Laxmi

Students can Download Maths Chapter 3 Algebra Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.2

Question 1.
Fill in the blanks.

(i) Unit digit of 124 × 36 × 980 is ______
(ii) When the unit digit of the base and its expanded form of that number is 9, then the exponent must be _______ power.
Answers:
(i) 0
(ii) Odd

Question 2.
Match the following:

Answer:
(i) – c
(ii) – d
(iii) – b
(iv) – f
(v) – a
(vi) – e

Question 3.
Find the unit digit of expanded form.
(i) 25²³
(ii) 11¹⁰
(iii) 46¹⁵
(iv) 100¹²
(v) 29²¹
(vi) 19¹²
(vii) 24²⁵
(viii) 34¹⁶
Solution:
(i) 25²³
Unit digit of base 25 is 5 and power is 23. Thus the unit digit of 25²³ is 5.

(ii) 11¹⁰
Unit digit of base 11 is 1 and power is 10. Thus the unit digit of 11¹⁰ is 1.

(iii) 46¹⁵
Unit digit of base 46 is 6 and power is 15. Thus the unit digit of 46¹⁵ is 6.

(iv) 100¹²
Unit digit of base 100 is 0 and power is 12. Thus the unit digit of 100¹² is 0.

(v) 29²¹
Unit digit of base 29 is 9 and power is 21 (odd power).
Therefore, unit digit of 29²¹ is 9.

(vi) 19¹²
Unit digit of base 19 is 9 and power is 12 (even power).
Therefore, unit digit of 19¹² is 1.

(vii) 24²⁵
Unit digit of base 24 is 4 and power is 25 (odd power).
Therefore, unit digit of 24²⁵ is 4.

(viii) 34¹⁶
Unit digit of base 34 is 4 and power is 16 (even power).
Therefore, unit digit of 34¹⁶ is 6.

Question 4.
Find the unit digit of the following numeric expressions.
(i) 114²⁰ + 115²¹ + 116²²
(ii) 10000¹⁰⁰⁰⁰ + 11111¹¹¹¹¹
Solution:
(i) 114²⁰ + 115²¹ + 116²²
In 114²⁰ unit digit of base 114 is 4 and power is 20 (even power).
∴ Unit digit of 114²⁰ is 6.
In 115²¹ unit digit of base 115 is 5 and power is 21 (Positive Integer).
∴ Unit digit of 115²¹ is 5.
In 116²² unit digit of base 116 is 6 and power is 22 (Positive Integer).
∴ Unit digit of 116²² is 6.
∴ Unit digit of 114²⁰ + 115²¹ + 116²² can be obtained by adding 6 + 5 + 6 = 17.
Unit digit of 114²⁰ + 115²¹ + 116²² is 7.

(ii) 10000¹⁰⁰⁰⁰ + 11111¹¹¹¹¹
In 10000¹⁰⁰⁰⁰ the unit digit of base 10000 is 0 and power is 10000.
Unit digit of 10000¹⁰⁰⁰⁰ is 0.
In 11111¹¹¹¹¹ the unit digit of base 11111 is 1 and power is 11111.
Unit digit of 11111¹¹¹¹¹ is 1.
Unit digit of 10000¹⁰⁰⁰⁰⁰ + 11111¹¹¹¹¹ is 0 + 1 = 1

Objective Type Question

Question 5.
Observe the equation (10 + y)⁴ = 50625 and find the value of y.
(i) 1
(ii) 5
(iii) 4
(iv) 0
Answer:
(ii) 5

Question 6.
The unit digit of (32 × 65)⁰ is
(i) 2
(ii) 5
(iii) 0
(iv) 1
Answer:
(iv) 1

Question 7.
The unit digit of the numeric expression 10⁷¹ + 10⁷² + 10⁷³ is
(i) 0
(ii) 3
(iii) 1
(iv) 2
Answer:
(i) 0

Students can Download Science Term 3 Chapter 7 Crop Production and Management Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Science Solutions Term 3 Chapter 7 Crop Production and Management

Samacheer Kalvi 8th Science Crop Production and Management Text Book Exercise

I. Choose the best answer:

8th Science Crop Production And Management Question 1.
The process of placing seeds in the soil is called as ………………..
(a) ploughing
(b) sowing
(c) crop production
(d) crop rotation
Answer:
(b) sowing

8th Standard Science Crop Production And Management Question 2.
Organism that control insects and pests of plant crops is …………………
(a) bio-pesticides
(b) bio-fertilizers
(c) earthworms
(d) neem leaves
Answer:
(a) bio-pesticides

8th Crop Production And Management Question 3.
The method in which water flows over the soil surface and allow it to infiltrate is ………………..
(a) irrigation
(b) surface irrigation
(c) springier irrigation
(d) drip irrigation
Answer:
(b) surface irrigation

Crop Production And Management Class 8 Question 4.
Effective microorganisms preparation is not used in ………………..
(a) seed treatment
(b) foliar spray
(c) soil treatment
(d) bio-predators
Answer:
(a) seed treatment

8th Standard Crop Production And Management Question 5.
Which of the following is not present in Panchakavya?
(a) cow dung
(b) cow’s urine
(c) curd
(d) sugar
Answer:
(d) sugar

II. Fill in the blanks:

1. The process of actively growing seedling from one place and planting in the main field for further growth is called ………….
2. …………. is a plant growing where it is not wanted.
3. The chemicals used for killing the weeds or inhibiting their growth are called as ………….
4. …………. seed transfer its unique characteristics to the descents.
5. …………. centers serve as the ultimate link between ICAR and farmers.
6. Several popular high yielding varieties of major crops have been developed by ………….

Answer:

1. Transplantation
2. Weed
3. herbicides
4. Heirloom seeds
5. Krishi Vigyon Kendra
6. IARi (Indian Agricultural Research Institute)

III. Match the following:

1. Bio-pesticide – (a) Neem Leaves
2. Bio-predators – (b) Bacillus thuringiensis
3. Bio-fertilizer – (c) Control white flies
4. Bio-indicators – (d) Improve soil fertility
5. Bio-repellants – (e) Quality of environment

Answer:

1. b
2. c
3. d
4. e
5. a

IV. Answer briefly:

Crop Production And Management Answers Question 1.
Define ploughing.
Answer:
Ploughing or tilling is the process of loosening and turning the soil up and down to facilitate the availability of nutrients in the root zone of that cultivating crop.

Samacheer Kalvi 8th Science Guide Question 2.
Name the methods of sowing.
Answer:
The different methods of sowing are

1. Sowing by hand
2. Seed drill
3. Dibbling

Samacheer Kalvi Guru 8th Science Question 3.
What is foliar spray?
Answer:

1. Foliar feeding is a technique of feeding plants by applying liquid fertilizer directly to their leaves.
2. Plants are able to absorb essential elements through the stomata in their leaves. Give a brief account on Krishi Vigyon Kendra.

Samacheer Kalvi Guru 8th Question 4.
Give a brief account on Krishi Vigyon Kendra.
Answer:

1. Krishi Vigyon Kendra is a farm science centre.
2. This centre serve as a ultimate link between ICAR and farmers.
3. They operate small farms to test new technologies.
4. They also provide advice to farmers about weather and pricing of crops.

Samacheer Kalvi Guru 8 Science Question 5.
What is bio-indicator? How does it help human beings?
Answer:

1. A bio-indicator or biological indicator is any species or group of species whose function or status reveals the qualitative status of the environment.
2. Biological indicators are used to document and understand changes in Earth’s living systems especially changes caused by the activities of an expanding human population.

Question 6.
What do you mean by weeding?
Answer:
The undesirable plants growing naturally with crop plants are called weeds. The removal of weeds is called weeding.

Question 7.
What is crop rotation?
Answer:
Crop rotation is planting a series of different crops in the same field following a defined order. This helps to maintain fertility of the soil.

Question 8.
What is green manure?
Answer:

1. Some plants like Sun hemp or guar are grown in the field prior to the sowing of the crop seeds.
2. These plants gradually decompose and turn into green manure which helps in ensuring the soil in nitrogen and phosphorous.
3. Application of green manure always enhance the growth and yield of the crops.

V. Answer in detail:

Question 1.
Explain the agricultural practices.
Answer:

1. Soil preparation: (LoosenIng of top soil)

• Ploughing – Process of loosening and turning the soil up and down to facilitate nutrient availability.
• levelling – Done with leveller and helps in uniform distribution of water for irrigation
• Basal Nianuring – Increase soil fertility by manuring

2. Sowing of seeds:

• Hand – Scattering of seeds by hand.
• Seed Drill – Sowing seeds by iron drills attached to a tractor.
• Dibbling – Placing seed in furrow or pits or hole by hand.

3. Adding manure and fertilisers:

• Manure is a substance added to the soil in the form of Nutrients to enhance plant growth.
• Organic Sources – Plant and animal waste
• Synthetic Sources – Urea, Super phosphate.

4. Irrigation:
Supply of water to crops:

• Traditional method – Irrigation done manually.
• Modem method – Involve two systems. Sprinkler system Drip system

5. Harvesting of crops:

• Manual harvesting – Harvested without tools. Eg – Groundnut crop, green gram, black gram, house gram.
• Mechanical method – Harvesting by instruments small sized farms Eg – Sickle.
• Machine harvesting – Used for large sized farms.

6. Threshing and Winnowing:
Process of separating the grains from their chaffs or pods.

7. Storage:

• Supply of the produce has to be stored properly.
• Grains must be free from moisture to avoid growth of microorganism.
• Need to be dried in Sun before storing.
• Collected in Gunny bags and stored in go downs.

Question 2.
Give a detailed account on irrigation.
Answer:
Irrigation:

1. The supply of water to crops at regular intervals is called irrigation.
2. Source of irrigation – Wells, tube wells, ponds, lakes, rivers, dams and canal.

Traditional Methods:

1. Irrigation is done manually.
2. Here, a farmer pulls out water from wells or canals by himself or using cattle and carries to farming fields.
3. Pumps used for lifting water from various sources.
4. Diesel, biogas, electricity and solar energy are the sources of energy needed to run these pumps.

Modern Methods:

1. It helps to overcome the problems exist in the traditional methods.
2. It also facilitates the even distribution of moisture in the field.

Sprinkler system:

1. Sprinkles water over the crop and helps in an even distribution of water.
2. This method is advisable in areas facing water scarcity.
3. Pump is connected to pipes which generate pressure and water is sprinkled through the fine nozzles of pipes.

Drip System:

1. Here, water is released drop by drop exactly at root zone using a hose or pipe.
2. This method is effective one in regions where the availability of water is less.

Question 3.
What is weed? Explain the different methods of weed control.
Answer:

Weed:
The undesirable plants may grow naturally along with the main crop, and these undesirable plants are called weeds.

Weeding:
The removal of weeds is called weeding. It is an important process because weeds compete with the crop plants for the nutrients, sunlight, water, space and other resources. It results in the undernourishment of crops and it leads to low yield.

Mechanical methods:

1. Here, weeds are destroyed physically.
2. Hand pulling or weeding with the help of weeding hole is the oldest and most efficient method for controlling weeds.

Tillage methods:

1. It is the practical methods of destroying weeds of all categories.
2. Weeds are buried in the soil and also exposed to Sun heat by deep ploughing.

Crop Rotation:
Proper rotation of crops is followed for controlling crop associated and parasitic weeds.

Summer tillage:
Deep ploughing after harvest of Rabi crop and exposing underground parts of weeds to strong sunlight during summer months is useful for destroying many annual and perennial weeds.

Biological weed control:

1. Bio agents like insects and pathogens are used to control weeds.
2. The objectives are not eradication, but reduction and regulation of the weed population.

Chemical methods:

1. Very effective in certain cases and have great scope in weed control.
2. The chemicals used for killing the weeds or inhibiting their growth are called herbicides.
3. Chemicals are mixed with water and sprayed over the crops.

Integrated weed management:

1. Integrated weed management combines different agronomic practices and herbicides use to manage weeds, so that the reliance on any one weed control technique is reduced.
2. Mechanical, biological, cultural and chemical methods are included in integrated weed managements.

Samacheer Kalvi 8th Science Crop Production and Management Additional Questions

I. Choose the correct answer:

Question 1.
…………… is a ornamental crop.
(a) Hemp
(b) Euphorbia
(c) Sorghum
(d) sesame
Answer:
(b) Euphorbia

Question 2.
Lichen is a ……………
(a) Bio-pesticide
(b) Bio-indicator
(c) Bio-predator
(d) Bio-fervilizer
Answer:
(b) Bio-indicator

Question 3.
…………… is a bio predator.
(a) Trichoderma
(b) Aphids
(C) Chrysopa spp
(d) Earthworm
Answer:
(c) Chrysopa spp

Question 4.
Neem is a good ……………
(a) Bio-fertilizer
(b) green manure
(c) insect repellant
(d) Bio predator
Answer:
(c) insect repellant

Question 5.
Royal Botanical garden is located in ……………
(a) Chennai
(b) Mumbai
(c) New Delhi is a method of
(d) Kolkatta
Answer:
(d) Kolkatta

Question 6.
…………… is a method of sowing seeds.
(a) Tillage
(b) Winnowing
(c) Weeding
(d) Dibbling
Answer:
(d) Dibbling

II. Fill in the blanks:

1. Crops sown in rainy season are called ……………
2. The summer crops are also called …………… crops.
3. …………… is an example of Rabi crop.
4. India is the largest producer of …………… in the world.
5. …………… is a fodder crop.
6. Placing a seed in a pit or furrow is called ……………
7. The process of separating the grains from their chaffs is called ……………
8. Heirloom seeds are also called …………… seeds.
9. …………… is commonly called Pusa institute.
10. The first KVK was established in 1974 in ……………
11. Vermiwash is used as a …………… for crops.
12. …………… can be used for seed treatment.
13. …………… is the main source of bio-fertilizer.

Answer:

1. Kharif
2. zaid
3. Wheat
4. Banana/ Mangoes
5. Sorghum
6. Dibbling
7. Threshing
8. Organic
9. IARI
10. Pondicherry
11. foliar spray
12. Pachgavya
13. Cyanobacteria

III. Match the following:

Question 1.

1. Paddy – (a) Fodder
2. Muskmelon – (b) Oil crop
3. Millet – (c) Zaid crop
4. Sesame – (d) Food crop

Answer:

1. d
2. c
3. a
4. b

Question 2.

1. Nostoc – (a) Pest
2. Bacillus – (b) Cyanobacteria
3. Cotton bollworm – (c) Legume
4. Rhizobium – (d) Bio-pesticide

Answer:

1. b
2. d
3. a
4. c

IV. Answer briefly:

Question 1.
How are crops classified based an seasons?
Answer:

1. Kharif crops – Sown in rainy season eg. Paddy.
2. Rabi crops – Grown in winter season eg. Wheat.
3. Zaid Crops – Grown in summer season eg. Muskmelon.

Question 2.
What is dibbling?
Answer:
It is a method of sowing seeds in which the seed material is placed in a furrow, pit or hole , spacing with a dibble and covering the same with soil.

Question 3.
Mention the difference between manure and fertilizer.
Answer:
Manure:

1. It is organic in origin
2. It does not cause pollution

Fertiliser:

1. It is synthetic in origin
2. It causes water and air pollution

Question 4.
Define threshing.
Answer:
The process of separating the grains from their chaffs or pods is called threshing.

Question 5.
Expand the following abbreviations. FCI, ICAR, KVK, IARI
Answer:

1. FCI – Food Corporation of India.
2. ICAR – Indian Council of Agricultural Research
3. KVK – Krishi Vigyon Kendra
4. IARI – Indian Agricultural Research Institute.

Question 6.
What are seed balls?
Answer:

1. Seed balls are a mixture of soil, compost and plant seeds.
2. These balls are thrown into land areas and is a step towards conserving the natural ecosystems.

Question 7.
What are heirloom seeds?
Answer:

1. An heirloom seed is the seed of plant that has been carefully cultivated and passed down through many generations.
2. Heirloom seed are harvested, dried and stored so that one can replant them in the following season.

Question 8.
What is bio-control?
Answer:

1. Bio-control or biological control is a method of controlling pests such as insects, mites, weed and plant diseases using other organisms.
2. Eg – Bio-predators, bio-pesticides, bio-repellents and bio-fertilizers.

Question 9.
What are bio-pesticides?
Answer:

1. Bio-pesticides are living organism or their derived parts which are used as bio-control agents to protect crops against insect pests.
2. Eg. Entomopathegenic viruses.

Question 10.
What are bio-fertilizer?
Answer:

1. Bio-fertilizers are organisms which bring about soil nutrient enrichment.
2. Nitrogen fixing microorganisms have the capability of converting free nitrogen into nitrogenous compounds and make the soil fertile.
3. Eg – Nostoc.

Question 11.
What is Panchgavya?
Answer:

1. A growth promoter with a combination of five products obtained from the cow, includes cow dung, cow’s urine, milk, curd and ghee.
2. It is used in seed treatment.

Question 12.
What is a seed bank?
Answer:
Seed bank is a place where seeds are stored in order to preserve genetic diversity.

Are you searching for the Samacheer Kalvi 12th Chemistry Chapter Wise Solutions PDF? Then, get your Samacheer Kalvi 12th Chapter Wise Solutions PDF for free on our website. Students can Download Chemistry Chapter 4 Transition and Inner Transition Elements Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements

All concepts are explained in an easy way on our website. So, students can easily learn and practice Tamilnadu State Board 12th Chemistry Chapter 4 Transition and Inner Transition Elements Question and Answers. You can enjoy the digitized learning with the help of the Tamilnadu State Board Chemistry Online Material.

Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements Text Book Evalution

I. Choose the correct answer.

12th Chemistry Chapter 4 Book Back Answers Questions 1.
Sc ( Z=21) is a transition element but Zinc (z=30) is not because ……………..
(a) both Sc³⁺ and Zn²⁺ ions are colourless and form white compounds.
(b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled
(c) last electron as assumed to be added to 4s level in case of zinc
(d) both Sc and Zn do not exhibit variable oxidation states
Answer:
(c) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled

12th Chemistry Unit 4 Book Back Answers Question 2.
Which of the following d block element has half filled penultimate d sub shell as well as half filled valence sub shell?
(a) Cr
(b) Pd
(c) Pt
(d) none of these
Answer:
(a) Cr
Hint: Cr ⇒ [Ar]3d⁵ 4s¹

12th Chemistry Lesson 4 Book Back Answers Question 3.
Among the transition metals of 3d series, the one that has highest negative (M²⁺/ M) standard electrode potential is ……………..
(a) Ti
(b) Cu
(c) Mn
(d) Zn
Answer:
(a) Ti

12th Chemistry Chapter 4 Question 4.
Which one of the following ions has the same number of unpaired electrons as present in V³⁺?
(a) Ti³⁺
(b) Fe³⁺
(c) Ni²⁺
(d) Cr³⁺
Answer:
(c) Ni²⁺

Transition And Inner Transition Elements Class 12 Question 5.
The magnetic moment of Mn²⁺ ion is ……………..
(a) 5.92BM
(b) 2.80BM
(c) 8.95BM
(d) 3.90BM
Answer:
(a) 5.92BM
Hint: Mn²⁺ ⇒ 3d contains 5 unpaired electrons
n = 5
= \(\sqrt { n(n+2) } BM\)
= \(\sqrt { 5(5+2) }\) = \(\sqrt { 35 }\) = 5.92 BM

Samacheer Kalvi 12th Chemistry Question 6.
Which of the following compounds is colourless?
(a) Fe³⁺
(6) Ti⁴⁺
(c) CO²⁺
(d) Ni²
Answer:
(b) Ti⁴⁺
Hint: Ti⁴⁺ contains no unpaired electrons in d orbital, hence no d-d transition.

12th Chemistry Samacheer Kalvi Question 7.
The catalytic behaviour of transition metals and their compounds is ascribed mainly due to ……………..
(a) their magnetic behaviour
(b) their unfilled d orbitals
(c) their ability to adopt variable oxidation states
(d) their chemical reactivity
Answer:
(c) their ability to adopt variable oxidation states

Samacheer Kalvi 12th Chemistry Solutions Question 8.
The correct order of increasing oxidizing power in the series ……………..
(a) \({ VO }_{ 2 }^{ + }\) < \({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\) < \(Mn{ O }_{ 4 }^{ – }\)
(b) \({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\) < \({ VO }_{ 2 }^{ + }\) < \(Mn{ O }_{ 4 }^{ – }\)
(c) \({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\) < \(Mn{ O }_{ 4 }^{ – }\) < \({ VO }_{ 2 }^{ + }\)
(d) \(Mn{ O }_{ 4 }^{ – }\) < \({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\) < \({ VO }_{ 2 }^{ + }\)
Answer:
(a) \({ VO }_{ 2 }^{ + }\) < \({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\) < \(Mn{ O }_{ 4 }^{ – }\)
Hint:  greater the oxidation state, higher is the oxidising power.

12th Chemistry Solutions Samacheer Kalvi  Question 9.
The alloy of copper that contain Zinc is ……………..
(a) Monel metal
(b) Bronze
(c) bell metal
(d) brass
Answer:
(d) brass

Class 12 Chemistry Samacheer Kalvi Question 10.
Which of the following does not give oxygen on heating?
(a) K₂Cr₂O₇
(b) (NH₄)₂Cr₂O₇
(c) KClO₃
(d) Zn(ClO₃)₂
Answer:
(b) (NH₄)₂Cr₂O₇
Hint:

12 Chemistry Samacheer Kalvi Question 11.
In acid medium, potassium permanganate oxidizes oxalic acid to ……………..
(a) Oxalate
(b) Carbon dioxide
(c) acetate
(d) acetic acid
Answer:
(b) Carbon dioxide
Hint: \({ 5(COO) }_{ 2 }^{ 2- }\) + \({ 2MnO }_{ 4 }^{ – }\) + 16H⁺ \(\underrightarrow { \triangle }\) 2Mn²⁺ + 10CO²⁺\(\uparrow\) + 8H²O

12th Chemistry 4th Lesson Question 12.
Which of the following statements is not true?
(a) on passing H₂S, through acidified K₂Cr₂O₇ solution, a milky colour is observed.
(b) Na₂Cr₂O₇ is preferred over K₂Cr₂O₇ in volumetric analysis
(c) K₂Cr₂O₇ solution in acidic medium is orange in colour
(d) K₂Cr₂O₇ solution becomes yellow on increasing the pH beyond 7
Answer:
(b) Na₂Cr₂O₇ is preferred over K₂Cr₂O₇ in volumetric analysis

Transition And Inner Transition Elements Class 12 Exercise Question 13.
Permanganate ion changes to in acidic medium ……………..
(a) \({ MnO }_{ 4 }^{ 2- }\)
(b) Mn²⁺
(c) Mn³⁺
(d) MnO₂
Answer:
(b) Mn²⁺
Hint: \({ MnO }_{ 4 }^{ – }\) + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O

Samacheer Kalvi Chemistry 12th Question 14.
A white crystalline salt (A) react with dilute HCl to liberate a suffocating gas (B) and also forms a yellow precipitate . The gas (B) turns potassium dichromate acidified with dil H₂SO₄ to a green coloured solution(C). A,B and C are respectively ……………..
(a) Na₂SO₃, SO₂, Cr₂(SO₄)₃
(b) Na₂S₂O₃, SO₂, Cr₂(SO₄)₃
(c) Na₂S, SO₂, Cr₂(SO₄)₃
(d) Na₂SO₄, SO₂, Cr₂(SO₄)₃
Answer:
(a) Na₂SO₃, SO₂, Cr₂(SO₄)₃

Class 12 Chemistry Chapter 4 Notes Question 15.
\({ MnO }_{ 4 }^{ – }\) react with Br in alkaline P^H to give ……………..
(a) \({ BrO }_{ 3 }^{ – }\), MnO₂
(b) Br₃, \({ MnO }_{ 4 }^{ 2- }\)
(c) Br₃, MnO₃
(d) BrO^–, \({ MnO }_{ 4 }^{ 2- }\)
Answer:
(a) \({ BrO }_{ 3 }^{ – }\), MnO₂
Hint: \({ 2MnO }_{ 4 }^{ – }\) + Br- + H₂O → 2OH- + 2MnO₂ + \({ BrO }_{ 3 }^{ – }\)

Class 12 Chemistry Chapter 4 Question 16.
How many moles of I₂ are liberated when 1 mole of potassium dichromate react with potassium iodide?
(a) 1
(b) 2
(c) 3
(d) 4
Answer
(c) 3
Hint: K₂Cr₂O₇ + 6KI + 7H₂SO₄ → 4K₂SO₄ + Cr₂ (SO₄)₃ + 7H₂O + 3I₂

Chapter 4 Chemistry Class 12 Notes Question 17.
The number of moles of acidified KMnO₄ required to oxidize 1 mole of ferrous oxalate(FeC₂O₄) is …………..
(a) 5
(b) 3
(c) 0.6
(d) 1.5
Answer:
(c) 0.6

Question 18.
When a brown compound of Mn (A) ids treated with HCl, it gives a gas (B). The gas (B) taken in excess reacts with NH₃ to give an explosive compound (C). The compound A, B and C are ……………
(a) MnO₂, Cl₂, NCl₃
(b) MnO, Cl₂, NH₄Cl
(c) Mn₃O₄, Cl₂, NCl₃
(d) MnO₃, Cl₂, NCl₂
Answer:
(a) MnO₂, Cl₂, NCl₃

Question 19.
Which one of the following statements related to lanthanons is incorrect?
(a) Europium shows +2 oxidation state.
(b) The basicity decreases as the ionic radius decreases from Pr to Lu.
(c) All the lanthanons are much more reactive than aluminium.
(d) Ce⁴⁺ solutions are widely used as oxidising agents in volumetric analysis.
Answer:
(c) All the lanthanons are much more reactive than aluminium.
Hint: As we move from La to Lu , their metallic behaviour because almost similar to that of aluminium.

Question 20.
Which of the following lanthanoid ions is diamagnetic?
(a) Eu²⁺
(b) Yb²⁺
(c) Ce²⁺
(d) Sm²⁺
Answer:
(6) Yb²⁺
Hint: Yb²⁺ – 4f¹⁴ – no unpaired electrons – diamagnetic

Question 21.
Which of the following oxidation states is most common among the lanthanoids?
(a) 4
(b) 2
(c) 5
(d) 3
Answer:
(d) 3

Question 22.
Assertion: Ce⁴⁺ is used as an oxidizing agent in volumetric analysis.
Reason: Ce⁴⁺ has the tendency of attaining +3 oxidation state.
(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:
(a) Both assertion and reason are true and reason is the correct explanation of assertion.

Question 23.
The most common oxidation state of actinoids is ……………
(a) +2
(b) +3
(c) +4
(d) +6
Answer:
(c) +4

Question 24.
The actinoid elements which show the highest oxidation state of +7 are ……………
(a) Np, Pu, Am
(b) U, Fm, Th
(c) U, Th, Md
(d) Es, No, Lr
Answer:
(a) Np, Pu, Am

Question 25.
Which one of the following is not correct?
(a) La(OH)₂ is less basic than Lu(OH)₃
(b) In lanthanoid series ionic radius of Ln³⁺ ions decreases
(c) La is actually an element of transition metal series rather than lanthanide series
(d) Atomic radii of Zr and Hf are same because of lanthanide contraction
Answer:
(a) La(OH)₂ is less basic than Lu(OH)₃

II. Answer the following Questions:

Question 1.
What are transition metals? Give four examples.
Answer:

1. Transition metal is an element whose atom has an incomplete d sub shell or which can give rise to cations with an incomplete d sub shell.
2. They occupy the central position of the periodic table, between s-block and p-block elements.
3. Their properties are transitional between highly reactive metals of s-block and elements of p-block which are mostly non metals.
4. Example – Iron, Copper, Tungsten, Titanium.

Question 2.
Explain the oxidation states of 3d series elements.
Answer:
1. The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by losing electrons from (n-l)d orbital and ns orbital as the energy difference between them is very small.

2. At the beginning of the 3d series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.

3. The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases. For example, in 3d series, first element Sc has only one oxidation state +3; the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.

4. Mn²⁺ (3d⁵) is more stable than Mn⁴⁺ (3d³) is due to half filled stable configuration.

Question 3.
What are inner transition elements?
Answer:
1. The inner transition elements have two series of elements.

• Lanthanoids
• Actinoids

2. Lanthanoid series consists of 14 elements from Cerium (₅₈Ce) to Lutetium (₇₁Lu) following Lanthanum (₅₇La). These elements are characterised by the preferential filling of 4f orbitals. .

3. Actinoids consists of 14 elements from Thorium (₉₀Th) to Lawrencium (₁₀₃Lr) following Actinium (₈₉Ac). These elements are characterised by the preferential filling of 5f orbital.

Question 4.
Justify the position of lanthanides and actinides in the periodic table.
Answer:
1. In sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these 14 elements following lanthanum show similar chemical properties. Therefore these elements are grouped together and placed at the bottom of the periodic table. This position can be justified as follows.

• Lanthanoids have general electronic configuration [Xe] 4f^2-14 5d^0-1 6s²
• The common oxidation state of lanthanoids is +3
• All these elements have similar physical and chemical properties.

2. Similarly the fourteen elements following actinium resemble in their physical and chemical properties.

3. If we place these elements after Lanthanum in the periodic table below 4d series and actinides below 5d series, the properties of the elements belongs to a group would be different and it would affect the proper structure of the periodic table.

4. Hence a separate position is provided to the inner transition elements at the bottom of the periodic table.

Question 5.
What are actinoides? Give three examples.
Answer:

1. The fourteen elements following actinium ,i.e., from thorium (Th) to lawrentium (Lr) are called actinoids.
2. All the actinoids are radioactive and most of them have short half lives.
3. The heavier members being extremely unstable and not of natural occurrence. They are produced synthetically by the artificial transformation of naturally occuring elements by nuclear reactions.
4. Example – Thorium, Uranium, Plutonium, Californium.

Question 6.
Why Gd³⁺ is colourless?
Answer:
Gd – Electronic Configuration: [Xe] 4f⁷ 5d¹ 6s²
Gd³⁺ – Electronic Configuration: [Xe] 4f⁷
In Gd³⁺, no electrons are there in outer d-orbitals. d-d transition is not possible. So it is colourless.

Question 7.
Explain why compounds of Cu²⁺ are coloured but those of Zn²⁺ are colourless.
Answer:
Cu (Z = 29) Electronic configuration is [Ar] 3d¹⁰ 4s¹
Cu²⁺: Electronic configuration is [Ar] 3d⁹.
In Cu²⁺, promotion of electrons take place in outer d-orbital by the absorption of light form visible region involves d-d transition. Due to this Cu²⁺ compounds are coloured. Where in Zn²⁺ electronic configuration is [Ar]3d¹⁰. It has completely filled d-orbital. So there is no chance of d – d transition. So Zn²⁺ compounds are colourless.

Question 8.
Describe the preparation of potassium dichromate.
Answer:
Preparation of potassium dichromate:
1. Potassium dichromate is prepared from chromite ore. The ore FeO. Cr₂O₃ is concentrated by gravity separation process.

2. The concentrated ore is mixed with excess sodium carbonate and lime and roasted in a reverbratory furnace.
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ \(\underrightarrow { 900-{ 1000 }^{ 0 }C }\) 8 Na₂CrO₄ + 2Fe₂O₃ + 8CO₂\(\uparrow\)

3. The roasted mass is treated with water to separate soluble sodium chromate from insoluble iron oxide. The yellow solution of sodium chromate is treated with concentrated sulphuric acid which converts sodium chromate into sodium dichromate.

4. The above solution is concentrated to remove less soluble sodium sulphate. The resulting solution is filtered and concentrated. It is cooled to get the crystals of Na₂SO₂.2H₂O.

