Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 16 Consumerism Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 16 Consumerism

12th Commerce Guide Consumerism Text Book Back Questions and Answers

I. Choose The Correct Answer.

Questions 1.
The term ‘consumerism’ came into existence in the year …………
a) 1960
b) 1957
c) 1954
d) 1958
Answer:
a) 1960

Question 2.
Who is the father of consumer movement?
a) Mahatma Gandhi
b) Mr.John F. Kennedy
c) Ralph Nader
d) Jawaharlal Nehru
Answer:
c) Ralph Nader

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 3.
Sale of goods act was passed in the year?
a) 1962
b) 1972
c) 1930
d) 1985
Answer:
c) 1930

Question 4.
The main objective of all business enterprises is ………….
a) Providing service
b) Providing better standard of life
c) Providing necessities to the society
d) Earn profit
Answer:
d) Earn profit

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 5.
The consumer protection act came into force with effect from
a) 1.1.1986
b) 1.4.1986
c) 15.4.1987
d) 15.4.1990
Answer:
c) 15.4.1987

Question 6.
……………………… of every year is declared as a consumer protection day to educate the public about their rights and responsibilities.
a) August 15
b) April 15
c) March 15
d) September 15
Answer:
c) March 15

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 7.
Any person who buys any goods or avails services for personal use, for a consideration is called as …………..
a) Customer
b) Consumer
c) Buyers
d) User
Answer:
b) Consumer

Question 8.
The General Assembly of United Nations passed resolution of consumer protection guidelines on
a) 1985
b) 1958
c) 1986
d) 1988
Answer:
a) 1985

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

II. Very Short Answer Questions

Question 1.
Who is a consumer?
Answer:
A consumer is one who consumes goods manufactured and sold by others or created (air, water, natural resources) by nature and sold by others.

Question 2.
Define consumerism.
Answer:
“Consumerism is an attempt to enhance the rights and powers by buyers in relation to sellers”. L. Massie.

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 3.
Give two examples of adulteration.
Answer:

  1. Powdered rice/wheat is adulterated with starch.
  2. Coffee powder is adulterated with tamarind seed.

Question 4.
What is caveat emptor?
Answer:
‘Caveat emptor’ is a Latin term that means “let the buyer beware.” Similar to the phrase”sold as is” this term means that the buyer assumes the risk that a product fails to meet expectations or have defects.

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 5.
What is Caveat Venditor?
Answer:

  • “Caveat Venditor” or “let the seller Beware” means by which goods are covered by an implied warranty of merchantability.
  • Unless or otherwise advertised [sold as is ] with the buyer nearly all consumer products are guaranteed to work if used for their intended purpose.

Question 6.
Write short notes on Consumer Protection Act, 1986
Answer:
The Central Government enacted a comprehensive law called the Consumer Protection Act in 1986. This Act came into force with effect from 15.04.1987. This Act was further amended in 1993. The Act is referred to in short as ‘COPRA’.

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

III. Short Answer Questions

Question 1.
What are the important legislations related to consumerism in India?
Answer:

  •  Sale of goods act -1930
  •  The agricultural products grading and marketing act -1937
  •  The drugs and cosmetics act -1940
  •  The prevention of food Adulteration act -1954
  •  The essential commodities act -1955
  •  Weight and measures act -1958

Question 2.
What is meant by artificial scarcity?
Answer:

  • Even in times of plenty of stock, the sellers create artificial scarcity in the market to push up the prices to earn more profit.(e.g) “No stock” – Board in front of shops. “House full” – Board in front of theaters.
  • But goods and cinema tickets are available at a higher price in the black market.

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 3.
Write the importance of consumerism. [APP DEAD]
Answer:
Consumer Legislation:

  1. The Indian Contract Act, 1982
  2. The Sale of Goods Act, 1982
  3. The Essential Commodities Act, 1955
  4. The Agricultural Products Grading and Marketing Act, 1937
  5. The Prevention of Food Adulteration Act, 1954
  6. Weights and Measures Act, 1958
  7. The Trademark Act, 1999

IV. Long Answer Questions

Question 1.
How consumers are Exploited? [FD PASS]
Answer:
False advertisement:
Producers make false claims and bogus presentations about their products in the media, [quality, price, grade etc.]
With a view to mislead the consumers [eg] A ‘pinch of Rin’ ad ’23 components of complaint ad

Duplication [Spurious goods]:

  •  Duplicates of popular brands of products are produced and sold.
  •  A consumer is not in a position to differentiate the duplicate from the original thus consumer is exploited.
  •  [eg] Watches – TITAN – TITTAN
  •  Shoes – ADIDAS – ADITAS

Product risk:

  •  Harmful and banned drugs, Electrical appliances with inadequate safety provisions are sold in the market.
  •  Such goods are dangerous and injurious to the lives of the customers.
  •  eg: Electrical Gadgets with inadequate insulation.

Adulteration:

  •  Adulteration is quite common in food articles.
  •  It means mixing or substituting undesirable materials in food
  •  It spoils the health of the consumer
  •  It is a crime which cannot be pardoned
  •  eg: Mixing of stone with rice
  •  Mixing of water with milk
  •  Mixing of brick powder with chilli powder
  •  Papaya seeds added to black pepper.

Selling at a higher price:

  • The price charged by the seller for a product or service is not proportionate to the quality but at times it is more than the fair price.
  • eg: Price printed in the book X 200
  • But it is sold for X 240

Sub-standard:

  •  On opening a packed and sealed container one finds the contents with poor quality and quantity [damaged and low weight]
    eg: Ice creams -Zandubalm – Colgate paste.

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 2.
Explain the role of business in consumer protection?
Answer:
Business enterprises should do the following towards protecting consumers.

  1. Avoidance of Price Hike: Business enterprises should desist from hiking the price in the context of acute shortage of goods.
  2. Avoidance of Hoarding: Business enterprises should not indulge in hoarding and black marketing to earn maximum profit.
  3. Guarantees for Good Quality: Business enterprises should not give false warranty for the products.
  4. Product Information: Business enterprises should disclose correct, complete and accurate information about the product viz. size, quality, quantity, weight etc.
  5. The in advertising: Business enterprises should not convey false, untrue, bogus information relating to the product through the advertisement.
  6. Money Refund Guarantee: Where the product becomes defective, business enterprises should replace it with new one or refund the purchase price.

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 3.
What are the needs for consumer protection?
Answer:
Though consumer is said to be the king of modern marketing his interests are virtually

  •  Shortage of goods makes the consumers to be content with whatever is offered for sales, v Quality is sacrificed.
  •  Warranty of performance has no meaning. v Health hazard is never considered.
  •  The main aim of business concerns is to earn more profit.
  •  In such a context consumer protection remains a vital importance.

Question 4.
Explain the role of consumers in Consumer Protection.
Answer:

  • The consumer who alone can put an end to all their unethical trade practices.
  • Consumer action councils established at village levels should educate consumers of the
  • It should regulate business enterprises according to the rules of the industry.
  • It should organize movement against the mal practice of manufacturers and traders.
  • It should contact the legislators to raise consumer issues in Assembly and Parliament,
  • VCO should publicise the malpractices of manufacturers and traders.
  • Consumers cooperatives need to be strengthened.

Question 5.
What are the objectives of the Consumer Protection Act, 1986?
Answer:
The Central Government enacted a comprehensive law called the Consumer Protection Act in 1986. This Act came into force with effect from 15.04.1987. It is in short, called as ‘COPRA’.
Objectives:

  1. Consumer Protection Act protects the interests of the consumers.
  2. This Act provides safeguards against defective goods and deficient services, untrade practices.
  3. It also gives settlement of consumer disputes.
  4. It is applicable to the public sector, financial and co-operative enterprises.

Question 6.
Write about five important consumer legislation.
Answer:
To protect the consumers from unfair traders, the government passed various legislative Acts.
They are as follows:

  1. The Indian Contract Act, 1982 was passed to bind the people on the promise made in the contract.
  2. The Essential Commodities Act, 1955 protects the consumers against artificial shortages created by the sellers by hoarding the goods.
  3. The Prevention of Food Adulteration Act, 1954 checks the adulteration of food articles and ensures the purity of goods supplied.
  4. Weights and Measures Act, 1958 protects the consumer against malpractices of underweight or under measurement.

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

12th Commerce Guide Consumerism Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
In the practical world, a ……………………………….. is a slave and not the sovereign as described.
a) Consumer
b) Producer
c) Wholesaler
d) Retailer
Answer:
a) Consumer

Question 2.
The consumer protection act was amended in the year …………………
a) 1986
b) 1987
c) 1993
d) 2000
Answer:
c) 1993

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 2.
The term ‘Caveat emptor’ is a Latin term, which means _________
(a) Let the seller beware.
(b) Let the buyer beware
(c) Consumer
(d) Marketer
Answer:
(b) Let the buyer beware

Question 4.
Which one of the following is not the correct method?
a) The trademark act – 1999
b) The essential commodities act – 1955
c) ISI act – 1952
d) The drugs and cosmetics act – 1900
Answer:
d) The drugs and cosmetics act – 1900

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 5.
Pick the odd one out.
a) Adulteration
b) Sub-standard
c) False advertisement
d) Sold as is
Answer:
d) Sold as is

Question 6.
Pick the odd one out.
a) The drugs and magic remedies act -1954
b) The Indian companies act -1956
c) Public liability insurance act -1991
d) The Narcotic drugs and psychotropic substances act -1985
Answer:
b) The Indian companies act-1956

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 7.
Choose the correct statement.
i) Significance of consumerism – Awakening and uniting consumers
ii) Encouraging unfair trade practices
iii) Awakening the government.
a) (i) is correct
b) (i) and) (ii) are correct
c) (i) and (iii) are correct and (ii) is incorrect
d) (i) (ii) (iii) are correct
Answer:
c) (i) and (iii) are correct and (ii) is incorrect

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

II. Match The Following

Question .1

List-I

List-II

(i) Indian contract act1. 1930
(ii) Sales of goods act2. 2009
(iii) Metrology act3. 2002
(iv) Competition act4. 1982

a) i-4, ii-1, iii-2, iv-3
b) i-4, ii-2, iii-1, iv-3
c) i-4, ii-3, iii-2, iv-1
d) i-3, ii-4, iii-1, iv-2
Answer:
a) i-4, ii-1, iii-2, iv-3

Question 2.

List – I

List – II

i. Caveat Emptor1. Malpractices
ii. Caveat Venditor2. House full
iii. Artificial Scarcity3. Let the buyer beware
iv. Weights and Measures act4. Let the seller beware

a) i-3, ii-4, iii-2, iv-1
b) i-4, ii-3, iii-1, iv-2
c) i-1, ii-2, iii-4, iv-3
d) i-2, ii-1, iii-3, iv-4
Answer:
a) i-3, ii-4, iii-2, iv-1

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

III. Assertion and Reason

Question 1.
Assertion (A) : Duplicate products of popular products are illegally produced and sold.
Reason (R) : A consumer is not in a position to distinguish duplicates from the originals
a) (A) and (R) are true
b) (A) and (R) false
c) (A) is true (R) is false
d) (A) is false (R) is true
Answer:
a) (A) and (R) are true

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

IV. Short Answer Questions

Question 1.
Write a short note on False Advertisements.
Answer:
The main motive of advertisements is to educate the consumers regarding various aspects of the products and services. Sometimes it makes a false representation of the quality, price, grade, composition, and utility of products. Often the products are not as attractive as shown in the advertisement by the sellers. Consumers who buy the products on the faith of claims made in advertisements are cheated.

Question 2.
Which are the three constituent elements of business?
Answer:

  •  Government
  •  Business
  •  Consumer

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 3.
What is the role of the Government in consumer protection?
Answer:

  • The government should assure an active role in safeguarding the consumers.
  • Governments [central and state] have brought out more legislation to protect the consumers in the country.
  • Most of the consumers [ including educated] are illiterate about their rights and passive
  • Mere statute book [law book] will not address the problems.
  • Law enforcement authorities should sting the offenders mercilessly.

V. Long Answer Questions

Question 1.
Explain the Evolution and growth of consumerism
Answer:
Evolution:
The major causes of consumerism in India have been identified as rising prices, poor product performance and service quality product shortages, and deceptive advertising and inflation.

Growth:

  • Consumerism or consumer movement is an outcome of the sufferings and exploitations of consumers.
  • It intends to shield the consumers from commercial terrorism and exploitative practices.
  • Its aim is to safeguard consumer interests by establishing their rights and powers in relation to products and sellers.
  • Ralph Nader fought against the monopoly and unethical trade practices of large companies in the USA.
  • He is considered to be the father of the consumer movement.
  • He initially fought against the automobile industry for violating safety standards and pollution control norms
  • Countries like Britain, Sweden, Demark, Netherlands, and even in Kenya have stringent laws against consumer exploitation in their respective countries.
  • The united nations general assembly has adopted a set of general guidelines for consumer protection.

Samacheer Kalvi 12th Commerce Guide Chapter 16 Consumerism

Question 2.
What are the salient features of the Consumer Protection Act, 1986?
Answer:

  • Protecting consumers against products and services which are harmful to the health of consumers.
  • Protecting consumers from the breach of contract by sellers /manufacturers.
  • Ensuring consumers with a supply of goods at fair quality.
  • Ensuring that consumers are charged a fair price.
  • Ensuring an uninterrupted supply of goods.
  • Protecting the consumers against unfair trade practices of unscrupulous traders.
  • Protecting the consumers against pollution of various kinds
  • Protecting consumers against the effect of competition.

Question 3.
What are the objectives of the united nations Guidelines for consumer protection?
Answer:

  • To facilitate production and distribution patterns responsive to the needs and desires of consumers.
  • To facilitate the development of independent consumer groups.
  • To encourage high levels of ethical conduct for those engaged in the production and distribution of goods and services to the consumers.
  • To encourage the development of market conditions that provide consumers with greater choice at lower prices.
  • To further international co-operative in the field of consumer protection.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 11 Hydroxy Compounds and Ethers Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 11 Hydroxy Compounds and Ethers

12th Chemistry Guide Hydroxy Compounds and Ethers Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

Question 1.
An alcohol (x) gives blue colour in victormayer’s test and 3.7g of X when treated with metallic sodium liberates 560 mL of hydrogen at 273 K and 1 atm pressure what will be the possible structure of X?
(a) CH3 CH (OH) CH2CH3
(b) CH3 – CH(OH) – CH3
(c) CH3 – C (OH) (CH3)2
(d) CH3 – CH2 – CH (OH) – CH2 – CH3
Answer:
(a) CH3 CH (OH) CH2CH3
Hint:
2R – OH + Na → 2RONa + 2H2 ↑ 2 moles of alcohol gives 1 mole of H2 which occupies
22.4L at 273K and 1 atm
number of moles of alcohol = \(\frac{2 \text { moles of } \mathrm{R}-\mathrm{OH}}{22.4 \mathrm{L} \text { of } \mathrm{H}_{2}}\) x 560 mL = 0.05 moles
number of moles = \(\frac{\text { mass }}{\text { molar mass }}\)
= molar mass = \(\frac{3.7}{0.05}\) = 74 g mol-1
General formula for
R – OH Cn H2n+1 – OH
n(12) + (2n+1) (1) + 16 +1 = 74
14n = 74 – 18
14n = 56
n = \(\frac { 56 }{ 4 }\) = 4
The 2° alcohol which contains 4 carbon is CHn CH(OH)CH2 CH3

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 2.
Which of the following compounds on reaction with methyl magnesium bromide will give tertiary alcohol.
(a) benzaldehyde
(b) propanoic acid
(c) methyl propanoate
(d) acetaldehyde
Answer:
(c) methyl propanoate
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 1

Question 3.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 2
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 3
Answer:
a
Solution:
hydro boration – Anti markownikoff product
i.e CH3 – CH2 – CH – CH2 – CH2 – OH

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 4.
In the reaction sequence, Ethane
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 4
Ethan – 1, 2 – diol. A and X respectively are ………….
(a) Chioroethane and NaOH
(b) ethanol and H2SO4
(c) 2 – chloroethan – 1 – ol and NaHCO3
(d) ethanol and H2O
Answer:
(c) 2 – chloroethan – 1 – ol and NaHCO3\
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 5
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 6

Question 5.
Which one of the following is the strongest acid ………..
(a) 2 – nitrophenol
(b) 4 – chlorophenol
(c) 4 – nitrophenol
(d) 3 – nitrophenol
Answer:
(c) 4 – nitrophenol

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 6.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 7
on treatment with Con. H2SO4, predominately gives ……………..
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 8
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 9
Answer:
b
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 10

Question 7.
Carbolic acid is …………..
(a) Phenol
(b) Picric acid
(c) benzoic acid
(d) phenylacetic acid
Answer:
(a) Phenol

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 8.
Which one of the following will react with phenol to give salicyladehyde after hydrolysis …………..
(a) Dichioro methane
(b) trichioroethane
(c) trichloro methane
(d) CO2
Answer:
(c) trichloro methane (Riemer Tiemann reaction)

Question 9.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 11
(a) (CH3)3 CCH = CH2
(b) (CH3)2 C = C (CH3)2
(c) CH2 = C(CH3)CH2 – CH2 – CH3
(d) CH2 = C (CH3) – CH2 – CH2 – CH3
Answer:
(b) (CH3)2 C = C (CH3)2
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 12

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 10.
The correct IUPAC name of the compound,
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 13
(a) 4 – chloro – 2, 3 – dimethyl pentan – 1 – ol
(b) 2, 3 – dimethyl – 4 – chloropentan – 1 – ol
(c) 2, 3, 4 – trimethyl – 4 – chiorobutan – 1 – ol
(d) 4 – chioro – 2, 3, 4 – trimethyl pentan – 1 – ol
Answer:
(a) 4 – chloro – 2, 3 – dimethyl pentan – 1 – ol

Question 11.
Assertion: Phenol is more acidic than ethanol
Reason: Phenoxide ion is resonance stabilized
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false.
Answer:
if both assertion and reason are true and reason is the correct explanation of assertion.

Question 12.
In the reaction Ethanol
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 14
is ………………
(a) ethane
(b) ethoxyethane
(c) ethylbisuiphite
(d) ethanol
Answer:
(d) ethanol
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 15

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 13.
The reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 16
can be classified as
(a) dehydration
(b) Williams on alcohol synthesis
(c) Williamson ether synthesis
(d) dehydrogenation of alcohol
Answer:
(c) Williamson ether synthesis
Solution:
Cyclic alcohol → sodium cyclic alkoxide → Williamson ether synthesis

Question 14.
Isoprophylbcnzene on air oxidation in the presence of dilute acid gives …………
(a) C6H5COOH
(b) C6H5COCH3
(c) C6H5COC6H5
(d) C6H5 – OH
Answer:
(a) C6H5 – OH (phenol)

Question 15.
Assertion: Phenol is more reactive than benzene towards electrophilic substitution reaction
Reason: In the case of phenol. the intermediate arenium ion is more stabilized by resonance.
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false,.
Answer:
(a) if both assertion and reason are true and reason is the correct explanation of assertion.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 16.
HO CH2 CH2 – OH on heating with periodic acid gives ………..
(a) methanoic acid
(b) Glyoxal
(c) methanol
(d) CO2
Answer:
(c) methanol

Question 17.
Which of the following compound can be used as artireeze in automobile radiators?
(a) methanol
(b) ethanol
(c) Neopentyl alcohol
(d) ethan -1, 2-diol
Answer:
(d) ethan -1, 2-diol

Question 18.
The reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 17
is an example of …………..
(a) Wurtz reaction
(b) cyclic reaction
(c) Williamson reaction
(d) Kolbe reactions
Answer:
(c) Kolbe reactions

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 19.
One mole of an organic compound (A) with the formula C3H8O reacts completely with two moles of HI to form X and Y. When Y is boiled with aqueous alkali it forms Z. Z answers the iodoform test. The compound (A) is ……………
(a) propan – 2 – ol
(b) propan- 1- ol
(c) ethoxy ethane
(d) methoxy ethane
Answer:
(d) methoxy ethane
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 18

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 20.
Among the following ethers which one will produce methyl alcohol on treatment with hot HI?
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 19
Answer:
a
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 20

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 21.
Williamson synthesis of preparing dimethyl ether is a / an
(a) SN1 reactions
(b) SN2 reaction
(c) electrophilic addition
(d) electrophilic substitution
Answer:
(b) SN2 reaction

Question 22.
On reacting with neutral ferric chloride, phenol gives
(a) red colour
(b) violet colour
(c) dark green colour
(d) no colouration
Answer:
(b) violet colour

II. Short Answer

Question 1.
IdentIfy the product (s) is/are formed when 1 – methoxy propane is heated with excess HI. Name the mechanism involved in the reaction.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 21
Ethers having primary alkyl group undergo \(\mathrm{S}_{\mathrm{N}}^{2}\) reaction

Question 2.
Draw the major product formed when 1 – ethoxyprop – 1 – ene is heated with one equivalent of HI
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 22

Question 3.
Suggest a suitable reagent to prepare secondary alcohol with an identical groups using a Grignard reagent.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 23

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 4.
What is the major product obtained when two moles of ethyl magnesium bromide is treated with methyl benzoate followed by acid hydrolysis
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 24

Question 5.
Predict the major product, when 2-methyl but – 2 – ene is converted into alcohol in each of the following methods.

  1. Acid-catalyzed hydration
  2. Hydroboration
  3. Hydroxylation using bayers reagent

Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 25

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 6.
Arrange the following in the increasing order of their boiling point and give a reason for your ordering

  1. Butan – 2 – ol, Butan – 1 – SI, 2 – methylpropan – 2 – ol
  2. Propan – 1 – ol, propan – 1, 2, 3 – triol, propan – 1, 3 – diol, propan – 2 – ol

Answer:
1. Boiling points increases regularly as the molecular mass increases due to a corresponding increase in their Van der Waal’s force of attraction. Among isomeric alcohols, 2° – alcohols have lower boiling points than 1° – alcohols due to a corresponding decrease in the extent

of H-bonding because of steric hindrance. Thus the boiling point of Butan – 2 – ol is lower than that of Butan – 1 – ol. Overall increasing order of boiling points is, 2 – methyl propane – 2 – ol < Butan – 2 – ol < Butan – 1 – ol

2. 2°-alcohols have lower boiling points than 1° – alcohols due to a corresponding decrease in the extent of H – bonding because of steric hindrance. Therefore Propan – 1 – ol has higher boiling point than Propan – 2 – ol. The hydrogen group increases, boiling point also increases. Overall increasing order of boiling points is, propan – 2 – ol < Propan – 1 – ol < propan – 1, 3 – diol < propan -1, 2, 3 – triol

Question 7.
Can we use nucleophiles such as NH3, CH3O for the Nucleophilic substitution of alcohols
Answer:
1. Increasing order of nucleophilicity,
NH3 < – OH < CH3O⊖-

2. Higher electron density will increase the nucleophilicity.

3. Negatively charged species are almost always more nucleophiles than neutral species.

4. RO has an alkyl group attached, allowing a greater amount of polarizability. This means oxygen’s lone pairs will be more readily available to reach in RO than in OH. Hence CH3O – is the better nucleophile for the nucleophilic substitution of alcohols. NH3 cannot act as nucleophiles for the nucleophilic substitution of alcohols.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 8.
Is it possible to oxidise t – butyl alcohol using acidified dichromate to form a carbonyic compound.
Answer:
3° – alcohols do not undergo oxidation reaction under normal conditions, but at elevated temperature, under strong oxidising agent cleavage of C – C bond takes place to give a mixture of carboxylic acid.

Yes, it is possible. t – butyl alcohol is readily oxidsing in acidic solution (K2Cr2O7 / H2SO4) to a mixture of a ketone and an acid each containing lesser number of carbon atoms than the original alcohol. The oxidation presumably occur via alkenes formed through dehydration of alcohols under acidic conditions.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 26

Question 9.
What happens when 1 – phenyl ethanol is treated with acidified KMnO4.
Answer:
1 – phenyl ethanol reacts with acidified KMnO4 to give Acetophenone.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 27

Question 10.
Write the mechanism of acid catalysed dehydration of ethanol to give ethene.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 28

Question 11.
How is phenol prepared form

  1. chloro benzene
  2. isopropyl benzene

Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 29

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 12.
Explain Kolbe’s reaction
Answer:
Kolbe’s (or) Kolbe’s Schmitt reaction:
In this reaction, phenol is first converted into sodium phenoxide which is more reactive than phenol towards electrophilic substitution reaction with CO2. Treatment of sodium phenoxide with CO2 at 400K, 4 -7 bar pressure followed by acid hydrolysis gives salicylic acid.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 30

Question 13.
Writes the chemical equation for Williamson synthesis of 2 – ethoxy – 2 – methyl pentane starting from ethanol and 2 – methyl pentan – 2 – ol
Answer:
A tertiary alkoxide and primary alkyl halide easily undergo williamson ether synthesis
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 31

Question 14.
Write the structure of the aldehyde, carboxylic acid and ester that yield 4 – methylpent – 2 – en – 1 – ol.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 32

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 15.
What is metamerism? Give the structure and IUPAC name of metamers of 2 – methoxy propane
Answer:
Metamerism:
It is a special type of isomerism in which molecules with same formula, same functional group, but different only in the nature of the alkyl group attached to oxygen.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 33

Question 16.
How are the following conversions effected

  1. benzyl chlorjde to benzyl alcohol
  2. benzyl alcohol to benzoic acid

Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 34

Question 17.
Complete the following reactions
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 35
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 36

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 18.
O.44g of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP 112 cm3 of methane with PCC the same alcohol form a carbonyl compound that answers silver mirror test. Identify the compound.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 37
CnH2n+1+OH
⇒ n x 12+ (2n + 1) x 1 + 1 x 16 + 1 x 1 = 88
12n + 2n + 1 + 16 + 1 = 88
14n + 18 = 88
14n = 88 – 18
14n = 70
n = 70/14 = 5
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 38

Question 19.
Complete the following reactions
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 39
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 40 Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 41
(According to Saytzeff s rule, during intramolecular dehydration, if there is a possibility to form C = C bond at different locations, the preferred location is the one that gives the more substituted alkene je, the stable alkene).

Question 20.
Phenol is distilled with Zn dust gives (A) followed by Friedel – crafts alkylation with propyl chloride to give a compound B, B on oxidation gives (C). Identify A,B and C.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 42
Note:
Carbon directly attached to the aromatic riñg is called benzylic carbon.
If there is hydrogen attached to benzylic carbon it will undergo oxidation.

Question 21.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 43
Identify A, B, C, D and write the complete equation.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 44 Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 45

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 22.
What will be the product for the following reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 46
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 47

Question 23.
How will you convert acetylene into n – butyl alcohol.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 48

Question 24.
Predict the product A, B, X and Y in the following sequence of reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 49
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 50

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 25.
3,3 – dimethyl butane – 2 – ol on treatment with conc. H2SO4 to give tetramethyl ethylene as a major product. Suggest suitable mechanisms.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 51

III. Evaluate yourself

Question 1.
Classify the following alcohols as 10, 20, and 30 and give their IUPAC Names.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 52
Answer:

a 2° alcohol5- bromo -5- methyl -3 hexanol
b 3° alcohol3 – ethyl -3- pentanol
c 2° alcohol3- chiorobut -3- en – 1 – o!
d 2°alcohol6-bromohept-3-en-2-ol
e 3° alcohol3- phenyl -3 – pentanol

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 2.
Write all the possible isomers of alcohol having the molecular formula C15H12O and their IUPAC names.
Answer:
Eight isomers are possible for C15H12O. They are,
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 53 Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 54

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 3.
Suggest a suitable carbonyl compound for the preparation of pent – 2 – en – 1 ol using LiAlH4.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 55

Question 4.
2 – methylpropan – 2 – ene
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 56
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 57

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 5.
How will you prepare the following using a Grignard reagent?

  1. t – butyl alcohol
  2. allyl alcohol

Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 58

Question 6.
Identify the products in the following reactions. Write their IUPAC names and mention the mechanism involved in the reactions.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 59
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 60

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 7.
What is the major product obtained when 2, 3 – dimethyl pentan – 3 – ol is heated in the presence of H2SO4
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 61 Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 62

Question 8.
Which of the following set of reactants will give 1 – methoxy – 4 – nitrobenzene.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 63
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 64

Question 9.
What happens when m – cresol is treated with an acidic solution of sodium dichromate?
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 65

Question 10.
When phenol is treated with propan – 2 – ol in the presence of HF, Friedel – Craft reaction takes place. Identify the products.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 66

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 11.
Given the IUPAC name for the following ethers and classify them as simple or mixed.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 67

Question 12.
1. Which of the following reaction will give 1 – methoxy – 4 – nitrobenzene.

  1. 4 – nitro – 1 – bromobenzene + sodium methoxide.
  2. 4 – nitrosodium phenoxide + bromomethane

Answer:
4-nitrosodium phenoxide + bromo methane → 1 – methoxy – 4 – nitrobenzene
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 68
Reason: 4 – nitro -1- brornobnzene + sodium methoxide can not be used, because bromine is strongly bonded to carbon atom of hcnzcne ring and it can not be replaced easily.

Question 13.
Arrange the following compounds in the increasing order of their acid strength. propan – 1 – ol, 2, 4, 6 – trinitroptienol, 3 – nitrophenol, 3,5 – dinitrophenol, phenol, 4 – methyl phenol.
Answer:
Phenols are stronger acids than alcohols because the phenoxide ion left after the removal of proton is stabilized by resonance while the alkoxide ion left after the removal of a proton from alcohol is not stabilized. Thus propan – 1 – ol is much weaker acid than any phenol.

Thus propan- 1 – ol is a much weaker acid than any phenol. We know that electron-donating groups decrease the acidic character and stronger is the electron-donating group, weaker is the phenol.

Compare to propan – 1 – ol, 4 – methyl phenol is a stronger acidic character. But comparing phenol and 4-methyl phenol, phenol is stronger acidic. Since electron-withdrawing groups increase the acidic character of phenols and the effect is more pronounced at the para position than at the meta position.

Therefore 4 – nitrophenol is a stronger acid than 3 – nitro phenol. Further as the number of electron-withdrawing groups increases the acidic strength further increases. Therefore 2, 4, 6 – trinitrophenol is a stronger acid than 3, 5 – dinitrophenol. It may be noted here that although the two nitro groups in 3, 5 – dinitrophenol are at m – position with respect to OH group,

their combined effect is however greater than one nitro group at p – position. Therefore 3, 5 – dinitrophenol is a stronger acid than 4-nitro phenol. Thus, the overall increasing order of acid strength is. Propan – 1 – 01 < 4 – methyl phenol < phenol < 3 – nitrophenol < 3, 5 – dinitrophenol < 2, 4, 6 – trinitrophenol.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 14.
1 mole of HI is allowed to react with t – butyl methylether. Identify the product and write down the mechanism of the reaction.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 69
The reaction involves the protonation of oxygen which is followed by the SN1 mechanism.
The products are t-butvl iodide and methyl alcohol.

12th Chemistry Guide Hydroxy Compounds and Ethers Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

Question 1.
In ethanol the – OH group is attached to hybridised carbon
a) sp
b) sp2
c) sp3
d) dsp2
Ans :
c) sp3

Question 2.
In Vinyl alcohol the -OH group is attached to hvbridised carbon
a) sp
b) sp2
c) sp3
d) dsp2
Ans :
b) sp2

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 3.
Sorbitol is
a) monohydric alcohol
b) dihydric alcohol
c) trihydric alcohol
d) hexa hydric alcohol
Answer:
d) hexa hydric alcohol

Question 4.
The IUPAC name of sorbitol is
a) Ethan -1, 2 – diol
b) Propan – 1, 2, 3 – triol
c) Hexan – 1, 2, 3, 4, 5, 6 – hexol
d) Ethenol
Answer:
c) Hexan – 1, 2, 3,4, 5, 6 – hexol

Question 5.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 70
a) prop – 2 – en – 1 – ol
b) 2 – methyl but – 3 – en – 2 ol
c) 3 – methyl but – 1 – en – 3 – ol
d) 1,1 – di methyl prop – 2 – en – 1 – ol
Answer:
b) 2 – methyl but – 3 – en – 2 ol

Question 6.
The C – O – H bond angle in methanol is
a) 107°
b) 109°.5°
c) 108.9°
d) 104°
Answer:
c) 108.9°

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 7.
CH3-CH2-a + NaOH → CH3-CH2-OH + NaCl This reaction follows mechanism
a) \(S_{N}^{1}[latex]
b) [latex]S_{N}^{2}\)
c) E1
d) E2
Answer:
b) \(S_{N}^{2}\)

Question 8.
Conversion of isobutyl chloride and isopropyl chloride into isobutyl alcohol and isopropyl alcohol respectively using dilute aquous NaOH follow the mechnisms
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 71
Answer:
c)
Reason : Isobutylchloride – Primary alkyl halide – \(S_{N}^{2}\) mechanism.
Iso propyl chloride – Secondary alkyl halide – Sjsj mechanism.
\(S_{N}^{1}\)

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 9.
Addition of water across the double bond
of an unsymmetrical alkene follows
a) Saytzeffs rule
b) Markownikoff’s rule
c) anti Markownikoff’s rule
d) Popoff’s rule
Answer:
b) Markownikoff’s rule

Question 10.
Hydroboration of an alkene follows
a) Saytzeff’s rule
b) Markownikoff’s rule
c) anti Markownikoff’s rule
d) Popoff’s rule
Answer:
c) anti – Markownikoff’s rule

Question 11.
Addition of water to 2-methyl propene in
presence of cone, sulphuric acid gives
a) 2-propanol
b) 2 – methyl – 2 – propanol
c) 2 – methyl – 1 – propanol
d) 2 – butanol
Answer:
b) 2 – methyl – 2 – propanol
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 72

Question 12.
2 – methyl propene reacts with diborane followed by H2O2 in presence of NaOH gives
a) 2 – proponol
b) 2 – methyl – 2 – propanol
c) 2 – methyl – 1 – propanol
d) 2 – butanol
Answer:
c) 2 – methyl -1 – propanol
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 73

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 13.
To prepare a primary alcohol a Grignard reagent must be reacted with
a) HCHO
b) RCHO
c) RCOR
d) RNH2
Answer:
a) HCHO

Question 14.
To prepare a secondary alcohol a
Grignard reagent must be reacted with
a) HCHO
b) RCHO
c) RCOR
d) RNH2
Answer:
b) RCHO

Question 15.
To prepare a tertiary alcohol a Grignard reagent must be reacted with
a) HCHO
b) RCHO
c) RCOR
d) RNH2
Ans:
b) RCOR

Question 16.
Phenyl magnesium bromide reacts with formaldehyde followed by hydrolysis gives
a) phenol
b) Benzyl alcohol
c) Benzaldehyde
d) Benzoic acid
Answer:
b) Benzyl alcohol
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 74

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 17.
Name the product obtained when 1 mole of methyl magnesium bromide reacts with ethyl methanoate
a) propan -2- ol
b) propan – 1 – ol
c) ethanal
d) ethanol
Answer:
c) ethanal
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 75

Question 18.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 76
The reducing agent X is
a) Ramey Ni
b) NaBH4
c) LiA/H4
d) Na – Hg
Answer:
b) NaBH4

Question 19.
The best reducing agent to prepare unsaturated alcohols from unsaturated carbonyl compounds is
a) Raney Ni
b) NaBH4
c) LiA/H4
d) both (b) & (c)
Answer:
d) both (b) & (c)

Question 20.
Ethene is converted into ethylene glycol by treating with
a) acidified potassium dichromate
b) alkaline potassium permanganate
c) Chromyl chloride
d) nitric acid
Answer:
b) alkaline potassium permanganate

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 21.
Cold dilute alkaline potassium permanganate is called as
a) Fenton’s reagent
b) Tollen’s reagent
c) Baeyer’s reagent
d) Schiff’s reagent
Answer:
c) Baeyer’s reagent

Question 22.
Glyceryl palmitate on alkaline hydrolysis gives
a) glycerol and soap
b) glycerol and sodium palmitate
c) glycerol and sodium hexadentate
d) all the above
Answer:
d) all the above

Question 23.
The preparation of glycerol and soap from oils and fats is known as
a) esterification
b) saponification
c) hydroboration
d) transesterification
Answer:
b) saponification

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 24.
Lucas reagent is
a) dil HCl & ZnCl2
b) Cone. HCl & anhydrous ZnCl2
c) dil H2SO4 & AlCl3
d) Conc. H2SO4 & anhydrous AlCl3
Answer:
b) Cone. HCl & anhydrous ZnCl2

Question 25.
Which among the following will give immediate turbidity with Lucas reagent?
a) propan – 2 – ol
b) propan – 1 – ol
c) 2 – methyl propan – 2 – ol
d) 2 – methyl propan – 1 – ol
Answer:
c) 2 – methyl propan – 2 – ol
Hint: A teritary alcohol gives immediate turbidity

Question 26.
In Victor Meyer’s test, the alcohol which
gives blue colour is
a) propan – 2 – ol
b) propan – 1 – ol
c) 2 – methyl propan – 2 – ol
d) 2 – methyl propan – 1 – ol
Answer:
a) propan – 2 – ol
Hint: A secondary alcohol gives blue colour.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 27.
In the given equation
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 77
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 78
Answer:
c

Question 28.
Among butyl alcohols which one have the lowest boiling point?
a) n-butyl alcohol
b) isobutyl alcohol
c) sec-butyl alcohol
d) ter – butyl alcohol
Answer:
d) ter – butyl alcohol Hint: Increasing order of boiling point is 30 < 2° < 1°

Question 29.
Which among the alcohol has a higher boiling point?
a) n-butyl alcohol
b) n-propyl alcohol
c) ethyl alcohol
d) methyl alcohol
Answer:
a) n-butyl alcohol
Hint: As molecular weight of the alcohol increases, boiling point increases.

Question 30.
Which among the following is less soluble in water?
a) n-butyl alcohol
b) n-propyl alcohol
c) ethyl alcohol
d) methyl alcohol
Answer:
a) n-butyl alcohol Hint: As the molecular weight increases, the solubility of alcohol decreases.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 31.
Conversion of alcohols into alkyl halide is an example of
a) Nucleophilic addition
b) Nucleophilic substitution
c) Electrophilic addition
d) Electrophilic substitution
Answer:
b) Nucleophilic substitution

Question 32.
Conversion of 2 – methyl – 1 – propanol into 2 – methyl – 1 – bromopropane is …………. reaction
a) SN2
b) SN1
c) E2
d) E1
Answer:
a) SN2

Question 33.
Primary alcohols undergo dehydration by mechanism
a) SN2
b) SN1
c) E2
d) E1
Ans:
d) E1

Question 34.
2 – methyl – 2 – propanol when reacted with cone. H2SO4 gives 2 – methyl propene. This reaction follows mechanism
a) SN2
b) SN1
c) E2
d) E1
Answer:
d) E1

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 35.
During intramolecular dehydration of 3, 3 – dimethyl – 2 – butanol the major product obtained is
a) 2, 3 – dimethyl but – 1 – ene
b) 2, 3 – dimethyl but – 2 – ene
c) 3, 3 – dimethyl but – 1 – ene
d) 3, 3 – dimethyl but – 2 – ene
Answer:
b) 2, 3 – dimethyl but – 2 – ene

Question 36.
The major product obtained when phenol reacts with con.H2SO4 at 280 K is
a) Salicyclic acid
b) picric acid
c) o-phenol sulphonic acid
d) p-phenol sulphonic acid
Answer:
c) o-phenol sulphonic acid

Question 37.
During intramolecular dehydration of alcohols say tzeff’s rule favours the formation of
a) unstable alkenes
b) less substituted alkenes
c) more substituted alkenes
d) none of the above
Answer:
c) more substituted alkenes

Question 38.
To stop the oxidation reaction of alcohols at the aldehyde / ketone stage…………..is used as an oxidising agent
a) Pottassium permanganate
b) Py ridinium chloro chromate (PCC)
c) Potassium di chromate
d) Sodium di chromate
Answer:
b) pyridinium chloro chromate (PCC)

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 39.
In Swern oxidation of alcohols into aldehydes/ketones the oxidising agent used is
a) Pyridinium chlorochromate (PCC)
b) Dimethyl sulfoxide (DMSO)
c) Alkaline potassium permanganate (Baeyer’s reagent)
d) Ferrous sulphate / H2O2 (Fenton’s reagent)
Answer:
b) Dimethyl sulfoxide (DMSO)

Question 40.
To detoxify the alcohol produced in animals by the fermentaion of food, the oxidising agent used is
a) ADH
b) ADP
c) NAD
d) ATP
Answer:
c) NAD

Question 41.
The catalyst which catalyses the oxidation of toxic alcohols into non-toxic aldehyde in animals is
a) ADH
b) ADP
c) NAD
d) ATP
Answer:
a) ADH

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 42.
When ethan-1, 2-diol is heated with anhydrous ZnCl2 under pressure it gives
a) Ethanol
b) Ethanal
c) Ethanoic acid
d) Ethene
Ans:
b) Ethanal

Question 43.
Glycerol can be oxidised to meso oxalic acid by
a) dii HNO3
b) HIO4
c) Bismuth nitrate
d) Fenton’s reagent
Answer:
c) Bismuth nitrate

Question 44.
The alcohol used in the manufacture of dynamite is
a) methanol
b) ethanol
c) ethv lene glycol
d) glycerol
Answer:
d) glycerol

Question 45.
The correct order of acidic nature of alcohols is
a) Ethanol < propan – 2 – ol < 2 – methyl propan – 2- ol
b) Propan – 2 – ol < Ethanol < 2 – methyl propan — 2 — ol
c) 2 – methyl propan – 2 – ol < Ethanol < propan – 2 – ol
d) 2 – methyl propan – 2 — ol Answer:
d) 2 – methyl propan – 2 – ol

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 46.
1, 3 – dihydroxy benzene is commonly
known as
a) Cresol
b) Catechol
c) Resorcinol
d) Quinol
Answer:
c) Resorcinol

Question 47.
The JUPAC name of pyrogallol is
a) 1, 2, 3 — trthydroxy benzene
h) 1, 2, 4 — trihydroxy beneze
c) 1, 3, 5 — trihydroxv benzene
d) 1, 4, 5 — tri hydroxy benzene .
Answer:
a) 1, 2, 3 — trihydroxy benzene

Question 48.
Oricinol is
a) 1, 2 – dihydroxy benzene
h) 3 — methyl phenol :
c) 3, 5 — dihydroxy toluene
d) 3, 5 — dimethyl toluene
Answer:
c) 3, 5 – dihydroxy toluene

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 49.
Which of the following reaction will give ether?
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 79
Ans:
d) All of the above

Question 50.
Phenol on oxidation with acidified K2Cr2O7gives
a) 1, 4 – dihydroxy benzene
b) 1,4- benzoquinone
c) cyclohexanol
d) cumene
Answer:
b) 1,4- benzoquinone

Question 51.
Which among the following is a simple ether?
a) 1 – methoxy propane
b) 2 – methoxy propane
c) methoxy methane
d) methoxv benzene
Answer:
c) methoxy methane

Question 52.
The oxygen atom of ether is
hybridised
a) sp
b) sp2
c) sp3
d) dsp2
Answer:
c) sp3

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 53.
Williamson’s synthesis is an example of
a) nucleophilic addition
b) nucleophilic substitution
c) electrophilic addition
d) electrophilic substitution
Answer:
b) nucleophilic substitution

Question 54.
Which among the following is more reactive
towards ethers?
a) HF
b) HCl
c) HBr
d) HI
Answer:
d) HI

Question 55.
The products obtained when methoxy
ethane reacted with one mole of HI are
a) methanol & iodoethane
h) iodomethane & ethanol
c) iodomethane & jodo ethane
d) Methanol & ethanol
Answer:
b) iodomethane & ethanol

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 56.
Which is used as a precursor to the synthesis of perfumes and insecticide pheromones?
a) Phenol
b) Phenoxy methane
c) methoxy benzene
d) Ethoxy benzene
Answer:
c) methoxy benzene

Question 57.
According to Lewis concept of acids and bases an ether is
a) acidic
b) basic
c) neutral
d) amphoteric
Answer:
b) basic

Question 58.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 80
a) CH3COCH3
b) C2H5OH
c) CH3-CH(OH)CH3
d) CH3CHO
Answer:
c) CH3-CH(OH)CH3
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 81

Question 59.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 82
a)Propan-1,3-diol
b) Ethan -1,2- diol
c) Propan -1,2,3- Triol
d) Ethanal
Answer:
b) Ethan -1,2- diol
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 83

Question 60.
Which of the following has lowest boiling point?
a) phenol
b) O-nitro phenol
c) m-nitro phenol
d) p-nitro phenol
Answer:
b) O-nitro phenol
Hint: O – nitrophenol has intramolecular hydrogen bonding while others intermolecular hydrogen bonding.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

II. Assertion and Reason

Question 1.
Assertion : Tertiary alcohols undergo dehydration more readily than primary alcohol.
Reason : Tertiary alcohols are less acidic than primary alcohol.
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:
b) Both assertion and reason are true but reason is not the correct explanation of assertion

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 2.
Assertion (A) : Alcohols can act as Bronsted acids
Reason (R) : It is due to the presence of unshared electron pairs on oxygen which make them proton acceptors.
a) Both (A) and (R) are correct, R explain A.
b) Both (A) and (R) are correct, R does not explain A.
c) (A) is correct but R is correct.
d) (A) is wrong but R is correct.
Answer:
d) (A) is wrong but R is correct Correct Reason (A) : Alcohols can act as Bronsted bases.

Question 3.
Assertion (A) : Phenol is more acidic than aliphatic alcohols
Reason (R) : Alkyl substituted phenols show a decreased acidity due to the electron releasing +1 effect of alkyl group.
a) Both (A) and (R) are correct, R explain A.
b) Both (A) and (R) are correct, R does not explain A.
c) (A) is correct but R is correct.
d) (A) is wrong but R is correct.
Answer:
b) Both (A) and (R) are correct, R does not explain A.
Correct Reason (R) : The phenoxide ion is more stabilised by resonance than phenol.

Question 4.
Assertion (A): Orthonitro phenol is slightly soluble in water and more volatile whereas p – nitro phenol is more soluble in water and less volatile.
Reason (R) : Orthonitro phenol forms intramolecular hydrogen bonding and p-nitro phenol forms intermolecular hydrogen bonding.
a) Both (A) and (R) are correct, R explains A.
b) Both (A) and (R) are correct, R does not explain A.
c) (A) is correct but R is correct.
d) (A) is wrong but R is correct.
Answer:
a) Both (A) and (R) are correct, R explains A.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

III. Pick out the correct statements

Question 1.
(i) The structure of – OH group attached to a sp3– hybridised carbon in alcohol is similar to the structure of – OH group attached to a hydrogen in water.
(ii) Due to lone pair – lone pair repulsion, the – C – O – H bond angle in methanol is reduced to 104° from the regular tetrahedral bond angle of 109.5°.
(iii) With Grignard reagent formaldehyde gives primary alcohol and other aldehydes give secondary alcohols.
(iv) Reaction of Grignard reagent with aldehydes and ketones to form alcohols is an example for nucleophilic substitution reaction.
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (i) & (iv)
Answer:
c) (i) & (iii)
Correct statements : (ii) Due to lone pair – lone pair repulsion, the – C – O – H bond anlge in methanol is reduced to 108.9° from the regular tetrahedral bond angle of 109.5°.
(iv) Reaction of Grignard reagent with adlehydes and ketones to form alcohols is an example for nucleophilic addition reaction.

