atom = 6 – \(\left[ 4+\frac { 4 }{ 2 } \right]\) = 6 – 6 = 0

Question 3.
Explain the ionic bond formation in MgO and CaF2
Magnesium oxide (MgO):
Answer:
Electronic configuration of Mg – 1s2 2s2 2p5 3s2
Electronic configuration of O – 1s2 2s2 2p6 3s6 3p4.
1. Magnesium has two electrons in its valence shell and oxygen has six electrons in its valence shell.

2. By losing two electrons, Mg acquires the inert gas configuration of Neon and becomes a dipositive cation Mg2+
Mg → Mg2+ + 2e

3. Oxygen accepts the two electrons to become a dinegative oxide anion, O2- thereby attaining the inert gas configuration of Neon.
O + 2e → O2-

4. These two ions, Mg2+ and O2- combine to form an ionic crystal in which they are held together by electrostatic attractive forces.

5. During the formation of magnesium oxide crystal 601.6 kJ mol-1 energy is released . This favours the formation of magnesium oxide (MgO) and its stabilisation.

CaF2, Calcium fluoride
1.  Calcium, Ca: [Ar] 4s2, Fluorine F: [He] 2s2 2p5

2. Calcium has two electrons in its valence shell and fluorine has seven electrons in its valence shell.

3. By losing two electrons, calcium attains the inert gas configuration ofArgon and becomes a dipositive cation, Ca2+.

4. Two fluorine atoms, each one accepts one electron to become two unincgative fluoride ions (F) thereby attaining the stable configuration of Neon.

5. These three ions combine to form an ionic crystal in which they are held together by electrostatic attractive force.

6. During the formation of calcium fluoride crystal 1225.91 kJ mol-1 of energy is released. This favours the formation of calcium fluoride, CaF2 and its stabilisation.

Question 4.
Write the resonance structures for

  1. Ozone molecule
  2. N2O

Answer:
1. Ozone molecule, O3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-57

2. Nitrous oxide, N2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-58

Question 5.
Of the two molecules OCS and CS2 which one has higher dipole moment value? Why?
Answer:
OCS and CS2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-59

The dipole moment µOCS = 0.7 149 ± 0.0003 Debye.
CS2:
S = C = S
In CS2, the bond dipoles of 2C = S have same values and bond dipoles cancel each other so dipole moment of CS2 is zero. Among OCS and CS2, OCS has a higher dipole moment because in OCS oxygen is more electronegative than sulphur and C = S and C = O bonds in OCS molecules do not cancel each other.

On the other hand CS2, due to linear structure, the bond dipole of two C = S bonds cancel each other. On the other hand, CS2 due to linear structure, the bond dipoles of two C = S bonds cancel each other and thc recultant dipo’c moment value is zero. So OCS has a higher dipole moment than CS2.

Question 6.
Arrange the following in the decreasing order of Bond angle

  1. CH4, H2O, NH3
  2. C2H2, BF3, CCl4

Answer:
1. CH4, H2O, NH3:
NH3 = 107°, Water= 104.5°, CH4 = 109.5°
Decreasing order of bond angle: H2O < NH3 < CH4

2. C2H2, BF3, CCl4:
C2H2 = 1800, BF3 = 120°, CCl4 = 109.5°
Decreasing order of bond angle: CCl4 < BF3 < C2H2

Question 7.
Bond angle in PH4+ is higher than in PH3. Why?
Answer:
Phosphorous in both PH3 and PH4+ is sp3 hybridised. Due to the absence of lone pair – bond pair repulsion and presence of four identical bond pair – bond pair interactions, PH4+ assumes tetrahedral geometry with a bond angle of 109° 28’.

But PH3 has three bond pairs and one lone pair around P. Due to greater lone pair-bond pair repulsion than bond pair-bond pair repulsion, the tetrahedral angle decreases from 109° 28’ to 93.6°. As a result PH3 is pyramidal.
PH3 – Pyramidal with bond angle of 93.6°. PH4+ Tetrahedral with bond angle of 109° 28’.

Question 8.
Explain the bond formation in SF4 and CCl4 using hybridisation concept.
Answer:
In SF4, the central atom is sp3d hybridised.
1. The molecule SF3 will have a total 34 valence electron 6 form sulphur, 7 each from four fluorine atoms.

2. Sulphur atom will from 4 single bonds with fluorine atoms. These bonds account for the 8 electrons out of the 34 valence electrons. Each fluorine atom will have 3 lone pair of electrons in order to have a complete octet structure.

These lone pairs will use up 24 valence electrons. So the total used valence electrons, are 32. The remaining 2 electrons will be placed on the sulphur atom as a lone pair. Sulphur atom gets a total of 10 electrons 8 from the bonds and 2 as lone pair.

3. This is quite Sulphur atom gets a total of 10 electrons 8 from the bonds and 2 as lone pair. This is quite possible for sulphur because it has easy access to its 3d orbital which means that it can expand its octet and accommodate more than 8 electrons.

4. Sulphur forms 4 single bonds and has 1 lone pair which means that its steric number is equal to 5. In this case sulphur will use five hybrid orbitais, such as one 3s orbital three 3p orbitais and one 3d orbital. So the central atom is sp3d hybridised.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-60

CCl4:
1. It is not necessary to invoke hybridisation especially in CCl4. It must be invoked for all tetrahedral bonds of carbon and other atoms.

2. The electronic configuration of an isolated carbon atom in its ground state is 1s2 s2 2p2.

3. CCl4 is a tetrahedral molecule comprising of four single bonds known as a bonds between the carbon atom and the chlorine atoms. In this type of bonding, the 2s orbital and three 2p orbitals of carbon atoms are mixed to produce four identical orbitals, a process known as sp3 hybridisation.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-61

Question 9.
The observed bond length of N2+ is larger than N2 while the bond length in NO+ is less than in NO. Why?
Answer:
(a)
(1) By molecular orbital theory, the bond order of both N2+ is 2.5 whereas N2 is 3.

2. N2 has 5e in the antibonding molecular orbital whereas N2+ has 4e in the antibonding molecular orbital. So N2+ will make a stronger and shorter bond length.

3. More the bond order and bond strength, and lesser will be the bond length.

4. So we can easily conclude N2 has more bond length than N2
Bond order in Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-62
Bond order inSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-63
So, N2 is more stable than N2+. But bond length N2+ is greater than N2.

(b) NO+ & NO
Bond order of NO = 2.5
Bond order of NO+ = 3
Due to lesser bond order in NO, the bond length is greater than NO+ So, NO+ bond length is shorter than NO bond length.

Question 10.
Draw the MO diagram for acetylide ion C22- and calculate its bond order.
Answer:
Acetylide ion, C22- in acetylene
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-64
Electronic configuration of C2 ion is:
σ1s2 σ*1s2 σ1s2 π2px2 π2py2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-65
Bond order
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-66

Samacheer Kalvi 11th Chemistry Chemical Bonding Additional Questions Solved

Samacheer Kalvi 11th Chemistry Chemical Bonding 1 Mark Questions and Answer:

I. Choose the correct answer.

Question 1.
Which is the correct Lewis structure of Helium?
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-67
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-68
Solution:
Helium has only two electrons in its valence shell which is represented as a pair of dots (duplet).