5. The saturated solution of sodium dichromate in water is mixed with KCl and then concentrated to get crystals of NaCl. It is filtered while hot and the filtrate is cooled to obtain K₂Cr₂O₇ crystals.

Question 9.
What is lanthanide contraction and what are the effects of lanthanide contraction?
Answer:
As we move across 4f series, the atomic and ionic radii of lanthanoids show gradual decrease with increase in atomic number. This decrease in ionic size is called lanthanoid contraction.
Effects (or) Consequences of lanthanoid contraction:
1. Basicity differences: As we move from Ce³⁺ to Lu³⁺ , the basic character of Ln³⁺ ions decrease. Due to the decrease in the size of Ln³⁺ ions, the ionic character of Ln – OH bond decreases (covalent character increases) which results in the decrease in the basicity.

2. Similarities among lanthanoids – In the complete f-series only 10 pm decrease in atomic radii and 20 pm decrease in ionic radii is observed. Because of this very small change in radii of lanthanoids, their chemical properties are quite similar.

The elements of second and third transition series resemble each other more closely than the elements of first and second transition series due to lanthanoid contraction. For example,

• 4d series – Zr – Atomic radius 145 pm
• 5d series – Hf – Atomic radius 144 pm

Question 10.
Complete the following
Answer:

Question 11.
What are interstitial compounds?
Answer:

1. An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice.
2. They are usually non-stoichiometric compounds.
3. Transition metals form a number of interstitial compounds such as TiC, ZrH_1.92, Mn₄N etc.
4. The elements that occupy the metal lattice provide them new properties.
   • They are hard and show electrical and thermal conductivity
   • They have high melting points higher than those of pure metals
   • Transition metal hydrides are used as powerful reducing agents
   • Metallic carbides are chemically inert.

Question 12.
Calculate the number of unpaired electrons in Ti³⁺, Mn²⁺ and calculate the spin only magnetic moment.
Answer:
Ti³⁺:
Ti (Z = 22). Electronic configuration [Ar] 3d² 4s²
Ti³⁺ – Electronic configuration [Ar] 3d¹
So, the number of unpaired electrons in Ti³⁺ is equal to 1.
Spin only magnetic moment of Ti³⁺ = \(\sqrt { 1(1+2) }\) = \(\sqrt { 3 }\) = 1.73µ_B

Mn²⁺:
Mn (Z = 25). Electronic configuration [Ar] 3d⁵ 4s²
Mn²⁺ – Electronic configuration [Ar] 3d⁵
Mn²⁺ has 5 unpaired electrons.
Spin only magnetic moment of Mn²⁺ = \(\sqrt { 5(5+2) }\) = \(\sqrt { 35 }\) = 5.91µ_B

Question 13.
Write the electronic configuration of Ce⁴⁺ and CO²⁺.
Answer:
Ce (Z = 58) → Ce⁴+4e^–
Ce⁴⁺ – Is² – 2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶ 5s²4d¹⁰ 5p⁶
CO²⁺ – Is²2s²2p⁶3s²3p⁶4s²3d⁵.

Question 14.
Explain briefly how +2 states becomes more and more stable in the first half of the first row transition elements with increasing atomic number.
First transition series.
Answer:

Question 15.
Which is more stable? Fe³⁺ or Fe²⁺ – explain.
Answer:
Fe (Z = 26)
Fe → Fe²⁺ + 2e^–
Fe → Fe³⁺ + 3e^–
Fe²⁺ [Number of electrons 24]
Electronic configuration = [Ar]3d⁶
Fe³⁺ [Number of electrons 23]
Electronic configuration = [Ar]3d⁵
Among Fe³⁺ and Fe²⁺, Fe³⁺ is more stable due to half filled d-orbital. This can be explained by Aufbau principle. Half filled and completely filled d-orbitals are more stable than partially filled d-orbitals. So Fe³⁺ is more stable than Fe²⁺.

Question 16.
Explain the variation in E⁰_{M²⁺/M³⁺}3d series.
Answer:
1. In transition series, as we move down from Ti to Zn, the standard reduction potential E⁰_M²⁺/M³ value is approaching towards less negative value and copper has a positive reduction potential, i.e. elemental copper is more stable than Cu²⁺.

2. E⁰_M²⁺/M value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d⁵ configuration in Mn²⁺ and completely filled d¹⁰ configuration in Zn²⁺.

3. The standard electrode potential for the M³⁺ / M²⁺ half cell gives the relative stability between M³⁺ and M²⁺.

4. The high reduction potential of Mn³⁺ / Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺.

5. For Fe³⁺ / Fe²⁺ the reduction potential is 0.77 V, and this low value indicates that both Fe³⁺ and Fe²⁺ can exist under normal condition.

6. Mn³⁺ has a 3d² configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d sub-shell makes the reduction of Mn³⁺ very feasible [E° = +1.51 V]

Question 17.
Compare lanthanides and actinides.
Answer:
Lanthanoids:

1. Differentiating electron enters in 4f orbital
2. Binding energy of 4f orbitals are higher
3. They show less tendency to form complexes
4. Most of the lanthanoids are colourless
5. They do not form oxo cations
6. Besides +3 oxidation states lanthanoids show +2 and +4 oxidation states in few cases.

Actinoids:

1. Differentiating electron enters in 5f orbital
2. Binding energy of 5f orbitals are lower
3. They show greater tendency to form complexes
4. ost of the actinoids are coloured.
   E.g : U³⁺ (red), U⁴⁺ (green), \({ UO }_{ 2 }^{ 2+ }\)(yellow)
5. They do form oxo cations such as \({ UO }_{ 2 }^{ 2+ }\) \({ NpO }_{ 2 }^{ 2+ }\) etc.
6. Besides +3 oxidation states actinoids show higher oxidation states such as +4, +5, +6 and +7.

Question 18.
Explain why Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidizing.
Answer:
Cr²⁺ is strong reducing while Mn³⁺ is strongly oxidising.
E⁰_{Cr³⁺/Cr²⁺} is -0.41 V
Cr²⁺ + 2e^– → Cr E° = – 0.91V.
If the standard electrode potential E° of a metal is large and negative, the metal is a powerful reducing agent because it loses electrons easily.
Mn → Mn³⁺ + 3e^–
Mn³⁺ + e^– → Mn²⁺

Mn³⁺ [Ar]3d⁴
E° = + 1.51 V
If the standard electrode potential E° of a metal is large and positive, the metal is a powerful oxidising agent because it gains electrons easily.

Question 19.
Compare the ionization enthalpies of first series of the transition elements. Ionization enthalpies of first transition series:
Answer:

1. Ionization energy of transition element is intermediate between those of s and p block elements.
2. As we move from left to right in a transition metal series, the ionization enthalpy increases as expected. This is due to increase in nuclear charge corresponding to the filling of d electrons.
3. The increase in first ionisation enthalpy with increase in atomic number along a particular series is not regular. The added electron enters (n-l)d orbital and the inner electrons act as a shield and decrease the effect of nuclear charge on valence ‘ns’ electrons. Therefore, it leads to variation in the ionization energy values.

Question 20.
Actinoid contraction is greater from element to element than the lanthanoid contraction, why?
Answer:

1. Actinoid contraction is greater from element to element than lanthanoid contraction. The 5f orbitals in Actinoids have a very poorer shielding effect than 4f orbitals in lanthanoids.
2. Thus, the effective nuclear charge experienced by electron in valence shells in case of actinoids is much more than that experienced by lanthanoids.
3. In actinoids, electrons are shielded by 5d, 4f, 4d and 3d whereas in lanthanoids, electrons are shielded by 4d, 4f only.
4. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.

Question 21.
Out of LU(OH)₃ and La(OH)₃ which is more basic and why?
Answer:

1. As we move from Ce³⁺ to Lu³⁺, the basic character of Lu³⁺ ions decreases.
2. Due to the decrease in the size of Lu³⁺ ions, the ionic character of Lu – OH bond decreases, covalent character increases which results in the decrease in the basicity.
3. Hence, La(OH)³ is more basic than Lu(OH)³.

Question 22.
Why europium (II) is more stable than Cerium (II)?
Answer:
Eu (Z = 63) – Electronic configuration – [Xe] 4f⁷ 5d° 6s²
Eu²⁺ – Electronic configuration Electronic 6s¹
Ce (Z = 58) – configuration – [Xe] 4f²⁺ 5d° 6s²⁺
Ce²⁺ – Electronic confluration – [Xe] 4f² 5d°
According to Aufbau principle, half filled and completely filled d (or) f orbitals are more stable than partially filled f orbitals.
Hence Eu²⁺ [Xe] 4f⁷ 5d° is more stable than Ce²⁺ [Xe] 4f² 5d°

Question 23.
Why do zirconium and Hafnium exhibit similar properties?
Answer:

1. The element of second and third transition series resemble each other more closely than the elements of first and second transition series due to lanthanoid contraction.
2. e.g., Zr – 4d series -Atomic radius 145 pm Hf – 5d series – Atomic radius 144 pm
3. The radii are very similar even though the number of electrons increases.
4. Zr and Hf have very similar chemical behaviour, having closely similar radii and electronic configuration.
5. Radius dependent properties such as lattice energy, solvation energy are similar.
6. Thus lanthanides contraction leads to formation of pair of elements and those known as chemical twins, e.g., Zr – Hf

Question 24.
Which is stronger reducing agent Cr²⁺ or Fe²⁺ ?
Answer:
Cr²⁺ and Fe²⁺
Cr (Z = 24) – Electronic configuration – [Ar] 3d⁵ 4s¹
Cr²⁺ Electronic configuration – [Ar] 3d⁴ 4s⁰
Fe (Z = 26) – Electronic configuration – [Ar] 3d⁶ 4s²
Fe²⁺ Electronic confimiration – [Ar] 3d⁶ 4s⁰

If standard electrode potential (E°) of a metal is large and negative, the metal is a powerful reducing agent

Cr²⁺ + 2e^– → Cr
Fe²⁺+2e^– → Fe
E° = – 0.91V
E°= – 0.44V

By comparing the above equation, Cr²⁺ is a powerful reducing agent.

Question 25.
The E⁰_M²⁺/M value for copper is positive. Suggest a possible reason for this.
Answer:

1. Copper has a positive reduction potential. Elemental copper is more stable than Cu²⁺.
2. Copper having positive sign for electrode potential merely means that copper can undergo reduction at faster rate than reduction of hydrogen.
3. The electron giving reaction (oxidation) of copper is slower than that of hydrogen. It is determined from the result of S.H.E (Standard Hydrogen Electrode) potential value experiment.

Question 26.
Predict which of the following will be coloured in aqueous solution Ti²⁺ , V³⁺, Sc⁴⁺, Cu⁺, SC³⁺, Fe³⁺, Ni²⁺ and CO³⁺
Answer:
Among Ti²⁺ , V³⁺, Sc⁴⁺, Cu⁺, Sc³⁺, Fe³⁺, Ni²⁺ and CO³⁺ in aqueous solution state.
Ti (Z = 22) – Ti²⁺ – Electronic configuration is [Ar] 3d²
V (Z = 23) – V³⁺ – Electronic configuration is [Ar] 3d²
Sc (Z = 21) – SC⁴⁺ – Electronic configuration is [Ar] 1s² 2s² 2p⁶ 3s² 3p⁵
Cu (Z = 29) – Cu⁺ – Electronic configuration is [Ar] 3d¹⁰
Sc (Z = 21) – SC³⁺ – Electronic configuration is [Ar] 3d°4s°
Fe (Z = 26) – Fe³⁺ – Electronic configuration is [Ar] 3d⁵
Ni (Z = 28) – Ni²⁺ – Electronic configuration is [Ar] 3d⁵
CO (Z = 27) – CO³⁺ – Electronic configuration is [Ar] 3d⁶
A transition metal ion is coloured if it has one or more unpaired electrons in (n-l)d orbital, i.e. 3d orbitals in the case of first transition series, when such species are exposed to visible radiation, d – d transition take place and the species are coloured.

Question 27.
Describe the variable oxidation state of 3d series elements.
Answer:
1. The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by loosing electrons from (n-l)d orbital and ns orbital as the energy difference between them is very small.

2. At the beginning of the 3d series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.

3. The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases. For example, in the 3d series, first element Sc has only one oxidation state +3 the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.

4. Mn²⁺ (3d⁵) is more stable than Mn⁴⁺ (3d³) is due to half filled stable configuration.

Question 28.
Which metal in the 3d series exhibits +1 oxidation state most frequently and why?
Answer:
1. The first transition metal copper exhibits only +1 oxidation state.

2. It is unique in 3d series having a stable +1 oxidation state.
Cu (Z = 29) Electronic configuration is [Ar] 3d¹⁰ 4s²

3. So copper element only can have +1 oxidation state.

Question 29.
Why first ionization enthalpy of chromium is lower than that of zinc?
Answer:
The first ionization enthalpy of chromium is lower than that of zinc. Cr (Z = 24) Electronic configuration [Ar] 3d⁵ 4s¹. In the case of Cr, first electron has to be removed easily from 4s orbital to attain the more stable half filled configuration. So Cr has lower ionization enthalpy. But in the case of Zinc (Z = 30), electronic configuration [Ar] 3d¹⁰ 4s². The first electron has to be removed from the most stable fully filled electronic configuration becomes difficult and it requires more energy.

Question 30.
Transition metals show high melting points why?
Answer:

1. All the transition metals are hard.
2. Most of them are hexagonal close packed, cubic close packed (or) body centered cubic which are characteristics of true metals.
3. The maximum melting point at about the middle of transition metal series indicates that d⁵ configuration of favourable for strong inter atomic attraction.
4. Due to the strong metallic bonds, atoms of the transition elements are closely packed and held together. This leads to high melting point and boiling point.

Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements Evaluate yourself

Question 1.
Compare the stability of Ni⁴⁺ and Pt⁴⁺ from their ionisation enthalpy values
Answer:

Pt⁴⁺ compounds are stable than Ni⁴⁺ compounds because the energy needed to remove 4 electrons in Pt is less than that of Ni.

Question 2.
Why iron is more stable in +3 oxidation state than in +2 and the reverse is true for Manganese?
Answer:
Fe (Z = 26). Electronic configuration [Ar] 3d⁶  4s²
Fe → Fe³⁺ + 3e^–
Fe³⁺ Electronic configuration p° [Ar]3d⁵.
If’d’ orbital is half filled, it is more stable than . Fe²⁺ where it is [Ar]3d⁶.
Mn (Z = 25). Electronic configuration [Ar]3d⁵ 4s²
Mn → Mn²⁺ + 2e^– . By the loss of 2e^–, Mn²⁺ is more stable due to half filled configuration. Mn → Mn³⁺ + 3e^–.
Mn³⁺ Electronic configuration [Ar]3d⁴ 4s°.
Among this Fe³⁺ is more stable than Fe²⁺ and the Mn²⁺ is more stable than Mn³⁺.

Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements Additional Questions

Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements 1 Mark Questions and Answers

I. Choose the correct answer.

Question 1.
The elements whose atom has incomplete d sub-shell are called …………..
(a) s-block element
(b) Alkali metals
(c) transition elements
(d) Representative elements
Answer:
(c) transition elements

2. Which one of the following is the other name of d-block elements?
(a) Chalcogens
(b) Halogens
(c) Inner-transition elements
(d) Transition elements
Answer:
(d) Transition elements

Question 3.
Which metals play an important role in the development of human civilization?
(a) Al and Mg
(b) Na and K
(c) Fe and Cu
(d) Mn and Ni
Answer:
(c) Fe and Cu

Question 4.
Which transition element is used in light bulb filaments?
(a) Al
(b) Ni
(c) W
(d) Fe
Answer:
(c) W

Question 5.
Which metal is used in manufacturing artificial joints?
(a) Molybdenum
(6) Titanium
(c) Tungsten
(d) Iron
Answer:
(b) Titanium

Question 6.
Which transition metal is applied in the manufacturing of boiler plants?
(a) Iron
(b) Copper
(c) Aluminium
(d) Molybdenum
Answer:
(d) Molybdenum

Question 7.
Identify the transition metal present in Hemoglobin …………..
(a) Cobalt
(b) Iron
(c) Manganese
(d) Copper
Answer:
(b) Iron

Question 8.
Which of the following transition metal is present in Vitamin B₁₂?
(a) Cobalt
(b) Platinum
(c) Copper
(d) Iron
Answer:
(a) Cobalt

Question 9.
The metal cobalt is present in
(a) Vitamin-A
(b) Vitamin-B
(c) Vitamin-B₁₂
(d) Vitamin-B₆
Answer:
(c) Vitamin-B₁₂

Question 10.
Consider the following statements.
Answer:
(i) Transition metals occupy group-3 to group-12 of the modem periodic table.
(ii) Representative elements occupy group-3 to group-12 of the modem periodic table.
(iii) Except group-11 elements of all transition metals are hard.
(iv) d-block elements are mostly non-metals.

Which of the above statements is/are incorrect?
(a) (ii) and (iv)
(b) (i) and (iii)
(c) (iii) only
(d) (i) only
Answer:
(a) (ii) and (iv)

Question 11.
Consider the following statements.
Answer:
(i) d-block elements composed of 3d series, Sc to Zn (4^th period).
(ii) 4d series composed of Y to Cd.
(iii) 5d series composed of La, Hf to Mercury.
(iv) d-block elements composed of 4d series Y to Cd.

Which of the above statements is/are incorrect
(a) (i) and (iv)
(b) (i), (ii) and (iii)
(c) (iii) and (iv)
(d) (iv) only
Ans.
(b) (i), (ii) and (iii)

Question 12.
Which of the following is the correct electronic configuration of Sc (Z = 21)?
(a) [Ar] 3d³
(b) [Ar] 3d’ 4s²
(c) [Ar] 3d² 4s¹
(d) [Ar] 4s² 4p’
Answer:
(b) [Ar] 3d’ 4s²

Question 13.
The correct electronic configuration of Cr is …………..
(a) [Ar] 3d⁴ 4s²
(b) [Ar] 3d⁵
(c) [Ar] 3d⁵ 4s¹
(d) [Ar] 3d⁶
Answer:
(c) [Ar] 3d⁵ 4s¹

Question 14.
Which of the following is the correct electronic configuration of copper?
(a) [Ar] 3d⁵ 4s¹
(b) [Ar] 3d¹⁰ 4s¹
(c) [Ar] 3d⁹ 4s²
(d) [Ar] 3d⁸ 4s² 4p¹
Answer:
(b) [Ar] 3d¹⁰ 4s¹

Question 15.
Which one of the following is the general electronic configuration of transition elements?
(a) [Noble gas] ns² np⁶
(b) [Noble gas] ( n – 2 ) f^1-14(n-l)d^1-10 ns²
(c) [Noble gas] ( n – 1 ) d^1-10 (n-l)f^1-14 ns²
(d) [Noble gas] ( n – 1 ) d^1-10 ns²
Answer:
(d) [Noble gas] ( n – 1 ) d^1-10 ns²

Question 16.
Which of the following d-block elements has the highest electrical conductivity at room temperature?
(a) Copper
(b) Silver
(c) Aluminium
(d) Tungsten
Answer:
(b) Silver

Question 17.
Consider the following statements.
(i) The melting point decreases from Scandium to Vanadium in 3d series.
(ii) In 3d transition series, atomic radius decreases from Sc to V and upto copper atomic radius nearly remains the same.
(iii) As we move down in 3d transition series, atomic radius increases.

Which of the above statements is/are incorrect?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i), (ii) and (iii)
Answer:
(a) (i) only

Question 18.
Which of the following transition element exhibit only +3 oxidation state?
(a) Cu
(b) Sc
(c) Mn
(d) Cr
Answer:
(b) Sc

Question 19.
Which one of the following transition element has maximum oxidation states?
(a) Manganese
(b) Copper
(c) Scandium
(d) Titanium
Answer:
(a) Manganese

Question 20.
Consider the following statements.
(i) In 3d series, the middle element Mn has +2 to +7 oxidation states.
(ii) The oxidation state of Ru and Os is +8.
(iii) Scandium has six different oxidation states.

Which of the above statements is/are not correct?
(a) (i) and (ii)
(b) (ii) only
(c) (i) only
(d) (iii) only
Answer:
(d) (iii) only

Question 21.
Which one of the following elements show high positive electrode potential?
(a) Ti⁺
(b) Mn²⁺
(c) CO²⁺
(d) Cr²⁺
Ans.wer:
(c) CO²⁺

Question 22.
Which one of the following elements show high negative electrode potential?
(a) Copper
(b) Manganese
(c) Cobalt
(d) Zinc
Answer:
(d) Zinc

Question 23.
Which one of the following is diamagnetic in nature?
(a) Ti³⁺
(b) Cu²⁺
(c) Zn²⁺
(d) V³⁺
Answer:
(c) Zn²⁺

Question 24.
Which one of the following is paramagnetic in nature?
(a) Sc³⁺
(b) Ti⁴⁺
(c) V⁵⁺
(d) Cu²⁺
Answer:
(d) Cu²⁺

Question 25.
Which of the following pair has maximum number of unpaired electrons?
(a) Mn²⁺, Fe³⁺
(b) CO³⁺, Fe²⁺
(c) Cr³⁺, Mn⁴⁺
(d) Ti²⁺, V³⁺
Answer:
(a) Mn²⁺, Fe³⁺

Question 26.
Which of the following pair has d¹⁰ electrons?
(a) Ti³⁺, V⁴⁺
(b) CO³⁺, Fe²⁺
(c) Cu⁺ , Zn²⁺
(d) Mn²⁺, Fe³⁺
Answer:
(c) Cu⁺, Zn²⁺

Question 27.
Which of the following is used as a catalyst in the manufacture of sulphuric acid form SO₃.
(a) V₃O₅
(b) Rh-Ir
(c) Ni
(d) Fe
Answer:
(a) V₅O₅

Question 28.
Which one of the following is Zeigler – Natta catalyst?
(a) CO₂(CO)₈
(b) Rh/Ir complex
(c) TiCl₄ + Al(C₂H₅)₃
(d) Fe / Mo
Answer:
(c) TiCl₄ + Al(C₂H₅)₃

Question 29.
Which one of the following is used as a catalyst in the polymerisation of propylene?
(a) V₂O₅
(b) Pt
(c) TiCl₄ + Al(C₂H₅)₃
(d) Fe / Mo
Answer:
(c)TiCl₄ + Al(C₂H₅)₃

Question 30.
Consider the following statements.
Answer:
(i) Transition metal hydrides are used as powerful oxidising agents.
(ii) Metallic carbides are chemically active.
(iii) Interstitial compounds are hard and show electrical and thermal conductivity.

Which of the above statements is/are incorrect?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) only
(d) (i) only
Answer:
(a) (i) and (ii)

Question 31.
Which one of the following oxide is covalent?
(a) Cr₂O₃
(b) CrO
(c) Mn₂O₇
(d) Na₂O
Answer:
(c) Mn₂O₇

Question 32.
Which one of the following oxide is amphoteric in nature?
(a) CrO
(b) Cr₂O₃
(c) Mn₂O₇
(d) MnO
Answer:
(b) Cr₂O₃

Question 33.
The oxidation state of Chromium in \({ CrO }_{ 4 }^{ 2- }\) and in \({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\) are …………..
(a) +3, +6
(b) +7, +4
(c) +6, +6
(d) +4, +6
Answer:
(c) +6, +6

Question 34.
Which one of the following is used to identify chloride ion in inorganic qualitative analysis?
(a) Barium chloride test
(b) Chromyl chloride test
(c) Brown ring test
(d) Ammonium molybdate test
Answer:
(b) Chromyl chloride test

Question 35.
Which one of the following is the formula of chromyl chloride?
(a) CrOCl₂
(b) CrCl₃
(c) CrO₂ Cl₂
(d) CrCl
Answer:
(c) CrO₂ Cl₂

Question 36.
Which ore is used to prepare potassium permanganate?
(a) Pyrolusite
(b) Chromite
(c) Argentite
(d) Cuprite
Answer:
(a) Pyrolusite

Question 37.
Which one of the following geometry is possesed by permanganate ion?
(a) Pyramidal
(b) Tetrahedral
(c) Octahedral
(d) linear
Answer:
(b) Tetrahedral

Question 38.
The hybridisation state of Mn⁺⁷ is permanganate ion is …………..
(a) sp² hybridisation
(b) dsp² hybridisation
(c) d²sp³ hybridisation
(d) sp³ hybridisation
Answer:
(d) sp³ hybridisation

Question 39.
Which one of the following is known as Baeyer’s reagent?
(a) Cold dilute alkaline KMnO⁴
(b) Chromyl Chloride
(c) Acidified potassium dichromate
(d) Acidified potassium manganate
Answer:
(a) Cold dilute alkaline KMnO₄

Question 40.
Which reagent is used in the conversion of ethylene into ethylene glycol?
(a) Chromyl chloride
(b) Zeigler – Natta catalyst
(c) Cold dilute alkaline KMnO₄
(d) Acidified K₄Cr₂O₇
Answer:
(c) Cold dilute alkaline KMnO₄

Question 41.
Baeyer’s reagent is used to detect unsaturation in an organic compound.
(a) Chloride ion
(b) unsaturated organic compound
(c) Sulphate ion
(d) Chromate ion
Answer:
(b) unsaturated organic compound

Question 42.
What is the equivalent weight of KMnO₄ in acid medium?
(a) 158
(b) 52.67
(c) 31.6
(d) 392
Answer:
(c) 31.6

Question 43.
What is the equivalent weight of KMnO₄ in basic medium?
(a) 158
(b) 52.67
(c) 392
(d) 31.6
Answer:
(a) 158

Question 44.
Which one of the following is used for the estimation of ferrous salts, oxalates, hydrogen peroxide and iodides?
(a) K₂MnO₄
(ft) KMnO₄
(c) K₂Cr₂O₇
(d) CrO₂ Cl₂
Answer:
(b) KMnO₄

Question 45.
Which of the following is the general electronic configuration of lanthanoids?
(a) [Xe] 4f⁷ 3d^1-10 5s²
(b) [Xe] 4f^1-14 3d¹⁰ 6s²
(c) [Xe] 5f^2-14 4d¹⁰ 6s²
(d) [Xe] 4f^2-14 5d^1-10 6s²
Answer:
(d) [Xe] 4f^2-14 5d^1-10 6s²

Question 46.
The most common oxidation state of Lanthanoids is …………..
(a) + 4
(b) + 3
(c) + 6
(d) +5
Answer:
(b) + 3

Question 47.
The expected electronic configuration of lanthanum (La) (Z = 57) is …………..
(a) [Xe] 4f¹ 5d° 6s²
(b) [Xe] 4f⁰ 5d¹ 6s²
(c) [Xe] 4f³
(d) [Xe] 4f⁰ 5d³
Answer:
(a) [Xe] 4f¹ 5d° 6s²

Question 48.
The actual electronic configuration of La (Z = 57) is …………..
(a) [Xe] 4f¹ 5d° 6s²
(b) [Xe] 4f³
(c) [Xe] 4f⁰ 5d¹ 6s²
(d) [Xe] 4f⁰ 5d³
Answer:
(c) [Xe] 4f⁰ 5d¹ 6s²

Question 49.
Which of the following lanthanoids have half-filled 4f orbital?
(a) Gd
(b) Tb
(c) Lu
(d) La
Answer:
(a) Gd

Question 50.
Which one of the following lanthanoids have completely filled 4f orbital?
(a) Gd and Eu
(b) La and Ce
(c) Yb and Lu
(d) Pr and Pm
Answer:
(c) Yb and Lu

Question 51.
Which one of the following is the main cause of lanthanoid contraction?
(a) Poor shielding effect of 5f sub-shell
(b) More shielding effect of 4f sub-shell
(c) Poor shielding effect of 4f sub-shell
(d) More shielding effect of 5f sub-shell
Answer:
(c) Poor shielding effect of 4f sub-shell

Question 52.
Which of the following pair has more or less same atomic radius due to lanthanide contraction?
(a) Ti and V
(b) Fm and Md
(c) No and Lr
(d) Zr and Hf
Answer:
(d) Zr and Hf

Question 53.
Consider the following statement.
(i) All the actinoids are non radioactive.
(ii) Neptunium and other heavier elements are produced by artificial transformation of naturally occurring elements by nuclear reactions.
(iii) Most of the actinoids have long half lives.

Which of the above statements is/are not correct.
(a) (i) only
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (i) and (iii)
Answer:
(d) (i) and (iii)

Question 54.
The general valence shell electronic configuration of actinoids is …………..
(a) [Xe] 4f^2-14 5d^0-2 6s²
(b) [Rn] 4f^2-14 5d^0-2 6s²
(c) [Rn] 5f^2-14 6d^0-1 7s²
(d) [Rn] 4f^0-7 5d^0-1 s²
Answer:
(c) [Rn] 5f^2-14 6d^0-2 7s²

Question 55.
Which pair of actinoids show +2 oxidation state?
(a) Am and Th
(b) Pa and U
(c) Pu and Cm
(d) No and Lr
Answer:
(a) Am and Th

Question 56.
Neptunium and Plutonium exhibit the maximum oxidation state as …………..
(a) + 8
(b) + 7
(c) + 6
(d) + 4
Answer:
(b) + 7

Question 57.
Consider the following statement.
Answer:
(i) Most of the actinoids are coloured.
(ii) Actinoids show greater tendency to form complexes.
(iii) Most of the actinoids are non-radioactive.

Which of the above statements is/are correct.
(a) (i) only
(b) (i) and (iii)
(c) (i) and (ii)
(d) (ii) and (iii)
Answer:
(c) (i) and (ii)

Question 58.
Consider the following statement.
Answer:
(i) Lanthanoids do not form oxo cations.
(ii) Most of the lanthanoids are colourless.
(iii) Binding energy of 4f orbitals are lower.

Which of the above statement is/are not correct.
(a) (i) and (ii)
(b) (iii) only
(c) (i) and (iii)
(d) (i), (ii) and (iii)
Answer:
(b) (iii) only

Question 59.
Which one of the following is more basic in nature?
(a) La(OH)₃
(b) Ce(OH)₃
(c) Gd(OH)₃
(d) Lu(OH)₃
Answer:
(a) La(OH)₃

Question 60.
Which one of the following is less basic in nature?
Answer:
(a) La(OH)₃
(b) Gd(OH)₃
(c) Lu(OH)₃
(d) Ce(OH)₃
Answer:
(c) LU(OH)₃

II. Fill in the blanks.

1. Transition elements occupy the central position of the periodic table between …………. elements.
2. Except …………. elements, all transition metals are hard and have very high melting point.
3. The metal present in Vitamin – B₁₂ is ………….
4. …………. metal is used in manufacture of artificial joints.
5. The extra stability of Cr and Cu is due to …………. of electrons and exchange energy.
6. Of all the known elements …………. has the highest electrical conductivity at room temperature.
7. The maximum melting point at about the middle of transition metal series indicates that …………. configuration is favourable for strong attraction.
8. The atomic radius of 5d elements and 4d elements are nearly same due to ………….
9. Ni (II) compounds are thermodynamically …………. than Pt (II) compounds.
10. The first transition metal …………. exhibits only +3 oxidation state.
11. The middle transition element …………. has six different oxidation states.
12. The oxidation state of Ru and Os is ………….
13. …………. is unique in 3d series having a stable +1 oxidation state.
14. The substance which is oxidised is a …………. agent and the one which is reduced is an …………. agent.
15. The oxidising and reducing power of an element is measured in terms of ………….
16. If the E° of a metal is large and negative, the metal is a ………….
17. The species with all paired electrons exhibit ………….
18. The magnetic moment of an ion is given by ………….
19. Many industrial processes use …………. or their as catalyst.
20. In the catalytic hydrogenation of an alkene …………. is used as catalyst.
21. In the preparation of acetic acid from acetaldehyde the catalyst used in ………….
22. The catalyst used in the hydroformylation of olefins is ………….
23. …………. catalyst is used in polymerization of propylene.
24. Metallic carbides are chemically ………….
25. Except Scandium all other 3d series transition elements form …………. metal oxides.
26. Cr₂O₃ …………. is and CrO is …………. in nature.
27. Mn₂O₇ dissolves in water to give ………….
28. On heating potassium dichromate, it decomposes to give …………. and molecular oxygen.
29. Potassium dichromate is a powerful …………. agent in acidic medium.
30. ………….is used in leather tanneries for chrome tanning.
31. Potassium dichromate is used in quantitative analysis for the estimation of …………. and ………….
32. Permanganate ion has …………. geometry in which Mn⁺⁷ is …………. hybridised.
33. Cold dilute alkaline KMnO₄ is known as ………….
34. …………. is used for the treatment of skin infections and fungal infections of the foot.
35. Baeyer’s reagent is used for detecting …………. in an organic compounds.
36. The equivalent weight of KMnO₄ in neutral medium is ………….
37. The common oxidation state of lanthanoids is ………….
38. Due to the decrease in the size of Ln³⁺ ions, the ionic character of Ln – OH bond decreases which results in the ………….
39.  All the actinoids are …………. and most of them have …………. half lives.
40. ………….do not form oxo cations.