Question 2.
(i) Lower alcohols are waxy solids and the higher members are colourless liquids.
(ii) Due to the presence of intermolecular hydrogen bonding alcohols have higher boiling point than the corresponding alkanes, aldehydes, ethers etc.
(iii) Due to the formation of intermolecular hydrogen bonding with water lower alcohols are highly soluble in water.
(iv) Among isomeric alcohols primary alcohols have lower boiling point and tertiary alcohols have higher boiling points.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statements : (i) Lower alcohols are colourless liquids and the higher members are waxy solids.
(iv) Among isomeric alcohols primary alcohols have higher boiling point and the tertiary alcohols have lower boiling points.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 3.
(i) In Swern Oxidation DMSO, oxalyl chloride and triethylamine are used to convert alcohols into aldehydes / ketones.
(ii) Vapours of primary alcohol passed over heated copper at 573 K undergoes dehydration. .
(iii) Methanol reacts with ethanoic acid in presence of an acid to form ethylethanoate.
(iv) Vapours of tertiary alcohols react with heated copper at 573 K to give alkenes.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)
Correct statements : (ii) Vapours of primary alcohol passed over heated copper at 573 K undergoes dehydrogenation.
(iii) Methanol reacts with ethanoic acid in presence of an acid to form methyl ethanoate. (or)
Ethanol reacts with ethanoic acid in presence of an acid to form ethyl ethanoate.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 4.
(i) Glycerol contains two primary alcoholic group and one secondry alcoholic group.
(ii) Oxidation of glycerol with cone. HNO3 gives mainly formic acid.
(iii) Glycerol is used in making printing inks, and stamp pad ink.
(iv) Glycerol is used as a preservative for biological specimens
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (i) & (iv)
Answer:
c) (i) & (iii)
Correct statements : (ii) Oxidation of glycerol with cone. HNO3 gives mainly glyceric acid
(iii) Ethanol is used as a preservative for biological specimens

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

IV. Pick out the incorrect statements

Question 1.
(i) The overall reaction of hydroboration is hydration of an alkene.
(ii) Hydroboration reaction occurs according to Markownikoff’s rule.
(iii) Raney Ni does not reduce the Carbon – Carbon double bond present in the Carbonyl compound to form unsaturated alcohols.
(iv) When two or more functional groups are present in a molecule sodium borohydride to used as a reducing agent to reduce the more reactive group
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statements : (ii) Hydroboration reaction occurs against Markownikoff’s rule.
(iii) Lithium aluminium hydride does not reduce the Carbon – Carbon double bond present in the Carbonyl compound to form unsaturated alchols.

Question 2.
(i) Ethanol forms a turbidity with Lucas reagent within 10 minutes.
(ii) 2 – Methyl propan – 2 – ol forms immediate turbidity with Lucas reagent due to the formation of insoluble 2 – chloro – 2 – methyl propane.
(iii) Ehanol forms red colour in Victor Meyer’s test.
(iv) 2 – 2 – dimethyl propan – 1 – ol gives a colourless solution in victor Meyer’s test,
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)
Correct statements : (i) Ethanol forms no tubridity with lucas reagent at room temperature.
(iv) 2,2 – dimethyl propan -1 – ol gives red colour in victor mayer’s test.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 3.
(i) Alcohols undergo nucleophilic substitution reaction with hydrohaloacids to form alkyl halides.
(ii) Alkyl halide formation from primary alchols follow E2 mechanism.
(iii) Alkyl halide formation from tertiary alcohols follow SN2 mechanism.
(iv) The conversion of methanol into chloromethane with thionyl chloride in the presence of pyridine follows SN2 mechanism.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statements : (ii) Alkyl halide formation from primary alcohols follow \(\mathrm{S}_{\mathrm{N}}^{2}\) mechanism.
(iii) Alkyl halide formation from tertiary alcohols follow \(\mathrm{S}_{\mathrm{N}}^{1}\) mechanism.

Question 4.
(i) In phenol the carbon bearing the – OH group is sp2 hybridised.
(ii) Unlike alcohols phenol reactswith sodium hydroxide to form sodium phenoxide.
(iii) In substituted phenols, the electron with drawing groups decreases the acidic nature of phenol.
(iv) Alkyl substituted phenols show increased acidity due to electron releasing +1 effect of alkvl group.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct statements: (iii) In substituted phenols, the electron withdrawing groups increases the acidic
nature of phenol.
(iv) Alkyl substituted phenols show a decreased acidity due to electron releasing + I effect of alkyl
group.

V. Match the Following

Question 1.

Oxidising agent

Oxidation product of glycerol

i) dil HNO3a meso oxalic acid
ii) Conc. HNO3b formaldehyde & formic acid
iii) Bismuth nitratec oxalic acid
iv) Fenton’s reagentd glyceric acid & tartronic acid
v) HIO4e glyceric acid
vi) Acidified KMnO4f glycerose

Answer:
i) d glyceric acid & tartronic acid
ii) e glyceric acid
iii) a meso oxalic acid
iv) f glycerose
v) b formaldehyde & formic acid
vi) c oxalic acid

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 2.

ExampleType of Phenol
i) Orcinola Monohydric pheno
ii) Catecholb Trihydric phenol
iii) Cresolc Dihydric phenol
iv) Phioroglucinold Substituted phenol

Answer:
i) d Substituted phenol
ii) c Dihydric phenol
iii) a Monohydric phenol
iv) b Trihydric phenol

Question 3.

CompoundUses
i) Phenola Antifreeze
ii) Glycerolb Perfumery
iii Glycolc Carbolic soaps
iv Anisoled Refrigerant
v) Diethyl ethere Cordite

Answer:
i) c Carbolic soaps
ii) e Cordite
iii) a Antifreeze
iv) b Perfumery
v) d Refrigerant

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 4.

Type of AlcoholExample
i 1° alcohola Glycerol
ii 2° alcoholb Sorbitol
iii 3° alcoholc Ethylene glycol
iv Dihydric alcohold Isopropyl alcohol
v Trihydric alcohole Neo pentyl alcohol
vi Polyhydric alcoholf 2-phenyl propan-2-ol

Answer:
i) e Neo pentyl alcohol
ii) d Isopropyl alcohol
iii) f 2-phenyl propan-2-ol
iv) c Ethylene glycol
v) a Glycerol
vi) b Sorbitol

VI. Two Marks Questions

Question 1.
Illustrate Markownikoff’s rule in the addition of water to an alkene
Answer:
Markownikoff’s rule states that the negative part of the adding molecules gets attached with the carbon containing least number of hydrogen atoms across the double bond.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 84

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 2.
Write a note on hydroboration reaction
Answer:
Diborane reacts with an alkene to form trialkyl borane which on treatment with H2O2 in presence of NaOH gives an alcohol.
The overall reaction is hydration of an alkene.
This reaction gives an anti MarkowniKoff’s product
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 85

Question 3.
Convert crotonaldehyde into crotylalcohol
Answer:
LiAlH4 does not reduce C = C bond present in the carbonyl compound.
Hence it is the best reagent to prepare unsaturated alcohols.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 86

Question 4.
An organic compound (A) – C3H8O3 used as sweetening agent, which on oxidation with Fenton’s reagent gives a mixture of compounds B and C. Identify A, B & C. Write possible reactions.
Answer:
An organic compound (A) – C3H8O3 is glycerol.
Oxidation of glycerol with Br2/H2O (or) NiaOBr (or) Fenton’s reagent (FeSO4 + H2O2) gives a mixture of glyceraldehvde and dihydroxv acetone ( This mixture is named as glvcerose)
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 87

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 5.
What is saponification?
Answer:
The alkaline hydroysis of naturally occurring oils and fats into glycerol and soap is called saponification.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 88

Question 6.
What is Saytzeff’s rule
Answer:
During intramolecular dehydration of an alcohol if there is a possibility to form a carbon- carbon double bond at different locations, the preferred location is the one that gives the more substituted alkene ie., the stable alkene.

Question 7.
Convert propan -1 – ol to propanal.
Answer:
To stop the oxidation reaction of alcohols at the aldehyde / ketone stage, pyridinium chloro chromate (PCC) is used as an oxidising agent.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 89

Question 8.
Write a note on Swern Oxidation
Answer:
In Swern oxidation dimethvl sulfoxide (DMSO) is used as the oxidising agent.
It converts alcohols to ketones / aldehydes.
Alcohol is treated with DMSO and oxah I chloride followed by the addition of triethylarnine.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 90

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 9.
Write note on biological oxidation that occurs in animals.
Answer:

  • The fermentation of the food consumed by an animal produces alcohol.
  • To detoxify the alcohol, liver produces an enzyme called alcohol dehydrogenase (ADH)
  • Nicotinamide adenine dinucleotide (NAD) present in animals acts as an oxidising agent.
  • ADH catalyses the oxidation of toxic alcohols into non-toxic aldehyde.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 91

Question 10.
What happens when ethylene glycol is reacted with PI3 ?
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 92

Question 11.
What happens when ethylene glycol is heated with conc. HNO3 and conc. H2SO4?
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 93

Question 12.
How is nitroglycerine prepared
Answer:
Glycerol reacts with nitric acid in the presence of cone. H2SO4 to form nitroglycerine (TNG)
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 94

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 13.
Write about the dehydration of glycerol
Answer:
When glycerol is heated with dehydrating agents such as H2SO4, KHSO4 etc., it undergoes dehvration to form acrolein.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 95

Question 14.
Convert aniline into phenol
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 96

Question 15.
What is Schotten – Baumann reaction
Answer:
Benzovlation of phenol with benzoyl chloride in presence of NaOH is known as Schotten Baumann reaction.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 97

Question 16.
Write a note on williamson ether synthesis?
Answer:
Methyl iodide undergoes nucleophilic substitution by phenoxide ion to form anisole
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 98

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 17.
Convert phenol into (i) 1, 4 – benzoquionone (ii) cyclo hexanol
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 99

Question 18.
How is picric acid prepared?
Answer:
Phenol undergoes nitration with cone HNO3 and cone H2SO4 at 298 K to form 2, 4/ 6 – trinitrophenol known as picric acid
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 100

Question 19.
Differentiate phenol and ethanol
Answer:

Reaction

Phenol

Ethanol

With benzene diazonium chlorideforms a red orange dyeNo reaction
With neutral ferric chlorideforms a purple colorationdoes not form purple colouration
WithNaOHgives sodium phenoxidedoes not react

Question 20.
Give four uses or diethyl ether.
Answer:

  1. Diethyl ether is used as a surgical anaesthetic agent in surgery.
  2. It is a good solvent for organic reactions and extraction.
  3. It is used as a volatile starting fluid for diesel and gasoline engine.
  4. It is used as a refrigerant.

Question 21.
Write the mechanism of intermolecular dehydration of alcohols?
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 101

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 22.
Write the mechanism of nucleophilic substitution of ethers
Answer:
Mechanism:
Ethers having primary alkyl group undergo SN2 reaction while tertiary alkyl ether undergo SN1 reaction.Protonation of ether is followed by the attack of halide ion.The halide ion preferentially attacks the less sterically hindered of the two alkyl groups which are attached to etherial oxygen.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 102

When excess HBr or HI is used, the alcohol formed will further react with HBr or HI to form alkyl halides.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 103

Question 23.
What is autooxidation of ethers?
Answer:

  • Ethers stored in the presence of atmospheric oxygen slowly oxidise to form hydroperoxides and dialkyl peroxides.
  • These are explosive in nature.
  • This spontaneous oxidation by atmospheric oxygen is called auto oxidation

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 104

Question 24.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 105
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 106

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 25.
Complete the reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 107
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 108
The bond of oxygen with phenyl ring is strong, hence it can not be broken and iodo benzene is not formed.

VII. Three Mark Questions

Question 1.
Explain the structure of the functional group of alcohol
Answer:

  • The structure of – O – H group attached to a SP3 hybridised carbon is similar to the structure of – O – H group attached to a hydrogen in water, ie., ‘V! shaped.
  • In such alcohols, one of the sp3 hybridised orbital of oxygen linearly overlap with the sp3 hybridised orbital of carbon to form a C – O, ‘σ’ bond.
  • Another SP3 hybridised orbital linearly overlap with IS orbital of hydrogen to form a O-H, a’ bond.
  • Remaining two sp3 hybridized orbitals of oxygen are occupied by two lone pairs of electrons.
  • Due to lone pair – lone pair repulsion, the C – O – H bond angle in methanol is reduced to 108.9C from the regular tetrahedral bond angle of 109.5°.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 109

Question 2.
Convert acetaldehyde into crotyl alc0hol
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 110
LiAlH4 selectively reduces the carbonyl group leaving the C = C bond.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 3.
Explain Lucas test of differentiating three types of alcohols
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 111
The alkyl chloride formed in this reaction is insoluble in water, hence it appears as turbidity.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 112

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 4.
Write a note on catalytic dehydrogenation of three types of alcohols.
Answer:
Catalytic dehydrogenation :
When the vapours of a primary or a secondary alcohol are passed over heated copper at 573K, dehydrogenation takes place to form aldehyde or ketone.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 113 Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 114

Question 5.
Write about the oxidation of ethylene glycol
Oxidation of glycol
Answer:
On oxidation, glycol gives a variety of products depending on the nature of oxidizing agent and
other reaction conditions.
i) When nitric acid (or) alkaline potassium permanganate is used as the oxidizing agent, the following products are obtained.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 115

ii) Oxidation of glycol with periodic acid :
Ethylene glycol on treatment with periodic acid gives formaldehyde. This reaction is selective for vicinal 1,2 – diols and it proceeds through a cyclic periodate ester intermediate.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 116

Question 6.
Why is C-O-C bond angle in ether slightly greater than the tetrahedral bond angle.
Answer:

  • Structure of ethereal oxygen attached to two alkyl groups is similar to the structure of – O – H group of alcohol.
  • Oxygen atom is sp3 hybridised.
  • Two sp3 hybridised orbitals of oxygen linearly overlap with two sp3 hybrid orbitals of the two carbon atoms attached directly to the oxygen forming two C – O ‘o ‘ bonds.
  • C-O-C bond angle is slightly greater than the tetrahedral bond angle due to the repulsive interaction between the two bulkier alkyl group.

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 117

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 7.
For the preparation of mixed ether having primary and tertiary alkyl group, primary alkyl halide and alkoxide are used. Why?
Answer:
Primary alkyl halides are more susceptible for SN2 reaction.
Hence for the preparation of mixed ether having primary and tertiary alkyl group, primary alkyl halide
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 118

On the other hand, if tertiary’ alkyl halide and primary alkoxide are used elimination reaction dominates and succeeds over substitution reaction to form an alkene.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 119

Question 8.
How do you convert anisole into (i) Methoxy toluene (ii) methoxy acetophenone
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 120

Question 9.
375 mg of an alcohol reacts with required amount of methyl magnesium bromide and releases 140 mL of methane gas at STP. What is the formula of the alcohol?
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 121
(1 mole) 22,400 mL of methane can be produced from 1 mole of alcohol.
140 ml of methane is liberated from 1/22400 x 140 mole of alcohol = 6.25 x 10-3 mole of alcohol
w = 375 mg = 375 x 10-3 g; n = 6.25 x 10-3
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 122
Molar mass of alcohol is 60 gmol-1
CnH2n+1 -OH=>nxl2 + (2n + 1)xl + lxi6+1xl=60
12n + 2n + 1 + 16 + 1 = 60 14n + 18 = 60
14n = 60 – 18
Hence the formula of the alcohol is C3H7OH .
n = 42/14 ⇒ n = 3

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 10.
How will you convert methanol into ethanol?
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 123

VIII. Five Mark Questions

Question 1.
Convert methyl magnesium iodide into
(i) ethanol
(ii) propan – 2 – ol
(iii) 2 – methyl propan – 2 – ol
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 124

Question 2.
Bring about the following conversion using Grignard reagents
i) Methanal into phenyl methanol
ii) Ethanal into butan – 2 – ol
iii) Propanone into 2 – methyl hexan – 2 -ol
iv) Ethyl methanoate into propan – 2 – ol
v) Formaldehyde into propan-l-ol
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 125

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 3.
A ether (A) C5H12O when heated with excess of hot concentration HI, produced two alkyl halides, which on hydrolysis forms compound (B) and (C). Oxidation of (B) gives an acid (D) where as oxidation of (C) gives ketone (E). Identify A, B, C, D and E and write the chemical equation.
Answer:
A ether (A) C5H12O when heated with excess of hot concentration HI, produced two alkyl halides.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 126 Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 127

CompoundName
A2- ethoxy propane
Bethanol
C2- propanol
DAcetic acid
EPropanone

Question 4.
Explain E1mechanism involved in the dehydration of tertiary butyl alcohol
Answer:
Tertiary alcohols undergo dehydration by E1 mechanism. It involves the formation of a carbocation.
Protonation of alcohol
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 128 Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 129

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 5.
An organic compound C2H6O (A) heated with Cone. H2SO4 at 443K to give and unsaturated hydrocarbon C2H4 (B), which on treatment with Baeyer’s reagent to give compound C2H6O2 (C). Which is used as antifreeze in automobile radiator. Compound (C) distilled with cone. H2SO4 to give cyclic compound (C4H8O2) (D). Compound (A) is heated with cone H2SO4 at 413K to give compound C4H10O (E). Identify compounds (A) to (E) and write equations Compound (A) is Ethanol heated with Cone. H2SO4 at 443K to give compound (B) is Ethene
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 130

Compound (B) on treatment with Baeyer’s reagent to give compound (C) is Ethylene glycol
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 131

Compound ( C) distilled with cone. H2SO4 to give cyclic compound (D) 1,4- dioxene
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 132

Compound (A) is heated with cone. H2SO4 at 413K to give compound (E) is diethylether
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 133

CompoundName
AEthanol
BEthene
CEthane 1,2-diol
D1,4 dióxane
EDiethylether

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 6.
Explain the oxidation of glycerol
Answer:

Oxidising agentProduct formed
i) dil HNO3glyceric acid & tartronic acid
ii) Cone HNO3glyceric acid
iii) bismuth nitratemeso oxalic acid
iv) BD/H2O (or) NaOBr
(or) Fenton’s reagent (FeSO4 + H2O2)
a mixture of glyceraldehyde and dihydroxyacetone knew as glycerose
v) HIO4 (or) Lead tetraacetate (LTA)formaldehyde & formic acid?
vi) acidified KMnO4Oxalic acid

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 134

Question 7.
Write a note on (i) Riemer Tiemann reaction (ii) Phthalein reaction (iii) coupling reaction
Answer:
i) Riemer Tiemann Reaction:
On treating phenol with CHCl3 /NaOH, a -CHO group is introduced at ortho position This reaction proceeds through the formation of substituted benzal chloride intermediate.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 135

(ii) Phthalein reaction:
On heating phenol with phthalic anhydride in presence of con.H2SO4 phenolphthalein is obtained.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 136

iii) Coupling reaction:
Phenol couples with benzene diazonium chloride in an alkaline solution to form p-hydroxy azobenzene (a red-orange dye).
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 137

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 8.
Write any three methods of preparing ethers
Answer:
(i) Intermolecular dehydration of alcohol
Ethanol undergoes intermolecular dehydration with conc. H2SO4 at 413 K to form diethyl ether. Conc. H2SO4
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 138

(ii) Williamson ether synthesis
When an alkyl halide is heated with an alcoholic solution of sodium alkoxide, the corresponding ether is obtained. This involves SN2 mechanism.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 139

(iii) Methylation of alcohol
Methyl ethers can be prepared by treating alcohol with diazomethane in presence of catalyst fluoroboric acid
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 140

Question 9.
How does diethylether react with the following
(i) Cl2/light
(ii) PCl5
(iii) dilH2SO4
(iv) CH3COCl/anhy.ZnCl2
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 141

Question 10.
Convert
(i) methanol into tertiary butyl alcohol
(ii) ethanol into 1- butanol
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 142 Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 143

Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers

Question 11.
Complete the reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 144
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 145
(y) Reduction to pinacols: Ketones, on reduction with magnesium amalgam and water, are reduced to symmetrical diols known as pinacol.
Samacheer Kalvi 12th Chemistry Guide Chapter 11 Hydroxy Compounds and Ethers 146

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 15 Recent Trends in Marketing Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions 15 Recent Trends in Marketing

12th Commerce Guide Recent Trends in Marketing Mix Text Book Back Questions and Answers

I. Choose The Correct Answer.

Question 1.
Selling goods/ services through internet is
a) Green marketing
b) E- business
c) Social marketing
d) Meta marketing
Answer:
b) E-business

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

Question 2.
Which is gateway to internet?
a) Portal
b) CPU
c) Modem
d) Webnaire
Answer:
c) Modem

Question 3.
Social marketing deals with:
a) Society
b) Social Class
c) Social change
d) Social evil
Answer :
c) Social change

Question 4.
Effective use of Social media marketing increase conversion rates of –
a) Customer to buyers
b) Retailer to customers
c) One buyer to another buyers
d) Direct contact of marketer
Answer:
a) Customer to buyers

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

Question 5.
A company’s products and prices are visually represented by
a) Shopping cart
b) Web portal
c) Electronic catalogue
d) Revenue model
Answer:
c) Electronic catalogue

Question 6.
Pure play retailers are called
a) Market creators
b) Transaction brokers
c) Merchants
d) Agents
Answer:
b) Transaction brokers

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

II. Very Short Answer Questions.

Question 1.
What is E-Business?
Answer:
If all the business transactions are carried out through the internet and other online tools, it is called E-business.

Question 2.
What is green marketing?
Answer:

  • It refers to a holistic marketing concept with growing awareness about the implications of Global warming, non -biodegradable solid waste, the harmful impact of pollutants, etc.
  • It is also known as “Environmental marketing”, “Ecological marketing” Eco-friendly marketing and sustainable marketing”.

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

Question 3.
What is service marketing?
Answer:
Service marketing denotes the processing of selling service goods like telecommunication, banking, insurance, car rentals, healthcare, tourism, professional services, repairs etc.

Question 4.
Define E-Marketing.
Answer:
“E-Marketing refers to the application of a broad range of information technology for creating more customer value through more effective segmentation, targeting differentiation and positioning strategies, planning more efficiently and executing the conception distribution, promotion and pricing of goods, services and ideas and creating exchanges that satisfy individual consumer and organization’s consumer objectives”. _ JUDY STRAUSS

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

Question 5.
What is E-Tailing?
Answer:
E-tailing or electronic retailing refers to the selling of goods and services through a shopping website (internet) or through a virtual Store to the ultimate consumer.

Question 6.
What is social marketing?
Answer:

  • Social marketing is a new marketing tool
  • It is the systematic application of marketing philosophy and techniques to achieve specific behavioural goals that ensure social good
  • [Eg] Asking people- Not to smoke in a public place.

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

III. Short Answer Questions.

Question 1.
What are the advantages of E-Marketing?
Answer:

  • Any time market: E-Marketing provides 24 hours and 7 days “24/7” Service to its users. So consumer can shop or order the product anytime from anywhere.
  • Direct contact of end consumer by the manufacturer cuts down the substantially intermediation cost. Thus products bought through e-marketing become cheaper.
  • Customers can buy whatever they want / need just by browsing the various sites.

Question 2.
Discuss the objectives E-Marketing (EARS)
Answer:

  • Expansion of market share.
  • Achieving higher brand awareness.
  • Reduction of distribution and promotional expenses
  • Strengthening database.

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

Question 3.
Explain in detail Niche marketing. [Nitch] [Neesh]
Answer:

  • “Niche marketing “ denotes a strategy of directing all marketing efforts towards one well-defined segment of the population.
  •  Actually, there is no market in a niche market.
  •  It is found by the company, by identifying the need of customers who are not served or underserved by the competitors.
  • It involves a specific target audience with a specialized offering.
  •  It aimed at being a big fish in a small pond instead of being a small fish in a big pond.
  •  [eg] Not everybody wants to watch the movie by paying 5x-6xtimes the cost of the normal tick.

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

IV. Long Answer Questions.

Question 1.
Explain in detail how traditional marketing differ from E-marketing [FEETS].
Answer:

Basic of differenceE-MarketingTraditional marketing
FasterIt is a faster way to promote the productIt takes more time to promote the products
EconomicalIt is very economicalIt is too expensive
ExpandIt can expand its operation with minimum manpowerIt needs more manpower to expand its operation
TimeIt is quite easier for promoting a product globally in a short timeIt requires more time
Sold and Bought Productcan be bought and sold around the yearThat is not possible

Question 2.
Explain the advantages and disadvantages of E-marketing.
Answer:
E-tailing or electronic retailing refers to the selling of goods and services through a shopping website:
Advantages:

  1. Customer can buy the product at any time from anywhere.
  2. Direct contact of end consumer by the manufacturer cuts down the cost.
  3. Customer can buy whatever they want by browsing the various sites.

Disadvantages:

  1. E-tailing needs a strong advertisement and for which it has to spend large amount.
  2. It is not suitable for small size business.

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

Question 3.
Discuss any two new methods of marketing with their advantages.
Answer:|
1. Rural Marketing:
Rural marketing is a process of developing pricing, promoting, and distributing rural specific goods and services leading to exchange with rural customers. There is an inflow of goods into rural markets for production and consumption and there is also an outflow of products to urban areas. the rural to urban flow consists of agricultural products like rice, wheat and sugar, etc.

2. Service Marketing:
Service marketing is a special branch of marketing. It denotes the processing of selling service goods like telecommunication, banking, insurance, healthcare, tourism, and professional services. The service products are mostly intangible. The unique feature of services marketing warrants different strategies compared with the marketing of physical goods.

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

12th Commerce Guide Recent Trends in Marketing Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
Pick the odd one out:
a) E-Commerce
b) E-Marketing
c) E-Tailing
d) E-Banking
Answer:
d) E-Banking

Question 2.
Find out which is not suitable? The products marketed in commodity exchange are ______
(a) Crude oil
(b) Rice
(c) Copper
(d) Gold
Answer:
(b) Rice

Question 3.
…………… is marketing knowns as environmental marketing or Ecological marketing or Eco-friendly marketing or sustainable marketing
a) Green
b) Social
c) Service
d) Rural
Answer:
a) Green

Question 4.
…………….. Marketing is holistic with growing awareness about implications of Global warming Non-Bio degradable solid waste harmful impact of pollutants etc.
a) Rural
b) Traditional
c) Service
d) Green
Answer:
d) Green

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

Question 5.
Which one of the following is not correctly matched?
a) Referral marketing – Reference
b) Content marketing – Communicating
c) E-Tailing – Online
d) Service marketing – Anti-drug
Answer:
d) Service marketing – Anti-drug

Question 6.
Select the odd one out.
a) Electronic marketing
b) Online marketing
c) Web marketing
d) Traditional marketing
Answer:
d) Traditional Marketing

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

Question 7.
Which one represents a cluster of manufacturers, content providers and online retailers organised around an activity?
a) Virtual mall
b) Association
c) Metomediary
d) Portal
Answer:
c) Metomediary

Question 8.
Green Shelter concept was introduced by group:
a) ACME
b) Tata
c) Reliance
d) ICI
Answer:
a) ACME

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

II. Match The Following

1. Match list -1 with list 2

LIST -1

LIST-2

(i) Green marketing1.Road safety
(ii) Service marketing2. Fast moving consumer good
(iii) Social marketing3. Banking
(iv) Rural marketing4.Friendly products

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing 1
Answer:
a) (i) 4, (ii) 3, (iii) 1, (iv) 2

Question 2.

LIST -1

LIST-2

i Niche marketing1. Without paying sponsoring expenses
ii Viral marketing2. Defined segment of populations
iii Ambush marketing3. Virus from person to person
iv Guerilla marketing4. Un convention-able system of promotions

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing 2

Answer:
c) (i) 2,(ii) 3,(iii) 1,(iv) 4

Question 3.

LIST-1

LIST-2

i Multilevel marketing1. Buyers accept delivery
ii Referral marketing2. Art of communicating
iii Content marketing3. Spreading the word
iv Commodity Exchange4. Recruit new distributors

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing 3

Answer:
a) (i) 4, (ii) 3, (iii) 2, (iv) 1

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

III. Assertion and Reason

Question 1.
Assertion [ A] : Actually there is no market in niche market
Reason [B]: It is found by identifying the need of customers which are not served and under served by the competitions .
a) A is true R is false
b) A is false R is true
c) A and R are true
d) A and R are false
Answer:
c) A and R are true

Question 2.
Assertion [A]: “Offer more – for less “ is a strategy to satisfy the consumers
Reason [R ]: They use computers, laptops tablets smart-phone android phone to access different websites.
a. A is true R is false
b. A and R are true
c. A and R are false
d. A and R are false
Answer:
b) A and R are true

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

IV. Short Answer Questions.

Question 1.
What is B2B and B2C type of E-Commerce?
Answer:
B2B-Business to Business:

  • Under this model .Business concerns transact with one another through internet.
  • [eg] Flipkart

B2C- Business to consumer:

  • Under this model Business concerns sell directly to the consumers.
  • [eg] Shopping zone .

Question 2.
Explain the importance of social marketing ?
Answer:

  • It helps to eradicate social evils that affect the society and quality of life
  • It promotes the consumption of socially desirable products and develops health consciousness.
  • It is a new marketing tool by using philosophy and techniques to achieve specific behavioural goals which ensure social goal.

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

Question 3.
Elucidate how E-Commerce differs from E-Business.
Answer:

  • E-Commerce simply refers to the buying and selling of products and services through online.
  • E-Business goes away beyond the simple buying and selling goods and services and much wider range of business processes.
  • E-Commerce and E-Business is used interchangeably in its broader meaning just as trade and commerce.

V. Long Answer Questions.

Question 1.
Describe the various strategies pursued in recent day’s marketers.
Answer:
Viral Marketing:

  • “Viral marketing “ is a marketing technique that impels the users to pass on a marketing message to other users, creating a potentially exponential in the message visibility and effect.
  •  It is able to generate interest and potential sale of a brand or product through messages that
    spread quickly like a virus from person to person.
  • [eg] Emergency needs through you tube, face book, twitter etc… [Need of blood -to save the life]

Guerrilla marketing:
“ Guerilla marketing “ represents an advertisement strategy to promote or services on streets or other public places with monkey like malls, parks, beach etc.

Referral marketing:
“Referral marketing is the method of promoting products or services to new customers through existing customers.

Ambush marketing:

  • It is a new technique where by a particular advertiser seeks to connect his product to the event in the mind of polentional customers without paying sponsoring expenses to the event.
  • [eg] X has sponsored a cricket match to promote his brand. A group of people sitting on the gallery wearing Y’ brand name T- shirts come in to focuses, in this case ‘Y’ promotes his brand at ‘X’ expense.

Multilevel marketing:

  • It is the marketing strategy where in the direct sales companies encourage its existing distributors
  • ID recruit new distributors to facilitate the sale of goods and services.

Content marketing:
It is said to be the art of communicating with customers and prospects without selling.

Samacheer Kalvi 12th Commerce Guide Chapter 15 Recent Trends in Marketing

Question 2.
Compare the concept of social marketing with service marketing. (BEEP)
Answer:

Basis of difference

Social marketing

Service marketing

BranchSocial marketing is a new marketing tooService marketing is a specialized branch of marketing
Ensures social goodIt ensures social good and by using marketing philosophy and techniques to achieve specific behavioural goals.It denotes the processing of selling service goals like banking, Insurances, Transport etc.
Eradicate social evils.It helps to eradicate social evils that affect the society and quality of life.It is tangible and in seperable from the service provider.  It does not eradicate social evils
Primary aimThe primary aim of the social marketing is social good like anti -tobacco, anti-drugs, anti-dowry, Road safety etc.Deliver service directly from the service provider without any inter median.

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 14 Marketing and Marketing Mix Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions 14 Marketing and Marketing Mix

12th Commerce Guide Marketing and Marketing Mix Text Book Back Questions and Answers

I. Choose The Correct Answer.

Question 1.
The initial stage of Marketing system is ……………
a) Monopoly system
b) Exchange to money
c) Barter system
d) Self producing
Answer:
c) Barter system

Question 2.
Who is supreme in the market?
a) Customer
b) Seller
c) Wholesaler
d) Retailer
Answer:
a) Customer

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 3.
In the following variables which one is not the variable of the marketing mix?
a) Place the variable
b) Product variable
c) Program variable
d) Price variable
Answer:
c) Program variable

Question 4.
Marketing mix means a marketing program that is offered by a firm to its target ………………… to earn profits through the satisfaction of their wants.
a) Wholesaler
b) Retailer
c) Consumer
d) Seller
Answer:
c) Consumer

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 5.
Which one is the example of an Intangible product?
a) Education
b) Mobiles
c) Garments
d) Vehicles
Answer:
a) Education

II. Very Short Answer Questions.

Question 1.
What is Marketing?
Answer:
Marketing is one of the business functions that all activities that take place in relation to markets for actualise potential exchanges for the purpose of satisfying human needs and wants.

Question 2.
Define marketing mix.
Answer:
“Marketing mix is a pack of four sets of variables namely product variable, price variable, promotion variable, and place variable”

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 3.
What is meant by Grading?
Answer:
Grading means the classification of standardized products into certain well-defined classes.

Question 4.
Define product.
Answer:
“A product is anything that can be offered to a market for attention, acquisition, use to consumption that might satisfy a want or a need” – PHILIP KOTLER

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

III. Short Answer Questions.

Question 1.
What are the objectives of marketing?
Answer:
Baker and Anshen say, “The end of all the marketing activities is the satisfaction of human wants”. The following are the objectives of marketing:

  1. Intelligent and capable application of modem marketing policies.
  2. To develop the marketing field.
  3. To develop guiding policies and their implementation for a good result.

Question 2.
What is need for market and explain the concept of marketing?
Answer:
Market is needed for the following:

  • To the movement of goods from the manufacturer to the consumer.
  • To increase the standard of living of the people.
  • To increase the national income
  • To increase employment opportunities.

Concept of marketing:
What I can sell?

  • Make what you can sell, but do not try to sell what you can make Shall I first create products?
  • No, first create a consumer, then create products.

Shall I love my products?
No, love your customers not the products.

Who is supreme in the market?
Customer is supreme or king.
Who will shape my decision?
Customers’ preferences shape your decisions.

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 3.
What are the factors affecting the Price of the Product? COMMO
Answer:
Factors affecting Price of product/service:
(a) Internal Factors:

  1. Marketing Objectives
  2. Marketing Mix Strategy
  3. Organizational considerations
  4. Costs
  5. Organization Objectives

(b) External Factors:

  1. The market after demand
  2. Competition
  3. Customers
  4. Suppliers
  5. Legal factors
  6. Regulatory factors..

Question 4.
What do you mean by marketing mix? Describe any two elements.
Answer:

  • Marketing mix means a marketing programme that is offered by a firm to its target consumers to earn profits through the satisfaction of their wants.
  • It is a mixture of four ingredients namely 4Ps.
    • Product
    • Price
    • Place
    • Promotion.
    • Product
  • “Product” is the main element of marketing.
  • Without a product, there can be no marketing.
  • Price is the value of a product expressed in monetary terms.
  • t is the amount charged for the product.

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

IV. Long Answer Questions.

Question 1.
Discuss the evolution of marketing.
Answer:
Barter system:
Before the invention of money, goods were exchanged for goods.

Production orientation:

  •  Producers more cared about the mass production of goods for the purpose of profit.
  •  They cared very little about the customers.

Sales orientation:

  • The stage witnessed a major change in all the spheres of economic life.
  • Sales become the dominant factor, neglect the consumer needs and satisfaction.

Marketing orientation:
Customer’s importance was realized, but only as a means of disposing of goods aggressive advertising is used to compete for the stiff competition.

Consumer orientation:
Under this stage only such products are brought forward to the market which are capable of satisfying the tastes, preferences and expectations of consumer satisfaction.

Management orientation:
It assumes a managerial role to co-ordinate the business activities with the objective of planning, promoting and services to the present and potential consumers.

Question 2.
Why marketing is important to society and individual firms? Explain.
Answer:
To the Society:

  •  Marketing is a connecting link between the consumers and the producers
  •  Helps in increasing the living standard of people.
  •  Helps in increasing the nation’s income.
  •  Helps to increase employment opportunities.
  •  II creates a modern cultivator.
  •  Reduction in the cost of marketing is a direct benefit to society.
  •  It includes all activities in the creation of Form, place, time, and possession utilities.

To the individual firms:

  • It generates revenue [profit] to the firm.
  • It helps the development of business and creates employment opportunities for people.
  • It provides information to the top management for taking an overall decision on production
  • Marketing and innovation are two basic functions of all business, the world is dynamic.

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 3.
Narrate the elements of the marketing mix.
Answer:

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix 1

Marketing mix means a marketing programme that is offered by a firm for the satisfaction of human wants. There are four elements of the marketing mix:

  1. Product: A Product is the main element of marketing. Without a product, there can be no marketing.
  2. Price: Price is the value of a product expressed in monetary terms. It is the amount charged for the product.
  3. Place (Physical Distribution): An excellent quality product, with a good price, will be waste, if it is not transferred from the production place to the consumption place.
  4. Promotion: An excellent product with a competitive price cannot achieve a desired success and acceptance in the market, with special features are conveyed to the consumers.

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

12th Commerce Guide Marketing and Marketing Mix Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
…………………………………….. Creates place utility.
a) Transport
b) Warehouse
c) Bank
d) Insurance
Answer:
a) Transport

Question 2.
Warehouse creates …………………………… utility.
a) Form
b) Place
c) Time
d) Possession
Answer:
c) Time

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 3.
Before the invention of money, goods were exchanged for goods known as …………..
a) Trade
b) Commerce
c) Business
d) Barter
Answer:
d) Barter

Question 4.
Pick the odd one out:
a) Grading
b) Branding
c) Packing
d) Warehousing
Answer:
d) Warehousing

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 5.
Pick the odd one out:
a) Price
b) Promotion
c) Place
d) Power
Answer:
d) Power

Question 6.
Choose the correct statement
i. Transport creates place utility.
ii. Were house creates time utility.
iii. Storage creates from utility

a) (i) and (ii) correct
b) (ii) and (iii) correct
c) All are correct
d) (i) and (ii) correct (iii) incorrect
Answer:
d) (i) and (ii) correct (iii) incorrect

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 7.
Which one is not the correct method?
a) Promotion – Advertisement
b) Branding – Name
c) Packaging – Wrapping
d) Grading – Value
Answer:
d) Grading – value

Question 8.
Which of the following elements is the related marketing mix?
a) Place
b) production
c) consumption
d) trade
Answer:
a) place

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

II. Match The Following

Question 1.
Match list-1 with list-2

List-1

List-2

(i) Branding1. Movement of goods
(ii) Pricing2. Preservation of goods
(iii) Storage3. Sum of the values
(iv) Transportation4. Symbol

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix 2

Answer:
a) (i) 4, (ii) 3, (iii) 2, (iv) 1

Question 2.
Match list-1 with list-2

List-1

List-2

i. Price1. Advertisement
ii. Product2. Terms and condition
iii. Place3. Distribution
iv. Promotion4. Innovation

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix 3

Answer:
a) (i) 2, (ii) 4, (iii) 3, (iv) 1

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

III. Assertion and Reason.

Question 1.
Assertion [A]: Movement of goods from the place of production to the place of consumption by means of transport
Reason [R]: It creates place utility,
a) (A) is false (R) is true
b)(A) is true (R) is true
c) (A) is false (R) is false
d) (A) is true (R) is false
Answer:
b) (A) is True (R) is True

Question 2.
Assertion [ A]: Product is the main element of marketing.
Reason [R]: There can be no market without a product.
a) Both (A) and (R) are correct
b) Both (A) and (R) are incorrect
c) (A) is correct
d) (R) is correct
Answer:
a) Both (A) and (B) is correct

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

IV. Very Short Answer Questions.

Question 1.
Define marketing.
Answer:
“Marketing is concerned with the people and the activities involved in the flow of goods and services from the producer to the consumer”. -AMERICAN MARKETING ASSOCIATION.

Question 2.
What are the functions of exchange? BAS
Answer:

  •  Buying
  •  Assembling
  •  Selling

Question 3.
Give the traditional marketing mix. 4Ps
Answer:

  • Price
  • Product
  • Place
  • Promotion

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 4.
Give the modern marketing mix. 4Ps
Answer:

  • People
  • Performance
  • Programme
  • Process

Question 5.
What are the functions of marketing? G.B. GILES.
Answer:

  •  Market research
  •  Planning
  •  Product development
  •  Promotion
  •  Distribution
  •  Service offer sales

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 6.
What are the functions of marketing? CLARK AND CLARK.
Answer:

  •  Functions of exchange
  •  Functions of physical supply
  •  Facilitating functions.

Question 7.
Define price.
Answer:
“Price is the amount of money charged for a product or service or the sum of the values that consumers exchange for the benefits of having or using the product or service”. – PHILIP KOTLER

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 8.
Give any two internal factors affecting the price of product/service. COMMO
Answer:

  •  Costs
  • Organizational objectives
  •  Marketing objectives
  •  Marketing mix strategy
  •  Organizational consideration

V. Long Answer Questions.

Question 1.
Explain the functions of physical supply.
Answer:
Transportation:

  •  Transport means carrying goods and services from one place to another. [Goods- men.]
  •  It creates place utility by moving goods from the places when they are available in plenty to places where they are needed.
  •  Types of transport – land, water, and air.

Storage:

  •  Generally, there is a time gap between production and consumption of goods.
  •  It involves the holding and preservation of goods from the time of production to the time of consumption.

Warehousing:
It creates time utility by storing the goods throughout the year and releasing them as and when they are needed.

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 2.
What are the facilitating functions of marketing?
Answer:

  • Financing: Long term – medium term and short term financing.
  • Risk bearing: Time – place – competition – human and political risks.
  • Grading: It means the classification of the standardized product into certain well-defined classes.
  • Branding: It means giving a name or symbol to a product to differentiate it from a competitive product
  • Packing: It means wrapping and crating of goods before distribution.
  • Pricing: Price must be determined only to offer to take all the relevant factors into consideration.

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 3.
What is marketing?
Answer:

  • Marketing is one of the business functions that all activities that take place in relation to markets for actualizing potential exchanges [from the producer to the consumer] for the purpose of satisfying human needs and wants.
  • Selling is basically concerned with putting the goods into the hands of the buyers for a price.
  • But marketing is much wider than selling.
  • It is an evolution rather than revolutionary.
  • Marketing is what a marketer does. It is not clear to understand the meaning.
  • It is one of the oldest professions in the world.
  • The traditional objective of marketing had been to make the goods available at places where they are needed.
  • This idea was later on changed by shifting the emphasis from exchange to satisfaction of human wants.
  • Marketing must first find out what customers want and then plan a product to satisfy the wants.

Question 4.
State to advantages of warehousing.
Answer:
Storage:

  • There is a time gap between production [supply] and consumption [demand].
  • During the time gap, the goods may be damaged or destroyed or perished.
  • For that goods need storage or warehouse.
  • It creates time utility.

Equalization of demand and supply:

  • It equalizes the demand and supply of goods.
  • Storing the goods- when there is no demand.
  • Releasing the goods – when there is a demand.

Price stabilization:

  • It ensures price stabilization.
  • It controls the price fluctuations.
  • The price of commodities more or less uniform throughout the year by storing the goods, if there is no demand and releasing the goods if they are demanded.

Risk bearing:
When the goods are in the warehouse, in case of damage due to flood fire, theft, etc., the warehouse keepers compensate the loss of the owner of the goods.

Finance:
By pledging warehouse receipt or warehouse warrant, a depositor can get loans from financial institutions.

Samacheer Kalvi 12th Commerce Guide Chapter 14 Marketing and Marketing Mix

Question 5.
How market information is helpful to the invention of a new product in the market?
Answer:

  •  Strong knowledge of customers.
  •  Marketing research – process for marketing opportunities- problem-solving collecting- analyzing marketing information.
  • Control over fact – the responsibility of the marketing research director.
  • Design research carefully and supervise its execution.
  • Various combinations of people, technologies for managing marketing information.
  • Broader management and strategy decisions.
  • Research consumer performance to contribute to improving decision-making performance.
  • It is helpful to accurate needs and decision-making by the marketer to be a regular source of information about marketing and trading brand awareness.

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 13 Concept of Market and Marketer Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 13 Concept of Market and Marketer

12th Commerce Guide Concept of Market and Marketer Text Book Back Questions and Answer

I. Choose The Correct Answer.

Question 1.
One who promotes (or) Exchange of goods or services for money is called as ……………
a) Seller
b) Marketer
c) Customer
d) Manager
Answer:
b) Marketer

Question 2.
The marketer initially wants to know in the marketing is …………………….
a) Qualification of the customer
b) Quality of the product
c) Background of the customers
d) Needs of the customers
Answer:
d) Needs of the customers

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 3.
The Spot market is classified on the basis of ………………..
a) Commodity
b) Transaction
c) Regulation
d) Time
Answer:
b) Transaction

Question 4.
Which one of the market deals in the purchase and sale of shares and debentures?
a) Stock Exchange Market
b) Manufactured Goods Market
c) Local Market
d) Family Market
Answer:
a) Stock Exchange Market

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 5.
Stock Exchange Market is also called …………
a) Spot Market
b) Local Market
c) Security Market
d) National Market
Answer:
c) Security Market

II. Very Short Answer Questions.

Question 1.
What is the Market?
Answer:
The word market is derived from the Latin word ‘Marcatus’ which means trade, commerce, merchandise, a place where the business is transacted.

Question 2.
Define Marketer.
Answer:
“A person whose duties include the identification of the goods and services desired by a set of consumers, as well as the marketing of those goods and services on behalf of a company” – BUSINESS DICTIONARY

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 3.
What is meant by Regulated Market?
Answer:
Very Short Period Market: Markets that deal in perishable goods like, fruits, milk, vegetables, etc., are called a very short period market.

Question 4.
What is meant by Spot Market?
Answer:
In such markets, goods are exchanged and the physical delivery of goods takes place immediately.

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 5.
What is meant by Commodity Market?
Answer:
A commodity market is a place where produced goods or consumer goods are bought and sold.

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

III. Short Answer Questions.

Question 1.
What can be marketed in the Market? [S.GOPI PEEPI]
Answer:

  • Services
  • Goods
  • Organization
  • Persons
  • Ideas
  • Places
  • Events
  • Experiences
  • Properties
  • Information

Question 2.
Mention any three Roles of the Marketer.
Answer:
Instigator:
As an Instigator, He keenly watches the developments taking place in the market and identifies marketing opportunities emerging in the ever-changing market.

Innovator:

  • Marketer seeks to distinguish his product or services by adding additional features or functionalities to the existing products.
  • Modifying the pricing structure, introducing new delivery patterns, creating new business models, introducing changes in the production process, and so on.
  • Implementer: Marketer plays the role of implementer when he or she actually converts marketing opportunities into marketable products with the help of several functional teams put in place in the organization.

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 3.
The marketer is an innovator? Do you agree?
Answer:
Marketer seeks to distinguish his products/services by adding additional features or functionalities to the existing product, modifying the pricing structure, introducing new delivery patterns, creating new business models, introducing change in production process and so on.

Question 4.
Explain the types of markets on the basis of time.
Answer:
Very Short Period Market:
Markets which deal in perishable goods like milk, Fish, Fruits, Vegetables, etc. are called as “Very Short Period Market”.

Short Period Market:
In certain goods, supply is adjusted to meet the demand. The demand is greater than supply. (Rice, wheat, etc) such markets are known as “Short Period Marker.

Long Period Market:
This type of market deals in durable goods. [TV – Fridge – computer etc] where the goods and services are dealt for longer period usages.

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 5.
List down the functions of Marketer? [GM] [5-P] [BTSS]
Answer:

  1. Very Short Period Market: Markets which deal in perishable goods like, fruits, milk, vegetables, etc., are called a very short period market. There is no change in the supply of goods.
  2. Short Period Market: In certain goods, supply is adjusted to meet the demand. The demand is greater than supply. Such markets are known as Short Period Market.
  3. Long Period Market: This type of market deals in durable goods, where the goods and services are dealt with for a longer period of usages.

IV.Long Answer Questions.

Question 1.
How the Market can be classified? GTTV RICE
Answer:

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer 1

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 2.
How the market can be classified on the basis of Economics?
Answer:
The Market can be classified on the basis of Economics as follows:
(a) Perfect Market: A market is said to be a perfect market if it satisfies the following conditions:

  1. A large number of buyers and sellers are there.
  2. Prices should be uniform throughout the market.
  3. Buyers and sellers have a perfect knowledge of the market.
  4. Goods can be moved from one place to another without restrictions.