Question 2.
Which one of the following form only covalent bonds?
(a) Alkali metals
(b) Metals
(c) Non metals
(d) Metalloids
Answer:
(c) Non metals

Question 3.
In which one of the following molecule triple bond is present?
(a) O2
(b) H2
(c) CO2
(d) N2
Answer:
(d) N2
Solution:
N ≡ N

Question 4.
Which of the following is the lewis structure of water?
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-69
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-70

Question 5.
Which one of the following element forms only one bond?
(a) Carbon
(b) Oxygen
(c) Fluorine
(d) Nitrogen
Answer:
(c) Fluorine

Question 6.
Which one is the preferred structure of CO2?
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-71
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-189

Question 7.
Which is the correct lewis structure of BF3?
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-72
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-73

Question 8.
Statement I: In sulphur hexafluoride, the central atom has more than eight valence electrons.
Statement II: The central atom can accommodate additional electron pairs by using outer vacant d orbitals.
(a) Statements I and II are correct and statement II is the correct explanation of statement I.
(b) Statements I and II are correct but statement II is not the correct explanation of statement I.
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(a) Statements I and II are correct and statement II is the correct explanation of statement I.

Question 9.
Which one of the following molecule has complete octet?
(a) BF3
(b) BeCl2
(c) BCl3
(d) CCI4
Answer:
(d) CCI4

Question 10.
Which one of the following does not have electrovalent bond?
(a) KCI
(b) NaI
(c) MgO
(d) CCI4
Answer:
(d) CCI4

Question 11.
Which one of the following has an ionic bond?
(a) CO2
(b) CH4
(c) CaF2
(d) BeCI2
Answer:
(c) CaF2

Question 12.
During the formation of 1 mole of KCI crystal, the amount of energy released is ………….
(a) 418.81 kJ
(b) 348.56 kJ
(c) 718 kJ
(d) 70.25 kJ
Answer:
(c) 718 kJ

Question 13.
Which one of the following has coordinate covalent bond?
(a) CaF2
(b) MgO
(c) [Fe(CN)6]4-
(d) KCI
Answer:
(c) [Fe(CN)6]4-

Question 14.
The distance between the nuclei of the two covalently bonded atoms is called …………….
(a) bond order
(b) bond length
(c) bond angle
(d) bond enthalpy
Answer:
(b) bond length

Question 15.
The length of a bond can be determined by ……………
(a) spectroscopic method
(b) x – ray diffraction method
(c) electron-diffraction method
(d) all the above
Answer:
(d) all the above

Question 16.
The value of carbon-carbon single bond length is ………..
(a) 1.43A
(b) 1.54Å
(c) 1.33A
(d) 1.20A
Answer:
(b) 1.54Å

Question 17.
The value of carbon – carbon double bond length is …………..
(a) 1.43Å
(b) 1.20Å
(c) 1.54A
(d) 1.33A
Answer:
(d) l.33A

Question 18.
The value of carbon – carbon triple bond length is ……………..
(a) 1.33A
(b) l.20Å
(C) 1.54A
(d) 1.43A
Answer:
(b) 1.20Å

Question 19.
Among the following which one has bond order as 3?
(a) N2
(b) O2
(c) HCHO
(d) CH4
Answer:
(a) N2

Question 20.
Which one of the flowing has bond order as 2?
(a) N2
(b) C2 – H4
(c) CH4
(d) HCN
Answer:
(b) C2H4

Question 21.
Identify the molecule with bond order 1.
(a) N2
(b) O2
(c) H2
(d) C2H4
Answer:
(c) H2

Question 22.
Which one of the following has zero dipole moment?
(a) HF
(b) H2
(c) CO
(d) NO
Answer:
(b) H2

Question 23.
Which one of the following is called polar molecule?
(a) H3
(b) O2
(c) F2
(d) NO
Answer:
(d) NO

Question 24.
Statement I: CuCl is more covalent than NaCI.
Statement II: As compared to Na+. Cu+ is small and have 3s2 3p6 3d10 configuration and show greater polarisation.
(a) Statement I & II are correct and II is the correct explanation of I
(b) Statement I & II are correct but II is not the correct explanation of I
(c) Statement I & II are correct but II is wrong
(d) Statement I & II are wrong and II is the correct.
Answer:
(a) Statement I & II are correct and II is the correct explanation of I

Question 25.
Which of the following has see saw shape?
(a) PCl5
(b) IO2F2
(c) SOF4
(d) ClO33
Answer:
(b) IO2F2

Question 26.
Which one of the following has trigonal bipyramidal shape?
(a) SF6
(b) IF4+
(c) AsF5
(d) SF4
Answer:
(c) AsF5

Question 27.
Which one of the following does not have tetrahedral shape?
(a) NH4+
(b) ClO4
(c) HCHO
(d) CH4
Answer:
(c) HCHO

Question 28.
Which one of the following has linear shape?
(a) O3
(b) CO32-
(c) NO3
(d) BCl3
Answer:
(a) O3

Question 29.
Which of the following has linear shape?
(a) PCI5
(b) SnBr2
(c) BeCl2
(d) CCl2F2
Answer:
(c) BeCl2

Question 30.
Which one of the following has tetrahedral shape?
(a) HCHO
(b) BeCl2
(c) PbCl2
(d) CF2Cl2
Answer:
(d) CF2CI2

Question 31.
Which one of the following pair has T – shapcd structure?
(a) BrF3, CIF3
(b) SF4, IF4+
(c) PCl5, AsF5
(d) NH3, PF3
Answer:
(a) BrF3, CIF3

Question 32.
Which one of the following has pentagonal bipyramidal shape?
(a) XeF4
(b) XeOF4
(c) IF7
(d) IOF5
Answer:
(c) IF7

Question 33.
Which one of the following has linear shape?
(a) I3
(b) ICI4
(c) BrF5
(d) IOF5
Answer:
(a) I3

Question 34.
Which one of the following is the correct increasing order of bond angle?
(a) H2O < CH4 < BF3 < BeCI2
(b) BeCI2 < BF3 < CH4 < H2O
(c) BF3 < CH4 < BeCI2 < H2O
(d) CH4 < BeCI2 < H2O < BF3
Answer:
(a) H2O < CH4 < BF3 < BeCI2

Question 35.
Which one of the following hybridisation takes place in the formation of BeCI2?
(a) sp2
(b) sp
(c) sp3
(d) dsp2
Answer:
(b) sp

Question 36.
Which hybridisation is possible in BF3?
(a) sp2
(b) sp
(c) sp3
(d) sp3d
Answer:
(a) sp2

Question 37.
Which one of the following has bond order as 2.5?
(a) O2
(b) NO
(c) CO
(d) H2
Answer:
(b) NO

Question 38.
Which one of the following is an electron deficient compound?
(a) Al2Cl6
(b) AlBr3
(c) SF6
(d) BF3
Answer:
(d) BF3

Question 39.
Apply the VSEPR model to XeF4, which of the following molecular shape is consistent with the model?
(a) Square planar
(b) Tetrahedral
(c) Square pyramidal
(d) Octahedral
Answer:
(a) Square planar

Question 40.
On the basis of molecular orbital theory, select the most appropriate option.
(a) The bond order of O2 is 2.5 and it is paramagnetic
(b) The bond order of O2 is 1.5 and it is paramagnetic
(c) The bond order of O2 is 2 and it is diamagnetic
(d) The bond order of O2 is 2 and it is paramagnetic
Answer:
(d) The bond order of O2 is 2 and it is paramagnetic

Question 41.
Which of the following molecule does not exist due to its zero bond order?
(a) H2
(b) He2+
(c) He2
(d) H2+
Answer:
(c) He2

Question 42.
Which of the following molecules have bond order equal to 1?
(a) NO, HF, HCl, Li2, CO
(b) H2, Li2, HF, Br2, HCI
(c) Li2, B2, CO, NO, He2+
(d) B2, CO, He2+, NO, HF
Answer:
(b) H2, Li2, HF, Br2, HCI
Solution:
Bond order of He2+ = 0.5
Bond order of NO = 2.5
Bond order of CO = 3

Question 43.
Arrange the following molecules in decreasing order of bond length.
(a) O2 >O2 > O2+ > O22-
(b) O22- > O2 > O2 > O2
(c) O22- > O2 > O2+ > O2
(d) O2+ > O2+ > O22- > O2
Answer:
(b) O22- > O2 > O2 > O2
Solution: Since the bond length is inversely proportional to the bond order so option ‘b’ is correct.