Answer:

1. sandpblock
2. group-II
3. cobalt
4. Titanium
5. symmetrical distribution
6. silver
7. d⁵, inter atomic
8. lanthanoid contraction
9. more stable
10. Scandium
11. Manganese
12. + 8
13. Copper
14. reducing, oxidising
15. Standard electrode potential
16. powerful reducing agent
17. diamagnetism
18. μ=\(g\sqrt { S(S+1) } \)μ_B
19. transition metals, compounds
20. Nickel
21. Rh/Ir complex
22. CO₂(CO)₈
23. Zeigler – Natta (or) TiCl₄ + Al(C₂H₅)₃
24. inert
25. ionic
26. amphoteric, basic
27. permanganic acid (HMnO₄)
28. Chromium (III) oxide – Cr₂O₃
29. Potassium dichrom.ate
30. iron compounds, iodides
31. tetrahedral,sp³
32. Baeyer’s reagent
33. Potassium permanganate
34. unsaturation
35. 52.67
36. + 3
37. decrease in the basicity
38. radioactive. shoit
39. Lanthanoids

III. Match the following using the code given below.

Question 1.
A. Tungsten – 1. Development of human civilization
B. Titanium – 2. Light bulb filament
C. Molybdenum – 3. Artificial joint
D. Copper – 4. Boiler plants

Answer:
(a) 2, 3, 4, 1

Question 2.
A. Iron – 1. Artificial joints
B. Platinum – 2. Hemoglobin
C. Cobalt – 3. Catalysis
D. Titanium – 4. Vitamin – B₁₂

Answer:
(b) 2, 3, 4, 1

Question 3.
A. Sc to Zn – 1. 5d series
B. Y to Cd – 2. Actinoids
C. LatoHg – 3. 3dseries
D. Ac to Lr – 4. 4d series

Answer:
(a) 3, 4, 1, 2

Question 4.
A. Cr – 1. [Ar] 3d¹⁰ 4s²
B. Cu – 2. [Ar] 3d⁵ 4s¹
C. Zn – 3. [Ar]3d¹ 4s²
D. Sc – 4. [Ar] 3d¹⁰ 4s¹

Answer:
(c) 2, 4, 1, 3

Question 5.
A. Scandium – 1. + 7
B. Manganese – 2. + 2
C. Copper – 3.+3
D. Titanium – 4. + I

Answer:
(a) 3, 1, 4, 2

Question 6.
A. Sc³⁺ – 1. 3d¹
B. Ti³ – 2. 3d⁰
C. Mn²⁺ – 3. 3d¹⁰
D. Zn²⁺ – 4. 3d⁵

Answer:
(b) 2, 1, 4, 3

IV. Assertion and Reason.

Question 1.
Assertion (A) – Cr and Cu having [Ar] 3d⁵ 4s¹ and [Ar] 3d¹⁰ 4s¹ are more stable.
Reason (R) – The extra stability of elements Cr and Cu is due to symmetrical distribution of electrons and exchange energy.
(a) Both (A) and (R) are correct and (R) explains (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) explains (A).

Question 2.
Assertion (A) – In 3d transition elements, the expected decrease in atomic radius is observed from Sc to V, thereafter upto Cu, the atomic radius nearly remains the same.
Reason (R) – As we move from Sc to V, the added 3d electrons only partially shield the increased nuclear charge but upto Cu, the extra electron added to 3d sub-shell repel the 4s electrons and the slight increase in nuclear charge operated in opposite direction and it leads to constancy in atomic radii.
(a) Both (A) and (R) are correct and (R) explains (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) explains (A).

Question 3.
Assertion (A) – In transition metal series, the ionization enthalpy increases.
Reason (R) – This is due to increase in nuclear charge corresponding to the filling of d electrons.
(a) Both (A) and (R) are correct and (R) explains (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) explains (A).

Question 4.
Assertion (A) – Ni (II) compounds are thermodynamically more stable than Pt (II) compounds.
Reason (R) – The energy required to form Ni^2- is less than that of Pt²⁺.
(a) Both (A) and (R) are correct and (R) explains (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) explains (A).

Question 5.
Assertion (A) – Except Scandium all 3d series, transition elements exhibit variable oxidation states.
Reason (R) – By loosing electrons from (n – l)d orbital and ns orbital as the energy difference between them is very small.
(a) Both (A) and (R) are correct and (R) explains (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) explains (A).

Question 6.
Assertion (A) – Mn²⁺ is more stable than Mn⁴⁺.
Reason (R) – Mn²⁺(3d⁵) is more stable than Mn⁴⁺(3d³) is due to extra stability of half-filled electronic configuration.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A)

Question 7.
Assertion (A) – Copper is unique in 3d series having a stable +1 oxidation state.
Reason (R) – Copper is prone to disproportionate to the +2 and 0 oxidation states.
(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) explains (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) explains (A).

Question 8.
Assertion (A) – Transition metals form large number of complexes.
Reason (R) – Transition metals are small and highly charged and they have vacant low energy orbitals to accept an electron pair donated by other groups.
(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 wrong
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation

V. Find the odd one out.

Question 1.
(a) Sc
(b) Ti
(c) Y
(d) Cr
Answer:
(c) Y
Reason: Yttrium be1ons to 4d series whereas others are 3d series

Question 2.
(a)Ru
(b)Rh
(c)Pd
(d) Pt
Answer:
(d) Pt
Reason: Pt belongs to 5d series whereas others are 4d series.

Question 3.
(a) Th
(b) La
(c) Ce
(d) Lu
Answer:
(a) Th
Reason: Th belongs to actinoids whereas others are lanthanoids.

Question 4.
(a) La
(b) Pr
(c) Am
(d) Lu
Answer:
(c) Am
Reason: Am belongs to actinoids whereas others are lanthanoids.

Question 5.
(a) Ce
(b) Th
(c) U
(d) Pu
Answer:
(a) Ce
Reason: Ce belongs to lanthanoids whereas others are actinoids.

Question 6.
(a) Ac
(b) U
(c) Pa
(4) Np
Answer:
(a) Ac
Reason: Except Actinium all the remaining elements are synthetically prepared.

Question 7.
(a) Ti²⁺
(b) V²⁺
(c) Zn²⁺
(d) Cu²⁺
Answer:
(d) Cu²⁺
Reason: It has positive reduction potential whereas others have negative reduction potential.

Question 8.
(a) CO³⁺
(b) Cr³⁺
(c) V³⁺
(4) Ti³⁺
Answer:
(a) CO³⁺
Reason: It has positive reduction potential whereas others have negative reduction potential.

Question 9.
(a) Mn³⁺
(b) Fe³⁺
(c) Cr³⁺
(d) CO³⁺
Answer:
(c) Cr³⁺
Reason: It has negative reduction potential whereas others have positive reduction potential.

Question 10.
(a) SC³⁺
(b) Ti⁴⁺
(c) V⁵⁺
(d) Cu²⁺
Answer:
(d) Cu²⁺
Reason: It is paramagnetic whereas others are diamagnetic.

Question 11.
(a) Cr³⁺
(b) Mn⁴⁺
(c) V²⁺
(d) Zn²⁺
Answer:
(d) Zn²⁺
Reason: It is diamagnetic whereas others are paramagnetic

VI. Find Out the correct pair.

Question 1.
(a) \({ CrO }_{ 4 }^{ – }\) and Cr₂\({ O }_{ 7 }^{ 2- }\)
(b) \({ MnO }_{ 4 }^{ – }\) and \({ MnO }_{ 4 }^{ 2- }\)
(c) H₂CrO₄ and HMnO₄
(d) Cr₂O₃ and Mn₂O₇
Answer:
(a) \({ CrO }_{ 4 }^{ – }\) and Cr₂\({ O }_{ 7 }^{ 2- }\)
In this pair Cr has +6 oxidation states whereas in others the metal has different oxidation state.

Question 2.
(a) Zn, Cu
(b) Hf, Zr
(c) Ag , Au
(d) Ti, Cu
Answer:
(b) Hf, Zr
This pair has same atomic radius whereas others have different atomic radius.

Question 3.
(a) Ru , Os
(b) Mn , Cu
(c) Sc , Cu
(d) Ni, Co
Answer:
(a) Ru , Os
Both has +8 as oxidation state whereas others have different oxidation state.

Question 4.
(a) Ti²⁺, CO²⁺
(b) Cr²⁺, Mn³⁺
(c) Fe²⁺ and CO³⁺
(d) CO³⁺ and Cu²⁺
Answer:
(d) CO³⁺ and Cu²⁺
Both have positive electrode potential whereas others have different value.

Question 5.
(a) Cu²⁺, Zn²⁺
(b) CO³⁺, Cr³⁺
(c) Ti³⁺, V³⁺
(d) Mn³⁺, Cr³⁺
Answer:
(c) Ti³⁺, V³⁺
have negative electrode potential whereas others have different values.

VII. Find out the incorrect pair.

Question 1.
(a) Sc³⁺, Ti⁴⁺
(b) Ti³⁺, Ti²⁺
(c) Cr²⁺, Mn³⁺
(d) Cu⁺, Zn²⁺
Answer:
(b) Ti³⁺, Ti²⁺
have d¹ and d¹ configuration whereas others have same configuration.

Question 2.
(a) Sc and Zn
(b) Y and Cd
(c) Ag and Au
(d) Na and K
Answer:
(d) Na and K. They belong to alkaline metals whereas others are d-block elements.

Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements 2 Mark Questions and Answers

Question 1.
d-block elements are called transition elements. Justify this statement.
Answer:
1. d-block elements occupy the central position of the periodic table, between s and p block elements.

2. Their properties are transitional between highly reactive metals of s-block and elements of p-block which are mostly non-metals. That is why d-block elements are called transitional elements.

Question 2.
How many series are in d-bloclc elements? What are they?
Answer:

• There are 4 series in d-block e1ements They are,
• 3d series – 4th period – Scandium to Zinc
• 4d series – 5th period – Yttrium to Cadmium
• 5d series – 6th period – Lanthanum to Mercury
• 6d series – 7th period – Actinium to Californium

Question 3.
Zn, Cd, Hg belong to d-block elements even though they do not have partially filled d-orbitals. Give reason.
Answer:

1. Zn, Cd, Hg belong to d-block elements even though they do not have partially filled d-orbitals either in their elemental state or in their normal oxidation states.
2. However they are treated as transition elements, because their properties are an extension of the properties of the respective transition elements.

Question 4.
Applying Aufbau principle, write down the electronic configuration of Sc (Z = 21) and Zn (Z = 30).
Answer:

1. According to Aufbau principle, the electron first fills the 4s orbital before 3d orbital.
2. Sc (Z = 21) Is² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹
   • Zn (Z = 30) Is² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰

Question 5.
Write a note about atomic radius of Zinc.
Answer:
At the end of 3d series, d-orbitals of Zinc contain 10 electrons in which the repulsive interaction between the electrons is more than the effective nuclear charge and hence the orbitals slightly expand and atomic radius slightly increases.

Question 6.
Write a not about oxidation state of 3d series.
Answer:

1. The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases.
2. Hence, the first and last elements show less number of oxidation states and the middle elements with more number of oxidation states.
3. For example, the first element Sc has only one oxidation state +3, the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.

Question 7.
Mn²⁺ is more stable than Mn⁴⁺. Why?
Answer:

1. The relative stability of different oxidation states of 3d metals is correlated with the extra stability of half-filled and fully filled electronic configurations.
2. Example – Mn²⁺ (3d⁵) is more stable than Mn⁴⁺ (3d³)

Question 8.
Ru and Os have highest oxidation state in which compounds? Explain with example.
Answer:

1. Ru and Os have +8 as the highest oxidation state.
2. The highest oxidation state of 4d and 5d elements are found in their compounds with the higher electronegative elements like O, F and Cl. For example: RuO₄, OsO₄

Question 9.
Copper is unique in 3d series. Prove this statement.
Answer:
Copper is unique in 3d series having a stable +1 oxidation state. It is prone to disproportionate to the +2 and 0 oxidation states.

Question 10.
Define – Standard electrode potential.
Answer:
Standard electrode potential is the value of the standard emf of a cell in which molecular hydrogen under standard pressure (latm) and temperature (273K) is oxidised to solvated protons at the electrode.

Question 11.
Which metal is used to reduce Cr³⁺ ion? Why?
Answer:
A stable Cr³⁺ ion, strong reducing agent which has high negative value for reduction potential like metallic zinc (E° = – 0.76 V) is required. Metallic zinc is a powerful reducing agent due to its large negative values.

Question 12.
Sc³⁺, Ti⁴⁺, V⁵⁺ are diamagnetic. Give reason.
Answer:

1. Sc³⁺, Ti⁴⁺, V⁵⁺ have d° electronic configuration, n = 0
2. µ = \(\sqrt { 0(0+2) }\) = 0 µ_B So they are diamagnetic.

Question 13.
Calculate the magnetic moment of Ti³⁺ and V⁴⁺.
Answer:
Ti (Z = 22) Ti³⁺ 3d¹
V (Z = 23) V⁴⁺ 3d¹
µ = \(\sqrt { 1(1+2) }\) = \(\sqrt { 35 }\) = 1.73 µ_B. So they are paramagnetic.

Question 14.
Cr³⁺, Mn⁴⁺, V²⁺ are paramagnetic. Calculate their magnetic moment values.
Answer:
Cr³⁺, Mn⁴⁺, V²⁺ Configuration is d³. Due to 3 unpaired electrons, they are paramagnetic.  µ = \(\sqrt { 3(3+2) }\) = \(\sqrt { 35 }\) = 3.87 µ_B

Question 15.
Mn²⁺, Fe³⁺ have high magnetic moment. Prove it.
Answer:

1. Mn²⁺, Fe³⁺ configuration is d⁵.
2. µ = \(\sqrt { 5(5+2) }\) = \(\sqrt { 35 }\) = 5.916 µ_B
   Among 3d series, Mn²⁺, Fe³⁺ have high magnetic moment as 5.916 µ_B .

Question 16.
How many unpaired electrons are present in CO³⁺, Fe²⁺? Calculate their magnetic moment.
Answer:
Answer:
CO (Z = 27) CO³⁺ [Ar] 3d⁶
Fe (Z = 26) Fe²⁺ [Ar] 3d⁶
The number of unpaired electrons are 4 as follows:

Their magnetic moment is µ = \(\sqrt { 4(4+2) }\) = \(\sqrt { 24 }\) = 4.89 µ_B

Question 17.
Calculate the magnetic moment and the number of unpaired electrons in Cu²⁺.
Answer:
Cu (Z = 29) Electronic configuration [Ar] 3d¹⁰ 4s¹
Cu²⁺ Electronic configuration [Ar] 3d⁹
The number of unpaired electrons
is 1.
Magnetic moment µ = \(\sqrt { 1(1+2) }\) = \(\sqrt { 3 }\) = 1.732 µ_B

Question 18.
Cu⁺, Zn²⁺ are diamagnetic. Prove it.
Answer:
Cu⁺, Zn²⁺ electronic configuration [Ar] 3d¹⁰
The number of unpaired electron is 0.
µ = \(\sqrt { 0(0+2) }\) = 0 µ_B. Cu⁺, Zn²⁺ are diamagnetic.

Question 19.
Most of the transition metals act as catalyst. Justify this statement.
Answer:
1. Many industrial processes use transition metals or their compounds as catalysts. Transition metal has energetically available d orbitals that can accept electrons from reactant molecule or metal can form bond with reactant molecule using its ‘d’ electrons.

2. For example, in the catalytic hydrogenation of an alkene, the alkene bonds to an active site by using its n electrons with an empty d orbital of the catalyst.

Question 20.
Explain the catalytic hydrogenation of alkene to alkane with equation.
Answer:
The σ bond in the hydrogen molecule breaks, and each hydrogen atom forms a bond with a d electron on an atom in the catalyst Nickel. The two hydrogen atoms then bond with the partially broken π -bond in the alkene to form an alkane.

Question 21.
Which catalyst is used in the hydroformylation of olefins? Give equation
Answer:

Question 22.
Which catalyst is used in the conversion of acetaldehyde to acetic acid? Give equation
Answer:

Question 23.
What is Zeigler -Natta catalyst? In which reaction it is used? Give equation.
Answer:
A mixture of TiCl₄ and trialkyl aluminium is Zeigler – Natta catalyst. It is used in the polymerization

Question 24.
d-block elements readily form complexes. Give reason.
Answer:
1. Transition elements (d-block elements) have a tendency to form coordination compounds (complexes) with a species that has an ability to donate an electron pair to form a coordinate

2. Transition metal ions are small and highly charged and they have vacant low energy orbitals to accept an electron pair donated by other groups. Due to these properties, transition metals form large number of complexes.

3. Examples – [Fe(CN)₆]^4-, [CO(NH₃)₆]³⁺

Question 25.
Prove that acidified potassium dichromate is a powerful oxidising agent.
Answer:
K₂Cr₂O₇ act as power oxidising agent in acidic medium. In the presence of H⁺ ions, the oxidation state of Cr froms Cr⁶⁺ is changed to Cr³⁺
Cr₂\({ O }_{ 7 }^{ 2- }\) + 14H⁺ + 6^– → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

Example:
Acidified K₂Cr₂O₇ oxidises Ferrous salts to Ferric salts.
Cr₂\({ O }_{ 7 }^{ 2- }\) + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.

Question 26.
What are the uses of potassium dichromate?
Answer:
K₂Cr₂O₇ is used

1. as a strong oxidising agent
2. in dyeing and printing
3. in leather tanneries for chrome plating
4. in quantitative analysis for the estimation of iron compounds and iodides

Question 27.
Draw and explain about the structure of permanganate ion.
Answer:

Permanganate ion has tetrahedral geometry in which the central Mn⁷⁺ is sp³ hybridised.

Question 28.
Explain the action of heat on potassium permanganate.
Answer:

Question 30.
What happens when thiosuiphate ion is treated with permanganate ion?
Answer:
Permanganate ion oxidises thiosulphate into sulphate.
\({ 8MnO }_{ 4 }^{ – }\) + 3S₂\({ O }_{ 3 }^{ 2- }\) → \({ 6SO }_{ 4 }^{ 2- }\) + 8MnO₂ + 2OH^–

Question 31.
What is Baeyer’s reagent? Where it is used?
Answer:

1. Cold dilute alkaline KMnO₄ is known as Baeyer’s reagent. It is used to oxidise alkene into diols.
2. For example, ethylene can be converted into ethylene glycol and this reaction is used as a test for unsaturation.

Question 32.
Acidified KMnO₄ is a very strong oxidising agent. Prove it.
Answer:
1. In the presence of dilute sulphuric acid, potassium permanganate acts as a very strong oxidising agent. Permanganate ion is converted into Mn²⁺ ion.
\({ MnO }_{ 4 }^{ – }\) + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O

2. Permanganate oxidises ferrous salt to ferric salt.
\({ 2MnO }_{ 4 }^{ – }\) + 10Fe²⁺ + 16H⁺ → 2Mn²⁺ + 10Fe³⁺ + 8H²O

Question 33.
KMnO₄ does not act as oxidising agent in the presence of HCl. Why?
Answer:
HCl cannot be used for making acidified KMnO₄ as oxidising agent, since it reacts with KMnO₄ as follows.
\({ 2MnO }_{ 4 }^{ – }\) + 10Cl^– + 16H⁺ → 2Mn²⁺ + 5Cl₂ + 8H₂O

Question 34.
HNO₃ cannot be used as an acid medium along with KMnO₄. Why?
Answer:
HNO₃ cannot be used since it is a good oxidising agent and it reacts with reducing agents in the reaction.

Question 35.
Among HCl, HNO₃ and H₂SO₄, which is the suitable medium for KMnO₄ in oxidising reaction?
HCl and HNO₃ cannot be used. HCl react with KMnO₄. HNO₃ is itself a good oxidising agent. However, H₂SO₄ is found to be most suitable since it does not react with potassium permanganate.

Question 36.
Explain about the causes of lanthanide contraction.
Answer:
1. As we move from one element to another in 4f series ( Ce to Lu) the nuclear charge increases by one unit and an additional electron is added into the same inner 4f sub-shell.

2. 4f sub-shell have a diffused shapes and therefore the shielding effect of 4f electrons are relatively poor. Hence, with increase of nuclear charge, the valence shell is pulled slightly towards nucleus.

3. As a result, the effective nuclear charge experienced by the 4f elelctoms increases and the size of Ln³⁺ ions decreases.

Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements 3 Mark Questions and Answers

Question 1.
Cr and Cu are more stable. Give reason.
Answwer:
1. The electronic configuration of Cr and Cu are [Ar] 3d³ 4s¹ and [Ar] 3d¹⁰ 4s¹ respectively. The extra stability of half filled and fully filled d orbitals, as already explained in XI STD, is due to symmetrical distribution of electrons and exchange energy.

2. The extra stability of half filled and fully filled d orbitals is due to symmetrical distribution of electrons and exchange energy.

3. When the d orbitals are considered together, they will constitute a sphere. So the half filled and fully filled configuration leads to complete symmetrical distribution of electron density.

4. On the other hand, an unsymmetrical distribution of electron density will result in building up of a potential difference. To decrease this and to achieve a tension free state with lower energy, a symmetrical distribution is preferred.

Question 2.
Explain about the metallic behaviour of d-block elements.
Answer:
1. All the transition elements are metals. They are good conductors of heat and electricity. Of all the known elements, silver has the highest electrical conductivity at room temperature.

2. Most of the transition elements are hexagonal close packed, cubic close packed or body centered cubic which are the characteristics of true metals.

Question 3.
Explain about the variation of melting point among the transition metal series.
Answer:
1. As we move from left to right along the transition metal series, melting point first increases as the number of unpaired d electrons available for metallic bonding increases, reach a maximum value and then decreases, as the d electron pairs up and become less available for bonding.

2. For example, in the first series the melting point increases from Scandium to a maximum of 2183 K for Vanadium, which is close to 2180K for chromium.

3. Manganese in 3d series and has low melting point. The maximum melting point at about the middle of transition metal series indicates that d⁵ configuration is favorable for strong interatomic attraction.

Question 4.
Explain about the variation of atomic radius along a period of 3d series.
Answer:
1. In general, atomic radius decreases along a period. But for the 3d transition elements, the expected decrease in atomic radius is observed from Sc to V , thereafter upto Cu the atomic radius nearly remains the same.

2. As we move from Sc to Zn in 3d series, the extra electrons are added to the 3d orbitals, the added 3d electrons only partially shield the increased nuclear charge and hence the effective nuclear charge increases slightly.

3. However, the extra electrons added to the 3d sub shell strongly repel the 4s electrons and these two forces are operated in opposite direction and as they tend to balance each other, it leads to constancy in atomic radii.

Question 5.
Ni (II) compounds are more stable than Pt (II) compounds. Give reason.
1. The ionisation enthalpy values can be used to predict the thermodynamic stability of their compounds. .

2. For Nickel I.E₁ + I.E₂ = 737+ 1753
= 2490 kJ mol^-1
For platinum, I.E₁ + I.E₂ = 864+ 1791
= 2655 kJ mol^-1
Since, the energy required to form Ni²⁺ is less than that of Pt²⁺ , Ni(II) compounds are thermodynamically more stable than Pt(II) compounds.

Question 6.
Compare the reduction potentials of Mn³⁺ / Mn²⁺ and Fe³⁺ / Fe²⁺ .
Answer:
1. Mn³⁺ + e^– → Mn²⁺
E° = + 1.51V
Fe³⁺ + e^– → Fe²⁺
E° = + 0.77V

2. The high reduction potential of Mn³⁺ / Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺. For Fe³⁺ / Fe²⁺ the reduction potential is 0.77V, this low value indicates that both Fe³⁺ and Fe²⁺ can exist under normal conditions.

3. The drop from Mn to Fe is due to the electronic structure of the ions concerned. Mn³⁺ has 3d⁴ configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d sub-shell makes the reduction of Mn³⁺ very feasible.

Question 7.
How alloys are formed in d-biock elements?
Answer:
1. An alloy is formed by blending a metal with one or more other elements. The elements may be metals or non-metals or both.

2. The bulk metal is named as solvent, and the other elements in smaller portion is called solute.

3. According to Hume – Rothery rule to form an alloy, the difference between the atomic radii of the solvent and solute is less than 15%. Both the solvent and solute must have the same crystal structure and valence and their electro negativity difference must be close to zero.

4. Since their atomic sizes are similar and one metal atom can be easily replaced by another metal atom from its crystal lattice to form an alloy. The alloys are hard and have high melting points. Examples – Gold – copper alloy.

Question 8.
What are interstitial compounds? Give their properties.
Answer:
An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They are usually non-stoichiometric compounds, e.g., TiC, ZrH_1.92, Mn₄N.

1. They are hard and show electrical and thermal conductivity.
2. They have high melting points.
3. Transition metal hydrides are used as powerful reducing agents.
4. Metallic carbides are chemically inert.

Question 9.
Explain the action of heat on potassium dichromate.
Answer:

Question10.
Draw and explain about the structure of chromate and dichromate ion.
Answer:

1. Both chromate and dichromate ions are oxo anions of chromium and they are moderately strong oxidising agents.
2. In both structures, chromium is in +6 oxidation state.
3. In an aqueous solution, chromate and dichromate ions can be inter convertible, and in an alkaline solution, chromate ion is predominant, whereas dichromate ion becomes predominant in acidic solutions. .

Question 11.
Explain the action of acidified K₂Cr₂O₇ – with

1. Iodide
2. Sulphide

Answer:
1. Acidified K₂Cr₂O₇ oxidises iodide ions to iodine.
Cr₂\({ O }_{ 7 }^{ 2- }\) + 6I^– + 14H⁺ → 2Cr³⁺ +3I₂ + 7H₂O

2. Acidified K₂Cr₂O₇ oxidises Sulphide ions to Sulphur.
Cr₂\({ O }_{ 7 }^{ 2- }\) + 3S^2- + 14H⁺ → 2Cr³⁺ + 3S + 7H²O

Question 12.
Explain the action of acidified K₂Cr₂O₇ with

1. Sulphur dioxide
2. Alcohols.

Answer:
1. Acidified K₂Cr₂O₇ oxidises SO₂ to sulphate ion.
Cr₂\({ O }_{ 7 }^{ 2- }\) + 3SO² + 2H⁺ -> 2Cr³⁺ + \({ 3SO }_{ 4 }^{ 2- }\) + H₂O

2. Acidified K₂Cr₂O₇ oxidises alcohol to acid. .

Question 13.
Explain about chromyl chloride test.
Answer:
1. When potassium dichromate is heated with any chloride salt in the presence of Conc.H₂SO₄, orange red vapours of chromyl chloride (CrO₂Cl₂) is evolved. This reaction is used to confirm the presence of chloride ion in inorganic qualitative analysis.

2. The chromyl chloride vapours are dissolved in sodium hydroxide solution and then acidified with acetic acid and treated with lead acetate. A yellow precipitate of lead chromate is obtained.

Question 14.
Explain the action of Conc.H₂SO₄ on potassium permanganate.
Answer:
1. On treating with cold conc.H₂SO₄, it decomposes to form manganese heptoxide, which subsequently decomposes explosively.

2. With hot Conc.H₂SO₄, potassium permanganate give MnSO₄ [Manganese(II) sulphate]

Question 15.
What are the uses of Potassium permanganate?
Answer:
Potassium permanganate is used,

1. as a strong oxidising agent.
2. for the treatment of various skin infections and fungal infections of foot.
3. used in water treatment industries to remove iron and hydrogen sulphide from well water.
4. as a Baeyer’s reagent for detecting unsaturation in an organic compound.
5. in quantitative analysis for the estimation of ferrous salts, oxalates, hydrogen peroxide (H₂O₂) and iodides.

Question 16.
Calculate the equivalent weight of KMnO₄ in

1. Acidic medium
2. Basic medium
3. Neutral medium

Answer:
1. Equivalent weight of KMnO₄ in acidic medium

2. Equivalent weight of KMnO₄ in basic medium

3. Equivalent weight of KMnO₄ in neutral medium

Question 17.
Explain about the oxidation state of actinoids.
Answer:

1. The most common oxidation state of actinoids is +3.
2. In addition to that actinoids show variable oxidation states such as +2, +3, +4, +5, +6 and +7.
3. The elements Americium (Am) and Thorium (Th) show +2 oxidation state in some compounds.
4. Th, Pa, U, Np, Pu and Am show +5 oxidation states.
5. Np and Pu exhibit +7 oxidation state.

Question 18.
Write the electronic configuration of

1. Ac (Z = 89)
2. Am (Z = 95)
3. Lr (Z = 103)

Answer:

1. Ac (Z = 89): [Rn] 5f⁰ 6d¹ 7s²
2. Am (Z = 95) : [Rn] 5f⁷ 6d⁰ 7s²
3. Lr (Z = 103): [Rn] 5f¹⁴ 6d¹ 7s²

Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements 5 Mark Questions and Answers

Question 1.
Explain about the magnetic properties of transition elements.
Answer:
1. Most of the compounds of transition elements are paramagnetic.

2. Materials with no elementary magnetic dipoles are diamagnetic. In other words a species with all paired electrons exhibits diamagnetism.

3. Paramagnetic solids having unpaired electrons possess magnetic dipoles which are isolated from one another.

4. Ferromagnetic materials have domain structure and in each domain the magnetic dipoles are arranged. But the spin dipoles of the adjacent domains are randomly oriented. Some transition elements or ions with unpaired d electrons show ferromagnetism.

5. 3d transition metal ions in paramagnetic solids often have a magnetic dipole moments corresponding to the electron spin contribution only. So the magnetic moment of the ion is given by
µ = \(g\sqrt { S(S+1) }\) µ_B
Where g = 2, S is the total spin quantum number of the electrons.
µ_B = Bohr Magneton.
For an ion with ‘n’ unpaired electrons S = \(\frac { n }{ 2 } \)
Therefore the spin only magnetic moment is µ = \(2\sqrt { \left( \frac { n }{ 2 } \right) \left( \frac { n }{ 2 } +1 \right) }\)µ_B
µ = \(2\sqrt { \left( \frac { n\left( n+2 \right) }{ 4 } \right) }\)µ_B
µ = \(2\sqrt { n(n+2) }\)µ_B

Question 2.
How will you prepare potassium permanganate from pyrolusite ore?
Answer:
Potassium permanganate is prepared from pyrolusite (MnO₂) ore. The preparation involves the following steps.
1. Conversion of MnO₂ to potassium manganate:
Powdered ore is fused with KOH in the presence of air or oxidising agents like KNO₃. A green coloured potassium manganate is formed.