(b) Imperfect Market: A market is said to be imperfect when

  1. Products are similar but not identical.
  2. Prices are not uniform.
  3. There is a lack of communication.

12th Commerce Guide Concept of Market and Marketer Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
One who promotes (or) Exchange of goods or services for money is called as _____
(a) Seller
(b) Marketer
(c) Customer
(d) Manager
Answer:
(b) Marketer

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 2.
Mercatus Mean ……….
a) A place where the business is transacted
b) A place where goods are produced
c) A place where manufacturers assembled
d) All of these
Answer:
a) A place where the business is transacted

Question 3.
The Need for the market is ……………………..
a) To Exchange
b) To adjust the price mechanism
c) To improve the quality
d) All the above
Answer:
d) All the above

Question 4.
Which one of the market deals in the purchase and sale of shares and debentures?
(a) Stock Exchange Market
(b) Manufactured Goods Market
(c) Local Market
(d) Family Market
Answer:
(a) Stock Exchange Market

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 5.
In …………………. market physical delivery of goods takes place immediately.
a) Future market
b) Spot market
c) Capital market
d) Commodity market
Answer:
b) Spot market

Question 6.
Short term securities are exchanged in ……………… market.
a) Stock
b) Finance
c) Money
d) All of the above
Answer:
c) Money

Question 7.
Stock Exchange Market is also called _____
(a) Spot Market
(b) Local Market
(c) Security Market
(d) National Market
Answer:
(d) National Market

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 8.
Which one is not correctly matched?
a) Very short period market – Vegetables
b) Future market – Delivery
c) Regulated market – No statutory measures
d) Bullion market – Gold
Answer:
c) Regulated market – No statutory measures

Question 9.
Which of the following statement correct?
i) In imperfect market products are similar but not identical.
ii) Prices are not uniform.
iii) There is a lack of communication.

a) (i) is correct
b) (ii) and (iii) are correct
c) (i) and (ii) are correct
d) All (i), (ii) and (iii) are correct
Answer:
d) All (i), (ii) and (ii) are correct

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

II. Match The Following.

Question 1.
Match List-I with List-II

List-I

List-II

i. Perfect market1. Prices are not uniform
ii. Imperfect market2. Large number of buyers and sellers
iii. Regulated market3. Short term securities
iv. Money market4. Statutory measures

a) (i) – 2 (ii) – 1 (iii) – 4 (iv) – 3
b) (i) – 4 (ii) – 3 (iii) – 2 (iv) – 1
c) (i) – 3 (ii) – 2 (iii) – 1 (iv) – 4
d) (i) – 1 (ii) – 4 (iii) – 3 (iv) – 2
Answer:
a) (i) – 2 (ii) – 1 (iii) – 4 (iv) – 3

Question 2.
Match List-I with List-II

List -I

List – II

i. Stock market1. Semi-finished good
ii. Bullion market2. Fruits – Fish
iii. Very short period market3. Gold – Silver
iv. Secondary market4. Shares – Debentures

a) (i) – 4 (ii) – 3 (iii) – 2 (iv) – 1
b) (i) – 3 (ii) – 2 (iii) – 1 (iv) – 4
c) (i) – 2 (ii) – 3 (iii) – 4 (iv) – 1
d) (i) – 1 (ii) – 4 (iii) – 3 (iv) – 2
Answer:
a) (i) – 4 (ii) – 3 (iii) – 2 (iv) – 4

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

III. Assertion and Reason.

Question 1.
Assertion (A): A market is said to be a perfect market.
Reason (R): There is a Large number of buyers and sellers.
a) (A) is True (R) is False
b) (A) is False (R) is True
c) Both (A) and (R) are True
d) Both (A) and (R) are False
Answer:
c) Both (A) and (R) are True

IV. Short Answer Questions.

Question 1.
Mention any three Roles of the Marketer.
Answer:

  1. Instigator: As an instigator, the marketer keenly watches the developments taking place in the market and identifies marketing opportunities emerging in the ever-changing market.
  2. Integrator: The marketer plays the role of the integrator in the sense that he collects feedback or vital inputs from channel members and consumers.
  3. Implementer: Marketer plays the role of implementer when he/she actually converts marketing opportunities into marketable products.

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 2.
Classify the market on the basis of Importance.
Answer:
Primary market:
The primary producers of firms sell their output or products through this type of market to wholesales or consumers.

Secondary market:

  • In this market, semi-finished goods are marketed.
  • The dealings commonly between wholesales and retailers.

Terminal market:
It is a central place that serves as an assembly and trading place for commodities in a
metropolitan area.

Question 3.
List down the functions of Marketer.
Answer:

  1. Gathering and Analysing market information
  2. Market planning
  3. Product Designing and development
  4. Standardization and Grading
  5. Packaging and Labelling
  6. Branding
  7. Customer Support Services
  8. Pricing of Products
  9. Promotion and Selling
  10. Physical Distribution
  11. Transportation
  12. Storage and Warehousing

Question 4.
Define Market.
Answer:
“Market includes both place and region in which buyers and sellers are in free competition with one another” – PYLE.

Question 5.
Explain the Need for marketing.
Answer:

  •  To Exchange [Barter] Goods and Services
  •  To Adjust Demand and Supply by the price mechanism.
  •  To improve the quality of life of the Society.

Question 6.
Mention any four differences between Wholesale Market and Retail Market? [LOQS]
Answer:

No Basis of difference

Wholesale Market

Retail Market

1. LinkIt is a link between manufacture and retailer.It is a link between a wholesaler and the ultimate consumer
2. One line ProductsHe deals in one line productsHe deals in different types of products.
3. QuantitiesHe buys in a large Quantity from the manufacturesHe buys in small Quantity from the wholesaler.
4. SituatedIt is situated in different areasIt is situated in a particular place nearer to the consumer.

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 7.
Why Customer support is needed to Market?
Answer:

  •  To exchange (Barter) goods and services.
  •  To adjust demand and supply by the price mechanism.
  •  To improve the quality of life of the society, [standard of living]
  •  To introduce new modes of life.
  •  To develop by enhancing the market segment.

V. Long Answer Questions.

Question 1.
How the market can be classified?
Answer:
On the basis of different approaches markets can be classified as follows:
I. On the basis of Geographical Area:

  • Family Market
  • Local Market
  • National Market
  • International Market or World Market

II. On the Basis of Commodities / Goods:
(a) Commodity Market

  • Produce Exchange Market
  • Manufactured Goods Market
  • Bullion Market

(b) Capital Market:

  • Money Market
  • Foreign Exchange Market
  • The Stock Market

III. On the Basis of Economics:

  • Perfect Market
  • Imperfect Market

IV. On the basis of transaction:

  • Spot Market
  • Future market

V. On the Basis of Regulation:

  • Regulated Market
  • Unregulated Market

VI. On the Basis of Time:

  • Very Short Period Market
  • Short Period Market
  • Long Period Market

VII. On the Basis of Volume of Business:

  • Wholesale Market
  • Retail Market

VIII. On the Basis of Importance:

  • Primary Market
  • Secondary Market
  • Terminal Market

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 2.
How the market can be classified on the basis of commodities [Goods]?
Answer:
A) Commodity Market:
A commodity market is a place where produced goods or consumer goods are bought and sold.

  1. Produce Exchange Market: It is an organised market where commodities or Agricultural produce are Bought and sold on a wholesale basis.
  2. Manufactured Goods Market: This market deals with [Leather goods, machinery, etc] manufactured goods.
  3. Bullion Market: This market deals with the purchase and sale of Gold and Silver.

B) Capital Markets:

  1. Money Market: It is a type of market where short term securities are exchanged. [Banks, – Finance to industries] .
  2. Foreign Exchange Market: This type of market, help Exporters and Importers, in converting their currencies into Foreign currencies and vice versa.
  3. Stock Market: This is a market where the purchase and sales of shares and Debentures, Bonds, etc. of companies are dealt with.

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 3.
What is your contribution to promote the market in modern society?
Answer:
A market is a place where buyers and sellers gather for purchase and sale. The market may be of Local market, national market, and international or global market. To develop and promote the market the following are needed:

  1. Eligible and satisfied and customer is needed.
  2. Quality and durable goods are to be marketed.
  3. Recent trends like E-marketing, online marketing are to be encouraged.
  4. After-sales service is to be provided to durable goods.
  5. Customers are to be financed for buying costly articles.
  6. New Innovations and marketing research are to be introduced to develop the market.

Samacheer Kalvi 12th Commerce Guide Chapter 13 Concept of Market and Marketer

Question 4.
Apart from goods and services, explain any five items that can be marketed,
Answer:
i) Experiences:
The unique and varied experiences pertaining to a place or a park or an event can be marketed under this concept. For eg. Amusement Park, Theme Park, Mountaineering, etc.

ii) Events:
The event marketing aims at promoting and marketing special events, shows, exhibitions, fairs, performances, sports events like the World Cup, Olympics, T20, etc.

iii) Properties:
Properties include both real property and financial property which involve the transfer of ownership on sale or purchase of the same. For eg. Constructed house, Land, Plot, Building.

iv) Organisations:
An organisation can market itself in order to build an image for it by sponsoring events, adoption of villages, donations for charitable causes, corporate social responsibility activities, etc.

v) Ideas:
An idea may change the world. Innovative and attractive ideas are desirable products for marketers to sell. Social media marketing thrives on new ideas in a consistent manner.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 10 Surface Chemistry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

12th Chemistry Guide Surface Chemistry Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

Question 1.
For freudlich isotherm a graph of log \(\frac{x}{m}\) is plotted against log P. The slope of the line and its y – axis intercept respectively corresponds to
(a) \(1 / n\), k
(b) log \(1 / n\), k
(c) \(1 / n\), log k
(d) log \(1 / n\), log k
Answer:
(c) \(1 / n\), log k
\(\frac{x}{m}\) = \(\mathrm{k} \cdot \mathrm{p}^{1 / \mathrm{n}}\)
log\((\frac{x}{m})\) = log k + \(\frac { 1 }{ n }\)log p
y = c + mx
m = \(\frac { 1 }{ n }\) and c = log k

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 2.
Which of the following is incorrect for physisorption?
(a) reversible
(b) increases with increase in temperature
(c) low heat of adsorption
(d) increases with increase in surface area
Answer:
(b) increases with increase in temperature

Question 3.
Which one of the following characteristics are associated with adsorption?
(a) ∆G and ∆H are negative but ∆S is positive
(b) ∆G and ∆S are negative but ∆H is positive
(c) ∆G is negative but ∆H and ∆S are positive
(d) ∆G. AH and ∆S all are negative.
Answer:
(d) ∆G, ∆H and ∆S all are negative.
Adsorption leads to decrease in randomness (entropy).i.e. ∆S < 0 for the adsorption to occur, ∆G should be – ve. We know that ∆G = ∆H – T∆S if ∆S is – ve, T∆S is + ve. It means that ∆G will become negative only when ∆H is – ve and ∆H > T∆S

Question 4.
Fog is colloidal solution of ……………..
(a) solid in gas
(b) gas in gas
(c) liquid in gas
(d) gas in liquid
Answer:
(c) liquid in gas
dispersion medium-gas, dispersed phase-liquid

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 5.
Assertion: Coagulation power of Al3+ is more than Na.
Reason: greater the valency of the flocculating ion added, greater is its power to cause precipitation
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(a) if both assertion and reason are true and reason is the correct explanation of assertion. (Hardy-Sechuize nile)

Question 6.
Statement: To stop bleeding from an injury, ferric chloride can be applied. Which comment about the statement is justified?
(a) It is not true, ferric chloride is a poison.
(b) It is true, Fe3+ ions coagulate blood which is a negatively charged sol
(c) It is not true; ferric chloride is ionic and gets into the blood stream.
(d) It is true, coagulation takes place because of formation of negatively charged sol with Cl.
Answer:
(b) It is true, Fe3+ ions coagulate blood which is a negatively charged sol

Question 7.
Hair cream is …………..
(a) gel
(b) emulsion
(c) solid sol
(d) sol.
Answer:
(b) emulsion
Emulsion dispersed phase, Dispersion medium -liquid

Question 8.
Which one of the following is correctly matched?
(a) Emulsion – Smoke
(b) Gel – butter
(c) foam – Mist
(d) whipped cream – sol
Answer:
(b) Gel – butter

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 9.
The most effective electrolyte for the coagulation of As2S3 Soils
(a) NaCl
(b) Ba(NO3)2
(c) K3[Fe(CN)6]
(d) AI2(SO4)3
Answer:
(d) AI2(SO4)3

Question 10.
Which one of the is  not a surfactant?
(a) CH3 – (CH2)15 – N – (CH3)2CH2Br
(b) CH3 – (CH2)15 – NH2
(c) CH3 – (CH2)16 – CH2OSO2 – Na+
(d) OHC – (CH2)14 – CH2 – COONa+
Answer:
(b) CH3 – (CH2)15 – NH2

Question 11.
The phenomenon observed when a beam of light is passed through a colloidal solution is ………….
(a) Cataphoresis
(b) Electrophoresis
(c) Coagulation
(d) Tyndall effect
Answer:
(d) Tyndall effect-scattering of light

Question 12.
In an electrical field, the particles of a colloidal system move towards cathode. The coagulation of the same sol is studied using K2SO4
(i). Na3PO4
(ii). K4[Fe(CN)6]
(iii). and NaCI
(iv). Their coagulating power should be …………..
(a) II > I >IV > III
(b) III > II > I > IV
(c) I > II > III > IV
(d) none of these
Answer:
(b) III > II > I > IV

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 13.
Collodion is a 4% solution of which one of the following compounds in alcohol – ether mixture?
(a) Nitroglycerine
(b) Cellulose acetate
(c) Glycoldinitrate
(d) Nitrocellulose
Answer:
(a) Nitrocellulose
pyroxylin (nitro cellulose)

Question 14.
Which one of the following is an example for homogeneous catalysis?
(a) manufacture of ammonia by Haber’s process
(b) manufacture of sulphuric acid by contact process
(c) hydrogenation of oil
(a) Hydrolysis of sucrose in presence of all HCI
Answer:
(a) Hydrolysis of sucrose in presence of all HCl
Both reactant and catalyst are in same phase. i.e. (1)

Question 15.
Match the following.

A)V2 O5i) High density
polyethylene
B) Ziegler – Nattaii) PAN
C) Peroxideiii) NH3
D) Finely divided Feiv) H2SO4

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 1
Answer:
(a) (iv) (i) (ii) (iii)

Question 16.
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of AS2S3 are given below
(I) (NaCl) = 52
(II) (BaCl) = 0.69
(III) (MgSO4) = 0.22
The correct order of their coagulating power is ……….
(a) III > II > I
(b) I > II > III
(c) I >III > II
(d) II > III > I
Answer:
(a) III > II > I

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 17.
Adsorption of a gas on solid metal surface is spontaneous and exothermic, then ……………
(a) ∆H increases
(b) ∆S increases
(c) ∆G increases
(d) ∆S decreases
Answer:
(a) ∆S decreases – ∆S is -ve

Question 18.
If x is the amount of adsorbate and m is the amount of adsorbent, which of the following relations is not related to adsorption process?
(a) x/m = f(P) at constant T
(b) x/m = f(T) at constant P
(c) P = f(T) at constant x/m
(d) x/m = PT
Answer:
(d) x/m = mPT

Question 19.
On which of the following properties does the coagulating power of an ion depend?
(a) Both magnitude and sign of the charge on the ion.
(b) Size of the ion alone
(c) the magnitude of the charge on the ion alone
(d) the sign of charge on the ion alone.
Answer:
(a) Both magnitude and sign of the charge on the ion.

Question 20.
Match the following.

A) Pure nitrogeni) Chlorine
B) Haber processii) Sulphuric acid
C) Contact processiii) Ammonia
D) Deacons processiv) Sodium azide (or) Barium azide

II. Short Answer Questions

Question 1.
Give two important characteristics of physisorption.
Answer:
Important characteristics of physisorption:

  1. It is reversible
  2. It has low heat of adsorption
  3. It has weak van der Waals forces of attraction with the adsorbent.
  4. It increases with an increase in pressure.
  5. It forms multi molecular layer.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 2.
Differentiate physisorption and chemisorption.
Answer:
Chemical adsorption or Chemisorption or Activated adsorption

  1. It is very slow
  2. It is very specific depends on nature of adsorbent and adsorbate.
  3. chemical adsorption is fast with increase pressure, it can not alter the amount.
  4. When temperature is raised chemisorption first increases and then decreases.
  5. Chemisorption involves transfer of electrons between the adsorbent and adsorbate, Heat of adsorption is high i.e., from 40 – 400kJ/mole.
  6. Monolayer of the adsorbate is formed.
  7. Adsorption occurs at fixed sites called active centres. It depends on surface area.
  8. Chemisorption involves the formation of activated complex with appreciable activation energy.
  9. Physical adsorption or van der Waals adsorption or Physisorptlon
  10. It is irreversible.

Physical adsorption or van der Waals adsorption or physisorption.

  1. It is instantaneous
  2. It is non-specific
  3. In Physisorption. when pressure increases the amount of adsorption increases.
  4. Physisorption decreases with increase in temperature.
  5. No transfer of electrons
  6. Heat of adsorption is low in the order of 40kJ/mole.
  7. Multilayer of the adsorbate is formed on the adsorbent.
  8. It occurs on all sides.
  9. Activation energy is insignificant.
  10. It is reversible.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 3.
In case of chemisorption, why adsorption first increases and then decrease with temperature?
Answer:
1. Chemisorption involves high activation energy so it is also referred to as activated adsorption.

2. It is found in chemisorption that it first increases and then decreases with an increase in temperature. When adsorption is plotted, the graph first increases and then decreases with temperature.

3. The initial increase illustrates the requirement of activation of the surface for adsorption is due to fact that the formation of activated complex requires certain energy. But later it decreases at high temperature is due to desorption as the kinetic energy of the adsorbate increases (exothermic nature)

Question 4.
Which will be adsorbed more readily on the surface of charcoal and why; NH3 or CO2?
Answer:
1. The gases having low critical temperature are adsorbed slowly, while gases with high critical temperature are adborbed readily.

2. Among CO2, and NH3, NH3 will be more readily adsorbed on the surface of the charcoal. This is because the critical temperature of ammonia gas is quite high than the CO2. Hence, it easily combines with the materials than the CO2 whether it is solid, liquid or any gases.

Question 5.
Heat of adsorption is greater for chemisorptions than physisorption. Why?
Answer:
Chemisorption has higher heat of adsorption. because in chemisorption the chemical bonds are much stronger. In adsorbed state the adsorbate is hold on the surface of adsorbent by attractive forces (bond). And chemisorption is irreversible one. Therefore, heat of adsorption is greater for chenil sorptions than physisorption. Chemisorption, heat of adsorption range 40 – 400kJ/mole.

Question 6.
In a coagulation experiment 10 mL of a colloid (X) is mixed with distilled water and 0.1M solution of an electrolyte AB so that the volume is 20 mL. It was found that all solutions containing more than 6.6 mL of AB coagulate with in 5 minutes. What is the flocculation values of AB for sol (X)?
Answer:
A minimum of 6.6mL of AB is required to coagulate the sol. The moles of AB in the sol is
\(\frac{6.6 \times 0.01}{20}\) = 0.033 moles
This means that a minimum of 0.033 moles or 0.0033 x 1000 = 3.3 milli moles are required for coagulating one litre of sol. Flocculation value of AB for X = 3.3

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 7.
Peptising agent is added to convert precipitate into colloidal solution. Explain with an example.
Answer:
1. Ions either positive or negative of peptizing agent (electrolyte) are adsorbed on the particles of precipitate. They repel and hit each other and break the particles of the precipitate into colloidal size.

2. For example, when we add a small volume of very dilute hydrochloric acid solution peptising agent to a fresh precipitate of a silver chloride, it leads to formation of silver chloride colloidal solution,
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 3

Question 8.
What happens when a colloidal sol of Fe(OH)3 and As2S3 are mixed?
Answer:
On mixing Fe(OH)3 positive sol and As2S3 negative sol, mutual coagulation occurs which causes precipitation. When this sol got mixed with each other, due to Fe3+ and S2- ions neutralisation of charges will happen, and precipitate will be formed.
Fe(OH)3 + As2S3 → Fe2S3 + As(OH)3

Question 9.
What is the difference between sol and gel?
Answer:
Sol

  1. The liquid state of a colloidal solution is called a sol.
  2. The sol does not have a definite structure.
  3. The dispersion medium of the sol may be water.
  4. The sol can be converted to gel by cooling The sol can be easily dehydrated.
  5. The viscosity of the sol is very low.
  6. Sol is categorized into lyophobic and lyophilic sols.
  7. Example: Blood

Gel

  1. The solid or semi-solid state of a colloidal solution is called gel.
  2. The gel possesses a honeycomb-like structure.
  3. The dispersion medium of gel will be hydrated colloid particles.
  4. The gel can be converted to sol by heating.
  5. The gel cannot be dehydrated.
  6. The viscosity of the gel is very high.
  7. There is no such classification of gel.
  8. Example: Fruit jelly, cooked gelatin jelly.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 10.
Why are lyophillic colloidal sols are more stable than lyophoblic colloidal sol?
Answer:
1. A lyophilic colloidal sols are stable due to the charge and the hydration of sol particles.

2. Lyophilic sols are more stable than lyophobilc sols because they are highly hydrated in the solution. And since more is the hydration more will be its stability.

3. Lyophilic sols are stabilized by electrostatic charge and hydration whereas lyophobic sols are only stabilized by charge, so they easily gets coagulated and requires a stabilising agent. Hence, lyophilic sols are more stable than lyophobilc sols.

Question 11.
Addition of Alum purifies water. Why?
Answer:
Purification of drinking water is activated by coagulation of suspended impurities in water using alums containing Al3+. That is why we are adding to purify water.

Question 12.
What are the factors which influence the adsorption of a gas on a solid?
Answer:
Factors that influence the adsorption of a gas on a solid is as follows:
1. Nature of the gas:
Easily liquefiable gases such as NH3, HCl etc are adsorbed to a great extent in comparison to gases such as H2, O2 etc. This is because van der Waal’s forces are stronger is easily liquifiable gases.

2. Surface area of the solid:
The greater the surface area of the adsorbent, the greater is the adsorption of gas on the solid surface.

3. Effect of pressure:
Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.

4. Effect of temperature:
Adsorption is an exothermic process. Thus in accordance with Le – Chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 13.
What are enzymes? Write a brief note on the mechanism of enzyme catalysis.
Answer:
Enzymes are complex protein molecules with three-dimensional structures. They catalyse the chemical reaction in a living organism. They are often present in colloidal state and extremely specific in catalytic action.

Each enzyme produced in a particular living cell can catalyse a particular reaction in the cell. Mechanism of enzyme catalysis: Mechanism of enzyme catalysed reaction is known as lock and key mechanism.
1. Enzymes arc highly specific in their action.

2. This specificity is due to the pressure of active sites. The shape of active site of any given enzyme is like cavity such that only a specific substrate can fit into it.

In the same way a key fits into lock. The specific binding needs to the formation of an enzyme-substrate complex which accounts for high specificity of enzyme catalysed reactions.

3. Once the proper orientation is attained the substrate molecules reacts to form the product in two steps.

4. Since product molecules do not have any affinity for the enzyme they leave the enzyme surface making room for fresh substrate.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 4

step 1: Formation of the enzyme-substrate complex
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 5
Step 2: Dissociation of the enzyme-substrate complex to form product
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 6
The rate of the formation of the product depends upon the concentration of ES.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 14.
What do you mean by activity and selectivity of catalyst?
Answer:
1. Activity of Catalyst:
The activity of a catalyst is its ability to increase the rate of a particular reaction, Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants in the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.

2. Selectivity of the catalyst:
The ability of the catalyst to direct a reaction to yield a particular product is referred to as the selectivity of the catalyst. For example, by using different catalysts, we can get different products for the reaction between H2 and CO.
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 7

Question 15.
Describe some features of catalysis b Zeoliles.
Answer:
1. Zeolites are microporous, crystalline, hydrated aluminosilicates, made of silicon and aluminium tetrahedra.

2. There are about 50 natural zeolites and 1 50 synthetic zeolites. As silicon is tetravalent and aluminium is trivalent, the zeolite matrix carries extra negative charge. To balance the negative charge, there are extra framework cations for example, H+ or Na+ ions.

3. Zeolites earring protons are used as solid acids, catalysis and they are extensively used in the petrochemical industry for cracking heavy hydrocarbon fractions into gasoline, diesel,etc.

4. Zeolites earring Na ions are used as basic catalysis.

5. One of the most important applications of zeolites is their shape selectivity. In zeolites, the active sites namely protons are lying inside their pores. So, reactions occur only inside the pores of zeolites.

Question 16.
Give three uses of emulsions.
Answer:

  1. The cleansing action of soap is due to emulsions.
  2. It is used in the preparation of vanishing cream.
  3. It is used in the preparation of cold liver oil.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 17.
Why does the bleeding stop by rubbing moist alum?
Answer:
Blood is a colloidal sol. When we nib the injured part with moist alum then coagulation of blood takes place. Hence main reason is coagulation, which stops the bleeding. Therefore bleeding stop by rubbing moist alum.

Question 18.
Why is desorption important for a substance to act as good catalyst?
Answer:
Desorption is important for a substance to act as a good catalyst, so that after the reaction, the products found on the surface separate out (desorbed) to create free surface again for other reactant molecules to approach the surface and react. If desorption does not occur then other reactants are left with no space on the surface of the catalyst for adsorption and the reaction will stop.

Question 19.
Comment on the statement: Colloid is not a substance but it is a state of the substance.
Answer:
The statement is true. Because the same substance may exist as a colloid under certain conditions and as a crystalloid under certain other conditions. For example. NaCl in water behaves as a crystalloid while in benzene, it behaves as a colloid. Similarly, dilute soap solution behaves

like a crystalloid while concentrated solutions behave as a colloid. It is the size of the particles which matters. That is the state in which the substance exists. If the size of the particles lies in the range 1 nm to 1oo nm, it is in the colloidal state.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 20.
Explain any one method for coagulation
Answer:
The flocculation and setting down of the soil particles is called coagulation. Various method of coagulation are given below:

  1. Addition of electrolytes
  2. Electrophoresis
  3. Mining oppositely charged sols
  4. Boiling.

Addition of electrolytes
A negative ion causes the precipitation of positively charged sol and vice versa. When the valency of ion is high, the precipitation power is increased. For example, the precipitation power of some cations and anions varies in the following order
Al3+ > Ba2+ > Na+, Similarly [Fe(CN)6]-3 > SO4-2 > Cl
The precipitation power of electrolyte is determined by finding the minimum concentration (millimoles / lit) required to cause precipitation of a sol in 2hours. This value is called flocculation value. The smaller the flocculation value greater will be precipitation.

Question 21.
Write a note on electroosmosis.
Answer:
Electro osmosis:
A sol is electrically neutral. Hence the medium carries an equal but opposite charge to that of dispersed particles. When sol particles are prevented from moving, under the influence of electric field the medium moves in a direction opposite to that of the sol particles. This movement of dispersion medium under the influence of electric potential is called electro-osmosis.
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 8

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 22.
Write a note on catalytic poison
Answer:
Catalytic poison:
Certain substances when added to a catalysed reaction, decreases or completely destroys the activity of catalyst and they are often known as catalytic poisons. For example, In the reaction,
2SO2 + O2 → 2SO3 with a Pt catalyst, the poison is AS2O3.
i.e., AS2O3 destroys the activity of pt. AS2O3 blocks the activity of the catalyst. So, the activity is lost.

Question 23.
Explain intermediate compound formation theory of catalysis with an example.
Answer:
The intermediate compound formation theory:
A catalyst acts by providing a new path with low energy of activation. in homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the catalyst is regenerated.

Consider the reactions:
A + B → AB ……………(1)
A + C → AC (intermediate) ………….(2)
C is the catalyst
AC + B → AB + C …………(3)
Activation energies for the reactions (2) and (3) are lowered compared to that of (1). Hence the formation and decomposition of the intermediate accelerate the rate of the reaction.
Example:
The mechanIsm of Fridel crafts reaction is given below
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 9
The action of catalyst is explained as follows .
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 10

This theory describes,

  1. The specificity of a catalyst.
  2. The increase in the rate of the reaction with increase in the concentration of a catalyst.

Limitations

  1. The intermediate compound theory fails to explain the action of catalytic poison and activators (promoters).
  2. This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 24.
What is the difference between homogenous and hetrogenous catalysis?
Answer:
Hornogenous Catalysis:

  1. In a catalysed reaction the reactants, products and catalyst are present in the same phase.
  2. For example.
    Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 11
    Hence NO act as catalyst.
  3. Homogeneous catalysis explained by intermediate compound formation theory.

Heterogeneous Catalysis:

  1. In a reaction, the catalyst is present in a different phase. i.e., catalyst is not present in the same phase as that of reactants and products.
  2. For example.
    Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 12
    Hence Pt(s) act as catalyst.
  3. Hetenogeneous catalysis explained by adsorption theory.

Question 25.
Describe adsorption theory of catalysis.
Answer:
Langmuir explained the action of catalyst in heterogeneous catalysed reactions based on adsorption.
This theory explains heterogeneous catalysis.
The reactant molecules are adsorbed on the catalytic surface, so heterogeneous catalysis is also called as contact catalysis.
When the reactants are adsorbed on the catalytic surface, an activated complex is formed.
This activated complex decomposes to give the product.
Steps involved in heterogeneous catalysis.

  1. Reactant molecules diffuse from bulk to the catalyst surface.
  2. Reactant molecules are adsorbed on the surface of the catalyst.
  3. Adsorbed reactant molecules are activated to form activated complex which is decomposed to form the products.
  4. Product molecules are desorbed.
  5. Product molecules diffuse away from the surface of the catalyst.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 13

12th Chemistry Guide Surface Chemistry Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

Question 1.
Adsorption is a
a) Bulk phenomenon
b) Surface phenomenon
c) Both (a) and (b)
d) None of the above
Answer:
b) Surface phenomenon

Question 2.
Absorption is a
a) Bulk phenomenon
b) Surface phenomenon
c) Both (a) and (b)
d) None of the above
Answer:
a) Bulk phenomenon

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 3.
Which among the following is an adsorbent?
a) N2
b) SO2
c) Ni
d) NH3
Answer:
c) Ni

Question 4.
In adsorption, if the concentration of a substance in the interface is high, then it is called?
a) Desorption
b) Positive adsorption
c) Negative adsorption
d) Absorption
Answer:
b) Positive adsorption

Question 5.
Adsorption process is
a) Spontaneous
b) Non-spontaneous
c) Slow
d) Bulk phenomenon
Answer:
a) Spontaneous

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 6.
Adsorption is always accompanied by
a) Increase in entropy
b) Increase in free energy
c) Decrease in free energy
d) No change in entropy
Answer:
c) Decrease in free energy

Question 7.
The force of attraction exist between adsorbent and adsorbate in physical adsorption is
a) Vanderwaal’s force
b) Dipole-dipole interaction
c) Dispersion forces
d) All the above
Answer:
d) All the above

Question 8.
Total amount of the gas adsorbed increases as the ……………. of the adsorbent increases.
a) Volume
b) Density
c) Surface area
d) Surface tension
Answer:
c) Surface area

Question 9.
The critical temperature of the gas which is readily adsorbed is
a) lower
b) higher
c) zero
d) none of the above
Answer:
b) higher

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 10.
Which among the following gas is adsorbed slowly?
a) SO2
b) NH3
c) N2
d) CO2
Answer:
c) N2
Reason: Critical temperature of N2 is low hence adsorbed slowly.

Question 11.
The process of adsorption is
a) Exothermic
b) Endothermic
c) Both (a) and (b)
d) None of the above
Answer:
a) Exothermic

Question 12.
Multi molecular layers are formed in
a) Absorption
b) Physisorption
c) Chemisorption
d) None of the above
Answer:
b) Physisorption

Question 13.
The rate constant of a reaction at temperature 200K is 10 times less than the rate constant at 400K. What is the activation energy of the reaction? (R=gas constant)
a) 1842.4R
b) 921.2 R
c) 460.6 R
d) 230.3 R
Answer:
b) 921.2 R

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 14.
In adsorption isobar the amount of adsorption is plotted against
a) Pressure
b) Temperature
c) Volume
d) Mass
Answer:
b) Temperature

Question 15.
In physical adsorption isobar x /m, …………………………. with increase in temperature.
a) Increases
b)Decreases
c) First increases then decreases
d) Does not change
Answer:
b) Decreases

Question 16.
In chemisorption isobar x /m …………….. with increase in temperature.
a) Increases
b)Decreases
c) First increases then decreases
d) Does not change
Answer:
c) First increases then decreases

Question 17.
Which of the following is not an equation for Freundlich isotherm?
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 14
Answer:
d

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 18.
Sugar prepared from molasses in decolourised by adding.
a) Silica gel
b) Permutit
c) Animal Charcoal
d) Activated charcoal
Answer:
c) Animal Charcoal

Question 19.
Which of the following is not an application of adsorption?
a) Heterogeneous catalysis
b) Gas masks
c) Froth floatation process
d) Softening of water by boiling
Answer:
d) Softening of water by boiling

Question 20.
The change of W/O emulsion into O/W emulsion is called …………..
a) Coagulation
b) Emulsification
c) Decomposition
d) Invension of phase
Answer:
d) Invension of phase

Question 21.
2SO2 + O2 + [NO] → 2SO3 + [NO] is an example for
a) Positive catalysis
b) Negative catalysis
c) Homogeneous catalysis
d) Both (a) and (c)
Answer:
d) Both (a) and (c)

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 22.
Which is not a heterogeneous catalysis?
a) Contact process
b) Haber’s process
c) Ester hydrolysis
d) Ostwald’s process
Answer:
c) Ester hydrolysis

Question 23.
Intermediate compound formation theory explains.
a) Homogeneous catalysis
b) Heterogeneous catalysis
c) Autocatalysis
d) Negative catalysis
Answer:
a) Homogeneous catalysis

Question 24.
In intermediate compound formation, the intermediate complex formed has
a) More activation energy than uncatalysed complex
b) Less activation energy than uncatalysed complex
c) Less kinetic energy than the reactants
d) Less kinetic energy than the products
Answer:
b) Less activation energy than uncatalysed complex

Question 25.
Adsorption theory explains
a) Homogeneous catalysis
b) Heterogeneous catalysis
c) Autocatalysis
d) Negative catalysis
Answer:
b) Heterogeneous catalysis

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 26.
The catalytic activity of a catalyst is increased by a promoter by
a) Increasing the number of active centres
b) Decreasing the number of active centres.
c) Blocking the number of active centres
d) Desorbing the active centres
Answer:
a) Increasing the number of active centres

Question 27.
Enzymes are often present as
a) Crystalloids
b) Suspension
c) Colloids
d) True solutions
Answer:
c) Colloids

Question 28.
In the conversion of 1 – chloro octane into 1 – cyano octane, tetra alkyl ammonium cation acts as a
a) Enzyme catalyst
b) Phase transfer catalyst
c) Zeolite catalyst
d) Nano catalyst
Answer:
b) Phase transfer catalyst

Question 29.
Which of the following is incorrect
a) Enzymes can be inhibited (poisoned)
b) Calalytic activity of enzymes is decreased by coenzvmes.
c) Enzyme catalysis is highly specific in nature
d) the rate of Enzyme catalysed reactions varies with the pH of the system.
Answer:
b) Calalytic activity of enzymes is decreased by coenzymes.

Question 30.
Nano catalysts can act as
a) Homogeneous catalysts
b) Heterogeneous catalysts
c) Phase transfer catalysts
d) Both (a) and (b)
Answer:
d) Both (a) and (b)

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 31.
Size of colloidal particle is
a) 1 – 200 Å
b) 1 – 200 nm
c) 1 – 200 pm
d) 1 – 200 pm
Answer:
b) 1 – 200 nm

Question 32.
In hydrosols the dispersion medium is
a) Benzene
b) Alcohol
c) Water
d) Ether
Answer:
c) Water

Question 33.
In lyophillic colloids which is true?
a) Definite attractive force exists between the dispersion medium and dispersed phase
b) More stable
c) Reversible sols
d) All the above
Answer:
d) All the above

Question 34.
In lyophobic colloids which is not true?
a) No attractive force exists between the dispersion medium and dispersed phase
b) Less stable
c) Can be produced again
d) Precipitated readily
Answer:
c) Can be produced again

Question 35.
Which among the following is a liquid aerosol?
a) Smoke
b) Fog
c) Shaving cream
d) Froth
Answer:
b) Fog

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 36.
Which one of the following is negatively charged colloid?
a) arsenic sulphide
b) Ferric hydroxide
c) Haemoglobin
d) Basic dyes
Answer:
a) arsenic sulphide

Question 37.
Butter is a colloid of
In butter the dispersed phase and the
dispersion medium are respectively [222EE3
a) Solid, gas
b) Liquid, solid
c) Solid, liquid
d) Gas, solid
Answer:
c) Solid, liquid

Question 38.
In butter the dispersed phase and the dispersion medium are respectively_______
a) Solid in liquid
b) Liquid in solid
c) Liquid in liquid
d) Liquid in gas
Answer:
b) Liquid in solid

Question 39.
Colloidal graphite can be prepared by
a) Peptisation
b) Mechanical dispersion
c) Ultrasonic dispersion
d) Double decomposition
Answer:
b) Mechanical dispersion

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 40.
Electro dispersion method is used to prepare the colloidal solution of
a) Copper
b) Silver
c) Gold
d) All the above
Answer:
d) All the above

Question 40.
Arsenic sulphide sol is prepared by the reaction. AS2O3 + 3H2S → As2S3 + 3H2O. This method is known as
a) Hydrolysis
b) Double decomposition
c) Oxidation
d) Reduction
Answer:
b) Double decomposition

Question 42.
Conversion of a colloid into a precipitate is known as
a) Peptisation
b) Dialysis
c) Coagulation
d) Electrophoresis
Answer:
c) Coagulation

Question 43.
Conversion of a precipitate into a colloid is known as
a) Peptisation
b) Dialysis
c) Coagulation
d) Electrophoresis
Answer:
a) Peptisation

Question 43.
The removal of electrolytic impurities from a colloidal solution is known as
a) Peptisation
b) Dialysis
c) Coagulation
d) Electrophoresis
Answer:
b) Dialysis

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 44.
The movement of dispersed phase under the influence of electric current is known as
a) Electro osmosis
b) Electrophoresis
b) Dialysis
d) Electrophoresis
Answer:
b) Dialysis

Question 45.
The movement of dispersed phase under the influence of electric current is known as
a) Electro osmosis
b) Electrophoresis
c) Electrodialysis
d) Ultra filteration
Answer:
b) Electrophoresis

Question 46.
The movement of dispersion medium under the influence of electric current is known as
a) Electro osmosis
b) Electrophoresis
c) Electrodialysis
d) Ultra filteration
Answer:
a) Electro osmosis

Question 47.
In ultra filteration, ultrafilters are made by using
a) Collodion
b) Cellophane
c) Both (a) and (b)
d) None of the above
Answer:
c) Both (a) & (b)

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 48.
Collodion is 4% solution of ……………… in a mixture of alcohol and water.
a) Cellulose acetate
c) Chioro cellulose
b) Cellulose sulphate
cl) Nitrocellulose
Answer:
d) Nitrocellulose

Question 49.
Blue colour of the sky. in nature is due to
a) Browniart movement
b) Tyndall effect
c) Both (a) and (b)
d) None of thé above
Answer:
b) Tyndall effect

Question 50.
Scattering of light by colloidal particles is known as
a) Brownian movement
b) Tyndall effect :
c) Electrophoresis
d) Electro osmosis
Answer:
b) Tyndall effect

II. Match the following

Question 1.

III
i) Auto CatalystHydrated alumino zilicates
ii) Nano catalystAl2O3 in Haber’s process
iii) Zeolite catalystFe / Pd
iv) Catalytic poisonUrease
v) PromoterAnhydrous AlCl3
vi) Enzyme catalystAs2O3 in contact process decomposition of arsine.

Answer:
i) Arsenic in the decomposition of arsine
i) Fe/Pd
ii) Hydrated aluminosilicates
iii) As2O3 in contact process
iv) Al2O3 in Haber’s process
v) Urease

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 2.

ProcessCatalyst
i) Haber’s processCupric chloride
ii) Contact processNitric oxide
iii) Lead chamber processFerric oxide
iv) Deacon’s processIron
v) Bosch’s processPlatinum

Answers :
i) Iron
ii) Platinum
iii) Nitric oxide
iv) Cupric chloride
v) Ferric oxide

Question 3.

Column AColumn B
i) Solid aerosolPaints
ii) EmulsionBread
iii) SolDust
iv) Solid foamAlloys
v) Solid solMilk

Answer:
i) Dust
ii) Milk
iii) Paints
iv) Bread
v) Alloys

III. Assertion and Reasoning

Question 1.
Assertion (A) : Physical adsorption occurs at low temperatures.
Reason (R) : The forces of attraction between the adsorbent and adsorbate are weak and heat of adsorption is low
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct.
Answer:
a) Both A and R are correct, R explains A

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 2.
Assertion (A) : In the reaction CH3COOC2H5 + H20 -> CH3COOH + C2H5OH is an example for autocatalysis.
Reason (R) : One of the product ethanol acts as a catalyst.
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct.
Answer:
c) A is correct but R is wrong.
Correct Reason: One of the product acetic acid acts as a catalyst.

Question 3.
Assertion (A) : 1- Chioro octane is converted into 1 – Cyano octane in I or 2 hours by reacting with sodium cyanide in presence of tetra alkyl ammonium chloride.
Reason (R) : The tetra alkyl ammonium cation transports CN from the aqueous phase to the
organic phase using its hydrophilic end.
a) Both A and R are correct, R explains A
h) Both A and R are correct, R does do not explain A
c) A is correct hut R is wrong
d) A is wrong but R is correct.
Answer:
a) Both A and R are correct, R explains A

Question 4.
Assertion (A): Measurements of osmotic pressure is used to find the molecular weight of colloidal particle.
Reason (R): Colloidal solutions show the kinetic property,
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain 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, R does not explain A Correct Reason: Colloidal solutions show colligative properties.

IV. Choose the correct statement

Question 1.
i) Chemisorption is instantaneous
ii) Chemisorption first increases and then decreases with temperature
iii) In chemisorption the heat of adsorption is high
iv) Chemisorption is independent of the surface area of the adsorbent
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement:
Chemisorption is slow,
Chemisorption depends on the surface area of the adsorbent.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 2.
i) A catalyst can initiate a reaction
ii) A solid catalyst will be more effective if it is taken in bulk
A catalyst does not affect the position of equilibrium and the value of the equilibrium constant
A catalyst is highly effective at a particular temperature called optimum temperature,
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)

Correct statement:
A catalyst cannot initiate a reaction,
A solid catalyst will be more effective if it is taken in a finely divided form.

Question 3.
i) When light passes through colloidal solution it is scattered in all directions.
ii) When a colloidal solution is observed through ultramicroscope, they showed a random, zigzag, ceaseless motion.
iii) Haemoglobin is a negatively charged colloid
iv) A sol is not electrically neutral.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
a) (i) & (ii)

Correct statement: iii) Haemoglobin is a positively charged colloid, iv) A sol is electrically neutral

Question 4.
i) The flocculation and setting down of the sol particles is called coagulation.
ii) The higher the flocculation value greater will be the precipitation.
iii) Lyophilic sols are precipitated readily even with small amount of electrolytes.
iv) When boiled due to increased collisions, the sol particles combine and settle down.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)

Correct statement:
ii) The smaller the flocculation value greater will be the precipitation,
iii) Lyophobic sols are precipitated readily even with small amount of electrolytes

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

V. Choose the incorrect statement

Question 1.
i) Physisorption decreases with an increase in pressure
ii) Physisorption decreases with an increase in temperature
iii) Activation energy of physisorption is significant
iv) No transfer of electrons occur in physisorption
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (iii) & (iv)
Answer:
c) (i) & (iii)

Correct statement:
i) Physisorption increases with increase in pressure,
iii) Activation energy of physisorption is insignificant.

Question 2.
i) Enzyme catalysis is highly specific in nature
ii) The rate of enzyme catalysed reactions does not vary with pH of the system
iii) Catalytic activity of enzymes is decreased by coenzymes.
iv) Enzyme catalysed reaction has maximum rate at optimum temperature.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)

Correct statement:
ii) The rate of enzyme catalysed reactions varies with pH of the system,
iii) Catalytic activity of enzymes is increased by coenzymes.

Question 3.
i) In zeolites the active sites namely protons are lying inside their pores.
ii) Reactions occur only outside the pores of zeolites.
iii) Bulkier reactant molecules are prevented from reaching the active sites within- the zeolite crystal.
iv) If the transition state of a reaction is large compared to the pore size of the zeolite, then product will be formed.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (ii) & (iv)
Answer:
d) (ii) & (iv)
Correct statement:
ii) Reactions occur only inside the pores of zeolites,
iv) If the transition state of a reaction is large compared to the pore size of the zeolite, then no product will be formed.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 4.
i) Colloidal solutions are heterogeneous in nature having two distinct phases.
ii) Colloidal solutions are unstable and they are affected by gravity.
iii) When the colloidal solution is dilute coagulation occurs.
iv) Unlike true solution, colloids diffuse less readily through membranes.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement:
ii) Colloidal solutions are quite stable and they are not affected by gravity,
iii) When the colloidal solution is dilute coagulation does not occur

VI. Two Mark Questions

Question 1.
What is adsorption?
Answer:

  • Adsorption is a phenomenon of concentration of a substance on the surface of a liquid or gas.
  • Adsorption is a surface phenomenon.

Question 2.
What is known as interface?
Answer:
The surface of separation of the two phases where the concentration of adsorbed molecule is high is known as the interface.

Question 3.
What is meant by positive adsorption and negative adsorption?
Answer:
If the concentration of a substance in the interface is high in an adsorption, then it is called positive adsorption. If it is less, then it is called negative adsorption.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 4.
What is desorption?
Answer:
The process of removing an adsorbed substance from the surface is called desorption.

Question 5.
What is meant by adsorbent and adsorbate?
Answer:
Absorbent is the material on which adsorption takes place.
Adsorbed substance is called an adsorbate.

Question 6.
Give some examples for adsorbates.
Answer:

  • Gaseous molecules like He, Ne, O2, SO2 and NH3
  • Solutions of NaCl, KCl

Question 7.
Give some examples for adsorbents.
Answer:
Silica Gel
Metals like Ni, Cu, Pi. Ag and Pd
Certain colloids.

Question 8.
What are the limitations of Freundlich isotherm?
Answer:

  • The equation is purely empirical
  • Valid over a limited pressure range.
  • Constants k and n values vary with temperature and for this no theoretical explanations were given.

Question 9.
How is adsorption principle used in the softening of hard water?
Answer:

  • Permutit adsorbs Ca2+ and Mg2+ ions present in hard water on its surface.
  • An ion exchange occurs on its surface. Na2Al2Si4O12 + CaCl2 → CaAl2 Si4 O12 + 2NaCl
  • Exhausted permutit is regenerated by adding a solution of common salt.
    CaAl2 Si4 O12 + 2NaCl → Na2Al2Si4O12 + CaCl2

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 10.
How is ion exchange resins work on the principle of adsorption in demineralising water?
Answer:

  • The process is carried out by passing water through two columns of cation and anion exchange resins.
  • Cation exchange:
    Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 15
  • Anion exchange:
    Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 16

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 11.
Define a catalyst.
Answer:
A catalyst is defined as a substance which alters the rate of chemical reaction without itself undergoing chemical change.

Question 12.
What is meant by positive catalysis and negative catalysis?
Answer:

  • In positive catalysis the rate of a reaction is increased by the presence of a catalyst.
  • In negative catalysis the rate of a reaction is decreased by the presence of a catalyst.

Question 13.
What are promoters?
Answer:

  • The substance which increases the activity of a catalyst in a reaction is called promoter.
  • (Eg-) In Haber’s process of synthesis of NH3, MO is the promoter of Fe catalyst.

Question 14.
What are the limitations of intermediate compound formation theory.
Answer:

  • Fails to explain the action of catalytic poison and promoters.
  • Unable to explain the mechanism of heterogeneous catalysis.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 15.
What are active centres?
Answer:

  • Surface of a catalyst is not smooth.
  • It bears steps, cracks and corners.
  • The atoms on such locations are coordinatively unsaturated.
  • So they have much residual force of attraction.
  • Such sites are called active centres.

Question 16.
What is a colloidal solution?
Answer:
When the diameter of the particles of a substance dispersed in a solvent ranges from 1 – 200 nm, the system is called a colloidal solution.

Question 17.
What is meant by dispersion medium and dispersed phase?
Answer:
In a colloid, the substance present in a larger amount is called dispersion medium and the substance present in less amount is called the dispersed phase.

Question 18.
What is the flocculation value?
Answer:
The precipitation power of electrolyte is determined by finding the minimum concentration(millimoles/lit ) required to cause precipitation of a sol in 2 hours. This value is called flocculation value.
The smaller the flocculation value greater will be precipitation.

Question 19.
Name the dispersion medium present in hydrosol, alcosol and benzosol.
Answer:

ColloidDispersion Medium
i HydrosolWater
ii AlcosolAlcohol
iii BenzosolBenzene

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 20.
How is a colloid prepared by exchange of solvent ?
Answer:

  • Colloidal solution of phosphorous or sulphur is obtained by preparing the solutions in alcohol and pouring them into water.
  • As they are insoluble in water, they form colloidal solution P in alcohol + water —> P Sol

Question 21.
What is Tyndall effect?
Answer:
The scattering of light by colloidal particles is called Tyndall effect.

Qu

estion 22.
What is Brownian movement?
Answer:

  • The random, zigzag, ceaseless motion of colloidal particles in dispersion medium is known as Brownian movement.
  • This is due to the continuous bombardment of dispersed phase bv the molecules of dispersion medium.

Question 23.
What is the significance of Brownian movement ?
Answer:

  • To calculate Avogadro number.
  • To confirm kinetic theory which considers the ceaseless rapid movement of molecules that increases with increase in temperature.
  • To understand the stability of colloids.
  • As the particles are in continuous rapid movement they do not come close and hence not get condensed. That is Brownian movement does not allow the particles to be acted by force of gravity.

Question 24.
Define ‘gold number’
Answer:

  • Gold number is defined as the number of milligrams of hydrophilic colloid that will just prevent the precipitation of 10 ml of gold sol on the addition of 1 ml of 10% NaCl solution.
  • Smaller the gold number greater the protective power.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 25.
What are emulsions?
Write their types.
Answer:

  • Emulsions are colloidal solution in which a liquid is dispersed in another liquid.
  • Two types of emulsions are
    (i) Oil in water (O/W)
    (ii) Water in oil (W/ O)

Question 26.
What is emulsification?
Answer:
The process of preparation of emulsion by the dispersal of one liquid in another liquid is called emulsification.

Question 27.
What is de emulsification?
Answer:
The process of separation of an emulsion into two separate layers is called de emulsification.

Question 28.
Write the uses of colloids in medicine?
Answer:

  • Antibodies such as penicillin and streptomycin are produced in colloidal form for suitable injections.
  • Colloidal gold and colloidal calcium are used as tonics.
  • Milk of magnesia is used for stomach troubles.
  • Silver sol protected by gelatine known as Argyrol is used as eye lotion.

Question 29.
Mention the shapes of the following colloidal particles,
i) AS2S3
ii) Blue gold sol
iii) Tungstic acid sol
Answer:
i) AS2S3 -Spherical
ii) Blue gold sol – Disc or plate like
iii) Tungstic acid sol – Rod like

VII. Three Mark Questions

Question 1.
Write the characteristics of adsorption
Answer:

  • Adsorption can occur in all interfacial surfaces (ie) between gas-solid, liquid-solid, liquid-liquid, solid-solid, gas-liquid
  • Adsorption is always accompanied by decrease in free energy. When ΔG = 0, the equilibrium is attained.
  • Adsorption is a spontaneous process.
  • Adsorption is accompanied by decrease in randomness (ie) decrease in entropy.
  • Adsorption is exothermic.
  • Adsorption is a quick process.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 2.
Write a note on Freundlich isotherm.
Answer:
A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorhate at constant temperature is called adsorption isotherm.

According to Freundlich x/m = kp1/n
x = amount of adsorbate
m = mass of adsorbent in gram
p = pressure
k, n = constants, value of n is always less than unity.

This equation is applicable for adsorption of gases on solid surfaces.
For adsorption in solutions of concentration C, the equation is x/m = kC1/n
These equations quantitatively predict the effect of pressure or concentration on the adsorption of gases or liquids at constant temperature
Taking log
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 18

Question 3.
Write a note on (i) auto catalysis , (ii) negative catalysis ?
Answer:
Auto catalysis:
When one of the products formed acts as a catalyst to the reaction it is called as auto catalysis and that product is-called as auto catalyst.
(Eg.) CH3COOC2H5 + H2O → CH3COOH + C2 + C2H5OH
Acetic acid acts as the auto catalyst.

Negative catalysis:
When certain substances decreases the rate of the reaction, it is called negative catalysis and the substance is called negative catalyst.
(Eg.) 2H2O2 → 2H2O + O2
Glycerol acts as a negative catalyst in the above reaction.

Question 4.
What is the role of adsorption in the heterogeneous catalysis.
Answer:
In heterogenous catalysis, the reactants are converted into products only when the reactants are adsorbed on the catalyst surface to form an activated complex which subsequently decomposes to give the product. Hence the role of adsorption is to form activated complex.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 5.
Give some examples for enzyme catalysis.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 19

Question 6.
Write a note on nano catalysis.
Answer:

  • Metallic nano particles, metal oxides are used as catalysts.
  • Nano catalysts can act as both homogeneous as well as heterogeneous catalysts.
  • Like homogeneous catalysts they give 100% selective transformations and excellent yield and show extremely high activity.
  • Like heterogeneous catalysts, they can be recovered and recycled.
  • Nano catalysts are actually soluble heterogeneous catalysts.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 20

Question 7.
How can you identify the two types of emulsion.
Answer:

TestOil in water (O/W) emulsion (aqueous emulsion)Water in oil (W/O) emulsion (oily emulsion)
i) Dye test: A small amount of dve soluble in oil is added.Does not take up the colourTakes up the colour.
ii) Viscosity testLower viscosityHigher viscosity
iii) Conductivity testHigher conductivityLower conductivity.
iv) Spreading testSpread less readily on an only surface.Spread readily on an only surface.

Question 8.
Name some of the de emulsification techniques.
Answer:

  • Distilling one component.
  • Adding an electrolyte to destroy the charge. Destroying the emulsifier using chemical methods.
  • Using solvent extraction to remove one component.
  • By freezing one of the components.
  • By applying centrifugal force.
  • Adding dehydrating agents for water in oil (W/O) type.
  • Using ultrasonic waves.
  • Heating at high pressure.

VIII. Five Mark Questions

Question 1.
What are the general characteristics of catalysts.
Answer:

  • Only a small quantity of catalyst is needed for a reaction.
  • There may be some physical changes, but the catalyst remains unchanged’ in mass and its chemical composition.
  • A catalyst cannot initiate a reaction, only it speeds up a slow reaction.
  • A solid catalyst is more effective in a finely divided form.
  • Catalysts are specific in nature.
  • A catalyst does not affect the position of equilibrium and the value of equilibrium constant but helps the quick attainment of equilibrium.
  • A catalyst is highly effective at a particular temperature called as optimum temperature.
  • Presence of a catalyst does not change the nature of products.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 2.
What are the special characteristics shown by enzyme catalysed reaction?
Answer:
Effective and efficient conversion An enzyme may transform a million molecules of reactant in a minute. (Eg.) 2H2O2 → 2H2O + O2
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 21
Enzyme catalysis has maximum rate at optimum temperature. The temperature at which the enzymic activity is high or maximum is called as optimum temperature.
Enzymes involved in human body have an optimum temperature 37°C.
The rate of enzyme catalysis varies with pH of the system. The pH at which the rate is maximum is called the optimum pH.
Enzymes can be inhibited or poisoned. Penicillin inhibits the action of bacteria and used for curing diseases like pneumonia, dysentery, cholera.
A small non protein (vitamin) called a coenzyme promotes the catalytic activity of enzyme.

Question 3.
Explain phase transfer catalysis.
Answer:

  • If the reactants of a reaction are present in two different solvents which are immiscible, the reaction between them is very slow.
  • As the solvents form separate phases, the reactants have to migrate across the boundary to react.
  • But migration across the boundary is not easy.
  • For such a situation a third solvent miscible with both is added.
  • So the phase boundary is eliminated and the reactants mix freely and react fast.
  • But for large scale production of any product, use of a third solvent is not convenient as it may be expensive.
  • Phase transfer catalysis gives a simple solution for such problems as it avoids the use of solvents.
  • Phase transfer catalyst facilitate the transport of a reactant in one solvent to the other solvent where the second reactant is present.
  • As the reactants are brought together, they rapidly react and form the product.
    R – Cl + NaCN → R-CN + NaCl
    Organic phase aqueous phase Organic phase aqueous phase
  • RC1 = 1 – chloro octane, RCN = 1 – cyano octane
    1- cyano octane is not obtained by directly heating organic phase, 1 – chloro octane ^ with aqueous phase sodium cyanide for^ ’ several days.
  • If a small amount of tetra alkyl ammonium chloride is added, 1 – cyano octane occurs in about 100% yield after 1 or 2 hours.
  • In this reaction, the tetra alkyl ammonium cation containing hydrophobic and hydrophilic ends, transports CN from the aqueous phase to the organic phase using its hydrophilic end.
    Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 22
  • Both in organic phase Organic phase
    It moves to aqueous phase, releases Cl— again picks up CN and transports it.
  • So phase transfer catalyst speeds up the reaction by transporting one reactant from one phase to another.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 4.
Explain dispersion methods of preparing colloids
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 23
Mechanical Dispersion:
The solid is ground to colloidal dimension using a colloid mill consists of two metal plates rotating in opposite direction at very high speed of nearlv 7000 revolution / minute.
Required colloidal si/.e is obtained bv adjusting the distance between the two plates.
(Eg.) Colloids of ink and graphite.

ii) Electro Dispersion :
An electric arc is struck between electrodes dispersed in water surrounded by ice.
When a current of l amp/100 v is passed, an arc produced forms vapours of metal which immediate!) condense to form colloidal solution
Alkali hvdroxide is added as an stabilising agent.
(Eg.) Colloids of metals like copper, silver, gold, platinum
Svedberg modified this method for the prepartion of non-aqueous inflammable liquids like pentane, ether and benzene using high frequency alternating current which prevents the decomposition of the liquid.

iii) Ultrasonic Dispersion:
Sound wraves of frequency more than 20 KHz transform coarse suspension to colloids. Claus obtained mercury sol by subjecting mercury to sufficiently high frequency ultrasonic vibrations.
Ultrasonic vibrations produced by generator spread through the oil and transfer the vibration to the vessel with mercury in water.

iv) Peptisation:
Conversion of a precipitate into colloid by the addition of electrolyte is called peptisation.
Added electrolyte is called peptising or dispersing agent.
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 24

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 5.
Explain condensation methods of preparing colloids.
Answer:
i) Oxidation:
Sols of non metals are prepared by this method
HIO3 + 5HI → 3H2O + I2 (sol)

ii) Reduction:
Gold sol is prepared by reducing auric chloride using organic reagent formaldehyde.
2AUCl3 + 3HCHO + 3H2 → 2Au (sol) + 6HCl + 3HCOOH

iii) Hydrolysis :
Sols of metal hydroxides are prepared by this method.
FeCl3 + 3H2O → Fe(OH)3 + 3HCl

Double decomposition:
Water insoluble sols are prepared by this method. Yellow coloured arsenic sulphide sol is prepared by passing hydrogen sulphide gas through arsenic oxide.
AS2O3+ 3H2S → AS2S3 + 3H2O

vi) Decomposition:

  • When a few drops of an acid is added to sodium thio sulphate, it decomposes to form insoluble free sulphur.
  • Insoluble free sulphur accumulates into small clusters which impart various colours blue, yellow and even red to the system.
  • This colour change is due to their growth within the size of colloidal dimensions.
    Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 25

Question 6.
Explain various methods of purification of colloids.
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 26
Answer:
Dialysis:

  • Colloidal solution is taken in a bag made up of semipermeable membrane.
  • It is suspended in a trough of flowing water.
  • Electrolytes diffuse out of the membrane and they are carried away by water.

Electro dialysis :

  • The presence of electric field increases the speed of removal of electrolytes from colloidal solution.
  • Colloidal solution with electrolytic impurity is placed between two dialysing membranes enclosed into two compartments filled with water.
  • When current is passed, the impurities pass into water compartment and get removed periodically.
  • This process is faster than dialysis, as the rate of diffusion of electrolytes is increased by the application of electricity.

iii) Ultra filtration:

  • The pores of ordinary filter papers allow the passage of colloidal particles.
  • In ultra filtration the membranes are made by using colloidion, cellophane or visiking.
  • When a colloidal solution is filtered using such a filter, colloidial particles are separated on the filter and the impurities are removed as washings.
  • This process is quickened by the application of pressure.
  • Collodion is 4% solution of nitrocellulose in a mixture of alcohol and water.
  • This method of separation of sol particles from electrolyte by filtration through an ultrafilter is called ultrafiltration.

Question 7.
Write a note on Helmholtz double layer.
Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 27

  • The surface of colloidal particle adsorbs one type of ion due to preferential adsorption
  • This layer attracts the oppositely charged ions in the medium.
  • Hence at the boundary separating the two, electrical double layers are set up.
  • This is called as Helmholtz electrical double layer.
  • As the particles nearby are having similar charges, they cannot come close and condense.
  • This explains the stability of a colloid.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry

Question 8.
Write a note on electrophoresis.
Answer:

  • The migration of sol particles towards an electrode under the influence of electric field is called electrophoresis or cataphoresis.
  • When electric potential is applied across two platinum electrodes dipped in a hydrophilic sol, the dispersed particles move toward one or other electrode.
  • If the sol particles migrate to the cathode then they possess positive charges.
  • If the sol particles migrate to the anode then thev have negative charges.
  • Thus from the direction of migration of sol particles, the charge on them can be determined.
  • Hence electrophoresis is used for the detection of presence of charges on the sol practicles.

Samacheer Kalvi 12th Chemistry Guide Chapter 10 Surface Chemistry 28

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 13 Organic Nitrogen Compounds Text Book Back Questions and Answers, Notes.

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

12th Chemistry Guide Organic Nitrogen Compounds Text Book Back Questions and Answers

Part – I Text Book Evaluation

I. Choose the Correct Answer

Question 1.
Which of the following reagent can be used to convert nitrobenzene to aniline
a) Sn/HCl
b) ZnHg/NaOH
c) LiAlH4
d) All of these
Answer:
a) Sn/HCl

Question 2.
The method by which aniline cannot be prepared is
a) degradation of benzamide with Br2/NaOH
b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.
c) Hydrolysis of phenylcyanide with acidic solution.
d) reduction of nitrobenzene by Sn/HCl
Answer:
b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.
 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

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

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

Question 5.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 1
a) bromomethane
b) α -Bromo sodium acetate
c) methanamine
d) acetamide
Answer:
c) methanamine
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 2

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 6.
Which one of the following nitro compounds does not react with nitrous acid?
a) CH3-CH2-CH2-N02
b) (CH3)2 CH-CH2 NO2
c) (CH3)3CNO2
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 3
Answer:
c) (CH3)3CNO2
Solution:
30 Nitroalkane

Question 7.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 4
a) Friedel – Crafts reaction
b) HVZ reaction
c) Schotten – Baumann reaction
d) None of these
Answer:
c) Schotten – Baurnann reaction

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

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 9.
Which of the following reaction is not correct
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 5
d) None of these
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 6
Solution:
p – nitrosation takes places, the product is im7
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 7

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

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

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

Question 12.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 9
a) H3PO2 and H2O
b) H+/H2O
c) HgSO4 / H2SO4
d) Cu2Cl2
Answer:
a) H3PO2 and H20

Question 13.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 10
a) C6H5-OH
b) C6H5-CH2OH
c) C6H5-CHO
d) C6H5NH2
Answer:
a) C6H5-OH
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 11

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 14.
Nitrobenzene on reaction with Con HNO3 / H2SO4 at 80- 100oC forms which one of the following products?
a) 1, 4 – dinitrobenzene
b) 2, 4, 6 – tirnitrobenzene
c) 1, 2 – dinitrobenzene
d) 1, 3 – dinitrobenzene
Answer:
d) 1, 3 -dinitrobenzene
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 12

Question 15.
C5H13N reacts with HNO2 to give an optically active compound – The compound is
a) pentan-1 -amine
b) pentan-2-amine
c) N, N – dimethyipropan -2-amine
d) N-methylbutan-2-amine
Answer:
d) N-methylbutan-2-amine

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

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

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

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

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 19.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 13
a) Ethanol, hydroxylamine hydrochloride
b) Ethanol, ammonium hydroxide
c) Ethanol, NH2OH
d) C3H5NH2, H2O
Answer:
a) Ethanol, hydroxylamine hydrochloride

Question 20.
UPAC name for the amine
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 14
a) 3 – Bimethy lamino – 3 – methyl pentane
b) 3(N,N – Triethyl) – 3 – amino pentane
c) 3-N,N – trimethyl pentanamine
d) 3 – (N,N – Dimethyl amino) – 3- methyl pentane
Answer:
d) 3 – (N,N – Dimethyl amino) -3- methyl pentane

Question 21.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 15
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 16
Answer:
b)Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 17

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 22.
Ammonium salt of benzoic acid is heated strongly with P2O5 and the product so formed is reduced and then treated with NaNO2/ HCl at low temperature. The final compound formed is
a) Benzene diazonium chloride
b) BenzYl alcohol
c) Phenol
d) Nitrosobenzene
Answer:
b) Benzyl alcohol
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 18

Question 23.
Identify X in the sequence given below.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 19
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 20
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 21

Question 24.
Among the following, the reaction that proceeds through an electrophilic substitution, is:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 22
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 23
Explanation:
a) Nucleophilic substitution
b) Electrophilic substitution
c) Addition Reaction
d) Nucleophilic substitution

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 25.
The major product of the following reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 24
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 25
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 26
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 27
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 28

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

II. Short Answer Questions

Question 1.
Write down the possible isomers of the C4H9NO2 and give their IUPAC names
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 29
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 30
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 31

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 2.
There are two isomers with the formula CH3 NO2. How will you distinguish between them?
Answer:

  • Primary and secondary nitroalkanes with a-H atom exhibit tautomerism.
  • Tertiary amines do not exhibit tautomerism due to CH =N the absence of a-H atom.
  • Nitromethane exists in two tautomeric forms namely \itro form Aci form nitroform and aciform.

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 32

Nitro FormAci – Form
1. Less AcidicMore acidic and also called pseudoacids (or) nitronic acids.
2. Dissolves in NaOH slowlyDissolves in NaOH instantly.
3. Decoiourises FeCl3 SolutionWith FeCl3, gives reddish-brown colour
4. Electrical conductivity is lowElectrical conductivity is high

Question 3.
What happens when
Answer:
i) 2 – Nitropropane boiled with HCl
ii). Nitrobenzene electrolytic reduction in strongly acidic medium.
iii). Oxidation of tert – butylamine with KMnO4
iv). Oxidation of acetone oxime with trifluoromethoxy acetic acid.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 33
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 34
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 35

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 4.
How will you convert nitrobenzene into
Answer:

  • 1, 3, 5 – trinitrobenzene
  • o and p – nitrophenol
  • m – nitro aniline
  • azoxvbenzcne
  • hvdrazohenzene
  • N – phenvl hyd roxylamine
  • aniline

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 36
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 37

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 5.
Identify compounds A, B, and C in the following sequence of reactions.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 38
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 39
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 40
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 41
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 42
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 43
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 44

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 6.
Write short notes on the following
Answer:

  • Hofmann’s bromide reaction
  • Ammonolysis
  • Gabriel phthalimide synthesis
  • Schotten – Baurnann reaction
  • Carhvlamine reaction
  • Mustard oil reaction
  • Coupling reaction
  • Diazotisation
  • Gorenberg reaction

I Hofmann’s bromamide reaction:

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 45
In Hofnianns degradation acid amide is converted into an amine with one carbon less by Br2/ KOH.

ii. Ammonolysis:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 46
In Hoffmann’s ammonolysis alkyl halides are heated with alcoholic ammonia in a sealed tube, mixtures of 1o, 2o and 3o amines and quarternary ammonium salts are obtained.

iii. Gabriel phthalimide synthesis
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 47
Phthalimide on treatment with alcoholic KOH forms potassium phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine. Aniline can not be prepared by this method.

iv. Schotten – Baumann reaction:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 48
Benzoylation of amines to give N-alkyl benzamide in presence of NaOH is known as Schotten Baumann reaction.

v. Carbylamine reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 49
Primary’ amines react with chloroform and alcoholic KOH to form isocyanides called carbylamines with unpleasant smell. This is used to identify primary amines.

vi. Mustard oil reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 50When primary amines are treated with carbon disulphide, N-alkyl dithiocarbamic acid is formed which on treatment with HgCl2 gives an alkyl isothiocyanate.

vii. Coupling reaction:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 51
Benzenediazonium chloride reacts with electron-rich aromatic compounds like phenol, undergoing
electrophilic substitution at para position.

viii. Diazotisation:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 52
Aniline reacts with nitrous acid at 273-278K to form benzene diazonium chloride.

ix. Gomberg reaction:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 53
Benienediazonium chloride reacts with benzene in presence of NaOH to give biphenyl.
 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 7.
How will you distinguish between Primary, Secondary, and tertiary aliphatic amines
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 54
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 55

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 8.
Account for the following
Answer:

  • Aniline does not undergo Friedel – Crafts reaction
  • Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
  • pKb of aniline is more than that of methylamine
  • Gabriel phthalimide synthesis is preferred for synthesising_primary amines
  • Ethvlamine is soluble in water whereas aniline is not
  • Amines are more basic than amides
  • Although amino group is o – and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m – nitroaniline.

i) Aniline does not undergo Friedel – crafts reaction, because Aniline is basic in nature
Aniline donates its lone pair to the lewis acid Alcl3 to form an adduct which inhibits further electrophilic substitution reaction

ii) The stability of diazonium salts of aromatic amines is due to the dispersal of the positive charge over the benzene ring.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 56

iii) In aniline the NH
group is directly attached to the benzene ring.
The lone pair of electron on nitrogen atom in aniline gets delocalised over the benzene ring.
Hence it is less available for protonation.
So aniline is less basic than methylamine and pKb of aniline is more than that of methylamine.

iv) Gabriel phthalimide synthesis results in the formation of primary amines only. Secondary and tertiary amines are not formed in this synthesis. Thus a pure primary amine can be obtained.
Therefore, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

v) Ethvlamine forms intermolecular hydrogen bonds with water. Hence it is soluble in water.
But aniline does not form hydrogen bond with water to a very large extent due to the presence of a large hydrophobic C6H5 – group. Hence aniline is insoluble in water

vi) In amides the carbonyl group (C = 0) is highly electronegative and draws the electrons towards it.
This makes the lone pair of amide nitrogen less available to accept a proton.
Hence amides are less basic than amines.
But in amines the alkyl groups being electron releasing groups. the electron pair on amine nitrogen is readily available to accept a proton.

vii) Hence amines are more basic than amides.
In strong acid medium aniline is protonated to form anilinium ion which is rn-directing.
Hence a substantial amount of rn-nitroaniline is formed during nitration of aniline.

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 9.
Arrange the following
Answer:
i) In increasing order of solubility in water C6H5NH2,(C2H5)2NH,C2H5NH2
ii) In increasing order of basic strength
a) aniline, p – toludine and p – nitroaniline
b) C6H5NH2 C6H5NHCH3, C6H5NH2, p-Cl-C6H4-NH2
iii) In decreasing order of basic strength in gas phase
(C2H5)NH2, (C2H5)NH, (C2H5)3N and NH3
iv) In increasing order of boiling point C6H5OH, (CH3)2NH,C2H5NH2
v) In decreasing order of the pKb values C2H5NH2, C6H5NHCH3, (C2H5)2NH and CH3NH2
vi) Increasing order of basic strength C6H5NH2, C6H5N(CH3) H53)2, (C2H5)2 NH and CH3 NH2
vii) In decreasing order of basic strength
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 153

i) Increasing order of solubility in water C6H5NH2 < (C2H5) NH < C2H5 NH2
Hint
Aromatic amine
aliphatic amine
Solubility α \(\frac{1}{\text { Molarmass }}\)
ii) Increasing order of basic strength
a) p-nitro aniline
aniline
p-toludine
b) p – Cl- C6H4– NH2 < C6H5NH2 < C6H5NHCH3 < C6H5NH2
Hint > p-nitrogroup (-I) electron with drawing, hence decreases the basic strength.
Hint > p-methyl group (+I) electron releasing, hence increases the basic strength.
iii) Decreasing order of basic strength in gas phase
(C2H5)3N > (C2H5)2 NH > C2H5 NH2> NH3
Hint > In the gas-phase there is no solvation effect. As a result the basic strength mainly depends on the + I effect. Higher the number of alkyl groups higher will be the +1 effect, stronger is the base.
iv) Increasing order of boiling point.
(CH3)2NH <C2H5NH2 <C6H5OH
Hint > Amines generally have lower boiling point than alcohols of comparable molar mass. Since amines have weaker H-bonds.
Hint
Secondary amines
primary amines
v) Decreasing order of pK values
C6H5NHCH3> C2H5NH2> CH3 NH2> (C2H5)2 NH
Hint > Basic character increases. pKb value decreases. Higher the basic nature, lower will be the pkb’ value.
vi) Increasing order of basic strength
C6H5NH2 < C6H5-N(CH3)2 < CH3 NH2 < (C2H5)2NH
vii) Decreasing order of basic strength

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 58

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 10.
How will you prepare propan – 1 – amine from
Answer:
i) butane nitrile
ii) propanamide
iii) 1-nitropropane
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 59

Question 11.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 60
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 61

CompoundName
AMethanamirie
BN – ethyl N – methyL ethanamine
CDiethyl methyl ammonium hydrogen sulphate

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 12.
How will you convert diethylamine into
Answer:
i) N,N – dimethylacetamide
ii) N – nitrosódiethylamine
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 62

Question 13.
Identify A, B and C
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 63

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 14.
Identify A,B,C and D
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 64

Question 15.
Complete the following reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 65
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 66

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 16.
Predict A,B, C and D for the following reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 67
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 68
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 69

CompoundName
APhthaliniide
BIsopropyl bromide
CN-isopropy1ththalimide
DPhthalic acid

Question 17.
A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO2 . (B) on heating with liquid ammonia followed by treating with Br2 / KOH gives (c) which on treating with NaNO2 and HC1 at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D.
Answer:
D is a monobasic and with molecular mass 74.
M.F of D is CnH2n+1COOH
MolarMassis 12 × n + 2n × 1 + 1 × 1 + 12 + 2 × 16 + 1 × 1 = 74
14n + 46 = 74
14n=28
n=2
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 70
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 71

CompoundName
A1, 1- dibromopropane
BButanoic acid
C1 – amino propane
DPropanoic acid

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 18.
Identify A to E in the following frequency of reactions
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 72
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 73

CompoundName
Ao – nitro toluene
Bo – amino toluene
Co – amino toluene
Do – methyl benzene diazonium chloride
Eo – cyano toluene

III. Evaluate Yourself

Question 1.
Write all possible isomers for the following compounds.
Answer:
i) C 2H 5 – NO 2
ii) C 3H 7 – NO 2
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 74
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 75

Question 2.
Find out the product of the following reactions.
Answer:

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 3.
Predict the major product that would be obtained on nitration of the following compounds?
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 77
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 78

Question 4.
Draw the structure of the following compounds
i. Neopentylaniine
ii. Tert – butylamine
iii. α- amino propionaldehyde
iv. Tribenzylamine
v. N – ethyl – N – methyihexan – 3- amine
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 79

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 5.
Give the correct IUPAC names for the following amines
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 80

12th Chemistry Guide Organic Nitrogen Compounds Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

Question 1.
Nitro ethane and ethyl nitrite are
a) chain isomers
b) position isomers
c) functional isomers
d) tautomers
Answer:
c) functional isomers

Question 2.
1-nitrobutane and 2-nitrobutane are
a) chain isomers
b) Position isomers
c) functional isomers
d) tautomers
Answer:
b) Position isomers

Question 3.
1-nitrobutane and 2- methyl-1-nitro propane are
a) chain isomers
b) Position isomers
c) funtional isomers
d) tautomers
Answer:
a) chain isomers

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 4.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 81
a) chain isomers
b) position isomers
c) functional isomers
d) tautorners
Answer:
d) tautomers

Question 5.
Which among the following is a tertiai nitroalkane?
a) 2-nitro propane
b) 2-methyl-.1—nitro propane
C) 2-methyl – 2-nitropropane
d) nitro ethane
Answer:
c) 2-methyl – 2-nitropropan

Question 6.
Which among the following does not sho tautomerism?
a) 2-nitro propane
b) 2-methyl-1-nitropropane
c) 2-methyl-2-nitropropane
d) nitro ethane
Answer:
c) 2-methyl-2-nitropropane
Hint : Tertiary Nitroalkanes do not exhibit tautomerism due to the absence of Oc-H atom.

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 7.
Which nitro alkane does not dissolve in NaOH?
a) 2-nitro propane
b) 2-methyl – 1 – nitro propane
c) 2 – methvl-2-nitropropane
d) nitroethane
Answer:
c) 2 – methyl- 2 – nitropropane
Hint : Tertiary mitroalkanes do not react with NaOH due to the absence of ∝-H atom.

Question 8.
Which among the following does not react with Nitrous acid.
a) 2 – nitro propane
h) 2-methyl-1-nitropropane
c) 2 – methyl – 2 – nitro propane
d) nitro ethane
Answer:
c) 2 – methyl – 2 – nitro propane
Hint : Tertiary mitroalkanes do not react with nitrous acid due to the absence of ∝- H atom.

Question 9.
The correct order of acidic nature of nitro alkanes is
a) nitro propane> nitroethane > nitro methane
b) nitro propane c) nitro methane < nitro ethane < nitro propane d) nitro ethane> nitro methane> nitro propane
Answer:
b) nitro propane < nitroethane < nitromethane
Hint: Length of alkyl group attached to the ∝- carbon increases, acidic nature decreases

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 10.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 82
The above reaction follows the mechanism
a) E1
b) E2
c) SN1
d) SN2
Answer:
d) SN2

Question 11.
Acetaldoxime is converted into nitroethane by reaction with
a) HNO2
b) CF3COOOH
c) KMnO4
d) HNO3/H2SO4
Answer:
b) CF3COOOH

Question 12.
Oil of mirbane is
a) nitro ethane
b) nitro propane
c) nitro benzene
d) nitro aniline
Answer:
c) nitrobenzene

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 13.
Direct nitration of nitro benzene gives
a) o – dinitro benzene
b) m – dinitro benzene
c) p – dinitro benzene
d) All the above
Answer:
b) m – dinitro benzerte

Question 14.
P – diamino benzene can be converted to p – dinitro benzene by
a) Caros acid
b) persuiphuric acid
c) peroxy trifluoro acetic acid
d) all the above
Answer:
d) all the above

Question 15.
Nef carbonyl synthesis given by
a) C6H5CHO
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 83
c) C6H5CH2NO2
d) All of these
Answer:
a) C6H5CHO

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 16.
In electrophilic substititution reaction of nitro benzene, the – NO2 group acts as
a) activating group and m – directing
b) deactivating group and rn-directing
c) activating group and o,p – directing
d) deactivating group and o,p -directing
Answer:
b) deactivating group and m-directing

Question 17.
Identify X in the following sequence of reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 84
a) (CH3)2CHCOCl
b) CH3CH2CH(OH)CH2Cl
c) CH3CH2CH2COCl
d) ClCH2CH2CH2CHO
Answer:
c) CH3CH2CH2COCl

Question 18.
The reducing agent used to reduce alkylcyanides to primary amines is
a)H2/Ni
b)LiAlH4
c) Na/C2H5OH
d) all the above
Answer:
d) all the above

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 19.
The reducing agent used in Mendius reaction is
a) H2/Ni
b) LiAlH4
c) Na/C2H5OH
d) NaBH4
Answer:
c) Na/C2H5OH

Question 20.
Reduction of alkyl iso cyanides give
a) primary amines
b) secondary amines
c) tertiary amines
d) none of the above
Answer:
b)secondary amines

Question 21.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 85
a) CH3NH2, CH3CH2NH2
b) CH3OH, CH2CH2OH
c) CH3CH2NH2, CH3NH2
d) CH3CN, CH3CH2CN
Answer:
c) CH3CH2NH2, CH3NH2

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 22.
Gabriel phthalimide synthesis is used for the preparation of
a) aliphatic primary amines
b) aromatic primary amines
c) both (a) & (b)
d) none of the above
Answer:
a)aliphatic primary amines

Question 23.
Which among the following can not be prepared by Gabriel phthalimide synthesis?
a) Methanamine
b) ethanamine
c) benzenamine
d) propananiine
Answer:
c)benzenamine

Question 24.
Two molecules of propannitrile in the presence of Na/Ether to form 3-imino- 2-methylpentanenitrile. This reaction is known as
a) Baltz – schiemann reaction
b) Thorpe nitrile condensation
c) Gomberg reaction
d) hotten – Baurnann reaction
Answer:
b) Thorpe nitrile condensation

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 25.
In Hoffmann’s ammonolysis, if excess alkyl halide is taken, the final product is
a) primary amine
b) secondary amine
c) tertiary amine
d) quaternary ammonium salt
Answer:
d) quaternary ammonium salt

Question 26.
Hoffmann’s ammonolysis reaction is an example of
a) nucleophilic addition
b) nucleophilic substitution
c) electrophilic addition
d) electrophilic substitution
Answer:
b) nucleophilic substitution

Question 27.
In Hoffmann’s ammonolysis the order of reactivity of alkyihalides with amines is
a) RI > RBr> RCl
b) RI < RBr < RCl c) RI > RBr < RCl
d) RI < RBr> RCl
Answer:
a)RI>RBr>RCl

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 28.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 86
A and B are respectively
a) CH3Na, CH3CN
b) CH3NH2,CH3N3
c) CH3N3 CH3NH2
d) CH3NH2,CH3CN
Answer:
c) CH3N3 CH3NH2

Question 29.
The correct ascending order of basic strength of alkylamines in aqueous solution is.
a) R2NH > RNH2 > R3N > NH3
b) NH3 > R3N > RNH2 > R2NH
c) NH3 < R3N < RNH2 < R2NH
d) R2NH < RNH2 < R3N < NH3
Answer:
c) NH3 < R3N < RNH2 < R2NH

Question 30.
Among the three types of amines, secondary amine is more basic due to
a) +1 effect
b) steric effect
c) hydration effect
d) all the above
Answer:
d) all the above

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 31.
In case of sustituted aniline, the groups which increase the basic strength are
(i) -CH3
(ii) -NO2
(iii) -OCH3
(iv) -Cl
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
b)(i) & (iii)

Question 32.
The correct order of basic strength of amines is
a) C6 H5 – NH2 > C6 H5 CH2 NH2> NH3 > C6 H5 CH2 NHCH3 > C2 H5 NH2

b) C6 H5 CH2 NH2 > C6 H5 CH2 NHCH3 > NH3 > C6 H5 NH2 > C2H5NH2

c) C2 H5 NH2 > C6H5 CH2 NH2 > NH3 > C6 H5 CH2 NHCH3 > C6 H5 NH2

d) C6 H5CH2 NH CH3 > C6 H5 CH2 NH2 > NH3 > C2 H5 NH2 > C6 H5 NH2
Answer:
c) C2 H5 NH2 > C6H5 CH2 NH2 > NH3 > C6 H5 CH2 NHCH3 > C6 H5 NH2
Hint : Alkylamine > Aralkvlamine > Ammonia > N – Aralkylamine > Arylamine

Question 33.
Schotten Baumann reaction is an example of
a) nucleophilic substitution
b) nucleophilic addition
c) electrophilic substitution
d) electropholic addition
Answer:
a)nucleophilic substitution

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 34.
The product ‘D’ of the reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 87
a) CH3CH2NH2
b) CH3CN
c) HCONH2
d) CH3CONH2
Answer:
d) CH3CONH2

Question 35.
The test used to identify a primary amine is
a) iodoform test
b) silver mirror test
c) libermann’s nitroso test
d) carbylamine test
Answer:
d) carbylamine test

Question 36.
The test used to identify a secondary amine is
a) iodo form test
b) silver mirror test
c) Libermann’s nitroso test
d) carbylamine test
Answer:
c) Libermann’s nitroso test

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 37.
When ethylamine is treated with chloroform and alcoholic KOH an unpleasant smell is evolved due to the formation of
a) ethylcyanide
b) ethylisocyanide
c) ethylisocyanate
d) ethylcyanate
Answer:
b) ethylisocyanide

Question 38.
C6H5N + 2Cl + H3PO2 + H2O → C6H6 + H3PO3 + HCl + N2 This reaction proceeds through
a) SN1 mechanism
b) SN2 mechanism
c) free radical mechanism
d) elimination reaction
Answer:
c) free radical mechanism

Question 39.
Benzene diazonium chloride is converted into chloro benzene by
(i) Sandmeyer reaction
(ii) Gatter mann reaction
(iii) Gomberg reaction
a) (i) &(ii)
b) (i.) & (iii)
c) (ii) & (iii)
d) (i) (ii) & (iii)
Answer:
a)(i) &(ii)

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 40.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 88
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 89
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 90

Question 41.
Hydrogen cyanide and hydrogen isocyanide are
a) chain isomers
b) position isomers
c) tautomers
d) Geometrical isomers
Answer:
c) tautomers

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 42.
The IUPAC name of C6H5CN is
a) phenyl cyanide
b) benzene carbonitrile
c) benzonitrile
d) benzene cyanide
Answer:
b) benzene carbonitrile

Question 43.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 91
a) CH3CN
b) CH3NH2
c) CH3CONH2
d) CH3CH2CONH2
Answer:
d) CH3CH2CONH2

Question 44.
Methyl magnesium chloride is converted into ethanenitrile by treating with
a) cyanogen
b) cyanogen chloride
c) methyl cyanide
d) hydrogen cyanide
Answer:
b) cyanogen chloride

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 45.
Which one of the following compound is a strong base?
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 92
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 93

Question 46.
Which among the following will not undergo diazotisation?
a) m- toluidine
b) aniline
c) p – amino phenol
d) benzylamine
Answer:
d) benzylamine
Hint: Aromatic amines in which – NH2 group is directly attached to the benzene ring undergo diazotisation

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 47.
An organic compound (A) C3H9N when treated with nitrous acid, gave an alcohol (B) and N2 gas. (A) on warming with CHCl3 and Caustic potash gave (C) which on reduction gave isopropyl methylamine. Organic compound (A) is
a) CH3CH2NH CH3
b) CH3 CH2 CH2 NH2
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 94
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 95
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 96

Question 48.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 97
a) Ethanamide, Ethanamine, Ethanol
b) Ethanarnine, Ethanol, Ethanoic acid
c) Ethanol, Ethanal, Ethanoic acid
d) Ethanamine, Ethanal, Ethanoic acid
Answer:
b) Ethanamine, Ethanol, Ethanoic acid
Hints:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 98

Question 49.
Which of the following nitro compounds when reacted with nitrous acid followed by treatment with alkali produces blue colour?
a) 2-methyl-2-nitropropane
b) 2-methyl-1-nitropropane
c) 2-nitropropane
d) nitrobezene
Answer:
c) 2-nitropropane
Hunt: Nitro group attached to a secondary carbon atom will give pseudonitrol with HNO2 which on further reaction with alkali gives blue colour.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 99

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
II. Assertion and reason

Question 1.
Assertion (A) : Aniline when exposed to air becomes coloured
Reason (R) : Aniline when exposed to air undergoes oxidation
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct
Answer:
a) Both A and R are correct, R explians A

Question 2.
Assertion (A) : Aniline is more basic than ammonia
Reason (R) : The lone pair of electron on nitrogen atom in aniline gets delocalised over the benzene ring.
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct
Answer:
d) A is wrong but R is correct.
Correct Assertion A: Aniline is less basic than ammonia.

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 3.
Assertion (A) : Nitroalkanes are more acidic than aldehvdes, ketones and cyanides.
Reason (R) : a-H atom of 10 and 20 nitroalkanes, show acidic character because of the electron releasing effect of NO2 group.
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct
Answer:
c) A is correct but R is wrong.
Correct Reason (R) : a- H atom of 1o and 2o nitroalkanes show acidic character because of the electron withdrawing effect of NO2 group.

Question 4.
Assertion (A) : Nitro benzene does not undergo Friedel crafts reactions:
Reason (R) : NO2 group present in the benzene ring is strongly deactivating in nature.
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct
Answer:
a) Both(A) and (R) are correct, R explains A.

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

III. Pick out the correct statement

Question 1.
(i) In amines the nitrogen atom is Sp3 hybridised.
(ii) In amines, the fourth Sp3 hybridised orbital contains a lone pair of electron.
(iii) Due to the presence of lone pair of electron C- N-C bond angle is more than the normal tetrehedral bond angle.
(iv) The C-N-C bond angle of trimethylamine is 1090 which is greater than tetrahedral angle,
(a) (i) & (ii)
(b) (ii)&(iii)
(c) (iii)&(iv)
(d) (I&(iv)
Answer:
(a)(i) & (ii)
Correct statement:
(iii) Due to the presence of lonepair of electron C-N-C bond angle is less than the normal tetrahedral bond angle.
(iv) The C-N-C bond angle of trimethylamine is 1080 which is lower than tetrahedral angle.

Question 2.
(i) Tertiary amines form inter molecular hydrogen bond.
(ii) Amines have higher boiling points than alcohols.
(iii) Nitrogen being less electronegative than oxygen, the N-H bond in amines is less polar than O-H bond in alcohols
(iv) Lower aliphatic amines are soluble in water, since they form hydrogen bond with water,
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer:
(c) (iii) & (iv)
Correct Statement:
(i) Tertiary amines do not form intermolecular hydrogen bond.
(ii) Amines have lower boiling point than alcohols.

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 3.
(i) In Hofmann’s degradation reaction primary amines with one carbon less than the parent amides are formed.
(ii) Gabriel phthalimide synthesis is useful for the preparation of three types of amines.
(iii) Alkyl halides can be converted to primary amines by treating with sodium azide followed by reduction using LiAlH4
(iv) When phenol reacts with ammonia in presence of anhydrous ZnCl2 gives p -amino phenol.
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (i) & (iv)
Answer:
c)(i) & (iii)
Correct statement:
(ii) Gabriel phthalimide synthesis is useful for the preparation of aliphatic primary amines
(iv) When phenol reacts with ammonia in presence of anhydrous ZnCl2 gives aniline.

Question 4.
(i) Alkyl cyanides on reduction with LiAlH4 give secondary amines.
(ii) Alkyl cyanides on hydrolysis with alkali or dilute mineral acid give carboxylic acids.
(iii) Nitriles containing a- H atom undergo condensation with esters in presence of sodamide in ether to form ketonitriles.
(iv) Alkycyanides have lower boiling points than analogous acetylenes
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d (i) & (iv)
Answer:
b)(ii) & (iii)
Correct statement:
(i) Alkylcyanides on reduction with LiAlH4 give primary amines.
(iv) Alkylcyanides have higher boiling points than analogous acetylenes.

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

IV. Pick out the incorrect statement

Question 1.
(i) Bromobenzene can be converted into nitrobenzene by reacting with KNO2
(ii) Nitroethane is suspected to cause genetic damage and be harmful to the nervous system.
(iii) 2-methyl – 2-nitro propane reacts with HCl undergoing hydrolysis to form 2-methyl – 2 – propanol.
(iv) Ethyl nitrite on reduction with Sn / HCl gives ethanol.
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (iii) & (iv)
Answer:
c) (i) & (iii)
Correct statement:
(i) Bromo benzene can not be converted into nitrobenzene by reacting with KNO2
(iii) 2-methyl -2-nitropropane does not undergo hydrolysis with HCl to form 2-methyl – 2 – propanol

Question 2.
(i) Nitroalkanes exhibit functional isomerism with alkyl nitrites.
(ii) Tertiary nitroalkanes exhibit tautomerism due to the absence of a-H atom.
(iii) The electrical conductivity of aciform of nitro alkanes is high.
(iv) When the number of alkyl group attached to the a- carbon of nitroalkane increases, its acidity increases.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
c) (ii) & (iv)
Correct statemet:
(ii) Tertiary nitroalkanes do not exhibit tautomerism due to the absence of a- H atom.
(iv) When the number of alkyl group attached to the a- carbon of nitro alkane increases its acidity decreases.

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 3.
(i) Benzoylation of aniline is known as Schotten. Baumann reaction
(ii) The reaction of ethylamine with nitrous acid is known as diazotisation.
(iii) The reaction of primary amine with nitrous acid is known as Libermann’s nitroso test.
(iv) The reaction of primary amine with CS2 and HgCl2 is known as Mustard oil reaction.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement:
(ii) The reaction of aniline with nitrous acid is known as diazotisation.
(iii) The reaction of secondary amine with nitrous acid is known as Libermann’s nitroso test.

Question 4.
(i) Diazo compounds obtained from the coupling reactions of diazonium salts are coloured and are used as dyes.
(ii) Aryl fluorides and iodides can not be prepared by direct halogenation but can be prepared by using benzene diozonium chloride.
(iii) In Sandmeyer reaction and Gattermann reaction the catalyst used are Cu/HCl and Cu2Cl2 / HCZ respectively .
(iv) Benzene diazonium chloride when added with boiling water gives benzene.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct statement:
(iii) In Sandmeyer reaction and Gattermann reaction the catalyst used are Cu2Cl2/ HCl and Cu / HCl respectively
(iv) Benzene diazonium chloride when added with boiling water gives phenol.

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

V. Match the following

Question 1.

Reducing agentReduced product of nitrobenzene
1. Zn / NH4ClHydrazobenzene
2. SnCl2 / KOHAniline
3. Sn /HClNitrosobenzene
4. Zn / NaOHPhenylhydroxylamine
5. Fe / steamAzobenzene

Answers:
1. – Phenylhydroxylamine
2. – Azobenzene
3. – Aniline
4. – Hydrazo benzene
5. – Nitrosobenzene

Question 2.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 100
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 101
Answer:
1. – Prop – 2-en-1-amine
2. – N – methylpropan – 2 – amine
3. – Hexan -1, 6 – diamine
4. – Phenylmethanamine
5. – N, N – dimethylben zenamine

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 3.

ReactionProduct
1. Schotten Baumann reactionMethyl isothiocyanate
2. DiazotistationP-hvdroxy azobenzene
3. Carbylamine testN – phenylbenzamide
4. Mustard oil reactionBenzene diazonium chloride
5. Coupling reactionMethylisocyanide

Answer:
1. – N-phenylbenzamide
2. – Benzene diazonium chloride
3. – Methyl isocyanide
4. – Methylisothiocyanate
5. – p-hydroxyazobenzene

Question 4.

CompoundUse
ChloropicrinFuel additive
Ethyl nitriteSolvent in perfume industry
Nitrobenzenediuretic
Nitro ethaneTNT
Nitrilesinsecticide

Answer:
1. – insecticide
2. – diuretic
3. – TNT
4. – fuel additive
5. – Solvent in perfume industry

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

VI. Two Mark Questions

Question 1.
Write a note on acidic nature of nitroalkanes.
Answer:

  • α-H atom of 10and 20nitroalkanes show acidic character.
  • This is due to the electron withdrawing effect of – NO2group.
  • Nitroalkanes are more acidic than aldehydes, Ketones, esters and cyanides.
  • Due to acidic nature nitroalkanes dissolve in NaOH to form a salt.
  • Aci-nitro form is more acidic than nitro form.
  • As the number of alkyl group attached to a- Carbon increases, acidity decreases due to + I effect of alkyl groups.

Question 2.
How is the oil of mirbane prepared?
Answer:
When benzene is heated at 330K with the nitrating mixture (Con HNO3 + Con H2S04), electrophilic substitution takes place to form nitrobenzene (oil of mirbane)
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 102

Question 3.
How is P – dinitrobenzene prepared?
Answer:

  • Direct nitration is nitrobenzene gives m-dinitro benzene.
  • The following indirect method is adopted to prepare p-dinitrobenzene.
    Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 103

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 4.
Convert p-diamino benzene into p-dinitro benzene.
Answer:
Amino group can be directly converted into nitro group using Caro’s acid (H2SO5) or persulphuric acid (H2S2O8) or peroxytrifluro acetic acid (CF3COOOH) as oxidising agent
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 104

Question 5.
Write the various reduction stages of nitromethane.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 105

Question 6.
Write about the reduction of nitro methane in acid medium and neutral medium.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 106

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 7.
What happpens when ethylnitrite is reduced by Sn / Hcl?
Answer:

  • Ethyl nitrite on reduction with Sn / HCl gives ethanol.

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 107

Question 8.
What happens when ethyl nitrite undergoes hydrolysis with acid or base?
Answer:

  • Ethyl nitrite undergoes hydrolysis with acid or base to give ethanol.

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 108

Question 9.
What is Nef Carbonyl Synthesis?
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 109

Question 10.
Write about the electrolytic reduction of nitro benzene?
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 110

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 11.
Convert m- dinitrobenzene into m – nitroaniline
Answer:
Selective reduction of rn – dinitrobenzene with ammonium suiphide gives m – nitroaniline
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 111

Question 12.
What happens when
(i) methyl cyanide and
(ii) methyl isocyanide are reduced?
Answer:
(i) Methyl cyanide on reduction gives a primary amine ethanamine.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 112
(ii) Methylisocyanide on reduction gives a secondary amine N-methylmcthanamine
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 113

Question 13.
How can you convert acetamide into
(i) ethanamine
(ii) methanamine.
Answer:
Acetamide on reduction with LiAlH4 gives ethanamine.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 114

Acetamide undergoes Hofmanns degradation with Br2/KOH to form methanamine
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 115

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 14.
Write about Sabatier – Mailhe method of preparing three types of Amines
Answer:

  • Vapour of an alcohol and ammonia are passed over alumina, W2O5 or Silica as 400o Call the types of amines are formed.
  • This method is known as Sabatier-Mailhe method.

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 116

Question 15.
Write a note on basic strength of aniline.
Answer:

  • Aniline is less basic than alkylamines and ammonia.
  • In aniline, the NH2 group is directly attached to the benzene ring.
  • The lone pair of electron in nitrogen atom in aniline gets delocalised over the benzene ring.
  • Hence it is less available for protonation.
  • In case of subtituted aniline, election releasing groups like – CH3, -OCH3, -NH2 increase the basic strength.
  • Electron with drawing groups like -NO2, -X, -COOH decrease the basic strength.

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 16.
How is sulphanilic acid prepared?
Answer:

  • Aniline reacts with conc H2SO4to form anilinium hydrogen sulphate.
  • This on heating with H2SO4 at 453-473K gives p-aminobenzene suiphonic acid known as
    sulphanilic acid

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 107 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 117

Question 17.
How will you convert aliphatic carboxylic acid into aromatic carboxylic acid?
Answer:

  • When benzene diazonium fluoroborate is heated with acetic acid, benzoic acid is obtained.
  • This reaction is used to convert aliphatic carboxylic acid into aromatic carboxylic acid.

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 118

Question 18.
Write about the isomerisation of alkylisocyanides.
Answer:
When alkyl isocyanides are heated at 250oC, they change into the more stable isomeric cyanides.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 119

Question 19.
Write the uses of nitroalkanes.
Answer:

  • They are good solvents for a large number of organic compounds including vinyl polymers, cellulose esters, synthetic rubbers, oils, fats, waxes and dyes.
  • Used in organic synthesis.
    Nitromethane reacts with halogen in presence of alkali to form trihalogen derivative, (e.g.,) with chlorine it forms chloropicrin, CCl3NO2 which is used as soil sterilizing agent.
  • Nitroethane is also used as a fuel additive and precursor to Rocket propellants.

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds

Question 20.
How is chloropicrin prepared ?
Answer:
Nitromethane on treatment with Cl2 in presence of NaOH gives chloropicrin.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 120
Chloropicrin (Trichloronitro methane)

VII. Three mark questions

Question 1.
How are nitro compounds classified?
Answer:
Classification of nitrocompounds
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 121

Question 2.
Convert (i) acetaldelyde into nitro ethane
(ii) acetone into 2-nitropropane
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 122

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 3.
Write about the hydrosysis of nitro alkanes
Answer:

  • Hydrolysis can be effected using conc: HCI or conc H2S04
  • Primary nitro alkanes give carboxylic acids.
  • Secondary nitro alkanes give ketones.
  • Tertiary nitro alkanes have no reaction.

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 123

Question 4.
Write about the classification of amines
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 124
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 125

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 5.
Write notes on structure of amines
Answer:

  • Like ammonia, nitrogen atom of amines is trivalent.
  • It carries a lone pair of electron.
  • Nitrogen atom is sp3 hybridised.
  • Out of four sp3 hybridised orbitals, three sp3 orbitals overlap with hydrogen or carbon of alkyl groups.
  • Fourth sp3 orbital contains a lone pair of electron.
  • Amines posses pyramidal geometry.
  • Due to the presence of lone pair of electron the C-N-H or C-N-C bond angle in less than the normal
    tetra hedral bond angle 109.5O
  • In trimethyl amine the C-N-C bond angle is 108O which is lower than the tetrahedral angle and higher than the H-N-H bond angle of 107O in ammonia.
  • This increase is due to the repulson between the bulky methyl groups.
    Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 126

Question 6.
Prepare aniline from
(i) chloro benzene
(ii) Phenol
(iii) nitro benzene
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 127

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 7.
Derive an expression for basic strength of amines.
Answer:

  • In aqueous solutions the following equilibrium exists.
  • It lies far to the left, hence amines are weak bases compared to NaOH.

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 128

  • Basicity constant K bis a measure of the extent to which the amine accepts the hydrogen – ion H+ from water.
  • Larger the value of Kb,, smaller is the value of pKb and stronger is the base.

Question 8.
An organic compound (A) on reduction gives compound (B). (B) on treatment with CHCl3 and alcoholic KOH gives (C) . (C) on catalytic reduction gives N-methyl aniline. Identify A, B, C and write its equation.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 129

CompoundName
ANitrobenzene
BAniline
CPhenyl Isocyanide

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 9.
Convert aniline into
(i) p-bromo aniline
(ii) p-nitro aniline
Answer:
(i) To get mono bromo derivatives, -NH2 is first acylated to reduce its activity
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 130
(ii) To get para product, the -NH2 group is protected by acetylation with acetic anhydride. Then the nitrated product is hydrolysed to form the product
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 131

VIII. Five mark questions

Question 1.
Explain structural isomerism exhibited by nitro alkanes
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 132

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 2.
Explain the reduction reactions of nitro benzene in various medium.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 133

Question 3.
An Organic compound (A) – C7H7NO on treatment with Br2 and KOH gives an amine (B), which gives carbylamine test. (B) upon diazotization to give (C). (C) on coupling_with p-cresol to give
compound (D). Identify A, B, C and D with necessary reaction.
Hofmann’s degradation reaction
Compound (A is Benzamide
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 134Compound B is Aniline (primary amine)
Diazotization reaction:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 135

 

CompoundName
ABenzamide
BAniline
CBenzene diazoniurn chloride
D2- phenylazo- 4- Methyl phenol

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 4.
Write a short note an Libermann’s nitroso test ?
Answer:
Alkyl and aryl secondary amines react with nitrous acid to give N-nitroso amine
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 136

Question 5.
An Organic compound (A) – CNCl react with methyl magnesium Bromide to give, compound B – (C2H3N). B – upon catalytic reduction to give compound C – (C2H7N). C gives carbylamine test, Identify compound A, B and C and write the reactions.
Answer:
An organic Compound A is cyanogon chloride react with methylmagnesium Bromide to give compound B – ethanenitrite – (C2H3N)
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 137
B – upon catalytic reduction to give compound C – Ethananxuie (C2H7N)
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 138

Compound

Name

Acyanogen chloride
Bethanenitrile
CEthanamine

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 6.
Explain the action of nitrous acid on three types of amine.
Answer:
Three classes of amines react differently with nitrous acid which is prepared in situ from a mixture of NaNO2 and Hcl.
(a) Primary amines:
Ethylamine reacts with nitrous acid to form ethyldiazonium chloride which is unstable. It is converted into ethanol by liberating N2.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 139

  • Aniline reacts with nitrous acid at low temperature (273 – 278K) to give benzene diazonium chloride.
  • Benzene diazonium chloride is stable for short time and slowly decomposes even at low
    temperatures.

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 140
b) Secondary amines:
Alkyl and aryl secondary amines react with nitrous acid to give N-nitroso amine, which is insoluble in water.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 141
c) Tertiary amines: :
Aliphatic tertiary amine reacts with nitrous acid to form trialkyl ammonium nitrite salt which is
insoluble in water.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 142
Aromatic tertiary amine reacts with nitrous acid at 273K to give p-nitroso compound
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 143

Question 7.
Explian:
(i) Sand mayer reaction
(ii) Gaftermann reaction
(iii) Baltz – schiemann reaction
Answer:
(i) Sand mayer reaction:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 144
(ii) Gattermann reaction:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 145
(iii) Baltz – Schiemann Reaction:
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 146

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 8.
Convert benzene diazonium chloride into:
(i) benzene
(ii) iodo benzene
(iii) phenol
(iv) nitrobenzene
(v) phenyl hydrazine
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 147

Question 9.
Explain coupling reactions of benzene diazonium chlorid
Answer:

  • Benzene diazonium chloride reacts with electron rich aromatic compounds like phenol, aniline to form brightly coloured azo compounds.
  • Coupling generally occurs at the para position.
  • If para position is occupied then coupling occurs at the ortho position.
  • Coupling tendency is enhanced if an electron donating group is present at the para-position to – N2 Cl group.
  • This is an electrophilic substitution.

Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 148

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 10.
Write notes on
(i) Thrope nitrile condensation
(ii) Levine and Hauser acetylation
Answer:
(i) Thrope nitrile condensation :
Self condensation of two molecules of alkylnitrile containing α-H atom in the presence of sodium to form iminonitrile.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 149
Nitriles containing α-H atom undergo condensation with esters in the presence of sodamide in ether to form ketonitriles,
This is known as Levine – Hauser acetylation.
This reaction involves the replacement of ethoxy (OC2H5) group by methylnitrile (-CH2 CN) group and is called as cyanomethylation reaction.
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 150

 Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds
Question 11.
An aromatic nitro compound (A) on reduction with Sn/HCl gives compound (B) C6H7N which on treatment with Benzoyl chloride in the presence of pyridine to give compound (C). Compound (B) on treatment with CH3Br to give compound (D) which further reacts with NaNO2 /HCl to give compound (E) with yellow oil liquid. Identify (A) to (E) and write the
reactions.
Compound (A) is nitro benzene on reduction with Sn/HCl gives compound (B) is Aniline
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 151
Compound (B) is Aniline reacts with Benzoyl chloride in the presence of pyridine to give compound
(C) is N – phenyl benzamide
Schotten- Baumann reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 13 Organic Nitrogen Compounds 152

CompoundName
ANitro benzene
BAniline
CN-phenyl benzamide
DN-Methylaniline
EN-Nitroso methyl phenylamine

Samacheer Kalvi 12th Commerce Guide Chapter 12 Employee Training Method

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 12 Employee Training Method Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 12 Employee Training Method

12th Commerce Guide Employee Employee Training Method Text Book Back Questions and Answers

I. Choose The Correct Answer.

Question1.
Off the job training is given
a) In the classroom
b) On off days
c) Outside the factory
d) In the playground
Answer:
c) Outside the factory

Question 2.
Vestibule training is provided
a) On the job
b) In the classroom
c) In a situation similar to the actual working environment
d) By the committee
Answer:
c) In a situation similar to the actual working environment

Samacheer Kalvi 12th Commerce Guide Chapter 12 Employee Training Method

Question 3.
Improves Skill Levels of employees to ensure better job performance
a) Training
b) Selection
c) Recruitment
d) Performance appraisal
Answer:
a) Training

Question 4.
When trainees are trained by a supervisor or by a superior at the job is called
a) Vestibule training
b) Refresher training
c) Roleplay
d) Apprenticeship training
Answer:
d) Apprenticeship training

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

Question 5.
……… is useful to prevent skill obsolescence of employees
a) Training
b) Job analysis
c) Selection
d) Recruitment
Answer:
a) Training

Question 6.
Training methods can be classified into training …….. training.
a) Job rotation and Job enrichment
b) On the Job and Off the Job
c) Job analysis and Job design
d) Physical and mental
Answer:
b) On the job and Off the Job

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

Question 7.
Elaborate discussion on specific topic comes under ……………. method of training.
a) Understudy
b) Coaching
c) Conferences
d) Counselling
Answer:
c) Conferences

II. Very Short Answer Questions.
Question 1.
What is meant by training?
Answer:
Training is the act of increasing/enhancing the new skill of problem-solving activity and technical knowledge of an employee for doing the jobs themselves.

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

Question 2.
What is Mentoring training method?
Answer:

  • Mentoring is the process of sharing knowledge and experience of an employee.
  • The focus in this training is on the development of attitude of trainees.
  • It is always done by senior person, it is also one-to-one interaction like coaching.
  • The mentor is responsible for the providing necessary support to trainees, and feedback

Question 3.
What is Roleplay?
Answer:
Under this method, trainees have explained the situation and assigned roles. They have to act out the roles assigned to them without any rehearsal.

Question 4.
State e-learning method.
Answer:

  • E-learning is the use to technological process of a traditional classroom or office.
  • It refers to online learning or web-based training.
  • Web-based training is any where and any time information can pass over the internet.

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

III. Short Answer Questions.

Question 1.
What is vestibule training?
Answer:
Vestibule training is the training of employees in an environment similar to the actual work environment. This type of training is given to avoid any damage or loss to machinery in the actual place by trainees.

Question 2.
What do you mean by on the job training?
Answer:

  • On the job training refers to the training which is given to the employee at the workplace by his immediate supervisor.
  • It is based on the principle of “Learning by Doing and Learning while Earning”.
  • It is suitable for imparting skills that can be learnt in a relatively short period of time.

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

Question 3.
Write down various steps in a training programme.
Answer:

Samacheer Kalvi 12th Commerce Guide Chapter Chapter 12 Employee Training Method 1Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process
Question 4.
Write short note on trainer and trainee.
Answer:
A person who is learning and practicing the skills of a particular job is called a trainee. Trainees should be selected on the basis of self-interest and recommendation by the supervisor.

A trainer is a person who teaches skills to the employee and prepares them for job activity. Trainers may be supervisor, co-workers, HR staffs, specialists in the other parts of the company, outside consultants.

IV. Long Answer Questions

Question 1.
Discuss various types of training.
Answer:
According to Edwin B. Flippo” Training is the act of increasing the knowledge and skills of an employee for doing particular jobs”. Training may be mainly divided into:

  • On the job training
  • Off the job training

(a) On the Job Training: On the job training refers to the training which is given to the employee at the workplace. The following are the on the job training methods.

  1. Coaching Method: In this method of training, the superior teaches or guides the new employee about the knowledge and skills.
  2. Mentoring Method: Mentoring is the process of sharing the knowledge and experience of an employee.
  3. Internship Training Method: A superior gives training to subordinates or understudy like an assistant to a manager.

(b) Off the Job Training: It is the training method wherein the workers learn the job role away from the actual workplace. The following are types of off the job training:

  1. Lecture Method: Under this method, trainees are educated about concepts, theories, principles in any particular area.
  2. E-learning Method: E-learning is the use of technological processes to access a traditional classroom or office.

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

Question 2.
What are the differences between on the job training and off the job training? [MALS] [CDM]
Answer:

Basis of difference

On the Job Training

Off the Job Training

1.MeaningThe employee learns the job in the actual work environmentIt involves the training of employees outside the actual workplace.
2. ApproachPractical ApproachTheoretical Approach
3. LocationAt the workplaceAway from the workplace
4. Suitable forImparted in case of Manufacturing or production-related jobsImparted for managerial and non-production related jobs.
5. CostIt is cheapest to carry outIt is the costliest to carry out. It requires expenses like separate training rooms, specialist, projectors etc.
6. Deals withDealing with the day-to-day requirement of job.It can more easily deal with groups of employees at the same time.
7. MethodsCoaching – Job rotation understudy Job instruction etcRole play – seminar – Field Trip – Lecture – Case study Demonstration etc.

Question 3.
Explain the benefits of training
Answer:
1. Benefits to the Organization:

  • It improves the skill of employees and increases productivity.
  • It reduces wastages of materials and idle time.
  • It minimizes the time for supervision.
  • It reduces the frequent accidents at the workplace and consequent payment of compensation.

2. Benefits to the Employees:

  • It increases the knowledge, skill of the employees.
  • It enables him to gain promotion in a shorter time.
  • It improves the employee’s productivity.
  • Employees get higher earnings through incentives and rewards.

3. Benefits of Customer:

  • Customers get better quality of product and service.
  • Customers get innovative products or value-added or feature-rich products.

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

12th Commerce Guide Employee Training Method Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
…………………………… training refers to the training which is given to the employee at the workplace by his immediate superior.
a) On the job
b) Off the job
c) PMKVY
d) NOTA
Answer:
a) On the job

Question 2.
…………………………… training is the training wherein the workers learn the job role away from the actual work floor.
a) Off the job
b) On the job
c) STEP
d) All of these
Answer:
a) Off the job

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

Question 3.
On the job training is given to the employee at the ____________
(a) college
(b) home
(c) workplace
(d) none of the above
Answer:
(c) workplace

Question 4.
Pick the odd one out:
a) Job Rotation
b) Job Instruction Techniques
c) Apprenticeship
d) Field Trip
Answer:
d) Field Trip

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

Question 5.
Which one of the following is not correctly matched?
a) Conference method – Seminar
b) Role play method – Assigned Roles
c) Coaching method – Subordinate Teaches Superior
d) Committee Assignment – Particular issue assigned
Answer:
c) Coaching method – Subordinate Teaches Superior

Question 6.
Which is the correct statement?
i) Training improves the employee’s productivity
ii) Employees get rewards and incentives.
iii) It decrease the morale of the employees
a) (i) is correct
b) (ii) is correct
c) (i) and (ii) are correct
d) (ii) and (iii) are correct

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

II. Match The Following.

Question 1.
Match List-I with List-II

List-I

List-II

(i) Job Rotation1. Guides the new employee about the skills.
(ii)  Job Instruction Techniques2. Sharing Knowledge and experience.
(iii)  Coaching3. How to perform his job
(iv)  Mentoring4. Periodical shifted

a) (i) – 4 (ii) – 3 (iii) -1 (iv) – 2
b) (i) – 3 (ii) – 2 (iii) – 4 (iv) -1
c) (i) – 2 (ii) – 4 (iii)-3 (iv) – 1
d) (i) – 1 (ii) – 3 (iii) – 2 (iv) – 4
Answer :
a) (i) – 4 (ii) – 3 (iii) -1 (iv) – 2

Question 2.

Match List-I with List-II

List-I

List – II

i. Roleplay1. Listen to the Lectures
ii. Field Trip2. Explained the situation
iii. Seminar3. undertake a journey
iv. E-learning4.  to access technological process

a) (i) – 3 (ii) – 4 (iii) – 2 (iv) – 1
b) (i) – 2 (ii) – 3 (iii) – 1 (iv) – 4
c) (i) – 4 (ii) – 3 (iii) – 2 (iv) – 1
d) (i) – 1 (ii) – 2 (iii) – 3 (iv) – 4
Answer :
b) (i) – 2 (ii) – 3 (iii) -1 (iv) – 4

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

III. Assertion and Reason.

Question 1.
Assertion (A): Under this method, trainees have explained the situation and assigned roles
Reason (R): They have to act out the roles assigned to them without any rehearsal
a) (A) is True (R) is False
b) (A) is False (R) is True
c) Both (A) and(R) are False
d) Both (A) and (R) are True
Answer :
d) Both (A) and (R) are True

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

IV. Very Short Answer Questions.

Question 1.
What is the Job Rotation method?
Answer:

  • Job Rotation is an important method of training for broadening the knowledge of executives.
  • Under this method, a trainee is periodically shifted from one work (Department) to another work (Department).

Question 2.
Write a note on internship training.
Answer:
When a superior gives training to subordinates, it is called internship training. The subordinates learn through experience and observation by participating in handling day-to-day problems.

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

Question 3.
What is the Internal ship [under study] Training method?
Answer:

  • A superior gives training to subordinates or understudy like an assistant to a manager or director.
  • The subordinate learns through experience and observation by participating in handling day-to-day problems.

Question 4.
What is the case study method?
Answer:

  • Trainees have described a situation which stimulate their interest to find solution.
  • They have to use their theoretical knowledge and practical knowledge to find solution to the problems presented.

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

Question 5.
What is the Group Discussion method?
Answer:

  • Under this method participants are divided into various groups.
  • Each group has to prepare solutions (issue provided to them) after deep discussion with their group members.

Question 6.
Demonstration Training method – Explain.
Answer:

  • It is a visual display of how something works or how to do something.
  • Demonstration involves showing by reason or proof explaining or making clear by use of examples or experiments.

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

V. Long Answer Questions.

Question 1.
Explain any three purposes of training.
Answer:
The purposes of training are as follows:

  1. Improved Quality of Work- Training helps to focus on a specific area and enables the employees to increase the quality of work carried out by them.
  2. Prevention of obsolescence- Training helps to learn more knowledge regarding the latest techniques and trends.
  3. Improved Safety Measures- Employees gain awareness about the risks involved in the job and safety measures to be adopted through a proper training program.

Samacheer Kalvi 12th Commerce Guide Chapter 11 Employee Selection Process

Question 2.
What are the steps in Designing a Training Programme?
Answer:
1) Whom to Train?

  • Training Department has to determine the candidates few whom Training should be imparted.

2) Who is a Trainee?

  • A person who is learning and practicing the skills of a particular job is called a “Trainee”.

3) Who are the Trainers?

  • A trainer is a person who teaches skills to the employee and prepare them for a job.

4) What method will be used?

  • The training segment should decide the appropriate method of training among the various methods of training available.

5) What should be the level of training?

  • Training Department should decide the level of training to be imparted to the employees.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 9 Electro Chemistry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 9 Electro Chemistry

12th Chemistry Guide Electro Chemistry Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

Question 1.
The number of electrons that have a total charge of 9650 coulombs is ………..
(a) 6.22 × 1023
(b) 6.022 × 1024
(c) 6.022 × 1022
(d) 6.022 × 10-34
Answer:
(c) 6.022 × 1022
Hint: IF = 96500 C = 1 mole of e = 6.023 × 1023 e
9650 C = \(\frac{6.22 \times 10^{23}}{96500} \times 9650\) = 6.022 × 1022

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
Consider the following half cell reactions:
Mn2+ + 2e → Mn E° = – 1.18V
Mn2+ → Mn3+ + e E = – 1.51V
The E for the reaction 3Mn2+ → Mn+2Mn3+, and the possibility of the forward reaction are respectively.
(a) 2.69V and spontaneous
(b) – 2.69 and non spontaneous
(c) 0.33V and Spontaneous
(d) 4.18V and non spontaneous
Answer:
(b) – 2.69 and non spontaneous
Hint: Mn+ + 2e → Mn(E0red) = 1.18V
2[Mn2+ → Mn3+ + e] (E0ox) = – 1.51V
3Mn2++ → Mn3+ + 2Mn3+ + (E0cell) = ?
E0red = (E0ox) + (E0cell)
= – 1.51 – 1.18 and non spontaneous
= – 2.69 V
Since E° is – ve ∆G is +ve and the given forward cell reaction is non – spontaneous.

Question 3.
The button cell used in watches function as follows
Zn(s) + Ag2O(s) + H2O(1) \(\rightleftharpoons\) 2Ag(s) + Zn2+(aq) + 2OH(aq) the half cell potentials are
Ag2O(s) + H2O(1) + 2e →2Ag(S) + 2OH(aq) E° = 034V. The cell potential will be
(a) 0.84V
(b) 1.34V
(c) 1.10V
(d) 0.42V
Answer:
(c) 1.10V
Hint: Anodic oxidation: (Reverse the given reaction)
(E0ox ) = 0.76V cathodic reduction
E0cell = (E0ox) + (E0red) = 0.76 + 0.34 = 1.1V

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 4.
The molar conductivity of a 0.5 mol dm-3 solution of AgNO3 with electrolytic conductivity of 5.76 × 10-3S cm-1at 298 K is ………….
(a) 2.88 S cm2 mo1-1
(b) 11.52 S cm2 mol-1
(c) 0.086 S cm2 mol-1
(d) 28.8 S cm2 mol-1
Answer:
(b) 11.52 S cm2 mol-1
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 1

Question 5.
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 2
Calculate A0HOAC using appropriate molar conductances of the electrolytes listed above at infinite dilution in water at 25°C.
(a) 517.2
(b) 552.7
(c) 390.7
(d) 217.5
Answer:
(c) 390.7
Hint:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 3

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 6.
Faradays constant is defined as
(a) charge carried by I electron
(b) charge carried by one mole of electrons
(c) charge required to deposit one mole of substance
(d) charge carried by 6.22 × 1010 electrons
Answer:
(b) charge carried by one mole of electrons
Solution:
IF = 96500 C = 1 charge of mole of e = charge of 6.022 × 1023 e

Question 7.
How many faradays of electricity are required for the following reaction to occur
MnO4 → Mn2+
(a) 5F
(b) 3F
(C) IF
(d) 7F
Answer:
(a) 5F
Hint:
7MnO4 + 5e → Mn2+ + 4H2O
5 moles of electrons i.e., 5F charge is required.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 8.
A current strength of 3.86 A was passed through molten Calcium oxide for 41 minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of Ca is 40g / mol and IF = 96500C).
(a) 4
(b) 2
(c) 8
(d) 6
Answer:
(b) 2
Solution:
m = ZIt
41mm 40sec = 2500 seconds
= \(\frac { 40 × 3.86 × 2500 }{ 2 x 96500 }\)
Z = \(\frac { m }{ n × 96500 }\) = \(\frac { 40 }{ 2 × 96500 }\)
= 2g

Question 9.
During electrolysis of molten sodium chloride, the time required to produce 0.1 mol of chlorine gas using a current of 3A is ………..
(a) 55 minutes
(b) 107.2 minutes
(c) 220 minutes
(d) 330 minutes
Answer:
(b) 107.2 minutes
Solution:
\(\frac { m }{ ZI }\) (mass of 1 mole of Cl2 gas = 71)
t = \(\frac { m }{ ZI }\) mass of 0.1 mole of Cl2 gas = 7.1 g mol-1)
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 4

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 10.
The number of electrons delivered at the cathode during electrolysis by a current of 1 A in 60 seconds is (charge of electron = 1.6 × 10-19C)
(a) 6.22 × 1023
(b) 6.022 × 1020
(c) 3.75 × 1020
(d) 7.48 × 1023
Answer:
(c) 3.75 × 1020
Solution:
Q = It
= 1A × 60S
96500 C charge 6.022 × 1023 electrons
60 C charge = \(\frac{6.022 \times 10^{23}}{96500} \times 960\)
= 3.744 × 1020 electrons

Question 11.
Which of the following electrolytic solution has the least specific conductance?
(a) 2N
(b) 0.002N
(c) 0.02N
(d) 0.2N
Answer:
(b) 0.002N
Solution:
In general, specific conductance of an electrolyte decreases with dilution. SO, 0.002N solution has least specific conductance.

Question 12.
While charging lead storage battery
(a) PbSO4 on cathode is reduced to Pb
(b) PbSO4 on anode is oxidised to PbO4
(c) PbSO4 on anode is reduced to Pb
(d) PbSO4 on cathode is oxidised to Pb
Answer:
(c) PbSO4 on anode is reduced to Pb.
Solution:
Charging: anode: PbSO4(s) + 2e → Pb (s) + SO4-2 (aq)
Cathode: PbSO4(s) + 2H2O (1) → PbO2 (s) + SO4-2 (aq) + 2e

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 13.
Among the following cells
I. Leclanche cell
II. Nickel – Cadmium cell
III. Lead storage battery
IV. Mercury cell
Primary cells are …………
(a) I and IV
(b) I and III
(c) III and IV
(d) II and III
Answer:
(a) I and IV

Question 14.
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
(a) Zinc is lighter than iron
(b) Zinc has lower melting point than iron
(c) Zinc has lower negative electrode potential than iron
(d) Zinc has higher negative electrode potential than iron
Answer:
(d) Zinc has higher negative electrode potential than iron
Solution:
E0Zn+|Zn = – 0.76V and E0Fe2+|Fe = 0.44V. Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc.

Question 15.
Assertion: pure iron when heated in dry air is converted with a layer Of rust.
Reason: Rust has the composition Fe3O4
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false.
Answer:
(d) both assertion and reason are false.
Solution:
Both are false

  1. Dry air has no reaction with iron
  2. Rust has the composition Fe2O3 x H2O

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 16.
In H2 – O2 fuel cell the reaction occur at cathode is ……….
(a) O2(g) + 2H2O (l) + 4e → 4OH(aq)
(b) H+(aq) + OH(aq) → H2O (l)
(c) 2H2(g) + O2(g) → 2H2O (g)
(d) H+ + e → 1/2 H2
Answer:
(a) O2(g) + 2H2O (l) + 4e → 4OH(aq)
Solution:
(a) O2(g) + 2H2O (l) + 4e → 4OH(aq)

Question 17.
The equivalent conductance of M/36 solution of a weak monobasic acid is 6mho cm2 and at infinite dilution is 400 mho cm2. The dissociation constant of this acid is ………….
(a) 1.25 x 10-16
(b) 6.25 x 10 -6
(c) 1.25 x 10-4
(d) 6.25 x 10-5
Answer:
(b) 6.25 x 10 -6
Hint: α = \(\frac { 6 }{ 400 }\)
Ka = α2C = \(\frac { 6 }{ 400 }\) x \(\frac { 6 }{ 400 }\) x \(\frac { 1 }{ 36 }\)
= 6.25 x 10-6

Question 18.
A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific conductance (K = 1.25 x 10-3 S cm-1 ) in the cell and the measured resistance was 800Ω at 250 C . The cell constant is,
(a) 10-1 cm-1
(b) 10-1 cm-1
(c) 1 cm-1
(d) 5.7 x 10-12
Answer:
(c) 1 cm-1
Hint: R = p.\(\frac { 1 }{ A }\)
Cell constant = \(\frac { R }{ ρ }\) = k.R \((\frac { 1 }{ ρ } =k)\) = 1.25 x 10-3 f-1cm-1 x 800Ω = 1cm-1

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 19.
Conductivity of a saturated solution of a sparingly soluble salt AB (1:1 electrolyte) at 298K is 1.85 x 10-5 S m-1. Solubility product of the saltAB at 298K (Λ0m)AB = 14 x 10-3 S m2 mol-1.
(a) 5.7 x 10-2
(b) 1.32 x 1012
(c) 7.5 x 10-12
(d) 1.74 x 10-12
Answer:
(d) 1.74 x 10-12
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 5

Question 20.
In the electrochemical cell: Zn|ZnSO4 (0.01M)||CuSO4 (1.0M)|Cu , the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that CuSO4 changed to 0.0 1M, the emf changes to E2. From the followings, which one is the relationship between E1 and E2?
(a) E1 < E2
(b) E1 > E2
(c) E2 = 0↑E1
(d) E1 = E2
Answer:
(b) E1 > E2
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 6
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 7

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 21.
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 8
Then the species undergoing disproportional is …………..
(a) Br2
(b) BrO4
(c) BrO3
(d) HBrO
Answer:
(d) HBrO
Hint:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 9
(Ecell)A = – 1.82 + 1.5 = – 0.32V
(Ecell)B = – 1.5 + 1.595 = + 0.095V
(Ecell)C = 1.595 + 1.0652 = – 0.529V
The species undergoing disproportionation is HBrO

Question 22.
For the cell reaction
2Fe3+(aq) + 2I(aq) → 2Fe2+ (aq) + I2(aq)
EC0cell = 0.24V at 298K. The standard Gibbs energy (∆ G0 ) of the cell reactions is …………
(a) – 46.32 KJ mol-1
(b) – 23.16 KJ mol-1
(c) 46.32 KJ mol-1
(d) 23.16 KJ mor-1
Answer:
(a) – 46.32 KJ mol-1
Solution:
n = 2; E0cell = 0.24V; ∆G = ?; F = 96500C
∆G0 = – nFE°
∆G0= – 2 × 96500 × 0.24
∆G0 = – 46320 J mol-1
∆G0 = – 46.32 KJ mol-1

Question 23.
A certain current liberated 0.504gm of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?
(a) 31.75
(b) 15.8
(c) 7.5
(d) 63.5
Answer:
(b) 15.8
Solution:
m1 = 0.504 g m2 = ?
e1 = 1.008 e2 = 31.77
\(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{e}_{1}}{\mathrm{e}_{2}}\)
∴ \(\mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{\mathrm{e}_{1}} \cdot \mathrm{e}_{2}=\frac{0.504 \times 31.77}{1.008}=15.885 \mathrm{~g}\)

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 24.
A gas X at 1 atm is bubble through a solution containing a mixture of 1MY and 1MZ-1 at 25°C . If the reduction potential of Z > Y> X, then
(a) Y will oxidize X and not Z
(b) Y will oxidize Z and not X
(c) Y will oxidize both X and Z
(d) Y will reduce both X and Z
Answer:
(a) Y will oxidize X and not Z
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 10
Zoxidises Yand X
Y oxidises X and reduces Z (does not oxidise Z)
X nduces Y and Z

Question 25.
Cell equation: A2+ + 2B → A2+ + 2B
A2+ + 2e → AE° = + 0.34V and log10 K = 15.6 at 300K for cell reactions find E° for
B1 + e → B
(a) 0.80
(b) 1.26
(c) – 0.54
(d) – 10.94
Answer:
(a) 0.80
Solution:
A2+ + 2B → A2+ + 2B E°cell
Half reaction anode A → A2+ + 2e
ox = -.034V [Given : A2+ + 2e → A E° = +0.34V]
Cathode 2B+ + 2e → 2B E°red = ?]
log10K = 156; T = 300K; n = 2;
F = 96500C;
R 8.314 JK-1 mol-1
∆G = – 2.303 RT log K
:. -nFE° = – 2.303 RT log K
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 11
∴ E°red = E°cell – E°oxid
= 0.4643 – (-0.34)
= 0.4643 + 0.34
E° 0.8043 V = 0.80V

II. Short Answer Questions

Question 1.
Define anode and cathode
Answer:

  1. Anode: The electrode at which the oxidation occur is called anode.
  2. Cathode: The electrode at which the reduction occur is called cathode.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
Why does conductivity of a solution decrease on dilution of the solution?
Answer:
Conductivity always decreases with decrease in concentration (on dilution of the solution) both for weak as much as for strong electrolytes. ¡t is because the number of ions per unit volume that carry the current is a solution decreases on dilution.

Question 3.
State Kohlrausch Law. How is it useful to determine the molar conductivity of weak electrolyte at infinite dilution.
Answer:
Kohlrausch’s law:
It is defined as, at infinite dilution the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions.

Determination of the molar conductivity of weak electrolyte at infinite dilution.
It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally. However, the same can be calculated using Kohlraushs Law. For example, the molar conductance of CH3COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCI, NaCI and CH3COONa.
Λ°CH3COONa = λ°Na+ + λ°CH3COONa …….(1)
Λ°HCl = λ°H+ + λ°Cl ………………(2)
Λ°NaCl = λ°Na+ + λ°Cl …………….(3)
Equation (1) + Equation (2) – Equation (3) gives,
(Λ°CH3COONa) + (Λ°HCl) – (Λ°NaCl) = λ°H+ + λ°CH3COONa = Λ°CH3COONa

Question 4.
Describe the electrolysis of molten NaCI using inert electrodes.
Answer:
1. The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they are connected to an external DC power supply via a key.

2. The electrode which is attached to the negative end of the power supply is called the cathode and the one is which attached to the positive end is called the anode.

3. Once the key is closed, the external DC power supply drives the electrons to the cathode and at the same time
pull the electrons from the anode.

Cell reactions:
Na+ ions are attracted towards cathode, where they combines with the electrons and reduced to liquid sodium.
Cathode (reduction)
Na+(I) + eNa(1)
E0 = – 2.7 1V
Similarly, Cl ions are attracted towards anode where they losses their electrons and oxidised to chlorine gas. Anode (oxidation)
2Cl(1) Cl2(g) + 2e
E° = – 1 .36V
The overall reaction is,
2Na+(l) + 2Cl(l) → 2Na(l) + Cl2(g) (g)
E0 = 4.07 V
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 12
The negative E° value shows that the above reaction is a non spontaneous one. Hence, we have to supply a voltage greater than 4.07V to cause the electrolysis of molten NaCI. In electrolytic cell, oxidation occurs at the anode and reduction occur at the cathode as in a galvanic cell, but the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode is -ve and anode is +ve.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 5.
State Faraday’s Laws of electrolysis.
Answer:
Faraday’s laws of electrolysis:
1. First law:
The mass of the substance (M) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell. M α Q

2. Second law:
When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents. M α Z

Question 6.
Describe the construction of Daniel cell. Write the cell reaction.
Answer:
The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.

Oxidation half cell:
The metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.

Reduction half cell:
A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.

Joining the half cells :
The zinc and copper strips are externally connected using a wire through a switch (K) and a load (ex : volt meter)

The electrolytic solution in the cathodic and anodic compartment are connected using an inverted U tube containing agar-agar gel mixed with an inert electrolyte like Kcl, Na2SO4 etc.,
This acts as the salt bridge.
When the switch (K) closes the circuit, the electrons flow from zinc strip to copper strip.
This is due to the following redox reactions.

Anodic oxidation:
Zinc strip acts as the anode.
Here oxidation occurs.
Zinc is oxidised to Zn2+ ions and electrons.
Zn2+ ions enter the solution and electrons enter the zinc metal strip; then flow through the external wire and enter the copper strip.
Electrons are liberated at zinc electrode and hence it is negative.
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 14

Cathodic reduction:
Copper strip acts as the cathode.
Here reduction occurs. .
Cu2+ ions are reduced to copper metal.
Cu2+ ions in the solution accepts electrons flowing through the circuit from zinc to copper strip and gets deposited on the electrode as copper metal.
Here electrons are consumed and hence it is positive.
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 13

Salt bridge:

The electrolytes present in two half cells are connected using salt bridge. To maintain electrical neutrality Cl- ions (from KCl) move from the salt bridge.
Due to cathodic reduction, cathodic compartment contains more number of SO42- ions in the solution and becomes
negatively charged. To maintain electrical neutrality K+ ions (from KCl) move from the salt bridge.

Completion of circuit:

Electrons flow from negatively charged zinc anode into positively charged copper cathode through the external wire.
At the same time anions move towards anode and cations move towards cathode.
This completes the circuit.

Consumption of electrodes:

As the Daniel cell operates, the mass of zinc electrode gradually decreases, while the mass of copper electrode increases.
Hence the cell will function until the entire metallic zinc is converted into Zn2+ ions or the entire Cu2+ ions are converted into metallic copper.
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 15

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 7.
Why is anode in galvanic cell considered to be negative and cathode positive electrode?
Answer:
A galvanic cell works basically in reverse to an electrolytic cell. The anode is the electrode where oxidation takes place, in a galvanic cell, it is the negative electrode, as when oxidation occurs, electrons are left behind on the electrode.

The anode is also the electrode where metal atoms give up their electrons to the metal and go into solution. The electron left behind on it render ¡t effectively negative and the electron flow goes from it through the wire to the cathode.

Positive aqueous ions in the solution are reduced by the incoming electrons on the cathode. This why the cathode is a positive electrode, because positive ions are reduced to metal atoms there.

Question 8.
The conductivity of a 0.01%M solution of a 1:1 weak electrolyte at 298K is 1.5 x 10-4 S cm-1.

  1. molar conductivity of the solution
  2. degree of dissociation and the dissociation constant of the weak electrolyte

Given that
λ0cation = 248.2 S cm2 mol-1
λ0anion = 51.8 S cm2 mol-1
Answer:
1. Molar conductivity
C = 0.01M
k = 1.5 × 10-4 S cm-1
(or)
K = 1.5 × 10-2 S m-1
\(\frac{\kappa \times 10^{-3}}{\mathrm{C}}\) S m-1 mol-1 m3 = \(\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01}\) S m2 mol-1
Λm = 1.5 × 10-3 s m-1

2. Degree of dissociation
α = \(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\infty}^{\circ}}\) (or) α = \(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\)
= (248.2 + 51.8)S cm2 mol-1
= 300 S cm2 mol-1
Ka = \(\frac{\alpha^{2} C}{1-\alpha}\)
= \(\frac{(0.05)^{2}(0.01)}{1-0.05}\)
Ka = 2.63 × 10-5

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 9.
Which of 0.1M HCl and 0.1 M KCl do you expect to have greater molar conductance and why?
Answer:
Compare to 0.1M HCl and 0.1 M KCl, 0.1M HCl has greater molar conductance.

  1. Molar conductance of 0.1M HCl = 39.132 × 10-3 S m2 mol-1.
  2. Molar conductance of 0.1 M KCl = 12.896 × 10-3 S m2 mol-1

Because, H+ ion in aqueous solution being smaller size than K+ ion and H+ ion have greater mobility than K ion. When mobility of the ion increases, conductivity of that ions also increases. Hence, 0. 1M HCI solution has greater molar conductance than 0.1 M KCI solution.

Question 10.
Arrange the following solutions in the decreasing order of specific conductance.

  1. 0.01M KCI
  2. 0.005M KCI
  3. 0.1M KCI
  4. 0.25 M KCI
  5. 0.5 M KCI

Answer:
0.005M KCl > 0.01M KCI > 0.1M KCI > 0.25KCl > 0.5 KCI.
Specific conductance and concentration of the electrolyte. So if concentration decreases, specific conductance increases.

Question 11.
Why is AC current used instead of DC in measuring the electrolytic conductance?
Answer:
1. AC current to prevent electrolysis of the solution.

2. If we apply DC current to the cell the positive ions will be attracted to the negative plate and the negative ions to the positive plate. This will cause the composition of the electrolyte to change while measuring the equivalent conductance.

3. So DC current through the conductivity cell will lead to the electrolysis of the solution taken in the cell. To avoid such a electrolysis, we have to use AC current for measuring equivalent conductance.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 12.
0.1M NaCI solution is placed in two different cells having cell constant 0.5 and 0.25cm-1 respectively. Which of the two will have greater value of specific conductance.
Answer:
The specific conductance values are same. Because the reaction (cation) of cell constant does not change.

Question 13.
A current of 1 .608A is passed through 250 mL of 0.5M solution of copper sulphate for 50 minutes. Calculate the strength of Cu2+ after electrolysis assuming volume to be constant and the current efficiency is 100%.
Answer:
Given, I = I .608A
t = 50 min (or) 50 × 60 = 3000 S
V = 250 mL
C = 0.5M
η = 100%
The number of Faraday’s of electricity passed through the CuSO4 solution
Q = It
= Q = 1.608 × 3000
Q = 4824C
Number of Faraday’s of electricity = \(\frac { 4824C }{ 96500C }\) = 0.5F
Electrolysis of CuSO4
Cu2+(aq) + 2e → Cu(s)
The above equation shows that 2F electricity will deposit 1 mole of Cu2+
0.5F electicity will deposit \(\frac { 1mol }{ 2F }\) × 0.5F = 0.025 mol
Initial number of molar of Cu2+ in 250 ml of solution = \(\frac { 1mol }{ 250mL }\) × 250mL = 0.125 mol
Number of molar of Cu2+ after electrolysis 0.125 – 0.025 = 0.1 mol
Concentration of Cu2+ = \(\frac { 0.1mol }{ 250mL }\) × 1000 mL = 0.4 M

Question 14.
Can Fe3+ oxidises Bromide to bromine under standard conditions?
Given: \(\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^{\mathrm{o}}=0.771\)
\(\mathrm{E}_{\mathrm{Br}_{2} \mid \mathrm{Br}}^{\mathrm{O}}=1.09 \mathrm{~V}\)
Answer:
Required half cell reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 16
E0cell = (E0ox) + (E0red) = – 1.09 + 0.771 = – 0.319V
We know that ∆G° = – nFE0cell
If E0cell is -ve; ∆G is +ve and the cell reaction is non-spontaneous.
Hence, Fe3+ cannot oxidise Bromide to Bromine.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 15.
Is it possible to store copper sulphate in an iron vessel for a long time?
Given:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 17
Answer:
E0cell = (E0ox) + (E0red) = 0.44 V + 0.34V = 0.78V
These +ve E0cell values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

Question 16.
Two metals M1 and M2 have reduction potential values of – xV and + yV respectively. Which will liberate H2 in H2SO4?
Answer:
Metals having negative reduction potential acts as powerful reducing agent. Since M1 has – xV, therefore M1 easily liberate H2 in H2SO4. Metals having higher oxidation potential will liberate H2 from H2SO4. Hence, the metal M1 having + xV, oxidation potential will liberate H2 from H2SO4.

Question 17.
Reduction potential of two metals M1 and M2 are
\(E_{\mathrm{M}_{1}^{2+}}^{0}=-2.3 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{M}_{1}^{2+}}^{0}=0.2 \mathrm{~V}\)
Predict which one Is better for coating the surface of iron.
Given:
\(\mathbf{E}_{\mathrm{Fe}^{2+} \mid \mathbf{F e}}^{0}=-\mathbf{0 . 4 4} \mathbf{V}\)
Answer:
Oxidation potential of M1 is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.

Question 18.
Calculate the standard emf of the cell: Cd | Cd2+|| Cu2+ | Cu and determine the cell reaction. The standard reduction potentials of Cu2+ | Cu and Cd2+ | Cd are 0.34V and – 0.40 volts respectively. Predict the feasibility of the cell reaction.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 18
emf is +ve, so ∆G is (-)ve, the reaction is feasible.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 19.
In fuel cell H2 and O2 react to produce electricity. In the process, H2 gas is oxidised at the anode and O2 at cathode. If 44.8 litre of H2 at 25°C and also pressure reacts in 10 minutes, what is average current produced? If the entire current is used for electro deposition of Cu from Cu2+, how many grams of Cu deposited?
Answer:
Oxidation at anode:
2H2(g) + 4OH (aq) → 4H2O (1) + 4e
1 mole of hydrogen gas produces 2 moles of electrons at 25°C and 1 atm pressure, 1 mole of hydrogen gas occupies = 22.4 litres
∴no. of moles of hydrogen gas produced = \(\frac{1 \mathrm{mole}}{22.4 \text { litres }}\) x 44.8 litres = 2 moles of hydrogen
∴2 of moles of hydrogen produces 4 moles of electro i.e., 4F charge. We know that Q = It
I = \(\frac { Q }{ t }\) = \(\frac{4 \mathrm{F}}{10 \mathrm{mins}}\) = \(\frac { 4×96500 }{ 10x60s }\)
I = 643.33 A
Electro deposition of copper
Cu2+(aq) + 2e → Cu(s)
2F charge is required to deposit
1 mole of copper i.e., 63.5 g
If the entire current produced in the fuel cell i.e., 4 F is utilised for electrolysis, then 2 x 63.5 i.e., 127.0 g copper will be deposited at cathode.

Question 20.
The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate and chromium nitrate respectively. If 2.935g of Ni was deposited in the first cell. The amount of Cr deposited in the another cell? Given: molar mass of Nickel and chromium are 58.74 and 52gm-1 respectively.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 19

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 21.
0.1M copper sulphate solution in which copper electrode is dipped at 25C. Calculate the electrode potential of copper.
Given : \(\left.\mathrm{E}_{\mathrm{Cu}^{2+}}^{\mathrm{o}}\right|_{\mathrm{Cu}}=0.34\)
Answer:
Given that
[Cu2+] = 0.1 M
E0Cu2+|Cu = 0.34
Ecell = ?
Cell reaction is Cu2+(aq) + 2e → Cu (s)
Ecell = E0 – \(\frac { 0.0591 }{ n }\) log \(\frac{[\mathrm{Cu}]}{\left[\mathrm{Cu}^{2+}\right]}\) = 0.34 – \(\frac { 0.0591 }{ 2 }\) log \(\frac { 1 }{ 0.1 }\)
= 0.34 – 0.0296 = 0.31 V

Question 22.
For the cell Mg(s) | Mg2+(aq) || Ag+ (aq) | Ag (s), calculate the equilibrium constant at 25°C and maximum work that can be obtained during operation of cell. Given:
\(\mathrm{E}_{\mathrm{Mg}^{2+} \mid \mathrm{Mg}}^{\mathrm{o}}=-2.37 \mathrm{~V}\) and \(\mathrm{E}_{\mathrm{Ag}^{2+} \mid \mathrm{Ag}}^{\mathrm{o}}=0.80 \mathrm{~V}\)
Answer:
Oxidation at anode
Mg → Mg2+ + 2e……………(1) (Eooxi = 2.37 V
Reduction at cathode
Ag+ + e → Ag……………(2) (Eored = 0.80 V
E0cell = (E0ox) + (E0red) = 2.37 + 0.80 = 3.17 V
Overall reaction
Mg + 2Ag+ → Mg2+ + 2Ag
∆G° = -nFE°
= – 2 × 96500 × 3.17
= – 6.118 × 105 J
We know that Wmax = ∆G°
Wmax = + 6.118 × 105 J
Relationship between ∆G° and Keq is,
∆G = – 2.303 RT logKeq
∆G = – 2.303 × 8.314 × 298 log Keq [25°C = 298 K]
log Keq = \(\frac{6.118 \times 10^{5}}{2.303 \times 8.314 \times 298}\) = \(\frac{6.118 \times 10^{5}}{5705.84}\)
log Keq = 107.223
Keq = Antilog (107.223)
Kc = 1.5849

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 23.
8.2 × 1012 litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of 2 × 106 Cs-1 at an appropriate voltage. How many years would it like to completely electrolyse the water in the lake. Assume that there is no loss of water except due to electrolysis.
Answer:
Hydrolysis of water:
At anode: 2H2O → 4H+ + O2 + 4e …………..(1)
At cathode: 2H2O + 2e → H2 + 2OH
Overall reaction: 6H2O → 4H + 4OH +2H2 + O2
(or)
Equation (1) + (2) × 2
= 2H2O → 2H2 + O2
According to Faraday’s Law of electrolysis, to electrolyse two mole of Water
(36g ≃ 36 mL. of H2O), 4F charge is required alternatively, when 36 mL of water is electrolysed,
the charge generated = 4 × 96500 C.
When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 20
Given that in 1 second, 2 × 106 C is generated therefore, the time required to generate
96500 × 1015 C is = \(\frac{1 \mathrm{S}}{2 \times 10^{6} \mathrm{C}}\) × 96500 × 1015C = 48250 × 109 S
Number of year = \(\frac{48250 \times 10^{9}}{365 \times 24 \times 60 \times 60}\) 1 year = 365 days
= 1.5299 × 106
= 365 × 24 hours
= 365 × 24 × 60 min
= 365 × 24 × 60 × 60 sec

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 24.
Derive an expression for Nernst equation.
Answer:
Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction.
Let us consider an electrochemical cell for which the overall redox reaction is,
Answer:
xA + yB = lC + mD
The reaction quotient Q is,
\(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\)
We know that,
∆G = ∆G0 + RT ln Q
∆G = – nFEcell
∆G0 = -nFE0cell
equation (1) becomes
– nFEcell = -nFE0cell + RT ln Q
Subsitute the Q value in equation (2)
– nFEcell = – nFE0cell + RT ln \(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\). ………..(3)
Divide the whole equation (3) by – nF
Ecell = E°cell – \(\frac { RT }{ nF }\) ln \(\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)\)
Ecell = E°cell – \(\frac { 2.303RT }{ nF }\) log \(\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)\) ……………(4)
This is called the Nernst equation.
At 25°C (298 K) equation (4) becomes,
Ecell = E°cell – \(\frac { 2.303×8.314×298 }{ nx96500 }\) log \(\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)\)
Ecell = E°cell – \(\frac { 0.0591 }{ n }\) log \(\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)\)

Question 25.
Write a note on sacrificial protection.
Answer:
In this method, the metallic structure to be protected is made cathode by connecting it with more active metal (anodic metal). So that all the corrosion will concentrate only on the active metal. The artificially made anode thus gradually gets corroded protecting the original metallic structure. Hence this process is otherwise known as sacrificial anodic protection. Al, Zn and Mg are used as sacrificial anodes.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 26.
Explain the function of H2 – O2 fuel cell.
Answer:
In this case, hydrogen act as a fuel and oxygen as an oxidant and the electrolyte is aqueous KOH maintained at 200°C and 20 – 40 atm. Porous graphite electrode containing Ni and NiO serves as the inert electrodes. Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively.
Oxidation occurs at the anode:
2H2(g)+ 4OH-(aq) → 4H2O(1) + 4e
Reduction occurs at the cathode O2(g) + 2 H2O(1) + 4e → 4 OH (aq)
The overall reaction is 2H2(g) + O2(g) → 2H2O(1)
The above reaction is the same as the hydrogen combustion reaction, however, they do not react directly ie., the oxidation and reduction reactions take place separately at the anode and cathode respectively like H2 – O2 fuel cell. Other fuel cell like propane – O2 and methane O2 have also been developed.
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 21

Question 27.
Ionic conductance at infinite dilution of Al3+ and SO42- are 189 and 160 mho cm2 equiv-1. Calculate the equivalent and molar conductance of the electrolyte Al2(SO4) at infinite dilution.
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 22
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 23

III. Evaluate Yourself

Question 1.
Calculate the molar conductance of 0.01M aqueous KCI solution at 25°C . The specific conductance of KCl at 25°C is 14.114 x 10-2 Sm-1.
Answer:
Concentration of KCI solution = 0.01 M.
Specific conductance (K) = 14.114 × 10-2 S m-1
Molar conductance (Λm) = ?
Λm = \(\frac{\kappa \times 10^{-3}}{M}\) = \(\frac{14.114 \times 10^{-2} \times 10^{-3}}{0.01}\)
S m-1 mol-1 m3
Λm = 14.114 × 10-5 × 102 = 14.114 × 10-3 Sm2 mol-1

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
The emf of the following cell at 25°C is equal to 0.34v. Calculate the reduction potential of copper electrode.
Pt(s) | H2(g,1atm) | H+ (aq,1M) || Cu2+(aq,1M) | Cu(s)
Answer:
SHE Value is zero
cell = E°R – E°L
= 0.34 – 0 = 0.34V
The reduction potential of copper electrode = 0.34V

Question 3.
Using the calculated emf value of zinc and copper electrode, calculate the emf of the following cell at 25°C.
Zn (s) | Zn2+ (aq, 1M) || Cu2+(aq, 1M) | Cu (s)
Answer:
cell = E°R – E°L
Ezn/zn2+ = 0.76V
ECu/Cu2+ = 0.76V
cell = 0.76 – (- 0.34V)
cell = 0.76 – (- 0.34)
cell = + 1.1 V

Question 4.
Write the overall redox reaction which takes place in the galvanic cell,
Pt(s) | Fe2+(aq),Fe2+(aq) || MnO4(aq), H+(aq), Mn2+(aq) || Pt(s)
Answer:
At Anode half cell – 5Fe2+(aq) → 5Fe3+(aq) + 5e
At cathode half cell – MnO4(aq) + 8H+(aq) + 5e → Mn2+(aq) + 4H2O(1)
Overall redox reaction – 5Fe2+(aq) + MnO4(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(1)

Question 5.
The electrochemical cell reaction of the Daniel cell is
Zn (s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
What is the change in the cell voltage on increasing the ion concentration in the anode compartment by a factor 10?
Answer:
Zn (s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
ln the case E°cell = 1.1V
Reaction quotient Q for the above reaction is, Q = \(\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\)
Ecell = E°cell – \(\frac { 0.0591 }{ n }\) log \(\frac{\left[Z n^{2+}\right]}{\left[C u^{2+}\right]}\)
If suppose concentration of Cu2+ is 1 .OM then the concentration of Zn2+ is 10M (why because,
ion concentration in the anode compartment increased by a 10 factor)
Ecell = 1.1 – \(\frac { 0.0591 }{ n }\) log \((\frac { 10 }{ 1 })\)
= 1.1 – 0.02955             ……………….(1)
= 1.070 V (cell voltage decreased)
Thus, the initial voltage is greater than E° because Q < 1. As the reaction proceeds, [Zn2+] in the anode compartment increases as the zinc electrode dissolves, while [Cu2+] in the cathode compartment decreases as metallic copper is deposited on the electrode.

During this process, the Q = [Zn2+] [Cu2+] steadily increases and the cell voltage therefore steadily decreases. [Zn2+] will continue to increase in the anode compartment and [Cu2+] will continue to decrease in the cathode compartment. Thus the value of Q will increase further leading to a further decrease in value.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 6.
A solution of a salt of metal was electrolysed for 150 minutes with a current of 0.15 amperes. The mass of the metal deposited at the cathode is 0.783g. calculate the equivalent mass of the metal.
Answer:
Given,
I = 0.15 amperes
t = 150 mins
= t = 150 × 6Osec
= t = 9000sec
Q = It
= Q = 0.15 × 9000 coulombs
= Q = 1350 coulombs
Hence, 135 coulombs of electricity deposit is equal to 0.783g of metal.
96500 coulombs of electricity, \(\frac { 0.783 × 96500 }{ 1350 }\) = 55.97 gm of metal
Hence equivalent mass of the metal is 559.7

12th Chemistry Guide Electro Chemistry Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

Question 1.
When electric current is passed through an electrolytic solution, charge is carried by
a) electrons
b) ions
c) atoms
d) molecules
Answer:
b) ions

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
Ohm’s law
a) I = VR
b) I = \(\frac{R}{V}\)
c) I = \(\frac{V}{R}\)
d) V = \(\frac{I}{R}\)
Answer:
c) I = \(\frac{V}{R}\)

Question 3.
The cell constant of a conductivity cell with platinum electrodes at a distance of 0.5 cm and area of cross section 5 cm2 is
a) 10 cm-1
b) 0.1 cm-1
c) 2.5 cm-1
d) 0.25 cm-1
Answer:
b) 0.1 cm-1
Cell constant = \(\frac{l}{A}=\frac{0.5}{5}\) = 0.1cm-1

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 4.
Specific conductance (or) conductivity =
a) 1/ρ
b) \(\frac{1}{R} \frac{A}{l}\)
c) \(\mathrm{R} \frac{\mathrm{A}}{l}\)
d) both (a) and (b)
Answer:
d) both (a) & (b)

Question 5.
For 1 : 1 electrolyte like NaCl equivalent conductance is
a) less than molar conductance
b) greater than molar conductance
c) equal to molar conductance
d) zero
Answer:
c) equal to molar conductance

Question 6.
The relationship between molar conductance and equivalent conductance of 1M H2SO4 is
\(a) \Lambda_{\mathrm{m}}=\frac{\Lambda}{2}
b) \Lambda_{\mathrm{m}}=2 \Lambda
c) \Lambda_{\mathrm{m}}=\Lambda
d) \frac{\Lambda_{\mathrm{m}}}{\Lambda}=0\)
Reason :
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 24
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 25

Question 7.
As concentration of the electrolyte decreases the specific conductance of the solution
a) decreases
b) increases
c) remains the same
d) becomes zero
Answer:
a) decreases

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 8.
As concentration of the electorlyte decreases the molar conductance and equivalent conductance of the solution
a) decreases
b) increases
c) remains the same
d) becomes zero
Answer:
b) increases

Question 9.
In the measurement of conductivity of an electrolyte using wheatstone bridge arrangement which is correct?
a) PQ = RS
\(b) \frac{Q}{P}=\frac{R}{S}
c) \frac{P}{Q}=\frac{R}{S}
d) \frac{P}{Q}=\frac{S}{R}\)
Answer:
c) \(\frac{P}{Q}=\frac{R}{S}\)

Question 10.
Which among the following solution of NaCl has the maximum molar conductance value?
a) 10-4 M
b) 10-3 M
c) 10-2 M
d) 10-1 M
Answer:
a) 10-4 M
Reason : As dilution increases, molar conductance increases.

Question 11.
Molar conductance of an electrolyte approaches a limiting value in
a) highly concentrated solution
b) concentrated solution
c) very dilute solution
d) dilute solution
Answer:
c) very dilute solution

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 12.
For a weak electrolyte there is a sudden increase in molar conductance as the concentration approaches
a) infinity
b) maximum value
c) zero
d) negative value
Answer:
c) zero

Question 13.
Limiting molar conductivity values of strong electrolytes can be determined by
a) Kohlrausch’s law
b) Faradays’ law
c) Nernst equation
d) extrapolating the straight line
Answer:
d) extrapolating the straight line

Question 14.
Limiting molar conductivity values of weak electrolytes can be determined by
a) Kohlrausch’s law
b) Faradays’ law
c) Nernst equation
d) extrapolating the straight line
Answer:
a) Kohlrausch’s law

Question 15.
Debye and Huckel derived an equation for the conductivity of strong electrolytes by assuming
a) partial dissociation
b) incomplete dissociation
c) complete dissociation
d) negligible dissociation
Answer:
c) complete dissociation

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 16.
The basis for Kohlrausch’s law is
a) molar conductance
b) specific conductance
c) specific resistance
d) limiting molar conductance
Answer:
d) limiting molar conductance

Question 17.
Degree of dissociation of weak electrolytes can be calculated using the expression
\(a) [latex]\alpha=\frac{\mathrm{K}_{\mathrm{m}}}{\mathrm{K}_{\mathrm{m}}^{\mathrm{o}}}
b) \quad \alpha=\frac{\mathrm{K}_{\mathrm{m}}^{\mathrm{O}}}{\mathrm{K}_{\mathrm{m}}}
c) \alpha=\frac{\Lambda_{\mathrm{m}}^{\mathrm{o}}}{\Lambda_{\mathrm{m}}}
d) \alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\mathrm{O}}}\)
Answer:
d) \(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\mathrm{O}}}\)

Question 18.
The device which converts chemical energy into electrical energy is known as
a) battery
b) Galvanic cell
c) Voltaic cell
d) all the above
Answer:
d) all the above

Question 19.
The device which converts electrical energy into chemical energy is known as
a) Galvanic cell
b) Voltaic cell
c) electrolytic cell
d) all the above
Answer:
c) electrolytic cell

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 20.
In Daniel cell zinc undergoes
a) oxidation
b) reduction
c) hydrolysis
d) galvanisation
Answer:
a) oxidation

Question 21.
In Daniel cell copper ions undergo
a) oxidation
b) reduction
c) hydrolysis
d) galvanisation
Answer:
b) reduction

Question 22.
In Daniel cell the anode and cathode half cells are respectively
a) Cu | Cu2+, Zn | Zn2+
b) ZZn | Zn2+, Cu | Cu2+
c) Zn | Zn2+, Cu2+ | Cu
d) Cu2+ | Cu, Zn2+ | Zn
Answer:
c) Zn | Zn2+, Cu2+ | Cu

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 23.
In Daniel cell, the number of electrons transfered in the redox reaction
Zn + CuSO4 → ZnSO4+ Cu is
a) 1
b) 2
c) 3
d) 4
Answer:
b) 2

Question 24.
Which of the following statement is correct with respect to electrolytic conductance? ULffi
a) Conductivity increases with the decrease ’ in Viscosity
b) Conductivity increases with increase in temperature
c) Molar conductance of a solution decreases with increase in dilution
d) Conductance decrease with increase in temperature.
Answer:
c) Molar conductance of a solution decreases with increase in dilution

Question 25.
In Daniel cell the Electronic solutions in the two half cells are connected using a
a) wire
c) copper strip
b) zinc strip
d) salt bridge
Answer:
d) salt bridge

Question 26.
A salt bridge is containing an inverted U tube
a) agar – agar gel with Kcl
b) agar – agar gel with Na2SO4
c) both (a) & (b)
d) none of the above
Answer:
c) both (a) & (b)

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 27.
The general representation of a fuel cell is
a) Fuel / Electrode / Electrolyte / Electrode / Oxidant
b) Oxidant /Electrode / Electrolyte / Fuel
c) Fuel / Electrode / Electrolyte / Electrode / Reductant
d) Oxidant /Electrode / Electrolyte /Reductant
Answer:
a) Fuel / Electrode / Electrolyte / Electrode / Oxidant

Question 28.
Which of the following is correct?
a) 1 C = 1J × 1V
b) 1V = 1C × 1J
c) 1J = 1C × 1V
d) 1V = \(\frac{1 C}{1 \mathrm{~J}}\)

Question 29.
Ecell is equal to
a) (Eox)anode + (Ered)cathode
b) (Ered)cathode – (Ered)anode
c) (Ered)anode + (Ered)cathode
d) both (a) & (b)
Answer:
d) both (a) & (b)

Question 30.
The emf of standard hydrogen electrode is
assigned an arbitrary value of
a) zero
b) one
c) infinity
d) negative
Answer:
a) zero

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 31.
Electrical charge carried by one electron is
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 26
Answer:
d) Both (a) & (b)

Question 32.
A cell reaction is spontaneous when AG° and ocell E are respectively
a) + ve and – ve
b) – ve and + ve
c) 1 and 0
d) 0 and 1
Answer:
b) – ve and + ve

Question 33.
Electrolysis is a
a) photochemical reaction
b) spontaneous reaction
c) non-spontaneous reaction
d) substitution reaction
Answer:
c) non-spontaneous reaction

Question 34.
If the atomic mass of an ion Mn+ is A, its electrochemical equivalent is
a) \(\frac{\mathrm{nA}}{\mathrm{F}}\)
b) \(\frac{\mathrm{F}}{\mathrm{nA}}\)
c) \(\frac{\mathrm{nF}}{\mathrm{A}}\)
d) \(\frac{A}{n F}\)
Answer:
d) \(\frac{A}{n F}\)
Reason Z = \(\frac{\text { Atomic mass }}{\text { Valency } \times \text { Faraday constant }}\)

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 35.
When Q coulomb electric charge is passed through a series of aqueous solution of AgNO3, AlCl3 and CUSO4, the increasing order of mass of the metals liberated is
a) Ag > Cu > Al
b) AZ < Cu < Ag
c) Cu > Ag > Al
d) Ag < Cu < Al
Ans:
b) AZ < Cu < Ag
Reason: m a equivalent mass eAl < eCu < eAg \ mAl < mCu < mAg

Question 36.
Leclanche cell is a
a) primary battery
b) secondary battery
c) rechargeable battery
d) electrolytic cell
Answer:
a) primary battery

Question 37.
Lithium ion battery is a
a) primary battery
b) secondary battery
c) non – rechargeable battery
d) none of the above
Answer:
b) secondary battery

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 38.
In cellular phones/ laptop Computers the battery used is
a) Lechlanche cell
b) Lead storage battery
c) Lithium – ion battery
d) H2 – O2 fuel cell
Answer:
c) Lithium – ion battery

Question 39.
Rusting of iron is
a) a hydrolysis reaction
b) an electrochemical redox reaction
c) an oxidation reaction
d) a reduction reaction
Answer:
b) an electrochemical redox reaction

Question 40.
The formula of rust is
a) PeO
b) Fe3O4
c) Fe (OH)3
d) Fe2O3. × H2O
Answer:
d) Fe2O3. × H2O

Question 41.
The most convenient method to protect the bottom of ship made of iron is
a) coating it with red lead oxide
b) white tin plating
c) connecting it with Mg block
d) connecting it with Pb block
Answer:
c) connecting it with Mg block
Reason: As Mg is more reactive metal than iron, it gets corroded in preference to iron. This is known as cathodic protection.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 42.
Standard reduction potential of three metals A, B and C are – 1.5 V, + 1 V and – 2 V respectively. The decreasing order of reducing power of these metals is
a) B > C > A
b) A > B > C
c) C > A > B
d) C > B > A
Answer:
c) C > A > B
– 2 V – 1.5 V + 1 V
Reason: lower the E° value higher is the reducing power.

Question 43.
Limiting molar conductivity of NH4OH (A?n)NH4OH is equal to
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 27
Answer:
d

Question 44.
A hydrogen gas electrode in made by dipping platinum wire in a solution of pH = 10 and by passing hydrogen gas around the platinum wire at one atmosphere pressure. The oxidation potential of the electrode would be
a) 0.59 V
b) 0.118 V
c) 1.18 V
d) 0.059 V
Answer:
a) 0.59 V

Reason: H2 —> 2H+ + 2e
Concentration 1 atm
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 28

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 45.
A button cell used in watches functions as following:
Zn(s) + Ag2O + H2O(l) ⇌ 2Ag(s)+ Zn2+(aq) + 2OH(aq)
If half cell potentials are :
Zn2+(aq) + 2e(aq) → Zn(s) ; E° = -0.76 V
Ag2O + H2O(l) + 2e → 2Ag(s) + 2OH(aq) E° = – 0.34 V
The cell potential will be
a) 0.42 V
b) 0.84 V
c) 1.34 V
d)1.10V
Answer:
d) 1.10 V
Reason : Ecell – (Eox)anode + (Ered)cathode
= + 0.76 + 0.34 = + 1.10V

Question 46.
How many grams of cobalt metal will be deposited when a solution of Cobalt (II) chloride is electrolysed with a current of 10 amperes for 109 minutes? (1 Faraday = 96500 C; Atomic mass of (Co = 59 u)
a) 0.66
b) 4.0
c) 20.0
d) 40.0
Answer:
c) 20.0
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 29

Question 47.
Consider the half-cell reactions
Mn2+ + 2e → Mn; E° = – 1.18 V
Mn2+ → Mn3+ + e ; E° = – 1.51 V
The E° for the reaction 3Mn2+ → Mn + 2Mn3+ and the possibility of the forward reaction are respectively.
a) – 2.69 and not possible
b) – 4.18 V and possible
c) + 0.33 V and possible
d) + 2.69 V and not possible
Answer:
a) – 2.69 and not possible
Reason : E° = E°oxid + E°red
= – 1.51 +(-1.18) E°= – 2.69V
= 1.51 -1.18 ΔG°= -nFE°,
E° is negative, ΔG° is positive and hence the forward reaction is not possible.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 48.
The electrolyte used in Lechlanche cell is
a) Paste of KOH and ZnO
b) 38 % solution of H2SO4
c) Moist of paste NH4Cl and ZnCl2
d) Moist sodium hydroxide
Answer:
c) Moist of paste NH4Cl and ZnCl2

Question 49.
During the electrolysis of fused NaCl, which reaction occurs at anode.
a) chloride ions are oxidised
b) chloride ions are reduced
c) Sodium ions are oxidised
d) sodium ions are reduced
Answer:
a) chloride ions are oxidised

Question 50.
Amount of electricity that can deposit 108 gm of silver from AgNO3 solution is
a) 1 ampere
b) 1 coulomb
c) 1 Faraday
d) 1 Volt
Answer:
c) 1 Faraday
Reason: 1 g eq.mass of Ag (108 g) will be liberated by 1 Faraday

II. Match the following:

Question 1.

AB
i) Specific conductancea) Sm2 mol-1
ii) Conductanceb) m-1
iii) Cell constantc) Sm2g equiv-1
iv) Molar conductanced) Sm-1
v) Equivalent conductancee) S

Answer:
d) Sm-1
e) S
b) m-1
a) Sm2 mol-1
c) Sm2g equiv-1

III. Pick out the correct statements

Question 1.
i) For strong electrolyte the plot ∧mVs√C is not a linear one.
ii) For a strong electrolyte, at high concentration the number of constituent ions in a given volume is high.
iii) At high concentration the ions experience a viscous drag due to greater solvation.
iv) At infinite dilution the ions are so close and the interaction between them becomes significant.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)

Correct Statement:
i) For strong electrolytes the plot ∧mVs√C is a straight line.
iv) At infinite dilution the ions are so apart and the interaction between them becomes insignificant. ‘

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
i) In Daniel cell electrons are liberated at : inc electrode and hence it is negative.
ii) Electrons flow from copper cathode to zinc anode.
iii) Through salt bridge ions can move into (or) out of the half cells.
iv) The zinc and copper strips are connected externally through salt bridge.

a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (i) & (iv)
Answer:
c) (i) & (iii)

Correct Statement:
(ii) Electrons flow from, zinc anode to copper cathode.
(iv) The zinc and copper strips are connected externally through a wire.

Question 3.
i) Electrolysis is carried out in an electrolytic cell by connecting it to an AC power supply.
ii) The device which is used to carry out the electrolysis is called the electrolytic cell.
iii) In the electrolytic cell cathode is + ve and anode is – ve.
iv) The electrochemical process occuring in the electrolytic cell and galvanic cell are the reverse of each other.
a) (i) & (ii) b) (ii) & (iii)
c) (iii) & (iv) d) (ii) & (iv)
Answer:
d) (ii) & (iv)

Correct Statement :
(i) Electrolysis is carried out in an electrolytic cell by connecting it to a DC power supply.
(iii) In the electrolytic cell cathode is – ve and anode is + ve.

Question 4.
i) The amount of substance deposited when 1 ampere current passed for 1 second is its equivalent mass.
ii) When same quantity of electric charge is passed through different solutions the masses liberated are inversely proportional to their equivalent masses.
iii) One gram equivalent mass of any substance can be liberated by passing one Faraday electricity.
iv) One Faraday of electric charge is carried by Avogadro number of electrons.

a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)

Correct Statement :
(i) The amount of substance deposited when 1 ampere current passed for 1 second is its electrochemical equivalent.
(ii) When same quantity of electric charge is passed through different solutions the masses liberated are directly proportional to their equivalent masses.

IV. Pick out the incorrect statements

Question 1.
i) At constant temperature, the current flowing through the cell (I) is directly proportional to the resistance of the cell.
ii) Resistance is the opposition that a cell offers to the flow of electric current through it.
iii) The conductivity of the electrolyte is measured using a conductivity cell.
iv) Resistance of an electrolytic solution is directly proportional to the cross sectional area and
inversely proportional to the length.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)

Correct Statement :
(i) At constant temperature, the current flowing through the cell (I) is directly proportional to the voltage across the cell (V).
(iv) Resistance of an electrolytic solution is directly proportional to the length and inversely proportional to the cross sectional area.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
i) The conductance of lm3 electrolytic solution is called the specific conductance.
ii) Conductivity increases with increase in viscosity.
iii) Specific conductivity increases with increase in dilution.
iv) Conductivity of a cell can be measured using wheatstone bridge arrangement.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)

Correct Statement:
(ii) Conductivity increases with decrease in viscosity.
(iii) Specific conductivity decreases with increase in dilution.

Question 3.
i) Molar conductivity is due to the independent migration of cations in one direction and anions in the opposite direction.
ii) At infinite dilution each constituent ion of the electrolyte makes a definite contribution towards the molar conductance.
iii) It is possible to determine the molar conductance at infinite dilution for weak electrolytes experimentally.
iv) The solubility product of a sparingly soluble salt can not be determined using conductivity measurements.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)

Correct Statement:
(iii) It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally.
(iv) The solubility product of a sparingly soluble salt can be determined using conductivity measurements.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 4.
i) Corrosion of iron is called rusting.
ii) Coating of iron over zinc is called galvanisation.
iii) The standard reduction potential (E°) is a measure of the oxidising tendency of the species.
iv) Greater the E° values in the spectrochemical series greater is the tendency of the species to donate electrons and undergo oxidation.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
c) (ii) & (iv)

Correct Statement:
(ii) Coating of zinc over iron is called galvanisation.
(iv) Greater the E° values in the spectrochemical series greater is the tendency of the species to accept electrons and undergo reduction.

V. Assertion and reason

Question 1.
Assertion (A): If the temperature of the electrolytic solution increases, conductance also increases.
Reason (R): As temperature increases, the kinetic energy of the ions decreases and the attractive force between the ions increases.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
c) A is correct, but R is wrong. Correct Reason R: As temperature increases, the kinetic energy of the ions increases and the attractive force between the ions decreases.

Question 2.
Assertion (A): In the wheatstone bridge arrangement AC power supply is used in the measurement of conductivity of an electrolyte in conductivity cell.
Reason (R): DC power supply will lead to the electrolysis of the solution taken in the cell.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
a) Both A& R are correct, R explains A

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 3.
Assertion (A): In lead acid storage battery, when, a potential greater than 2V is applied across electrodes the cell reactions of discharge process are reverse.
Reason (R): The electrochemical reactions which take place in galvanic cell may be reversed by applying a potential slightly greater than the emf generated by the cell.

a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
a) Both A & R are correct, R explains A

Question 4.
Assertion (A): Magnesium is used as a sacrificial anode to protect iron from rusting. Reason (R): The reduction potential of magnesium is higher than iron hence magnesium has lesser tendency to undergo corrosion in preference to iron.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
c) A is correct, but R is wrong Correct Reason R: The reduction potential of magnesium is less than iron and hence magnesium has higher tendency to undergo corrosion in preference to iron.

VI. Two Mark Question

Question 1.
State Ohm’s law.
Answer:
At constant temperature, the current flowing through the cell (1) is directly proportional to the voltage across the cell (V).
ie) I ∝ V (or) I = V/R (or) V = IR

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
Define resistivity.
Answer:
Resistivity is defined as the resistance of an electrolyte confined between two electrodes having unit cross sectional area (1m2) and are separated by unit distance (1m).

Question 3.
Define specific conductance (or) conductivity.
Answer:
Specific conductance is defined as the conductance of a cube of an electrolytic solution of unit dimension, (lm3).

Question 4.
Define molar conductivity (∧m)
Answer:
Conductivity of a cell in which the electrodes are lm apart containing 1 mole of electrolyte in V m3 of electrolytic solution is called as molar conductivity.
Relation between specific conductance and molar conductance.
\(\Lambda_{\mathrm{m}}=\frac{\mathrm{K}\left(\mathrm{sm}^{-1}\right) \times 10^{3}}{\mathrm{M}} \mathrm{mot}^{-1} \mathrm{~m}^{3}\)

Question 5.
Define equivalent conductance (∧)
Answer:
Conductivity of a cell in which the electrodes are lm apart and containing 1 gram equivalent of electrolyte in Vm3 of electrolytic solution is called as equivalent conductance (∧)
\(\Lambda=\frac{\kappa \times 10^{-3}}{N} \mathrm{Sm}^{2} \mathrm{~g} \text { equiv }^{-1}\)
K = Specific conductance
N Normality

Question 6.
Define Electrochemical equivalent.
Answer:
Electrochemical equivalent is defined as the amount of substance deposited at the electrode by the passage of 1 coulomb charge.
Electrochemical equivalent
Z = \(\frac{\text { Equivalent mass }}{96500}\)

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 7.
What is the role of salt bridge in Galvanic cell?
Answer:
To maintain the electrical neutrality in both the compartments, the non reactive anions Cl (from KCl taken in the salt bridge) move from the salt bridge and enter into the anodic compartment, at the same time some of the K+ ions move from the salt bridge into the cathodic compartment.

Question 8.
What are the applications of Kohlrausch’s law?
Answer:
It is used to calculate the molar conductance of a weak electrolyte at infinite dilution.
\((\mathrm{eg}) \Lambda_{\mathrm{CH}_{\mathrm{COOH}}}^{\mathrm{o}}=\Lambda_{\mathrm{CH}_{2} \mathrm{COONa}}^{\mathrm{o}}+\Lambda_{\mathrm{HCl}}^{\mathrm{o}}-\Lambda_{\mathrm{NaCl}}^{\circ}\)
It is used to calculate the degree of dissociaton of weak electrolytes.
\(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\mathrm{o}}}\)
It is used to calculate the solubility product of sparingly soluble salts.
For 1:1 electrolyte
\(K_{S P}=\left(\frac{K \times 10^{-3}}{\Lambda_{m}^{o}}\right)^{2}\)

Question 9.
Define emf of a-cell.
Answer:

  • The force that pushes the electrons away from the anode and pulls them towards the cathode is called the electromotive force (emf) or the cell potential.
  • SI unit of emf is volt (V).

Question 10.
Define Joule.
Answer:
When there is one volt difference in electrical potential between the anode and cathode, the energy released for each coulomb of charge that moves between them is defined as one joule of energy.
1 J = 1 C × 1 V.

Question 11.
Write the factors affecting cell voltage.
Answer:

  • Nature of the electrodes.
  • Concentration of the electrolytes.
  • Temperature at which the cell is operated.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 12.
What is the IUPAC definition of electrode potential (E)?
Answer:
Electrode potential is the electromotive force of a cell in which the electrode on the left is a standard hydrogen electrode and the electrode on the right is the electrode in question.

Question 13.
What is the IUPAC definition of standard electrode potential (E°)?
Answer:
Standard electrode potential (E°) is the value of the standard emf of a cell in which molecular hydrogen under standard pressure is oxidised to solvated protons at the left hand electrode.

Question 14.
Define electrolysis.
Answer:

  • Electrolysis is a process in which the electrical energy is used to cause a non- spontaneous chemical reaction to occur.
  • The energy is often used to decompose a compound into its elements.

Question 15.
What is galvanisation?
Answer:

  • Coating of zinc over iron is called galvanisation.
  • Galvanisation of iron prevents rusting.

Question 16.
How cathodic protection helps to protect the metal corrosion
Answer:

  • Unlike galvanisation in this method the entire surface of iron is not covered with zinc.
  • Metals like Mg or Zn which corrode more easily than iron can be used as a sacrificial anode.
  • Iron material acts as a cathode.
  • So iron is protected but Mg or Zn is corroded during rusting.
  • This is known as cathodic protection.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 17.
How metals can be protected from corrosion?
Answer:

  • Coating metal surface by paint.
  • Galvanisation.
  • Cathodic protection.
  • Passivation.
  • Alloy formation.

Question 18.
What is electrochemical series?
How is it useful to predict corrosion?
Answer:

  • The series in which the standard aqueous electrode potential at 298 K for various metal – metal ion electrodes are arranged in the decreasing order of their standard reduction potential values is called electrochemical series.
  • Standard reduction potential E° is a measure of the oxidising tendency of The species.
  • Greater the E° value, greater is the tendency shown by the species to accept electrons and undergo reduction.
  • Higher the E° value lower is the tendency to undergo corrosion..

VII.Three Mark Questions

Question 1.
The molar conductivity of a strong electrolyte and a weak electrolyte increases with dilution, why?
Answer:
Strong electrolytes:

  • When dilution increases, the ions are far apart.
  • Attractive force between the ions decreases.
  • The interaction between the ions becomes insignificant.
  • Hence molar conductivity increases and reaches a maximum value at infinite dilution.

Weak electrolytes:

  • According to Ostwald’s dilution law, as dilution increases the dissociation of weak electrolyte increases.
  • As dissociation increases the number of ions increases.
  • Hence molar conductivity increases with dilution.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
Write a note on Debye-Huckel and onsagar equation.
Answer:

  • At infinite dilution, the interaction between the ions in the electrolyte solution is negligible.
  • Except this condition, electrostatic interaction between the ions alters the properties of the solution from those expected from the free – ions – value.
  • The influence of ion-ion interactions on the conductivity of strong electrolytes was studied by Debye and Huckel.
  • They considered that each ion is surrounded by an ionic atmosphere of opposite sign.
  • They derived an expression relating the molar conductance of strong electrolytes with the concentration by assuming complete dissociation.
  • Later, this equation was further developed by onsagar.
  • For a uni-univalent electrolyte the Debye, Huckel and onsagar equation is given as,
    Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 30
    A and B are constants which depend only on the nature of the solvent and temperature.
    D – dielectric constant of the medium, r) – viscosity of the medium.
    T – temperature in Kelvin.

Question 3.
From the following data, prove that each constituent ion of the electrolyte makes a definite contribution towards the molar conductance irrespective of the nature of other ion with which it is associated.
Answer:

Electrolyte∧°m at 298 K
KCl
NaCl
149.86
126.45
KBr
NaBr
151.92
128.51
KNO3
NaNO3
114.96
121.55

Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 31

  • This value 23.41 for the difference between the (λ°m) values of K+ and Na+ is constant irrespective of the anion (Cl, Br or NO3) with which they are associated.
  • This proves that at infinite dilution each constituent ion of the electrolyte makes a definite contribution towards the molar conductance of the electrolyte irrespective of the nature of the other ion with which it is Associated
  • Thus Kohlraush’s law can be proved.
  • Similarly we can conclude that

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 4.
What is known as intercalation?
Answer:

  • In lithium-ion battery, during discharge, the Li+ ions produced at the anode move towards cathode through the non-aqueous electrolyte.
  • When a potential greater than the emf produced by the cell, is applied across the electrode, the cell reaction is reversed.
  • Now the Li+ ions move from cathode to anode where they become embedded on the porous electrode.
  • This is known as intercalation.

Question 5.
Write a note on standard Hydrogen Electrode (SHE).
Answer:

  • Standard Hydrogen Electrode is used as a reference electrode.
  • Its emf is assigned an arbitrary value of zero volt.
  • It consists of a platinum electrode in contact with 1 M HC1 and 1 atm hydrogen gas.
  • The hydrogen gas is bubbled through the solution at 25°C.
  • SHE can act as a cathode as well as an anode. If SHE is used as a cathode, the reduction reaction is
    Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 32
    2H+(aq,1M) + 2e → H2 (g, latm) E° = 0 volt
  • If SHE is used as an anode, the oxidation reaction
    is H2, (g, 1atm) →2H+(aq,1M) + 2e E°= 0 volt

Question 9.
Write a note on lithium – ion battery.
Answer:

  • Lithium – ion battery is a secondary battery (rechargeable)
  • Anode – Porous graphite
  • Cathode – transition metal oxide such as Co02.
  • Electrolyte – Lithium salt in an organic solvent.
  • Oxidation at anode :
    Li(s) → Li+(aq) + e ………………(1)
  • Reduction at cathode :
    Li++ CoO2(s) + e → LiCoO2(s) ……………..(2)
  • Equation (1) + (2) gives the overall redox reaction.
    Li(s) + CoO2(s) – LiCoO2(s) ………………. (3)
  • Both electrodes allow Li+ ions to move in and out of their structures.
  • Discharge process – Li+ ions produced at the anode move towards cathode through the non-aqueous electrolyte.
  • Recharge process – A potential greater than
    the emf produced by the cell is applied across the electrode, the cell reaction is reversed. .
  • Li+ ions move from cathode to anode where they become embedded on the porous electrode. This is known as intercalation.
  • Uses – cellular phones, laptop computers, digital cameras.

VIII. Five Mark Questions.

Question 1.
What are the factors affecting electrolytic conductance
Answer:

  • As inter ionic attraction increases, conductance decreases.
  • Solvent of high dielectric constant show high conductance in solution.
  • As viscosity of the medium increases, conductance decreases.
  • An temperature of the electrolytic solution increases, conductance increases because kinetic energy of ions increase and the attractive force between them decrease.
  • As dilution increases specific conductance decreases as the number of ions for unit volume decreases.
  • As dilution increases, molar conductance and equivalent conductance of a strong electrolyte increases, because inter ionic attractive force between the ions decreases with dilution.
  • As dilution increases, molar conductance and equivalent conductance of a weak electrolyte increases, because degree of dissociation increases with dilution.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
How is the conductivity of an electrolytic solution determined using a Wheatstone bridge?
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 33
Answer:

  • Conductivity of an electrolytic solution is determined by using a Wheatstone bridge.
  • Here one of the resistance is replaced by a conductivity cell containing the electrolytic solution of unknown conductivity.
  • If DC power supply is used, it will lead to the electrolysis of the electrolytic solution.
  • So, AC power supply is used to prevent electrolysis.
  • A Wheatstone bridge is constituted using conductance and the curve is almost parallel to Am axis.
  • This is because as the dilution increases dissociation of the weak electrolyte also increases.
  • ∧°m values for strong electrolytes can be obtained by extrapolating the straight line.
  • But this is not applicable for weak electrolytes, as the plot is not a linear one.
  • ∧°m values of the weak electrolytes can be determined using Kohlrausch’s law.

Question 4.
Explain the thermodynamics of cell reactions.
Answer:
In a galvanic cell, the chemical energy is converted into electrical energy.
Electrical energy produced by the cell = Total charge of electrons x emf of the cell.
If ‘n’ is the number of moles of electrons exchanged between the oxidising and reducing agents in the overall cell reaction, the electrical energy produced by cell is Electrical energy = charge of ‘n’ mole of electrons x Ecell …………… (1)

Charge of 1 mole of electrons = 1 F
∴ Charge of n mole of electrons = n F
∴ Equation (1) becomes
Electrical energy = nFEcellE0u……………(2)

This energy is ueto do work
∴ Maximum work obtaJed from a galvanic cell is
(Wmax))cell nFEcell ………………….(3)
The (-). sign indicates work is done by the system on the surroundings.

According to ‘second aw of thermodynamics
Maximum work done by the system =
change -in-Gibbs free energyof the.’system
(Wcell = ∆G ………………..(4)
Combining’ (3) and (4)
∆G = -nFEcell ………………(5)

If Ecell is positive, ∆G is negative and the cefi reaction is spontaneous.

When all the cell components are m their
standard state, tbe above equation becomes
∆G° = -nFE°cell ……………..(6)
Also ∆G° = – RT In Keq ……………..(7)
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 34

Question 5.
Write about Lechlanche cell
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 35

  • Lechlanche cell is a primary battery (non – rechargeable)
  • Anode : Zinc container
  • Cathode : Graphite rod in contact with MnO2.
  • Electrolyte : NH4Cl & ZnCl2 in water.
  • EMF: 1.5 V

Oxidation at anode :
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 36
Equation (1) + (2) +(3) gives the overall redox reaction
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 37

Ammonia produced at the cathode combines with Zn2+ to form a complex ion [Zn (NH3)4]2+.
As the reaction proceeds, the concentration of NH4+ will decrease and the aqueous NH3 will increase which lead to the decrease in the emf of the cell.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 6.
Write about mercury button cell.
Answer:

  • Mercury button cell is a primary battery (non – rechargeable)
  • Anode : Zinc amalgamated with mercury.
  • Cathode : HgO mixed with graphite.
  • Electrolyte : Paste of KOH and ZnO.
  • EMF :1.35 V
  • Oxidation at anode :
    Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 38
  • It has higher capacity and longer life.
  • Uses : In pacemakers, electronic watches, cameras, etc.,

Question 7.
Write about lead storage battery.
Answer:

  • Electrochemical reactions which take place in a galvanic cell may be reversed by applying a potential slightly greater than the emf . generated by the cell.
  • This principle is used in secondary batteries to regenerate the original reactants.
  • Lead storage battery is a secondary battery (reachargeable)
  • Anode: Spongy lead.
  • Cathode : Lead plate bearing PbO2.
  • Electrolyte : 38 % by mass of H2SO4 with density 1.2 g / mL.
  • EMF : emf of a single cell is 2 V. Usually six cells are combined in series to produce 12 volt.

Oxidation at anode:
\(\mathrm{Pb}_{(\mathrm{s})} \rightarrow \mathrm{Pb}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \)………..1
\(\mathrm{Pb}^{2+} ions combine with \mathrm{SO}_{4}^{2-} to form PbSO4 precipitate.
\mathrm{Pb}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4}^{2} \underset{(\mathrm{aq})} \rightarrow \mathrm{PbSO}_{4(\mathrm{~s})}\) ………………..(2)

Reduction at cathode;
\(\mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}+2 \mathrm{e} \rightarrow \mathrm{Pb}_{(\mathrm{aq})}^{2+}+2 \mathrm{H}_{2} \mathrm{O}_{(1)}\) …………(3)
Pb2+ ions combine with \(\mathrm{SO}_{4}^{2-}\) from H2SO4 to form PbSO4 precipitate..
\(\mathrm{Pb}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \rightarrow \mathrm{PbSO}_{4(\mathrm{~s})}\)………………..(4)

Equation (1) + (2) + (3) + (4) gives the overall redox reaction.
\(\begin{array}{r}
\mathrm{Pb}_{(s)}+\mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{SO}_{4(\mathrm{aq})}^{2-} \rightarrow \\
2 \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{H}_{2} \mathrm{O}_{(1)}
\end{array}\)
Emf of the cell depends on the concentration of H2SO4
As the cell reaction uses \(\mathrm{SO}_{4}{ }^{2}\) ions, the concentration of H2SO4 decreases.
When the cell potential decreases to about 1.8 V, the cell has to be recharged.

Recharge of the cell:
A potential greater than 2 V is applied across the electrodes, the cell reactions of the discharge process are reversed.
During recharge, the role of anode and cathode is reversed and H2S04 is regenerated.
Oxidation occurs at the cathode (now act as anode)

\(\begin{aligned}
\mathrm{PbSO}_{4(s)}+& 2 \mathrm{H}_{2} \mathrm{O}_{(1)} \rightarrow \\
& \mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+2 \mathrm{e}
\end{aligned}\)
Reduction occurs at the anode (now act as cathode)
\(\mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}_{(s)}+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\)
Overall redox reaction is
\(\begin{array}{c}
2 \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{H}_{2} \mathrm{O}_{(1)} \rightarrow \\
\mathrm{Pb}_{(\mathrm{s})}+\mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}+2 \mathrm{SO}_{4}^{2-}
\end{array}\)

  • Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 39

Thus the overall cell reaction is exactly the reverse of the redox reaction which takes place while discharging. .
Uses : In automobiles, trains, inverters etc.,

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 8.
Explain the electrochemical mechanism of corrosion.
Answer:

  • The redox process which causes deterioration of metal by oxygen and moisture is called corrosion.
  • Rusting or corrosion of iron is an electrochemical process.
  • Rusting requires both oxygen and water.
  • Since it is an electrochemical redox process, it requires an anode and cathode in different places on the iron.
  • Iron surface and a droplet of water on the surface form a tiny galvanic cell as shown in the figure.
  • The region enclosed by water is exposed to low amount of oxygen and it acts as anode.
  • The remaining area has high amount of oxygen and it acts as cathode.
  • Based on the oxygen content, an electrochemical cell is formed. Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 40

Corrosion occurs at the anode ie, the region covered by water.
Oxidation at anode : Iron dissolves in the anode region.
\(2 \mathrm{Fe}_{(\mathrm{s})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+4 \mathrm{e}^{-} \quad \mathrm{E}^{\circ}=0.44 \mathrm{~V}\) ……………………. (1)
Electrons move through the iron metal from the anode to the cathode area.
At the cathode oxygen dissolved in water is reduced to water.

Reduction at cathode:
Atmospheric carbon dioxide and water react to give carbonic acid which furnishes the H+ions for reduction.
O2(g) + 4HM++(aq) + 4e → 2H2O(l)
E° = 1.23V ……………..(2)
Electrical circuit is completed by the migration of ions through water droplet.
The overall redox reaction is, (1) + (2)
\(2 \mathrm{Fe}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+} \rightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
E° = 0.44 + 1.23 = 1.67 V
The positive emf value shows that the reaction is spontaneous.
Fe2+ ions are further oxidised to Fe3+, which on further reaction with oxygen form rust.
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 41

VIII. Additional problems:

Problems based on conductivity :

Question 1.
A conductivity cell has two platinum electrodes separated by a distance 1.5cm and the cross sectional area of each electrode is 4.5sq cm. Using this cell, the resistance of 0.5 N electrolytic solution was measured as 15Ω. Find the specific conductance of the solution.
Solution:
l = 1.5 cm = 1.5 × 10-2 m
A = 4.5 cm2 = 4.5 × (10-4)m2
R = 15Ω

\(K=\frac{1}{R}\left(\frac{\ell}{A}\right)\)
\(K=\frac{1}{15 \Omega} \times \frac{1.5 \times 10^{-2} \mathrm{~m}}{4.5 \times\left(10^{-4}\right) \mathrm{m}^{2}}\)
= 2.22 Sm-1

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
0.04 N solution of a weak add has a specific conductance 4.23 × 10-4 Scm-1. The degree of dissociation of acid at this dilution is 0.0612. Calculate the equivalent conductance of weak acid at infinite dilution.
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 42

Problems based on Kohlrausch’s Law

Question 1.
Equivalent conductances of NaCl, HCl and CH3COONa at infinite dilution are 126.45,426.16 and 91 S cm2g equiv-1. Calculate the equivalent conductance of CH3COOH at infinite dilution.
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 43
= 91 + 426.16 – 126.45
= 517.16 -126.45
∧°CH3COOH = 390.71Sm2gequ-1

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
The equivalent conductance of M/36 solution of a Weak monobasic acid is 6 mho cm2 equivalent and infinite dilution is 400 mho cm2 equivalent-1. Calculate the dissociation constant of this acid.
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 44

Problems based on Faraday’s law

Question 1.
If 50 milli ampere of current is passed through copper coulometer for 60 minutes, calculate the amount of copper deposited.
Solution:
I = 50 mA = 50 × 10-3 A;
t = 60 min = 60 × 60 = 3600 seconds
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 45
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 46

Question 2.
0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minuets. What is the electro chemical equivalent of copper?
Solution:
m = 0.1978 g; I = 0.2 A;
t = 50 mins = 50 × 60 = 3000 sec
Z = ?
m = ZIt
∴ \(Z=\frac{m}{I t}\)
\(=\frac{0.1978}{0.2 \times 3000}\)
Z = 3.297 × 10-4gC-1

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 3.
What current strength in amperes will be required to liberate 10 g of iodine from potassium iodide solution in one hour?
Solution:
l = ?m = 10g;
t = 1 hr = 60 × 60 = 3600 sec
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 47

Question 4.
An electric current is passed through three cells in series containing respectively solutions of copper sulphate, silver nitrate and potassium iodide. What weight of silver and iodine will be liberated while 1.25 g copper is being deposited?
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 48

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 5.
On passing a current of 1 ampere for 16 min 5 sec through 1 litre solution of CuCl2 all copper of the solution was deposited at cathode. Calculate the normality of CuCl2 solution.
Solution:
1 = 1 ampere
t = 16 min 5 sec = 16 * 60 + 5 = 965 sec
N = ? m = ZIt
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 49

Problems based on emf

Question 1.
The reaction \(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Co}^{2+} \rightleftharpoons \mathrm{Co}_{(\mathrm{s})}+\mathrm{Zn}^{2+}\) occurs in a cell Compute the standard emf of the cell.
Solution:
Given that
\(E^{0} z_{n} / Z n^{2+}\) = +0.76 V and \(E_{\mathrm{Co} / \mathrm{Co}}^{2+}\) = +0.28 V
Zn(s) + Co2+ Co(s) +Zn2+
Oxidation potential of Co ion is change into reduction potential
+0.28 V = -0.28 V
E° Cell = E° oxidation (2n/Zn2+) + E° reduction(Co2+/Co)
E° cell = + 0.76 + (-0.28)
E°cell = +0.48 V
This reaction is feasible.

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 2.
What is the E°el, of the reaction
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{Sn}_{(\mathrm{aq})}^{2+} \rightleftharpoons \mathrm{Cu}_{(s)}+\mathrm{Sn}_{(\mathrm{aq})}^{4+}\) at 25°C, If the equilibrium constant for the reaction is 1 × 106.
Solution :
Keq= 1 × 106 T = 25° C + 273 = 298 K
R = 8.314 JK-1 mol-1
Cu2+ + 2e → Cu
Sn2+ → Sn4+ + 2e
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 50

Question 3.
Cu+(aq) | is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction.
\(2 \mathrm{Cu}_{(\mathrm{aq})}^{+} \rightleftharpoons \mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(s)}\)
Calculate E° for the above reaction
\({ }^{\mathrm{E}^{\mathrm{O}}} \mathrm{Cu}^{2+} \mid \mathrm{Cu} = 0.34\) and \(\mathrm{E}_{\mathrm{Cu}^{2+} \mathrm{Cu}^{+}}^{\mathrm{o}}=0.15 \mathrm{~V}\)
Solution:
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 51
Equation (1) = Equation (2) + Equation (3)
ΔG°1 = ΔG°2 + ΔG°3
– 0.68 F= 0.15F—1E°3F
-0.68 = -0.15 – E°3
3 = 0.68 — 0.15 = 0.53 V
\(2 \mathrm{Cu}_{(\mathrm{aq})}^{+} \rightleftharpoons \mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}\)
Half reactions are
Cu+ → Cu2+ + e
(Oxidation at anode)
E = 0.15V ……………….(4)
(Reduction at cathode)
Cu+ + e→ Cu
red =E°3= 0.53V ………………..(5)
Overall reaction is
2Cu+ → Cu2+ + Cu
:.E° = (E°oxid)Anode + (E°red)cathode
= – 0.15 + 0.53
cell = 0.38V

Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry

Question 4.
For the redox reaction
\(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.1 \mathrm{M}) \rightleftharpoons \mathrm{Zn}^{2+}(1 \mathrm{M})+\mathrm{Cu}_{(\mathrm{s})}\)
if E°cell = 1.1V calculate Ecell of the reaction
Solution:
E°ell = 1.1V; n = 2; Ecell=?
[Cu2+] = 0.1 M; [Zn2+] = 1M
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 52

Question 5.
Calculate the emf of the cell Zn | Zi2+ (0.001 M | Ag+ (0.1 M) | Ag. The standard potential of Ag+ | Ag half cell is + 0.80 V and Zn | Zn2+ is + 0.76 V.
Anode : Zn → Zn2+ + 2e E° = + 0.76 V
Cathode : 2Ag+ + 2e → 2Ag E° = + 0.80 V
Cell reaction : Zn + 2Ag+ ⇌ Zn2+ + 2Ag
E° = + 0.76 + 0.80 = + 1.56 V
Samacheer Kalvi 12th Chemistry Guide Chapter 9 Electro Chemistry 53
Ecell = 1.56 – 0.02955 log 10-1
Ecell = 1.56 + 0.02955 = 1.58955 V

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 3 Magnetism and Magnetic Effects of Electric Current Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

12th Physics Guide Magnetism and Magnetic Effects of Electric Current Text Book Back Questions and Answers

Part – 1

Text Book Evaluation:

I. Multiple choice questions:

Question 1.
The magnetic field at the center O of the following current loop is
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 1
a) \(\frac{\mu_{0} I}{4 r} \otimes\)
b) \(\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \mathrm{r}} \odot\)
c) \(\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}} \otimes\)
d) \(\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}} \odot\)
Answer:
a) \(\frac{\mu_{0} I}{4 r} \otimes\)
Solution:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 3

Question 2.
An electron moves straight line inside a charged parallel plate capacitor of uniform charge density σ. The time taken by the electron to cross the parallel plate capacitor undeflected when the plates of the capacitor are kept under constant magnetic field of induction B is
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 2
a) \(\varepsilon_{0} \frac{\mathrm{elB}}{\sigma}\)
b) \(\varepsilon_{\mathrm{O}} \frac{\mathrm{lB}}{\sigma l}\)
c) \(\varepsilon_{\mathrm{o}} \frac{\mathrm{lB}}{\mathrm{e} \sigma}\)
d) \(\varepsilon_{0} \frac{\mathrm{B}}{\sigma}\)
Answer:
\(\varepsilon_{0} \frac{\mathrm{B}}{\sigma}\)
Solution:
\(\mathrm{E}=\frac{\sigma}{\varepsilon_{\mathrm{o}}}\)
[∴Parallel plate capacitors]
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 5

Question 3.
A particle having mass m and charge q accelerated through a potential difference V. Find the force experienced when it is kept under perpendicular magnetic field B is
a) \(\sqrt{\frac{2 q^{3} B V}{m}}\)
b) \(\sqrt{\frac{q^{3} B^{2} V}{2 m}}\)
c) \(\sqrt{\frac{2 q^{3} B^{2} V}{m}}\)
d) \(\sqrt{\frac{2 q^{3} B V}{m^{3}}}\)
Answer:
\(\sqrt{\frac{2 q^{3} B^{2} V}{m}}\)
Solution:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 5

Question 4.
A circular coil of radius 5 cm arid has 50 turns carries a current of 3 amperes. The magnetic dipole moment of the coil is
a) 1.0 Am2
b) 1.2 Am2
c) 0.5 Am2
d) 0.8 Am2
Answer:
b) 1.2 Am2
Solution:
r = 5 cm = 5 × 10-2m
N = 50 turns
I = 3A
M = NIA = 50 × 3 × 3.14 × 25 × 104
M = 1.2amp – m2

Question 5.
A thin insulated wire forms a plane spiral of N=100 tight turns carrying a current I = 8mA (milliampere). The radii of inside and outside turns are a = 50mm and b = 100 mm, respectively. The magnetic induction at the center of the spiral is
a) 5 µT
b) 7 µT
c) 8 µT
d) 10 µT
Answer:
b) 7 µT
Solution:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 6

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 6.
Three wires of equal lengths are bent in the form of loops. One of the loops is a circle, another is a semi-circle and the third one is a square. They are placed in a uniform magnetic field and the same electric current is passed through them. Which of the following loop configuration will experience greater torque?
a) circle
b) semi-circle
c) square
d) all of them
Answer:
a) Circle
Solution:
Circle has greater torque because torque is directly proportional to area.

Question 7.
Two identical coils, each with N turns and radius R are placed coaxially at a distance R as shown in the figure. If I is the current passing through the loops in the same direction, then the magnetic field at a point P at a distance of R/2 from the center of each coil
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 7
a) \(\frac{8 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{\sqrt{5} \mathrm{R}}\)
b) \(\frac{8 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{5^{3 / 2} \mathrm{R}}\)
c) \(\frac{8 \mathrm{~N} \mu_{\mathrm{O}} \mathrm{I}}{5 \mathrm{R}}\)
d) \(\frac{4 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{\sqrt{5} \mathrm{R}}\)
Answer:
b) \(\frac{8 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{5^{3 / 2} \mathrm{R}}\)
Solution:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 8
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 9

Question 8.
A wire of length l carrying a current I along the Y direction is kept in a and magnetic field is given by
\(\mathrm{B}=\frac{\beta}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k}) T\).
The magnitude of Lorentz force acting on the wire is
a) \(\sqrt{\frac{2}{\sqrt{3}}} \beta \mathrm{I} l\)
b) \(\sqrt{\frac{1}{\sqrt{3}}} \beta I l\)
c) √2βIl
d) \(\sqrt{\frac{1}{2}} \beta \mathrm{I} l\)
Answer:
a) \(\sqrt{\frac{2}{\sqrt{3}}} \beta \mathrm{I} l\)
Solution:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 10

Question 9.
A bar magnet of length l and magnetic moment pm is bent in the form of an arc as shown in Figure. The new magnetic dipole moment will be
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 11
a) pm
b) \(\frac{3}{\pi} p_{m}\)
c) \(\frac{2}{\pi} \mathrm{p}_{\mathrm{m}}\)
d) \(\frac{1}{2} \mathrm{p}_{\mathrm{m}}\)
Answer:
b) \(\frac{3}{\pi} p_{m}\)
Solution:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 12

Question 10.
A non-conducting charged ring of charge of q, mass m and radius r is rotated about its axis with constant angular speed to. Find the ratio of its magnetic moment with angular momentum is
a) \(\frac{q}{m}\)
b) \(\frac{2 q}{m}\)
c) \(\frac{\mathrm{q}}{2 \mathrm{~m}}\)
d) \(\frac{q}{4 m}\)
Answer:
c) \(\frac{\mathrm{q}}{2 \mathrm{~m}}\)
Solution:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 13

Question 11.
The BH curve for a ferromagnetic material is shown in the Figure. The material is placed inside a long solenoid which contains 1000 turns/cm. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 14
a) 1.00 mA
b) 1.25 mA
c) 1.50 mA
d) 1.75 mA
Answer:
c) 1.50 mA

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 12.
Two short bar magnets have magnetic moments 1.20 Am2 and 1.00 Am2, respectively. They are kept on a horizontal table parallel to each other with their north poles pointing towards the south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid¬point O of the line joining their centers is (Horizontal components of Earth’s magnetic induction is 3.6 × 10-5 Wbm-2)
a) 3.60 × 10-5 Wbm-2
b) 3.5 × 10-5 Wbm-2
c) 2.56 × 10-4 Wbm-2
d) 2.2 × 10-4 Wbm-2
Answer:
c) 2.56 × 104-5 Wbm2
Solution:
\(B=\frac{\mu_{0}}{4 \pi} \frac{M}{\left(d^{2}+L^{2}\right)^{3 / 2}}\)
M1 = 1.2 Am2
M2 = 1 Am2
2L = 0.01m, 2d = 0.2m
\(B_{1}=\frac{4 \pi x 10^{-7} \times 1.2}{4 \pi\left[(0.1)^{2}+(0.005)^{2}\right]^{3 / 2}}\)
B1 = 1.1955 × 10-4 Wbm-2
\(\mathrm{B}_{2}=\frac{4 \pi \times 10^{-7} \times 1}{4 \pi\left[(0.10)^{2}+(0.005)^{2}\right]^{3 / 2}}\)
B2 = 1.1963 × 10-4 Wbm-2
B = B1 + B2 + Bearth
= (1.1955 + 1.1963 + 0.36) × 10-4
B = 2.56 × 10-4 Wbm-2

Question 13.
The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of the angle of dip at this place?
a) 30°
b) 45°
c) 60°
d) 90°
Answer:
b) 45°
Solution:
BH = BE cos I
BV = BE sin I
BE cos I = BE sin I
\(\frac{\sin I}{\cos I}=1\)
tan I = 1
I = 45°

Question 14.
A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is σ. The disc rotates about an axis perpendicular to its plane passing through the center with angular velocity ω. Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field whose strength is B which is directed perpendicular to the axis of rotation
a) \(\frac{1}{4}\)σωπBR
b) \(\frac{1}{4}\)σωπBR2
c) \(\frac{1}{4}\)σωπBR3
d) \(\frac{1}{4}\)σωπBR4
Answer:
d) \(\frac{1}{4}\)σωπBR4
Solution:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 83

Question 15.
The potential energy òf magnetic dipole whose dipole moment is \(\vec{p}_{m}=(-0.5 \hat{i}+0.4 \hat{j}) \mathrm{Am}^{2}\) Am2 kept in uniform magnetic field \(\vec{B}=0.2 \hat{i} \mathrm{~T}\)
a -0.1 J
b) -0.8 J
c) 0.1 J
d) 0.8 J
Answer:
c) 0.1 J

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

II. Short Answer Questions:

Question 1.
What is meant by magnetic induction?
Answer:
The magnetic induction (total magnetic field) inside the specimen \(\vec { B } \) is equal to the sum of the magnetic field \(\vec { B } \)0 produced in vacuum due to the magnetising field and the magnetic field \(\vec { B } \)m due to the induced magnetisation of the substance.
\(\vec { B } \) = \(\vec { B } \)0 + \(\vec { B } \)m = µ0\(\vec { H } \) + µ0\(\vec { I } \) = µ0 (\(\vec { H } \) + \(\vec { I } \)) = (\(\vec { H } \) + \(\vec { I } \))

Question 2.
Define magnetic flux.
Answer:

  1. The number of magnetic field lines crossing per unit area.
  2. φB = \(\overrightarrow{\mathrm{B}}\) . \(\overrightarrow{\mathrm{A}}\) = BA cos θ
  3. S.I unit: weber (Wb); Dimensional formula [ML2T-2A-1]

Question 3.
Define magnetic dipole moment.
Answer:
The magnetic dipole moment is defined as the product of its pole strength and magnetic length.
\(\vec { P } \) = qm \(\vec { d } \)

Question 4.
State Coulomb’s inverse law.
Answer:
The force of attraction or repulsion between two magnetic poles is directly proportional to the product of their pole strengths and inversely proportional to the square of distance between them,
\(\overrightarrow{\mathrm{F}} \alpha \frac{\mathrm{q}_{\mathrm{m}_{\mathrm{A}}} \mathrm{q}_{\mathrm{m}_{\mathrm{B}}}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\) ;
\(\mathrm{F}=\frac{\mathrm{K} \mathrm{q}_{\mathrm{m}} \mathrm{A} \quad \mathrm{q}_{\mathrm{m}} \mathrm{B}}{\mathrm{r}^{2}}\)
qmA & qmB – pole strengths of two poles,
r – distance between two magnetic poles.

Question 5.
What is magnetic susceptibility?
Answer:
It is defined as the ratio of the intensity of magnetisation (\(\vec { M } \)) induced in the material due to the magnetising field (\(\vec { H } \))
\(\chi_{m}=\left|\frac{\overrightarrow{\mathrm{M}}}{\overrightarrow{\mathrm{H}}}\right|\)

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 6.
State Biot-Savart’s law.
Answer:
The magnitude of magnetic field d\(\overrightarrow{\mathrm{B}\) at a point P at a distance r from the small length on a current-carrying conductor varies,
i) directly as the strength of the current I
ii) directly as the magnitude of the length element dI and r
iii) directly as the sine of the angle between dI and r̂
iv) inversely as the square of the distance between the point P and length element dl
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 15

Question 7.
What is magnetic permeability?
Answer:
Magnetic permeability: Magnetic permeability can be defined as the measure of the ability of the material to allow the passage of magnetic field lines through it or a measure of the capacity of the substance to take magnetisation or the degree of penetration of the magnetic field through the substance.

Question 8.
State Ampere’s circuital law.
Answer:
The line integral of magnetic field over a closed loop is μ0 times net current enclosed by loop.
\(\oint_{C} \vec{B} \cdot \vec{l}=\mu_{o} I_{\text {enclosed }}\)
Ienclosed → net current in the closed loop C.

Question 9.
Compare dia, para and ferro-magnetism.
Answer:

DiamagnetismParamagnetismFerro magnetism
A force tends to move the material from the stronger part to the weaker part of the external field.Weak magnetism in the direction of the applied field.Strong magnetism in the direction of the applied field.
Each electron has finite orbital magnetic dipole moments.Each atom has a net magnetic dipole moment.Have net dipole moment as in a paramagnetic material.
Since the orbital planes are oriented in a random manner, the vector sum of magnetic moments is zero.Due to the random orientation of the magnetic moments, the net magnetic moment of the material is zero.Within each domain, the magnetic moments are spontaneously aligned in a direction.
The resultant magnetic moment for each atom is zero.There is a net magnetic dipole moment induced in the direction of the applied field.Since the direction of magnetisation varies from domain to domain, the net magnetisation of the specimen is zero.

Question 10.
What is meant by hysteresis?
Answer:
The phenomenon of lagging of magnetic induction behind the magnetising field is called hysteresis. Hysteresis means Tagging behind’.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 11.
Define magnetic declination and inclination?
Answer:
Magnetic declination:

  1. The angle between magnetic meridian at a point and geographical meridian.
  2. Higher altitude – Greater declination
  3. Equator – Smaller declination.
  4. Magnetic inclination: The angle subtended by the Earth’s total magnetic field \(\overrightarrow{\mathrm{B}}\) with the horizontal direction in the magnetic meridian.

Question 12.
What is the resonance condition in a cyclotron?
Answer:

  1. In cyclotron operation when the frequency f at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator fosc. This is called resonance condition.
    fosc = \(\frac{q B}{2 \pi m}\)
  2. The time period of oscillation is T = \(\frac{2 \pi \mathrm{m}}{\mathrm{qB}}\)
  3. The kinetic energy of charged particle is \(\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{q}^{2} \mathrm{~B}^{2} \mathrm{r}^{2}}{2 \mathrm{~m}}\)

Question 13.
Define ampere.
Answer:
Current passed through each of the two infinitely long parallel straight conductors kept at a distance of one meter apart in vacuum causes each conductor to experience a force of 2 × 10-7 newton per meter length of the conductor.

Question 14.
State Fleming’s left-Hand rule.
Answer:

  1. Stretch forefinger, the middle finger, and the thumb of the left hand such that they are in mutually perpendicular directions.
  2. The forefinger points the direction of the magnetic field.
  3. The middle finger points in the direction of the electric current.
  4. Thumb will point the direction of the force experienced by the conductor.

Question 15.
Is an ammeter connected in series or parallel in a circuit? Why.
Answer:

  1. An ammeter is a low resistance instrument and it is always connected in series to the circuit.
  2. Such that it will change the current passing through it.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 16.
Explain the concept of the selector.
Answer:

  1. For a given magnitude of E field and B field, the forces act only on the particle moving with particular speed p.
    \(v=\frac{E}{B}\)
  2. This speed is independent of mass and charge.

Question 17.
Why is the path of a charged particle not a circle when its velocity is not perpendicular to the magnetic field?
Answer:

  1. If a charged particle moves in a region of a uniform magnetic field such that its velocity is not perpendicular to the magnetic field, then the velocity of the particle is split up into two components.
  2. One component is parallel to the field while the other component perpendicular to the field.
  3. The component of velocity parallel to find remains unchanged and the component perpendicular to the field keeps changing due to Lorentz force.
  4. Hence the path of the particle is not a circle; it is helical around the field lines.
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 16
    Helical path of the electron in a uniform.

Question 18.
Give the properties of dia/para/ferromagnetic materials.
Answer:

PropertiesDiaParaFerro
Magnetic susceptibilityNegativePositive and smallPositive and large
Relative permeabilitySlightly less than unityGreater than unityLarge
Nature of magnetic field linesRepelled or expelled from diamagnetic materialsAttracted into the paramagnetic materialsStrongly attracted to the ferromagnetic materials
SusceptibilityTemperature independent.Inversely proportional to temperature.Inversely proportional to temperature.
ExamplesBismuth, copper, and water, etc.Aluminum, platinum, chromium, etc.Iron, Nickel, and Cobalt

Question 19.
What happens to the domains in a ferromagnetic material in the presence of an external magnetic field?
Answer:

  1. The domains having magnetic moments parallel to the field glow bigger in size.
  2. The other domains (not parallel to the field) are rotated so that they are aligned with the field.

Question 20.
How is a galvanometer converted into (i) an ammeter and (ii) a voltmeter?
Answer:

  1. A galvanometer can be converted into an ammeter of a given range by connecting of suitable low resistance S called shunt in parallel to the galvanometer.
  2. A galvanometer can be converted into suitable high resistance. Rh in series with the given galvanometer.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

III. Long Answer Questions:

Question 1.
Discuss Earth’s magnetic field in detail.
Answer:

  1. William Gilbert proposed that Earth behaves like a gigantic powerful bar magnet.
  2. The temperature inside the earth is very high, so it is not possible for a magnet to retain its magnetism.
  3. Gover suggested that the Earth’s magnetic field is due to hot rays coming out from the sun.
  4. The branch of physics which deals with the Earth’s magnetic field is called Geomagnetism or Terrestrial magnetism.
  5. Three quantities required to specify the Earth’s magnetic field are
    • magnetic declination (D)
    • magnetic dip or inclination (I)
    • the horizontal component of the Earth’s magnetic field (BH)
  6. BE → Earth’smagneticfield.
    BH → Horizontal component
    BV → Vertical component
    Horizontal component BH = BE cos I
    Vertical component BV= BE sin I;
    \(\tan I=\frac{B_{V}}{B_{H}}\)
    i) At magnetic equator BH = BE [I = 0°]
    BV = 0
    ii) At magnetic poles BH = 0 [I = 90°]
    BV = BE

Question 2.
Deduce the relation for the magnetic field at a point due to an infinitely long straight conductor carrying current.
Answer:

  1. Let NM be a long straight wire carrying a current I.
  2. Let dl be the elemental length at a distance l from O.
  3. Let P be a point at a distance ‘a’ from ‘o’.
  4. According to Biot savart law, Magnetic Induction at P is
    \(\overrightarrow{\mathrm{dB}}=\frac{\mu_{\mathrm{o}} \mathrm{I\overrightarrow{\mathrm{dl}}}}{4 \pi \mathrm{r}^{2}} \sin \theta\) ………….(1)

The direction of the field is perpendicular to the plane of the paper and going into it.

From the figure, In △PAO,
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 17
tan(π – θ) = \(\frac{a}{l}\)
∵ tan(π – θ) = -tan θ
-tan θ = \(\frac{a}{l}\)
l = \(\frac{-a}{\tan \theta}\)
l = -a cot θ ………………(2)

From △PAO
sin θ = \(\frac{a}{r}\)
\(r=\frac{a}{\sin \theta}\)
∴r = a cosec θ …………….(3)
Differentiating (2)
dl = a cosec2θ dθ …………..(4)
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 18

Question 3.
Obtain a relation for the magnetic field at a point along the axis of a circular coil carrying current.
Answer:

  1. Let R be the radius of a current-carrying circular loop.
  2. I be the current flowing through the wire.
  3. Let P be a point on the axis of the circular coil at a distance z from its center ‘o’.
  4. Take two diametrically opposite element d\(\overrightarrow{\mathrm{I}\) at C and D.
  5. Let \(\overrightarrow{\mathrm{r}}\) be the vector joining the current I\(\mathrm{d} \overrightarrow{\mathrm{l}}\) at C.
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 19
    Magnetic field due to current carrying circular loop using Biot-Savart’s law.
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 20
  6. PC = PD = r = \(\sqrt{R^{2}+Z^{2}}\)
    ∠CPO =∠DPO = θ
  7. According to Biot – Savart’s law,
    \(\mathrm{d} \overrightarrow{\mathrm{B}}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{Id} \overrightarrow{\mathrm{l}} \times \hat{\mathrm{r}}}{\mathrm{r}^{2}}\)
  8. \(\mathrm{d} \overrightarrow{\mathrm{B}}\) is resolved into two components
    dB sin θ – along y – direction
    dB cos θ – along z – direction
  9. Horizontal components cancel each other vertical component contribute to total magnetic field.
  10. Integrating d\(\overrightarrow{\mathrm{I}\) around the ioop, d\(\overrightarrow{\mathrm{B}\) sweeps out a cone.
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 21
    (Using pythagorous)
    ∵\(\cos \theta=\frac{R}{\sqrt{R^{2}+Z^{2}}}\)
    r2 = R2 + Z2
    \(B=\frac{\mu_{0} I}{4 \pi} \int \frac{d l}{R^{2}+Z^{2}} \frac{R}{\sqrt{R^{2}+Z^{2}}} \hat{k}\)
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 22

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 4.
Compute the torque experienced by a magnetic needle in a uniform magnetic field.
Answer:

  1. Consider a magnetic needle or bar magnet of length 2l.
  2. Pole strength → qm.
  3. Each pole experience a force qmB, but in opposite direction.
  4. The net force is zero.
  5. The magnetic needle will rotate and constitute a couple.
  6. Force experienced by the north pole
  7.  Force experienced by south pole

\(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=\mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}}\)
\(\begin{array}{l}
\overrightarrow{\mathrm{F}}_{\mathrm{S}}=-\mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}} \\
\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{\mathrm{N}}+\overrightarrow{\mathrm{F}}_{\mathrm{S}}
\end{array}\)
\(\overrightarrow{\mathrm{F}}\) = 0
\(\begin{aligned}
\text { Torque } \vec{\tau} &=\mathrm{O} \overrightarrow{\mathrm{N}} \times \overrightarrow{\mathrm{F}}_{\mathrm{N}}+\mathrm{O} \overrightarrow{\mathrm{S}} \times \overrightarrow{\mathrm{F}}_{\mathrm{S}} \\
&=\mathrm{O} \overrightarrow{\mathrm{N}} \mathrm{x} \mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{OS}} \times\left(-\mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}}\right)
\end{aligned}\)
Using right hand cork screw rule, total torque points into the paper.
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 23
Magnetic dipole kept in a uniform magnetic field

\(\begin{array}{l}
|\mathrm{ON}|=|\mathrm{OS}|=l \\
\left|\mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}}\right|=\mid-\mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}}
\end{array}\)
∴\(\overrightarrow{\mathrm{τ}}\) = l × qmBsin θ + l × qmBsin θ

Question 5.
Calculate the magnetic field at a point on the axial line of a bar magnet.
Answer:

  1. Consider a bar magnet NS.
  2. N – Northpole, S – Southpole
  3. Pole strength – qm.
  4. Distance – 2l
  5. C be a point along the axis of the magnet.
  6. By Coulomb’s law of magnetism,
  7. The force of repulsion
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 24
    Magnetic field at a point along the axial line due to magnetic dipole
    \(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{(\mathrm{r}-l)^{2}} \hat{\mathrm{i}}\) …………..(1)
  8. r – l → is the distance between north pole of the bar magnet and unit north pole at C.
  9. The force of attraction,
    \(\overrightarrow{\mathrm{F}}_{\mathrm{S}}=\frac{-\mu_{0}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{(\mathrm{r}+l)^{2}} \hat{\mathrm{i}}\) ………………(2)
  10. r + l → is the distance between south pole of the bar magnet and unit north pole at C.
  11. Net force at point C.
    \(\overrightarrow{\mathrm{F}}\) = \(\overrightarrow{\mathrm{FN}}\) + \(\overrightarrow{\mathrm{FS}}\)
    \(\overrightarrow{\mathrm{F}}\) = \(\overrightarrow{\mathrm{B}}\)
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 25
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 26

Question 6.
Obtain the magnetic field at a point on the equatorial line of a bar magnet.
Answer:

  1. Consider a bar magnet NS.
  2. N – Northpole, S – Southpole
  3. Pole strength – q.
  4. Distance between two poles – 2l
  5. C be a point along the equatorial line
  6. By Coulomb’s law of magnetism,
  7. The force of repulsion,

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 27
Magnetic field at a point along the equatorial line due to a magnetic dipole

\(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=-\mathrm{F}_{\mathrm{N}} \cos \theta \hat{\mathrm{i}}+\mathrm{F}_{\mathrm{N}} \sin \hat{\theta} \hat{\mathrm{j}}\)
where,
\(\mathrm{F}_{\mathrm{N}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{\mathrm{r’}^{2}}\) …………(1)

The force of attraction,
\(\overrightarrow{\mathrm{F}}_{\mathrm{S}}=-\mathrm{F}_{\mathrm{S}} \cos \theta \hat{\mathrm{i}}-\mathrm{F}_{\mathrm{S}} \sin \hat{\theta} \hat{\mathrm{j}}\)
where,
\(\mathrm{F}_{\mathrm{S}}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{\mathrm{r}^{\prime 2}}\) ……………(2)

The net force at point C,
\(\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{\mathrm{N}}+\overrightarrow{\mathrm{F}}_{\mathrm{S}}\)
\(\overrightarrow{\mathrm{B}}\) = – (FN + FS) cosθî
Here,
\(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=\overrightarrow{\mathrm{F}}_{\mathrm{S}}\)
∴ \(\begin{array}{l}
\overrightarrow{\mathrm{B}}=-\frac{2 \mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{\mathrm{r}^{\prime 2}} \cos \theta \hat{\mathrm{i}} \\
\overrightarrow{\mathrm{B}}=-\frac{2 \mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{\left(\mathrm{r}^{2}+1^{2}\right)} \cos \theta \hat{\mathrm{i}}
\end{array}\)
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 28

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 7.
Find the magnetic field due to a long straight conductor using Ampere’s circuital law.
Answer:

  1. I be the current flowing in a straight conductor of infinite length.
  2. The wire is geometrically cylindrical in shape.
  3. Symmetrical about an axis.
  4. Amperian loop is constructed in the form of a circular shape at a distance r from the centre of
  5. From Ampere’s law, \(\oint\) \(\overrightarrow{\mathrm{B}}\) . \(\overrightarrow{\mathrm{dl}}\) = μ0I

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 29
Amerian loop for current-carrying straight line

\(\overrightarrow{\mathrm{dl}}\) is the line element along the loop.
The angle between magnetic field and line element is zero.
∴\(\oint\) B.dl = μ0I
B is uniform over the loop
∴ B\(\oint\)dl = μ0I
The circumference of the loop = 2πr
∴ \(\mathrm{B} \int_{0}^{2 \pi \mathrm{r}} \mathrm{dl}\) = μ0I
B.2πr = μ0I
\(\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi \mathrm{r}}\)
In vector form,
\(\vec{B}=\frac{\mu_{0} I}{2 \pi r} \hat{n}\)
n̂ – unit vector along the tangent to the loop.

Question 8.
Discuss the working of a cyclotron in detail.
Answer:
Cyclotron:

  1. A device used to accelerate the charged particles to gain large kinetic energy.
  2. It is also called as high energy accelerator.

Principle:
When a charged particle moves normally to the magnetic field, it experiences magnetic Lorentz force.

Construction:

  1. The particles are allowed to move in between two semi-circular metal containers called Dees.
  2. Uniform magnetic field is controlled by an electromagnet.
  3. The direction of magnetic field is normal to the plane of the Dees.
  4. Source ‘S’ is kept between the two Dees.
  5. Dees are connected to high frequency alternating potential difference.

Working:

  1. The ion ejected from the source is positively charged.
  2. It is accelerated towards a Dee-1 which has negative potential at that time.
  3. The ion undergoes a circular path.
  4. After one semi-circular path, the ion reaches the gap between the Dees.
  5. The polarities of the Dees are reversed, so that ion accelerated towards Dee-2 with greater velocity.
  6. The centripetal force is provided by Lorentz force 0 ; Bqv
    ∴ \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{Bqv}\)
    \(r=\frac{m}{B q} v\)
    i.e. r α v
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 30
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 31
  7. If the radius of the circular path increases, velocity also increases.
  8. Particles undergo a spiral path with increasing radius.
  9. Particles took out with the help of the deflector plate and hit the target T.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 32

Limitations :

  1. The speed of the ion is limited.
  2. An electron cannot be accelerated.
  3. Unchanged particles cannot be accelerated.

Question 9.
What is tangent law? Discuss in detail.
Answer:
Tangent law:-
When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.
B = BH tan θ
Construction:

  1. A circular coil of wire wound over brass or wood.
  2. The circular turn table is adjusted through levelling screws.
  3. Compass box at the center
  4. Pivoted magnet and aluminium pointer are found.
  5. The circular scale graduated with four quadrants.

Theory:

  1. Two magnetic fields are perpendicular to each other.
  2. Magnetic induction due to the current in the coil acting normal to the plane of the coil.
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 33
    BH – horizontal component of Earth’s magnetic field.
    Magnetic induction at the centre of a coil
    B = \(\mathrm{B}=\mu_{\mathrm{o}} \frac{\mathrm{NI}}{2 \mathrm{R}}\) …………(2)
    ∴ Substituting (2) in (1)
    \(\begin{aligned}
    \mu_{0} \frac{\mathrm{NI}}{2 \mathrm{R}} &=\mathrm{B}_{\mathrm{H}} \tan \theta \\
    \mathrm{B}_{\mathrm{H}} &=\mu_{\mathrm{o}} \frac{\mathrm{NI}}{2 \mathrm{R}} \frac{1}{\tan \theta}
    \end{aligned}\) in tesla.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 10.
Derive the expression for the torque on a current-carrying coil in a magnetic field.
Answer:
Consider a rectangular loop PQRS carrying current I is placed in a uniform magnetic field B. Let a and b be the length and breadth rectangular loop respectively. The unit vector n̂ normal to the plane of the loop makes an angle θ with the magnetic field.
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 34
Rectangular coil placed in a magnetic field

1. The magnitude of the magnetic force acting on the current-carrying arm PQ is FPQ = IaBsin (π/2) = IaB.
2. The magnitude of the force on the arm QR is FQR =IbBsin \(\left(\frac{\pi}{2}-\theta\right)\) = IbB cos θ and its direction.
3. The magnitude of the force on the arm RS is FRS =IaBsin( π/2) = IaB and its direction is downwards.
4. The magnitude of the force acting on the arm SP is FSP = lbB\(\left(\frac{\pi}{2}-\theta\right)\) = IbB cos θ and its direction.
5. Since the forces FQR and FSP are equal, opposite, and collinear, they cancel each other. But the forces FPQ and FRS.
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 35
Side view of the current loop

6. The magnitude of torque acting on the arm PQ about AB is τPQ = (1/2 sinθ) and points in the direction of AB. The magnitude of the torque acting on the arm. RS about AB is τRS = (b/2 sinθ) IaB and points also in the same direction AB.
7. The total torque acting on the entire loop about an axis AB is given by
\(\tau=\left(\frac{b}{2} \sin \theta\right) F_{P Q}+\left(\frac{b}{2} \sin \theta\right) F_{R S}\)
= Ia(bsinθ)B
τ = IABsinθ along the direction B.
In vector form,
\(\overrightarrow{\mathrm{τ}}\) = (\(\overrightarrow{\mathrm{IA}}\)) × \(\overrightarrow{\mathrm{B}}\).
8. The above equation can also be written in terms of magnetic dipole moment.
\(\overrightarrow{\mathrm{τ}}\) = \(\overrightarrow{\mathrm{pm}}\) × \(\overrightarrow{\mathrm{B}}\) where \(\overrightarrow{\mathrm{pm}}\) = \(\overrightarrow{\mathrm{IA}}\)
9. If there are N turns in the rectangular loop, the torque is given by
τ = NIAB sinθ
Special cases:
a) When θ =90° or the plane of the loop is parallel to the magnetic field, the torque on the current loop is maximum.
τmax = IAB
b) When θ = 0°/180° or the plane of the loop is perpendicular to the magnetic field, the torque on the current loop is zero.

Question 11.
Discuss the conversion of a galvanometer into an ammeter and also a voltmeter.
Answer:
Galvanometer to an ammeter:
1. The ammeter must offer low resistance, no change in current.
2. Galvanometer is converted to an ammeter by connecting low resistance in parallel.
3. Low resistance – Shunt resistance.
4. I – current in the circuit.
5. Ig – Galvanometer current.
6. Rg – Galvanometer resistance.
7. Current through the shunt, Is = I – Ig
8. Vgalvanometer = Vshunt
Ig Rg = (I – Ig)S
\(\mathrm{S}=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}\) ………..(1)
\(\mathrm{I}_{\mathrm{g}}=\frac{\mathrm{S}}{\mathrm{S}+\mathrm{R}_{\mathrm{g}}} \mathrm{I}\)
Ig α I
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 36

9. The deflection in the galvanometer is proportional to current θ α Ig
\(\begin{array}{l}
\frac{1}{\mathrm{R}_{\text {eff }}}=\frac{1}{\mathrm{R}_{\mathrm{g}}}+\frac{1}{\mathrm{~s}} \\
\frac{1}{\mathrm{R}_{\text {eff }}}=\frac{\mathrm{R}_{\mathrm{g}} \mathrm{S}}{\mathrm{R}_{\mathrm{g}}+\mathrm{S}}=\mathrm{R}_{\mathrm{a}}
\end{array}\)
10. Ra is very low.
11. An ideal ammeter has zero resistance.

Galvanometer to a voltmeter:

  1. The voltmeter must have high resistance.
  2. A galvanometer is converted to a voltmeter by connecting high resistance in series.
  3. Rg – Galvanometer resistance.
  4. Ig – Galvanometer current.
  5. Rh – High resistance
  6. Since it is connected in series, the current is the same.
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 37
    i.e. I = Ig
    Ig = \(\frac{\text { Potential difference }}{\text { total resistance }}\)
    Rv = Rg + Rh
    \(\begin{array}{l}
    \mathrm{Ig}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{g}}+\mathrm{R}_{\mathrm{h}}} \\
    \mathrm{R}_{\mathrm{h}}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{R}_{\mathrm{g}}
    \end{array}\)
    ∴Ig α V
  7. An ideal voltmeter has infinite resistance.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 12.
Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law.
Answer:

  1. Consider a solenoid of length L having N turns.
  2. Solenoid’s diameter is much smaller compared to its length.
    The magnetic field inside the solenoid:
  3. Consider a rectangular loop abcd.
  4. From Ampere’s circuital law,
  5. Elemental length bc and da are perpendicular to the magnetic field.
    ∴ \(\int_{b}^{c} \vec{B} \cdot d \vec{l}=\int_{b}^{c}|\vec{B}| d \vec{l} \mid \cos 90^{\circ}=0 \quad \int_{d}^{a} \vec{B} \cdot d \overrightarrow{1}=0\)
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 38
    Amperian loop for solenoid.
    \(\begin{array}{l}
    \oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{l}}=\mu_{\mathrm{O}} \mathrm{I}_{\text {enclosed }} \\
    \mathrm{C} \\
    \oint_{\mathrm{C}} \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{l}}=\int_{\mathrm{a}}^{b} \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{1}+\int_{\mathrm{b}}^{c} \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{l}}+\int_{\mathrm{c}}^{\mathrm{d}} \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{l}}+\int_{\mathrm{d}}^{\mathrm{a}} \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{1}
    \end{array}\)
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 39

Question 13.
Derive the expression for the force between two parallel, current-carrying conductors.
Answer:
1. Two long parallel current-carrying conductors separated by a distance V in air.
2. I1, I2 – current passing through the conductors A and B in the same direction.
3. Magnetic field due to current I2 of elemental length dl in A
\(\begin{array}{l}
\overrightarrow{\mathrm{B}}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi \mathrm{r}}(-\hat{\mathrm{i}}) \\
\overrightarrow{\mathrm{B}}_{1}=\frac{-\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi \mathrm{r}}(\hat{\mathrm{i}}
\end{array}\)
4. The direction of magnetic field is perpendicular to the plane of the paper and outwards.
5. A force on a small elemental length dl in B, experience magnetic field B1.
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 40
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 41
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 42
6. Attractive force – direction of electric current is same.
7. Repulsive force – direction of electric current is opposite.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 14.
Give an account of magnetic Lorentz force.
Answer:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 43

Direction of the Lorentz force on
(a) positive charge
(b) negative charge
1. An electric charge q is moving with velocity \(\overrightarrow{\mathrm{v}}\) in the magnetic field \(\overrightarrow{\mathrm{B}}\), experience a force called magnetic force \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) deduced by Lorentz.
\(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) = q (\(\overrightarrow{\mathrm{v}}\) × \(\overrightarrow{\mathrm{B}\)) …………(1)
\(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) = qvBsin θ ……………(2)
Equations (1) & (2) implies

2. \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) is directly proportional to the magnetic field \(\overrightarrow{\mathrm{B}}\).

3. \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) is directly proportional to the velocity \(\overrightarrow{\mathrm{V}}\).

4. \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) is directly proportional to sine of the angle between the velocity and magnetic field.

5. \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) is directly proportional to the magnitude of the charge q.

6. \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) is always perpendicular to \(\overrightarrow{\mathrm{V}}\) and \(\overrightarrow{\mathrm{B}}\).

7. The direction of \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) on negative charge is opposite to the direction of \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) on positive charge.

8. If velocity \(\overrightarrow{\mathrm{v}}\) of the charge is along magnetic field, \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) = 0.

Question 15.
Compare the properties of soft and hard ferromagnetic materials.
Answer:

PropertiesSoft ferromagnetic materialsHard ferromagnetic materials
1. Absence of external field.Magnetisation disappearMagnetisation remains
2. Area of the loopSmallLarge
3. RetentivityLowHigh
4. CoercivityLowHigh
5. Susceptibility and magnetic permeabilityHigh

 

Low

 

6. Hysteresis lossLessMore
7. UsesSolenoid core, transformer core, and electromagnetsPermanent magnet
8. ExamplesSoft iron, Mumetal, Stelloy, etc.Steel, Alnico, Lodestone, etc.

Question 16.
Derive an expression for the force acting on a current-carrying conductor placed in a magnetic field.
Answer:

  1. The force experienced is equal to the sum of Lorentz forces on the individual charge carriers in the wire.
  2. Consider a small length of wire ‘dl’, area – A, Current – I.
  3. The free electrons drift opposite to the direction of the current.
  4. The relation between I & Vd is given by
    I = neAVd
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 44
  5. In magnetic field, the force experienced is given by
    \(\overrightarrow{\mathrm{F}}\) = -e(\(\overrightarrow{\mathrm{v}}_{\mathrm{d}}\) × \(\overrightarrow{\mathrm{B}}\))
    n → number of free electrons per unit volume.
    n = \(\frac{\mathrm{N}}{\mathrm{V}}\)
    V = A dl
  6. Lorentz force = n A dl × – e(\(\overrightarrow{\mathrm{v}}_{\mathrm{d}}\) × \(\overrightarrow{\mathrm{B}}\))
    dF = -enA dl (\(\overrightarrow{\mathrm{v}}_{\mathrm{d}}\) × \(\overrightarrow{\mathrm{B}}\))
  7. Current element, I\(\overrightarrow{\mathrm{dl}}\) = – enA\(\overrightarrow{\mathrm{v}}_{\mathrm{d}}\)
    ∴ d\(\overrightarrow{\mathrm{F}}\) = (I\(\overrightarrow{\mathrm{dl}}\) × \(\overrightarrow{\mathrm{B}}\))
    \(\overrightarrow{\mathrm{F}}\) = (I \(\overrightarrow{\mathrm{l}}\) × \(\overrightarrow{\mathrm{B}}\))
    F = BIl sin θ
    Case (i):
    If the conductor is along the magnetic field; θ = 0° ; F = 0.
    Case (ii):
    If the conductor is perpendicular to the magnetic field; θ = 90°; F = BIl.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

IV. Numerical problems:

Question 1.
A bar magnet having a magnetic moment pm is cut into four pieces i.e., first cut in two pieces along the axis of the magnet and each piece is further cut into two pieces. Compute the magnetic moment of each piece.
Answer:
If cut along the axis of the magnet of length ‘l’ into 4 pieces,
New pole strength M’= \(\frac{\mathrm{m}}{4}\)
New length l’ = l
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 45

Question 2.
A conductor of linear mass density 0.2 g m1 suspended by two flexible wire as shown in figure. Suppose the tension in the supporting wires is zero when it is kept inside the magnetic field of I T whose direction is into the page. Compute the current inside the conductor and also the direction for the current.
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 46
Answer:
To have zero tension in the wires, the magnetic force per unit length must be upwards and equal to the weight per unit length.
∴\(\left|\frac{F_{\mathrm{m}}}{\mathrm{L}}\right|=\mathrm{BI}=\frac{\mathrm{mg}}{\mathrm{L}}\)
I = \(\frac{\left(\frac{\mathrm{m}}{\mathrm{L}}\right) \mathrm{g}}{\mathrm{B}}\)
\(\frac{\mathrm{m}}{\mathrm{L}}\) = 0.2 gm-1
= 0.2 × 10-3 kgm-1
B = It
g = 10ms-2
∴I = \(\frac{0.2 \times 10^{-3} \times 10}{1}\) = 2 × 10-3 A
I = 2 mA

Question 3.
A circular coil with a cross-sectional area of 0.1 cm2 is kept in a uniform magnetic field of strength 0.2 T. If the current passing in the coil is 3A and the plane of the loop is perpendicular to the direction of magnetic field. Calculate
a) total torque on the coil
b) total force on the coil
c) average force on each electron in the coil due to the magnetic field. The free electron density for the material of the wire is 1028 m-3.
Answer:
N= 1
A = 0.1 × 10-4 m2
B = 0.2 T
I = 3A
θ = 0° [Plane is perpendicular to the field]
n = 1028 m-3
a) Torque, τ = NIB A sinθ
= l × 3 × 0.2 × 0. 1 × 10 × sin 0°
Torque = 0
b) Total force on a current loop is always zero in a magnetic field.
c) for free electron, drift velocity,
vd = q(\(\overrightarrow{\mathrm{v}}\) × \(\overrightarrow{\mathrm{B}}\))
= qvdB sin 90°
F = Bqvd
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 47
= \(\frac{0.6}{0.1}\) × 1028 × 104
= 6 × 10-24
Average Force, F = 0.6 × 10-23 N.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 4.
A bar magnet is placed in a uniform magnetic field whose strength is 0.8 T. If the bar magnet is oriented at an angle 30° with the external field experiences a torque of 0.2 Nm. Calculate
i) the magnetic moment of the magnet
ii) the work is done by the magnetic field in moving it from the most stable configuration to the most its configuration and also compute the work done by the applied magnetic field in this case.
Answer:
i) τ = 0.2 Nm
τ = PmB sinθ
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 48

ii) W = Uf – Ui
Uf = μB cos 180°
Ui = -μB cos 0°
∴W = μB cos 180° – (μB cos 0°)
= μB + μB
W = 2μB;
W = 2 PmB
W = 2 × 0.5 × 0.8
W = 0.85 J and
Wmag = -0.85 J

Question 5.
A non-conducting sphere has a mass of 100g and a radius of 20cm. A flat compact coil of wire with turns 5 is wrapped tightly around it with each turn concentric with the sphere. This sphere is placed on an inclined plane such that the plane of the coil is parallel to the inclined plane. A uniform magnetic field of 0.5 T exists in the region in a vertically upward direction. Compute the current I required to rest the sphere in equilibrium.
Answer:
The sphere is in translational equilibrium,
fa – Mg sinθ = 0 ……………..(1)
The sphere is in rotational equilibrium.
torque = μB sinθ (Produces by magnetic field clockwise)
Frictional force (anticlockwise torque) = fsR
R – radius of the sphere
fsR – μB sinθ = 0 ……………..(2)
Substitute (1) in (2)
fs = mg sinθ
∴ mg sinθ R – μB sinθ = 0
mg sinθ R = μB sinθ
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 49
μB = mgR ……………(3)
μ = NIA
μ = NIπr² …………….(4)
NIπr² = mgR
I = \(\frac{\mathrm{mgR}}{\mathrm{BN} \pi \mathrm{R}^{2}}\)
I = \(\frac{\mathrm{mg}}{\mathrm{BN} \pi \mathrm{R}}\)
m = 100g = 100 ×10-3 kg
= 0.1 kg
R = 20 cm = 0.2m
B = 0.5 T
N = 5 turns
g = 10 m/s2
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 50

Question 6.
Calculate the magnetic field at the center of a square loop which carries a current of 1.5 A, length of each side being is 50cm.
Answer:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 51
Similarly for four sides BC, CD, DA
B = \(\frac{4 \mu_{0} I}{\sqrt{2} \pi L}\)
Here,
I = 1.5A
L = 50 cm
L = 0.5 m
B = \(\frac{4 \mu_{0} I}{\sqrt{2} \pi L}\)
\(B=\frac{4 \times 4 \pi \times 10^{-7} \times 1.5}{\sqrt{2} \pi \times 0.5}\)
= \(\frac{24 \times 10^{-7}}{70.7 \times 10^{-2}}\)
= 0.3394 × 10-7 × 102
B = 3.39 × 10-6 T
B = 3.4 × 10-6 T

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Part II:

12th Physics Guide Magnetism and Magnetic Effects of Electric Current Additional Questions and Answers

I. Choose the incorrect pair:

Question 1.
1. Aurora borealis – northern lights
2. Aurora australis – southern lights
3. Bluish-Orange – collision
4. Purplish – red – Nitrogen molecules
Answer:
Bluish orange – collision

Question 2.
1. Anticlockwise current – North pole
2. Clockwise current – South pole
3. Angular momentum – \(\frac{\mathrm{nh}}{2 \pi}\)
4. Bohr magneton – \(\frac{4 \pi \mathrm{m}}{\mathrm{eh}}\)
Answer:
Bohr magneton – \(\frac{4 \pi \mathrm{m}}{\mathrm{eh}}\)

Question 3.
1. Tangent law – BH tan θ
2. Tangent galvanometer – Joule’s law
3. Magnetic polarity – End rule
4. Ampere’s – circuital law – Biot – Savart law
Answer:
Tangent galvanometer – Joule’s law

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

II. Choose the odd man out:

Question 1.
High permeability, high retentivity, low coercivity, thin hysteresis loop.
Answer:
High retentivity – property of permanent magnet. Others are properties of electromagnets.

Question 2.
William Gilbert, Gover, Hans Christian Oersted, George Simon ohm.
Answer:
George Simon Ohm – Electricity. Others explain the magnetic effect.

Question 3.
Aluminium, Platinum, Chromium, Bismuth
Answer:
Bismuth – Diamagnetic. Others are paramagnetic.

Question 4.
Phosphor bronze, suspension strip, torsion head, Dees
Answer:
Dees – Parts of the cyclotron. Others are parts of moving coil galvanometer.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

III. Fill in the blanks:

Question 1.
When a charged particle enters in a uniform magnetic field, its kinetic energy ________________
Answer:
remain constant.

Question 2.
Work done by the magnetic Lorentz force is ________________
Answer:
zero.

Question 3.
The direction of the force on a current-carrying conductor placed in a magnetic field is given by ________________
Answer:
Fleming’s left-hand rule.

Question 4.
The tangent galvanometer is most sensitive at a deflection of ________________
Answer:
45°

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

IV. Match the following:

Question 1.

III
i. Magnetic dipole momenta. NA-1 m-1
ii. Magnetic fieldb. weber metre-2
iii. Magnetic fluxc. Am2
iv. Magnetic flux densityd. weber

Answer:
i. c
ii. a
iii. d
iv. b

Question 2.

III
i. Charge of an electron (e)a. 9.27 x 10-24 Am2
ii. Gyromagnetic ratio (e/2m)b. 9.11 x 10-31 kg
iii. Bohr magneton (μB)c. 1.60 x 10-19 C
iv. Mass of an electron (m)d. 8.78 x l010 Ckg-1

Answer:
i. c
ii. d
iii. a
iv. b

Question 3.

III
i. Diamagnetica. Aluminium
ii. Paramagneticb. Steel
iii. Soft Ferromagneticc. Copper
iv. Hard Ferromagneticd. Stelloy

Answer:
i. c
ii. a
iii. d
iv. b

Question 4.

III
i. Curie’s lawa. \(\frac{\mathrm{C}}{\mathrm{T}-\mathrm{T}_{r}}\)
ii. Coulomb lawb. \(\frac{\mathrm{Kq}_{\mathrm{m}_{\mathrm{A}}} \mathrm{q}_{\mathrm{m}_{\mathrm{B}}}}{r^{2}}\)
iii. Tangent lawc. C/T
iv. Curie – Weiss lawd. BH tan θ

Answer:
i. c
ii. b
iii. d
iv. a

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

V. Choose the correct pair:

Question 1.
1) Right-hand thumb rule – a. Direction of the magnetic moment
2) Gauss’s law – b. Electrostatics
3) Biot – Savart law – c. Direction of current
4) Maxwell’s right-hand cork screw rule – d. The direction of the magnetic field.
Answer:
Biot – Savart law – Direction of current

Question 2.
1. Radius (r) = \(\frac{\mathrm{qB}}{\mathrm{mV}}\)
2. Frequency (fosc) = \(\frac{2 \pi \mathrm{m}}{\mathrm{qB}}\)
3. Time period (T) = \(\frac{\mathrm{qB}}{2 \pi \mathrm{m}}\)
4. Kinetic energy (K.E) = \(\frac{q^{2} B^{2} r^{2}}{2 m}\)
Answer:
Kinetic energy (K.E) = \(\frac{q^{2} B^{2} r^{2}}{2 m}\)

Question 3.
1. Figure of merit – deflection of two-scale division
2. Current sensitivity – Deflection produced per unit time
3. Voltage sensitivity – Deflection produced per unit voltage
4. Phosphor – bronze wire – Large couple per unit twist
Answer:
Voltage sensitivity – Deflection produced per unit voltage

Question 4.
1. Ammeter – measures the potential difference
2. Voltmeter – measure current
3. Shunt resistance – Parallel connection with a galvanometer
4. Ideal Ammeter – infinite resistance
Answer:
Shunt resistance – Parallel connection with galvanometer

Question 5.
1. Gyro – magnetic ratio – 8.78 × 1010 Ckg -1
2. Mass of an electron – 9.11 × 1031 kg
3 Planck’s constant – 6.63 × 10-24 Js
4 Bohr magneton – 9.27 × 10-34 Am2
Answer:
Gyromagnetic ratio – 8.78 × 1010 Ckg1

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

VI. Assertion and Reason:

a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
b) If both assertion and reason are true but the reason is not the correct explanation of the assertion.
c) If an assertion is true but the reason is false.
d) If the assertion and reason both are false.

Question 1.
Assertion:
Cyclotron does not accelerate electrons.
Reason:
The mass of the electron is very small.
Answer:
a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

Question 2.
Assertion:
The poles of the magnet cannot be separated by breaking them into two pieces.
Reason:
The magnetic moment will be reduced to half when a magnet is broken into two equal pieces.
Answer:
b) If both assertion and reason are true but reason is not the correct explanation of the assertion.

Question 3.
Assertion:
When a magnet is brought near iron nails, only translatory force acts on it.
Reason:
The field due to a magnet is generally uniform.
Answer:
d) If the assertion and reason both are false.

Question 4.
Assertion:
The reduction factor of a tangent galvanometer helps in reducing deflection to current.
Reason:
Reduction factor increases with increase of current.
Answer:
c) If assertion is true but reason is false.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

VII. Choose the correct statements:

Question 1.
(i) A freely suspended magnet always comes to rest in south-north direction.
(ii) Earth itself behaves like a gigantic powerful bar magnet.
(iii) Earth’s magnetic field is due to hot rays coming out from the sun.
(iv) The north pole of a magnetic compass needle is attracted towards the magnetic north pole of the earth.
Answer:
Statements (ii) and (iii) are correct.

Question 2.
(i) When \(\overrightarrow{\mathrm{B}}\) is parallel to the surface, θ = 90°, the magnetic flux, φB = 0.
(ii) Magnetic flux is a vector quantity.
(iii) The S.l Unit for magnetic flux is maxwell.
(iv) The CGS unit for magnetic flux is weber.
Answer:
Statement (i) is correct.

Question 3.
(i) T.G consists of a copper coil wounded on a magnetic circular frame.
(ii) T.G is provided with only one coil of a different number of turns.
(iii) A thin aluminium pointer is attached to the magnetic needle.
(iv) The frame is made up of silver.
Answer:
Statement (iii) is correct.

Question 4.
(i) For a fixed current I, the magnetic field outside the solenoid is constant.
(ii) The magnetic field of the solenoid demagnetizes the iron bar.
(iii) A solenoid is a long coil of wire closely wound in the form of a ring.
(iv) Solenoids can be used as electromagnets.
Answer:
Statement (iv) is correct.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

VIII. Choose the incorrect statements:

Question 1.
i) Tangent galvanometer is a device for measuring current.
ii) The unit of reduction factor of T.G. is ampere.
iii) The tangent galvanometer is most sensitive at a deflection of 45°.
iv) The allowed deflection in T.G is between 45° and 55°.
Answer:
Statement (iv) is incorrect.

Question 2.
i) Permeability of a medium is μ0r
ii) Permeability of free space is 4π × 10-7 henry / meter.
iii) Relative permeability of a medium is μ /μ0
iv) The unit of magnetic induction is tesla.
Answer:
Statement (i) is incorrect.

Question 3.
i) The magnetic Lorentz force on a charged particle in the absence of an electric field is \(\overrightarrow{\mathrm{F}}\) = q(\(\overrightarrow{\mathrm{V}}\) × \(\overrightarrow{\mathrm{B}}\))
ii) The magnetic lorentz force on a charged particle in the presence of an electric field is F = q(\(\overrightarrow{\mathrm{V}}\) x \(\overrightarrow{\mathrm{B}}\)) + qE
iii) Work is done by the magnetic Lorentz force is infinite.
iv) The Lorentz force on a charged particle moving perpendicular to the magnetic field is maximum.
Answer:
Statement (iii) is incorrect.

Question 4.
i) In a circuit, the ammeter should be connected in series.
ii) Shunting means connecting a high resistance in parallel with the galvanometer.
iii) In a circuit, a voltmeter should be connected in parallel.
iv) The resistance of an ideal ammeter is zero.
Answer:
Statement (ii) is incorrect.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

IX. Choose the best answer:

Question 1.
Magnetic induction at the center of the ring is shown in the diagram.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 52
a) 0.2 T
b) 0.5 T
c) 0 T
d) 1.0 T
Answer:
c) 0 T

Question 2.
Magnetic field lines-
(a) cannot intersect at all
(b) intersect at infinity
(c) intersect within the magnet
(d) intersect at the neutral points
Answer:
(a) cannot intersect at all

Question 3.
When the number of turns in a galvanometer is doubled.
a) both current sensitivity and voltage sensitivity are doubled
b) current sensitivity does not change but voltage sensitivity is doubled
c) voltage sensitivity does not change but current sensitivity is doubled
d) both current sensitivity and voltage sensitivity do not change
Answer:
a) both current sensitivity and voltage sensitivity are doubled.

Question 4.
The magnetic declination angle for Chennai is
a) 1°8′
b) 2°8′
c) -1°8′
d) -2°8′
Answer:
c) -1°8′

Question 5.
Earth’s magnetic field always has a horizontal component except for at-
(a) equator
(b) magnetic pole
(c) a latitude of 60°
(d) a latitude of 50°
Answer:
(b) magnetic pole

Question 6.
The direction of magnetic moment is from ________to________
a) South pole, north pole
b) North pole, north pole
c) North pole, South pole
d) South pole, the south pole
Answer:
a) Southpole, north pole

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 7.
A simple pendulum with charged bob is oscillating with time period T and lets O be the angular displacement. if the uniform magnetic field is switched on in a direction perpendicular to the plane of oscillation then
a) time period will decrease but θ will remain constant
b) time period will remain constant but θ will decrease
c) both time period and θ remain constant
d) both time period and θ will decrease
Answer:
c) both time period and θ remain constant

Question 8.
At magnetic poles the angle of dip is-
(a) 45°
(b) 30°
(c) 0°
(d) 90°
Answer:
(d) 90°

Question 9.
The magnitude and direction of Lorentz force is _______
a) \(\overrightarrow{\mathrm{F}}=q(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{V}})\)
b) \(F=q(\vec{B} \times \vec{V})\)
c) \(\overrightarrow{\mathrm{F}}=q(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{E}})\)
d) \(\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{V}} \cdot \overrightarrow{\mathrm{B}})\)
Answer:
b) \(F=q(\vec{B} \times \vec{V})\)

Question 10.
1 Weber = _______.
a) 106 maxwell
b) 10-6 maxwell
c) 108 maxwell
d) 10-8 maxwell
Answer:
c) 108 maxwell

Question 11.
Magnetic field at any point at a distance R due to a long straight conductor carrying current varies is
a) R2
b) R
c) \(\frac{1}{R^{2}}\)
d) \(\frac{1}{R}\)
Answer:
d) \(\frac{1}{R}\)

Question 12.
Demagnetization of a magnet can be done by-
(a) rough handling
(b) magnetizing in the opposite direction
(c) heating
(d) all the above
Answer:
(d) all the above

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 13.
For free space, the relative permeability is ______
a) unity
b) zero
c) greater than one
d) constant
Answer:
a) unity

Question 14.
An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current I along the same direction is shown in fig. Magnitude of force per unit length on the middle wire B is given by:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 53
a) \(\frac{21^{2} \mu_{0}}{\pi d}\)
b) \(\frac{1^{2} \mu_{0}}{\sqrt{2 \pi d}}\)
c) \(\frac{\sqrt{21}^{2} \mu_{0}}{\pi \mathrm{d}}\)
d) \(\frac{1^{2} \mu_{0}}{2 \pi d}\)
Answer:
b) \(\frac{1^{2} \mu_{0}}{\sqrt{2 \pi d}}\)

Question 15.
The magnetic susceptibility of Aluminium is ______.
a) 2.3 × 105
b) 2.3 × 10-5
c) 2.2 × 105
d) 2.2 × 10-5
Answer:
b) 2.3 × 10-5

Question 16.
Which of the following is most suitable for the core of an electromagnet?
(a) air
(b) soft iron
(c) steel
(d) Co-Ni alloy
Answer:
(b) soft iron

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 17.
Dimension of Resistance is ______.
a) ML2 T-3 A-2
b) ML2 T-1 A-1
c) ML2 T2 A-3
d) ML2 T-1 A-2
Answer:
a) ML2 T-3 A-2

Question 18.
Photo has spin _______.
a) S = \(\frac{1}{2}\)
b) S = –\(\frac{1}{2}\)
c) S = 1
d) S = 2
Answer:
c) S = 1

Question 19.
At Curie point, a ferromagnetic material becomes-
(a) non magnetic
(b) diamagnetic
(c) paramagnetic
(d) antiferromagnetic
Answer:
(c) paramagnetic

Question 20.
The variation of magnetic susceptibility (k) with magnetizing field (H) for a paramagnetic substance is _______
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 54
Answer:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 55

Question 21.
The force per unit length between two conductors carrying one ampere current each and separated by a distance of 1 m is
a) 2π × 10-7 N
b) 2 × 10-7 N
c) 4π × 107 N
d) 2π × 107 N
Answer:
b) 2π × 10-7 N

Question 22.
A cyclotron cannot accelerate _______.
a) electrons
b) protons
c) deuterons
d) α – particles
Answer:
a) electrons

Question 23.
\(\oint \overrightarrow{\mathbf{B}} \cdot \overrightarrow{\mathbf{d}}l\) is equal to
a) μI0
b) μ0I0
c) \(\frac{\mu_{\mathrm{o}}}{\mathrm{I}_{\mathrm{o}}}\)
d) \(\frac{\mathrm{I}_{\mathrm{o}}}{\mu_{\mathrm{o}}}\)
Answer:
b) μ0I0

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 24.
A magnetic needle is placed in a uniform magnetic field. It experience-
(a) a force and a torque
(b) a force but not a torque
(c) a torque but not a force
(d) neither a force nor a torque
Answer:
(c) a torque but not a force

Question 25.
In a galvanometer, the magnetic field used is _____
a) radial
b) parallel
c) perpendicular
d) tangential
Answer:
a) radial

Question 26.
In a tangent galvanometer, a current 1A produces a deflection of 30° the current required to produce a deflection of 60° is
a) 3 A
b) 2 A
c) √2A
d) √3A
Answer:
a) 3A

Question 27.
A magnetic needle suspended freely-
(a) orients itself in a definite direction
(b) remains in any direction
(c) become vertical with N-pole up
(d) become vertical with N-pole down
Answer:
(a) orients itself in a definite direction

Question 28.
The voltmeter resistance is
a) Rg = Rv +Rh
b) Rv = Rg+ Rh
c) Rh = Rv + Rg
d)Rg = Rh – Rv
Answer:
b) Rv = Rg+ Rh

Question 29.
Kirchoff’s I an II laws are based on conservation of
a) charge and energy
b) energy and charge
c) energy and voltage
d) energy and current
Answer:
a) charge and energy

Question 30.
A moving charge produces-
(a) an electric field only
(b)a magnetic field only
(c) both electric and magnetic fields
(d) neither an electric nor a magnetic field
Answer:
(c) both electric and magnetic fields

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

X. Two Mark Questions:

Question 1.
Define magnetic declination (D).
Answer:
The angle between magnetic meridian at a point and geographical meridian is called the declination or magnetic declination (D).

Question 2.
What is a magnetizing field?
Answer:

  1. The magnetic field is used to magnetize a sample or specimen.
  2. Vector quantity and denoted by \(\overrightarrow{\mathrm{H}}\).
  3. Unit – Am-1

Question 3.
What is relative permeability?
Answer:

  1. The ratio between absolute permeability of the medium to the permeability of free space.
  2. \(\mu_{\mathrm{r}}=\frac{\mu}{\mu_{\mathrm{o}}}\)
  3. For free space, μr = 1
  4. Isotropic medium, μ → scalar.
  5. Non-isotropic medium, μ → tensor.

Question 4.
What is the intensity of magnetisation?
Answer:

  1. The net magnetic moment per unit volume of material.
  2. \(\overrightarrow{\mathrm{M}}=\frac{\text { magnetic moment }}{\text { volume }}=\frac{1}{\mathrm{~V}} \overrightarrow{\mathrm{p}}_{\mathrm{m}}\)
  3. Unit – Am-1

Question 5.
State Curie’s law.
Answer:

  1. The magnetic susceptibility decreases with increase in temperature.
  2. \(\begin{array}{l}
    \mathrm{X}_{\mathrm{m}} \alpha \frac{1}{\mathrm{~T}} \\
    \mathrm{X}_{\mathrm{m}}=\frac{\mathrm{C}}{\mathrm{T}}
    \end{array}\)
  3. C – curie constant, T – Kelvin.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 6.
Define in terms of magnetising field.
Answer:
The magnetic field which is used to magnetize a sample or specimen is called the magnetising field. Magnetising field is a vector quantity and it denoted by \(\vec { H } \) and its unit is A m-1.

Question 7.
What kind of magnetic properties electromagnet has?
Answer:
Materials should possess

  • High initial permeability
  • Thin hysteresis loop.
  • Low retentivity
  • Low coercivity
  • Examples: Soft iron and Mumetal (Nickel Iron alloy)

Question 8.
What is needed to design transformer cores?
Answer:

  1. Large magnetic induction
  2. High initial permeability
  3. Thin hysteresis loop with a small area.

Question 9.
State ‘One ampere’.
Answer:
One ampere is defined as that current when it is passed through each of the two infinitely long parallel straight conductors kept at a distance of one meter apart in a vacuum causes each conductor to experience a force of 2 x 10-7 newton per meter length of the conductor.

Question 10.
Define magnetic dipole moment of the current loop.
Answer:
The product of the current and area of the loop.

Question 11.
What is a solenoid?
Answer:
A long coil of wire closely wound in the form of a helix.

Question 12.
What is toroid?
Answer:
Both the ends of a solenoid are joined together to form a closed ring shape.

Question 13.
Define one tesla.
Answer:
Unit charge moving with unit velocity experiences unit force.
\(1 \mathrm{~T}=\frac{1 \mathrm{Ns}}{\mathrm{Cm}}=1 \frac{\mathrm{N}}{\mathrm{Am}}=1 \mathrm{NA}^{-1} \mathrm{~m}^{-1}\)

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 14.
State Lorentz force.
Answer:

  1. The total force experienced by the charge.
  2. \(\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{q}} \mathrm{E}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})\)

Question 15.
Give the expression for cyclotron frequency or gyro frequency.
Answer:
f = \(\frac{\mathrm{qB}}{2 \pi \mathrm{m}}\)
ω = 2πf = \(\frac{\mathrm{qB}}{\mathrm{m}}\)

Question 16.
Give the uses of the mass spectrometer.
Answer:
A mass spectrometer is used in many areas

  1. In sciences; especially in medicine, space science, geology, etc.
  2. In medicine, anaesthesiologists use it to measure respiratory gases.
  3. Biologists use it to determine the reaction mechanisms in photosynthesis.

Question 17.
How is a galvanometer is converted into an ammeter and a voltmeter?
Answer:

  1. A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.
  2. A galvanometer is converted into a voltmeter by connecting a high resistance Rh in series with the galvanometer.

Question 18.
What is the reason for the hemispherical magnetic poles used in moving coil galvanometer?
Answer:

  1. The hemispherical magnetic poles produce a radial magnetic field.
  2. The plane of the coil is parallel to the magnetic field in all its positions.

Question 19.
Why did we use the coils in different thicknesses in the Tangent galvanometer?
Answer:
The coil consists of 2, 5, and 50 turns of different thicknesses to measure the current of different strengths.

Question 20.
The electrons cannot be accelerated by a cyclotron, why?
Answer:

  1. At high frequencies, a relativistic variation of the mass of the electron is appreciable.
  2. Hence electrons cannot be accelerated by a cyclotron.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 21.
Why Phosphor bronze wire is used as suspension wire in moving coil galvanometer?
Answer:

  1. Because phosphor bronze wire produces a small couple per unit twist.
  2. A small couple per unit twist increases the sensitivity of the galvanometer.

Question 22.
Why freely suspended bar magnets in your lab experience only torque but not any translatory motion even though Earth has a non-uniform magnetic field?
Answer:
Earth’s magnetic field in the lab is uniform. Therefore, it experiences torque and not a translatory motion.

Question 23.
List the differences between coulomb force and Lorentz force.
Answer:

Coulomb forceLorentz force
1. Force between two charged particles separated by a distance.Force experienced by current carrying conductor placed in a uniform magnetic field.
2. It decreases with an increase in distance.It remains constant.
3. It depends upon the medium.It doesn’t depend upon the medium.

Question 24.
What is the Meissner effect?
Answer:
The expulsion of magnetic flux from a superconductor during its transition to the superconducting state.

Question 25.
Difference between Tangent Galvanometer and Moving coil galvanometer.
Answer:

Tangent GalvanometerMoving coil galvanometer
1. It works on the principle of Tangent law.It works on the principle that when a current carrying coil suspended in a uniform magnetic field
2. It consists of a non-magnetic circular frame.It consists of a rectangular coil of fine insulated copper wire.
3. It is used to measure magnetic induction.It is used to measure potential.

Question 26.
Difference between the natural magnet and artificial magnet.
Answer:

Natural magnetArtificial magnet
1. Formed in nature.Man-made magnets.
2. It has no regular size and shape.It has the perfect size and shape.
3. Ex: Magnetite or lode stoneEx: Bar magnet.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

XI. Three Mark Questions:

Question 1.
State Curie-Weiss law.
Answer:

  1. The susceptibility of the material above the curie temperature is given by
  2. \(X_{m}=\frac{C}{T-T_{C}}\)
    This relation is called the Curie – Weiss law.
  3. C – Curie constant
    T – Temperature
    Tc – Curie temperature

Question 2.
Define in terms of the Current Sensitivity of a galvanometer.
Answer:
It is defined as the deflection produced per unit current flowing through it.
Is = \(\frac { θ }{ I }\) = \(\frac { N A B }{ K }\) ⇒ Is = \(\frac { 1 }{ G }\)

Question 3.
What are the applications of hysteresis loop?
Answer:

  1. It provides information such as retentivity, coercivity, permeability, susceptibility, and energy loss.
  2. It helps to select and suitable material for a given purpose.

Question 4.
State Maxwell’s right-hand corkscrew rule.
Answer:

  1. If a right-handed corkscrew is rotated, the direction of current is the same as the direction in which the screw advances and the direction of rotation of the screw gives the direction of the magnetic field.
  2. This rule is used to determine the direction of the magnetic field.

Question 5.
Write short notes on MRI (Magnetic Resonance Imaging).
Answer:

  1. It helps to diagnose or monitor treatment for a variety of abnormal conditions in the head, chest, abdomen, and pelvis.
  2. It consists of a superconducting wire to produces a strong magnetic field.
  3. It uses radiofrequency pulses and a computer to produce pictures of organs.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 6.
Difference between Coulomb’s law and Biot-Savart’s law.
Answer:

Coulomb’s lawBiot -Savart’s law
1. It is produced by a scalar source, ie. an electric charge q.It is produced by a vector source ie) current element I \(\overrightarrow{\mathrm{dl}}\).
2. It is directed along the position vector joining the source and the point at which the field is calculated.It is directed perpendicular to the position vector \(\overrightarrow{\mathrm{r}}\) and the current element I \(\overrightarrow{\mathrm{dl}}\).
3. It does not depend on the angle.It depends on the angle between the position vector \(\overrightarrow{\mathrm{r}}\) and the current element I \(\overrightarrow{\mathrm{dl}}\).

Question 7.
List out the deflection based on the velocity – velocity selector.
Answer:

VelocityDeflection of charged particle
1. V >V0In the direction of Lorentz force.
2. V < V0In the direction of Coulomb force.
3. V = V0No deflection and the particle moves straight.

Question 8.
Define the current sensitivity of a galvanometer.
Answer:

  1. The deflection produced when unit current passes through the galvanometer.
  2. The galvanometer produces a large deflection for a small current.
  3. Current sensitivity.
    \(\frac{\theta}{\mathrm{I}}=\frac{\mathrm{NBA}}{\mathrm{K}}\)

Question 9.
How can the current sensitivity of the galvanometer increase?
Answer:

  1. By increasing the number of turns.
  2. By increasing the magnetic induction.
  3. By increasing the area of the coil.
  4. By decreasing the couple per unit twist of the suspension wire.

Question 10.
Define voltage sensitivity of a galvanometer?
Answer:

  1. The deflection per unit voltage.
  2. Voltage sensitivity
    \(\frac{\theta}{\mathrm{V}}=\frac{\theta}{\mathrm{IR}_{\mathrm{g}}}=\frac{\mathrm{NBA}}{\mathrm{KR}_{\mathrm{g}}}\)
  3. Rg – Galvanometer resistance.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 11.
Why does the voltage sensitivity remain constant even if the number of turns in increased?
Answer:

  1. When the number of turns is doubled, current sensitivity also doubled.
  2. If the number of turns (n) increases, the resistance (G) increases.
    Hence voltage sensitivity remains constants.
  3. Voltage sensitivity = \(\frac{\mathrm{NBA}}{\mathrm{KR}_{\mathrm{g}}}\)

Question 12.
In a galvanometer, the current sensitivity does not necessarily increase the voltage sensitivity. Give reason.
Answer:

  1. Current sensitivity does not depend on Resistance. It depends on the number of turns.
  2. Voltage sensitivity decreases with an increase in resistance. But this is made constant by increasing the number of turns.
  3. There is no increase in voltage sensitivity.

Question 13.
Derive an expression for a magnetic dipole in a current loop.
Answer:

  1. Consider a circular loop of radius R.
  2. The magnetic field along the axis is given by
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 56
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 57

Question 14.
Why magnetic field outside the solenoid is zero?
Answer:

  1. The magnetic field lines exist outside the solenoid.
  2. But the number of field lines per unit area outside the solenoid is much less compared to the number of field lines per unit area inside the solenoid.
  3. Therefore the magnetic field is very much small, so it is considered to be zero.

Question 15.
Why the resistance of the voltmeter be very large?
Answer:

  1. The resistance of the voltmeter should be very large compared to the resistance of the voltmeter which is connected across to measure the potential difference.
  2. The current through the circuit decreases and the voltmeter shows a low potential difference.
  3. The error is eliminated, when the voltmeter has a high resistance.
  4. An ideal voltmeter has infinite resistance.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 16.
Two wires of equal length are bent in the form of two loops. One loop is square whereas the other is circular. These are suspended in the same magnetic field and the same current is passed through them. Explain with the reason which will experience greater torque?
Answer:
For a given length, the circle has the greatest area.
τ = NIAB.
Torque is directly proportional to area. So, the circular current loop experiences greater torque.

Question 17.
What are the precautions to be taken while handling the Tangent galvanometer?
Answer:

  1. Keep away all the nearly magnets and magnetic materials.
  2. Using a spirit level, by adjusting the levelling screw, the small magnetic needle is exactly horizontal and the coil is exactly vertical.
  3. Make the coil remain in the magnetic meridian.
  4. By rotating the coil, the plane of the coil is parallel to the small magnetic needle.
  5. The compass is rotated such that the pointer reads O° – O°.

Question 18.
Define figure of merit of a galvanometer? Which has more resistance, a galvanometer or a millimeter?
Answer:

  1. The current produces a deflection of one scale division in the galvanometer.
  2. The millimeter can detect a very small amount of current.
  3. Its resistance is low compared to a galvanometer.
  4. ∴ A galvanometer has more resistance than a millimeter.

Question 19.
Mention the factors on which the direction of force experienced by a current-carrying conductor placed in a magnetic field.
Answer:
The direction of force experienced by the current-carrying conductor depends on

  • direction of current
  • the direction of the magnetic field.

Question 20.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is pushed into the coil?
Answer:

  1. The magnetic field lines linked with the coil change.
  2. Electric current to get induced in it.
  3. The needle of the galvanometer will move momentarily in one direction.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

XII. Five Mark Questions:

Question 1.
What are the properties of a magnet?
Answer:

  1. A freely suspended bar magnet will always point along the north-south direction.
  2. The attractive force is maximum near the end of the bar magnet.
  3. A magnet attracts another magnet or magnetic substance towards itself.
  4. When a magnet is broken into pieces, each piece behaves like a magnet with poles at its ends.
  5. Two poles of a magnet have pole strength equal to one another.
  6. The length of the bar magnet is called a geometrical length.
  7. The length between two magnetic poles in a bar magnet is called magnetic length.
  8. The ratio of magnetic length and geometrical length is \(\frac{5}{6}=0.833\)

Question 2.
List out the properties of magnetic field lines.
Answer:

  1. Continuous closed curves.
  2. The direction is from the North to South pole – outside the magnet.
  3. The direction is from the south to north pole – inside the magnet.
  4. The direction of the magnetic field at any point is known by the tangent to the magnetic line of force at that point.
  5. Magnetic field lines never intersect.
  6. The magnetic field is strong where the magnetic field lies crowd.
  7. The magnetic field is weak where magnetic field lines thin out.

Question 3.
Obtain an expression for the potential energy of a bar magnet in a uniform magnetic field.
Answer:

  1. A bar magnet has a dipole moment \(\overrightarrow{\mathrm{Pm}}\)
  2. It makes an angle θ with the uniform magnetic field B.
  3. The torque acting on the dipole
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 58
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 59

case i:
If θ = 0°
U = -PmB cos0 = -PmB ………….(6)
Potential energy is minimum when aligned along the external magnetic field.
case ii:
If θ=180°
U = -PmB(c0s 180°) = -PmB(-1)
U = Pm
Potential energy is maximum when aligned antiparallel to the external magnetic field.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 4.
Discuss the variation of magnetic induction \(\overrightarrow{\mathrm{B}}\) with magnetising field H in a ferromagnetic material, with the graph.
Answer:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 60

  1. A ferromagnetic material is magnetised slowly.
  2. The magnetic induction \(\overrightarrow{\mathrm{B}}\) increases from point A, then attains a saturated level.
  3. This is plotted as AC in the graph.

Saturation magnetization:

  1. The maximum point up to which the material can be magnetised.
  2. Further on reducing magnetising field, the magnetic induction also decreases.
  3. It does not retrace the original path CA. It takes a different path CD.
  4. When the magnetising field is zero, the magnetic induction is not zero. It has a positive value.

Remanence or Retentivity:

  1. The ability of the materials to retain the magnetism when \(\overrightarrow{\mathrm{H}}\) = 0 . (AD)
  2. To demagnetise the material, \(\overrightarrow{\mathrm{H}}\) gradually increases in reverse direction.
  3. \(\overrightarrow{\mathrm{B}}\) decreases along DE and at E, \(\overrightarrow{\mathrm{B}}\) = 0.
  4. AE brings the residual magnetism to zero.

Coercivity:

  1. For the reverse magnetising field, the residual magnetism vanishes.
  2. On increasing \(\overrightarrow{\mathrm{H}}\), \(\overrightarrow{\mathrm{B}}\) also increases along EF, till the saturation point at F.
  3. If \(\overrightarrow{\mathrm{H}}\) decreased or increased, it traces the path FGKC.
  4. This closed curve ACDEFGK is called hysteresis loop, (cycle of magnetisation)
  5. In this cycle, \(\overrightarrow{\mathrm{B}}\) lags behind \(\overrightarrow{\mathrm{B}}\)
  6. Lagging of \(\overrightarrow{\mathrm{B}}\) behind \(\overrightarrow{\mathrm{B}}\) is called hysteresis.
  7. Hysteresis means “lagging behind”.

Question 5.
Discuss the magnetic field around
i) a straight conductor carrying current
ii) circular coil carrying current.
Answer:
i) Current carrying straight conductor:

  1. When a magnetic compass is kept near a current-carrying conductor, it experiences a torque.
  2. It aligns in the direction of the magnetic field.
  3. The nature of the magnetic field is like concentric circles having their center at the axis of the conductor.
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 61
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 62
  4. The direction of a magnetic field is clockwise or anticlockwise depending on the direction of the current.
  5. If current increases, the magnetic field is also increased.
  6. If distance increases, the magnetic field decreases.

ii) Circular coil carrying:

  1. When a magnetic compass is kept near a current-carrying conductor, it experiences a torque.
  2. It aligns in the direction of the magnetic field.
  3. The magnetic field lines are circular at points A and B.
  4. The magnetic field lines are nearly parallel to each other near the center.
  5. If the current or number of turns increases, the magnetic field increases.
  6. The polarity depends on the direction of the current in the loop.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 6.
Deduce the magnetic dipole moment of a revolving electron.
Answer:
1. Electron revolves around a nucleus in a circular orbit of radius R.
2. Circulating electron is like a current in a circular loop.
3. The magnetic dipole moment
μL = IA …………….(1)
4. If T is the time period if an electron, then-current
\(I=\frac{-\mathrm{e}}{\mathrm{T}}\) ……………..(2)
-e → Charge of an electron
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 63
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 64
5. Angular momentum, L = Rp; L = mvR
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 65
\(\mu_{\mathrm{L}}=\frac{\text { neh }}{4 \pi \mathrm{m}}\)

6. On substituting known values,
μL = n × 9.27 × 10-24 Am2
7. When n=1, μL = n × 9.27 × 10-24 Am2 (or)
μL = 9.27 × 10-24 JT-1
8. (μL)min = μB (Bohr magneton)
μB = \(\frac{\mathrm{eh}}{4 \pi \mathrm{m}}\) = 9.27 × 10-24 Am2

Question 7.
Deduce the magnetic field inside, interior, and exterior region in the toroid.
Answer:
i) Open space interior to the toroid:

  1. Consider Amperian loop 1 with radius r1
  2. To find the magnetic field Bp at P, the length of the loop is given by
    L1 = 2πr1(circumference) By Ampere’s circuital law,
    \(\begin{aligned}
    &\oint \overrightarrow{\mathrm{B}}_{\mathrm{P}} \cdot \overrightarrow{\mathrm{d}} l=\mu_{\mathrm{o}} \mathrm{I}_{\text {enclosed }}\\
    &\text { loop1 }
    \end{aligned}\)
  3. Loop 1 encloses no current, so Ienclosed = 0.
    \(\oint \overrightarrow{\mathrm{B}} \mathrm{p} \cdot \overrightarrow{\mathrm{d}} l=0\)
  4. This is possible only if, \(\overrightarrow{\mathrm{B}} p\) = 0
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 66

ii) Open space exterior to the toroid:

  1. Construct on Amperian loop 3 of radius r3 around the point Q.
  2. To find the magnetic field BQ at Q1 the length of the loop is given by,
    L3 = 2πr3
    By Ampere’s circuital law,
    \(\left.\oint_{\text {loop } 3} \overrightarrow{\mathrm{B}}_{\mathrm{Q}} \cdot \overrightarrow{\mathrm{d}}\right]=\mu_{0} \mathrm{I}_{\text {enclosed }}\)
  3. Current coming out and going into the plane of paper, Ienclosed = 0
    \(\oint_{\text {loop3 }} \overrightarrow{\mathrm{B}}_{\mathrm{Q}} \overrightarrow{\mathrm{d}} l=0\)
  4. This is possible only if, BQ = 0

iii) Inside the toroid:

  1. Construct an Amperian loop 2 of radius r2 around the point S.
  2. To find the magnetic field Bs at S, the length of the loop is given by
  3. By Ampere’s circuital law,
    \(\left.\oint_{\text {loop } 2} \overrightarrow{\mathrm{B}}_{\mathrm{S}} \cdot \overrightarrow{\mathrm{d}}\right]=\mu_{0} \mathrm{I}_{\text {enclosed }}\)
  4. Let I – current passing through the toroid,
    N – Number of turns of the toroid.
    Ienclosed = NI
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 67
  5. n → number of turns per unit length.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 8.
Discuss the motion of a charged particle in a uniform magnetic field.
Answer:

  1. Consider a charged particle ‘q’ having mass’m’ enters into uniform magnetic field \(\overrightarrow{\mathrm{B}}\) with velocity \(\overrightarrow{\mathrm{v}}\).
  2. Velocity is perpendicular to both magnetic field \(\overrightarrow{\mathrm{B}}\) and velocity \(\overrightarrow{\mathrm{v}}\).
  3. The charged particle moves in a circular orbit.
  4. The Lorentz force on the charged particle
    \(\overrightarrow{\mathrm{B}}\) = q(\(\overrightarrow{\mathrm{v}}\) × \(\overrightarrow{\mathrm{B}}\)) …………(1)
  5. The magnitude of the net force is
    Fm = qvB …………….(2)
  6. This Lorentz force exhibits as centripetal force to execute circular motion.
    qvB = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) …………….(3)
  7. The radius of circular path
    \(r=\frac{m v}{B q}\) ……………(4)
    ∴ \(r=\frac{p}{B q}\) ……………..(5)
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 68
    Circular motion of a charged particle in a perpendicular uniform magnetic field.
  8. p = mv is the linear momentum of the particle.
  9. Let T be the time taken to complete one circular motion.
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 69
  10. Eqns., (9) & (10) are called as cyclotron frequency or gyro frequency.
  11. Time period and frequency depend only on charge-to-mass ratio and independent of Velocity or radius.

Question 9.
Discuss the motion of a charged particle under a crossed electric and magnetic field.
Answer:

  1. Consider a charged particle ‘q having mass’m’ enters into uniform magnetic field \(\overrightarrow{\mathrm{B}}\) with velocity \(\overrightarrow{\mathrm{v}}\).
  2. Velocity is not perpendicular to both magnetic field \(\overrightarrow{\mathrm{B}}\) and velocity \(\overrightarrow{\mathrm{v}}\).
  3. The path of the particle is a helix.
  4. The Lorentz force is balanced by the Coulomb force by electric and magnetic fields.
  5. The Coulomb force acts along the direction of a magnetic field.
  6. The Lorentz force is perpendicular to the direction of magnetic field.
  7. To balance these forces, both electric and magnetic fields are perpendicular to each other.
  8. Perpendicular electric and magnetic fields are known as cross fields.
  9. In a parallel plate capacitor, the magnetic field is perpendicular to the electric field.
  10. The net force \(\overrightarrow{\mathrm{F}}\) = q(\(\overrightarrow{\mathrm{E}}\) + \(\overrightarrow{\mathrm{v}}\) × \(\overrightarrow{\mathrm{B}}\))
  11. For a positive charge, an electric force acts downward, the Lorentz force acts upward.
    Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 70                                        Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 71
  12. Two fields balance each other qE = – qv0B
    \(\mathrm{v}_{\mathrm{o}}=\frac{\mathrm{E}}{\mathrm{B}}\)
    Force can experience only for the particle moving with v0
  13. There are three possibilities of deflection for the charged particle.
    • v > v0 – Direction of Lorentz force
    • v < v0 – Direction of Coulomb force
    • v = v0 – No deflection, moves straight.

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

XIII. Additional Problems (one mark):

Question 1.
A current of 10A is flowing in a wire of length 1.5 M. A force of 15 N acts on it when it is placed in a uniform magnetic field of 2 teslas. The angle between the magnetic field and the direction of current is
a) 30°
b) 45°
c) 60°
d) 90°
Answer:
a) 30°
Solution:
F = BIL sin θ
sin θ = \(\frac{\mathrm{F}}{\mathrm{BIl}}\)
= \(\frac{15}{2 \times 10 \times 1.5}\)
sin θ = 1/2
θ = 30°

Question 2.
3 A of current is flowing in a linear conductor having a length of 40 cm. The conductor has placed in a magnetic field of
strength 500 gauss and makes an angle of 30° with the direction of the field. It experiences a force of magnitude _____.
a) 3 × 104 newton
b) 3 × 102 newton
c) 3 × 10-2 newton
d) 3 × 10-4 newton
Answer:
c) 3 × 10-2 newton
Solution :
F = Bil sin θ
= 500 × 10-4 × 3 × (40 × 10-2) × sin 30°
= 500 × 10-4 × 3 × 40 × 10-2 × \(\frac{1}{2}\)
F = 3 × 10-2 N

Question 3.
A circular coil of a radius of 4 cm has 50 turns. In this coil, a current of 2A is flowing. It is placed in a magnetic field of 0.1 weber/m2. The amount of work done in rotating it through 180° from its equilibrium position will be ______.
a) 0.1 J
b) 0.2 J
c) 0.4 J
d)0.3 J
Answer:
a) 0.1 J
Solution:
M = NIA
= NI(πr²)
W = MB(1 – cosθ)
= MB(1 – cos(180°))
= MB(1 – (-1))
= MB(2)
W = 2MB
W = 2(NIπr²)B
= 2 × (50 × 2 × 3.14 × 16 × 10-4) × 0.1
W =0.1 J.

Question 4.
Two straight parallel wires, both carrying current lOA in the same direction attract each other with a force of 1 x 103 N. If both currents are doubled, the force of attraction will be
a) 1 × 10-3N
b) 2 × 10-3N
c) 4 × 10-3N
d) 0.25 × 10-3N
Answer:
c) 4 × 10-3N
Solution:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 72

Question 5.
A wire of length I meter carrying a current of I ampere is bent in the form of a circle. Its magnitude of the magnetic moment will be ______.
a) \(\frac{\mathrm{IL}}{4 \pi}\)
b) \(\frac{\mathrm{IL}^{2}}{4 \pi}\)
c) \(\frac{\mathrm{I}^{2} \mathrm{~L}^{2}}{4 \pi}\)
d) \(\frac{\mathrm{I}^{2} \mathrm{~L}}{4 \pi}\)
Answer:
b) \(\frac{\mathrm{IL}^{2}}{4 \pi}\)
Solution:
If the radius of the circle is r, then
2πr = L
r = \(\frac{\mathrm{L}}{2 \pi}\)
A = πr²
= \(\frac{\pi \mathrm{L}^{2}}{4 \pi^{2}}\)
A = \(\frac{\mathrm{L}^{2}}{4 \pi}\)
Magnetic moment M = IA
M = \(\frac{\mathrm{IL}^{2}}{4 \pi}\)

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 6.
Two materials X and Y are magnetised whose intensity of magnetization is 500Am-1 and 2000 Am-1 respectively. The magnetising field is 1000 Am-1. What is the ratio between the susceptibilities of the two materials?
Answer:
The susceptibility of material χ is
∴ Areas,
\(\begin{array}{l}
\chi_{m, x}=\frac{|\overrightarrow{\mathrm{M}}|}{|\overrightarrow{\mathrm{H}}|}=\frac{500}{1000}=0.5 \\
\chi_{\mathrm{m}, \mathrm{y}}=\frac{|\overrightarrow{\mathrm{M}}|}{|\overrightarrow{\mathrm{H}}|}=\frac{2000}{1000}=2
\end{array}\)
Since, the susceptibility of material Y is greater than that of material X, which implies that material Y Can be easily magnetized. The ratio between the susceptibilities of the two materials = 0.5 : 2.

Question 7.
An electron has a mass of 9 x 10-31 kg and charges 1.6 x 10-19 C is moving with a velocity of 106 m/s. enters a region where a magnetic field exists. if it describes a circle of radius 0.10 m, the intensity of the magnetic field must be______
a) 1.8 × 10-4 T
b) 5.6 × 10-5 T
c) 14.4 × 10-5 T
d) 1.3 × 10-6 T
Answer:
b) 5.6 × 10-5 T
Solution:
B = \(\frac{\mathrm{mv}}{\mathrm{qr}}\)
= \(\frac{9 \times 10^{-31} \times 10^{6}}{1.6 \times 10^{-19} \times 0.1}\)
B = 5.6 × 10-5 tesla

Question 8.
A galvanometer coil of resistance 12Ω gives full-scale deflection for 3 mA. How will you convert it into a voltmeter of range 0 to 18 V and an ammeter of range 0 to 6A?
Answer:
Given:
G = 12Ω
Ig = 3 × 10-3 A
V = 18V
I = 6A
i) R = \(\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}\) – Galvanometer
= \(\frac{18}{3 \times 10^{-3}}\) – 12
= 6 × 10-3 – 12
= 6000 – 12
R = 5988Ω
A resistance of 5988Ω should be connected in series with the galvanometer.

ii)Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 73

Question 9.
An alpha particle travels in a circular path of radius 0.45 m in a magnetic field B=1.2 wb/m2 with a speed of 2.6 x 107 m/s. The period of revolution of an alpha particle is _____.
0 = 4π × 10-7weber / ampere-metre)
a) 4π × 10-5 tesla
b) 8π × 10-5 tesla
c) 4 × 10-5 tesla
d) 2 × 10-5 tesla
Answer:
a) 4π × 10-5 tesla
Solution:
B = \(\frac{\mu_{\mathrm{o}} \mathrm{NI}}{2 \mathrm{r}}\)
= \(\frac{4 \pi \times 10^{-7} \times 100 \times 0.1}{2 \times 5 \times 10^{-2}}\)
B = 4π × 10-5 tesla

Question 10.
The strength of the magnetic field at a point r near a long straight current-carrying wire is B. The field at a distance \(\frac{r}{2}\) will be _____.
a) \(\frac{\mathrm{B}}{2}\)
b) \(\frac{\mathrm{B}}{4}\)
c) 2B
d) 4B
Answer:
c) 2B
Solution:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 74

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

XIV. Additional Problems (5 marks):

Question 1.
A circular coil carrying current has a radius R. Find the distance from the center of the coil on the axis, where the magnetic induction will be \(\frac{1}{8}\)th to its value at the center of the coil?
Answer:
\(\frac{B_{\text {centre }}}{B_{\text {axis }}}=\left(1+\frac{x^{2}}{R^{2}}\right)^{3 / 2}\)
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 75

Question 2.
A coil having N turns is wound tight in the form of a spiral with inner and outer radii a and b respectively when a current I passes through the coil, Find the magnetic field at the center.
Answer:
Number of turns per unit width = \(\frac{\mathrm{N}}{\mathrm{b}-\mathrm{a}}\)
Consider an elemental ring of radius x, thickness dx.
Number of turns in the ring dN = \(\frac{\mathrm{Ndx}}{\mathrm{b}-\mathrm{a}}\)
The magnetic field at the center due to the ring element.
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 76

Question 3.
An electron moving perpendicular to a uniform magnetic field 0.5 T undergoes circular motion of radius 2.5 mm. What is the speed of the electron?
Answer:
Charge of an electron q = 1.6 × 10-19 C
Magnitude of magnetic field, B = 0.5 T
Mass of an electron, m = 9.11 × 10-31 Kg
Radius of the orbit, r = 2.5 mm =2.50 × 10-3 m
Velocity of the electron, v = \(\frac{\mathrm{qrB}}{\mathrm{m}}\)
v = \(\frac{1.6 \times 10^{-19} \times 2.50 \times 10^{-3} \times 0.5}{9.11 \times 10^{-31}}\)
v = 2.195 × 108 ms-1

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 4.
A circular coil of radius 20 cm has 100 turns wire and it carries a current of 5 A. Find the magnetic induction at a point along its axis at a distance of 20 cm from the center of the coil.
Answer:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 77
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 78

Question 5.
Three tangent galvanometers have turned 2 : 3 : 5. When connected in series in a circuit, they show deflections of 30°, 45°, and 60° respectively. Find the ratio of their radii.
Answer:
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 79

Question 6.
Show that the magnetic field at any point on the axis of the solenoid having n turns per unit length is
\(\mathbf{B}=\frac{1}{2} \mu_{0} \mathrm{nI}\left(\cos \theta_{1}-\cos \theta_{2}\right)\)
A solenoid is a long cylindrical coil having number of circular turns.
Answer:
n → number of turns per unit length
R → radius of a solenoid.
The current carrying element dx at a distance x from origin and distance r from point P.
\(r=\sqrt{R^{2}+\left(x_{0}-x\right)^{2}}\)
The magnetic field due to current carrying circular coil is
\(\mathrm{dB}=\frac{\mu_{\mathrm{o}} \mathrm{IR}^{2}}{2 \mathrm{r}^{3}} \mathrm{xN}\)
N = ndx
\(\mathrm{dB}=\frac{\mu_{0}}{2} \frac{\mathrm{nIR}^{2}}{\mathrm{r}^{3}} \mathrm{dx}\)
sin θ = \(\frac{R}{r}\)
r = R cosecθ

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 80

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 7.
Let I1 and I2 be the steady currents passing through a long horizontal wire XY and PQ, respectively. Suppose the wire PQ is fixed in a horizontal plane and the wire XY is allowed to move freely in a vertical plane. Let the wire XY is in equilibrium at a height d over the parallel wire PQ as shown in the figure.
Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 81
Show that if the wire XY is slightly displaced and released, it executes Simple Harmonic Motion (SHM). Also, compute the time period of oscillations.
Answer:
If XY is allowed to move freely in a vertical plane.
∴Vertical oscillation of the wire xy experience a force,
F = – ky
Applying Newton’s second law

Samacheer Kalvi 12th Physics Guide Chapter 3 Magnetism and Magnetic Effects of Electric Current 82