Question 44.
Among the following which shows the maximum covalent character?
(a) MgCI2
(b) FeCI2
(c) SnCI2
(d) AICI3
Answer:
(d) AICI3

Question 45.
Which of the following has maximum number of lone pairs associated with Xe?
(a) XeF2
(b) XeO3
(c) XeF4
(d) XeF6
Answer:
(a) XeF2

Question 46.
During the formation of a chemical bond …………………
(a) energy decreases
(b) energy increases
(c) energy remains zero
(d) energy remains constant
Answer:
(a) energy decreases

Question 47.
Using MO theory, predict which of the following species has the shortest bond length?
(a) O2+
(b) O2
(c) C22-
(d) O22+
Answer:
(d) O22+

Question 48.
Identify the incorrect statement regarding the molecule XeO4.
(a) XeO4 molecule is tetrahedral
(b) XeO4 molecule is square planar
(c) There are four pπ – dπ bonds
(d) There are four sp3 – p, s bonds
Answer:
(b) XeO4 molecule is square planar

Question 49.
Which of the following contains maximum number of lone pairs on the central atom?
(a) ClO3
(b) XeF4
(c) SF4
(d) I3
Answer:
(d) I3

Question 50.
Which one of the following is a correct set?
(a) H2O, sp3, bent
(b) H2O, sp2, linear
(c) NH4+, dsp2, square planar
(d) CH4+, dsp2 tetrahedal
Answwer:
(a) H2O, sp3, bent

II. Match the following

Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-76
Answer:
(b) 4 3 1 2

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-77
Answer:
(a) 3 4 1 2

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-78
Answer:
(a) 3 4 2 1

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-79
Answer:
(b) 3 4 1 2

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-80
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-80.1
Answer:
(c) 2 3 4 1

Question 6.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-81
Answer:
(b) 2 4 1 3

Question 7.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-82
Answer:
(c) 4 3 1 2

Question 8.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-83
Answer:
(d) 2 3 4 1

Question 9.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-84
Answer:
(a) 2 4 1 3

Question 10.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-85
Answer:
(a) 4 3 2 1

Question 11.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-86
Answer:
(a) 4 1 2 3

III. Fill in the blanks.

Question 1.
The electrovalent bond is present in ………..
Answer:
NaCI
Solution:
Na+ cation and Cl anion are held together by electrostatic attractive forces and this is called electrovalent bond.

Question 2.
The structure (or) shape of water molecule is …………
Answer:
inverted ‘V’ shape
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-87

Question 3.
The structure of CO2 is ………..
Answer:
linear
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-88

Question 4.
In the formation of a chemical bond between Na and C[, they attain the stable configuration of …………….
Answer:
Ne, Ar
Solution:
Na+: 1s2 2s2 2p6 = [Ne]
Cl: 1s2 2s2 3s2 3p6 = [Ar]

Question 5.
The mutual sharing of one or more pair of electrons between the two combining atoms results in the formation of …………..
Answer:
Covalent bond

Question 6.
Formal charge of an atom can be calculated by the formula ………….
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-89

Question 7.
The formal charge on the carbon atom in the following structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-90 is …………………
Answer:
zero
Solution:
Formal charge on carbon atom
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-91

Question 8.
The formal charge on both oxygen atoms in the structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-92 is …………..
Answer:
0
Solution:
Formal charge on oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-93

Question 9.
The formal charge on singly bonded oxygen atom in the structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-94 is …………..
Answer:
-1
Solution:
Formal charge on singly bonded oxygen atom
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-95

Question 10.
The formal charge on the triply bonded oxygen atom in the structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-96 is …………
Answer:
+ 1
Solution:
Formal charge on triply bonded oxygen atom
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-97

Question 11.
The complete transfer of one or more valence electron from one atom to another leads to the formation of ………….
Answer:
Ionic bond

Question 12.
The shape of the molecule is determined approximately by ……………
Answer:
bond angle

Question 13.
The unit of bond enthalpy is …………..
Answer:
kJ mol-1

Question 14.
The electronegativity of hydrogen and fluorine on Pauling’s scale are …………..
Answer:
2.1 and 4

Question 15.
The unit of dipole moment is ……………
Answer:
Coulomb-1 m2

Question 16.
The dipole moment of CO2 is ……………
Answer:
0
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-98

Question 17.
The shape of sulphur hexafluoride is …………
Answer:
Octahedral

Question 18.
The type of hybridisation takes place in methane is ………………
Answer:
sp3

Question 19.
The type of hybridisation takes place in SF6 is …………..
Answer:
sp3d2

Question 20.
The number of lone pair of electrons on C – atom present in CO2 are …………
Answer:
4

Qustion 21.
In SF6, the bond angle is …………….
Answer:
900

Question 22.
The ions have noble gas electronic configuration was suggested by ……………….
Answer:
Kossel

Question 23.
Tetrachlorornethane is a molecule …………..
Answer:
non polar

Question 24.
In C2H4, type of bonds present are ……………
Answer:
Covalent bonds only

Question 25.
Molecule with bond of shape trigonal pyramid is …………..
Answer:
BF3

Question 26.
When magnesium reacts with oxygen, nature of bond formed is …………….
Answer:
ionic

Question 27.
The number of lone pair of electrons in water molecule is …………..
Answer:
2

Question 28.
Double bonds as compared to single bonds are ……………….
Answer:
Shorter

Question 29.
Number of chlorine atoms which form equatorial bonds in PCI5 molecule are/is ………..
Answer:
3

Question 30.
The hybridisation of B in BF3 is ……………
Answer:
sp2

Question 31.
Bond order of O2, F3, N2 respectively are ……………
Answer:
2, 1, 3

Question 32.
Hybridisation which takes place in acetylene is …………….
Answer:
sp

Question 33.
Bond order of O2, F2, N2 respectively are ………….
Answer:
2, 1,3

Question 34.
Hybridisation which takes place in acetylene is …………..
Answer:
sp

Question 35.
The hybndisation of orbitais of N atom in NO3, NO3+ and NH4+ are respectively ……………….
Answer:
sp2, sp, sp3

Question 36.
Malleability and ductility of metals can be accounted due to the capacity of layers of …………………. to slide over one another.
Answer:
metal ions

Question 37.
For a stable molecule, the value of bond order must be ……………
Answer:
positive

Question 38.
In acetylene molecule between the carbon atoms there are ………….. σ and ……………. bonds.
Answer:
one, two

IV. Choose the odd one out.

Question 1.
(a) Hydrogen
(b) Chlorine
(c) Neon
(d) Argon
Answer:
(c) Neon. It is mono atomic whereas others are diatomic.

Question 2.
(a) NaCl
(b) CO2
(c) LiF
(d) MgO
Answer:
(b) CO2. It contains covalent bond whereas others have ionic bond.

Question 3.
(a) Methane
(b) Ceasium chloride
(c) Ammonia
(d) Boron trifluoride
Answer:
(b) Ceasium chloride. It is an ionic compound whereas others are covalent compounds.

Question 4.
(a) H2
(b) O2
(c) Cl2
(d) F2
Answer:
(b) O2. It’s bond order is 2 whereas in others bond order is 1.