2. Oxidation of potassium manganate to potassium permanganate:
Potassium manganate can be oxidised in two ways, either by chemical oxidation or electrolytic oxidation.

3. Chemical oxidation – In this method potassium manganate is treated with ozone (O₃) or chlorine to get potassium permanganate.
\({ 2MnO }_{ 4 }^{ 2- }\) + O₃ + H₂O → \({ 2MnO }_{ 4 }^{ 2- }\) + 2OH^– + O₂
\({ 2MnO }_{ 4 }^{ 2- }\) + Cl₂ → \({ 2MnO }_{ 4 }^{ 2- }\) + 2Cl^–

4. Electrolytic oxidation: In this method aqueous solution of potassium manganate is electrolyzed in the presence of little alkali.
K₂MnO₄ \(\equiv\) 2K⁺ + \({ 2MnO }_{ 4 }^{ 2- }\)
H₂O \(\equiv\) H⁺ + OH^–

Manganate ions are converted into permanganate ions at anode.

The purple coloured solution is concentrated by evaporation and forms crystals of potassium permanganate on cooling.

Question 3.
Explain acidified KMnO₄ is a powerful oxidising agent with 5 examples.
Answer:
1. In the presence of dilute sulphuric acid, potassium permanganate acts as a very strong oxidising agent. Permanganate ions is converted into Mn²⁺ ion.
\({ MnO }_{ 4 }^{ – }\) + 8H⁺ +5e^– → Mn²⁺ + 4H₂O
Example:
1. Potassium permanganate oxidises ferrous salts to ferric salts.

2. Potassium permanganate oxidises iodide ions to iodine.
\({ 2MnO }_{ 4 }^{ – }\) + 10I^– + 16H⁺ → 2Mn²⁺ + 5I² + 8H²O

3. Potassium permanganate oxidises sulphide ion to sulphur.
\({ 2MnO }_{ 4 }^{ – }\) + 5S^2- + 16H⁺ → 2Mn²⁺ + 5S + 8H²O

4. Potassium permanganate oxidises oxalic acid to CO₂.
\({ 2MnO }_{ 4 }^{ – }\) + 5(COO)^2- + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

5. Potassium permanganate oxidises alcohols to aldehyde.
2KMnO₄ + 3H₂SO₄ + 5CH₃CH₂OH → 2K₂SO₄ + 2MnSO₄ + 5CH₃CHO + 8H₂O

Question 4.
Explain about the oxidation state of Lanthanoids.
Answer:

1. The common oxidation state of lanthanoids is +3. In addition to that some of the lanthanoids also show either +2 or +4 oxidation states.
2. Gd³⁺ and Lu³⁺ ions have extra stability, it is due to half filled and completely filled f-orbitals.
3. Cerium and terbium attain 4f⁷ and 4f¹⁴ configurations respectively in the +4 oxidation states.
4. Eu²⁺ and Yb²⁺ ions have exactly half filled and completely filled f orbitals.
5. Lu shows only +3 oxidation state.
6. Ce, Pr, Nd, Tb and Dy exhibit +3 and +4 oxidation states.
7. Nd, Sm, Eu, Tm, Yb exhibit +2 oxidation states also.

Common Errors:
Atomic number of d-block elements may get confused.

Rectifications:
Atomic number of d-block elements in group wise jumped as 18, 18, 18, 32.
e.g. Sc (Z = 21)
Y (Z = 39)
La (Z = 57)
Ac (Z = 89)

Hope you love the Samacheer Kalvi 12th Chemistry Chapter Wise Material. Clearly understand the deep concept of Chemistry learning with the help of Tamilnadu State Board 12th Chemistry Chapter 4 Transition and Inner Transition Elements Questions and AnswersPDF. Refer your friends to and bookmark our website for instant updates. Also, keep in touch with us using the comment section.

Students can Download English Poem 5 Father to his Son Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 12th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th English Solutions Poem Chapter 5 Father to his Son

A Father To His Son Poem Summary Warm up

Every parent is anxious about the welfare of his / her children. Parents express their anxiety by advising them almost all the time. What kind of advice do you frequently receive from your parents? Fill in the bubbles. Tick the ones you like to follow implicitly and reasons for the ones you don’t like to follow.

Answer:

• Wash your fingers before eating.
• Don’t stay awake till late night.
• Who are you talking to over the phone for a long time?

1. I don’t like to tell my parents what I do with the phone. Why should they interfere with my privacy?
2. I don’t gossip; I just share what I enjoyed. My friends tell me what they like. How can my parents call it ‘gossip’?
Time spent in sharing our dreams is not wasted but invested. The other three agree.

Samacheer Kalvi 12th English Father to his Son Textual Questions

1. Fill in the blanks choosing the words from the box given and complete the summary of the poem.

A father sees his son nearing manhood.
What shall he tell that son?
“Life is hard; be steel; be a rock.”
And this might stand him for the storms
and serve him for humdrum monotony

and guide him among sudden betrayals
and tighten him for slack moments.
“Life is a soft loam; be gentle; go easy.”
And this too might serve him.
Brutes have been gentled where lashes failed.

The growth of a frail flower in a path up
has sometimes shattered and split a rock.
A tough will counts. So does desire.
So does a rich soft wanting.
Without rich wanting nothing arrives.

Tell him too much money has killed men
and left them dead years before burial:
the quest of lucre beyond a few easy needs
has twisted good enough men
sometimes into dry thwarted worms

Tell him time as a stuff can be wasted.
Tell him to be a fool ever so often
and to have no shame over having been a fool
yet learning something out of every folly
hoping to repeat none of the cheap follies

thus arriving at intimate understanding
of a world numbering many fools.
Tell him to be alone often and get at himself
and above all tell himself no lies about himself
whatever the white lies and protective fronts

he may use against other people.
Tell him solitude is creative if he is strong
and the final decisions are made in silent rooms.
Tell him to be different from other people
if it comes natural and easy being different.

Let him have lazy days seeking his deeper motives.
Let him seek deep for where he is born natural.
Then he may understand Shakespeare
and the Wright brothers, Pasteur, Pavlov,
Michael Faraday and free imaginations

Bringing changes into a world resenting change.
He will be lonely enough
to have time for the work
he knows as his own.

Lines 1-25

The poet Carl Sandburg gives a vivid description of a father’s worldly (1) _______ in directing a son who is at the threshold of his (2) _______ Here the father motivates his son to be like a hard (3) _______ and withstand life’s (4) _______ and sudden betrayals. (5) _______ is like a fertile soil. We can make our life fruitful if we are gentle, and take life as it comes. At times (6) _______ overtakes harshness. The growth of a (7) _______ can split a rock. One should have a (8) _______ and strong will to achieve. Greed for(9) _______ has left men dead before they really die. Good men also have fallen prey in quest for (10) _______ money. Time for (11) _______ is not a waste. When you seek knowledge never feel ashamed to be called a (12) _______ for not knowing, at the same time learn from your (13) _______ and never (14) _______ it.
Answers:

1. wisdom
2. manhood
3. rock
4. challenges
5. life
6. gentleness
7. tender-flower
8. deep desire
9.money
10. easy
11. leisure
12. fool
13.mistakes
14. repeat

Lines 26-44

Do (15) ______ often, and do not hesitate to accept your shortcomings, avoid (16) to protect self against other people. Solitude helps to be (17) and (18) are taken in silent rooms. Instead of being one among many, be (19) , if that is your nature. The son may need lazy days to find his (20) abilities, to seek what he is bom for. He will then know how free imaginations bring (21) to the world, which (22) change.
During such resentment, let him know that it is time for him to be on his own, and (23) to achieve like Shakespeare, the Wright brothers, Pasteur, Pavlov and Michael Faraday.
Answer:

15. introspect
16. white lies
17. creative
18. final decisions
19. different
20. inherent
21. changes
22. resents
23. work

2. Based on your understanding of the poem answer the following questions in one or two sentences.

A Father To His Son Poem Paragraph Question (a)
How would the poet’s advice help his son who is at the threshold of the manhood?
Answer:
The poet’s advice would help the son at the threshold of manhood, to grow as a positive individual and succeed in life like great scientists and dramatists.

12th English Poem A Father To His Son Question (b)
A tough will counts.’ Explain.
Answer:
One’s physical strength doesn’t matter. Any person with a strong will can achieve great feats. A frail flowering plant succeeds in splitting a rock due to its hard will. Thus a tough will (i.e.) “rich soft wanting” determines one’s success.

A Father To His Son Summary Question (c)
What happened to the people who wanted too much money?
Answer:
People who wanted too much money fell a prey to greed. They lost their reputation also.

Father To His Son Poem Summary Question (d)
What has twisted good men into thwarted worms?
Answer:
One’s desire or love for money should be limited to meeting one’s basic needs like food, clothing and shelter. Beyond that, in ordinate desire to possess a lot of money in a dishonourable way reduces even good men to “thwarted worms”.

A Father To His Son Question (e)
How would his being alone help the boy?
Solitude would help the boy to be creative.

A Father To His Son By Carl Sandburg Summary Question (f)
Where are the final decisions taken?
Answer:
In silent rooms, final decisions are taken.

A Father To His Son Poem Summary In Tamil Question (g)
What are the poet’s thoughts on ‘being different’?
Answer:
Instead of being one among many, one can be different if it is one’s nature to be so. One need not take conscious efforts to be different.

A Father To His Son Poem Explanation Question (h)
Why does the poet advise his son to have lazy days?
Answer:
Lazy days would help the boy to find his inherent abilities and goad him to discover the purpose for which he was bom.

A Father To His Son Poem Analysis Question (i)
The poet says
‘Without rich wanting nothing arrives’
but he condemns ‘the quest of lucre beyond a few easy needs.’Analyse the difference and write.
Answer:
Appreciating rich wanting and condemning ‘the quest for lucre beyond a few easy’ needs seems like a conundmm. One needs to take both the statements with a pinch of salt. One needs a strong will power to succeed. One can earn a lot too. Jack Ma has had strong will and earned beyond lucre but invested major part in community development and charitable works retaining some for his basic needs. Warren Buffet, Bill Gates and Ratan Tata are other such examples. The wealth earned should benefit a large number of people and not the individual who initiated the wealth.

3. Here are a few poetic devices used in the poem.

A Father To His Son Poem Question (a)
Antithesis – It is a literary device that em phasises the idea of contrast,
eg. The growth of a frail flower in a path up
has sometimes shattered and split
a rock. Brutes have been gentled where lashes failed.

Father To His Son Poem Question (b)
Transferred Epithet – It is a figure of speech in which an epithet grammatically qualifies a noun other than the person or a thing, it is actually meant to describe.
eg. and left them dead years before burial:
Let him have lazy days seeking his deeper motives.
Bringing changes into a world resenting change.

A Father To His Son Poem Figure Of Speech Question (c)
Repetition – It is a figure of speech.
eg. Tell him to be alone often and get at himself and above all tell himself no lies about himself

Appreciate The Poem

4. Read the lines given below and answer the questions that follow.

(a) “Life is hard; be steel; be a rock.’’

A Father To His Son By Carl Sandburg Question (i)
How should one face life?
Answer:
One should face life like a steel.

Question (ii)
Identify the figure of speech in the above line.
Answer:
Metaphor

(b) “Life is a soft loam; be gentle; go easy. ”
And this too might serve him.

Question (i)
Why does the poet suggest to take life easy?
Answer:
There are certain occasions one needs to treat life like wet clay very gently. Take life as it comes without hard resistance. The poet says, “Brutes have been gentled where lashes failed.”

Question (ii)
Identify the figure of speech in the above line.
Answer:
Metaphor

(c) “ Tell him solitude is creative if he is strong
and the final decisions are made in silent rooms.”

Question (i)
Can being in solitude help a strong human being? How?
Answer:
Yes, solitude helps the strong person to be creative. Solitude helps even a strong human being to introspect and analyse his own mistakes.

Question (ii)
Identify the figure of speech in the above line.
Answer:
Personification

(d) “Tell him time as a stuff can be wasted.
Tell him to be a fool every so often”

Question (i)
Why does the poet suggest that time can be wasted?
Answer:
Unless one wastes one’s time, one may not commit mistakes and learn from them. Besides learning not to repeat those mistakes would naturally make him wiser.

Question (ii)
Identify the figure of speech in the above line.
Answer:
Simile

Question (e)
“Tell him to be a fool ever so often
and to have no shame over having been a fool
yet learning something out of every folly
hoping to repeat none of the cheap follies”

Question (i)
Is it a shame to be a fool at times?
Answer:
No, everyone does commit funny mistakes in life. One might just laugh at them.

Question (ii)
What does one learn from every folly?
Answer:
Every folly teaches a person his limitations and vulnerabilities. By making conscious efforts to avoid them in future, one will become stronger and wiser.

(f)“______ Free imaginations
Bringing changes into a world resenting change ”

Question (i)
How does free imagination help the world?
Answer:
Free imagination brings changes in the world.

Question (ii)
Identify the figure of speech.
Answer:
Personification

Question (e)
Pick out the alliterated words from the poem and write.
“And this might stand him for the storms”
Answer:
The words stand and storms alliterate.

Additional Questions

(a) “And this might stand him for the storms
and serve him for humdrum monotony”

Question (i)
What does the poet mean by storms?
Answer:
The poet means life’s challenges by ‘storms’.

Question (ii)
What can help the son overcome “the boring routine” in life?
Answer:
One can overcome ‘the boring routine’ by keeping a strong steel/rock-like will power and face life boldly.

(b) “and guide him among sudden betrayals
and tighten him for slack moments.”

Question (i)
What could guide the son among unexpected betrayals?
Answer:
Rock/steel-like would guide the son among betrayals.

Question (ii)
What could happen to the boy during slack moments?
Answer:
During slack moments, the boy may be betrayed by his trusted friends.

(c) “Brutes have been gentled where lashes failed.’”

Question (i)
When does the Government or law use lashes?
Answer:
When a brute has committed a crime, the law uses lashes to punish him.

Question (ii)
What can transform brutes?
Answer:
Gentleness can transform brutes.

(d) “The growth of a frail flower in a path up has sometimes shattered and split a rock.”

Question (i)
What do you mean by ‘frail’?
Answer:
Frail means weak and delicate.

Question (ii)
How does a frail flower plant split a rock?
Answer:
The plant’s tough will to survive and grow forces the rock to split.

Question (e)
“the quest of lucre beyond a few easy needs
has twisted good enough men
sometimes into dry thwarted worms. ”

Question (i)
What do you mean by ‘quest for lucre ?
Answer:
‘Quest for lucre’ means seeking money in a dishonourable way.

Question (ii)
What is the result of such a quest?
Answer:
It results in frustrating and ruining good men.

5. Explain the following lines with reference to the context.

Question (а)
“and guide him among sudden betrayals
and tighten him for slack moments.”
Answer:
Reference: These lines are from the poem “A Father to his Son” written by Carl August Sandburg.
Context and Explanation: The poet says these words while explaining the need to stay strong like a rock or a steel. The steel will guide him when confronted with unexpected betrayals from trusted friends. It will also help him to be strong against future experimentations.

Question (b)
“Brutes have been gentled where lashes failed.”
Answer:
Reference: These lines are from the poem “A Father to his Son” written by Carl August Sandburg.
Context and Explanation: The poet highlights the importance of soft but firm will to melt even hardened criminals. Love can transform even criminals. Harsh punishment may harden them but gentleness and love may bring about a change of heart.

Question (c)
“ Yet learning something out of every folly hoping to repeat none of the cheap follies”
Answer:
Reference: These lines are from the poem “A Father to his Son” written by Carl August Sandburg.
Context and Explanation: The poet says these words while hinting at the possibilities of failures in the pursuit of knowledge. People may mock at a beginner as a fool. He should not feel ashamed of being called a fool. He must continue the pursuit of knowledge doggedly to achieve success.

Question (d)
“He will be lonely enough
to have time for the work”
Answer:
Reference: These lines are from the poem ‘A Father to his Son” written by Carl August Sandburg.
Context and Explanation: The poet says these words to explain how creative thinkers and those who strive to bring about changes are left alone to fend for themselves. The poet says his son must take advantage of this loneliness to pursue his creative imagination and succeed like Shakespeare, Wright brothers, Pasteur, Pavlov and Faraday.

Additional Questions

Question (a)
“and above all tell himself no lies about himself
whatever the white lies and protective fronts”
Answer:
Reference: These lines are from the poem ‘A Father to his Son” written by Carl August Sandburg.
Context and Explanation: The poet says these words to his son as to how to conduct himself after a folly is done. He wants him not to tell lies or defend himself or blame others for his predicament. He must learn from his follies and become wiser.

Question (b)
“Let him have lazy days seeking his deeper motives.
Let him seek deep for where he is born natural. ”
Answer:
Reference: These lines are from the poem ‘A Father to his Son” written by Carl August Sandburg.
Context and Explanation: The poet says these words while highlighting the advantages of spending leisure hours in introspection. When a person looks within in solitude he will realise his inner strength, innate abilities and also realise the purpose for which he has been created.

Question (c)
“Life is hard; be steel; be a rock.”
And this might stand him for the storms”
Answer:
Reference: These lines are from the poem ‘A Father to his Son” written by Carl August Sandburg.
Context and Explanation: The poet emphasises the need to have rock or steel-like determination to confront the challenges in life. He believes that such a stance would help his son to overcome boredom of the routine in life.

Question (d)
“A tough will counts. So does desire.
So does a rich soft wanting
Answer:
Reference: These lines are from the poem ‘A Father to his Son” written by Carl August Sandburg.
Context and Explanation: The poet says these words while stressing the need to have tough will to create a path for success even on a rocky terrain. One needs to have deep passionate desire and strong will to succeed in life.

Question (e)
“the quest of lucre beyond a few easy needs
has twisted good enough men’”
Answer:
Reference: These lines are from the poem “A Father to his Son” written by Carl August Sandburg.
Context and Explanation: The poet says these words while explaining how ill-gotten wealth suffocates the good men. The money earned through dishonest means frustrates good men and even reduces to the level of worms.

Question (f)
“Tell him too much money has killed men
and left them dead years before burial:”
Answer:
Reference: These lines are from the poem ‘A Father to his Son” written by Carl August Sandburg.
Context and Explanation: The poet says these words while dwelling on the evil effects of amassing wealth beyond basic needs of men. Oliver Goldsmith says, “Where wealth accumulates, man decays.” The poet says that a man who amasses wealth against ethical principles is spiritually dead.

6. Answer the following questions in about 100-150 words each.

Question (a)
Explain how the poet guides his son who is at the threshold of manhood, to face the challenges of life.
Answer:
The poet shares his wisdom with his son who is at the threshold of manhood. He persuades his son to be hard like steel or rock to withstand challenges and unforeseen betrayals in life. A person with soft heart will crumble before a breach of trust. Similarly he wants his son to be discerning enough to be soft when needed to grow like a frail flower plant splitting a rock. Occasionally one has to go with the current because life is at times fertile with a lot of opportunities to grow even among the harshest circumstances. ‘Rich soft wanting’ can help a person to win against all odds. He reiterates this’idea by explaining how gentleness can reform a hardened criminal when lashes would, in contrast, harden them further.

Question (b)
How according to the poet is it possible for his son to bring changes into a world that resents change?
Answer:
The poet advises his son to introspect often and not feel ashamed of being called a fool especially when he pursues knowledge. He can examine himself and remove his follies. He must enjoy the advantages of solitude. Solitude would help him to be creative. He would invariably leam that final decisions are always taken in silent rooms. Being alone, he can identify his innate potential and talents. His free imagination will bring about changes even if the world resents them. The zest to bring about changes will elevate him to the level of Shakespeare, Pasteur, Wright Brothers, Pavlov and Michael Faraday. Thus he will be remembered as one of the great men who changed the world.

Listening Activity

Listen to the poem read by the teacher or to the recorded version and write a synopsis in about 100 words. The teacher can choose any three stanzas.

Lincoln’s Letter to his Son’s Teacher:
He will have to learn, I know, that all men are not just, all men are not true. But teach him also that for every scoundrel there is a hero; that for every selfish politician, there is a dedicated leader.. .Teach him for every enemy there is a friend,

Steer him away from envy, if you can, teach him the secret of quiet laughter.

Let him learn early that the bullies are the easiest to lick… .Teach him, if you can, the wonder of books.. .But also give him quiet time to ponder the eternal mystery of birds in the sky, bees in the sun, and the flowers on a green hillside.

In the school teach him it is far honourable to fail than to cheat…Teach him to have faith in his own ideas, even if everyone tells him they are wrong .Teach him to be gentle with gentle people, and tough with the tough.

Try to give my son the strength not to follow the crowd when everyone is getting on the band wagon…Teach him to listen to all men.. .but teach him also to filter all he hears on a screen of truth, and take only the good that comes through.

Teach him if you can, how to laugh when he is sad…Teach him there is no shame in tears, Teach him to scoff at cynics and to beware of too much sweetness…Teach him to sell his brawn and brain to the highest bidders but never to put a price-tag on his heart and soul. Teach him to close his ears to a howling mob and to stand and fight if he thinks he’s right. Treat him gently, but do not cuddle him, because only the test of fire makes fine steel.

Let him have the courage to be impatient.. .let him have the patience to be brave. Teach him always to have sublime faith in himself, because then he will have sublime faith in mankind. This is a big order, but see what you can do.. .He is such a fine little fellow, my son!

Abraham Lincoln
Synopsis of the first three stanzas:
Abraham Lincoln’s letter to his son’s teacher is full of his optimism and ethical values he believed in. He asks the teacher to teach his son to have faith in humanity and teach him not to be discouraged by scoundrels, selfish politicians and enemies. For every selfish politician, there is a dedicated hero. He need not condemn all politicians as bad. There may be enemies but there is a friend for every enemy. Thus, the equations in life are wonderful. He must learn the dignity of labour and hard work. He should be taught the necessity of undergoing the pain of failure in order to appreciate success. He requests his son to stay away from envy and to enjoy quiet laughter. He wants him to teach his son that bullies accept defeat quickly. He also urges him to teach his son the wonders of reading books and deep quiet reflection on the beautiful mysteries of nature.

Given below is a well-known quotation.
“Cowards die many rimes before their death”.
Study the quotations and identify the adverse human qualities that are worse than ‘death’ and discuss the underlying message conveyed.
Greed and the craze to amass wealth through dishonourable methods is like dying years before one actually dies. Cowardliness is equally worse than death. Shakespeare has said, “the valiants die only once in their lifetime.”

Father to his Son About the Poet

“Poetry is the opening and closing of a door, leaving those who look through to guess about what was seen during a moment.” – Carl Sandburg

Sandburg was born in Galesburg, Illinois, to parents of Swedish ancestiy. Carl August Sandburg (January 6, 1878 – July 22, 1967) was an American poet, writer, and editor. He won three Pulitzer Prizes. During his lifetime, Sandburg was widely regarded as “a major

figure in contemporary literature”, especially for volumes of his collected verse, including Chicago Poems (1916), Comhuskers (1918), and Smoke and Steel (1920). He enjoyed ;unrivalled appeal as a poet in his day. At his death in 1967, President Lyndon B. Johnson observed that “Carl Sandburg was more than the voice of America, more than the poet of its strength and genius. He was America.”

Father to his Son Summary in English

Inheritance of wisdom
Poet Carl Sandburg wants to leave his wisdom to his son who is at the threshold of his manhood. He wants his son to be like steel and rock to confront the harsh challenges and betrayals. But he should use his discernment go easy because life can be like soft clay too. He explains lucidly how a tender flowering plant with a ‘tough will’ can split a rock. He makes his son understand how gentleness has reformed criminals when lashes failed. One can’t achieve much without a deep desire to achieve success. He advises his son to be cautious with money for greed of easy money has heralded the downfall of good men. One can’t acquire wisdom or knowledge overnight. One may commit mistakes in life and be called a fool. But one can overcome frailties by consciously avoiding such mistakes.

Value of introspection
One has to look within and analyse one’s shortcomings. One must not be defensive but accept one’s own drawbacks. While contemplating in solitude, one gains flashes of insight into one’s own self and becomes creative. The best decisions in life are usually taken in silent rooms. If it is natural on one’s part to be different, ignoring criticisms one can dare to be different.

Changes wrought by free imagination
Great geniuses like Shakespeare, Pasteur, Pavlov, and Michael Faraday achieved grand success in their pursuits because they wanted to use their free imagination to change the world in their own way The world is complacent with the existing order of things and may resent changes. But if one brushes aside the popular resentment and strives hard with free imagination one can achieve success like scientists and social thinkers who have transformed the world.

Conclusion
The greatest legacy a father can possibly leave his son or daughter is not wealth but wisdom and positive attitude to meet challenges in life.

Father to his Son Summary in Tamil

கவிஞர், கார்ல் ஆகஸ்ட் சான்ட்பர்க் தன்
மகனுக்கு கூறும் அறிவுரையாக இந்தக் கவிதை அமைந்துள்ளது. வாழ்க்கையில் சோதனைகளும், ஏமாற்றங்களையும் எதிர் கொள்ள தன் மகனை இரும்பை போல் அல்லது கல்லைப் போல் கலங்காமல் இருக்கவேண்டும் என்கிறார். வாழ்க்கை களிமண் போன்று மிருதுவாகவும் இருக்கக் கூடும் என்பதால் தன் பகுத்தறிவின்படி தேவைப்படும் போது மென்மையாகவும் நடந்து கொள்ளுமாறு அறிவுறுத்துகிறார். பாறையின் மேல் படர்ந்த மிருதுவான மலர்க்கொடி அப்பாறையைத் துளையிடும் வல்லமை கொண்டது என்பதை தெளிவாக கூறுகிறார். சவுக்கடி தந்தும் திருந்தாத முரடனை நல்ல வார்த்தைகள் திருத்தி மென்மையானவனாக்கி இருக்கிறது. ஒன்றினை அடைவதற்கு ஆழ்ந்த உந்துதல் இல்லையெனில் அதை அடைதல் அரிது. பணத்தின் பால் கவனம் செலுத்துமாறு மகனுக்கு அறிவுறுத்துகிறார். பணத்தின் மீதான் பேராசை காரணமாக பலர் விழுந்திருக்கிறார்கள் என்பதை பிரகடனம் செய்கிறார். ஒரு நாள் இரவில் ஞானத்தைப் பெற்று விட முடியாது. ஒருவன் வாழ்நாளில் தவறுகள் புரிந்து முட்டாள் என பெயர் வாங்கலாம். கவனமாக நடந்து கொண்டால் இந்த பலவீனத்தை தவிர்க்கலாம்.

தற்சோதனையின் மதிப்பு:
ஒருவன் தன்னையே பரிசோதித்து, தன் பலகீனத்தை ஆராய வேண்டும். ஒருவன் தனக்கு இருக்கும் குறைகளை மறுக்காமல் ஒப்புக் கொள்ள வேண்டும். தனியாகத் திட்டமிடும் போது அவனுள்ளே “ஞானம்” பிறந்து அவனை ஆக்கப் பூர்வமானவனாக ஆக்குகிறது. வாழ்க்கையின் முக்கியத் திட்டங்கள் தனிமையில் உதிக்கின்றன. மனிதன் தன் மேல் சாற்றப்படும் தவறான கூற்றுகளைக் கண்டு கொள்ளாதவனாய் அடுத்தவனிடத்திலிருந்து தனித்து நிற்க விரும்பினால் அவ்வாறு துணிந்து நிற்கலாம்.

சுயேட்சையாக சிந்திப்பதனால் உண்டான மாற்றங்கள்:
மேதாவிகளான சேக்ஸ்பியர், பாஸ்ச்சர், பாவ்லோவ் மற்றும் மைக்கேல் ஃபாரடே போன்றவர்கள் தாம் மேற்கொண்ட செயல்களில் சுயமாகச் சிந்தித்து உலகை தனக்கு ஏற்றவாறு மாற்ற முயன்றனர்.உலகம் தற்போதைய சூழ்நிலைக்கு ஏற்ற வண்ணம் செல்வதால் மாறுதலை ஆட்சேபிக்கலாம். ஆனால், எதிர்ப்பைத் தவிர்த்து சுயமாக சிந்தித்து செயல்பட்டால் | விஞ்ஞானிகள் மற்றும் சீர்திருத்தவாதிகள் போல் உலகை மாற்றி அமைக்கலாம்.

முடிவுரை:
தகப்பன் மகனுக்கு விட்டுச் செல்லும் பரம்பரைச் சொத்து பணம், பொருள் அல்ல. ஆனால், நல்ல அறிவுரைகளும் உலகை எதிர்கொள்ளத் தன்னம்பிக்கை அளிக்கும் முற்போக்கு சிந்தனைகளுமே ஆகும்.

Father to his Son Glossary

Textual:

Additional:

You can Download Samacheer Kalvi 6th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Science Solutions Term 3 Chapter 2 Water

Samacheer Kalvi 6th Science Water Textual Evaluation

I. Choose the appropriate answer :

Samacheer Kalvi 6th Science Question 1.
Around 97% of water available on earth is _______ water.
(a) fresh
(b) pure
(c) Salty
(d) polluted
Answer:
(c) Salty

Samacheer Kalvi Guru 6th Science Question 2.
Which of the following is not a part of water cycle?
a. evaporation
b. condensation
c. rain
d. distillation

Science Term 3 Question 3.
Which of the following processes add water vapour to the atmosphere?
i. Transpiration
ii. Precipitation
iii. Condensation
iv. Evaporation
(a) ii and iii
(b) ii and iv
(c) i and iv
(d) i and ii
Answer:
(c) i and iv

6th Standard Samacheer Kalvi Science 3rd Term Question 4.
About 30% of the freshwater is found in?
a. glaciers
b. groundwater
c. other sources of water
d. 0.3%
Answer:
b. groundwater

Samacheer Kalvi 6th Standard Science Third Term Question 5.
Using R.O. plant at home eliminates a lot of non-potable water. The best way to effectively use the expelled water of R.O. plant is _______
(a) make the expelled water go and seep near the bore well
(b) use it for watering plants
(c) to drink the expelled water after boiling and cooling
(d) to use for cooking as the water is full of many nutrients
Answer:
(b) use it for watering plants

II. Fill in the blanks :

1. Only _______ percent of natural water is available for human consumption.
2. The process of changing water into its vapour is called _______
3. _______ is built on rivers to regulate water flow and distribute water.
4. Water levels in rivers increase greatly during _______
5. Water cycle is also called as _______

Answer:

1. 0.3
2. Evaporation
3. Dam
4. Raining
5. Hydrological cycle

III. True or False. If False, give the correct statement:

Samacheer Kalvi Science Question 1.
Water present in rivers, lakes and ponds is unfit for use by human beings.
Answer:
False. Water present in rivers, lakes and ponds is fit for use by human beings.

Samacheer Kalvi 6th Science Term 3 Question 2.
Seas are formed when the water table meets the land surface.
Answer:
False. Ponds are formed when the water table meets the land surface.

Samacheer Kalvi 6th Standard Term 3 Question 3.
The evaporation of water takes place only in sunlight.
Answer:
True.

6th Science Term 3 Question 4.
Condensation results in the formation of dew on grass.
Answer:
True.

Samacheer Kalvi 6th Science Guide Question 5.
Sea water can be used for irrigation as such.
Answer:
False. Sea water cannot be used for irrigation as such.

IV. Match the following :

Answer:

V. Arrange the following statements in the correct sequence :

1. These vapours condense to form tiny droplets of water.
2. The water droplets come together to form large water droplets.
3. The heat of the sun causes evaporation of water from the surface of the earth, oceans, lakes, rivers and other water bodies.
4. The large water droplets become heavy and the air cannot hold them, therefore, they fall as rains.
5. Water vapour is also continuously added to the atmosphere through transpiration from the surface of the leaves of trees.
6. Warm air carrying clouds rises up.
7. Higher up in the atmosphere, the air is cool.
8. These droplets floating in the air along with the dust particles form clouds.