Question 5.
(a) BeCI2
(b) CS2
(c) BF3
(d) HCN
Answer:
(c) BF3. It is AB3 type whereas others are AB2 type.

Question 6.
(a) XeO2F2
(b) PCI5
(c) AsF5
(d) SOF4
Answer:
(a) XeO2F2. It is AB4L type whereas others are AB5 type.

V. Choose the correct pair.

Question 1.
(a) NaCI – ionic compound
(b) NH3 – coordinate compound
(c) BF3 – ionic compound
(d) H2 – ionic compound
Answer:
(a) NaCl – ionic compound

Question 2.
(a) O2 – Bond order 3
(b) H2 – Bond order 2
(c) N2 – Bond order 3
(d) Cl2 – Bond order 2
Answer:
(c) N2 – Bond order 3
[N = N] Bond order is 3.

Question 3.
(a) CH4 – ionic bond
(b) BF3 – dative bond
(c) NH3 – metallic bond
(d) CCI4 – covalent bond
Answer:
(d) CCI4 – covalent bond

Question 4.
(a) CH4 – 107° 18’
(b) H2O – 109°28’
(c) NH3 – 104°35’
(d) BF3 – 120°
Answer:
(d) BF3 – 120°

Question 5.
(a) AB3 – Linear
(b) AB3 – V-shape(or)bent
(c) AB4 – Trigonal planar
(d) AB5 – T-shape
Answer:
(a) AB3 – Linear

VI. Choose the incorrect pair.

Question 1.
(a) CS2 – Linear
(b) BF1 – Trigonal planar
(c) CH4 – T-shape
(d)NH3 – Pyramidal
Answer:
(c) CH4 – T-shape

Question 2.
(a) AB3 – Trigonal planar
(b) AB3L2 – T-shape
(c) AB5 – Trigonal bipyramidal
(d) AB3L – Bent
Answer:
(a) AB3L : Bent.
Actually AB3L is pyramidal.

Question 3.
(a) AB7 – IF7
(b) AB4L2 – ICI4
(c) AB6 – XeOF4
(d) AB5L – IF5
Ans.
(c) AB6 – XeOF4
Actually XeOF4 is AB5L type.

Question 4.
(a) Fluorine – Bond order 1
(b) Oxygen – Bond order 2
(c) Nitrogen – Bond order 2
(d) Cyanide – Bond order 3
Answer:
(c) Nitrogen – Bond order 2.
Actually N = N bond order is 3.

Question 5.
(a) CH4 – sp3
(b) PCI5 – sp3d
(c) BeCl2 – sp
(d) BF3 – sp3d2
Answer:
(d) BF3 – sp3d2
Actually BF4 is sp2 hybridised.

VII. Assertion and Reason.

Question 1.
Assertion (A): Diatomic molecules such as H2, O2, F2 are non-polar molecules.
Reason (R): H2, O2, F2 have zero dipole moment.
(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): HF, HCl, CO and No are polar molecules.
Reason (R): They have non zero dipole moments and so they are polar 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 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 3.
Assertion (A): H2, Li2, C2, N2 are diamagnetic.
Reason (R): All have no unpaired electrons and so they are diamagnetic.
(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): B2, O2, NO are paramagnetic in nature.
Reason (R): They have unpaired electrons and are paramagnetic.
(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) is 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) is the correct explanation of (A).

Question 5.
Assertion (A): Metals have high thermal conductivity.
Reason (R): Absence of bond gap is the main reason for high thermal conductivity.
(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:
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).

Question 6.
Assertion (A): Metals have high thermal conductivity.
Reason (R): Due to thermal excitation of many electrons from the valence band to the conductance band, metals have high thermal conductivity.
(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).

VIII. Choose the correct statement.

Question 1.
(a) The metallic luster is due to reflection of light by the electron cloud.
(b) Metals have low inciting point and low boiling point.
(c) Metals have low thermal conductivity.
(d) Electrical conductivity of metals is low.
Answer:
(a) The metallic luster is due to reflection of light by the electron cloud.

Question 2.
(a) NO molecules is diamagnetic
(b) O2 molecules is paramagnetic
(c) N2 molecules is paramagnetic
(d) Li2 molecules is paramagnetic
Answer:
(b) O2 molecules is paramagnetic

Question 3.
(a) BeCl2 undergoes sp3 hybridisation
(b) 8F3 undergoes sp3d hybridisation
(c) CH4 undergoes sp3d2 hybridisation
(d) PCl5 undergoes sp3d hybridisation
Answer:
(d) PCl5 undergoes sp3d hybridisation

Samacheer Kalvi 11th Chemistry Chemical Bonding 2 Mark Questions and Answers

I. Write brief answer to the following questions.

Question 1.
What are chemical bonds?
Answer:
The interatomic attractive forces which holds the constituent atoms/ions together in a molecule are called chemical bonds.

Question 2.
State octet rule.
Answer:
The atoms transfer or share electrons so that all the atoms involved in chemical bonding obtain eight electrons in their outer shell (valence shell). It is called octet rule.

Question 3.
What is meant by covalent bond?
Answer:
The mutual sharing of one or more pair of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond.

Question 4.
Draw the lewis structure of

  1. H2O
  2. SO3.

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-99

Question 5.
Calculate the formal charge on the carbon atom and oxygen atom in the structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-100
Answer:
Formal charge on carbon
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-101
Formal charge on oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-102

Question 6.
Calculate the formal charge on the carbon atom and oxygen atom in the structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-103
Answer:
Formal charge on carbon
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-104
Formal charge on singly bonded oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-105
Formal charge on triply bonded oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-106

Question 7.
AmongSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-107 and Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-108 which is a preferable structure for CO2 molecule why?
Answer:
Structure I of Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-109
Formal charge on carbon
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-110
Formal charge on oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-111

Structure II of Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-112
Formal charge on carbon
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-113
Formal charge on singly bonded oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-114
Formal charge on triply bonded oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-115
A structure in which all formal charges are zero is preferred over the one with non – zero charges. In case of CO2 structure , structure I is preferred over the structure II as it has zero formal charge for all the atoms.

Question 8.
Draw the lewis structures of a few molecules containing odd electrons.
Answer:
Few molecules have a central atom with an odd number of valence electrons. For example, in nitrogen dioxide and nitric oxide all the atoms does not have octet configuration.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-116

Question 9.
Draw the lewis structure of PCl5 and SF6
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-117

Question 10.
Define bond length.
Answer:
The distance between the nuclei of two covalently bonded atoms is called bond length. For e.g., in a covalent molecule A – B. the bond length is equal to the sum of the radii of bonded atoms. i.e., rA + rß = bond length.

Question 11.
Prove that bond order is inversely proportional to bond length.
Answer:
1. Bond order ∝ Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-118

2. An example for illustrating the above equation is Carbon – carbon single bond length (I .54Å) is longer than the carbon-carbon double bond length (1 .34Å) and the carbon- carbon triple bond length (1 .20Å).

Question 12.
Define Bond angle.
Answer:
Covalent bonds are directional in nature and are oriented in specific direction in space. This directional nature creates a fixed angle between two covalent bonds in a molecule and this angle is termed as bond angle.

Question 13.
Define Resonance.
Answer:
The similar structures in which the relative position of the atoms are same but they differ in the position of bonding and lone pair of electrons are called resonance structures and this phenomenon is called resonance.

Question 14.
What are polar and non-polar molecules?
Answer:
1. Diatomic molecules such as H2, O2, F2 have zero dipole moment and are called non polar molecules.

2. Molecules such as HF, HCl, CO, NO have non zero dipole moment values and are called polar molecules.

Question 15.
What is meant by polarisaion?
Answer:
The ability of a cation to polarise an anion is called its polarising ability and the tendency of the anion to get polarised is called its polarisability. This phenomenon is known as polarisation.

Question 16.
Among NaCI, MgCI2 and AICI3 which shows more covalent character? Why?
Answer:
Among, the ionic compounds NaCI, MgCl2 and AICI3 the charge of the cation increases in the order Na+ < Mg2+ < Al3+, thus the covalent character also follows the same order NaCl < MgCI2 < AlCI3. So AICI3 shows more covalent character.

Question 17.
Lithium chloride is more covalent than sodium chloride. Justify this statement.
Answer:
1. The smaller cation and larger anion shows greater covalent character due to greater extent of polarisation.

2. The size of Li+ ion is smaller than Na+ ion and hence the polarising power of Li+ ion is more. So lithium chloride is more covalent than sodium chloride.

Question 18.
Lithium iodide is more covalent than Lithium chloride. Give reason.
Answer:
Lithium iodide is more covalent than Lithium chloride as the size of I ion is larger than Cl ion. Hence I ion will be more polarised than Cl ion by the cation Li+. So LiI is more covalent than LiCl.

Question 19.
Draw the structure of AB4L and AB3L2 type of molecules with example.
Answer:
AB4L:
Bond pairs = 4
Lone pair = 1
Shape = see saw
e.g., SF4 Strucutre

Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-119

AB3L2:
Bond pairs = 3
Lone pair = 2
Shape = T shaped
e.g., CIF3 Strucutre
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-120

Question 20.
Draw the structure of AB4L2 and AB7 type of molecules with example
Answer:
1. AB4L2:
Bond pairs = 4
Lone pairs = 2
Shape = Square planare.g., XeF4
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-121

2. AB7:
Bond pairs = 7
Lone pairs = Nil
Shape = pentagonal bipyramidal
e.g., IF7
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-122

Question 21.
Explain the bond formation of hydrogen molecule.
Answer:
1. Electronic configuration of hydrogen atom is 1s1.

2. During the formation of H2 molecule, the 1 s orbitais of two hydrogen atoms containing one unpaired electron with opposite spin overlaps with each other along the internuclear axis.

This overlap is called s – s overlap. Such axial overlap results in the formation of a sigma (σ) covalent bond.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-123

Question 22.
Explain the bond formation of fluorine molecule.
Answer:
1. Valence shell electronic confIguration of fluorine atom is 2s2 2px2 2py2 2pz1

2. When the half filled pz orbitais of two fluorine atoms overlap along the z – axis, a σ covalent bond is formed between them.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-124

Question 23.
How is HF molecule formed?
Answer:

  1. Electronic configuration of hydrogen atom is 1s1.
  2. Valence shell electronic configuration of fluorine atom is 2s2 2px2 2py2 2pz1.
  3. When half filled is orbital of hydrogen linearly overlaps with a halt filled 2pz orbital of fluorine, a σ covalent bond is formed between hydrogen and fluorine.

Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-125

Question 24.
What is meant by metallic bond?
Answer:
The forces that keep the atoms of the metal so closely in a metallic crystal constitute what is generally known as the metallic bond.

Question 25.
Why metallic bonding is referred as electronic bonding?
Answer:
1. Metallic crystals are an assemblage of positive ions immersed in a gas of free electrons. The free electrons are due to the ionisation of the valence electrons of the atom of the metal.

2. As the valence electrons of the atoms are freely shared by all the ions in the crystal, the metallic bonding is also referred to as electronic bonding.

Question 26.
Metals have high density. Give reason.
Answer:
The electrostatic attraction between the metal ions and the free electrons yields a three dimensional close packed crystal with a large number of nearest metal ions. So metals have high density.

Question 27.
Metals are ductile in nature. why?
Answer:
In the close packed structure of metallic crystal. it contains many slip planes along which movement can occur during mechanical loading, so the metal acquires ductility.

Question 28.
Give reason behind the lustrous nature, high melting point and boiling point of metals?
Answer:
1. The metallic lustre is due to the reflection of light by the electron cloud.

2. As the metallic bond is strong enough. the metal atoms are reluctant to break apart into a liquid or gas, so the metals have high melting and boiling points.

Question 29.
Metals are very good electrical conductors. Why?
Answer:
1. The bonding in metal is explained by molecular orbital theory. As per this theory. the atomic orbitals of large number of atoms in a crystal overlap to form numerous bonding and anti-bonding molecular orbitals without any band gap.

2. The bonding molecular orbitals are completely filled with an electron pair in each, and the anti-bonding molecular orbitals are empty.

3. Absence of band gap accounts for high electrical conductivity of metals.

Question 30.
Metals have high thermal conductivity. Give reason.
Answer:
High thermal conductivity of metals is due to thermal excitation of many electrons from the valence band to the conduction band.

Question 31.
Except Cu, Ag and Au, most metals are black. Why?
Answer:
Most metals are black except copper, silver and gold. It is due to the absorption of light of all wavelengths. Absorption of light of all wavelengths is due to the absence of band gap in metals.

Question 32.
Write the favourable factors for the formation of ionic bond.
Answer:

  1. Low ionisation enthalpy of metal atoms.
  2. High electron gain enthalpy of non-metal atoms.
  3. High lattice enthalpy of compound formed.

Question 33.
Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-126
Because of two lone pair of electrons on O – atom, repulsion on bond pairs is greater in H2O in comparison to NH3. Thus, the bond angle is less in H2O molecules.

Question 34.
Write the significance/applications of dipole moment.
Answer:

  1. In predicting the nature of the molecules: Molecules with specific dipole moments are polar in nature and those with zero dipole moments are non-polar in nature.
  2. In the determination of shapes of molecules.
  3. In calculating the percentage of ionic character.

Question 35.
Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.
Answer:
In CO2, there are two C = O bonds. Each C = O bond is a polar bond. The net dipole moment of CO2 moleculc is zero. This is possible only if CO2 is a linear molecule. (O = C = O). The bond dipoles of two C = O bonds cancels the dipole moment of each other.

Whereas, H2O molecule has a net dipole moment (1.84 D). H2O molecule has a bent structure because here the O – H bonds are oriented at an angle of 104.5° and do not cancel the bond dipole moments of each other.

Question 36.
What is the total number of sigma and pi bonds in the following molecules?

  1. C2H2
  2. C2H4

Answer:

1. H – C = C – H
Sigma bond = 3
π bond = 2

2.Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-127
Sigma bond = 5
π bond = 1

Question 37.
Use molecular orbital theory to explain why the Be2 molecule does not exist
Answer:
E.C of Be = 1 s2 2s2
M.O.E.C of Be2 = σ1s2 σ2s2 σ*2s2
Bond order = \(\frac { 1 }{ 2 }\) (4 – 4) = 0
Hence, Be2

Question 38.
Compare the relative stabililty of the following species and indicate their magnetic properties O2, O2+, O2 (superoxide), O22-(peroxide)
Answer:
O2 – Bond order = 2 paramagnetic
O2+ – Bond order = 2.5, paramagnetic
O2 – Bond order = 1.5, paramagnetic
O22- – Bond order = 1, diagmagnetic
Order of relative stability is:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-128

Question 39.
Account for the following:

  1. Water is a liquid while H2S is a gas
  2. NH3 has higher boiling point than PH3.

Answer:

  1. In case of water, hydrogen bonding causes association of the H2O molecules. There is no such hydrogen bonding in H2S, that is why it is a gas.
  2. There is hydrogen bonding in NH3 but not in PH3.