Answer:

1. The heat of the sun causes evaporation of water from the surface of the earth, oceans, lakes, rivers and other water bodies.
2. Water vapour is also continuously added to the atmosphere through transpiration from the surface of the leaves of trees.
3. Higher up in the atmosphere, the air is cool.
4. These vapours condense to form tiny droplets of water.
5. These droplets floating in the air along with the dust particles form clouds.
6. Warm air carrying clouds rises up.
7. The water droplets come together to form large water droplets.
8. The large water droplets become heavy and the air cannot hold them, therefore, they fall as rains.

VI. Analogy:

6th Standard Science Water Lesson Question 1.
Population explosion : Water scarcity :: Recycle : __________
Answer:
Water Management.

6th Science Water Question 2.
Ground water : __________ :: Surface water : lakes
Answer:
Tube wells

VII. Give Very Short Answer:

6th Science Samacheer Kalvi Question 1.
Name four different sources of water.
Answer:
Rivers, wells, lakes, glaciers, ponds etc.

Samacheer Kalvi 6 Science Question 2.
How do people in cities and rural areas get water for various purposes?
Answer:
In city, people get water from water tanks, hand pipes and bore wells.
In rural area, people get water from wells, canals, ponds and rivers.

Samacheer Kalvi 6th Science Book Back Answers Question 3.
Take out of cooled bottle of water from refrigerator and keep it on a table. After some time you notice a puddle of water around it. Why?
Answer:
The cooled surface of bottle cools the air around it and the water vapour of the air condenses. So after some time a puddle of water can be noticed around the bottle.

Water Lesson 6th Standard Question 4.
We could see clouds almost every day. Why doesn’t it rain daily?
Answer:

1. The millions of tiny droplets do not collide with another to form larger droplets.
2. The air around the clouds is not cool.

Question 5.
Name the places where water is found as ice.
Answer:
Polar ice-caps, Ice sheets and glaciers in Artie region arid Antarctica.

Question 6.
How do aquatic animals manage to live in Arctic and Antarctic Circle?
Answer:
In Arctic and Antarctic circle, water in lakes and ponds will be frozen and a solid layer of ice is formed on the surface of water. Still aquatic animals living under the ice do not die. This is because the floating layer of ice acts as a protective coat and does not permit heat to escape from water. So as the surface water alone turns to ice, the aquatic animals manage it.

Question 7.
What are the types of rainwater harvesting?
Answer:
There are two types of rainwater harvesting.

1. Collecting water from where it falls.
2. Collecting rainwater by constructing bunds.

VIII. Give Short Answer:

Question 1.
Differentiate between surface water and groundwater.
Answer:
Surface water:

1. Water present on the surface of the earth.
2. Ex : River, lake, ponds, streams or freshwater.

Groundwater:

1. Water present beneath earth’s surface in soil.
2. Ex : open wells, tube wells (or) hand pumps, Spings etc.

Question 2.
Write a few slogans of your own on the topic “Save Water”?
Answer:

1. “To a thirsty man, a drop of water is worth more than a sack of gold”.
2. “Water covers 2/3 of the surface of the earth. But only 0.002% is drinkable. Save water”
3. “Save water to secure your future”
4. “Don’t make nature cry, keep your water clean”
5. “Store water for dry days”.

Question 3.
About 71% of earth’s surface is covered with water, then why do we face scarcity of water?
Answer:
About 71% of the earth’s surface is covered with water and even then we face scarcity of water. Reasons:

1. 97% of total water is found in seas and oceans, which is salty and not fit for human consumption.
2. Only 3% found is the freshwater and that too available in polar ice caps and glaciers.
3. Out of 3% fresh water, only 0.3% is available to us as surface water, in lakes, rivers, and swamps.

Question 4.
Give a reason for the following statement – Sewage should not be disposed of in rivers or oceans before treatment.
Answer:

1. Sewage should not be disposed of in rivers or oceans before treatment.
2. If we dispose of sewage before treatment the rivers and oceans will be polluted.
3. Aquatic animals and species will die due to pollution.
4. We can not use the river water for our day to day life.

Question 5.
The freshwater available on earth is only 3%. We cannot increase the amount of water. In that case, how can sustain the water level?
Answer:

1. The sewage water treatment is to be adopted.
2. Decrease the usage of pesticides, insecticides, and fertilizers in agriculture.
3. Protect forest and trees.
4. Adopt drip irrigation and sprinkler irrigation in agriculture.
5. Rainwater harvesting should be implemented in every building.
6. Create awareness about the impact of throwing wastes into the water bodies.

IX. Answer in detail:

Question 1.
What is potable water? List down its characteristics?
Answer:

1. Potable water is the water which is safe to drink.
2. On average the human body requires 2-3 litres of water per day for proper functioning.

Characteristics of potable water :

1. Cleaned of harmful contaminants.
2. It is transparent.
3. It is odourless and colourless.
4. It is harmless or free from disease-causing bacteria.

Question 2.
Who is known as Waterman of India? Browse the net and find the details about the award, the Waterman received for water management. State the findings by drafting a report.
Answer:
The ‘Waterman of India’ is Dr. Rajendra Singh. He is a well-known water conservationist and environmentalist from Alwar district, Rajasthan.

Awards:

1. He get Ramon Magasaysay award for community leadership in 2001.
2. In 2005, he got Jamnalal Bajaj award for Science and Technology for rural development.
3. In 2008, The Guardian, named him amongst its list of 50 people who could save the planet.
4. In 2015, he won Stockholm Water Prize. He runs an NGO called ‘Tarun Bharat Sangh’ (TBS), which was founded in 1975.
5. In 2016, he was bestowed with Ahimsa Award by Institute of Jainology based in UK.

Question 3.
What is rainwater harvesting? Explain in a few sentences how it can be used in houses.
Answer:

1. Direct collection and use of rain water is called rainwater harvesting.
2. The system is easy to install, operate and maintain for all types of houses.
3. Excellent and valuable source of water in emergencies.
4. Reduces rainwater runoff and solve drainage problems in houses.
5. Ideal solution for inadequacy of water.
6. Increase groundwater level.

X. Questions based on Higher Order Thinking Skills :

Question 1.
When there is no pond or lake in an area, will there be the formation of clouds possible in that area?
Answer:
Yes, the formation of clouds is possible at that area because plants also release water vapour by transpiration process. This water vapour will form clouds.

Question 2.
To clean the spectacles, people often breathe out on glasses to make them wet. Explain why do the glasses become wet.
Answer:
When we breathe out, the hot air comes out from our mouth which get condensed and changed into tiny water droplets and glasses become wet.

XI. Crossword :

DOWN:
1. A method of water conservation.
2. Process of getting water vapour from sea water.
6. Water stored in dams is used for generation of _____
ACROSS
3. _______ is a large body of non-potable water found in nature.
4. In summer, the body loses water as _______
5. Plants undergo _______ and contribute to water cycle.

Answer:
Down:
1. Recycling
2. Evaporation
6. Electricity

Across:
3. Ocean
4. Sweat
5. Transpiration

XI.

Question 1.
Observe the given graph carefully and answer the questions.

a. What percentage of water is seen in fish?
b. Name the food item that has maximum amount of water in its content.
c. Name the food item that has minimum amount of water in its content.
d. Human body consists of about _________ percentage of water.
e. Specify the food item that can be consumed by a person when he/she is suffering from dehydration.
Answer:
a. 70% of water is seen in fish.
b. Watermelon has maximum amount of water in its content.
c. Fish has minimum amount of water in its content.
d. 60%.
e. Watermelon is the food item which can be consumed by a person when he/she is suffering from dehydration.

Question 2.
Look at the map of Tamilnadu showing annual rainfall and answer the questions given below.

a. Identify the districts that get only low annual rainfall in Tamilnadu.
b. Identify the districts that get a medium annual rainfall in Tamilnadu.
c. State the districts that enjoy high annual rainfall in Tamilnadu.
Answer:
a. Dharmapuri, Erode, Karur, Trichy, Perambalur, Tanjore, Pudukottai, Dindugul, Madurai, Sivagangai, Ramanathapuram.
b. Thiruvallur, Chennai, Kanchipuram, Vellore, Krishnagiri, Thiruvannamalai, Salem, Namakkal, Vizhupuram, Ariyalur, Thiruvarur, Thiruppur, Virudunagar, Tuticorin.
c. Cuddalore, Nagappatinam, Nilgiri, Coimbatore, Theni, Thirunelveli, Kanyakumari.

Samacheer Kalvi 6th Science Water Intext Activities

Activity 1

Relative amount of water at various sources

Take a 20 litre bucket, a 500 ml mug, a 150 ml tumbler and a 1 ml spoon. If the capacity of the bucket is 20 litre, then it represents the total amount of water present on the Earth.

Now, transfer a mug of water from the bucket and it is 500 ml and then it represents the total amount of fresh water present in the Earth. The water left in the bucket represents seas and oceans. This water is not fit for human use.

The water present in the mug represents the freshwater which is present in frozen form on snow-covered mountains, glaciers and polar ice caps. This water is also not readily available for human use. Next, transfer 150 ml of water to the tumbler, then it represents the total amount of ground water. Finally, take one-fourth spoonful of water while the capacity of the spoon is 1 ml, then it represents the total amount of surface water (i.e) water seen in all the rivers, lakes and ponds of the world. It can be taken as potable water.

When such a small amount of potable water is available, then we should be more cautious in handling water. Is it not?
Answer:
Activity to be done by the students themselves

Activity 2
Conduct the activity with common salt, sand, chalk powder, charcoal powder and copper sulphate.

Fill up the following table.

From the above activity, we could observe that common salt and copper sulphate dissolve in water and contribute their properties like colour and other properties to water but sand, chalk powder and charcoal powder do not dissolve in water.
Answer:

Activity 3

Water contains dissolved salts

Take some tap water in a china dish and heat it. Continue heating till all the water gets dried up. Stop the heating and look at the china dish. What do you observe inside the china dish?

Deposits of some solid particles on the surface of china dish can be observed. The deposit is of salts that are dissloved in water. This shows that water has dissolved salts in it.

Note : Do not use distilled water or water from purifier or RO unit and the like for this activity.

Activity 4
Spread a piece of wet cloth in the sunlight. Observe after some time. Where has the water in the wet cloth gone?

The water evaporates into the atmosphere due to the heat of the sun.

Activity 5

Condensation of water vapour

Take a glass half filled with water. Wipe the outer surface of the glass with a clean piece of cloth. Add some ice into the water. Wait for one or two minutes. Observe the changes that take place on the outer surface of the glass.

From where do water drops appear on the outer side of the glass?

The cold surface of the glass containing icy water cools the air around it and the water vapour of the air condenses on the surface of the glass. This process is also the result of condensation of water vapour.

Activity 6

Estimation of water consumed by a family on a day

Answer:

Samacheer Kalvi 6th Science Water Additional Questions

I. Choose the appropriate answer:

Question 1.
_______ form of water is present in mountain and polar region.
(a) Solid
(b) Liquid
(c) Gaseous
(d) All these
Answer:
(a) Solid

Question 2.
The melting point of Ice is ……….
(a) 100°C
(b) 0°C
(c) 0.100°C
(d) 10°C
Answer:
(b) 0°C

Question 3.
_______ water contains 0.05% to 1% of salts.
(a) Brackish
(b) Fresh
(c) Sea
(d) Sewage
Answer:
(b) Fresh

Question 4.
The human body requires ………. litres of water per day for proper functioning.
(a) 2 – 3 litres
(b) 3 – 4 litres
(c) 4 – 5 litres
(d) 1 – 2 litres
Answer:
(b) 3 – 4 litres

Question 5.
In plants, the loss of water in the form of vapour from the aerial parts through _______
(a) Root
(b) Shoot
(c) Stomatal Pores
(d) Fruit
Ans:
(c) Stomatal Pores

Question 6.
_______ water is obtained through open wells, tube wells or hand pumps, springs.
(a) Surface
(b) Frozen
(c) Saline
(d) Ground
Ans:
(d) Ground

Question 7.
_______ of Asia’s largest rivers flow from the Himalayas.
(a) 10
(b) 9
(c) 11
(d) 15
Answer:
(a) 10

Question 8.
Volume of liquid is measured by _______
(a) Gallon
(b) Litre
(c) Cusec
(d) Estuaries
Answer:
(c) Mangrove

Question 9.
_______ forests are found in pichavaram near Chidambaram
(a) Green
(b) Grass lands
(c) Mangrove
(d) Estuaries
Answer:
(c) Mangrove

II. Fill up the blanks:

1. _______ plays a vital role in the evolution and survival of life.
2. _______ is present in the air around us.
3. In the distribution of total 0.3% of surface water, _______ have 87% surface water.
4. The molecular formula of water is _______
5. The oceanic volcanoes which are present inside also add _______ to the sea.
6. Every year _______ is observed as the world water day.
7. The water vapour gets cooled and changes into tiny water droplets that form _______ in the sky.
8. A larger portion of water is _______ % of the total available fresh water in frozen state.
9. Water level in the reservoirs is measured in _______
10. Adoption of _______ and irrigation in agriculture.

Answers:

1. Water
2. Vapour
3. Lakes
4. H₂O
5. Salt
6. March 22nd
7. Clouds
8. 68.7
9. Cubic feet per second (cusecs)
10. Drip, Sprinkler

III. True or False. If False, give the correct statement:

Question 1.
Mountains helps to regulate the temperature of our earth.
Answer:
False. Water helps to regulate the temperature of our earth.

Question 2.
Solid form of water is present in underground.
Answer:
False. Liquid form of water is present in underground.

Question 3.
Water while passing through layers of soil dissolves salt and minerals to a maximum extent.
Answer:
True.

Question 4.
Water freeze at 100° Celsius at normal pressure.
Answer:
False. Water freeze at 0° celsius at normal pressure.

Question 5.
When the air around the clouds is cool, these drops of water fall in the form of snow or rain.
Answer:
True.

Question 6.
Direct collection and use of rain water is called rain water harvesting.
Answer:
True.

Question 7.
Estuaries are harmful to unique plants and animal species.
Answer:
False. Estuaries are home to unique plants and animal species.

IV. Analogy:

Question 1.
Ice berg : Solid form :: Water vapour : _______
Answer:
Gaseous form.

Question 2.
Contain more than 3% of salt: _______ :: upto 3% of salt dissolved : Brackish water.
Answer:
Sea water.

Question 3.
Water bodies meet the sea : Estuaries :: Wet lands : _______
Answer:
Swamps.

Question 4.
Napier Bridge area : _______ :: Pallikkaranai: Wet land.
Answer:
Estuaries.

V. Give Short Answer:

Question 1.
Give the uses of water.
Answer:
Cooking food, washing cloths, cleaning utensils, bathing, agricultural etc.

Question 2.
What are the three forms of water?
Answer:
Water is available in solid, liquid and vapour forms.

Question 3.
Which places have fresh water?
Answer:
Ponds, pools, rivers, tube wells have fresh water.

Question 4.
How is 3% of fresh water distributed?
Answer:
Polar ice caps and glaciers – 68.7%
Groundwater – 30.1%
Another source of water – 0.9%
Surface water – 0.3%

Question 5.
Define – Saline water.
Answer:
Water with a large amount of dissolved solids is not potable or suitable for drinking. Such water is called saline water.

Question 6.
How water is classified based on its salinity?
Answer:
Based on salinity, water is classified into three main categories. Such as Fresh water, Brackish water and Sea water.

Question 7.
What are the salts dissolved in sea water?
Answer:
The salts are dissolved in sea water are sodium chloride, magnesium chloride and calcium chloride.

Question 8.
Define – Water cycle.
Answer:
The water on the earth evaporates into the atmosphere due to the heat of the sun. The water vapour in the atmosphere forms clouds. From the clouds, water falls on the earth in the form of rain or snow. By this natural process, water gets renewed. This is called water cycle.

Question 9.
What is Condensation?
Answer:
Water vapour which enters into the atmosphere by evaporation moves upward with air, gets cooled and changes into tiny water droplets that form clouds in the sky. This process is known as condensation.

Question 10.
What is transpiration?
Answer:
It is the process of loss of water from the aerial parts of a plant in vapour form. This is called transpriation.

Question 11.
What is frozen water?
Answer:
Water that is present in the frozen form as polar ice caps and glaciers are called frozen water. A large portion of water is 68.7% of the total available fresh water is in frozen state.

Question 12.
Define – Scarcity of water.
Answer:
There is no change in.the total quantity of water available on the earth. It remains the same. But the water useful for plants, animals and man is decreasing day by day. It is called Scarcity of water.

Question 13.
List the wet lands in Tamilnadu.
Answer:

1. Pichavaram Mangroves near Chidambaram.
2. Muthupet Mangrove wet-land.
3. Pallikkaranai wet-land in Chennai.
4. Chembarambakkam in Kancheepuram are a few examples of swamps in Tamilnadu.

Question 14.
What is meant by conservation of water?
Answer:
Saving water for the future generations by using water carefully and in a limited way is conservation of water.

VI. Answer in detail:

Question 1.
Draw and describe the total water on earth by pie chart.
Answer:
If the total water on earth be 100%, let’s see what percent would be the availability of fresh water. Look at the pie chart given below
From the pie chart, it can also be noted that 97% water is saline water. Only 3% found is the freshwater and that too in polar ice caps and glaciers. So this portion of water is not readily available for drinking.

The distribution of the totally available (3%) freshwater is as follows:
Polar ice caps and glaciers – 68.7%
Ground water – 30.1 %
Other source of water – 0.9%
Surface water – 0.3%
The distribution of total 0.3% of surface water is as follows:
Lakes – 87%
Rivers – 2%
Swamps – 11%
Thus the above pie chart explains that we have a very small amount of fresh water available for human usuage and so maintaining the water table and the conservation of water is very essential.

Question 2.
What are the methods of water conservation? Explain?
Answer:
1. Water management :

• Bringing awareness to the people.
• Recycling water by separating pollutants.
• Minimize the use of fertilizers.
• Controlling deforestation.
• Adopt drip and sprinkler irrigation in agriculture.

2. Rainwater harvesting:

• Collecting water from where it falls:
• Collecting water from the roof tops.
• Collecting flowing rain water :
• Collecting rain water by constructing dams.

Question 3.
Give the importance of water.
Answer:
The importance of water:

1. Our body uses water in all its cells, organs and tissues to help regulate its temperature and maintain other bodily functions.
2. On an average, the human body requires 2-3 litres of water per day for proper functioning.
3. Water helps in digestion of food and removal of toxins from the body.
4. Water is used in the domestic activities like cooking, bathing, washing clothes, washing utensils, keeping houses and common places clean, watering plants, etc.
5. It is also essential for the healthy growth of farm crops and farm stock and is used in the manufacture of many products.
6. Industry depends on water at all levels of production.

Question 4.
Describe water distribution and treatment system through flow chart.
Answer:

Question 5.
What are the reasons for scarcity of water?
Answer:
The main reasons for scarcity of water are:

1. Population explosion.
2. Uneven distribution of rainfall.
3. Decline of ground watertable.
4. Pollution of water.
5. Careless use of water.

Students can Download English Lesson 3 I Dream of Spices Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 6th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations

Tamilnadu Samacheer Kalvi 6th English Solutions Term 1 Poem Chapter 3 I Dream of Spices

I Dream Of Spices Poem Overview

I Dream Of Spices Questions And Answers Glossary

cinnamon – the bark of a tree that gives a delicious flavour to food
garlic – a small bulb with a strong taste used in cooking
rattle off – recite
sesame seeds – gingelly seeds
ouch – sound that expresses pain

Read and Understand

A. Answer the following questions.

I Dream Of Spices Poem Summary Question 1.
Who is Raj?
Answer:
Raj is a little boy.

Dream Of Spices Question 2.
Where did Raj’s mother send him?
Answer:
Raj’s mother sent him to a shop.

I Dream Of Spices Poem Question 3.
Who is Muthu?
Answer:
Muthu is the owner of a shop.

I Dream Of Spices Mind Map Question 4.
What did mother ask Raj to buy?
Answer:
His mother asked him to buy some cinnamon, betel leaves, ginger and garlic.

Samacheer Kalvi Guru 6th English Question 5.
What did Raj buy?
Answer:
Raj bought sesame seeds,onions, tomatoes and pickles.

B. Choose the correct answers.

1. Mother called _______
(a) Muthu
(b) Raj
(c) Ram
Answer:
(b) Raj

2. Mother did not ask for _______
(a) cinnamon
(b) cardamom
(c) betel leaves
Answer:
(b) cardamom

Spices Poem Question 3.
Raj did not buy _______
(a) onions and sesame
(b) ginger and garlic
(c) tomato and pickles
Answer:
(b) ginger and garlic

Dream About Spices Appreciating The Poem

C Find an example of alliteration in the poem.

sesame – seeds
what – want
ginger – garlic

D. Listen to the poem read by your teacher.

Read the poem aloud in pairs. One person reads out Raj’s words and the other reads the mother’s. Take turns and read.
(To be done by the students)

E. Tell the stoiy of the poem In three or four sentences with the help of the pictures given below.

Answer:

• Mother asks her son to go to shop and buy a few items.
• The boy goes to shop and buys different items.
• He returns home.
• Mother sees the items & punishes him.

Writing

F. Read the jumbled lines from the poem and rearrange them in correct order.

Answer:
4. My mother would say:
6. “Little boy Raj…
3. Go to Muthu’s
5. and get some
1. cinnamon, betel leaves
2. and ginger and garlic.”
7. And so I go to the shops
8. singing all the way
13. and when Muthu asks me
12. what I’d want
11. I rattle off a list:
10. “Sesame seeds, onions
9. tomatoes and pickles”
15. And back home,
14. My mother twists my ears
16. Ouch!

G. Fill in the blanks with different words and write, your own poem.

Your Title for the poem : Me and Mani!
My mom would say :
“Little boy / girl Mano
Go to Mani’s
and get some
Tomato, Brinjai
Carrot and Onion”
And so I go to the Mani’s
singing all the way
and when Mani asks me
what I want
I rattle off a list:
“Biscuit, Candies
Cakes and Wafers”
And back home,
My mom twists my ears
Ouch!

I Dream of Spices Additional Questions

I. Poem Comprehension.

1. My mother would say:
“Little boy Raj….
Go to Muthu s

Question a.
Who would say to go to shop?
Answer:
Raj’s mother.

Question b.
How does she call Raj?
Answer:
Little boy

2. get some
cinnamon, betel leaves
and ginger and garlic. ”

Question a.
Where will he get the items?
Answer:
Muthu’s shop.

Question b.
What should he buy?
Answer:
Cinnamon, betel leaves, ginger, garlic.

3. And so I go to the shops
singing all the way

Question a.
Who goes to shops?
Answer:
Raj goes to shops.

Question b.
How does he go?
Answer:
He goes singing all the way.

4. when Muthu asks me
what I’d want
I rattle off a list:
“Sesame seeds, onions
tomatoes and pickles ’’

Question a.
What does Muthu ask?
Answer:
Muthu asks Raj what he would want.

Question b.
List the things he will buy?
Answer:
Sesame seeds, onions, tomatoes and pickles.

5. And back home,
Mother twists my ears
Ouch!

Question a.
What does mother do?
Answer:
Mother twists Raj’s ears.

Question b.
Why does she do so?
Answer:
He forgot the list of items told by her.

II. Poetic Devices.

1. And back home
Mother twists my ears.

What Is the poetic devise used in the second line?
Answer:
Alliteration – Mother – my

2. Cinnamom, beetal leaves
and ginger and garlic

Pick out the Alliteration.
Answer:
Alliterati on – ginger – garlic.

III. Paragraph Questions.

Question 1.
Describe Raj’s experience in helping his mom at shopping?
Answer:
This is a simple and interesting poem by Raj Arumugam on a little boy’s memory while shopping for his Mom. One-day when Raj’s Mom asks him to buy cinnamon, betel leaves, ginger and garlic but Raj bought sesame seeds, onions, tomatoes and pickles. He gets lovingly punished by his Mom.

Question 2.
How did Raju react when his mother asked him to buy a list of items?
Answer:
Raju’s mother called him and gave him a list of items to be bought from Muthu’s shop. She told him to get some cinnamon, betel leaves, ginger and garlic. But, being a little boy, he goes to the shop, singing happily all the way. When the shop owner asks him what does he want, he forgets what his mother had told him to buy and recites different items like sesame seeds, onions, tomatoes and pickles. When he gets back home, his mother gets angry and twists his ears due to his poor memory.

I Dream of Spices Summary

This is a simple and interesting poem by Raj Arumugam on a little boy’s memory while shopping for his Mom. One-day when Raj’s Mom asks him to buy cinnamon, betel leaves, ginger and garlic but Raj bought sesame seeds, onions, tomatoes and pickles. He gets lovingly punished by his Mom.

You can Download Samacheer Kalvi 9th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Science Solutions Chapter 15 Carbon and its Compounds

Samacheer Kalvi 9th Science Carbon and its Compounds Textbook Exercises

I. Choose the correct answer.

Chapter 15 Carbon And Its Compounds Question 1.
A phenomenon in which an element exists in different modification in same physical state is called ……………….
(a) isomerism
(b) allotropy
(c) catenation
(d) crystallinity
Answer:
(b) allotropy

Carbon And Its Compounds 9th Class Question 2.
Number of free electron(s) in each carbon of graphite is ………………….
(a) one
(b) two
(c) three
(d) four
Answer:
(c) three

Carbon And Its Compounds Class 9 Samacheer Question 3.
Carbon forms large number of organic compounds due to ………………………
(a) Allotropy
(b) Isomerism
(c) Tetravalency
(d) Catenation
Answer:
(a) Allotropy

9th Science Carbon And Its Compounds Question 4.
Which of the following does not contain double bond?
(a) CO₂
(b) C₂H₄
(C) HCl
(d) O₂
Answer:
(C) HCl

Carbon And Its Compounds Class 9 Question 5.
Raagav brings his lunch every day to school in a plastic container which has resin code number 5. The container is made of …………….
(a) Polystyrene
(b) PVC
(c) Polypropylene
(d) LDPE
Answer:
(c) Polypropylene

Carbon And Its Compounds Samacheer Kalvi Question 6.
Plastics made of Polycarbonate (PC) and Acrylonitrile Butadiene Styrene (ABS) are made of resin code ………………
(a) 2
(b) 5
(c) 6
(d) 7
Answer:
(d) 7

Carbon And Its Compounds Question 7.
The lead pencil contains ………………
(a) graphite
(b) diamond
(c) lead
(d) charcoal
Answer:
(a) Graphite

Carbon And Its Compounds Solutions Question 8.
Graphene is one atom thick layer of carbon obtained from ……………..
(a) diamond
(b) fullerene
(c) graphite
(d) gas carbon
Answer:
(c) graphite

Class 9 Science Chapter 15 Notes Question 9.
The legal measures to prevent plastic pollution come under the ……………….. Protection Act 1988.
(a) forest
(b) wildlife
(c) environment
(d) human rights
Answer:
(c) Environment

II. Fill in the blanks.

1.  ………….. named carbon.
2. Buckminster Fullerene contains ……………. carbon atoms.
3. Compounds with same molecular formula and different structural formula are known as ………………
4. Different methods of formation of carbon is the main reason for its ………………..
5. There are …………….. plastic resin codes.

Answer:

1. Antoine Lavoisier
2. 60
3. isomer
4. Allotropy
5. 7

III. Match the following.

Answer:

1. (d) Triple bond
2. (c) Graphene
3. (a) Bucky Ball
4. (e) Polystyrene
5. (b) Oxidation

IV. Answer in brief.

Class 9 Science Chapter 15 Exercise Questions And Answers Question 1.
Differentiate graphite and diamond.
Answer:

Science Chapter 15 Class 9 Question 2.
What are saturated and unsaturated compounds?
Answer:
Saturated carbon compounds are called alkanes as they have single bond between carbon atoms. Unsaturated carbon compounds are called alkenes as they have one or more double bonds between carbon atoms.

Question 3.
Carbon do not form ionic compounds. Why?
Answer:
Carbon shares electrons only through covalent bonding, hence it does not form ionic compounds.

Question 4.
What is the valency of carbon in carbon monoxide?
Answer:
Valency of carbon in carbon monoxide is 2.

Question 5.
Why are one-time use and throwaway plastics harmful?
Answer:
Use and throwaway plastics cause short and long-term environmental damage as they are difficult to recycle. They stay in an environment for over 1000 years.

These block drains and pollute water bodies. One-time use plastic causes health problems for humans, plants and animals. Some examples are plastic carry bags, cups, etc.

V. Answer in detail.

Question 1.
What is catenation? How does carbon form catenated compounds?
Answer:
Catenation is the binding of an element to itself or with other elements through covalent bonds to form open chain or closed chain compounds. .
Carbon is the most common element which undergoes catenation and forms long chain compounds. Carbon atom links repeatedly to itself through covalent bond to form linear chain, branched chain or ring structure. This property of carbon itself is the reason for the presence of large number of organic carbon compounds. So organic chemistry essentially deals with catenated carbon compounds.

Question 2.
What are the chemical reactions of carbon?
Answer:
Elemental carbon undergoes no reaction at room temperature and limited number of reactions at elevated temperatures. But its compounds undergo large number of reactions even at room temperature.
Oxidation – (Reaction with oxygen)
Carbon combines with oxygen to form its oxides like carbon monoxide (CO) and carbon dioxide (CO₂) with evolution of heat. Organic carbon compounds like hydrocarbon also undergo oxidation to form oxides and steam with evolution of heat and flame. This is otherwise called Burning,
2C_(s) + O_2(g) ➝ 2CO_(g) + heat
C_(s) + O_2(g) ➝ CO_2(g) + heat

• Reaction with steam
  Carbon reacts with steam to form carbon monoxide and hydrogen. This mixture is called water gas.
  C_(s) + H_(s)O_(g) ➝ CO_(g) + H_2(g)
• Reaction with sulphur
  With sulphur, carbon forms its disulphide at high temperature.
  C_(s) + S_(s) ➝ CS_2(g)
• Reaction with metals
  At elevated temperatures, carbon reacts with some metals like iron, tungsten, Titanium, etc. to form their carbides.
  W_(s) + C_(g) ➝ WC_(s)

Question 3.
Name the three safer resin codes of plastics and describe their features.
Answer:

VI. HOTS

Question 1.
Why do carbon exist mostly in combined state?
Answer:
Carbon is an element that can form many different compounds, as each carbon atom can form 4 chemical bonds with other atoms and because the carbon atom is just the right size to fit in comfortably as parts of very large molecules.

Question 2.
When a carbon fuel burns in less aerated room, it is dangerous to stay there. Why?
Answer:
When carbon fuel bums in less aerated room, carbon monoxide is formed. When people are exposed to CO, it enters into human body through breathing and affects the function of haemoglobin. CO displaces oxygen from haemoglobin thereby stops its function (supply of oxygen to the parts of body) leading to death.

Question 3.
Explain how dioxins are formed? Which plastic type they are linked to and why they are harmful to humans?
Answer:
Burning PVC plastic releases dioxins which are one of the most dangerous chemicals known to humans.

Question 4.
Yugaa wants to bdy a plastic water bottle. She goes to the shop and sees four different kinds of plastic bottles with resin codes 1, 3, 5 and 7. Which one should she buy? Why?
Answer:
Yuga should by plastic bottle with resin code 5, as it is considered to be more safer.

Activity

Question 1.
With the help of your teacher, try to classify the following compounds and materials and, . fill in the table accordingly. HCN, CO₂, Propane, PVC, CO, Kerosene, LPG, Coconut oil, Wood, Perfume, Alcohol, Na₂CO₃, CaCO₃, MgO, Cotton, Petrol.
Answer:

Samacheer Kalvi 9th Science Carbon and its Compounds Additional Questions

I. Answer briefly.

Question 1.
Explain carbon cycle or biogeochemical cycle.
Answer:
The carbon cycle is a biogeochemical cycle by which carbon is exchanged among the biosphere, geosphere, hydrosphere and atmosphere of the Earth. Carbon is the main component of biological compounds as well as major component of many minerals such as lime stone.

Steps in carbon cycle:

1. Carbon enters the atmosphere as CO₂ from respiration and combustion.
2. CO₂ is absorbed by producers to make carbohydrates in photosynthesis.
3. Animals feed on the plant passing the carbon compounds along the food chain. Most of the carbon they consume is exhaled as CO₂, formed during respiration. The animals and plants eventually die.
4. The dead organisms are eaten by decomposers and the carbon in their bodies is returned to the atmosphere as CO₂. The plant and animal material may then be available as fossil fuel in the future for combustion.

Question 2.
What is isomerism? Explain with an example.
Answer:
Isomerism is a special feature of catenated organic compounds.
For example, if we consider the molecular formula of an organic compound to be C₂H₆O, we will not be able to name the compound. This is because the molecular formula of an organic compound represents only the number of different atoms present in that compound. It does not indicate the way in which the atoms are arranged and hence its structure.

The phenomenon in which the same molecular formula may exhibit different structural arrangements is called isomerism.
The given formula C₂H₆O can be represented with two kinds of arrangements.

Both the compounds have same molecular but different kind of arrangement. In compound ‘a’ the oxygen atom is attached to a hydrogen and a carbon. It is an alcohol. Whereas, compound ‘b’ the oxygen atom is attached to two carbon atoms and it is an ether. Both these compounds have different physical and chemical properties.

Question 3.
Write a short note on graphene.
Answer:
Graphene is most recently produced allotrope of carbon which consists of honeycomb-shaped hexagonal ring repeatedly arranged in a plane. Graphene is the thinnest compound known to man at one atom thick. It is the lightest material known (with 1 square metre weighing around 0.77 milligrams) and the strongest compound discovered (100-300 times stronger than steel).

It is the best conductor of heat at room temperature. Layers of graphene are stacked on top of each other form graphite, with an interplanar spacing of 0.335 nanometres. The separate layers of graphene in graphite are held together by Vander Waals forces.

Question 4.
What is carbon monoxide? Why is it toxic?
Answer:
Carbon monoxide is a toxic gas of carbon. When fuels undergo incomplete combustion, it results in the formation of carbon monoxide. It is released into the atmosphere from various sources like vehicle fuels, domestic fuels, industries, furnaces, etc, cigarette smoking also is a source of carbon monoxide.

Carbon monoxide is an odorless toxic gas. When CO enters the human body through breathing, it affects the function of hemoglobin. CO displaces oxygen from hemoglobin thereby stopping its function, i.e., a supply of oxygen is blocked to the parts of the body, thereby leading to death.

Question 5.
What are the drawbacks of plastics?
Answer:
Drawbacks of plastics

• Plastics take a long time to fully break down in nature.
• The microbes that breakdown plastics are too few and the quantity of plastics we produce is too many.
• A lot of plastics does not get recycled and ends up polluting the environment.
• Some plastics contain harmful chemical additives that are not good for human health.
• Burning of plastics releases toxic gases that are harmful to health and environment
• One time use and throwaway plastics end up littering and polluting the environment.

Question 6.
What is a resin code? What is the need to display them on plastic articles?
Answer:
The resin code is a set of symbols appearing on plastic products that is helpful in the identification of the type of polymer used to make that product.

Platic needs to be recycled or disposed of safely. Each plastic is composed of a different polymer or set of molecules. Different molecules do not mix when plastics are recycled. For this reason, plastics need to be segregated before recycling. These resin codes help in easy segregation of plastic products to be recycled.

Question 7.
A survey showed that one time use plastic items were among the top 10 plastic items found in garbage washed up from oceans. Can this be true? Explain how.
Answer:
Use and throwaway plastics cause short and long time environmental damage. These items like plastic carry bags, straws, plates, spoons, pouches, etc., take a few seconds to be made in a factory, but can stay in the environment forever a 1000 years. These items block drains and pollute water bodies. These plastics break down into pieces that are smaller than 5mm in diameter. These microplastics pollute the ocean and harm the marine life, who mistake them as their food.

These one time use plastics when disposed carelessly are carried to the ocean by winds, cause harm to marine life and are brought back to the shore by waves.

Students can Download English Lesson 2 The Mystery of the Cyber Friend Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 8th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th English Solutions Term 3 Supplementary Chapter 2 The Mystery of the Cyber Friend

The Mystery Of The Cyber Friend Summary Read And Understand

A. Say whether the following are ‘True’ or ‘False’.

1. Shree spends most of the time on T.V.
2. Shrees aunt stays with them.
3. Chaitra is Shrees school friend.
4. Chaitra gifted Shree a new camera phone.
5. Shree went alone to the train station to meet Chaitra.
6. A fraud middle aged man pretended to be Chaitra. Name the speaker.

Answer:

1. False
2. True
3. False
4. False
5. False
6. True

B. Name the speaker.
Answer:

C. Answer the following questions

The Mystery Of The Cyber Friend Questions And Answers Question 1.
Why did Shree’s parents buy her a computer?
Answer:
Shrees parents bought her a computer because they wanted her to learn computers.

The Mystery Of The Cyber Friend Question 2.
How did Shree make friends through computer?
Answer:
Two months ago, on her thirteenth birthday, Shree joined “Friends Net”. Through this net, she made friends.

The Mystery Of Cyber Friend Question 3.
What were the online activities given in this story?
Answer:
Chaitra tells lies to Shree, saying that she is studying in a school near to Shree’s school. She used a film stars picture as her profile. She asked Shree to come to the raiway station all alone.

The Mystery Of The Cyber Friend Book Back Answers Question 4.
How did Shree’s aunt save Shree from the man who pretended to be Chaitra?
Answer:
Shrees aunt accompanied her to the railway station. She informed the station manager about the new friend and asked him to help them. When she saw the man talking to Shree, she hit him with her handbag. He ran away into the crowded train.

Summary Of The Mystery Of The Cyber Friend Question 5.
How did the police find the man who pretended to be Chaitra?
Answer:
A Cyber crime officer went through Shrees computer within hours he found the man, who pretended to be Chaitra.

D. Based on your reading of the text, list out the merits and demerits of using a computer.
Answer:

The Mystery Of The Cyber Friend Story Step To Success

8th English The Mystery Of The Cyber Friend Question 1.
The following series is provided and you need to answer the question accordingly.
ABCDEFGHIJKLMN| |OPRSTUVWXYZ
In this series find the letter which is fifth to the left from the thirteenth letter from your right.
(a) M
(b) I
(c) H
(d) J
Answer:
(c) H

The Mystery Of The Cyber Friend Summary In English Question 2.
Based on the above series of English alphabets, if every alternate alphabet starting from C is deleted than which of the following alphabet will seventh from the left side of the series?
(a) H
(b) K
(c) I
(d) G
Answer:
(b) K

The Mystery Of The Cyber Friend Mind Map Question 3.
If the above series is written in reverse order than what is the eleventh letter of the fifteenth letter from your left?
(a) V
(b) W
(c) D
(d) X
Answer:
(b) W

The Mystery of the Cyber Friend Additional Questions

I. Choose the correct answer.

The Mystery Of The Cyber Friend Lesson Plan Question 1.
Shree lives in an apartment in a small town called ___________ Junction.
(a) Thirupathi
(b) Katpadi
(c) Gudur
(d) Thiruvallur
Answer:
(b) Katpadi

Mystery Of The Cyber Friend Question 2.
They want her to learn ___________
(a) arts
(b) singing
(c) music
(d) computers

The Mystery Of The Cyber Friend Pdf Question 3.
After tiffin, shree likes to ___________
(a) read books
(b) watch TV
(c) play
(d) sleep
Answer:
(c) play

The Mystery Of The Cyber Friend In Tamil Question 4.
The computer also has ___________ net.
(a) Friends
(b) Visitors
(c) Chat
(d) Vice
Answer:
(a) Friends

Question 5.
Shree says that she got extra sugar with her ___________
(a) coffee
(b) milk
(c) tea
(d) juice
Answer:
(c) tea

Question 6.
She chats with her ___________ friends through Friends Net.
(a) school
(b) college
(c) office
(d) best
Answer:
(a) school

Question 7.
Chaitra ___________ Shree to accept her as a new friend.
(a) pleaded
(b) begged
(c) requested
(d) ordered
Answer:
(c) requested

Question 8.
Soon Shree gets a ___________ from Chaitra.
(a) message
(b) warning
(c) threatening
(d) gift
Answer:
(a) message

Question 9.
Chaitra offers to give Shree her old ___________
(a) photos
(b) phone
(c) paintings
(d) laptop
Answer:
(b) phone

Question 10.
Best friends should be of the same ___________
(a) size
(b) race
(c) age
(d) status
Answer:
(c) age

II. dentify the character and speaker:

1. “Do you have many friends?”
2. “I lost a few friends today”
3. “Are you feeling unwell”.
4. “Too much of school work”
5. “Hello, Friend. How are you today?”
6. “Great. Now can you send me a selfie?”
7. “I have a new friend, Akka!”
8. “But that’s the film actress I like so much, Madhoo.”
9. “Don’t you dare come near my niece”
10. “You are clever to confide in a trusted adult.”

Answer:

1. Chaitra
2. Shree
3. Shree’s aunt
4. Shree
5. Chaitra
6. Chaitra
7. Shree
8. Shree’s aunt
9. Shree’s aunt
10. Policewoman

III. Give Very Short Answers :

Question 1.
Who was Shree’s best friend?
Answer:
The computer was Shree’s best friend.

Question 2.
What does Shree like to play on the computer?
Answer:
Shree likes to play games on the computer.

Question 3.
When did Shree join the Friends Net?
Answer:
She joined the friends Net two months ago on her thirteenth birthday.

Question 4.
How was Chaitra’s profile?
Answer:
Chaitra’s profile was very pretty. She looked like a film star.

Question 5.
Did Shree see Chaitra’s school? Why?
Answer:
No, Shree couldn’t see Chaitra’s school because it was not there.

Question 6.
Why didn’t Shree eat her tiffin?
Answer:
She couldn’t eat her tiffin because she was eagerly waiting to talk to Chaitra.

Question 7.
Where did Chaitra want to meet Shree ?
Answer:
Chaitra wanted to meet Shree at the Railway Station.

Question 8.
What did chaitra promise to give Shree?
Answer:
Chaitra promised to give her an old camera phone.

Question 9.
Who came to meet Shree at the Railway Station?
Answer:
A man, who was her father’s age, came to meet Shree.

IV. Write Short Answers :

Question 1.
Where did Shree live and what were his parents?
Answer:
Shree lived in an apartment in a small town called Katpadi Junction. Her mother works in a jewelery shop. Her father works as a taxi driver.

Question 2.
What does Shree do everyday?
Answer:
After snacks and tea, Shree chats with her school friends through Friend Net. They tell each other what they did everyday.

Question 3.
Who stays with Shree’s Family? What do everyone calls her?
Answer:
Shree’s aunt stays with their family. Everyone calls her “Akka”. Most of the time, she sleeps in front of the TV.

Question 4.
What did Shree’s aunt tell about Chaitra?
Answer:
Shree’s aunt tells Shree that Chaitra’s profile is not her. It is the picture of her favourite actress Madhoo.

Question 5.
What happend to the imposter at the Railway Station?
Answer:
Shree’s aunt hit the imposter with her handbag. The Station Manager rushed to catch the man. But he disappeared into the crowded train.

V. Answer in Detail:

Question 1.
How was the Cyber friend of Shree caught by the police?
Answer:
Shree went to meet her friend Chaitra at the railway station. Her aunt accompanied her. She informed about the imposter to the Station Manager and requested him to help them. When they were waiting at the station, a man came towards Shree. He was of Shree’s father’s age. He tried to convince Shree and requested her to be his friend. Seeing him talking to her neice, Shree’s aunt hit him with her handbag. The Station Manager also rushed forward to catch him. But he disappeared in the crowded train. Later the Cyber Crime officer went through Shree’s computer. He found the man and nabbed him in Bengaluru.

VI. Rearrange the jumbled sentences :

Question A.
1. So they are pleased that she likes to use it.
2. Appa works as a taxi driver.
3. They want her to learn computers.
4. Shree lives in an apartment in a small town called Katpadi Junction.
5. Amma works in a jewellery shop.
Answer:
4. Shree lives in an apartment in a small town called Katpadi Junction.
5. Amma works in a jewellery shop.
2. Appa works as a taxi driver.
3. They want her to learn computers.
1. So they are pleased that she likes to use it.

Question B.
1. She watched her favourite old films.
2. Shrees aunt also stays with them.
3. She gives Shree hot dosas stuffed with spicy noodles.
4. Everyone calls her Akka.
5. Most of the time, she naps in the front of the TV.
Answer:
2. Shrees aunt also stays with them.
4. Everyone calls her Akka.
5. Most of the time, she naps in the front of the TV.
1. She watched her favourite old films.
3. She gives Shree hot dosas stuffed with spicy noodles.

Question C.
1. Within hours, the police find the man who pretended to be Chaitra.
2. They nab him from his office in Bengaluru.
3. They find out that he has been trying to befriend many young girls and boys on social media.
4. Shree decides to only have friends from her own school.
5. The next day, a cyber crime officer goes through Shrees computer.
Answer:
5. The next day, a cyber crime officer goes through Shrees computer.
1. Within hours, the police find the man who pretended to be Chaitra.
2. They nab him from his office in Bengaluru.
3. They find out that he has been trying to befriend many young girls and boys on social media.
4. Shree decides to only have friends from her own school.

VII. Comprehension.

Question 1.
The next morning, Shree looks for school near hers. But she cannot see one. It is a little strange that she hasn’t heard of any other school nearby. But Chaitra is nicer than all her other friends. She is Shrees special friend and nobody else’s friend.

(a) What did Shree look for?
Answer:
Shree looked for Chaitra’s School.

(b) Did she see the school?
Answer:
No, She did not see the school.

(c) What did Shree think about Chaitra?
Answer:
She thought that Chaitra was nicer than all her other friends.

(d) Whose friend is Chaitra?
Answer:
Chaitra is Shrees special friend.

Question 2.
But Shree doesn’t do her homework. Instead, she starts the computer, goes online and waits for her new friend to ping her. Soon she gets a message from Chaitra.
“Hello, friend. How are you today?”

(a) Does Shree do her homework?
Answer:
No, Shree doesn’t do her homework.

(b) What does Shree do, instead doing of her homework?
Answer:
Shree starts the computer, goes online and waits for her new friend to ping her.

(c) What does Shree get from Chaitra?
Answer:
Shree gets a message from Chaitra.

(d) What was the message?
Answer:
The message was “Hello Friend. How are you today?”

The Mystery of the Cyber Friend Summary

Section – I

Shree lived in an apartment with her father, mother and aunt. Her parents wanted her to learn computers. So they were pleased with her, when she used it. She used to find information for school projects and send e-mails. Two months ago, on her thirteenth birthday, she joined the “Friend Net” on the computer. She used to chat with her friends everyday. One day, a new friend requested her to accept her. Shree accepted her and chatted with her. Her name was Chaitra. She told Shree that she was studying in a school, which was near to her school. Shree considered her to be her special friend. On a Sunday, Ghaitra asked her to meet her at the railway station.

Her aunt saw Chaitra’s profile picture. She told Shree that it was actually the picture of her favourite heroine Madhoo. It was not the picture of Chaitra. Shree could not find any school near her school. Chaitra had also asked Shree to come to the Railway Station all alone. All these things, created suspicion in Shree about Chaitra.

Section – II

Shree was very confused and couldn’t focus on her homework. Shree informed about Chaitra to her aunt. The next day Shree’s aunt accompanied Shree to the railway station. She informed the station manager and waited to see who would come to see Shree. A man who was of her father’s age met her. He tried to convince Shree that he was an uncle and he would like to be her friend. Shree’s aunt hit the man with her handbag. The station manager tried to catch him, but he disappeared into the crowded train. They went to the police station. The policewoman praised them and told Shree to put up a poster about cycber security in her school.

Next day, a cyber crime officer went through Shree’s computer. Within hours, the police found the man, who pretended to be Chaitra. They nabbed him from his office in Bengaluru. They found that he had been trying to befriend many young girls and boys on social media. Shree waited eagerly to tell her friends about her frightening adventure with her cyber friend.

Immediately get Chapter Wise Tamilnadu State Board Solutions for 11th Economics to gain more marks and start Quick Revision of all Chapters. All Chapters Pdf is provided along with the Questions and Answers. You Can Dowload Samacheer Kalvi 11th Economics Book Solutions Questions and Answers for Chapter wise are given with a clear explanation. Tamilnadu State Board Solutions for 11th Economics Chapter 3 Production Analysis Questions and Answers is for free of cost.

Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 3 Production Analysis

Tamilnadu State Board Solutions for 11th Economics Chapter 3 Production Analysis Questions and Answers PDF has all given in Chapter Wise Section. Check Out daily basis with Tamilnadu State Board Solutions 11th Economics PDF will help to improve your score. Improve your level of accuracy to answer a question by reading with Samacheer Kalvi 11th Economics Book Solutions Questions and Answers PDF.

Samacheer Kalvi 11th Economics Production Analysis Text Book Back Questions and Answers

Part – A

11th Economics Chapter 3 Book Back Answers Multiple Choice Questions

Samacheer Kalvi Guru 11th Economics Question 1.
The primary factors of production are:
(a) Labour and Organisation
(b) Labour and Capital
(c) Land and Capital
(d) Land and Labour.
Answer:
(d) Land and Labour.

Samacheer Kalvi Guru 11 Economics Question 2.
The man-made physical goods used to produce other goods and services are referred to as …………………………
(a) Land
(b) Labour
(c) Capital
(d) Organization
Answer:
(c) Capital

Samacheer Kalvi 11th Economics Solutions Question 4.
Which factor is called the changing agent of society?
(a) Labourer
(b) Land
(c) Organizer
(d) Capital
Answer:
(c) Organizer

Economics Samacheer Kalvi Question 5.
Who said, that one of the key of an entrepreneur is “uncertainty-bearing”.
(a) J.B.Clark
(b) Schumpeter
(c) Knight
(d) Adam Smith
Answer:
(c) Knight

Economics Class 11 Samacheer Kalvi Question 6.
The functional relationship between “inputs” and “outputs” is called………………………..
(a) Consumption Function
(b) Production Function
(c) Savings Function
(d) Investment Function
Answer:
(b) Production Function

Samacheer Kalvi Economics 11th Question 7.
In a firm 5 units of factors produce 24 units of the product. When the number of factor increases by one, the production increases to 30 units. Calculate the Average Product.
(a) 30
(b) 6
(c) 5
(d) 24
Answer:
(c) 5

Samacheer Kalvi 11th Solutions Question 8.
The short-run production is studied through ………………………
(a) The Laws of Returns to Scale
(b) The Law of Variable Proportions
(c) Iso – quants
(d) Law of Demand
Answer:
(b) The Law of Variable Proportions

11th Economics Samacheer Kalvi Question 9.
The long-run production function is explained by
(a) Law of Demand
(b) Law of Supply
(c) Returns to Scale
(d) Law of Variable Proportions
Answer:
(c) Returns to Scale

Economics Class 11 Chapter 3 Notes Question 10.
An Iso – quant curve is also known as …………………………
(a) Inelastic Supply curve
(b) Inelastic Demand Curve
(c) Equi – marginal Utility
(d) Equal Product Curve
Answer:
(d) Equal Product Curve

11 Th Samacheer Kalvi Economics Question 11.
Mention the economies reaped from inside the firm
(a) Financial
(b) Technical
(c) Managerial
(d) All of the above
Answer:
(d) All of the above

11th Economics Solutions Samacheer Kalvi Question 12.
Cobb – Douglas production function assumes ………………………….
(a) Increasing returns to scale
(b) Diminishing returns to scale
(c) Constant returns to scale
(d) All of the above
Answer:
(c) Constant returns to scale

Economics Class 11 Chapter 3 Question Answers Question 13.
Name the returns to scale when the output increases by more than 5%, for a 5% increase in the inputs,
(a) Increasing returns to scale
(b) Decreasing returns to scale
(c) Constant returns to scale
(d) All of the above
Answer:
(a) Increasing returns to scale

Question 14.
Which of the following is not a characteristic of land?
(a) It’s a limited supply
(b) It is mobile
(c) Heterogeneous
(d) Gift of Nature
Answer:
(b) It is mobile

Question 15.
Product obtained from additional factors of production is termed as
(a) Marginal product
(b) Total product
(c) Average product
(d) Annual product
Answer:
(a) Marginal product

Question 16.
Modem economists have propounded the law of ………………………
(a) Increasing returns
(b) decreasing returns
(c) Constant returns
(d) variable proportions
Answer:
(a) Increasing returns

Question 17.
Producer’s equilibrium is achieved at the point where:
(a) The marginal rate of technical substitution(MRTS) is greater than the price ratio
(b) MRTS is lesser than the price ratio
(c) MRTS and price ratio are equal to each other
(d) The slopes of Iso-quant and Iso-cost lines are different.
Answer:
(c) MRTS and price ratio are equal to each other

Question 18.
The relationship between the price of a commodity and the supply of a commodity is ……………………….
(a) Negative
(b) Positive
(c) Zero
(d) Increase
Answer:
(b) Positive

Question 19.
If average product is decreasing, then marginal product
(a) Must be greater than average product
(b) Must be less than average product
(c) Must be increasing
(d) Both a and c
Answer:
(b) Must be less than average product

Question 20.
A production function measures the relation between ……………………….
(a) Input prices and output prices
(b) Input prices and the quantity of output
(c) The number of inputs and the quantity of output
(d) The number of inputs and input prices.
Answer:
(c) The number of inputs and the quantity of output

Part – B

Answer the following questions in one or two sentences

Question 21.
Classify the factors of production.
Answer:
Factors of production are classified as land, labour, capital and organisation.

1. Land and Labour – Primary factors
2. Capital and organisation – Secondary factors.

Question 22.
Define labour?
Answer:

1. Labour is the active factor of production.
2. In common parlance, labour means manual labour or unskilled work. But in Economics the term ‘labour’ has a wider meaning.
3. It refers to any work undertaken for securing an income or reward. Such work may be manual or intellectual. For example, the work done by an agricultural worker or a cook or rickshaw puller or a mason is manual.
4. The work of a doctor or teacher or an engineer is intellectual.
5. In short, labour in economics refers to any type of work performed by a labourer for earning an income.

Question 23.
State the production function.
Answer:
Production function is the relationship between inputs of productive services and .outputs of product per unit of time.

Question 24.
Define Marginal Product of a factor?
Answer:
It is the addition or the increment made to the total product when one more unit of the variable input is employed. In other words, it is the ratio of the change in the total product to the change in the units of the input. It is expressed as
MP = ∆TP/∆N
Where MP = Marginal Product
∆TP = Change in total product
∆N = Change in units of input
It is also expressed as MP = TP [n] – TP [n-1]
MP = Marginal product
TP [n] = Total product of employing nth unit of a factor.
TP [n – 1] = Total product of employing the previous unit of a factor, that is, [n – 1]^th unit of a factor.

Question 25.
What is the Iso-cost line?
Answer:
An Iso-cost line represents different combinations of inputs which shows the same amount of cost.

Question 26.
What are the conditions for the producer’s equilibrium?
Answer:
The two conditions that are to be fulfilled for the attainment of producer equilibrium are:

1. The isocost line must be tangent to the isoquant curve.
2. At the point of tangency, the isoquant curve must be convex to the origin or MRTSLk must be declining.

Question 27.
What are the reasons for the upward-sloping supply curve?
Answer:
As the price of the commodity increases the quantum supplied also increases. So the supply curve has a positive slope.

Part – C

Answer the following questions in One Paragraph

Question 28.
What are the characteristics of land?
Answer:

1. Land is a primary factor of production
2. Land is a passive factor of production
3. Land is the free gift of nature
4. Land has no cost of production
5. Land is fixed in supply. It is inelastic in supply
6. Land is permanent
7. Land is immovable
8. Land is heterogeneous as it differs infertility
9. Land has alternative uses
10. Land is subject to Law of Diminishing Returns.

Question 29.
What are the factors governing elasticity of supply?
Answer:
1. Nature of the commodity:
The elasticity of supply of durable goods are high but perishables have low elasticity of supply.

2. Cost of production:
Under constant or increasing returns the elasticity of supply is greater, under diminishing returns elasticity is less.

3. Technical condition:
In large scale production supply cannot be adjusted easily. So elasticity of supply is lesser and vice versa.

4. Time factor:
During very short period, supply cannot be adjusted. In short period, variable factors can be changed so elasticity is more and in long period, supply is highly elastic.

Question 30.
What are the functions of entrepreneurs?
Answer:
Functions of an Entrepreneur:

• Initiation:
  An organizer is the initiator of the business, by considering the situation and availability of resources and planning the entire process of business of production.
• Innovation:
  A successful entrepreneur is always an innovator. He introduces new methods in the production process.
• Coordination:
  An organizer applies a particular combination of the factor of production to start and run the business or production.
• Control, Direction, and Supervision: An organizer controls so that nothing prevents the organization from achieving its goal. He directs the factors to get better results and supervises for the efficient functioning of all the factors involved in the process of production.
• Risk-taking and uncertainty-bearing: There are risk-taking and uncertainty-bearing obstacles. Risks may be insured but uncertainties cannot be insured. They reduce the profit.

Question 31.
State and explain the elasticity of supply.
Answer:
The elasticity of supply may be defined as the degree of responsiveness of change in supply to change in price on the part of sellers.
Mathematically:

Q – Supply, p – price, ∆ – change.

Question 32.
Bring out the Relationship among Total, Average and Marginal Products.
Answer:

Question 33.
Illustrate the concept of Producer’s Equilibrium.
Answer:
Producer equilibrium is the situation where the producer maximizes his output. It is also known as the optimum combination of the factors of production.
(Eg.) Maximum output at minimum cost.
Producer attains equilibrium where the Iso-cost line is tangent to the Iso-quant.

In the figure profit of the firm is maximised at the point of equilibrium.
At the point of equilibrium slope of the Iso-cost line = Slope of Iso-product curve At the point E, the firm employs OM units of labour and ON units of capital which is the least cost combination.

Question 34.
State the Cobb – Douglas Production Function?
Answer:
The Cobb – Douglas Production Function was developed by Charles W. Cobb and Paul H. Douglas, based on their empirical study of the American manufacturing industry. It is a linear homogeneous production function which implies that the factors of production can be substituted for one another up to a certain extent only.

The Cobb – Douglas production function can be expressed as follows.
Q = AL²K²
Where, Q = output; A = positive constant;
K = capital;
L = Labor α and β are positive fractions showing, the elasticity coefficients of outputs for the inputs labor and capital, respectively.
P = (1 – a) since a + p = 1. denoting constant returns to scale.
Factor intensity can be measured by the ratio β / α.

The sum of α + β shows the returns to scale.

1. (α + β) = 1, constant returns to scale.
2. (α + β) < 1, diminishing returns to scale.
3. (α + β) >1, increasing returns to scale.

• The production function explains that with the proportionate increase in the factors, the output also increases in the same proportion.
• Cobb – Douglas production function implies constant returns to scale.
  Cobb – Douglas Production Function is a specific standard equation applied to describe how much output can be made with capital and labour inputs.

Part – D

Answer the following questions in about a page

Question 35.
Examine the Law of Variable Proportions with the help of a diagram.
Answer:
According to G. Stigler, “As equal increments of one input are added, the inputs of other productive services being held constant, beyond a certain point the resulting increments of the product will decrease (ie) the marginal product will diminish”.
Assumptions:

1. Only one factor is variable.
2. All units of the variable factor are homogeneous
3. The product is measured in physical units.
4. There is no change in the state of technology and the price of the product.

Stage of production:

Units of variable factor (labour) are employed along with other fixed factors. There are three stages of production.

Stage I:

1. In the first stage MP_L increases upto third labourer and is higher than the average product.
2. So that total produt is increasing at an increasing rate.

Stage II:

1. MP_L decreases up to sixth unit of labour where MP_L curve intersects the x axis.
2. At fourth unit of labour MP_L = AP_L
3. After this, MP_L curve is lower than the AP_L. TP_L increases at a decreasing rate.

Stage III:

1. The seventh unit of labour is marked by negative MP_L the AP_L continues to fall but positive.
2. TP_L declines with the employment of more units of labour.

Question 36.
List out the properties of Iso-quants with the help of diagrams.
Answer:
Iso-quant:
“An Iso-quant is a curve showing all possible combinations of inputs physically capable of producing a given level of output” – Ferguson.

Properties of Iso-quant curve:
1. The Iso-quant curve has negative slope:

1. It slopes downwards from left to right indicating that the factors are substitutable.
2. In the diagram combination A refers to more of capital K₅ and less of labour L₂
3. As the producer moves to B, C and D more labour and less capital are used.

2. Convex to the origin:

1. In production, the capital substituted by 1 unit of labour goes on decreasing. It is called diminishing marginal rate of technical substitution.
2. If the marginal rate of technical substitution diminishes the Iso-quants will be convex to the origin.
3. Constant MRTS (straight line) and increasing MRTS (concave).

3. Non-inter-section of Iso-Quant curves:

1. Point A lie on the Iso-quants IQ₁ and IQ₂
2. But the point C shows a higher output and point B shows a lower level of output IQ₁
3. If C = A, B = A, then C – B. But C > B which is illogical.
   

4. An upper Iso-quant curve represents a higher level of output:

1. Higher IQ_s show higher outputs and lower IQ_s show lower output.
2. The arrow in the figure shows an increase in the output with a right and upward shift of an Iso-quant curve.

5. Iso-quant curve does not touch either X axis or Y axis:
No Iso-quant curve touches the X – axis or Y – axis because in IQ(, only capital is used and in IQ₂ only labour is used.

Question 37.
Elucidate the Laws of Returns to Scale. Illustrate.
Answer:
In the long-run all factors are variable. The laws of returns to scale explain the relationship between output and the scale of inputs in the long-run when all the inputs are increased in the same proportion.
Assumptions :

1. All the factors are variable except the organization.
2. There is no change in technology.
3. There is perfect competition in the market.
4. Outputs or returns are increased in physical quantities.

Three phases of returns to scale:
1. Increasing returns to scale:
If all inputs are increased by one percent, output increase by more than one percent.

2. Constant returns to scale:
In this case, if all inputs are increased by one percent, output increases by one percent.
Diagrammatic Illustration:

In the diagram, the movement from point a to point b represents increasing returns to scale. Between these two points, input has doubled but the output was tripled.

The law of constant returns is implied by the movement from point b to point c. Between these two points inputs have doubled and output also has doubled.

Decreasing returns to scale are denoted by the movement from the point c to point d since doubling the factors from 4 units to 8 units produce less than the increase in inputs, that is by only 33.33%.

Question 38.
Explain the internal and external economies of scale.
Answer:
The scale of production is an important factor affecting production. Every producer wishes to reduce the costs of production by using economies of scale.
Marshall classified economies into two.
1. Internal economies of scale: Internal economies refer to the reduction in the cost of production of the commodity. They are of various types.

• Technical economies: When the size and capital of the firm is large, up-to-date technologies can be used to improve the productivity of the firm. Here research and development strategies can be applied easily.
• Financial economies: Big firms can float shares in the market for capital expansion, while small firms cannot do so.
• Managerial economies: Large scale production facilitates specialization and delegation.
• Labour economies: Large scale production implies greater and minute division of labour. This leads to specialization which enhances the quality. This increases the productivity of the firm.
• Marketing economies: The producers can both buy raw-materials in bulk at a cheaper cost and can take the products to distant markets.
• Economies of survival: Product diversification is possible and it reduces the risk in production. Even if the market for one product collapses, the market for other commodities offsets it.