Question 40.
Why B2 is paramagnetic in nature while C2 is not?
Answer:
The molecular orbital electronic configuration of both B2 and C2 are:
B2:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-129
C2:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-130
Since, B2 has two unpaired electrons, therefore, B2 is paramagnetic C2 has no unpaired electron, therefore, C2 is diamagnetic.

Samacheer Kalvi 11th Chemistry Chemical Bonding 3 Mark Questions and Answers

Question 1.
Draw the lewis structure of

  1. Nitrogen
  2. Carbon
  3. Oxygen

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-131
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-132

Question 2.
Draw the lewis structure of

  1. Ammonia
  2. Methane
  3. Dinitrogen pentoxide.

Answer:
Lewis dot structures of Molecules
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-133

Question 3.
Calculate the bond enthalpy of OH bond in water.
Answer:
1. In the case of polyatomic molecules with two or more same bond types, the arithmetic mean of the bond energy value of the same type of bonds is considered as average bond enthalpy.

2. For e.g., in water, there are two OH bonds present and the energy needed to break them are not same.

3. H2O(g) → H(g) + OH(g)
∆H1 = 502 kJ mol-1
OH(g) → H(g) + O(g)
∆H = 427 kJ mol
The average bond enthalpy of OH bond in water = \(\frac { 502 + 427 }{ 2 }\) = 464.5 kJ mol-1

Question 4.
Explain how the ionic character in a covalent bond is related to electronegativity?
Answer:
1. The extent of ionic character in a covalent bond can be related to the electronegativity difference of the bonded atoms.

2. In a typical polar molecule Aδ- – Bδ+ the electronegativity difference (XA – XB) can be used to predict the percentage of the ionic character as follows

3. If the electronegativity difference XA – XB is equal to 1.7, then the bond A – B has 50% ionic character.

4. If it is greater than 1.7, then the bond XA – XB has more than 50% ionic character.

5. If it is greater than 1.7, then the bond A – B has more than 50% ionic character.

6. If it is lesser than 1.7, then the bond A – B has less than 50% ionic character.

Question 5.
CuCI is more covalent than NaCl. Give reason.
Answer:
1. Cations having ns2np6nd10 configuration show greater polansing power than the cations with ns2 np6 configuration. Hence they show greater covalent character.

2. CuCI is more covalent than NaCI. As compared to Na+ (1.13Å), Cu+(0.6Å)is small and has 3s23p63d10 configuration.

3. Electronic configuration of Cu+: [Ar] 3d10
Electronic configuration of Na+: [He] 2s22p6
So CuCI is more covalent than NaCI

Question 6.
Draw the structure of AB2, AB3, AB3L type of molecules with example.
Answer:
1. AB2
Number of bond pairs = 2
Shape = Linear
Example – BeCl2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-134

2. AB3
Number of bond pairs = 3
Shape = Tirgonal planar
Example – BF3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-135

3. AB2L
Number of bond pairs = 2
Number of lone pairs = 1
Shape = Bent (or) inverted V shape
Example – O3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-136

Question 7.
Give example and structure of

  1. AB3L
  2. AB5
  3. AB2L2

type of molecules with example.
Answer:
1. AB3L
Number of bond pairs = 3
Number of lone pairs = 1
Shape = Pyramidal
Example – NH3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-137

2. AB5
Number of bond pairs = 5
Number of lone pairs = 0
Shape = Trigonal bipyramidal
Example = PCIC5
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-138

3. AB2L2
Number of bond pairs = 2
Number of lone pairs = 2
Shape = Bent
Example – H2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-139

Question 8.
Draw the shape of

  1. XeF2
  2. IOF5
  3. XeOF4

Answer:
1. XeF2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-140

2. IOF5
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-141

3. XeOF4
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-142

Question 9.
Explain the bonding in oxygen molecule.
Answer:
1. Valence shell electronic configuration of oxygen atom is
2s2 2px2 2py12pz1
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-143

2. When the half filled pz orbitaIs of two oxygen atoms overlap along the z – axis a σ covalent bond is formed between them. Other two hail filled py orbitais of two oxygen atoms overlap laterally to form a π – cova1ent bond between the oxygen atoms.

3. Thus in oxygen molecule, two oxygen atoms are connected by two covalent bonds (double bond). The other two pair of electrons present in the 2s and 2px orbital do not involve in bonding and remains as lone pair on the respective oxygen atoms.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-144

Question 10.
Explain about the molecular orbital diagram of hydrogen molecule.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-145

  1. Electronic configuration of H atom 1s1
  2. Electronic configuration of H, molecule – σ1s1
    Bond order Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-146
  3. Molecule (H2) has no unpaired electrons, hence it is diamagnetic.

Question 11.
Draw and explain the M.O. diagram of lithium molecule.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-147

  1. Electronic configuration of Li atom – 1s1
  2. Electronic configuration of Li2 molecule is ais2 σ*1s2 σ*1s2 σs2
  3. Bondorder = Nb – Nb/2 = 4 – 2/2
  4. Li2 molecule has no unpaired electrons, hence it is diamagnetic.

Question 12.
Draw and explain the M.O. diagram of Boron molecule.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-148
Answer:

  1. Electronic configuration of B = 1s2 2s2 2p3
  2. Electronic configuration of B, = σ1s2 σ*1s2σ2s2 σ*2s2 π2px1 π 2pz1
  3. Bond order Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-149
  4. B2 molecule has two unpaired electrons hence it is paramagnetic.

Question 13.
Draw and explain the molecular orbital diagram of carbon molecule.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-150

  1. Electronic configuration of C atom – 1s2 2s2 2p2
  2. Electronic configuration of C2 molecule is σ1s2 σ*1s2 σ*2s2 σ*2s2 π 2px2 π 2py2
  3. Bond orderSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-151

Question 14.
Write Lewis dot symbols for atoms of the following elements: Mgq Naq B O, N, Br.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-152

Question 15.
write Lewis symbols for the following atoms and ions: S and S2-; Al and Al3+; H and OH
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-153

Question 16.
Draw the Lewis structures for the following molecules and ions H2S, SiCl4, BeF2, CO32-, HCOOH
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-154

Question 17.
Define Octet rule. Write its significance and limitations.
Answer:
Octet rule:
Atoms of elements combine with each other in order to complete their respective octet so as to acquire the stable nearest noble gas configuration.

Significance:
It helps to explain why dilfferent atoms combine with each other to form ionic compounds or covalent compounds.

Limitations of Octet rule:
1. According to octet rule, atoms take part in chemical combination to achieve the configuration of nearest noble gas elements.

However, some of noble gas elements like Xenon have formed compounds with fluorine and oxygen. For example: XeF2, XeF4, XeO3 etc. Therefore, validity of the octet rule has been challenged.

2. This theory does not account for the shapes of molecules.

Question 18.
Write the resonance structure for SO3, NO2 and NO3
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-155

Question 19.
What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Answer:
The electron pair involved in sharing between two atoms during covalent bonding is called shared pair or bond pair. At the same time, the electron pair which is not involved in sharing is called lone pair of electrons.
For example, in Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-156
there are only 4 bond pairs, but in Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-157 there are two bond pairs and two lone pairs.

Question 20.
Distinguish between a sigma bond and a pi bond
Answer:
Sigma (σ) Bond

  1. σ – bond is formed by the axial overlap of the atomic orbitais.
  2. The bond is quite strong.
  3. Only one lobe ofthep-orbitals is involved in the overlap.
  4. Electron cloud of the molecular orbital is symmetrkal around the internuclear axis.