Question 39.
External economies of scale:
Answer:
External economies of scale refer to changes in any factor outside the firm or industry causing an improvement in the production process.
Important external economies are:

1. Increased transport facilities.
2. Banking facilities
3. Development of townships
4. Development of information and communication

Samacheer Kalvi 11th Economics Production Analysis Additional Questions and Answers

Question 1.
Production refers to …………………….
(a) Destruction of utility
(b) Creation of utilities
(c) Exchange value
(d) None
Answer:
(b) Creation of utilities

Question 2.
_______ is a person who combines the factors of production in the production process to earn profit.
(a) Producer
(b) Entrepreneur
(c) Consumer
(d) Labours
Answer:
(b) Entrepreneur

Question 3.
Division of labour is introduced by ………………………
(a) Adam Smith
(b) Marshall
(c) Schumpeter
(d) Hawley
Answer:
(a) Adam Smith

Question 4.
Total product is
(a) TP = ∑AP
(b) TP = ∑MP
(c) TP = ∆N
(d) TP = TP / N
Answer:
(b) TP = ∑MP

Question 5.
The Initial supply price of land is ……………………..
(a) Zero
(b) Greater than one
(c) Less than one
(d) Equal to one
Answer:
(a) Zero

Question 6.
Cobb – Douglas production function depends on ______
(a) Land, Labour
(b) Land, Capital
(c) Labour, capital
(d) Labour, organisation
Answer:
(c) Labour, capital

Question 7.
Who is the primary factor of production?
(a) Organizer
(b) Capital
(c) Machine
(d) Man
Answer:
(a) Organizer

Question 8.
_______ is the man-made factor of production.
(a) Labour
(b) Capital
(c) Savings
(d) Land
Answer:
(c) Savings

Question 9.
Technological relationship between inputs and output is called ……………………….
(a) Production function
(b) Technical function
(c) Capital function
(d) Organiser function
Answer:
(a) Production function

Question 10.
In the production process _______ bears incertainity.
(a) Manager
(b) Labour
(c) Entrepreneur
(d) None
Answer:
(c) Entrepreneur

Question 11.
α + β = 1 refers _______
(a) Increasing returns to scale
(b) Constant return to scale
(c) Diminishing returns to scale
(d) None
Answer:
(b) Constant return to scale

Choose the correct statement

Question 1.
(a) Production refers to the creation of utility
(b) Production means the transformation of inputs into output
(c) Production creates economic well-being.
(d) The scale of production influences the revenue.
Answer:
(d) The scale of production influences the revenue.

Question 2.
(a) An entrepreneur is a risk-taker – Walker.
(b) Capital consists of all kinds of wealth other than the free gift of nature -Marshall.
(c) An entrepreneur innovates – Schumpeter.
(d) An entrepreneur is an uncertainty bearer – Walker.
Answer:
a. An entrepreneur is a risk-taker – Walker.

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

Question 3.

a. 3 4 1 2
b. 1 2 3 4
c. 4 1 2 3
d. 2 1 3 4
Answer:
(c) Active factor – (3) Cobb-Douglas

Question 4.

a. 2 4 1 3
b. 1 2 3 4
c. 3 4 1 2
d. 4 1 2 3
Answer:
(a) (α + β) =1 – (1) Diminishing returns.

Question 5.

Answer:
a. 1 2 3 4
b. 4 1 2 3
c. 3 1 2 4
d. 2 3 1 4
Answer:
(c) Perfectly inelastic supply = (3) Es < 1

Question 6.

a. 2 3 1 4
b. 3 4 1 2
c. 4 1 2 3
d. 3 4 2 1
Answer:
(b) Labour = (2) Initiator

Choose the odd one out

Question 7.
(a) Land
(b) Labour
(c) Production
(d) Organisation
Answer:
(c) Production

Question 8.
(a) Technical economics
(b) Financial economics
(c) Economics of survival
(d) Banking facility
Answer:
(d) Banking facility

Choose the correct statement

Question 9.
(a) Iso-quant has a positive slope.
(b) Iso-quants never intersect each other
(c) Iso-quants are concave to the origin
(d) Iso-quant curve touch either X or Y-axis.
Answer:
(b) Iso-quants never intersect each other

Question 10.
Cobb-Douglas production function.
(a) Implies constant returns to scale
(b) It considers all the factors of production.
(c) The elasticity of substitution between the factors is less than one.
(d) Is a non-linear homogeneous production function.
Answer:
(a) Implies constant returns to scale

Analyze the reason for the following

Question 11.
Assertion (A) : Labour is the human input into the production process.
Reason (R) : Labour cannot be separated from the labourer.
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 true but, (R) is false.
Answer:
b. Both (A) and (R) are true, (R) is not the correct explanation of (A)

Question 12.
Assertion (A) : Perishables have a low elasticity of supply.
Reason (R) : Perishables cannot be stored for a long time.
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. (A) is true but (R) is false.
d. (A) is false but (R) is true.
Answer:
a. Both (A) and (R) are true, (R) is the correct explanation of (A)

Choose the incorrect pair

Question 13.

Answer:
(a) Q = AL^α K^β

Question 14.

Answer:
(d) Q = f (N,L,K,T)

Fill in the blanks with the suitable option given below

Question 15.
Cobb-Douglas production function depends on _____
(a) Land, labour
(b) Land, capital
(c) Labour, capital
(d) Labour, organisation
Answer:
(c) Labour, capital

Question 16.
Total product is _____
(a) TP = ∑AP
(b) TP = ∆N
(c) TP = ∑MP
(d) TP = \(\frac { TP }{ N } \)
Answer:
(c) TP = ∑MP

Question 17.
In the production process __________ bears uncertainty.
(a) Manager
(b) Labour
(c) Worker
(d) Entrepreneur
Answer:
(d) Entrepreneur

Choose the best option

Question 18.
is the man-made factor of production.
(a) Labour
(b) Capital
(c) Savings
(d) Land
Answer:
(b) Capital

Question 19.
The labour exercised without expecting income is __________
(a) Service
(b) Physical labour
(c) Mental labour
(d) None of the above
Answer:
(a) Service

Question 20.
Marginal product is ________
(a) MP = \(\frac { AP }{ N } \)
(b) MP = \(\frac { TP }{ N } \)
(c) MP = \(\frac { ∆AP }{ ∆N } \)
(d) MP = \(\frac { ∆TP }{ ∆N } \)
Answer:
(d) MP = \(\frac { ∆TP }{ ∆N } \)

Part – B

Answer the following questions in one or two sentences

Question 1.
What are the characteristics of capital?
Answer:

1. Capital is a man-made factor
2. Capital is mobile between places and persons
3. Capital is a passive factor of production
4. Capital’s supply is elastic
5. Capital’s demand is a derived demand
6. Capital is durable

Question 2.
What is the land?
Answer:
Land means all gifts of nature owned and controlled by human beings which yield an income.

Question 3.
Explain the law of variable proportions assumptions?
Answer:

1. The Law of Variable Proportions is based on the following assumptions.
2. Only one factor is variable while others are held constant.
3. All units of the variable factor are homogeneous.
4. The product is measured in physical units.
5. There is no change in the state of technology.
6. There is no change in the price of the product.

Question 4.
What is the total product?
Answer:
Total product is the summation of marginal products.
TP = ∑MP

Question 5.
What is average product ?
Answer:
Average product refers to the output per unit of the input.

Question 6.
What is diseconomies of scale ?
Answer:
Diseconomies of scale are a disadvantage to a firm or industry or organisation. This necessarily increases the cost of production.

Question 7.
What are Iso-quants ?
Answer:
An Iso-quant curve can be defined as the locus of points representing various combinations of two inputs yielding the same output.

Part – C

Answer the following questions in One Paragraph

Question 1.
State the characteristics of labour.
Answer:

1. Labour is an active factor of production.
2. Labour may be manual or intellectual.
3. Labour is perishable.
4. Labour is inseparable from the labourer.
5. Labour is less mobile.
6. Labour is a means as well as an end.
7. Labour units are heterogeneous.
8. Labour has weak bargaining power.

Question 2.
Define Iso – quant, and Iso – quants Assumptions?
Answer:
Definition of Iso – quant:
According to Ferguson, “An Iso – quant is a curve showing all possible combinations of inputs physically capable of producing a given level of output”.

Assumptions:

1. It is assumed that only two factors are used to produce a commodity.
2. Factors of production can be divided into small parts.
3. The technique of production is constant.
4. The substitution between the two factors is technically possible. That is, the production function is of “variable proportion” type rather than a fixed proportion.
5. Under the given technique, factors of production can be used with maximum efficiency.

Question 3.
Who is an entrepreneur?
Answer:
An entrepreneur is a person who combines land, labour and capital in the production process to earn a profit. He not only runs the business but bears the risk of the business.

Question 4.
Differentiate the short period and long period.
Answer:
The short-run is the period where some inputs are variable. Another feature is that firms do not enter or exit the industry. The long-run is the period where all the inputs are variable. It is featured by the entry of new firms and the exit of existing firms from the industry.

Question 5.
Name the classification of a production function.
Answer:
Production function may be classified into two.

1. Short-run production function as illustrated by the law of variable proportions.
2. Long-run production function as explained by the laws of returns to scale.

Question 6.
What is an Iso-quant map?
Answer:
An Iso-quant map has different Iso-quant curves representing the different combinations of factors of production, yielding the different levels of output. An Iso-quant map is a family of Iso-quants..

Part – D

Answer the following questions in about a page

Question 1.
What are the characteristics of labour?
Answer:

1. Labour is the animate factor of production.
2. Labour is an active factor of production.
3. Labour implies several types: It may be manual [farmer] or intellectual [teacher, lawyer, etc]
4. Labour is perishable.
5. Labour is inseparable from the labourer.
6. Labour is less mobile between places and occupations.
7. Labour is a means as well as an end. It is both the cause of production and consumer of the product.
8. Labour units are heterogeneous.
9. Labour differs in ability.
10. Labour – supply determines its reward [wage]
11. Labour has weak bargaining power.

Question 2.
Explain the law of supply with a diagram.
Answer:
1. Law of supply:
The Law of supply describes a direct relationship between the price of a good and the supply of that good.

2. Definition:
“Other things remaining the same, if the price of a commodity increases its quantity supplied increases and if the price of a commodity decreases, quantity supplied also decreases”.

3. Supply function:
Q = f (P_x, P_r, P_r T, O, E)
Q_s – Quantity supplied of x commodity
P_x – Price of x commodity
P_r – Price of related goods
P_f – Price of factors of production.
T – Technology
O – Objective of the producer.
E – Expected price of the commodity

Assumptions:

1. There is no change in the prices of factors of production and capital goods.
2. Natural resources, technology, climate, political situations and tax policy remain unchanged.
3. Prices of substitutes are constant.

Supply schedule :
A supply schedule shows the different quantities of supply at different prices

Supply curve:
The quantum supplied is represented on X axis. And the price of the commodity is represent on the Y axis

As the price of the commodity increases, the quantum supplied of the commodity also increases. Thus the supply curve has a positive slope. The points such as e, d, c, b and a on the supply curve SS, represent various quantities of different prices.

Question 3.
Explain the factors determining supply.
Answer:
1. Price of the commodity:
Higher the price larger the supply Price is the incentive for the producers and sellers to supply more.

2. Price of other commodities:
The supply of a commodity not only depends upon its price but also prices of other commodities. For instance if the price of commercial crops increases, the supply of food crops may decrease.

3. Price of factors:
When the input prices go up, this results in rise in cost and so supply will be affected.

4. Price expectations:
The expectation over future prices determines present supply.

5. Technology:
With advancement in technology, production level improves, average cost declines and as a result supply level increases.

6. Natural factors:
In agriculture, natural factors like a monsoon, climate, etc., play a vital role in determining the production level.

7. Discovery of new raw materials:
The discovery of cheaper and high-quality raw materials tends to increase the supply of the product.

8. Taxes and subsidies:
Subsidies encourage the producers to produce more whereas taxes kill the ability and willingness to produce more.

9. The objective of the firm:
When the goal of the firm is sales maximization or improving market share, the supply of the product is likely to be higher.

Question 4.
Explain the characteristics of labour.
Answer:
According to Marshall, labour represents services provided by the factor labour, which helps in yielding an income to the owner of the labour-power.
Characteristics of labour:

1. Labour is the animate factor of production.
2. Labour is an active factor of production.
3. Labour implies several types it may be manual or intellectual.
4. Labour is perishable.
5. Labour is inseparable from the labourer.
6. Labour is less mobile between places and occupations.
7. Labour is a means as well as an end.
8. Labour units are heterogenous.
9. Labour-supply determines its reward.
10. Labour has weak bargaining power.

If you need some more information about Tamilnadu State Board Solutions for 11th Economics Chapter 3 Production Analysis Questions and Answers then visit our site frequently and get all the resources that you look for. Also, share this page with your friends and make them learn all the concepts thoroughly before the exams.

If you are looking for the best material to read Bio Botany Chapter 1 Asexual and Sexual Reproduction in Plants Questions and Answers, Notes then Download Samacheer Kalvi 12th Bio Botany Book Solutions Guide Pdf. You can learn all the topics and subtopics by referring to the Tamilnadu State Board 12th Bio Botany Chapter 1 Asexual and Sexual Reproduction in Plants Questions and Answers. Samacheer Kalvi12th Bio Botany Subject Material is the notes that you are looking for. Majority of students love to use Samacheer Kalvi Bio Botany Material while preparing for the exam.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 1 Asexual and Sexual Reproduction in Plants

Every concept of Tamilnadu State Board 12th Bio Botany Subject PDF is explained clearly in an understandable way. Make sure to answer all the Samacheer Kalvi 12th Questions on your own then look for our explanations to make it more easy. We have included all the topics and subtopics to help students for better learning. Best learning comes when you practice with Samacheer Kalvi Board Solutions for 12th Bio Botany Chapter 1 Asexual and Sexual Reproduction in Plants Questions and Answers.

Samacheer Kalvi 12th Bio Botany Asexual and Sexual Reproduction in Plants Text Book Back Questions and Answers

12th Bio Botany 1st Lesson Question 1.
Choose the correct statement from the following
(a) Gametes are involved in asexual reproduction
(b) Bacteria reproduce asexually by budding
(c) Conidia formation is a method of sexual reproduction
(d) Yeast reproduce by budding
Answer:
(c) Conidia formation is a method of sexual reproduction

Samacheer Kalvi Guru 12th Bio Botany Question 2.
An eminent Indian embryologist is ____________
(a) S.R. Kashyap
(b) P. Maheswari
(c) M.S. Swaminathan
(d) K.C. Mehta
Answer:
(b) P. Maheswari

12th Bio Botany Chapter 1 Question 3.
Identity the correctly matched pair ____________
(a) Tuber – Allium cepa
(b) Sucker – Pistia
(c) Rhizome – Musa
(d) Stolon – Zingiber
Answer:
(c) Rhizome – Musa

12th Botany 1st Lesson Question 4.
Pollen tube was discovered by ____________
(a) J.G. Kolreuter
(b) G.B. Amici
(c) E. Strasburger
(d) E. Hanning
Answer:
(b) G.B. Amici

Samacheer Kalvi Guru 12th Botany Question 5.
Size of pollen grain in Myosotis ____________
(a) 10 micrometer
(b) 20 micrometer
(c) 200 micrometer
(d) 2000 micrometer
Answer:
(a) 10 micrometer

Botany Class 12 Chapter 1 Question 6.
First cell of male gametophyte in angiosperm is ____________
(a) Microspore
(b) Megaspore
(c) Nucleus
(d) Primary Endosperm Nucleus
Answer:
(a) Microspore

Botany Chapter 1 Class 12 Question 7.
Match the following ____________
(I) External fertilization – (i) pollen grain
(II) Androecium – (ii) anther wall
(III) Male gametophyte – (iii) algae
(IV) Primary parietal layer – (iv) stamens
(a) I-(iv); II-(i); III-(ii); IV-(iii)
(b) I-(iii); II-(iv); III-(ii); IV-(i)
(c) l-(iii); II-(iv); III-(ii); IV-(i)
(d) I-(iii); II-(i); III-(iv); IV-(ii)
Answer:
(b) I-(iii); II-(iv); III-(ii); IV-(i)

Samacheer Kalvi 12 Bio Botany Solutions Question 8.
Arrange the layers of anther wall from locus to periphery
(a) Epidermis,middle layers, tapetum, endothecium
(b) Tapetum, middle layers, epidermis, endothecium
(c) Endothecium, epidermis, middle layers, tapetum
(d) Tapetum, middle layers endothecium epidermis
Answer:
(d) Tapetum, middle layers endothecium epidermis

Samacheer Kalvi 12th Botany Book Back Answers Question 9.
Identify the incorrect pair
(a) sporopollenin – exine of pollen grain
(b) tapetum – nutritive tissue for developing microspores
(c) Nucellus – nutritive tissue for developing embryo
(d) obturator – directs the pollen tube into micropyle
Answer:
(c) Nucellus – nutritive tissue for developing embryo

Samacheer Kalvi 12th Bio Botany Solutions Question 10.
Assertion : Sporopollenin preserves pollen in fossil deposits
Reason : Sporopollenin is resistant to physical and biological decomposition
(a) assertion is true; reason is false
(b) assertion is false; reason is true
(c) Both Assertion and reason are not true
(d) Both Assertion and reason are true.
Answer:
(b) assertion is false; reason is true

Samacheer Kalvi Guru 12th Biology Question 11.
Choose the correct statement(s) about tenuinucellate ovule
(a) Sporogenous cell is hypodermal
(b) Ovules have fairly large nucellus
(c) sporogenous cell is epidermal
(d) ovules have single layer of nucellus tissue
Answer:
(a) Sporogenous cell is hypodermal

Samacheer Kalvi 12th Bio Botany Guide Question 12.
Which of the following represent megagametophyte ____________
(a) Ovule
(b) Embryo sac
(c) Nucellus
(d) Endosperm
Answer:
(b) Embryo sac

12th Botany Samacheer Kalvi Question 13.
In Haplopappus gracilis, number of chromosomes in cells of nucellus is 4. What will be the chromosome number in Primary endosperm cell?
(a) 8
(b) 12
(c) 6
(d) 2
Answer:
(b) 12

Question 14.
Transmitting tissue is found in
(a) Micropylar region of ovule
(b) Pollen tube wall
(c) Stylar region of gynoecium
(d) Integument
Answer:
(c) Stylar region of gynoecium

Question 15.
The scar left by funiculus in the seed is ____________
(a) tegmen
(b) radicle
(c) epicotyl
(d) hilum
Answer:
(d) hilum

Question 16.
A Plant called X possesses small flower with reduced perianth and versatile anther. The probable agent for pollination would be ____________
(a) water
(b) air
(c) butterflies
(d) beetles
Answer:
(b) air

Question 17.
Consider the following statement(s)

1. In Protandrous flowers pistil matures earlier
2. In Protogynous flowers pistil matures earlier
3. Herkogamy is noticed in unisexual flowers
4. Distyly is present in Primula

(a) i and ii are correct
(b) ii and iv are correct
(c) ii and Hi are correct
(d) i and iv are correct
Answer:
(b) ii and iv are correct

Question 18.
Coelorhiza is found in ____________
(a) Paddy
(b) Bean
(c) Pea
(d) Tridax
Answer:
(a) Paddy

Question 19.
Parthenocarpic fruits lack ____________
(a) Endocarp
(b) Epicarp
(c) Mesocarp
(d) seed
Answer:
(d) seed

Question 20.
In majority of plants pollen is liberated at ____________
(a) 1 celled stage
(b) 2 celled stage
(c) 3 celled stage
(d) 4 celled stage
Answer:
(b) 2 celled stage

Question 21.
What is reproduction?
Answer:
Reproduction is the biological process of producing young ones of their own kind. It is a vital process for the existence of a species and it also brings suitable changes through variation in tlie offsprings for their survival on earth.

Question 22.
Mention the contribution of Hofmeister towards Embryology.
Answer:
Hofmeister described the structure of pollen tetrad

Question 23.
List out two sub-aerial stem modifications with example.
Answer:

1. Stolon – E.g.: Fragaria
2. Sucker – E.g.: Chrysanthemum

Question 24.
What is layering?
Answer:
Layering is a conventional propagation method, where the stem of a parent plant is allowed to develop roots while still intact. When the root develops, the rooted part is cut and planted to grow as a raw individual. E.g.: Jasminum.

Question 25.
What are clones?
Answer:
Individuals developed by asexual reproduction are morphologically and genetically identical. Such individuals are called as clones.

Question 26.
A detached leaf of Bryophyllum produces new plants. How?
Answer:
In Bryophyllum, the leaf is succulent and notched on its margin. Adventious buds develop at these notches and are called epiphyllous buds. They develop into new plants forming a root system and become independent plants when the leaf gets decayed.

Question 27.
Differentiate Grafting and Layering.
Answer:
Grafting:

1. In grafting, two different plants (stock & scion) are used to develop new plant.
2. The new plant will support to possess the characters of both the parents or new variation can be noticed.

Layering:

1. In layering, only one plant is used to develop new plant.
2. Variation cannot be expected. The new individual is exactly similar to parent plant.

Question 28.
Tissue culture is the best method for propagating rare and endangered plant species”- Discuss.
Answer:
Micropropagation of plants invitro through tissue culturing is a modem and alternative tool to conserve and safeguard rare plant species. Since the basic principle behind PTC is totipotency. With the help of a single explant it is possible to generate a huge population of plantlets within a short span of time. Conservation through micropropagation offers the possibility to rescue endangered and endemic species.

Question 29.
Distinguish mound layering and air layering
Answer:
Mound Layering:
In mound layering, lower flexible branch with leaves is bent to ground and a part of the stem is buried in the soil and the tip of branch is exposed above the soil. After the roots emerge from the buried stem, a cut is made in parent plant so that the buried plant grows into a new plant.

Air Layering:
In air layering, the stem is girdled at nodal part and hormones are applied and covered with moist soil using polythene sheet. Roots emerge in these branches after 2-4 months. Such branches are removed from parent plant and grown separately.

Question 30.
Explain the conventional methods adopted in vegetative propagation of higher plants.
Answer:
The common methods of conventional propagation are cutting, grafting and layering.
a. Cutting: It is the method of producing a new plant by cutting the plant parts such as root, stem and leaf from the parent plant. The cut part is placed in a suitable medium for growth. It produces root and grows into a new plant. Depending upon the part used it is called as root cutting (Malus), stem cutting (Hibiscus, Bougainvillea and Moringa) and leaf cutting (Begonia, Bryophyllum). Stem cutting is widely used for propagation.

b. Grafting: In this, parts of two different plants are joined so that they continue to grow as one plant. Of the two plants, the plant which is in contact with the soil is called stock and the plant used for grafting is called scion. Examples are Citrus, Mango and Apple. There are different types of grafting based on the method of uniting the scion and stock. They are bud grafting, approach grafting, tongue grafting, crown grafting and wedge grafting.

(i) Bud grafting: AT- shaped incision is made in the stock and the bark is lifted. The scion bud with little wood is placed in the incision beneath the bark and properly bandaged with a tape.

(ii) Approach grafting: In this method both the scion and stock remain rooted. The stock is grown in a pot and it is brought close to the scion. Both of them should have the same thickness. A small slice is cut from both and the cut surfaces are brought near and tied together and held by a tape. After 1-4 weeks the tip of the stock and base of the scion are cut off and detached and grown in a separate pot.

(iii) Tongue grafting: A scion and stock having the same thickness is cut obliquely and the scion is fit into the stock and bound with a tape.

(iv) Crown grafting: When the stock is large in size scions are cut into wedge shape and are
inserted on the slits or clefts of the stock and fixed in position using graft wax.

(v) Wedge grafting: In this method a slit is made in the stock or the bark is cut. A twig of scion is inserted and tightly bound so that the cambium of the two is joined.

c. Layering: In this method, the stem of a parent plant is allowed to develop roots while still intact. When the root develops, the rooted part is cut and planted to grow as a new plant.
Examples: Ixora mdJasminum Mound layering and Air layering are few types of layering.

(i) Mound layering: This method is applied for the plants having flexible branches. The lower branch with leaves is bent to the ground and part of the stem is buried in the soil and tip of ( the branch is exposed above the soil. After the roots emerge from the part of the stem buried in the soil, a cut is made in parent plant so that the buried part grow into a new plant.

(ii) Air layering: In this method the stem is girdled at nodal region and hormones are applied to this region which promotes rooting. This portion is covered with damp or moist soil using a ; polythene sheet. Roots emerge in these branches after 2-4 months. Such branches are removed-from the parent plant and grown in a separate pot or ground.

Question 31.
Highlight the milestones from the history of plant embryology.
Milestones in Plant Embryology
Answer:

1. 1682 – Nehemiah Grew mentioned stamens as the male organ of a flower.
2. 1694 – R.J. Camerarius described the structure of a flower, anther, pollen and ovule
3. 1761 – J.G. Kolreuter gave a detailed account on the importance of insects in pollination.
4. 1824 – G.B. Amici discovered the pollen tube.
5. 1848 – Hofmeister described the structure of pollen tetrad.
6. 1870 – Hanstein described the development of embryo in Capsella and Alisma.
7. 1878 – E. Strasburger reported polyembryony.
8. 1884 – E. Strasburger discovered the process of Syngamy.
9. 1899 – S.G. Nawaschin and L. Guignard independently discovered Double fertilization.
10. 1904- E. Hanning initiated embryo culture.
11. 1950 – D.A. Johansen proposed classification for embryo development.
12. 1964 – S. Guha and S.C. Maheswari raised haploids from Datura pollen grains
13. 1991 – E.S. Coen and E.M. Meyerowitz proposed the ABC model to describe the genetics of initiation and development of floral parts
14. 2015 – K.V. Krishnamurthy summarized the molecular aspects of pre and post fertilization reproductive development in flowering plants

Question 32.
Discuss the importance of Modern methods in reproduction of plants.
Answer:
Advantages of modern methods

1. Plants with desired characteristics can be multiplied rapidly in a short duration.
2. Plants produced are genetically identical.
3. Tissue culture can be carried out in any season to produce plants.
4. Plants which do not produce viable seeds and seeds that are difficult to germinate can be propagated by tissue culture.
5. Rare and endangered plants can be propagated.
6. Disease free plants can be produced by meristem culture.
7. Cells can be genetically modified and transformed using tissue culture.

Question 33.
What is Cantharophily?
Answer:
Pollination carried out by beetles is said to be cantharophily.

Question 34.
List any two strategy adopted by bisexual flowers to prevent self-pollination.
Answer:
(a) Protandry or protogyny
(b) Herkogamy

Question 35.
What is endothelium?
Answer:
In Asteraceae species, the inner layer of integument get specialized for nourishing the embryosac and this is called integumentary tapetum or endothelium.

Question 36.
“The endosperm of angiosperm is different from gymnosperm”. Do you agree. Justify your answer.
Answer:
Endosperm of Angiosperm:

1. Develops as a s result of double fertilization.
2. Endosperm is generally triploid (polyploid).

Endosperm of Gymnosperm:

1. Develops before the fertilization process.
2. Endosperm is haploid.

Question 37.
Define the term Diplospory.
Answer:
Diplospory is a condition where a diploid embryosac is formed from megaspore mother cells without a regular meiotic division.
E.g: Eupatorium.

Question 38.
What is polyembryony. How it can commercially exploited.
Answer:
Occurrence of more than one embryo in a seed is called polyembryony.

1. Embryos developed through polyembryony are found virus free.
2. The seedlings formed from nuclear tissue in citrus are found on better clones for orchards.

Question 39.
Why does the zygote divides only after the division of Primary endosperm cell.
Answer:
The primary endosperm nuclear (PEN) divides prior to zygotic division and form endosperm. Endosperm acts as a nutritive tissue and nourishes the developing embryo.

Question 40.
What is Mellitophily?
Answer:
Pollination carried out by Bees is said to be mellitophily.

Question 41.
“Endothecium is associated with dehiscence of anther” Justify the statement.
Answer:
The inner tangential wall develops bands (sometimes radial walls also) of a cellulose (sometimes also slightly lignified). The cells are hygroscopic. The cells along the junction of the two sporangia of an anther lobe lack these thickenings. This region is called stomium. This region along with the hygroscopic nature of endothecium helps in the dehiscence of anther at maturity.

Question 42.
List out the functions of tapetum.
Answer:

1. It supplies nutrition to the developing microspores.
2. It contributes sporopollenin through ubisch bodies thus plays an important role in pollen wall formation.
3. The pollenkitt material is contributed by tapetal cells and is later transferred to the pollen surface.
4. Exine proteins responsible for ‘rejection reaction’ of the stigma are present in the cavities of the exine. These proteins are derived from tapetal cells.

Question 43.
Write short note on Pollenkitt.
Answer:
Pollenkitt is contributed by the tapetum and coloured yellow or orange and is chiefly made of carotenoids or flavonoids. It is an oily layer forming a thick viscous coating over pollen surface. It attracts insects and protects damage from UV radiation.

Question 44.
Distinguish tenuinucellate and crassinucellate ovules.
Answer:
Tenuinucellate Ovule:

1.  Ovules with hypodermal sporogerous cell with unilayerd nucellus tissue is called tenuinucellate type.
2. They have very small nucellus

Crassinucellate Ovule:

1. Ovule with subhypodermal sporogenous cell is called crassinucellate type.
2. They have large nucellus

Question 45.
‘Pollination in Gymnosperms is different from Angiosperms’ – Give reasons.
Answer:
In gymnosperms, the ovules are exposed and the pollens are deposited directly on it. Hence the pollution is direct in gymnosperm. Whereas in angiosperms it is said to be indirect, as the pollens are deposited on stigma or the pistil.

Question 46.
Write short note on Heterostyly.
Answer:
Heterostyly: Some plants produce two or three different forms of flowers that are different in their length of stamens and style. Pollination will take place only between organs of the same lepgth.
E.g: Primula.

Question 47.
Enumerate the characteristic features of Entomophilous flowers.
Answer:
The characteristic features of entomophilous flowers are as follows:

1. Flowers are generally large or if small they are aggregated in dense inflorescence. Example: Asteraceae flowers.
2. Flowers are brightly coloured. The adjacent parts of the flowers may also be brightly coloured to attract insect. For example in Poinsettia and Bougainvillea the bracts become coloured.
3. Flowers are scented and produce nectar.
4. Flowers in which there is no secretion of nectar, the pollen is either consumed as food or used in building up of its hive by the honeybees. Pollen and nectar are the floral rewards for the visitors.
5. Flowers pollinated by flies and beetles produce foul odour to attract pollinators.
6. In some flowers juicy cells are present which are pierced and the contents are sucked by the insects.

Question 48.
Discuss the steps involved in Microsporogenesis.
Answer:
Microsporogenesis: The stages involved in the formation of haploid microspores from diploid microspore mother cell through meiosis is called Microsporogenesis. The primary sporogeneous cells directly, or may undergo a few mitotic divisions to form sporogenous tissue. The last generation of sporogenous tissue functions as microspore mother cells.

Each microspore mother cell divides meiotically to form a tetrad of four haploid microspores (microspore tetrad). Microspores soon separate from one another and remain free in the anther locule and develop into pollen grains.

Question 49.
With a suitable diagram explain the structure of an ovule.
Answer:
Structure of ovule(Megasporangium):

Ovule is also called megasporangium and is protected by one or two covering called integuments. A mature ovule consists of a Raphe stalk and a body. The stalk or the funiculus (also called funicle) is present at the base and it attaches the ovule to the placenta. The point of attachment of funicle to the body of the ovule is known as hilum. It represents the junction ovule and funicle. In an inverted ovule, the funicle is adnate to the body of the ovule forming a ridge called raphe. The body of the ovule is made up of a central mass of parenchymatous tissue called nucellus which has large reserve food materials. The nucellus is enveloped by one or two protective coverings called integuments. Integument encloses the nucellus completely except at the top where it is free and forms a pore called micropyle.