Pi (π) Bond

  1. π – bonnd is formed by the sidewise overlap of atomic orbitais.
  2. It is comparatively a weaker bond.
  3. Both lobes of the p-orbitais are involved in the overlap.
  4. The electron cloud is not symmetrical.

Question 21.
Write the Important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer:
1. The combining atomic orbitals should have comparable energies. For example, is orbital of one atom can combine with 1s atomic orbital of another atom, 2s orbitai can combine with 2s orbital and so on.

2. The combining atomic orbitals must have proper orientations so that they are able to overlap to a considerable extent.

3. The extent of overlapping should be large.

Question 22.
What are Lewis structures? Write the Lewis structure of H2, BeF2 and H2O.
Answer:
The outer shell electrons are shown as dots surrounding the symbol of the atom. These symbols are known as Lewis symbols or Lewis structures. The Lewis structure of
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-158

Question 23.
What are the main postulates of Valence Shell Electron Pair Repulsion (VSEPR) theory?
Answer:

  1. The shape of a molecule depends upon the no. of electron pairs around the central atom.
  2. There is a repulsive force between the electron pairs, which tend to repel one another.
  3. The electron pairs in space tend to occupy such positions that they arc at maximum distance, so that the repulsive force will be minimum.
  4. A multiple bond is treated as lilt is a single bond and the remaining electron pairs which constitute the bond may be regarded as single super pair.

Question 24.
Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with four H atoms at the corners of the square and C atom at its centre. Explain why CH4 is not square planar?
Answer:
Electronic configuration of carbon atom: C: σ1s22s22p2.
in the excited state, the orbital picture of carbon can be represented as:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-159
Hence, carbon atom undergoes sp3 hybridisation in CH4 molecule and takes tetrahedral shape.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-160
For a square planar shape, the hybridisation of the central atom has to be dsp3. However, an atom of carbon does not have d – orbitals to undergo dsp3 hybridisation. Hence, the structure of CH4 is tetrahedral.

Question 25.
Explain why BeH2 molecule has a zero dipole moment although the Be – H bonds are polar.
Answer:
The Lewis structure for BeH2 molecule is as follows:Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-161. There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, BeH2, is of the type AB2. It has a linear structure.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-162
Dipole moments of each Be – H bond are equal and opposite in direction. Therefore, they nullify each other. Hence, BeH2 has a net zero dipole moment.

IV. Answer the following questions in detail:

Question 1.
Explain about Kossel-Lewis approach to chemical bonding.
Answer:
1. Kossel and Lewis approach to chemical bonding is based on the inertness of the noble gases which have little or no tendency to combine with other atoms.

2. They proposed that noble gases are stable due to their completely filled outer electronic configuration.

3. Elements other than noble gases try to attain the completely filled outer electronic configuration by losing, gaining or sharing one or more electrons from their outer shell.

4. For e.g., sodium loses one electron to form Na ion and chlorine accepts that electron to give chloride ion, Cl.These two ions are held together by electrostatic attractive forces, a bond known as an electrovalent bond.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-163

5. In diatomic molecules such as nitrogen and oxygen, they achieve the stable noble gas electronic configuration by mutual sharing of electrons.

6. Lewis introduced a scheme to represent the chemical bond and the electrons present in the outer shell of the atom called Lewis dot structure.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-164

7. For example, the electronic configuration of nitrogen is 1s22s22p3. It has 5 electrons in its outer shell. The lewis structure of nitrogen isSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-165

8. In N, molecule, equal sharing of 3 electrons from each nitrogen atom takes place as fol lows
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-166

Question 2.
What is meant by covalent bond?
Explain the covalent bonding in H2, O2, N2.
Answer:
1. Mutual sharing of one or more pair of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond.

2. If two atoms share just one pair of electron, a single covalent bond is formed as in the çase of hydrogen molecule (H2).

3. If two or three electron pairs are shared benveen the two combining atoms, then the covalent bond is called double bond and triple bond respectively, as in the case of O2 and N2 molecules respectively.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-167

Question 3.
What is an ionic bond?
Explain about the formation of ionic bond with a suitable example.
Answer:
1. The complete transfer of electrons leads to the fomiation of a cation and an anion. Both these ions are held together by electrostatic attractive forces which is known as ionic bond.

2. KCl: Potassium chloride
Electronic configuration of K [Ar] 4s
Eleçtronic configuration of Cl = [Ne] 3s2 3p5

3. Potassium has 1 electron in its valence shell and chlorine has 7 electrons in its valence shell.

4. By losing one electron potassium attains the nearest inert gas configuration of Argon and becomes a unipositive cation (K) and chlorine accepts this electron to become uninegative chloride ion (CI) to attain the stable configuration of nearest noble gas, Argon.

5. These two ions combine to form an ionic crystal in which they are held together by electrostatic attractive forces.

6. During the formation of one mole of potassium chloride crystal from its constituent ions, 718 kJ of energy is released. This favours the formation of KCl and its stabilisation.

Question 4.
Define coordinate covalent bond. Illustrate the formation of coordinate covalent bond with a suitable example.
Answer:
1. In the bond formation, one of the combining atoms donates a pair of electrons i.e., two electrons which are necessary for the covalent bond formation and these electrons are shared by both the combining atoms, and the bond formed is called coordinate covalent bond.

2. The combining atom which donates the pair of electron is called the donor atom and the other atom is called the acceptor atom. This bond is denoted by an arrow starting from the donor atom pointing towards the acceptor atom.

3. For example, in ferricyanide ion [Fe(CN)6]4- each cyanide ion (CN) donates a pair of electrons to form a coordinate bond with iron (Fe2+) and these electrons are shared by Fe2+ and CN ions.

4. Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-168

5. Ammonia having a lone pair of electrons donates its pair to an electron deficient molecule such as BF3 to form a coordinate covalent bond
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-169

Question 5.
Explain about valence bound theory for the formation of H2 molecule.
Answer:
1. Two hydrogen atoms Ha and Hb are separated by infinite distance. At this stage, there is no interaction between these two atoms and the potential energy of this system is arbitrarly taken as zero.

2. As these two atoms approach each other, in addition to electrostatic attractive forces between the nucleus and its own electrons, the following new forces begins to operate
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-170

3. The new attractive forces Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-171 arise between:

  • nucleus of Ha and valence electron of Hb
  • nucleus of Hb and the valence electron of Ha

4. The new repulsive forces Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-172 arise between:

  • the nucleus Of Ha and Hb
  • the valence electrons of Ha and Hb

5. The attractive forces tend to bring Ha and Hb together whereas the repulsive forces tends to push them apart.

6. At the initial stage, as the two hydrogen atoms approach each other, the attractive forces are stronger than repulsive forces and the potential energy decreases.

7. A stage is reached where the net attractive forces are exactly balanced by repulsive forces and the potential energy of the system acquires a minimum energy.

8. At this stage, there is a maximum overlap between the atomic orbitals of Ha and Hb and atoms Ha and Hb are now said to be bonded together by a covalent bond.

Question 6.
What arc the salient features of Valence Bond (VB) theory?
Answer:
1. When half filled orbitals of two atoms overlap, a covalent bond will be formed between them.

2. The resultant overlapping orbitals are occupied by the two electrons with opposite spins. For example when H2 is formed, the two is electron of two hydrogen atoms get paired up and occupy the overlapped orbitals.

3. The strength of a covalent bond depends upon the extent of overlap of atomic orbitals. Greater the overlap, larger is the energy released and stronger will be the bond formed.

4. Each atomic orbital has a specific direction (excepts-orbital which is spherical) and hence orbital overlap takes place in the direction that maximises overlap.

5. Depending upon the nature of overlap, the bonds are classified as σ covalent bond and π it covalent bond.

6. When two atomic orbitals overlap linearly along the axis, the resultant bend is called a sigma (σ) bond. This overlap is also called or “axial overlap”.

7. When two atomic orbitals overlap sideways the resultant covalent bond is called a pi (π) bond.

Question 7.
Explain about sp hybridisation with suitable example.
Answer:

  1. bond rormation in Beryllium chloride takes place by sp hybridisation.
  2. The valence shell of Beryllium has the electronic configuration as follows:

Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-173

3. In BeCl2, both the Be – Cl bonds are equivalent and it was observed that the molecule is linear. VB theory explains this observed behaviour by sp hybridisation. One of the paired electrons in the 2s orbital gets excited to 2p orbital.

4. Now the 2s and 2p orbitals hybridise and produce two equivalent sp hybridised orbitals which have 50% s-character and 50% p-character. These sp hybridised orbitals are oriented in opposite direction.

5. Each of the sp hybridised orbitals linearly overlap with p orbital of the chlorine to form a covalent bond between Be and Cl atoms as follow:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-174

Question 8.
Explain the formation of methane using VB theory?
Answer:
1. Methane is formed by sp3 hybridisation. In CH4 molecule, the central carbon atom is bounded to four hydrogen atoms.

2. The ground state valence shell electronic configuration of carbon is [He] 2s22px2 2py1 2px0

3. En order to form four covalent bonds with the four hydrogen atoms, one of the paired electrons in 2s orbital of carbon is promoted to its 2Pz orbital in the excited state.

4. The one 2s orbital and three 2p orbitals of carbon atom mixes to give four equivalent sp3 hybridised orbitals. The angle between any of the two sp3 hybridised orbitals is 109°•28’
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-175

5. The Is orbital of the four hydrogen atoms overlap linearly with the four sp3 hybridised orbitais of carbon to form four C – H σ bonds in the methane molecule as follows
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-176

Question 9.
Explain sp3d hybridisation with a suitable example.
Answer:
1. In the PCl5 molecule, the central atom phosphorous is covalently bonded to five chlorine atoms. Here the atomic orbitals of phosphorous undergoes sp3d2 hybridisation which involves its one 3s orbital, three 3p orbitals and one vacant 3d orbital (dz2)

2. The ground state electronic configuration of phosphorous is [Ne] 3s23px23py13pz1
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-177

3. One of the paired electrons in the 3s orbital of phosphorous ¡s promoted to one of its vacant 3d orbital (dz2) in the excited state.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-178

4. The 3pz orbitals of the five chlorine atoms linearly overlap along the axis with the five sp3d hybridised orbitals of phosphorous to form the five P – CI bonds as follows.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-179

Question 10.
Explain about. sp3d2 hybridisation with an example.
Answer:
1. In sulphur hexafluoride SF6, the central atom sulphur extend its octet to undergo sp3d hybridisation to generate six sp3d2 hybridised orbitals which accounts for six equivalent S – F bonds.

2. The ground state electronic configuration of sulphur is [Ne] 3s23px13py13pz1
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-180

3. One electron each form 3s orbital and 3p orbital of sulphur is promoted to its two vacant 3d orbitals dz2 and dx2-y2 in the excited state.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-181

4. A total of six valence orbitals from sulphur (one 3s orbital, three 3p orbitals and two 3d orbitals) (dx2 and dx2-y2) which mixes to give six equivalent sp3d2 hybridised orbitals. The orbital geometry is octahedral.

5. The six sp3d2 hybridised orbitals of sulphur overlaps linearly with 2pz orbital of six fluorine atoms to form the six S – F bonds in sulphur hexa fluoride.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-182

Question 11.
Explain about the salient features of molecular orbital theory.
Answer:
1. When atoms combine to form molecules, their individual atomic orbitals lose their identity and form new orbitals called molecular orbitals.

2. The shape of molecular orbitals depend upon the shapes of combining atomic orbitals.

3. The number of molecular orbitals formed is the same as the number of combining atomic orbitals. half the number of molecular orbitals formed will have lower energy and are called bonding orbitals, while the remaining half molecular orbitals will have higher energy and are called anti-bonding molecular orbitals.

4. The bonding molecular orbitals are represented as σ (sigma), π (pi), δ (delta) and the corresponding anti-bonding orbitals are called σ*, π* and δ*.

5. The electrons in the molecule are accommodated in the newly formed molecular orbitals. The filling of electrons in these orbitals follow Autbau’s Principle, Pauli’s exclusion principle and Hund’s rule as in the case of filling of electrons in the atomic orbitals.

6. Bond order gives the number of covalent bonds between the two combining atoms. The bond order of a molecule can be calculated using
the foLlowing equation:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-183
Nb = Number of electrons in bonding molecular orbitals.
Na = Number of electrons in anti-bonding molecular orbitals.

7. A bond order of zero value indicates that the molecule does not exist.

Question 12.
Explain the MO diagram for NO molecule.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-184
Answer:

  1. Electronic configuration of N atom is 1s2 2s2 2p3
  2. Electronic configuration of O atom is 1s2 2s2 2p4
  3. Electronic configuration of NO molecule is
    Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-185
  4. Bond order Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-186
  5. NO molecule has one unpaired electron, hence it is paramagnetic.

Question 13.
Explain about metallic bonding.
Answer:
1. The forces that keep the atoms of the metal so closely in a metallic crystal constitute what is known as metallic bond.

2. According to Drude and Lorentz, metallic crystal is an assemblage of positive ions immersed in a gas of free electrons. The free electrons are due to ionisation of the valence electrons of the atoms of the metal.

3. As the valence electrons of the atoms are freely shared by all the ions in the crystal, the metallic bonding is referred to as electronic bonding.

4. The electrostatic attraction between the metal ions and the free electrons yield a three dimensional close packed crystal with a large number of nearest metal ions. So metals have high density.

5. As the close packed structure contains many slip planes along which movement can occur during mechanical loading, metal acquires ductility.

6. As metal ion is surrounded by electron cloud in all directions, the metallic bonding has no and thermal conductivity. The metallic lustre is due to the reflection of light by the electron cloud.As the metallic bond is strong enough, the metal atoms are reluctant to break apart into a liquid or gas, so the metals have high melting and boiling points.

7. High thermal conductivity of metals is due. to thermal excitation of many electrons from the valence bond to the conduction band.

Question 14.
Explain about the bonding in metals by molecular orbital theory.
Answer:
1. According to molecular orbital theory the atomic orbitals of large number of atoms in a crystal overlap to form numerous bonding and anti-bonding molecular orbitals without any band gap.

2. The bonding molecular orbitals are completely filled with an electron pair in each and the anti-bonding molecular orbitals are empty.

3. Absence of band gap accounts for high electrical conductivity of metals.

4. High thermal conductivity is due to thermal excitation of many electrons from the valence band to the conduction band.

5. With an increase in temperature, the electrical conductivity decreases due to vigorous thermal motion of lattice ions that disrupts the uniform lattice structure. that is required for free motion of electrons within the crystal.

Common Errors

  1. The number of bonds formed by elements may go wrong.
  2. When writing Lewis structure, electrons may be written in an irregular way.
  3. Coordinate covalent bond should not be written as a line

Rectifications

  1. Always hydrogen and fluorine form 1 bond Oxygen 2 bonds, Nitrogen 3 bonds, Carbon 4 bonds
  2. When writing lewis structure, each atom should be surrounded by eight electrons in such a way as 4 pairs of electrons.
  3. Coordinate covalent bond should be written from donor atom to acceptor atom as an arrow mark Donor→ Acceptor

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