The ovule with one or two integuments are said to be unitegmic or bitegmic ovules respectively. The basal region of the body of the ovule where the nucellus, the integument and the funicle meet or merge is called chalaza. There is a large, oval, sac-like structure in the nucellus toward the micropylar end called embryo sac or female gametophyte. It develops from the functional megaspore formed within the nucellus. In some species(unitegmic tenuinucellate) the inner layer of the integument may become specialized to perform the nutritive function for the embryo sac and is called as endothelium or integumentary tapetum (Example : Asteraceae).

Question 50.
Give a concise account on steps involved in fertilization of an angiosperm plant.
Answer:
Steps involved in fertilization of angiospermic plant:

1. Germination of pollen grain on stigma.
2. Formation of pollen tube in stigma.
3. Growth of pollen tube inside the style.
4. Direction of pollen tube towards the micropyle of ovule.
5. Entry of pollen tube into the synergid of embryo sac.
6. Discharge of male gametes from the pollen tube.
7. Fusion of male gamete with egg cell (syngany)
8. Fusion of second male gamete with polar nuclei (triple fusion/double

Question 51.
What is endosperm? Explain the types.
Answer:
1. Endosperm: The primary endosperm nucleus (PEN) divides immediately after fertilization but before the zygote starts to divide, into an endosperm. The primary endosperm nucleus is the result of triple fusion (two polar nuclei and one sperm nucleus) and thus has 3n number of chromosomes. It is a nutritive tissue and regulatory structure that nourishes the developing embryo. Depending upon the mode of development three types of endosperm are recognized in angiosperms. They are nuclear endosperm, cellular endosperm and helobial endosperm.

2. Nuclear endosperm: Primary Endosperm Nucleus undergoes several mitotic divisions without cell As, wall formation thus a free nuclear condition exists in the endosperm.
Examples: Coccinia, Capsella and Arachis.

3. Cellular endosperm: Primary endosperm nucleus divides into 2 nuclei and it is immediately followed by wall formation. Subsequent divisions also follow cell wall formation.
Examples: Adoxa, Helianthus and Scoparia.

4. Helobial endosperm: Primary Endosperm Nucleus moves towards base of embryo sac and divides into two nuclei. Cell wall formation takes place leading to the formation of a large micropylar and small chalazal chamber. The nucleus of the micropylar chamber undergoes several free nuclear division whereas that of chalazal chamber may or may not divide.
Examples : Hydrilla and Vallisneria.

5. Ruminate endosperm: The endosperm with irregularity and unevenness in its surface forms ruminate endosperm. Examples : Areca catechu, Passiflora and Myristica

Question 52.
Differentiate the structure of Dicot and Monocot seed
Answer:
Dicot Seed:

1. Possess two cotyledons
2. Absence of coleoptile and coleorhiza
3. Endosperm is scarce or absent

Monocot Seed:

1. Possess only one cotyledon
2. Presence of coleoptile and colerhiza surrounding plumule and radicle respectively.
3. Endosperm from the major storage tissue.

Question 53.
Give a detailed account on parthenocarpy. Add a note on its significance.
Answer:
In some plants, fruit like structures may develop from the ovary without the act of fertilization. Such fruits are called parthenocarpic fruits. Invariably they will not have true seeds. Many commercial fruits are made seedless.
Examples: Banana, Grapes and Papaya.Nitsch in 1963 classified the parthenocarpy into following types:

1. Genetic Parthenocarpy: Parthenocarpy arises due to hybridization or mutation.
   Examples: Citrus,Cucurbita.
2. Environmental Parthenocarpy: Environmental conditions like frost, fog, low temperature, high temperature etc., induce Parthenocarpy. For example, low temperature for 3-19 hours induces parthenocarpy in Pear. Chemically
3. induced Parthenocarpy: Application of growth promoting substances like Auxins and Gibberellins induces parthenocarpy.
4. Significance: The seedless fruits have great significance in horticulture.
   • The seedless fruits have great commercial importance.
   • Seedless fruits are useful for the preparation of jams, jellies, sauces, fruit drinks etc.
   • High proportion of edible part is available in parthenocarpic fruits due to the absence of seeds.

Samacheer Kalvi 12th Bio Botany Asexual and Sexual Reproduction in Plants Additional Questions and Answers

1 – Mark Questions

Question 1.
Match the following:
(1) Conidia – (i) Yeast
(2) Budding – (ii) Bacteria
(3) Gemma cups – (iii) Aspergilus
(4) Binary fission – (iv) Marchantia
(a) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)
(b) 1 – (ii), 2 – (iv), 3 – (iii), 4 – (i)
(c) 1 – (iv), 2 – (ii), 3 – (i), 4 – (iii)
(d) 1 – (i), 2 – (iii), 3 – (ii), 4 – (iv)
Answer:
(a) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)

Question 2.
The unit of reproductive structure used in vegetative propagation is called as
(a) Diplospores
(b) Aplanospores
(c) Diaspores
(d) Condiospores
Answer:
(c) Diaspores

Question 3.
Which of the following aquatic plant is popularly known as the “Terror of Bengal”?
(a) Eichornia crassipes
(b) Vallisneria spiralis
(c) Pistia stratiotes
(d) Zostera marina
Answer:
(a) Eichornia crassipes

Question 4.
Identify the incorrect statement regarding vegetative reproduction.
Answer:
(a) Only one parent is required for propagation.
(b) New individuals are genetically dissimilar.
(c) Easy mode of reproduction.
(d) Variation does not exist.
Answer:
(b) New individuals are genetically dissimilar.

Question 5.
The genetic ability of a plant cell to produce the entire plant is said to be
(a) Multipotency
(b) Totipotency
(c) Pleuripotency
(d) Differentiation
Answer:
(b) Totipotency

Question 6.
A typical anther is
(a) Bisporangiate
(b) Tetrasporangiate
(c) Unisporangiate
(d) Multisporangiate
Answer:
(b) Tetrasporangiate

Question 7.
Match the following:
Vegetative Reproductive structures

(a) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)
(b) 1 – (ii), 2 – (iv), 3 – (iii), 4 – (i)
(c) 1 – (iv), 2 – (ii), 3 – (i), 4 – (iii)
(d) 1 – (i), 2 – (iii), 3 – (ii), 4 – (iv)
Answer:
(a) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)

Question 8.
Innermost layer of anther wall is
(a) Endothecium
(b) Endothecum
(c) Endothelium
(d) Tapetum
Answer:
(d) Tapetum

Question 9.
Identify the mismatched pair:
(a) Epidermal layer – Protective infunction
(b) Endothecium layer – Helps in dehiscence of anther
(c) Middle layer – Persistent layer
(d) Tapetum – Nutritive in function
Answer:
(c) Middle layer – Persistent layer

Question 10.
Name the person who discovered the pollen tube?
(a) E.Strasburger
(b) Hofineister
(c) Nehemiah Grew
(d) G.B.Amici
Answer:
(d) G.B.Amici

Question 11.
Identify the mismatched pair
(i) Sucker – Chrysanthemum
(ii) Bulbils – Agave
(iii) Stolon – Fragaria
(iv) Runner – Lilium
(a) i only (b) ii only (c) iii only (d) iv only
Answer:
(d) iv only

Question 12.
Assertion (A): Epidermis is protective in function.
Reason (R): Epidermis is outermost unilayer of anther wall.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
(b) R explains A.
Answer:
(b) R explains A.

Question 13.
Assertion (A) : Microspores are the first cell of male gametophyte.
Reason (R) : Microspores undergo development and forms pollen grains.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(b) R explains A.

Question 14.
Assertion (A) : Carica papaya is a dioecious plant.
Reason (R): Both male and female flowers are borne on same plant.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(a) A is correct R is incorrect.

Question 15.
Assertion (A) : Anemophilous pollination occurs by animals.
Reason (R) : Pollen grains are sticky for easy attachment on animals.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(c) Both A and R are incorrect.

Question 16.
Assertion (A) : Fusion of male and female gametes results in zygote.
Reason (R): Product of triple fusion is PEN.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(d) Both A and R are correct. R does not explain A.

Question 17.
Assertion (A) : Zea mays is a monocotyledonous plant.
Reason (R) : Shield shaped cotyledon is called scutellum.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(b) R explain A.

Question 18.
Assertion (A) : In Bryophyllum, vegetative propagation occurs through leaf.
Reason (R) : Epiphyllous buds are noticed in Bryophyllum.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(b) R explain A.

Question 19.
Assertion (A): Androecium and Gynoecium are essential whorls of flower
Reason (R) : Androecium and Gynoecium assist the reproduction.
(a) A is correct R is incorrect
(b) R explains A
(c) Both A and R are incorrect
(d) Both A and R are correct. R does not explain A
Answer:
(a) A is correct R is incorrect

Question 20.
Identify the correct statement.
(a) Grafting is a modem method of artificial propagation.
(b) The plant which is used for graft is scion.
(c) In tongue grafting, the scion bud is placed inside the incision beneath bark.
(d) Grafting is usually carried out in monocot plants.
Answer:
(b) The plant which is used for graft is scion.

Question 21.
Statement 1: Flower is a highly condensed shoot for reproductive purpose.
Statement 2: A complete flower possess four whorls.
(a) Both the statements are incorrect.
(b) Statement 1 is correct and Statement 2 is incorrect.
(c) Both the statements are correct.
(d) Statement 1 is incorrect and statement 2 is correct.
Answer:
(c) Both the statements are correct.

Question 22.
Identify the incorrect statement.
(a) One seeded fruit of paddy is caryopsis.
(b) Primitive root is called coleorhiza.
(c) Sctellum is a part of monocot seed.
(d) Embryonic axis above the cotyledon is epicotyl.
Answer:
(b) Primitive root is called coleorhiza.

Question 23.
Cleavage polyembryony is noticed in
(a) Orchids
(b) Casuarina
(c) Balanophora
(d) Syzygium
Answer:
(a) Orchids

Question 24.
Pick out the non-spermous seed
(a) Wheat
(b) Sunflower
(c) Bean
(d) Orchids
Answer:
(c) Bean

Question 25.
The type of endosperm noticed in Hydrilla seed
(a) Ruminate endosperm
(b) Nuclear endosperm
(c) Cellular endosperm
(d) Helobial endosperm
Answer:
(b) Nuclear endosperm

Question 26.
Which is note a part of mature seed?
(a) Funiculus
(b) Testa and tegma
(c) hilm
(d) Chalaza
Answer:
(d) Chalaza

Question 27.
Select the wrong statement(s) regarding cross pollination.
(a) Pollination depends on external agent and so it is certain.
(b) New varieties are produced.
(c) Continuous cross pollination leads to weaker progeny.
(d) Germination capacity is highly declined.
(i) a and d
(ii) b and c
(iii) a, b and d
(iv) a, c and d
(iv) a, c and d
Answer:
(iv) a, c and d

Question 28.
Which of the following characters does not exist in Omithophilous flowers?
(a) Huge sized flowers
(b) Bright coloured
(c) Scented flowers
(d) Nectar is secreted in large
Answer:
(c) Scented flowers

Question 29.
Which of the following plant was introduced as a contaminant into India along with wheat?
(a) Parthenium hysterophorus
(b) Zea mays
(c) Rosaindica
(d) Mangifera indica
Answer:
(a) Parthenium hysterophorus

Question 30.
_____ is a carotenoid derivative of exine layer which provides resistance to pollen grains.
Answer:
Sporopollenin

Question 31.
The most common type of ovule noticed in dicots and monocots is ______
(a) Orthotropus
(b) Anatropous
(c) Campylotropus
(d) Amphitropous
Answer:
(b) Anatropous

Question 32.
Identity the incorrect statement.
(а) The stalk of the ovule is funiculus.
(b) Nucellus is composed of sclerenchymatous tissue.
(c) Basal region of the ovule is chalaza end.
(d) Micropyle is always oriented opposite to chalaza.
Answer:
(b) Nucellus is composed of sclerenchymatous tissue.

Question 33.
Generally the pollen grains are liberated from anther at
(a) 2-celled stage
(b) 4-celled stage
(c) 6-celled stage
(d) 8-celled stage
Answer:
(a) 2-celled stage

Question 34.
Assertion (A) : Self – pollination is certain in cleistogamous flowers.
Reason (R) : Flowers never open and do not expose reproductive organs.
(a) Both A and R are incorrect.
(b) A is correct R is incorrect.
(c) R explains A.
(d) Both A and R are correct. R is not correct explanation for A.
Answer:
(c) R explains A.

Question 35.
Assertion (A): Entamophily is the most common type of pollination.
Reason (R) : Birds and animals brings out effective pollination.
(a) Both A and R are incorrect.
(b) A is correct R is incorrect.
(c) R explains A.
(d) Both A and R are correct. R is not a correct explanation for A.
Answer:
(d) Both A and R are correct. R is not a correct explanation for A.

Question 36.
Statement 1: Primary sporogenous cell functions as megaspore mother cell.
Statement 2: Megaspore mother cell undergoes mitotic division producing megaspores.
(a) Statement 1 is correct and statement 2 is incorrect.
(b) Statement 1 is incorrect and statement 2 is correct.
(c) Both the statements 1 and 2 are correct.
(d) Both the statements 1 and 2 are incorrect.
Answer:
(a) Statement 1 is correct and statement 2 is incorrect.

Question 37.
Statement 1: Apomixis does not involve meiosis and syngamy.
Statement 2: The term Apomixis was introduced by Winkler.
(a) Statement 1 is correct and statement 2 is incorrect.
(b) Statement 1 is incorrect and statement 2 is correct.
(c) Both the statements 1 and 2 are correct.
(d) Both the statements 1 and 2 are incorrect.
Answer:
(c) Both the statements 1 and 2 are correct.

Question 38.
Statement 1: The pollen grains are deposited on the receptive surface of style.
Statement 2: After landing, the first visible change in pollen is hydration.
(a) Statement 1 is correct and statement 2 is incorrect.
(b) Statement 1 is incorrect and statement 2 is correct.
(c) Both the statements 1 and 2 are correct.
(d) Both the statements 1 and 2 are incorrect.
Answer:
(b) Statement 1 is incorrect and statement 2 is correct.

Question 39.
Which of the following post fertilization change is incorrectly matched?
(a) Secondary nucellus – Endosperms
(b) Antipodals – Degenerates
(c) Nucellus – Testa and tegma
(d) Funicle – Seed stalk
Answer:
(c) Nucellus – Testa and tegma Identify the parthenocarpic fruit

Question 40.
Identify the parthenocarpic
(a) Banana
(b) Pear
(c) Papaya
(d) More than one option is correct
Answer:
(d) More than one option is correct

Question 41.
A mature angiospermic embryo sac is
(a) 8 celled and 8 nucleated
(b) 7 celled and 8 nucleated
(c) 8 celled and 7 nucleated
(d) 7 celled and 8 nucleated
Answer:
(A) 7 celled and 8 nucleated

Question 42.
Identify the type of ovule, where the nucellus acquires a horse-shoe shaped structure.
(a) Anatropus
(b) Hemianatropus
(c) Campylotropus
(d) Amphitropus
Answer:
(d) Amphitropus

Question 43
(a) 1 egg cell and 2 anti
(b) 1 egg cell and 2 polar nuclei
(c) 1 egg cell and 1 secondary nuycleus
(d) 1 egg cell and 2 synergids
Answer:
(d) 1 egg cell and 2 synergids

Question 44.
Match the following :
(1) Hemianatropous
(2) Circinotropus
(3) Campylotropus
(4) Anatropous

Question 45.
Product of triple fusion is
(a) PEN
(b) PEG
(c) PVC
(d) PPT
Answer:
(a) PEN

Question 46.
Ex-albuminous seeds are
(a) Pea, castor, paddy
(b) Paddy, Coconut, Groundnut
(c) Beans, coconut, castor
(d) Groundnut, pea, beans
Answer:
(d) Groundnut, pea, beans

Question 47.
The white edible part of coconut is…
(a) Epicarp
(b) Endosperm
(c) Embryo
(d) Mesocarp
Answer:
(b) Endosperm

Question 48.
Observe the diagram and select the correct option mentioning the parts A,B,C and D.


Answer:
(b) (A) Plumule, (B) Cotyledon, (C) Testa, (D) Radicle

Question 49.


Answer:
(d) (A) Hemiamphitropus, (B) Campylotropus, (c) Amphitropus, (d) Circinotropus

Question 50.
Attractants and rewards are required for.
(а) Anemophily
(b) Entamophily
(c) Malacophily
(d) Cheiropterophily
Answer:
(b) Entamophily

Question 51.
Filiform apparatus is a special cellular thickening which is seen in
(a) Antipodals
(b) Polar nuclei
(c) Nucellus
(d) Synergids
Answer:
(d) Synergids

Question 52.
In anatropous ovule, the micropyle faces
(a) Right side
(b) Leftside
(c) Upward
(d) Downward
Answer:
(d) Downward

Question 53.
Observe the diagram and select the correct option mentioning the parts A,B,C and D.

Answer:
(A) Synergid, (B) Egg, (C) polar nuclei, (D) Antipodals

Question 54.
Identify the correct adaptation that checks autogamy
(a) Homogamy
(b) Cleistogamy
(c) Herkogamy
(d) None of the above
Answer:
(c) Herkogamy

Question 55.
In monoecious plants,
(a) Both autogamy and geitonogamy are prevented
(b) Both autogamy and geitonogamy are takes place
(c) Autogamy takes place preventing geitonogamy
(d) Autogamy is prevented whereas geitonogamy takes place
Answer:
(d) Autogamy is prevented whereas geitonogamy takes place

Question 56.
Antipodals are located at of embryo sac.
Answer:
Chalazal end

Question 57.
Identify the correct sequence of anther wall layers from periphery towards core part.
(a) Epidermis → endothelium → stomium → tapetum
(b) Epidermis → middle layer → endothecium → endothelium
(c) Epidermis → endothelium → middle layers → tapetum
(d) Epidermis → endothelium → endothecium → tapetum
Answer:
(c) Epidermis → endothelium → middle layers → tapetum

Question 58.
The proteins responsible for rejection reaction present in exine cavities of pollen is a- derivative of.
(a) Stomium
(b) Endothecium
(c) Tapetum
(d) Ubisch bodies
Answer:
(c) Tapetum

Question 59.
Pick out the mismatched pair:
(a) Entamophily – Insects
(b) Malacophily – Mammals
(c) Cheiropterophily – Bats
(d) Omithophily – Birds
Answer:
(b) Malacophily – Mammals

Question 60.
Which is the most common type of style seen in monocots?
(a) Open type
(b) Closed type
(c) Solid type
(d) Half closed type
Answer:
(a) Open type

2 – Mark Questions

Question 1.
Write the names of organisms that undergo the following types of asexual reproduction.
(a) Budding
(b) fragmentation
(c) Regeneration
(d) Gemma cup formation.
Answer:
(a) Budding – Hydra
(b) Fragmentation – Spirogyra
(c) Regeneration – Planaria
(d) Gemma cup formation – Marchantia

Question 2.
What are diaspores?
Answer:
The unit of reproductive structures used in vegetative propagation is called diaspore or reproductive propagules.

Question 3.
Name the vegetative propagules of the following plants.
(a) Allium cepa
(b) Zingiber officinalis
(c) Agave
(d) Colocasia
Answer:
(a) Allium cepa – Bulb
(b) Zingiber officinalis – Rhizome
(c) Agave – Bulbils
(d) Colocasia – Corm

Question 4.
Point out the advantages of natural vegetative reproduction.
Answer:
(a) Only one parent is required.
(b) New individuals are genetically similar
(c) Rapid spreading
(d) Large scale production

Question 5.
Mention any two conventional propagation techniques.
Answer:
(a) Cutting
(b) Grafting.

Question 6.
What do you mean by terms‘stock’and‘scion’in grafting technique?
Answer:
In Grafting, parts of two different plants are joined so that they continue to grow as one plant. Of the two plants, the plant which is in contact with the soil is called stock and the plant used for grafting is called scion.

Question 7.
Name any four types of grafting.
Answer:
(a) Bud grafting
(b) Tongue grafting
(c) Crown grafting
(d) Wedge grafting

Question 8.
Define totipotency.
Answer:
The genetic ability of a plant cell to produce the entire plant under suitable conditions is said to be totipotency.

Question 9.
What are the term micropropogation refers to?
Answer:
The regeneration of a whole plant from single cell, tissue or small pieces of vegetative structures through tissue culture is called micropropagation. It is one of the modem methods used to propagate plants.

Question 10.
Name the four whorls of a typical flower.
Answer:
(a) Calyx
(b) Corolla
(c) Androecium
(d) Gynoecium

Question 11.
Write any four valid points on Androecium
Answer:
(a) Androecium is the male part of a flower
(b) It is made up of stamens
(c) Each stamen possess anther and a filament
(d) Anthers bear pollen grains (male gametophyte)

Question 12.
What is poilinium? Give example.
Answer:
In some plants, all the microspores in a microsporangium remain held together called poilinium.
Example: Calotropis.

Question 13.
Name the four anther wall layers.
Answer:
(a) Epidermis
(b) Endothecium
(c) Middle layers
(d) Tapetum

Question 14.
Tapetum is dual in origin – Justify.
Answer:
Tapetum is derived partly from peripheral anther wall layer and partly from the connective tissue of anther lining the anther locule. Hence it is said to dual in origin.

Question 15.
Name the two types of tapetum. Mention any one function of tapetum.
Answer:
Tapetum are of two types. Secretory tapetum and Invasive tapetum. Tapetum nourishes the pollen grains.

Question 16.
Differentiate between Exine and Intine layers of pollen grain.
Answer:
Exine:

1. Thick outer layers
2. It is made of cellulose, sporopollenin and pollenkitt.
3. Ununiform layer

Intine:

1. Thin inner layer
2. It is made up of pectin, hemicellulose, cellulose and callose
3. Uniform layer

Question 17.
What are the chemical components that make up the wall layers of pollen grains?
Answer:
Pectin, cellulose, hemicellulose, callose, sporopollenin, pollenkitt and other proteins.

Question 18.
Draw and label the structure of a typical pollen grain
Answer:

Question 19.
At which cellular stage, does the pollen grains are usually liberated from anther? What happens to the generative cell if the pollen reaches the stigma?
Answer:
In general, the pollen grains are liberated at 2 – celled stage. On reaching the stigma at this stage, the generative cell divides meiotically and form two male gametes (male cells)

Question 20.
Write the equivalent botanical terms for the following words / sentences.
(a) Landing platform of pollen
(b) Ovarian cavity
(c) Megasporangium
(d) Basal swollen part of pistil
Answer:
(a) Stigma
(b) Locule
(c) Ovule
(d) Ovary

Question 21.
What is Nucellus?
Answer:
The body of the ovule is made up of a central mass of parenchymatous tissue called nucellus which has large reserve food materials.

Question 22.
What is integumentary tapetum?
Answer:
In Asteraceae species, the inner layer of integument get specialized for nourishing the embryosac and this is called integumentary tapetum or endothelium.

Question 23.
Mention the types of ovule seen in the members of
Answer:
(a) Cactaceae
(b) Leguminosae
(c) Polygonaceae
(d) Primulaceae.
Answer:
(a) Cactaceae – Circinotropous ovule
(b) Leguminosae – Campylotropous ovule
(c) Polygonaceae – Orthotropous ovule
(d) Primulaceae. – Hemianatropous ovule

Question 24.
What does the term ‘Bisporic development of embryo sac’ refers to? Give example.
Answer:
In gynoecium, out of four megaspores formed, if two are involved in embryo sac formation then it is said to be bisporic embryo sac. E.g. Peperomia

Question 25.
State the role of filiform apparatus found in embryo sac of angiosperm.
Answer:
(a) Filiform apparatus helps in absorption conduction of nutrients from nucellus to embryo sac
(b) It guides the pollen tube into the egg.

Question 26.
What is pollination? Mention its types.
Answer:
The process of transfer of pollen grains from the anther to a stigma of a flower is called pollination. The pollination is classified into two kinds, namely, self-pollination (Autogamy) and cross pollination(Allogamy).

Question 27.
Why pollination in gymnosperm is said to be direct?
Answer:
Pollination in gymnosperms is said to be direct as the pollens are deposited directly on the exposed ovules.

Question 28.
Define Homogamy with an example
Answer:
When the stamens and stigma of a flower mature at the same time it is said to be homogamy. It favours self-pollination to occur. Example: Mirabilis jalapa.

Question 29.
What is cross – pollination? What are its types?
Answer:
Cross – pollination refers to the transfer of pollens on the stigma of another flower.
The cross-pollination is of two types:

1. Geitonogamy
2. Xenogamy.

Question 30.
Distinguish between monoecious and dioecious plants.
Answer:
Monoecious:
Male & Female flowers develop on the same plant.
E.g: Maize

Dioecious:
Male and Female flowers develop on different plants.
E.g: Papaya

Question 31.
Define the following terms.
(a) Protandry
(b) Protogyny
Answer:
(a) Protandry: The stamens mature earlier than stigma. E.g: Helianthus
(b) Protogyny: The stigma mature earlier than stamens. E.g: Aristolochia bracteata

Question 32.
What is Heterostyly? Give example.
Answer:
Heterostyly: Some plants produce two or three different forms of flowers that are different in their length of stamens and style. Pollination will take place only between organs of the same length. E.g: Primula.

Question 33.
How self-pollination is avoided in Abutilon?
Answer:
In Abutilon, the self-pollination is avoided by self sterility or self-incompatibility, in which if the pollen grain reaches the stigma of the same flower, it will be prevented from germination. It is a genetic mechanism.

Question 34.
Name the agents of the following types of pollination
(a) Anemophily
(b) Ornithophily
(c) Cheiropterophily
(d) Malacophily
Answer:
(a) Winds
(b) Birds
(c) Bats
(d) Snails and slugs

Question 35.
Give any four unique characters exhibited by anemophilous flowers.
Answer:
(a) Flowers are small and inconspicuous
(b) Colourless
(c) Non-scented
(d) No nectar secretion

Question 36.
How the pollen grains of Vallisneria protect themselves?
Answer:
Vallisneria is an aquatic plant. Pollen grains of vallisneria are covered by mucilage coating which protects them from wetting.

Question 37.
Point out the differences between anemophilous flowers and ornithophilous flowers.
Answer:

Question 38.
Mention any two disadvantages of self-pollination.
Answer:

• Continuous self-pollination, generation after generation results in weaker progeny.
• Chances of producing new species and varieties are meager.

Question 39.
What is pollen-pistil interaction?
Answer:
The events from pollen deposition on the stigma to the entry of pollen tube in to the ovule is called pollen – pistil interaction. It is a dynamic process which involves recognition of pollen and to promote or inhibit its germination and growth.

Question 40.
What are the major post fertilization events in a flower?
Answer:
Endosperm development, embryo development, seed formation and fruit formation.

Question 41.
What is perisperm? Give example.
Answer:
In some plants, the nucellar tissue is not utilized by the embryo completely, a small portion will remain as a storage tissue in the seed, which is called as perisperm.
E.g: Black Pepper.

Question 42.
What happens to the following floral parts, after the fertilization process?
(a) Ovary
(b) Secondary nucleus
(c) Outer integument of ovule
(d) Funicle
Answer:
(a) Ovary → Fruit
(b) Secondary nucleus → Endosperm
(c) Outer integument of ovule → Outer seed coat (Testa)
(d) Funicle → Stalk of the seed

Question 43.
What is the product of triple fusion? Mention its ploidy.
Answer:
The product of triple fusion is Primary Endosperm Nucleus (PEN). It is triploid (3n) in condition.

Question 44.
Coconut is an albmunious seed. Why?
Answer:
Since coconut seed possess endosperm, it is called as albmunious seed. Endosperm nourishes the embryo during seed germination.

Question 45.
Name the various types of endosperms.
Answer:
(a) Nuclear Endosperm
(b) Helobial Endosperm
(c) Cellular Endosperm
(d) Ruminate Endosperm

Question 46.
What is scutellum?
Answer:
In monocot seeds, the embryo is small and consists of single shield-shaped cotyledon known as scutellum present towards lateral side of embryonal axis.

Question 47.
What is coleoptile and coleorhiza?
Answer:
In monocot embryo, the plumule is covered by a protective sheath called coleoptile and the radicle along with root cap is covered by a protective sheath called coleorhiza.

Question 48.
Who coined the term Apomixis? Define it.
Answer:
The term Apomixis was introduced by Winkler in the year 1908. It is defined as the substitution of the usual sexual system (Amphimixis) by a form of reproduction which does not involve meiosis and syngamy.

Question 49.
What are parthenocarpic fruits?
Answer:
In some plants, fruits develop from ovary without fertilization process. Such fruits are called parthenocarpic fruits. They lack true seeds.
E.g: Grapes.

3 – Mark Questions

Question 50.
How tongue grafting differs from wedge grafting?
Answer:
Tongue grafting:
In tongue grafting A scion and stock having the same thickness is cut obliquely and the scion is fit into the stock and bound with a tape.

Wedge grafting:
In wedge grafting a slit is made in the stock or the bark is cut. A twig of scion is inserted and tightly bound so that the cambium of the two is joined.

Question 51.
List any three advantages of micropropagation.
Answer:

1. Tissue culture can be carried out in any season to produce plants.
2. Plants which do not produce viable seeds and seeds that are difficult to germinate can be propagated by tissue culture.
3. Cells can be genetically modified and transformed using tissue culture.

Question 52.
Where the stomium is located? What is its role?
Answer:
In a mature anther, the cells along with junction or the two sporangia of an anther lobe lack cellulose and lignin thickening. This region is called stomium. Stomium along with hygroncopic nature of Endothecium helps in the dehiscence of anther at maturity.

Question 53.
Briefly explain about the types of tapetum.
Answer:
There are two types of tapetum based on its behaviour.
They are:
Secretory tapetum (parietal/glandular/ cellular): The tapetum retains the original position hnd cellular integrity and nourishes the developing microspores.

Invasive tapetum (periplasmodial): The cells loose their inner tangential and radial walls and the protoplast of all tapetal cells coalesces to form a periplasmodium.

Question 54.
Enumerate the functions of tapetum.
Answer:

1. It supplies nutrition to the developing microspores.
2. It contributes sporopollenin through ubisch bodies thus plays an important role in pollen wall formation.
3. The pollenkitt material is contributed by tapetal cells and is later transferred to the pollen surface.
4. Exine proteins responsible for ‘rejection reaction’ of the stigma are present in the cavities of the exine. These proteins are derived from tapetal cells.

Question 55.
State the significance of sporopollenin.
Answer:
The wall material sporopollenin is contributed by both pollen cytoplasm and tapetum. It is derived from carotenoids. It is resistant to physical and biological decomposition. It helps to withstand high temperature and is resistant to strong acid, alkali and enzyme action. Hence, it preserves the pollen for long periods in fossil deposits, and it also protects pollen during its journey from anther to stigma.

Question 56.
What do you know about bee pollen?
Answer:

1. Bee pollen is a natural substance with high proteins, carbohydrate and trace amount of vitamins and minerals.
2. It is used as dietary supplement as tablets.
3. It increases performance of athletes, race horses and also heals bum wounds.

Question 57.
Write a note on pollenkitt.
Answer:
Pollenkitt is contributed by the tapetum and coloured yellow or orange and is chiefly made of carotenoids or flavonoids. It is an oily layer forming a thick viscous coating over pollen surface. It attracts insects and protects damage from UV radiation.

Question 58.
Draw and label the structure of a mature embryo sac of angiosperm.
Answer: