Category: CBSE Class 12

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 15 Polymers. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 15 Important Extra Questions Polymers

Polymers Important Extra Questions Very Short Answer Type

Question 1.
Define the term ‘homopolymerisation’ giving an example. (CBSE Delhi 2012)
Answer:
The polymers formed from one type of monomers and having same repeating units are called homopolymers. For example polyvinyl chloride (PVC), polythene.

Question 2.
Give one example of condensation polymer. (CBSE 2013)
Answer:
Nylon-6, 6

Question 3.
Which of the following is a natural polymer? (CBSE 2014)
Buna-S, Proteins, PVC
Answer:
Proteins.

Question 4.
Based on molecular forces what type of polymer is neoprene? (CBSE 2014)
Answer:
Elastomer.

Question 5.
Which of the following is a fibre?
Nylon, Neoprene, PVC (CBSE 2014)
Answer:
Nylon

Question 6.
Draw structure for the polymer used for the manufacture of non-stick utensils. (CBSE2019C)
Answer:

Question 7.
What is the primary structural feature necessary for a molecule to make it useful in a condensation polymerisation reaction? (CBSE AI 2010)
Answer:
The monomers must be bifunctional i.e. contain two functional groups.

Question 8.
Arrange the following polymers in the increasing order of tensile strength. Nylon 6, Buna-S, Polythene. (CBSE Sample Paper 2010)
Answer:
Buna – S < Polythene < Nylon 6.

Question 9.
What type of reaction occurs in the formation of Nylon 6,6 polymer? (CBSE Sample Paper 2019)
Answer:
Condensation

Question 10.

Is a homopolymer or R copolymer? (CBSE 2019C)
Answer:
Homopolymer

Question 11.
Identify the Monomers of the following polymer:

Answer:
Buta-1,3-diene, styrene.
It is obtained by the polymerisation of buta-1, 3-diene and styrene in the ratio of 3: 1 in the presence of sodium.

Polymers Important Extra Questions Short Answer Type

Question 1.
Draw the molecular structures of the monomers of
(i) PVC
Answer:
PVC : CH₂ = CH – Cl (Vinyl chloride)

(ii) Teflon (CBSE 2010)
Answer:
Teflon : F₂C = CF₂ (Tetrafluoroethene)

Question 2.
Mention two important uses of each of the following : (CBSE Delhi 2011)
(i) Bakelite
Answer:
Uses of bakelite:

• Used for making combs, fountain pens, barrels, electrical switches and handles of utensils.
• Soft bakelites are used for making glue for binding wooden laminated planks in varnishes.

(ii) Nylon 6
Answer:
Uses of Nylon 6:

• It is used for making tyre cords.
• It is used for making fabrics and ropes.

Question 3.
Draw the structure of the monomer for each of the following polymers:
(i) Nylon 6
Answer:
The monomer of Nylon-6:

(ii) Polypropene (CBSE Delhi 2012)
Answer:
The monomer of polypropene:

Question 4.
Differentiate between thermoplastic and thermosetting polymers. Give one example of each. (CBSE Delhi 2010, 2012) Answer:
Thermoplastics when heated become soft and more or less fluid. These can be moulded into any desired shape. The thermoplastics ‘ can be processed again and again.

The intermolecular forces in thermoplastics are intermediate between those of elastomers and fibres.

On the other hand, thermosetting plastics on heating become hard and insoluble masses. These cannot be moulded into the desired shape and cannot be reprocessed. The intermolecular forces in thermosetting are strong and there are cross-links that hold the molecules in place so that heating does not allow them to move freely.

The common examples are:

1. Thermoplastics: Polythene, polystyrene and polyvinyl chloride.
2. Thermosetting: Bakelite and melamine formaldehyde.

Question 5.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Bakelite
Answer:
Phenol and formaldehyde

(ii) Neoprene
Answer:
2-Chloro-1,3-butadiene(chloroprene)

Question 6.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Terylene
Answer:
Ethylene glycol, terephthalic acid

(ii) Nylon-6, 6
Answer:
Hexamethylenediamine and adipic acid.

Question 7.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Teflon
Answer:
Tetrafluoroethylene

(ii) Buna-N
Answer:
1, 3-butadiene and acrylonitrile

Polymers Important Extra Questions Long Answer Type

Question 1.
Write the structures of monomers used for getting the following polymers:
(i) Nylon-6,6
(ii) Glyptal
(iii) Buna-S
OR
(i) Is it a homopolymer or copolymer? Give reason.
(ii) Write the monomers of the following polymer:

(iii) What is the role of the Sulphur in the vulcanization of rubber? (CBSE Delhi 2019)
Answer:
(i) Nylon-6,6
Hexamethylenediamine: NH₂ – (CH₂)₆ – NH₂ and Adipic acid

(ii) Glyptal

(iii) Buna – S
CH₂ = CH — CH = CH₂
1, 3 Butadiene

OR
(i) It is a homopolymer because it has only one monomer.

(iii) Sulphur makes the rubber hard, tough with greater tensile strength and non-sticky by forming sulphur crosslinks.

Question 2.
(a) Write the names of monomers of the following polymer:

Answer:
Ethylene glycol and terephthalic acid or Ethane-1,2- diol and Benzene 1, 4- dicarboxylic acid.

Terylene: It is a polymer of ethylene glycol (ethane-1,2-diol) and terephthalic acid (benzene-1, 4-dicarboxylic acid). It is obtained by heating a mixture of ethylene glycol and terephthalic acid at 420 to 460 K in the presence of zinc acetate-antimony trioxide, [Zn(OOCCH₃)₂ + Sb₂O₃] catalyst. It is known as terylene or dacron.

(b) What does part 6,6 mean in the polymer Nylon-6,6?
Answer:
It represents 6 carbon atoms present in both the monomer units.

Nylon-6,6: The monomer units of nylon-6,6 are hexamethylenediamine and adipic acid. It is prepared by condensation of hexamethylene diamine with adipic acid under high pressure and high temperature.

(c) Give an example of a Biodegradable polymer. (CBSE 2019C)
Answer:
Poly-p-hydroxybutyrate-co-p-hydroxy valerate (PHBV).

It is a copolymer of 3-hydroxybutyric acid and 3-hydroxypentanoic add, in which the monomer units are joined by ester linkages.

Question 3.
Explain the following terms giving a suitable example for each :
(i) Elastomers
Answer:
Elastomers: These are polymers in which the polymer chains are held together by weak intermolecular forces. Because of the presence of weak forces, the polymer can be easily stretched. However, a few cross-linked are also introduced in the chains which impart the property of regaining the original positions after the stretching force is released. A common example is vulcanised rubber.

(ii) Condensation polymers
Answer:
Condensation polymers: A polymer formed by the condensation of two or more than two monomers is called condensation polymer. In this type, each monomer generally contains two functional groups, and condensation takes place by the loss of molecules such as H₂0, HCl, etc. Common examples are nylon-6, nylon-6,6, bakelite, terylene and alkyl resins.

(iii) Addition polymers (CBSE 2012)
Answer:
Addition polymers: A polymer formed by the direct addition of repeated monomers is called an addition polymer. In this type, the monomers are unsaturated compounds and are generally derivatives of ethene. In addition, polymers have the same empirical formula as their monomers. For example, the addition of polymers polyethene or polypropylene are obtained as:

Question 4.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
Answer:
Buna-S:
1, 3-Butadiene: CH₂ = CH – CH == CH₂
Styrene: C₆H₅CH = CH₂

(ii) Neoprene
Answer:
Neoprene:
Chloroprene: CH₂ = C – CH = CH₂

(iii) Nyton-6, 6 (CBSE Delhi 2013)
Answer:
Nylon 6, 6:

• Adipic acid: HOOC (CH₂)₄ COOH
• Hexamethylenediamine: NH₂(CH₂)₆NH₂

Question 5.
Write the names and structures of the monomers of the following polymers:
(i) Bakelite
Answer:
Bakelite:

• Phenol:
  
• Formaldehyde: HCHO

(ii) Nylon-6
Answer:
Nylon-6:
Caprolactam

(iii) Polythene (CBSE Delhi 2013)
Answer:
Polythene: Ethene: CH₂ = CH₂.

Question 6.
(i) What is the role of benzoyl peroxide in the polymerisation of ethene?
(ii) Identify the monomers in the following polymer:

(iii) Arrange the following polymers in the increasing order of their intermolecular forces: Nylon-6, 6, Polythene, Buna-S
OR
Write the mechanism of free radical polymerisation of ethene. (CBSE 2016)
Answer:
(i) Benzoyl peroxide acts as a free radical generator. In the presence of benzoyl peroxide, phenyl free radical is formed which initiates the chain initiation step.

(iii) Buna-S < Polythene < Nylon 6, 6
OR
Mechanism
(i) Chain initiation step:

(ii) Chain propagating step:

(iii) Chain termination step:

Question 7.
Write the names and structures of the monomers of the following polymers:
(a) Neoprene
Answer:
Chloroprene CH₂ = C — CH = CH₂

(b) Bakelite
Answer:
C₆H₅OH(Phenol) and HCHO (formaldehyde)

(c) PVC (CBSE 2019C)
Answer:
Polyvinyl chloride CH₂ = CH – Cl

Question 8.
Write the structures of monomers used to obtain the following polymers:
(a) Natural rubber
(b) PVC
(c) Nylon-6,6
OR
Write the mechanism of free radical polymerisation of ethene. (CBSE AI 2019)
Answer:
(a) Natural rubber: 2-methyl-1,3-butadiene

(b) PVC: vinyl chloride
CH₂ = CH – Cl

(c) Nylon-6,6:
HOOC-(CH₂)₄ – COOH (Adipic acid)

NH₂ — (CH₂)₆ — NH₂
(Hexamethylenedlamine)
OR
Chain Initiation Step:

Chain propagation:

Chain termination step

Question 9.
Write the structures of monomers used to obtain the following polymers:
(a) Buna-S
(b) Glyptai
(c) Nylon-6
OR
(a) Arrange the following polymers in increasing order of their intermolecular forces: Polyvinylchloride, Neoprene, Terylene
(b) Write one example of each of
(i) Natural polymer
(ii) Thermosetting polymer
(c) What is the significance of numbers 6,6 in the polymer Nylon-6,6? (CBSE Al 2019)
Answer:
(a) Buna-S
CH₂ = CH – CH = CH₂ (1,3-Butadiene) and C₆H₅CH = CH₂ (Styrene)

(b) Glyptal
HO – CH₂ — CH₂ — OH (Ethylene glycol) and

(c) Nylon-6

OR
(a) Neoprene < Polyvinyl chloride (PVC) < Terylene
(b) (i) Isoprene
(ii) Bakelite
(c) The numbers ‘6, 6’ signifies that both the monomers of nylon-6, 6 have six carbon atoms.

Question 10.
What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester. (CBSE AI 2013)
Answer:
Polymers that are degraded by micro-organisms within a suitable period so that the polymers and their degraded products do not cause any serious effects on the environment are called biodegradable polymers. For example,

Poly β – hydroxybutyrate – co – β – hydroxy valerate (PHBV)

Question 11.
Write the names and structures of the monomers of the following polymers:
(i) Polystyrene
Answer:
Polystyrene: Styrene:

(ii) Dacron
Answer:
Dacron:

• Ethylene glycol: HOCH₂CH₂OH
• Terephthalic acid:
  

(iii) Teflon
Answer:
Teflon:
Tetrafluoroethene: F₂C = CF₂

Question 12.
Write the names and structures of the monomers of the following polymers:
(a) Teflon
Answer:
Tetrafluoroethylene, CF₂ = CF₂

(b) Terylene
Answer:

(c) Buna-N
Answer:

Question 13.
Write the structures of monomers used to obtain the following polymers:
(a) Neoprene
(b) PHBV
(c) Bakelite
OR
(a) Arrange the following polymers in decreasing order of their intermolecular forces:
Bakelite, Polythene, Buna-S, Nylon-6,6
(b) Write the monomers of the following polymer:

(c) What is the structural difference between high-density polythene (HDP) and low-density polythene (LDP)? (CBSE AI 2019)
Answer:
(a) Neoprene:
Chloroprene

(b) PHBV

(c) Bakelite

OR
(a) Buna-S < potythene < Bakelite < Nylon-6,6

(b) HO — CH₂ — CH₂ — OH and

(c) Low-density polythene (LDP) consists of highly branched chain molecules. Due to branching, the polythene molecules do not pack well and therefore, it has low density.

High-density polythene (HDP) consists of linear chains and therefore, the molecules can be closely packed. Hence, it has a high density.

Question 14.
What are high density and low-density polythene?
Answer:
High-density polythene is obtained by heating ethene at about 333-343 K under a pressure of 6-7 atm in the presence of a catalyst such as triethylaluminium and titanium tetrachloride (known as Zeigler-Natta catalyst)

This polymer consists of linear chains and therefore the molecules can be closely packed in space. It, therefore, has a high density (0.97g/cm3) and a higher melting point (403K). It is harder, tougher and has greater tensile strength than low-density polythene.

Low-density polythene (LDP) is prepared by free radical addition and hydrogen atom abstraction. It consists of highly branched chain molecules. Due to branching, the polythene molecules do not pack well and therefore it has low density (0.92 g/cm3) and low melting point (384 K). Low-density polythene is transparent. It has moderate tensile strength and high toughness. It is chemically inert.

Question 15.
Discuss the mechanism of free radical addition polymerisation.
Answer:
A large number of unsaturated compounds such as alkenes or dienes and their derivatives are polymerised by this process. The polymerisation takes place through the generation of an initiator, which is a molecule that decomposes to form free radicals. The commonly used initiator is t-butyl peroxide.

The addition occurs as

For example, most of the commercial addition polymers are obtained from alkenes and their derivatives,

The general model of polymerisation is:

Chain initiation step:

Chain Propagating Step:

Chain terminating step

Question 16.
List some important differences between natural rubber and vulcanised rubber.
Answer:
Differences between natural rubber and vulcanised rubber:

Natural rubber	Vulcanized rubber
1. Natural rubber is soft and sticky.	1. Vulcanized rubber is hard and non-sticky.
2. It has low tensile strength.	2. It has a high tensile strength.
3. It has low elasticity.	3. It has high elasticity.
4. It can be used over a narrow range of temperature (from 10°C to 60°C).	4. It can be used over a wide range of temperature (-40°C to 100°C).
5. It has low wear and tears resistance.	5. It has high wear and tears resistance.
6. It is soluble in solvents like ether, carbon- tetrachloride, petrol etc.	6. It is insoluble in all the common solvents.

Question 17.
(a) What is the difference between two notations: nylon-6 and nylon-6,6? Give their synthesis.
Answer:
Nylon-6 has only one compound having 6-carbon atoms while nylon-6,6 refers to a polymer obtained from 6-carbon atoms of dicarboxylic acid (adipic acid) and 6-carbon atoms of diamine (hexamethylene diamine).

Nylon-6 is synthesised from caprolactam as

Nylon-6,6 is synthesised from hexamethylenediamine and adipic acid.

(b) How is Buna-S prepared?
Answer:
Buna-S is obtained by copolymerisation of styrene and 1, 3-butadiene.

Question 18.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
Answer:
1, 3-Butadiene: CH₂ = CH — CH = CH₂
Styrene: C₆H₅CH = CH₂

(ii) Buna-N
Answer:
1, 3-Butadiene: CH₂ = CH – CH = CH₂
Acrytonitnie: CH₂ = CH — CN

(iii) Dacron
Answer:

• Ethylene glycol:
  
• Terephthalic acid:
  

(iv) Neoprene.
Answer:
Chloroprene:

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 12 Aldehydes, Ketones and Carboxylic Acids.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 12 Important Extra Questions Aldehydes, Ketones and Carboxylic Acids

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Very Short Answer Type

Question 1.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions: ethanal, propanal, propanone, butanone. (CBSE Delhi 2012)
Answer:
butanone < propanone < propanal < ethanal

Question 2.
Which of the following compounds would undergo the Cannizzaro reaction? Benzaldehyde, Cyclohexanone, 2- Methylpentanal. (CBSE Sample Paper 2019)
Answer:
Benzaldehyde

Question 3.
Draw the structure of semicarbazone of cyclopentanone. (CBSE 2019C)
OR
Draw the structure of the product formed when propanal is treated with zinc amalgam and concentrated hydrochloric acid.
Answer:

OR

Question 4.
Write the IUPAC name of the following: (CBSE AI 2012)

Answer:
Pent-2-enal.

Question 5.
Write the structure of 3-methylbutanal. (CBSE Delhi 2013)
Answer:

Question 6.
Write the structure of the p-Methylbenzal- dehyde molecule. (CBSE Delhi 2013, 2014)
Answer:

Question 7.
Draw the structure of the compound whose IUPAC name is 4-chloropentan 2-one. (CBSE Delhi 2008, CBSE AI 2011, 2013, 2014)
Answer:

Question 8.
Draw the structural formula of a 1-phenyl propane-2-one molecule. (CBSE 2010)
Answer:

Question 9.
Draw the structure of 3-methylbutanal. (CBSE 2011, CBSE Delhi 2013)
Answer:

Question 10.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions:
ethanal, propanal, propanone, butanone. (CBSE 2012)
Answer:
butanone < propanone < propanal < ethanal

Question 11.
Write the IUPAC name of (CBSE 2012)
Ph — CH = CH — CHO.
Answer:
3-Phenyl prop-2-en-al.

Question 12.
Rearrange the following compounds in the increasing order of their boiling points:
CH₃ – CHO, CH₃ – CH₂ – OH, CH₃ – CH₂ – CH₃ (CBSE 2013)
Answer:
CH₃CH₂CH₃ < CH₃CHO < CH₃CH₂OH

Question 13.
Ethanol is soluble in water. Why? (CBSE 2013)
Answer:
Ethanol is soluble in water because of hydrogen bonding between the polar carbonyl group of ethanal and water molecules:

Question 14.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Answer:
3-Hydroxybutanoic acid

Question 15.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Answer:

Question 16.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Answer:
3-Aminobutanal

Question 17.
Write the IUPAC name of the following compound: (CBSE AI2019)

Answer:
But-3-en-2-one

Question 18.
Write the IUPAC name of the following compound: (CBSE AI 2019)

Answer:
1-Phenylbutan-2-one.

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Short Answer Type

Question 1.
Write structures of compounds A and B in each of the following reactions: (CBSE Delhi 2019)

Answer:

Answer:

Question 2.
What is Tollen’s reagent? Write one use of this reagent. (CBSE 2010)
Answer:
The ammoniacal solution of silver nitrate is called Tollen’s reagent. It is used as an oxidizing reagent and helps to distinguish between aldehydes and ketones. Aldehydes are oxidized by Tollen’s reagent whereas ketones are not.

Question 3.
Name the reagents used in the following reactions: (CBSE Delhi 2015)

Answer:
Lithium aluminium hydride, LiAlH₄

Answer
Alkaline potassium permanganate (KMnO₄), KOH

Question 4.
Distinguish between
C₆H₅CH = CH – COCH₃ and C₆H₅CH = CH CO CH₂ CH₃ (CBSE AI 2016)
Answer:
Heat both the compounds with NaOH and l₂. C₆H₅CH = CHCOCH₃ gives yellow ppt of iodoform. C₆H₅CH = CHCOCH₂CH₃ does not give yellow ppt. of iodoform.

Question 5.
Arrange the following in the increasing order of their reactivity in nucleophilic addition reactions.
CH₃CHO, C₆H₅CHO, HCHO (CBSE AI 2016)
Answer:
C₆H₅CHO < CH₃CHO < HCHO

Question 6.
Write the product in the following reaction: (CBSE AI 2017)

Answer:

Question 7.
Write the structure of the products formed: (CBSE Sample Paper 2019)
(a) CH₃CH₂COOH
Answer:
CH₃ – CH(Cl) – C00H

(b) C₆H₅COCl
Answer:
C₆H₅CHO

Question 8.
Name the reagents in the following reactions: (CBSE AI 2015)

Answer:
Lithium aluminium hydride, LiAlH₄

Answer:
Alkaline potassium permanganate (KMnO₄), KOH

Answer:
CH₃ Mg Br, H₃O⁺

Answer:
Cl₂, P (Hell-Volhard-Zelinsky reaction)

Question 9.
Predict the organic products of the following reactions:

Answer:

Answer:

Answer:

Answer:

Question 10.
Describe how the following conversions are carried out:
(i) Toluene to benzoic acid
Answer:

(ii) Bromobenzene to benzoic acid (CBSE AI 2013)
Answer:

(iii) Ethylcyanide to ethanoic acid
Answer:

(iv) Butan-1-ol to butanoic acid (CBSE Delhi 2010)
Answer:

Question 11.
Write the structures of A, B, C, and D in the following reaction: (CBSE AI 2016)

Answer:

Question 12.
Aromatic carboxylic acids do not undergo Friedel Crafts reaction. Explain. (CBSE Al 2018)
Answer:
Aromatic carboxylic acids do not undergo Friedel Crafts reaction because -COOH group is deactivating and the catalyst aluminum chloride (Lewis acid) gets bonded to the carboxyl group.

Question 13.
pK_a value of 4-nitrobenzoic acid is lower than that of benzoic acid. Explain. (CBSE AI 2018)
Answer:
Due to the presence of a strong electron-withdrawing (-NO₂) group in 4-nitrobenzoic acid, it stabilizes the carboxylate anion and hence strengthens the acid. Therefore, 4-nitrobenzoic acid is more acidic than benzoic acid and its pK_a value is lower.

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Long Answer Type

Question 1.
Complete the following reactions:

Or
Write chemical equations for the following reactions:
(i) Propanone is treated with dilute Ba(OH)₂.
(ii) Acetophenone is treated with Zn(Hg)/Conc. HCI
(iii) Benzoyl chloride is hydrogenated in presence of Pd/BaSO₄. (CBSE Delhi 2019)
Answer:



Or

Question 2.
Predict the products of the following B reactions: (CBSE Delhi 2015)

Answer:

Answer:

Answer:

Question 3.
Predict the products of the following reactions: (CBSE 2015)

Answer:

Answer:

Answer:

Question 4.
What happens when
(a) Salicylic acid Is treated with (CH₃CO)₂O/H⁺?
Write a chemical equation in support of your answer. (CBSE AI 2019)
Answer:
Salicylic acid is treated with acetic anhydride In the presence of H⁺ to give aspirin.

(b) Phenol is oxidized with Na₂Cr₂O₇/H⁺?
Answer:
Phenol in the presence of acidified sodium dichromate gives benzoquinone.

(C) Anisole Is treated with CH₃Cl/anhydrous AlCl₃?
Answer:

Question 5.
Draw the structures of the following compounds:
(i) 3-Methylbutanal
Answer:

(ii) 4-Chlocopentan-2-one
Answer:

(iii) 4-Methylpent-3-en-2-one
Answer:

(iv) p-Methyl benzaldehyde (CBSE AI 2014)
Answer:

Question 6.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene (CBSE Delhi 2017)
Answer:
Propanone to propene

(ii) Propanal to butanone
Answer:
Propanal to butanone

(iii) Benzaldehyde to benzophenone
Answer:
Benzaldehyde to benzophenone

(iv) Benzaldehyde to 3-phenyl propan – 1-ol
Answer:
Benzaldehyde to 3- phenyl propan -1 -ol

(v) Benzaldehyde to α-hydroxyphenyl acetic acid
Answer:
Benzaldehyde to α-hydroxyphenyl acetic acid

(vi) Ethanol to 3-hydroxybutanal
Answer:
Ethanol to 3-hydroxybutanal

Question 7.
Convert the following:
(i) Ethanal to propanone. (CBSE AI 2018)
Answer:
Ethanal to propanone

(ii) Ethanal to lactic acid.
Answer:
Ethanal to lactic acid

(iii) Ethanal to 2-hydroxy-3-butenoic acid.
Answer:
Ethanal to 2-hydroxy-3-butenoic acid

(iv) Acetaldehyde to formaldehyde.
Answer:
Acetaldehyde to formaldehyde

(v) Formaldehyde to acetaldehyde.
Answer:
Formaldehyde to acetaldehyde

(vi) Acetaldehyde to crotonic acid.
Answer:
Acetaldehyde to crotonic acid

Question 8.
A compound ‘X’ (C₂H₄O) on oxidation gives ‘Y’ (C₂H₄O₂). ‘X’ undergoes haloform reaction. On treatment with HCN ‘X’ forms a product ‘V which on hydrolysis gives 2-hydroxy propanoic acid.
(i) Write down structures of ‘X’ and ‘Y’.
(ii) Name the product when ‘X’ reacts with dil NaOH.
(iii) Write down the equations for the reactions involved. (CBSE Sample Paper 2007)
Answer:
Compound ‘X’ (C₂H₄O) is oxidized to ‘Y’ (C₂H₄O₂). Since it undergoes a haloform reaction, it must be acetaldehyde.
(i) X = CHCHO Y = CH₃COOH
On treatment with HCN, X gives cyanohydrin which on hydrolysis gives 2-hydroxypropanoic acid.

(ii) When ‘X’ reacts with dil. NaOH, undergoes an aldol condensation reaction forming aldol which on heating gives but-2-enal.

(iii) Equations for reactions

Other equations are given above.

Question 9.
Complete the following reactions:

Answer:

Answer:

Answer:

Answer:

Answer:

(CBSE Sample Paper 2017 – 2018)
Answer:

(CBSE Delhi 2013)
Answer:

(CBSE AI 2013)
Answer:

Question 10.
(i) An organic compound (A) has a characteristic odor. On treatment with NaOH, it forms two compounds (B) and (C). Compound (B) has molecular formula C₇H₈0 which on oxidation gives back (A). The compound (C) is a sodium salt of an acid. When (C) is treated with soda lime it yields an aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D). Write the sequence of reactions involved.
Answer:
The compound A is C₆H₅CHO, benzaldehyde having a characteristic odor. The reactions are:

(ii) Complete each synthesis by filling the missing starting materials, reagents, or products. (X and Z).
(a) C₆H₅CHO + CH₃CH₂CHO NaOH X
Answer:

(b) CH₃(CH₂)₉ COOC₂H₅ CH₃(CH₂),CHO
Answer:

(iii) How will you bring about the following conversions in not more than two steps?
(a) Toluene to benzaldehyde
Answer:

(b) Ethylcyanide to 1-phenylpropanoid. (CBSE Sample Paper 2011)
Answer:

Question 11.
(i) How do you convert the following?
(a) Ethanal to propanone
(b) Toluene to benzoic acid
OR
(ii) (A), (B), and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C₄H₈O. isomers (A) and (C) give positive Tollens’ test whereas isomer (B) does not give Tollens’ test but gives positive iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCI gives the same product (D).
(a) Write the structures of (A), (B), (C), and (D).
(b) Out of (A), (B), and (C) isomers, which one is least reactive towards the addition of HCN? (CBSEAI 2018)
Answer:

OR
(ii) (a) Out of A, B, and C, (A) and (C) give positive Tollens’ test, and therefore, these are aldehydes. (B) does not give Tollens’ test and therefore, it is a ketone, with a group because it gives positive iodoform test. Thus, the three isomers are:

Question 12.
Predict the products of the following reactions:

(CBSE Delhi 2015)
Answer:

(CBSE Delhi 2015)
Answer:

(CBSE AI 2015)
Answer:

(CBSE AI 2015)
Answer:

(CBSE AI 2017)
Answer:

(CBSE AI 2017)
Answer:

Question 13.
Write the structures of A, B, C, D, and E in the following reactions: (C8SE Delhi 2016)

Answer:

Question 14.
(a) Give chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanone.
(ii) Pentan-2-one and Pentan-3-one.
(b) Arrange the following compounds in increasing order of their acid strength: Benzoic acid, 4- Nitrobenzoic acid, 3,4 -Dinitrobenzoic acid, 4- Methoxybenzoic acid.
OR
Compare the reactivity of benzaldehyde and ethanal towards nucleophilic addition reactions. Write the cross aldol condensation product between benzaldehyde and ethanal. (CBSE Sample Paper 2019)
Answer:
(a) (i)

Experiments	Ethanal	Propanone
1. Tollen’s Test: Warm the organic compound with freshly prepared ammoniacal silver nitrate solution (Tollen’s reagent).	A bright silver mirror is produced.	No silver mirror is formed.
2. Fehling’s Test: Heat the organic compound with Fehling’s reagent.	A reddish-brown precipitate is obtained.	No precipitate is obtained.
Anyone test

Acetaldehyde (Ethanal) and Acetone (Propanone)
(i) Acetaldehyde gives silver mirror with Tollen’s reagent.

Acetone does not give this test.

(ii) Acetaldehyde gives red ppt with Fehling solution.

(ii) Pentan-2-one and Pentan-3-one

Experiment	Pentan-2-one	Pentan-3-one
Iodoform Test: The organic compound is heated with iodine in presence of sodium hydroxide solution.	A yellow precipitate is obtained.	No yellow precipitate is obtained.

Pentan-3-one and Pentan-2-one
(i) Pentan-2-one forms yellow ppt with an alkaline solution of iodine (iodoform test), but pentane-3-one does not give iodoform test.
CH₃COCH₂CH₂CH₃ + 3I₂+ 4NaOH → CH₃CH₂CH₂COONa + CHI₃ + 3H₂0 + 3NaI

(ii) Pentan-2-one gives white ppt with sodium bisulphite while pentan-3-one does not.

Or any other suitable test.

(b) 4- Methoxybenzoic add < Benzoic acid < 4- Nitrobenzoic acid < 3,4-Dinitrobenzoic acid Effect of substituents on acidic strength of acids. The substituents have a marked effect on the acidic strength of carboxylic acids. The nature of substituents affects the stability of the conjugate base (carboxylate ion) and hence affects the acidity of the carboxylic acids.

In general, electron-withdrawing groups (EWG) increase the stability of the carboxylate ion by delocalizing the negative charge and hence increase the acidity of the carboxylic acid. Conversely, electron-donating groups (EDG) decrease the stability of the carboxylate ion by intensifying the negative charge and hence decrease the acidity of the carboxylic acid.
OR
The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than the carbon atom of the carbonyl group present in ethanol. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance hence less reactive than ethanal.

Aromatic Aldehydes and Ketones
In general, aromatic aldehydes and ketones are less reactive than the corresponding aliphatic analogs. For example, benzaldehyde is less reactive than aliphatic aldehydes. This can be easily understood from the resonating structures of benzaldehyde as shown below:

It is clear from the resonating structures that due to the electron releasing (+I effect) of the benzene ring, the magnitude of the positive charge on the carbonyl group decreases, and consequently it becomes less susceptible to the nucleophilic attack. Thus, aromatic aldehydes and ketones are less reactive than the corresponding aliphatic aldehydes and ketones.

However, amongst aromatic aldehydes and ketones, aromatic aldehydes are more reactive than alkyl aryl ketones which in turn are more reactive than diaryl ketones. Thus, the order of reactivity of aromatic aldehydes and ketones is:

Question 15.
(i) Illustrate the following name reaction giving a suitable example:
(a) Clemmensen reduction
(b) Hell-Volhard-Zelinsky reaction

(ii) How are the following conversions carried out?
(a) Ethylcyanide to ethanoic acid
(b) Butan-1-ol to butanoic acid
(c) Benzoic acid to m-bromobenzoic acid .
OR
(i) Illustrate the following reactions giving a suitable example for each.
(a) Cross aldol condensation
(b) Decarboxylation

(ii) Give simple tests to distinguish between the following pairs of compounds:
(a) Pentan-2-one and Pentan-3-one
(b) Benzaldehyde and Acetophenone
(c) Phenol and Benzoic acid (CBSE Delhi 2012)
Answer:
(i) (a) Clemmensen reduction

(b) Hell-Volhard-Zelinsky reaction

(ii)

OR
(i) (a) Cross Aldol condensation

(b) Decarboxylation

(ii) (a) Pentan-2-one and pentan-3-one can be distinguished by iodoform test

Pentan-3-one does not give this test.

(b) Benzaldehyde and acetophenone: Benzaldehyde forms a silver mirror with ammoniacal silver nitrate solution (Tollen’s reagent).

Acetophenone does not react with Tollen’s reagent.

(c) Phenol and Benzoic acid: Benzoic acid reacts with NaHC03 to give effervescence due to the evolution of CO₂.

Phenol does not give effervescence.

Question 16.
(i) Write a suitable chemical equation to complete each of the following transformations:
(a) Butan-1-ol to butanoic acid
(b) 4-Methylacetophenone to benzene-1, 4-dicarboxylic acid.

(ii) An organic compound with molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Toilen’s reagent, and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid. Identify the compound.
OR
(i) Give chemical tests to distinguish between (a) Propanol and propanone
(b) Benzaldehyde and acetophenone

(ii) Arrange the following compounds in increasing order of their property as indicated:
(a) Acetaldehyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)
(b) Benzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Answer:

(ii) The given compound forms a 2,4-DNP derivative. Therefore, it is an aldehyde or ketone. Since it reduces Tollen’s reagent, it must be an aldehyde. The compound undergoes Cannizzaro’s reaction, so it does not contain hydrogen. On vigorous oxidation, it gives 1,2-benzene dicarboxylic acid, which means that it must be containing an alkyl group at 2-position with respect to the CHO group on the benzene ring.
The molecular formula suggests it should be
(2- Ethytbenza(dehyde)

Answer:
(i) (a) Propanone gives iodoform test but propanol does not. When propanone is heated with aqueous sodium carbonate and iodine solution, a yellow precipitate of iodoform is obtained.

Propanol does not give this test.

(b) Acetophenone forms yellow ppt. of iodoform with an alkaline solution of iodine (iodoform test). Benzaldehyde does not react.

(ii) (a) Methyl tert-butyl ketone < Acetone < Acetaldehyde.
(b) 4-Methoxybenzoic acid < Benzoic add < 3, 4-Dinitrobenzoic acid.
(c) (CH₃)₂CHCOOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

Question 17.
(i) Although phenoxide ion has more resonating structures than carboxylate ion, the carboxylic acid is a stronger acid than phenol. Give two reasons.
(ii) How will you bring about the following conversions?
(a) Propanone to propane
(b) Benzoyl chloride to benzaldehyde
(c) Ethanal to but-2-enal
OR
(i) Complete the following reactions:

(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Ethanal and Propanal
(b) Benzoic acid and Phenol (CBSE AI 2013)
Answer:
(i) The carboxylic acids are stronger acids than phenols. The difference in the relative acidic strengths can be understood if we compare the resonance hybrid structures of carboxylate ion and phenoxide ion.

The resonance hybrids may be represented as:

(a) The electron charge in the carboxylate ion is more dispersed in comparison to the phenoxide ion since there are two electronegative oxygen atoms in the carboxylate ion as compared to only one oxygen atom in the phenoxide ion.

(b) Carboxylate ion is stabilized by two equivalent resonance structures in which negative charge is on more electronegative 0 atoms. But phenoxide ion has non-equivalent resonance structures in which the negative charge is also on a less electronegative carbon atom.

Therefore, the carboxylate ion is relatively more stable as compared to the phenoxide ion. Thus, the release of H⁺ ions from carboxylic acid is comparatively easier or it behaves as a stronger acid than phenol.

OR

(ii) (a) Ethanal gives a yellow ppt. of iodoform on heating with iodine and NaOH solution whereas propanal does not give.

(b) Benzoic acid reacts with NaHC03 to give effervescence due to the liberation of CO₂.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂0 + CO₂
Phenol does not give effervescence.
Phenol gives violet color with FeCl₃ solution but benzoic acid does not give such color.

Question 18.
(i) How will you convert the following:
(a) Propanone to Propan-2-ol
(b) Ethanal to 2-hydroxypropanoic acid
(c) Toluene to benzoic acid

(ii) Give a simple chemical test to distinguish between:
(a) Pentan-2-one and Pentan-3-one
(b) Ethanal and Propanal
OR
(i) Write the products of the following reactions:

(ii) Which acid of each pair shown here would you expect to be stronger? (CBSE AI 2013)
(a) F – CH₂ – COOH or Cl – CH₂ – COOH OH

Answer:

(ii) (a) Pentan-2-one forms yellow ppt. with an alkaline solution of iodine (iodoform test), but pentane-3-one does not give iodoform test.

(b) Ethanal gives a yellow ppt. of iodoform on heating with iodine and NaOH solution whereas propanal does not give.

OR

(ii) (a) FCH₂COOH
(b) CH₃COOH

Question 19.
(a) Predict the main product of the following reactions:

(b) Give a simple chemical test to distinguish between

(c) Why is alpha (a) hydrogen of carbonyl compounds acidic in nature?
OR
(a) Write the main product formed when propanal reacts with the following reagents:
(i) 2 moles of CH₃OH in presence of dry HCI
(ii) Dilute NaOH
(iii) H₂N – NH₂ followed by heating with KOH in ethylene glycol

(b) Arrange the following compounds in increasing order of their property as indicated:
(i) F – CH₂COOH, O₂N – CH₂COOH, CH₃COOH, HCOOH – acid character
(ii) Acetone, Acetaldehyde, Benzaldehyde, Acetophenone — reactivity towards the addition of HCN (CBSE AI 2019)
Answer:

(b) On adding NaOH/l2 and heat, acetophenone gives yellow ppt of iodoform but benzophenone does not.

(c) α – hydrogen of carbonyl compounds is acidic because of the electron-withdrawing nature of the carbonyl group.
OR

(b) (i) CH₃COOH < HCOOH < FCH₂COOH < NO₂-CH₃ COOH
(ii) Acetophenone < Benzaldehyde < Acetone < Acetaldehyde

Question 20.
(a) An organic compound with the molecular formula C₇H₆0 forms a 2,4-DNP derivative, reduces Tollen’s reagent, and undergoes the Cannizzaro reaction. On oxidation, it gives benzoic acid. Identify the compound and state the reactions involved.
(b) Give chemical tests to distinguish between the following pair of compounds:
(i) Phenol and propanol
(ii) Benzoic acid and benzene
OR
(a) Predict the products of the following:

(b) Arrange the following in increasing order of acidic character: (CBSE 2019C)
HCOOH, CF₃COOH, ClCH₂COOH, CCl₃COOH
Answer:
(a) Compound = Benzaldehyde or C₆H₅CHO
Reaction
Reaction with 2,4-DNP

With Tollens reagent
RCHO + 2[Ag(NH₃)₂ ]⁺ + 30H^– → RCOO^– + 2 Ag + 2H₂O + 4NH₃ (where R = – C₆H₅)
Cannizzaro

(b) (i) FeCl₃ test: Add 2 drops of neutral FeCl₃ to both the compounds separately. Phenol gives violet color.

(ii) Sodium bicarbonate test: Add NaHCO₃ to both the compounds: Benzoic acid reacts with NaHCO₃ to give effervescence due to liberation of CO₂.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂
OR
(a) A = CH₃COOH
B = CH₃COCl
C = CH₃CONH₂
D = CH₃NH₂

(b) HCOOH < ClCH₂COOH < CCl₃COOH < CF₃COOH

• the electron-withdrawing substituents disperse the negative charge on carboxylate ion and stabilize it and thus, increase acidity.
• the electron-releasing substituents intensify the negative charge of the carboxylate ion, destabilize it and thus, decrease the acidity.

Question 21.
(i) Give chemical tests to distinguish between:
(a) Propanal and propanone
(b) Benzaldehyde and acetophenone

(ii) How would you obtain:
(a) But-2-enal from the ethanal
(b) Butanoic acid from butanol
(c) Benzoic acid from ethylbenzene? (CBSE 2011)
Answer:
(i) (a) Propanal gives a silver mirror with Tollen’s reagent.

Propanone does not give this test.

(b) Benzaldehyde forms a silver mirror with ammoniacal silver nitrate solution (Tollen’s reagent). Acetophenone does not react.

Question 22.
(i) Describe the following giving linked chemical equations:
(a) Cannizzaro reaction
Answer:
Aldehydes that do not contain any a-hydrogen atom (e.g. benzaldehyde, formaldehyde) undergo self- oxidation and reduction reaction on treatment with cone, solution of caustic alkali. In this reaction, one molecule is oxidized to acid while another molecule is reduced to an alcohol.

(b) Decarboxylation
Answer:
The process of removal of a molecule of C02 from a carboxylic acid is called decarboxylation. It is usually carried out by heating a mixture of carboxylic acid or its sodium salt with soda lime (NaOH + CaO).

(ii) Complete the following chemical equations: (CBSE Delhi 2011)

Answer:

Answer:

Answer:

Question 23.
(i) Give chemical tests to distinguish between the following:
Benzoic acid and ethyl benzoate.
Answer:
(a) When treated with NaHCO₃ solution, benzoic acid gives brisk effervescence while ethyl benzoate does not

(b) Ethyl benzoate on boiling with an excess of NaOH gives ethyl alcohol which on heating with iodine gives yellow ppt. of iodoform.

(ii) Complete each synthesis by giving missing reagents or products in the following:

Answer:

Answer:

(CBSE 2011)
Answer:

Question 24.
(i) Write the products formed when CH₃CHO reacts with the following reagents:
(a) HCN
(b) H2N – OH
(C) CH₃CHO in the presence of dilute NaOH
(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Benzoic acid and Phenol
(b) Propanal and Propanone
OR
(i) Account for the following:
(a) Cl – CH₂COOH is a stronger acid than CH₃COOH.
(b) Carboxylic acids do not give reactions of the carbonyl group.
(ii) Write the chemical equations to illustrate the following name reactions:
Rosenmund reduction
(iii) Out of CH₃CH₂—CO—CH₂ – CH₃ and CH₃CH₂ — CH₂ — CO — CH₃, which gives iodoform test? (CBSE 2014)
Answer:

(ii) (a) Add neutral FeCl₃ in both the solutions, phenol gives violet color with FeCl₃ solution, but benzoic acid does not give such color.
(b) Add an ammoniacal solution of silver nitrate (Tollen’s reagent) in both the solutions, propanal gives silver mirror whereas propanone does not.

OR
(0) Chlorine is an electron-withdrawing atom (-I effect). It withdraws the electrons from carbon to which it is attached and therefore, this effect is transmitted throughout the chain. As a result, electrons are withdrawn more strongly towards oxygen of 0—H bond and promotes the release of the proton.

Consequently, ClCH₂COOH is a stronger acid than CH₃COOH.

(b) Carboxylic acids do not give the characteristic reactions of the carbonyl group (>C = 0) as given by aldehydes and ketones. In carboxylic acids, the carbonyl group is involved in resonance, as follows:

Therefore, it is not a free group. But no resonance is possible in aldehydes and ketones. They give the characteristic reactions of the group.

(ii) Rosenmund reduction. Acid chlorides are converted to corresponding aldehydes by catalytic reduction. The reaction is carried out by passing through a hot solution of the acid chloride in the presence of palladium deposited over barium sulfate (partially poisoned with sulfur or quinoline).

The poisoning of palladium catalyst decreases its activity and it does not allow the further reduction of an aldehyde into alcohol.

(iii) CH₃CH₂CH₂COCH₃ gives iodoform test

Question 25.
(i) Write the structures of A, B, C, and D in the following reactions:

(ii) Distinguish between:
(a) C₆H₅ – CH = CH – COCH₃ and C₆H₅ – CH = CH – CO CH₂CH₃
(b) CH₃CH₂COOH and HCOOH

(iii) Arrange the following in the increasing order of their boiling points:
CH₃CH₂OH, CH₃COCH₃, CH₃COOH
OR
(i) Write the chemical reaction involved in the Etard reaction.
(ii) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
CH₃ – CHO, C₆H₅COCH₃, HCHO
(iii) Why pKa of Cl-CH₂-COOH is lower than the pK_a of CH₃COOH?
(iv) Write the product in the following reaction.

(v) A and B are two functional isomers of compound C₃H₆O. On heating with NaOH and l₂, isomer A forms a yellow precipitate of iodoform whereas isomer B does not form any precipitate. Write the formulae of A and B. (CBSE 2016)
Answer:

(ii) (a) Heat both the compounds with NaOH and l₂, C₆H₅ CH = CHCOCH₃ gives yellow ppt. of iodoform. C₆H₅CH = CHCOCH₂CH₃ does not give yellow ppt. of iodoform.
(b) Add ammoniacal AgNO₃ solution (Tollen’s reagent), HCOOH gives silver mirror while CH₃CH₂COOH does not.

(iii) CH₃COCH₃ < CH₃CH₂OH < CH₃COOH
OR
(i) Etard’s reaction

(ii) C₆H₅COCH₃ < CH₃CHO < HCHO

(iii) Chlorine is an electron-withdrawing group (-I inductive effect).
It withdraws the electrons from the C to which it is attached and this effect is transmitted throughout the chain. As a result, the electrons are withdrawn more strongly towards 0 of the O-H bond and promote the release of the proton. Consequently, Cl — CH₂COOH is more acidic than acetic acid.

Therefore, pKa of ClCH₂COOH is less than that of CH₃COOH.

(iv) CH₃ CH₂ CH = CH CH₂ CN CH₃ CH₂ CH = CH CH₂ CHO

(v) Two isomers are CH₃COCH₃ and CH₃CH₂CHO
Since A forms yellow ppt. of iodoform on heating with NaOH and l₂, it is CH₃COCH₃.
CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3Nal + 3H₂O
B does not give iodoform, it is CH₃CH₂CHO.
A: CH₃COCH₃ B: CH₃CH₂CHO

Question 26.
Write the structures of A, B, C, D, and E in the following reactions:

OR
(i) Write the chemical equation for the reaction involved in the Cannizzaro reaction.
(ii) Draw the structure of the semicarbazone of ethanal.
(iii) Why pKa of F-CH₂-COOH is lower than that of Cl – CH₂ – COOH?
(iv) Write the product in the following reaction:
CH₃ – CH = CH – CH₂CN
(v) How can you distinguish between propanal and propanone? (CBSE Delhi 2016)
Answer:

OR
(i) Cannizzaro’s reaction: Aldehydes that do not contain a-hydrogen undergo self-oxidation and reduction on treatment with cone. KOH.


(iii) Fluorine is more electronegative than Cl and therefore, will have a greater electron-withdrawing effect (-I effect). As a result, the carboxylate ion will be stabilized to a larger extent and therefore fluoro acetic acid will be a stronger acid than chloroacetic acid. Hence, the pK_a of FCH₂COOH will be lower than that of ClCH₂COOH.

(iv) CH₃ — CH = CH — CH₂ CN CH₃ CH = CH CH₂ CHO

(v) Propanal gives red ppt with Fehling solution but propanone does not.

Question 27.
(i) Give reasons :
(a) HCHO is more reactive than CH₃-CHO towards the addition of HCN. (CBSE 2018C)
(b) pK_a of O₂N—CH₂—COOH is lower than that of CH₃—COOH.
(c) Alpha hydrogen of aldehydes and ketones is acidic in nature.

(ii) Give simple chemical tests to distinguish between the following pairs of compounds :
(a) Ethanal and Propanal
(b) Pentan-2-one and Pentan-3-one
OR
(i) Write the structure of the product(s) formed :
(a) CH₃ — CH₂ — COOH
(b) C₆H₅COCl
(c) 2HCHO

(ii) How will you bring the following conversions in not more than two steps :
(a) Propanone to propene
(b) Benzyl chloride to phenyl ethanoic acid
Answer:
(i) (a) Due to + I effect of the methyl group is CH₃CHO.
(b) Due to – I effect of the nitro group in nitroacetic acid.
(c) Due to the strong electron-withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base.

(ii) (a) Add NaOH and l₂ to both the compounds and heat, ethanal gives yellow ppt of iodoform.
(b) Add NaOH and l₂ to both the compounds and heat, pentan-2-one gives yellow ppt of iodoform.
OR

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 7 The p-Block Elements.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 7 Important Extra Questions The p-Block Elements

The p-Block Elements Important Extra Questions Very Short Answer Type

Question 1.
Which one of PCl₄⁺ and PCl₄^– is not likely to exist and why? (CBSE Delhi 2012)
Answer:
PCl₄^– is not likely to exist because lone pair of PCl₃ can be donated to Cl⁺ and not to Cl^–.

Question 2.
Of PH₃ and H₂S which is more acidic and why? (CBSE Delhi 2012)
Answer:
H₂S is more acidic because of weak S-H bond. PH₃ behaves as a Lewis base because of the presence of lone pair of electrons on P.

Question 3.
Which is a stronger reducing agent, SbH₃ or BiH₃ and why? (CBSE Al 2012)
Answer:
BiH₃ is a stronger reducing agent than SbH₃. This is because BiH₃ is less stable than SbH₃ because of larger size of Bi than Sb.

Question 4.
What is the covalency of nitrogen in N₂O₅? (CBSE Delhi 2013)
Answer:
Four.

Question 5.
Name two poisonous gases which can be prepared from chlorine gas. (CBSE AI 2013)
Answer:

1. Phosgene (COCl₂)
2. Mustard gas (ClCH₂CH₂SCH₂CH₂Cl)

Question 6.
What is the basicity of H₃PO₃ and why? (CBSE AI 2013)
Answer:
H₃PO₃ contains two P-OH bonds and therefore can give 2H⁺ ions. Its basicity is two.

Question 7.
Why does ammonia act as a Lewis base? (CBSE 2014)
Answer:
Nitrogen atom in NH₃ has one lone pair of electrons which is available for donation. Therefore,it acts as a Lewis base.

Question 8.
What is the oxidation number of phosphorus in H₃PO₂ molecule? (CBSE Delhi 2010)
Answer:
+1

Question 9.
Out of white phosphorus and red phosphorus, which one is more reactive and why? (CBSE 2015)
Answer:
White phosphorus is more reactive because of angular strain in the P₄ molecules where the angles are only 60°.

Question 10.
What is the basicity of H₃PO₄? (CBSE Delhi 2015)
Answer:
Basicity of H₃PO₄ is three because it has three ionisable hydrogen atoms.

Question 11.
On heating Pb(NO₃)₂ a brown gas is evolved which undergoes dimerisation on cooling. Identify the gas. (CBSE 2016)
Answer:
NO₂.

2NO₂ → N₂O₄

Question 12.
Solid PCl₅ is ionic in nature. Give reason. (CBSE AI 2016)
Answer:
PCl₅ is ionic in the solid state because it exists as [PCl₄]⁺ [PCl₆]^– in which the cation is tetrahedral and anion is octahedral.

Question 13.
Write the structural difference between white P and red P. (CBSE Delhi 2014)
Answer:
White P consists of P₄ units in which four P atoms lie at the corners of a regular tetrahedron with ∠PPP = 60°. Red P also consists of P₄ tetrahedra units but it has polymeric structure consisting of P₄ tetrahedra linked together by covalent bonds.

Question 14.
What is the basicity of H₃PO₂ acid and why? (CBSE AI 2012)
Answer:
H₃PO₂ has the structure:

It has only one ionisable hydrogen and therefore, its basicity is one.

Question 15.
Name the promoter used In Haber’s process. (CBSE Sample Paper 2017-18)
Answer:
Molybdenum.

Question 16.
NF₃ is an exothermic compound whereas NCl₃ is not. Explain. (CBSE AI 2011, 2012)
Answer:
NF₃ is an exothermic compound while NCl₃ is not because:

• The bond dissociation enthalpy of F₂ is lower than that of Cl₂.
• Size of F is small as compared to Cl and therefore F forms stronger bonds with nitrogen releasing large amount of energy.

Therefore, overall energy is released during the formation of NF₃ and energy is absorbed during the formation of NCl₃.

Question 17.
N-N single bond is weaker than P-P single bond. (CBSE Delhi 2014)
Answer:
N-N single bond is weaker than P-P single bond because of high interelectronic repulsions of non-bonding electrons due to small bond length.

Question 18.
What happens when orthophosphorous acid is heated? (CBSE Sample Paper 2017-18)
Answer:
Phosphoric acid and phosphine are formed.

Question 19.
Write the order of thermal stability of the hydrides of group 16 elements. (CBSE Delhi 2017)
Answer:
The thermal stability of the hydrides of group 16 elements decreases from H₂O to H₂Te as:
H₂O > H₂S > H₂Se > H₂Te

Question 20.
The two O – O bond lengths in ozone molecule are equal. Why? (CBSE AI 2013, CBSE Delhi 2014)
Answer:
Ozone is a resonance hybrid of two structures and therefore, the two O-O bond lengths are equal.

Question 21.
SF₆ is not easily hydrolysed. (CBSE Delhi 2005)
OR
SF₆ is kinetically inert substance. Explain. (CBSE Al 2011, CBSE Delhi 2011)
Answer:
SF₆ is chemically inert and therefore, does not get hydrolysed. Its inert nature is due to the presence of stearically protected sulphur atom which does not allow thermodynamically favourable hydrolysis reaction.

Question 22.
Which of the following compounds has a lone pair of electrons at the central atom? (CBSE Sample Paper 2011)
H₂S₂O₈, H₂S₂O₇, H₂SO₃, H₂SO₄.
Answer:
H₂SO₃

Question 23.
Ozone is thermodynamically unstable. Explain. (CBSE Sample Paper 2011)
Answer:
Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (∆H is -ve) and increase in entropy (∆S is +ve). These two factors reinforce each other resulting negative ∆G (∆G = ∆H – T∆S) for its conversion to oxygen.

Question 24.
Which neutral molecule would be isoelectronic with ClO^–? Is that molecule a Lewis base? (CBSE Delhi 2008)
Answer:
ClO^– is isoelectronic with ClF. Yes, it is a Lewis base.

Question 25.
Which compound of xenon has distorted octahedral shape? (CBSE Sample Paper 2012)
Answer:
XeF₆ has distorted octahedral shape.

Question 26.
Iodide ions can be oxidised by oxygen in acidic medium. Give chemical equation to support this. (CBSE Sample Paper 2011)
Answer:
Iodide can be oxidised by oxygen in acidic medium:
4I^–(aq) + O₂(g) + 4H⁺(oq) > 2l₂(s) + 2H₂O(l).

Question 27.
What happens when XeF6 undergoes complete hydrolysis? (CBSE Sample Paper 2017-18)
Answer:
XeO₃ is formed.
XeF₆ + 3H₂O → XeO₃ + 6HF

Question 28.
Which xenon compound is isostructural with lCl₄^–? (CBSE Sample Paper 2011)
Answer:
XeF₄

Question 29.
Why conductivity of silicon increases on doping with phosphorus? (CBSE Delhi 2019)
Answer:
When silicon is doped with phosphorus, an extra electron is introduced after forming four covalent bonds. This extra electron gets delocalised and serves to conduct electricity. Hence, conductivity increases.

The p-Block Elements Important Extra Questions Short Answer Type

Question 1.
What inspired N. Bartlett for carrying out reaction between Xe and PtF₆? (CBSE Delhi 2013)
Answer:
In 1962, N. Bartlett noticed that platinum hexafluoride PtF₆, is a powerful oxidising agent which combines with molecular oxygen to form ionic compound, dioxygenyl hexafluoroplatinate (V), O₂⁺ [PtF₆]^–.
O₂(S) + PtF₆(g) → O₂⁺ [PtF₆]^–
This indicates that PtF₆ has oxidised O₂ to O₂⁺. Now, oxygen and xenon have some similarities:

• The first ionisation energy of xenon gas (1170 kJ mol^-1) is fairly close to that of oxygen (1177 kJ mol^-1).
• The molecular diameter of oxygen and atomic radius of xenon are similar (4Å). These prompted Bartlett to carry out the reaction between Xe and PtF₆.

Question 2.
Calculate the number of lone pairs on central atom in the following molecule and predict the geometry. (CBSE Sample Paper 2019)
Answer:
XeF₄ has square planar structure. It involves sp³d² hybridisation of Xenon in which two positions are occupied by lone pairs.

Structure of XeF₄ (Square planar)

Question 3.
How are interhalogen compounds formed? What general compositions can be assigned to them? (CBSE AI 2013)
Answer:
The compounds containing two or more halogen atoms are called interhalogen compounds. They may be assigned general composition as XX’_n, where X is halogen of larger size and X’ is of smaller size and X’ is more electronegative than X, e.g. ClF, ICl₃, BrF₅, IF₇, etc.

Question 4.
Though nitrogen exhibits +5 oxidation state, it does not form pentahalides. Give reason. (CBSE 2013, CBSE Delhi 2015)
Answer:
Nitrogen belongs to second period (n = 2) and has only s and p-orbitals. It does not have d-orbitals in its valence shell and therefore, it cannot extend its octet. That is why nitrogen does not form pentahalides.

Question 5.
Why does NO₂ dimerise? Explain. (CBSE 2014, CBSE Delhi 2014)
Answer:
NO₂ contains odd number of valence electrons. It behaves as a typical molecule. In the liquid and solid state, it dimerises to form stable N₂O₄ molecule, with even number of electrons. Therefore, NO₂ is paramagnetic, while N₂O₄ is diamagnetic in which two unpaired electrons get paired.

Question 6.
Draw the structure of O₃ molecule. (CBSE Delhi 2010)
Answer:
Ozone has angular structure as:

It is resonance hybrid of the following structures:

Question 7.
Complete the following equations: (CBSE 2014)
(i) Ag + PCl₅ →
(ii) CaF₂ + H₂SO₄ →
Answer:
(i) 2Ag + PCl₅ → 2AgCl + PCl₃
(ii) CaF₂ + H₂SO₄ → CaSO₄ + 2HF

Question 8.
Write balanced chemical equations for the following processes:
(i) XeF₂ undergoes hydrolysis.
(ii) MnO₂ is heated with cone. HCl.
OR
Arrange the following in order of property indicated for each set:
(i) H₂O, H₂S, H₂Se, H₂Te – increasing acidic character
(ii) HF, HCl, HBr, HI – decreasing bond enthalpy (CBSE Delhi 2019)
Answer:
(i) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂
(ii) MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
OR
(i) H₂O < H₂S < H₂Se < H₂Te
(ii) HF > HCI > HBr > HI

Question 9.
Explain the following giving an appropriate reason in each case.
(i) O₂ and F₂ both stabilise higher oxidation states of metals but O₂ exceeds F₂ in doing so.
(ii) Structures of xenon fluorides cannot be explained by Valence Bond approach. (CBSE 2012)
Answer:
(i) This is because fluorine can show only -1 oxidation state, whereas oxygen can show -2 oxdiation state as well as positive oxidation states such as +2, +4 and +6.

(ii) Structures of xenon fluorides cannot be explained by Valence Bond approach because xenon is a noble gas and its octet is complete. Therefore, according to Valence Bond theory, xenon should not combine with fluorine to form fluorides due to its stable outermost shell configuration.

Question 10.
Draw the structures of the following molecules:
(i) XeOF₄
(ii) H₃PO₃ (CBSE 2013)
Answer:
(i)

Structure of XeOF₄

(ii)

Structure of H₃PO₃

Question 11.
How are interhalogen compounds formed? What general compositions can be assigned to them? (CBSE 2013)
Answer:
The compounds containing two or more halogen atoms are called interhalogen compounds. They may be assigned general composition as XX’_n, where X is halogen of larger size and X’ is of smaller size and X’ is more electronegative than X,
e.g. ClF, ICl₃, BrF₅, IF₇, etc.

Question 12.
Draw the structures of the following molecules: (CBSE 2013)
(i) N₂O₅
(ii) XeF₂
Answer:
(i)

(ii)

Structure of XeF₂ (Linear)

Question 13.
What happens when
(i) PCl₅ is heated?
(ii) H₃PO₃ is heated?
Write the reactions involved. (CBSE Delhi 2013)
Answer:
(i) On heating, PCl₅ sublimes and decomposes on strong heating:

(ii) On heating, H₃PO₃ decomposes into phosphoric acid and phosphine.
4H₃PO₃ → 3H₃PO₄ + PH₃

Question 14.
Complete the following chemical equations:
(i) Ca₃P₂ + H₂O →
(ii) Cu + H₂SO₄(conc.) →
OR
Arrange the following in the order of property indicated against each set:
(i) HF, HCl, HBr, HI – increasing bond dissociation enthalpy.
(ii) H₂O, H₂S, H₂Se, H₂Te – increasing acidic character. (CBSE Delhi 2014)
Answer:
(i) Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃
(ii) Cu + 2H₂SO₂₄;(conc.) → CuSO₄
+ SO₂ + 2H₂O
OR
(i) HI < HBr < HCl < HF
(ii) H₂O < H₂S < H₂Se < H₂Te

Question 15.
Complete the following equations:
(i) P₄ + H₂O →
(ii) XeF₄ + O₂F₂ → (CBSE 2014)
Answer:
(i) P₄ + H₂O → no reaction
(ii) XeF₄ + O₂F₂ → XeF₆ + O₆

Question 16.
Draw the structures of the following:
(i) XeF₂
(ii) BrF₃
Answer:
(i)

Structure of XeF₂ (Linear)

(ii)

Structure of BrF₃ (T Shape)

Question 17.
(a) Draw the structure of XeF₄.
(b) What happens when CaF₂ reacts with cone. H₂SO₄? Write balanced chemical equation. (CBSE 2019C)
Answer:
(a)

(b) HF is formed.
CaF₂ + H₂SO₄ → 2HF + CaSO₄

Question 18.
(a) Draw the structure of H₂S₂O₇.
(b) What happens when carbon reacts with cone. H₂SO₄? Write balanced chemical equation. (CBSE 2019C)
Answer:
(a)

(b) Carbon dioxide is formed.
C + 2H₂SO₄(conc.) → CO₂ + 2 SO₂ + 2H₂O

Question 19.
Draw the structures of the following molecules: (CBSE 2014)
(i) XeO₂
(ii) H₂SO₄
Answer:
(i)

(ii)

Question 20.
Write the structure of the following: (HPO₃)₃ (CBSE 2016)
Answer:
(HPO₃)₃

Question 21.
Why is N₂ less reactive at room temperature? (CBSE Al 2015, H.P.S.B. 2015, 2016, Meghalaya S.B. 2017)
Answer:
In molecular nitrogen, there is a triple bond between two nitrogen atoms (NsN) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Question 22.
Phosphine has lower boiling point than ammonia. Give reason. (CBSE Al 2008,
CBSE Delhi 2013)
Answer:
Ammonia exists as associated molecule due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike NH₃, phosphine (PH₃) molecules are not associated through hydrogen bonding in liquid state.
This is because of low electronegativity of P than N. As a result, the boiling point of PH₃ is lower than that of NH₃.

Question 23.
Draw structures of the following:
(a) XeF₄
(b) S₂O₈^2- (CBSE AI 2019)
Answer:
(a)

(b)

Question 24.
(i) Draw the structure of phosphinic acid (H₃PO₂).
(ii) Write a chemical reaction for its use as reducing agent. (CBSE Sample Paper 2011)
Answer:
(i) Phosphinic acid, H₃PO₂.

(ii) H₃PO₂ reduces Ag⁺ ion to Ag which shows its reducing nature.
H₃PO₂ + 4AgNO₃ + 2H₂O → 4Ag + 4HNO₃ + H₃PO₄

Question 25.
Among the hydrides of Group 15 elements, which have the
(a) lowest boiling point?
(b) maximum basic character?
(c) highest bond angle?
(d) maximum reducing character? (CBSE AI 2018)
Answer:
(a) PH₃
(b) NH₃
(c) NH₃
(d) BiH₃

Question 26.
The bond angles (O-N-O) are not of the same value in NO₂^– and NO₂⁺. Give reason. (CBSE Delhi 2012)
Answer:
In NO₂ there is one electron on N. Therefore, in NO₂⁺, there is no electron on N atom and there is a lone pair of electrons on N atom in NO₂^–.

Question 27.
H₂PO₂ is a stronger reducing agent than H₃PO₃. Give reason. (CBSEAI 2014, 2016)
Answer:
The reducing character of the acid is due to H atoms bonded directly to P atom. In H₃PO₂, there are two P-H bonds whereas in H₃PO₃, there is one P-H bond. Therefore, H₃PO₂ is a stronger reducing agent than H₂PO₃.

Question 28.
Bi(V) is a stronger oxidising agent than Sb(V). Give reason. (CBSE Delhi 2014)
Answer:
On moving down the group, the stability of +5 oxidation state decreases while the stability of +3 oxidation state increases due to inert pair effect. Therefore, +5 oxidation state of Bi is less stable than +5 oxidation state of Sb. Thus, Bi(V) is a stronger oxidising agent than Sb(V).

Question 29.
Dioxygen is a gas while sulphur is a solid at room temperature. Why?
(CBSE AI 2013, 2018)
Answer:
Oxygen exists as a stable diatomic molecule and is, therefore, a gas. On the other hand, sulphur exists in solid state as S8 molecules and have puckered ring structure. The main reason for this different behaviour is that oxygen atom has strong tendency to form multiple bonds with itself and forms strong O=O bonds rather than O-O bonds.

On the other hand, sulphur-sulphur double bonds (S=S) are not very strong. As a result, catenated -O-O-O- chains are less stable as compared to O=O molecule while catenated -S-S-S- chains are more stable as compared to S = S molecule.
Therefore, oxygen exists as a diatomic gas and sulphur exists as S₈ solid.

Question 30.
Draw structures of the following:
(a) H₂S₂O₇
(b) HClO₃ (CBSE Al 2019)
Answer:
(a) H₂S₂O₇

(b) HClO₃

Question 31.
Complete and balance the following equations:
(a) C + H₂SO₄ (cone.) →
(b) XeF₂ + PF₅ →
OR
Write balanced chemical equations involved in the following reactions:
(a) Fluorine gas reacts with water.
(b) Phosphine gas is absorbed in copper sulphate solution. (CBSE AI 2019)
Answer:
(a)

(b)

OR

(a) 2F₂ + 2H₂O → 4HF + O₂

(b)

Question 32.
Sulphuric acid has low volatility. Give chemical reactions in support of this. (CBSE Sample Paper 2011)
Answer:
Sulphuric acid has low volatility and decomposes the salts of volatile acids forming its own salts:
2MX + H₂SO₄ → 2HX + M₂SO₄
(M = metal, X = F, Cl, NO3)
e.g.,NaCl + H₂SO₄ → NaHSO₄ + HCl
KNO₃ + H₂SO₄ → KHSO₄ + HNO₃.

Question 33.
Electron gain enthalpies of halogens are largely negative. Why? (CBSE AI 2017)
Answer:
The halogens have the smallest size in their respective periods and therefore, high effective nuclear charge. Moreover, they have only one electron less than the stable noble gas configuration (ns²np⁶).

Therefore, they have strong tendency to accept one electron to acquire noble gas electronic configurations and hence have largely negative electron gain enthalpies.

Question 34.
Why does R₃P = 0 exist but R₃N = 0 does not (R = alkyl group). (CBSE AI 2014)
Answer:
R₃N = 0 does not exist because nitrogen cannot have covalency more than four. Moreover, R₃P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ-pπ bond whereas nitrogen cannot form dπ-pπ bond.

Question 35.
Write any two oxoacids of sulphur and draw their structures. (CBSE Al 2019)
Answer:
(i) Sulphuric acid

(ii) Peroxodisulphuric acid

Question 36.
Complete and balance the following equations:
(a) NH₃ (excess) + Cl₂ →
(b) XeF₆ + 2H₂O →
OR
Write balanced chemical equations involved in the following reactions:
(a) XeF₄ reacts with SbF₅.
(b) Ag is heated with PCl₅. (CBSE AI 2019)
Answer:
(a)
8NH₃ + 3Cl₂ → 6NH₄Cl + N₂
excess
(b) XeF₆ + 2H₂O → XeO₂F₂ + 4HF

OR

(a) XeF₄ + SbF₅ → [XeF₃]⁺[SbF₆]^–
(b) 2Ag + PCl₅ → 2AgCl + PCl₃

Question 37.
Complete and balance the following equations:
(a) S + H₂SO₄ (cone.) →
(b) PCl₃ + H₂O →
Write balanced chemical equations involved in the following reactions:
(a) Chlorine gas reacts with cold and dilute NaOH
(b) Calcium phosphide Is dissolved in water. (CBSE AI 2019)
Answer:
(a) S + 2H₂SO₄(conc.) → 3SO₂ + 2H₂O
(b) PCl₃ +3H₂O → H₃PO₃ + 3HCl
OR
(a) Cl₂ + ZNaOH → NaCl + NaOCl + H₂O
(b) Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃

Question 38.
Bond enthalpy of fluorine Is lower than that of chlorine. Why? (CBSE AI 2018)
Answer:
Fluorine atoms are smalL and internuclear distance between two F atoms In F₂ molecule is also small (1 .43Å). As a result, electron-electron repulsions among the lone pairs on two fluorine atoms in F₂ molecule are large. These Llarge electron-electron repulsions weaken the F-F bond. However, the electron-electron repulsions among lone pairs in Cl₂ molecule are less because of Larger Cl atoms and larger Cl-Cl distance (1.99Å). Therefore, the bond enthalpy of fluorine is lower than that of chlorine.

The p-Block Elements Important Extra Questions Long Answer Type

Question 1.
Give reasons for the following:
(a) Dioxygen is a gas but sulphur a solid.
(b) NO (g) released by jet aeroplanes is slowly depleting the ozone layer.
(c) Interhalogens are more reactive than pure halogens. (CBSE Al 2019)
Answer:
(a) Due to small size and high electronegativity, oxygen atom forms pπ-pπ double bond, O = O. The intermolecular forces in oxygen are weak van der Waal’s forces and therefore, oxygen exists as a gas. On the other hand, sulphur does not form stable pπ-pπ bonds and does not exist as S₂. It is linked by single bonds and forms polyatomic complex molecules having eight atoms per moledule (S₈) and has puckered ring structure. Therefore, S atoms are strongly held together and it exists as a solid.

(b) Nitrogen oxide (NO) gas emitted from the exhaust of jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen.
NO + O₃ → NO₂ + O₂
Since jet aeroplanes fly in the stratosphere, near ozone layer, these are responsible for the depletion of ozone layer.

(c) This is because covalent bond between dissimilar atoms in interhalogens (X- Y) is polar and weaker than between similar atoms (X-X and Y-Y). This is due to the fact that the overlapping of orbitals of dissimilar atoms is less effective than the overlapping between similar atoms.

Question 2.
Give reasons for the following:
(i) Where R is an alkyl group, R₃P = 0 exists but R₃N = 0 does not.
(ii) PbCl₄ is more covalent than PbCl₂.
(iii) At room temperature, N₂ is much less reactive. (CBSE AI 2013)
Answer:
(i) R₃N = 0 does not exist because nitrogen cannot have covalency more than four. But R₃P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ-pπ bonds whereas nitrogen cannot form dπ-pπ bond.

(ii) Pb⁴⁺ has smaller size than Pb²⁺. As a result, it can polarise the Cl^– ion to greater extent than Pb²⁺. Consequently, PbCl₄ is more covalent than PbCl₂.

(iii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Question 3.
Give reasons for the following:
(i) Oxygen is gas but sulphur is a solid,
(ii) O₃ acts as a powerful oxidising agent.
(iii) BiH₃ is the strongest reducing agent amongst all the hydrides of Group 15 elements. (CBSE Al 2013)
Answer:
(i) Oxygen exists as a stable diatomic molecule and is, therefore, a gas.
On the other hand, sulphur exists in solid state as S₈ molecules and have puckered ring structure. The main reason for this different behaviour is that oxygen atom has good tendency to form multiple bonds with itself and forms strong o = o bonds than o – o bonds. On the other hand, sulphur- sulphur double bonds (S = S) are not very strong. As a result, catenated -O-O-O- chains are less stable as compared to O = O molecule while catenated -S-S-S- chains are more stable as compared to S = S molecule. Therefore, oxygen exists as a diatomic gas and sulphur exists as S₈ solid.

(ii) Ozone acts as a powerful oxidising agent because it has higher energy content and decomposes readily to give atomic oxygen as:
O₃ → O₂ + O
Therefore, ozone can oxidise a number of non-metals and other compounds. For example,
PbS (s) + 4O₃ (8) → PbSO₄ (s) + 4O₂ (g)
2I^– (aq) + H₂O(l) + O₃(g) → 2OH^– (aq) + I₃₂(s) + O₂(g)

(iii) The stability of group 15 hydrides decreases from NH₃ to BiH₃. Therefore, BiH₃ is most unstable hydride among group 15 hydrides because of lowest M-H bond strength. As a result, BiH₃ is the strongest reducing agent.

Question 4.
Give reasons for the following:
(i) Though nitrogen exhibits +5 oxidation state, it does not form pentahalide.
(ii) Electron gain enthalphy with negative sign of fluorine is less than that of chlorine.
(iii) The two oxygen-oxygen bonds lengths in ozone molecule are identical. (CBSE Al 2013)
Answer:
(i) Nitrogen does not have vacant d-orbitals in its valence shell. Therefore, it cannot extend its valency beyond 3. Thus, it cannot form pentahalides.

(ii) The less negative electron gain enthalpy of fluorine as compared to chlorine is due to very small size (72 pm) of fluorine atom. As result, there are strong inter- electronic repulsions in the relatively small 2p sub-shell of fluorine and therefore, the incoming electron does not feel much attraction. Thus, its electron gain enthalpy is small.

(iii) Ozone is resonance hybrid of two structures and therefore, the two oxygen-oxygen bonds are identical.

Question 5.
How would you account for the following:
(i) H₂S is more acidic than H₂O.
(ii) The N-O bond in NO₂^– is shorter than the N-O in NO₃^–.
(iii) Both O₂ and F₂ stabilise higher oxidation states but the ability of oxygen to stabilise the higher oxidation states exceeds that of fluorine. (CBSE 2011)
Answer:
(i) Since size of S is more than that of O, the distance between central atom and hydrogen atoms is more in H₂S than in H₂O. Therefore, bond dissociation enthalpy of H₂S is less and bond cleavage is more easy than in H₂O. Therefore, H₂S is more acidic than H₂O.

(ii) In the resonance structures of NO₂^–, two bonds are sharing a double bond, while in NO₃^–, three bonds are sharing a double bond.

As a result, the bond in NO₂^– will be shorter than in NO₃^–.

(iii) The larger tendency of oxygen to stabilise the higher oxidation state is because of the tendency of oxygen to form multiple bonds with metal.

Question 6.
How would you account for the following:
(i) NF₃ is an exothermic compound but NCl₃ is not.
(ii) The acidic strength of compounds increases in the order:
PH₃ < H₂S < HCl
(iii) SF₆ is kinetically inert. (CBSE 2011)
Answer:
(i) NF₃ is an exothermic compound while NCl₃ is an endothermic compound because
(a) The bond dissociation enthalpy of F₂ is lower as compared to Cl₂.
(b) Size of F is small as compared to Cl and therefore, F forms stronger bond with N releasing large amount of energy.
Therefore, overall energy is released during the formation of NF₃ and energy is absorbed during the formation of NCl₃.

(ii) The bond dissociation enthalpy decreases in the order:
P-H > S-H > Cl-H
Therefore, acidic character follows the order:
PH₃ < H₂S < HCl

(iii) SF₆ is kinetically inert because sulphur is sterically protected by six F atoms and it is a coordinatively saturated compound. Therefore, it does not undergo thermodynamically favourable hydrolysis reaction.

Question 7.
(a) Draw the structures of the following molecules:
(i) XeOF₄
(ii) H₂SO₄
(b) Write the structural difference between white phosphorus and red phosphorus. (CBSE Delhi 2014)
Answer:
(a) (i) XeOF₄

(ii) H₂SO₄

(b) White phosphorus consists of P₄ units in which four P atoms lie at the corners of a regular tetrahedron with ΔPPP = 60°.
Answer:
Red phosphorus also consists of P₄ tetrahedra units, but it has polymeric structure consisting of P₄ tetrahedra linked together by covalent bond.

Question 8.
Account for the following:
(i) Bi(V) is a stronger oxidising agent than Sb(V).
(ii) N-N single bond is weaker than P-P single bond.
(iii) Noble gases have very low boiling points. (CBSE Delhi 2014)
Answer:
(i) On moving down the group, the stability of +5 oxidation state decreases while the stability of +3 oxidation state increases due to inert pair effect. Therefore, +5 oxidation state of Bi is less stable than +5 oxidation state of Sb. Thus, Bi(V) is a stronger oxidising agent than Sb(V).

(ii) N-N single bond is weaker than single P-P bond because of high interelectronic repulsions of non-bonding electrons due to small bond length.

(iii) Noble gases are monoatomic gases and are held together by weak vander Waals forces. Therefore, these are liquified at very low temperatures. Hence they have low boiling points.

Question 9.
Give reasons for the following: (CBSE 2014)
(i) (CH₃)₃ P = 0 exists but (CH₃)₃ N = 0 does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) H₃PO₂ is a stronger reducing agent than H₃PO₃.
Answer:
(i) (CH₃)₃ N = 0 does not exist because
nitrogen cannot have covalency more than four. Moreover, (CH₃)₃ P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ – pπ bonds whereas nitrogen cannot form dπ – pπ bonds.

(ii) This is due to small size of oxygen atom so that its electron cloud is distributed over a small region of space and therefore, it repels the incoming electron. Hence the electron gain enthalpy of oxygen is less negative than sulphur.

(iii) In H₃PO₃, there are two P-H bonds whereas in H₃PO₃, there is one P-H bond. Therefore, H₃PO₃ is stronger reducing agent than H₃PO₃.

Question 10.
Assign reason for the following:
(i) H₃PO₂ is a stronger reducing agent than H₃PO₄.
(ii) Sulphur shows more tendency for catenation than oxygen.
(iii) Reducing character increases from HF to HI. (CBSE 2016)
Answer:
(i) The reducing character of the acid is due to H atoms bonded directly to P atom. In H₃PO₂, there are two P-H bonds whereas in H₃PO₃, there is one P-H bond. Therefore, H₃PO₂ is a stronger reducing agent than H₃PO₃.

(ii) Sulphur shows more tendency for catenation than oxygen because of stronger S-S bonds than O-O bonds.

(iii) Due to decrease in bond enthalpy down the group, the thermal stability decreases from HF to HI.
As a result of decrease in stability from HF to HI, the reducing character increases down the group as:
HF < HCl < HBr < HI

Question 11.
What happens when
(i) (NH₄)₂Cr₂O₇ is heated?
(ii) PCl₅ is heated?
(iii) H₃PO₃ is heated? Write the equations involved. (CBSE AI 2015, CBSE Delhi 2008, 2013, 2017)
Answer:
(i) On heating (NH₄)₂Cr₂O₇, nitrogen gas is evolved.

(ii) On heating, PCl₅ first sublimes and then decomposes on strong heating:

(iii) On heating, H₃PO₃ disproportionates to give orthophosphoric acid and phosphine.

Question 12.
(a) Suggest a quantitative method for estimation of the gas which protects us from U.V. rays of the sun.
(b) Nitrogen oxides emitted from the exhaust system of supersonic jet aeroplanes slowly deplete the concentration of ozone layer in upper atmosphere. Comment. (CBSE Sample Paper 2011)
Answer:
(a) The gas which protects us from U.V. rays of the sun is ozone. It reacts with I^– ions to give iodine as:
O₃ + 2I^– + H₂O → O₂ + l₂ + 2OH^–
l₂ liberated is titrated against sodium thiosulphate solution and amount of O₃ can be estimated.

(b) The release of nitrogen oxides (NO_x) into stratosphere by the exhaust system of supersonic jet aeroplanes deplete the concentration of O₂ because NO reacts with O₃ to give O₂.
NO (g) + O₂ (g) → NO₂ (S) + O₂ (s)
Therefore, NO is slowly depleting the concentration of ozone.

Question 13.
Give reasons for the following:
(a) Acidic character decreases from N₂O₃ to Bi₂O₃.
(b) All the P-Cl bonds in PCl₅ are not equivalent.
(c) HF is a weaker acid than HCl in an aqueous solution. (CBSE Al 2019)
Answer:
(a) On moving down the group, metallic character increases. Therefore acidic character of oxides decreases from nitrogen to bismuth.

(b) PCl₅ has trigonalbipyramidal geometry.

In this geometry, axial bonds are larger than equatorial bonds because of large repulsions at axial positions. Therefore, all P-Cl bonds in PCl₅ are not equivalent.

(c) The H-F bond has high bond dissociation enthalpy than H-Cl bond. Therefore, H-F bond is stronger and can be dissociated into H⁺ and F^– ions with difficulty. Hence, it is weaker acid than HCl.

Question 14.
Give reasons for the following:
(a) O-O single bond is weaker than S-S single bond.
(b) Tendency to show -3 oxidation state decreases from Nitrogen (N) to Bismuth (Bi).
(c) Cl₂ acts as a bleaching agent. (CBSE AI 2019)
Answer:
(a) O-O single bond is weaker than S-S single bond because oxygen has smaller size and has high inter-electronic repulsion.

(b) Tendency to show -3 oxidation state decreases because metallic character and size increase down the group.

(c) Cl₂ acts as a bleaching agent in the presence of moisture
Cl₂ + H₂O → 2HCl + [O]
Coloured matter + [O] → Colourless matter
The nascent oxygen is responsible for the bleaching action of Cl₂.

Question 15.
(a) Draw the structures of the following molecules:
(i) XeOF₄
(ii) H₂SO₄
(b) Write the structural difference between white phosphorus and red phosphorus. (CBSE Delhi 2014)
Answer:
(a) (i) XeOF₄

(ii) H₂SO₄

(b) White phosphorus consists of P₄ units in which four P atoms lie at the corners of a regular tetrahedron with ΔPPP = 60°.
Red phosphorus also consists of P₄ tetrahedra units, but it has polymeric structure consisting of P₄ tetrahedra linked together by covalent bond.

Question 16.
Account for the following:
(i) PCl₅ is more covalent than PCl₃.
(ii) Iron on reaction with HCl forms FeCl₂ and not FeCl₃.
(iii) The two O-O bond lengths in the ozone molecule are equal. (CBSE Delhi 2014)
Answer:
(i) In PCl₅, the oxidation state of P is +5 while in PCl₃, it is +3. As a result of higher oxidation state of the central atom, PCl₅ has larger polarising power and can polarise the chloride ion (Cl^–) to a greater extent than in the corresponding PCl₃. Since larger the polarisation, larger is the covalent character; PCl₅ is more covalent than PCl₃.

(ii) Iron reacts with HCl to form ferrous chloride with the evolution of hydrogen gas.
Fe + 2HCl → FeCl₂ + H₂

(iii) The two O-O bond lengths in ozone molecule are equal and are intermediate between single and double bonds. This is because of resonance between two structures (a) and (b) and actual molecule is resonance hybrid of these two structures:

Question 17.
Account for the following:
(a) Moist SO₂ decolourises KMnO₄ solution.
(b) In general, interhalogen compounds are more reactive than halogens (except fluorine).
(c) Ozone acts as a powerful oxidising agent. (CBSE Sample Paper 2019)
Answer:
Moist sulphur dioxide behaves as a reducing agent. It reduces MnO₄^– to Mn²⁺.
5 SO₂ + 2MnO₄^– + 2H₂O → 5 SO₄^2- + 4H⁺ + 2Mn²⁺

(b) X-X’ bond in inter halogens is weaker than X-X bond in halogens except F-F bond. This is due to the fact that the overlapping of orbitals of two dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.

(c) Due to the ease with which it liberates atoms of nascent oxygen, Ozone acts as a powerful oxidising agent.
O₃ → O₂ + O

Question 18.
(a) What happens when
(i) chlorine gas is passed through a hot concentrated solution of NaOH?
(ii) sulphur dioxide gas is passed through an aqueous solution of Fe(lll) salt?
Answer:
(a) (i) Sodium chlorate and sodium chloride are formed.

(ii) Fe (III) salt is reduced to Fe (II) salt.
2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^{2- + 4H+}

(b) Answer the following:
(i) What is the basicity of H₃PO₃ and why?
(ii) Why does fluorine not play the role of a central atom in interhalogen compounds?
(iii) Why do noble gases have very low boiling points? (CBSE Delhi 2011)
Answer:
(i) H₃PO₃ is dibasic and has basicity of
two because it has two P-OH bonds which are ionisable. The third H atom is linked to P and is non-ionisable.

H₃PO₃ ⇌ HPO₃^2- + 2H⁺

(ii) Fluorine does not play the role of a central atom in interhalogen compounds because it is highly electronegative. Moreover, it has only one electron less than the octet and does not have vacant d-orbitals in its valence shell. Therefore, it can form only one bond with other halogen atoms and cannot act as central atom in interhalogen compounds.

(iii) Noble gases have very low boiling points because only weak van der Waals’ forces are present between the atoms of the noble gases in the liquid state.

Question 19.
(a) Give reasons for the following:
(i) Bond enthalpy of F₂ is lower than that of Cl₂.
(ii) PH₃ has lower boiling point than NH₃.
Answer:
(a) (i) Due to small size of F atom, there are strong repulsions between the non-bonding electrons of F atoms in the small sized F₂ molecule. Therefore, bond enthalpy of F₂ is lower than relatively larger Cl₂ molecule in which repulsions between non-bonding electrons are less.

(ii) Ammonia exists as associated molecules due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike NH₃, phosphine (PH₃) molecules are not associated through hydrogen bonding in liquid state. This is because of low electronegativity of P than N. As a result, the boiling point of PH₃ is lower than that of NH₃.

(b) Draw the structures of the following molecules:
(i) BrF₃
(ii) (HPO₃)₃
(iii) XeF₄
Answer:
(i)

BrF₃ (T shape)

(ii)

(HPO₃)₃

(iii)

XeF₄

OR

(a) Account for the following:
(i) Helium is used in diving apparatus.
(ii) Fluorine does not exhibit positive oxidation state.
(iii) Oxygen shows catenation behaviour less than sulphur.
Answer:
(a) (i) Helium alongwith oxygen (helium- oxygen mixture) is used by deep sea divers in preference to nitrogen-oxygen mixture because of its low solubility in blood.

(ii) Fluorine is the most electronegative element and therefore, it shows an oxidation state of -1 only. It does not show any positive oxidation state.

(iii) Sulphur has strong tendency for catenation, i.e. forming bonds with itself in comparison to oxygen. This is because of stronger S-S single bond than O-O single bond.

(b) Draw the structures of the following molecules.
(i) XeF₂
(ii) H₂S₂O₈ (CBSE Delhi 2013)
Answer:
(i)

XeF₂ (Linear)

(ii)

Question 20.
(a) Give reasons for the following:
(i) Sulphur in vapour state shows paramagnetic behaviour.
(ii) N-N bond is weaker than P-P bond.
(iii) Ozone is thermodynamically less stable than oxygen.
Answer:
(a) (i) In the vapour state, sulphur exists as S2 molecules. S2 molecule has two unpaired electrons in anti-bonding molecular orbitals (π^x and π^y) and hence shows paramagnetism.

(ii) The N-N bond is weaker than P-P bond because of high interelectronic repulsions of non-bonding electrons due to small N-N bond length (109 pm).

(iii) Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (ΔH is negative) and increase in entropy (ΔS positive). These two factors make large negative ΔG for its conversion to oxygen.

(b) Write the name of gas released when Cu is added to
(i) dilute HNO₃ and
(ii) conc. HNO₃
Answer:
(i) With dilute HNO₃; NO (nitric oxide)
3Cu + 8HNO₃(dil) → 3Cu(NO₃)₂ + 2NO + 4H₂O

(ii) With cone. HNO₃: NO₂ (nitrogen dioxide)
Cu + 4HNO₃(conc) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

OR

(a) (i) Write the disproportionation reaction of H₃PO₃.
(ii) Draw the structure of XeF₄.
Answer:
(i) 4H₃PO₃ → 3H₃PO₄ + PH₃

(ii)

(b) Account for the following:
(i) Although fluorine has less negative electron gain enthalpy, yet F₂ is strong oxidising agent.
(ii) Acidic character decreases from N₂O₃ to Bi₂O₃ in group 15.
Answer:
(i) Fluorine is the strongest oxidising agent, although it has less negative electron gain enthalpy. This is because of its smaller size, large bond dissociation enthalpy of F₂ and high exothermic hydration enthalpy of small F^– ion.

(ii) On moving down the group, the metallic character increases. Therefore, the acidic character of oxides decreases down the group from N₂O₃ to Bi₂O₃

(c) Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved. (CBSE Delhi 2019)
Answer:
SO₂ decolourises pink violet colour of acidified potassium permanganate solution.
2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄

Question 21.
(a) Complete the following chemical reaction equations:
(i) P₄ + SO₂Cl₄ →
(ii) XeF₆ + H₄O →
Answer:
(i) P₄ + 10SO₂Cl₂ → 4PCl₅ + 10SO₂

(ii) XeF₆ + H₂O → XeOF₄ + 2HF

(b) Predict the shape and the asked angle (90° or more or less) in each of the following cases:
(i) SO₃^2- and the angle O – S – O
(ii) ClF₃ and the angle F – Cl – F
(iii) XeF₂ and the angle F – Xe – F
Answer:
(i) SO₃^2-

The O-S-O angle is more than 90°.

(ii) ClF₃

The F-Cl-F bond angle is equal to 90°.

(iii)
XeF₂

F-Xe-F bond angle is equal to 180° (greater than 90°).

OR

(a) Complete the following chemical equations:
(i) NaOH + Cl₂ →
(hot and cone.)
(ii) XeF₄ + O₂F₂ →
Answer:
(i) 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O
(Hot and conc.)

(ii) XeF₄ + O₂F₂ → XeF₆ + O₂

(b) Draw the structures of the following molecules:
(i) H₃PO₂
(ii) H₂S₂O₇ (CBSE 2012)
Answer:
(i) H₃PO₂

(ii) H₂S₂O₇

Question 22.
(a) Draw the molecular structures of following compounds:
(i) XeF₆
(ii) H₂S₂O₈
Answer:
(i)

(ii)

(b) Explain the following observations:
(i) The molecules NH₃ and NF₃ have dipole moments which are of opposite directions.
(ii) All the bonds in PCl₅ molecule are not equivalent.
(iii) Sulphur in vapour state exhibits paramagnetism.
Answer:
(i) Both NH₃ and NF₃ have pyramidal shape with one lone pair on N atom.

The lone pair on N is in opposite direction to the N-F bond moments and therefore, it has very low dipole moment (about 0.234 D). But ammonia has high dipole moment because its lone pair is in the same direction as the N-H bond moments.

(ii) PCl₅ has trigonal bipyramidal structure in which there are three P-Cl equatorial bonds and two P-Cl axial bonds. The two axial bonds are being repelled by three bond pairs at 90° while the three equatorial bonds are being repelled by two bond pairs at 90°. Therefore, axial bonds are repelled more by bond pairs than equatorial bonds and hence are larger (219 pm).

(iii) In vapour state, sulphur partly exists as S₂ molecule and S₂ molecule like O₂ has two unpaired electrons in anti-bonding π* molecular orbitals. Therefore, it is paramagnetic.

OR

(a) Complete the following chemical equations:
(i) XeF₄ + SbF₅ →
(ii) Cl₂ + F₂ (excess ) →
Answer:
(i) XeF₄ + SbF₅ → XeF₄.SbF₅ → [XeF₃]⁺ [SbF₆]^–

(ii)

(b) Explain each of the following:
(i) Nitrogen is much less reactive than phosphorus.
(ii) The stability of +5 oxidation state decreases down group 15.
(iii) The bond angles (O – N – O) are not of the same value in NO₂^– and NO₂⁺. (CBSE Delhi 2012)
Answer:
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is non-polar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) The stablity of +5 oxidation state decreases down the group because of inert pair effect. Therefore, the +5 oxidation state of Bi is less stable than that of Sb.

(iii) In NO₂, there is one electron on N while in NO₂⁺ there is no electron and in NO₂⁺, there is a lone pair of electrons on N as shown below:

NO₂⁺ molecule is linear and has bond angle of 180°. The repulsion by single electron is less as compared to repulsion by a lone pair of electrons. Therefore, bond pairs in NO₂^– are forces more closer than in NO₂.

Question 23.
(a) Draw the structure of the following molecule: H₃PO₂
Answer:
(a)

(b) Explain the following observations:
(i) Nitrogen is much less reactive than phosphorus.
(ii) Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.
(iii) Sulphur has greater tendency for catenation than oxygen in the same group.
Answer:
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is nonpolar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) Though electronegativity of F is more than O, yet water forms more extensive hydrogen bonding than HF. In H₂O, each oxygen is tetrahedrally surrounded by two covalent bonds and two hydrogen bonds, (-O….H). In HF there is only one hydrogen bond (-F….H).

(iii) Sulphur has strong tendency for catenation, i.e. forming bonds with itself in comparison to oxygen. This is because of stronger S-S single bond than O-O single bond.

OR

(a) Draw the structures of the following molecules:
(i) N₂O₅
(ii) HClO₄
Answer:

(a)
(i)

(ii)

(b) Explain the following observations:
(i) H₂S is more acidic than H₂O.
(ii) Fluorine does not exhibit any positive oxidation state.
(iii) Helium forms no real chemical compound. (CBSE 2012)
Answer:
(i) The size of S is more than that of O. Therefore, the distance between S and H, i.e. S-H bond length, is more than O-H bond length. As a result, the bond dissociation enthalpy of S-H will be less and it will be easier to break the bond in H₂S than O-H bond in water. Therefore, H₂S will be more acidic than H₂O.

(ii) Fluorine is the most electronegative element and therefore, it shows oxidation state of – 1 only. It does not show any positive oxidation state.

(iii) Helium does not form compounds because it has very high ionisation enthalpy and smallest size. Its electron gain enthalpy is also almost zero.

Question 24.
Write balanced equations for the following reactions:
(a) P₄ + NaOH + H₂O → CBSE Delhi 2009)
(b) As₄ + Cl₂(excess) →
(c) P₄O₁₀ + H₂O →
(d) Ca₃P₂ + H₂O → (CBSE Delhi 2008, 2014)
(e) POCl₃ + H₂O →
(f) HgCl₂ + PH₃ → (CBSE AI 2010)
(g) Ag + PCl₅ → (CBSE AI 2014)
Answer:
(a) P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
(b) As₄ + 10Cl₄(excess) → 4AsCl₅
(c) P₄O₁₀ + 6H₄O → 4H₃PO₄
(d) Ca₃P₂ + 6H₂O → 2PH₃ + 3 Ca(OH)₂
(e) POCl₃ + 3H₂O → H₃PO₄ + 3HCl
(f) 3HgCl₂ + 2PH₃ → Hg₃P₂ + 6HCl
(g) 2Ag + PCl₅ → 2 AgCl + PCl₃

Question 25.
Complete the following reactions
(i) NaOH + Cl₂ →
hot and conc.
(ii) NH₃ + Cl₂ (excess) → (CBSE Delhi 2017)
(iii) NaNO₂ + HCl →
(iv) F₂(g) + H₂O(l) → (CBSE Delhi 2008)
(v) K₂CO₃ + HCl →
(vi) Cl₂ + H₂O → (CBSE Delhi 2017)
(vii) F₂ + 2Cl^– → (CBSE Delhi 2017)
Answer:
(i) 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O
hot and conc.
(ii) NH₃ + 3Cl₂ (excess) → NCl₃ + 3HCl
(iii) 2NaNO₂ + 2HCl → 2NaCl + H₂O + NO₂ + NO
(iv) 2F₂(g) + 2H₂O(l) → 4H⁺(aq) + 4F^– (aq) + O₂(g)
(v) K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O
(vi) 2Cl₂ + 2H₂O → 4HCl + O₂
(vii) F₂ + 2Cl^– → 2F^– + Cl₂

Question 26.
Complete the following reactions:
(i) XeF₄ + SbF₅ → (CBSE Delhi 2012)
(ii)

(CBSE Delhi 2012, CBSE Al 2012, 2014)
(iii) XeF₂ + H₂O →
(iv) XeF₄ + H₂O →
(v) XeF₆ + H₂O → (CBSE Delhi 2012)
(vi) XeF₆ + 2H₂O → (CfiSE Delhi 2017)
(vii) XeF₆ + 3H₂O → (CBSE Delhi 2017)
Answer:
(i) XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]^–
(ii)

(iii) XeF₂ + H₂O → 2Xe + 4HF + O₂
(iv) XeF₄ + H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂
(v) XeF₆ + H₂O → XeOF₄ + 2HF
(vi) XeF₆ + 2H₂O → XeO₂F₂ + 4HF
(vii) XeF₆ + 3H₂O → XeO₃ + 6HF

Question 27.
(a) Account for the following:
(i) Ozone is thermodynamically unstable.
(ii) Solid PCl₅ is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.
Answer:
(a) (i) Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (ΔH is -ve) and increase in entropy ΔS is +ve). These two factors reinforce each other resulting in negative ΔG (ΔG = ΔH – TΔS) for its conversion to oxygen.

(ii) PCl₅ has trigonal bipyramidal structure and is not very stable. It splits up into more stable tetrahedral and octahedral structures which are stable as
PCl₅ ⇌ [PCl₄]⁺ [PCl₆]^–
Therefore, it exists as ionic.

(iii) Due to small size and high electronegativity, fluorine cannot act as central atom in higher oxoacids.

(b) Draw the structure of
(i) BrF₅
(ii) XeF₄
Answer:
(i)

Square Pyramidal

(ii)

OR

(i) Compare the oxidising action of F₂ and Cl₂ by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
Answer:
F₂ is stronger oxidising agent than Cl₂. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:

The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F₂ than for Cl₂. Hence, F₂ is stronger oxidising agent than Cl₂.

(ii) Write the conditions to maximise the yield of H₂SO₄ by contact process.
Answer:
Conditions for maximum yield of H₂SO₄ by Contact Process:
(a) Low temperature (optimum temperature 720 K)
(b) High pressure (optimum pressure 2 bar)
(c) Presence of catalyst (V₂O₅ catalyst).

(iii) Arrange the following in the increasing order of property mentioned:
(a) H₃PO₃, H₃PO₄, H₃PO₂ (Reducing character)
(b) NH₃, PH₃, ASH₃, SbH₃, BiH₃ (Base strength) (CBSE 2016)
Answer:
(a) H₃PO₂ > H₃PO₃ > H₃PO₄
(b) NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

Question 28.
(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
Answer:
(a) (i) In gaseous state, hydrogen halides are covalent. But in aqueous solution, they ionise and behave as acids. The acidic strength of these acids decreases in the order:
HI > HBr > HCl > HF
Thus, HF is the weakest acid and HI is the strongest acid among these hydrogen halides.

The above order of acidic strength is reverse of that expected on the basis of electronegativity. Fluorine is the most electronegative halogen, therefore, the electronegativity difference will be maximum in HF and should decrease gradually as we move towards iodine through chlorine and bromine.

Thus, HF should be most ionic in nature and consequently it should be strongest acid. Although many factors contribute towards the relative acidic strengths, the major factor is the bond dissociation energy. The bond dissociation energy decreases from HF to HI so that HF has maximum bond dissociation energy and HI has the lowest value.

Since H-I bond is weakest, it can be dissociated into H⁺ and I^– ions readily while HF can be dissociated with maximum difficulty. Thus, HI is the strongest acid while HF is the weakest acid among the hydrogen halides.

(ii) Oxygen molecule is held by weak van der Waals forces because of the small size and high electronegativity of oxygen. On the other hand, sulphur molecules do not exist as S₂ but form polyatomic molecules having eight atoms per molecule (S₈) linked by single bonds. Therefore, S atoms are strongly held together by intermolecular forces and its melting point is higher than that of oxygen. Hence, there is large difference in melting and boiling points of oxygen and sulphur.

(iii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation enthalpy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(b) Draw the structures of the following:
(i) CIF₃
(ii) XeF₄
Answer:
(i)

T – Shaped molecule

(ii)

Square planar molecule

OR

(i) Which allotrope of phosphorus is more reactive and why?
Answer:
White phosphorus is most reactive of all the allotropes because it is unstable due to angular strain on P₄ molecule with bond angle of 60°.

(ii) How are the supersonic jet aeroplanes responsible for the depletion of ozone layers?
Answer:
Nitrogen oxide emitted from exhausts of supersonic jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen. Since supersonic jets fly in the stratosphere near ozone layer, they are responsible for the depletion of ozone layer.

(iii) F₂ has lower bond dissociation enthalpy than Cl₂. Why?
Answer:
Because of small size of fluorineatoms, there are strong electron-electron repulsions between the lone pairs of electrons on F atoms. Hence, bond dissociation enthalpy of F₂ is lower than that of Cl₂.

(iv) Which noble gas is used in filling balloons for meteorological observations?
Answer:
Helium is used for filling balloons for meteorological observations because it is non-inflammable.

(v) Complete the equation: (CBSE 2015)
XeF₂ + PF₅ →
Answer:
XeF₂ + PF₅ → [XeF]⁺ [PF₆]^–

Question 29.
(a) Account for the following:
(i) Interhalogens are more reactive than pure halogens.
(ii) N₂ is less reactive at room temperature.
(iii) Reducing character increases from NH₃ to BiH₃.
Answer:
(i) Interhalogen compounds are more reactive than component halogens.
This is because covalent bond between dissimilar atoms in interhalogen compounds is polar and weaker than between similar atoms in halogens (except F-F). This is due to the fact that the overlapping of orbitals of dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.

(ii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(iii) The reducing character of hydrides of group 15 depends upon the stability of the hydride. On going down the group, the size of the central atom increases and therefore, its tendency to form stable covalent bond with small hydrogen atom decreases.

The greater unstability of a hydride, the greater is its reducing character. Since the stability of the group 15 hydrides decreases from NH₃ to BiH₃, hence reducing character increases.

(b) Draw the structures of the following:
(i) H₄P₂O₇ (Pyrophosphoric acid)
(ii) XeF₄
Answer:
(i)

(ii)

OR

(a) Which poisonous gas is evolved when white phosphorus is heated with conc. NaOH solution? Write the chemical equation involved.
Answer:
Phosphine, PH₃
P₄ + 3NaOH + 3H₂O → 3NaH₂PO₃ + PH₃

(b) Which noble gas has the lowest boiling point?
Answer:
Helium

(c) Fluorine is a stronger oxidising agent than chlorine. Why?
Answer:
Fluorine has lower bond dissociation enthalpy of F-F bond than Cl-Cl bond of Cl₂ and high enthalpy of hydration because of smaller size of F ion. As a result it has greater tendency to accept electron in solution and is stronger oxidising agent than chlorine.

(d) What happens when H₃PO₃ is heated?
Answer:

(e) Complete the equation:
PbS + O₃ → (CBSE 2015)
Answer:
PbS + 4O₃ → PbSO₄ + 4O₂

Question 30.
(a) Account for the following observations:
(i) SF₄ is easily hydrolysed whereas SF₆ is not easily hydrolysed.
(ii) Chlorine water is a powerful bleaching agent.
(iii) Bi(V) is a stronger oxidising agent than Sb(V).
Answer:
(a) (f) S atom in SF₄ is not sterically protected as it is surrounded by only four F atoms, so attack of water molecules can take place easily. In contrast, S atom in SF₆ is protected by six F atoms. Thus attack by water molecules cannot take place easily.

(ii) Chlorine water produces nascent oxygen (causes oxidation) which is responsible for bleaching action.
Cl₂ + H₂O → 2HCl + O

(iii) Due to inert pair effect Bi(V) can accept a pair of electrons to form more stable Bi (III) (+3 oxidation state of Bi is more stable than its +5 oxidation state).

(b) What happens when
(i) White phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂.
(ii) XeF₆ undergoes partial hydrolysis.
(Give the chemical equations involved).
Answer:
(i) Phosphorus undergoes disproportionation reaction to form phosphine gas.
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂

(ii) On partial hydrolysis, XeF6 gives oxyfluoride XeOF4 and HF.
XeF₆ + H₂O → XeOF₄ + 2HF

OR

(a) What inspired N.Bartlett for carrying out reaction between Xe and PtF₆?
Answer:
N. Bartlett first prepared a red compound O₂⁺PtF₆^–. He then realised that the first ionisation enthalpy of molecular oxygen was almost identical with Xenon. So he carried out reaction between Xe and PtF₆.

(b) Arrange the following in the order of property indicated against each set:
(i) F₂, I₂, Br₂, Cl₂ (increasing bond dissociation enthalpy)
(ii) NH₃, ASH₃, SbH₃, BiH₃, PH₃ (decreasing base strength)
Answer:
(i) I₂ < F₂ < Br₂ < Cl₂
(ii) NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

(c) Complete the following equations:
(i) Cl₂ + NaOH (cold and dilute) →
(ii) Fe³⁺ + SO₂ + H₂O →
(CBSE Sample Paper 2018)
Answer:
(i) 2NaOH + Cl₂ → NaCl + NaOCl + H₂O
(ii) 2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2- + 4H⁺

Question 31.
(a) Give reasons:
(i) H₃PO₃ undergoes disproportionation reaction but H₃PO₄ does not.
(ii) When Cl₂ reacts with excess of F₂, ClF₃ is formed and not FCl₃.
(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.
Answer:
(i) In +3 oxidation state phosphorus tends to disproportionate to higher and lower oxidation states / Oxidation state of P in H₃PO₃ is +3 so it undergoes disproportionation but in H₃PO₄ it is +5 which is the highest oxidation state.

(ii) F cannot show positive oxidation state as it has highest electronegativity/ Because Fluorine cannot expand its covalency / As Fluorine is a small sized atom, it cannot pack three large sized Cl atoms around it.

(iii) Oxygen has multiple bonding due to pπ-pπ bonding whereas sulphur shows catenation. Oxygen is diatomic therefore held by weak intermolecular force while sulphur is polyatomic and held by strong intermolecular forces.

(b) Draw the structures of the following:
(i) XeF₄
(ii) HClO₃
Answer:
(i)

(ii)

OR

(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas Intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).
(i) Identify (A) and (B).
(ii) Write the structures of (A) and (B).
(iii) Why does gas (A) change to solid on cooling?
Answer:
(i) A = NO₂, B = N₂O₄
(ii)

(iii) NO₂ contains an odd electron. So it dimerises to give N₂O₄.

(b) Arrange the following In the decreasing order of their reducing character:
HF, HCl, HBr, HI
Answer:
HI > HBr > HCl > HF

Complete the following reaction:
XeF₄ + SbF₅ →, (CBSE 2018)
Answer:
(C) XeF₄ + SbF₅ → [XeF₃] [SbF₆]

Question 32.
(i) What happens when
(a) chlorine gas reacts with cold and dilute solution of NaOH?
(b) XeF₂ undergoes hydrolysis?
Answer:
(a) 2NaOH + Cl₂ → NaCl + NaOCl + H₂O
(cold and dilute)
(b) 2XeF₂(s) + 2H₂O (l) → 2Xe (g) + 4HF(aq) + O₂(S)

(ii) Assign suitable reasons for the following:
(a) SF₆ Is inert towards hydrolysis.
(b) H₃PO₃ is diprotic.
(C) Out of noble gases only Xenon is known to form established chemical compounds.
Answer:
(a) Sulphur is sterically protected by six F atoms, hence does not allow the water molecules to attack.

(b) It contains only two ionisable H-atoms which are present as -OH groups, thus behaves as dibasic acid.

(c) Xe has least ionisation energy among the noble gases and hence it forms chemical compounds particularly with O₂ and F₂.

OR

(i) Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and Cl₂.
Answer:
F₂ is stronger oxidising agent than Cl₂. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:

The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F₂ than for Cl₂.
Hence, F₂ is stronger oxidising agent than Cl₂.

(ii) Complete the following reactions :
(a) Cu + HNO₃(dilute) →
(b) Fe³⁺ + SO₂ + H₂O →
(c) XeF₄ + O₂F₂ → (CBSE 2018)
Answer:
(a) 3Cu + 8 HNO₃ (dilute) → 3Cu(NO₃)₂ + 2NO + 4H₃O
(b) 2Fe³⁺ + SO₃ + 2H₃O → 2 Fe²⁺ + SO₄^2- + 4H⁺
(c) XeF₄ + O₂F₂ → XeF₆ + O₂

Question 33.
A crystalline solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified KMnO₄ (aq.) solution and reduces Fe³⁺ to Fe²⁺. Identify ‘A’ and ‘B’ and write the reactions involved. (CBSE 2019C)
Answer:
A = S₈ / Sulphur
S₈ + 8 O₂ → 8SO₂ / S + O₂ → SO₂
B = SO₂

Decolourises KMnO₄
2KMnO₄ + 5 SO₂ + 2H₄O → 2H₂SO₄ + 2MnSO₄ + K₂SO₄ / 2MnO₄^– + 5SO₂ + 2H₂O → 4H⁺ + 2Mn²⁺ + 5SO₄^2-
Reduces Fe³⁺ to Fe²⁺
2Fe³⁺ + SO₂ + 2 H₂O → 2 Fe²⁺ + SO₄^2- + 4H⁺

OR

Answer the following:
(a) Arrange the following hydrides of Group 16 elements in the decreasing order of their acidic strength:
H₂O, H₂S, H₂Se, H₂Te
Answer:
H₂Te > H₂Se > H₂S > H₂O

(b) Which one of PCl₄⁺ and PCl₄^– is not likely to exist and why?
Answer:
PCl^4- is not likely to exist because lone pair on P in PCl₃ can be donated to Cl⁺ and not to Cl^–. Phosphorus has 10e^– which cannot be accommodated in sp³ orbitals.

(c) Which aliotrope of sulphur is thermally stable at room temperature?
Answer:
Rhombic sulphur is thermally stable at room temperature. Its melting point is 385.8 K. All other varieties of sulphur change into this form on standing. It has low thermal and electrical conductivity.

(d) Write the formula of a compound of phosphorus which is obtained when cone. HNO₃ oxidises P₄.
Answer:
HNO₃ oxidises phosphorus to phosphoric acid. H₃PO₄ is formed
P₄ + 2OHNO₃ → 4H₃PO₄ + 2ONO₂ + 4H₂O

(e) Why does PCl₃ fume in moisture?
Answer:
PCl₃ fumes in moist air and reacts with water violently to form phosphorus acid. It hydrolyses in the presence of moisture to give fumes of HCl.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 8 The d-and f-Block Elements. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 8 Important Extra Questions The d-and f-Block Elements

The d-and f-Block Elements Important Extra Questions Very Short Answer Type

Question 1.
How would you account for the increasing oxidising power in the series:
V0₂⁺ < Cr₂0₇ ^2-< Mn0₄^–? (CBSE Sample Paper 2010)
Answer:
This is due to the increasing stability of the lower species to which they are reduced.

Question 2.
Which metal in the first transition series exhibits a +1 oxidation state most frequently and why? (CBSE Delhi 2013)
Answer:
Copper has electronic configuration 3cf104s1. It can easily lose one (4s1) electron to give stable 3d10 configuration.

Question 3.
The magnetic moments of a few transition metal ions are given below:

Metal ion	Magnetic moment (BM)
Sc3+	0.00
Cr2+	4.90
Ni2+	2.84
Ti3+	1.73

(atomic no. Sc = 21, Ti = 22, Cr = 24, Ni = 28)
Which of the given metal ions:
(i) has the maximum number of unpaired electrons?
Answer:
Cr²⁺

(ii) forms colourless aqueous solution?
Answer:
Sc³⁺

(iii) exhibits the most stable +3 oxidation state? (CBSE Sample Paper 2017-18)
Answer:
Sc³⁺

Question 4.
Based on the data, arrange Fe²⁺, Mn²⁺ and Cr²⁺ in the increasing order of stability of +2 oxidation state:
E°_{Cr³⁺/Cr²⁺} = – 0.4 V, E°_{Mn³⁺/Mn²⁺} = 1.5 V, E°_{Fe³⁺/Fe²⁺} = 0.8 V (CBSE Sample Paper 2011)
Answer:
As the value of reduction potential increases, the stability of +2 oxidation state increases. Therefore, correct order of stability is Cr³⁺ | Cr²⁺ < Fe³⁺ | Fe²⁺ < Mn³⁺ | Mn²⁺.

Question 5.
(i) Name the element showing the maximum number of oxidation states among the first series of transition metals from Sc (Z = 21) to Zn (Z = 30).
Answer:
Manganese

(ii) Name the element which shows only +3 oxidation state. (CBSEAI2013)
Answer:
Scandium

Question 6.
Identify the oxoanion of chromium which is stable in an acidic medium. (CBSE Sample Paper 2017-18)
Answer:
Cr₂0₇^2-

Question 7.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (CBSE Delhi 2015)
Answer:
This is because of relatively poor shielding by 5f electrons in actinoids in comparison with shielding of 4f electrons in lanthanoids.

Question 8.
Name an important alloy that contains some of the Ianthanoid metals. (CBSE AI 2013)
Answer:
Misch metal

Question 9.
Identify the Ianthanoid element that exhibits a +4 oxidation state. (CBSE Sample Paper 2017-18)
Answer:
Cerium

Question 10.
Write the formula of an oxo-anion of chromium (Cr) in which it shows the oxidation state equal to its group number. (CBSE Delhi 2017)
Answer:
Cr₂0₇^2-

The d-and f-Block Elements Important Extra Questions Short Answer Type

Question 1.
Assign reasons for the following:
(i) Cu(I) Is not known In an aqueous solution.
Answer:
Because of the Lesser hydration enthalpy of Cu(I), It Is unstable In an aqueous solution and therefore, It undergoes disproportionation.

(ii) Actinolds exhibit a greater range of oxidation states than lanthanoids. (CBSE AI 2011)
Answer:
Lanthanoids show Limited number of oxidation states, such as + 2, + 3 and + 4 + 3 is the principal oxidation state). This is because of the large energy gap between 5d and 4f subshells. On the other hand, actinoids also show a principal oxidation state of + 3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of + 3, + 4, + 5, + 6 and + 7 and neptunium (Z = 94) shows oxidation states of + 3, +4, + 5, + 6 and + 7. This is because of the small energy difference between 5f and 6d orbitals.

Question 2.
Give reasons:
(a) MnO is basic whereas Mn₂O₇    is acidic in nature.
Answer:
When a metal is in a high oxidation state, its oxide Is acidic and when a metaL is in a low oxidation state its oxide is basic.
The oxides of manganese have the following behavior:


(b) Transition metals form alloys. (CBSE 2019C)
Answer:
The transition metals are quite similar in size and, therefore, the atoms of one metal can substitute the atoms of other metal in its crystal lattice. Thus, on cooling a mixture solution of two or more transition metals, solid alloys are formed which is shown in Fig.

Question 3.
Explain why the E^θ value for the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺. (CBSE AI 2008, 2010, 2018)
Answer:
Mn²⁺ has 3d5 electronic configuration. It is stable because of the half-filled configuration of the d-subshell. Therefore, Mn has a very high third ionization enthalpy for the change from d5 to d4 and it is responsible for a much more positive E^θ value for Mn³⁺/Mn²⁺ couple in comparison to Cr³⁺/Cr²⁺ and Fe³⁺/ Fe²⁺ couples.

Question 4.
Complete the following chemical equations:
(i) Cr₂O₇²⁺+ H⁺ + I^– →
Answer:
Cr₂O₇²⁺ + 6I^– + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O

(ii) MnO₄^– + NO₂^– + H⁺ → (CBSE Delhi 2012)
Answer:
2MnO₄^– + 5N0₂^– + 6H+ → 2Mn2+ + 5NO₃^–+ 3H₂O

Question 5.
(i) Which metal in the first transition series (3d series) exhibits +1 oxidatIon state most frequently and why?
Answer:
Copper, because it has [Ar] 3d¹⁰4s¹ electronic configuration. After losing one electron it gets completely filled electronic configuration (3d¹⁰).

(ii) Which of the following cations are colored in aqueous solutions and why?
Sc³⁺, V³⁺, Ti⁴⁺, Mn²⁺
(At. nos. Sc = 21, V = 23, Ti = 22, Mn = 25) (CBSE Delhi 2013)
Answer:
Sc³⁺: [Ar]  V³⁺: [Ar] 3d²
Ti⁴⁺: [Ar]  Mn²⁺: [Ar] 3d⁵
V³⁺ and Mn²⁺ cations are colored because they contain partially filled d-orbitals.

Question 6.
Why do transition elements exhibit higher enthalpies of atomization? (CBSE Delhi 2008, 2012)
Answer:
The high enthalpies of atomization are due to a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interactions and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.

Question 7.
(a) Actinoid contraction is greater than lanthanoid contraction. Give reason,
Answer:
This is due to poorer shielding by 5f electrons in actinoids as compared to shielding by 4f electrons in lanthanoids. In the Ianthanoid series, as we move from one element to another, the nuclear charge increases by one unit, and one electron are added. The new electrons are added to the same inner 4f-subshells. However, the 4f-electrons shield each other from the nuclear charge quite poorly because of the very diffused shapes of the f-orbitals.

The nuclear charge, however, increases by one at each step. Hence, with increasing atomic number and nuclear charge, the effective nuclear charge experienced by each 4f-electron increases. As a result, the whole of the 4f-electron shell contracts at each successive element, though the decrease is very small.

The actinoid contraction is due to the imperfect shielding of one 5felectron by another in the same subshell. Therefore, as we move along the series, the nuclear charge and the number of 5f-electrons increase by one unit at each step. However, due to imperfect shielding of 5f orbitals, the effective nuclear charge increases which results in contraction of the size.

(b) Out of Fe and Cu, which has a higher melting point and why? (CBSE 2019C)
Answer:
Fe has a higher melting point. The metallic bond is formed due to the interaction of electrons in the outermost orbitals. The strength of bonding is related to the number of unpaired electrons. Fe has more unpaired electrons leading to stronger metallic bonding. So, it has a higher melting point.

Question 8.
How would you account for the following:
(i) Cr²⁺ is reducing in nature while with the same d-orbital configuration (d⁴), Mn³⁺ is an oxidizing agent.
Answer:
Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g level. On the other hand, the change from Mn³⁺ to Mn²⁺ results in the half-filled (d⁵) configuration which has extra stability.

(ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series. (CBSE 2011)
Answer:
This is because, in the middle of the transition series, the maximum number of electrons are available for sharing with others. The small number of oxidation states at the extreme left side is due to the lesser number of electrons to lose or share. On the other hand, at the extreme right-hand side, due to a large number of electrons, only a few orbitals are available in which the electrons can share with others for higher valence.

Question 9.
When MnO₂ is fused with KOH in the presence of KNO₃ as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionates in an acidic solution to give a purple compound (B). An alkaline solution of compound (B) oxidizes Kl to compound (C), whereas an acidified solution of compound (B) oxidizes Kl to (D). Identify (A), (B), (C), and (D). (CBSE Delhi 2019)
Answer:
When MnO₂ is fused with KOH in the presence of KNO₃, as an oxidizing agent, it gives a dark green compound (A).

Question 10.
State reasons for the following:
(i) Cu(I) is not stable in an aqueous solution.
Answer:
Cu⁺ ion is not stable in an aqueous solution because of its less negative enthalpy of hydration than that of Cu²⁺ ion.

(ii) Unlike Cr³⁺, Mn²⁺, Fe³⁺ and the subsequent other M²⁺ ions of the 3d series of elements, the Ad and the 5d series metals generally do not form stable atomic species. (CBSE 2011)
Answer:
Because of lanthanoid contraction, the expected increase in size does not occur.

Question 11.
Assign reasons for each of the following:
(i) Transition metals generally form colored compounds.
Answer:
Most of the transition metal ions are colored both in the solid-state and in aqueous solutions. The color of these ions is attributed to the presence of an incomplete (n – 1) d-subshell. The electrons in these metal ions can be easily promoted from one energy level to another in the same d-subshell. The amount of energy required to excite the electrons to higher energy states within the same d-subshell corresponds to the energy of certain colors of visible light. Therefore, when white light falls on a transition metal compound, some of its energy corresponding to a certain color is absorbed causing the promotion of d-electrons. This is known as d-d transitions. The remaining colors of white light are transmitted and the compound appears colored.

(ii) Manganese exhibits the highest oxidation state of +7 among the 3d-series of transition elements. (CBSE Delhi 2011)
Answer:
Manganese has the electronic configuration [Ar] 3d⁵ 4s². It can lose seven electrons due to the participation of 3d and 4s electrons and therefore, exhibits the highest oxidation state of +7 in its compounds.

Question 12.
What are the transition elements? Write two characteristics of the transition elements. (CBSE Delhi 2015)
Answer:
Transition elements are those elements that have incompletely filled (partly filled) d-subshell in their ground state or in any one of their oxidation states.

Characteristics:

1. Transition elements show variable oxidation states.
2. They exhibit catalytic properties.

Question 13.
Though both Cr²⁺ and Mn³⁺ have d4 configurations, yet Cr²⁺ is reducing while Mn³⁺ is oxidizing. Explain why? (CBSE AI 2008, 2012, CBSE Delhi 2012)
Answer:
E° value for Cr³⁺/Cr²⁺ is negative (-0.41V), this means that Cr³⁺ ions are more stable than Cr²⁺. Therefore, Cr²⁺ can readily lose electrons to undergo oxidation to form Cr³⁺ ion, and hence Cr (II) is strongly reducing. On the other hand, the E° value for Mn³⁺ Mn²⁺ is positive (+1.57V), this means that Mn³⁺ ions can be readily reduced to Mn²⁺ and hence Mn (III) is strongly oxidizing.

Question 14.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state? (CBSE Delhi 2012)
Answer:
The electronic configuration of Mn²⁺ is [Ar] 3d⁵ which is half-filled and hence it is stable. Therefore, the third ionization enthalpy is very high, i.e. the third electron1 cannot be easily removed. In the case of Fe²⁺, the electronic configuration is 3d⁶. Therefore, Fe²⁺ can easily lose one electron to acquire a 3d⁵ stable electronic configuration.

Question 15.
The E°_M2+|M for copper is positive (0.34 V). Copper is the only metal in the first series of transition elements showing this behavior. Why? (CBSE Sample Paper 2012, CBSE AI2012)
Answer:
The E°_M2+|M value for copper is positive and this shows that it is the least reactive metal among the elements of the first transition series. This is because copper has a high enthalpy of atomization and enthalpy of ionization. Therefore, the high energy required to convert Cu(s) to Cu²⁺ (aq) is not balanced by its hydration enthalpy.

Question 16.
Chromium is a typical hard metal while mercury is a liquid. (CBSE Sample Paper 2011)
Answer:
In chromium, the M-M interactions are strong due to the presence of six unpaired electrons in the 3d and 4s subshell. On the other hand, in mercury, all the electrons in the 5d and 6s subshell are paired and therefore, the M-M interactions are weak. Therefore, chromium is a typical hard metal while mercury is a liquid.

Question 17.
Silver is a transition metal but zinc is not. (CBSE Sample Paper 2011)
Answer:
According to the definition, transition elements are those which have partially filled d-subshell in their elementary state or in one of the oxidation states. Silver (Z = 47) can exhibit a +2 oxidation state in which it has an incompletely filled d-subshell (4d9 configuration). Hence, silver is regarded as a transition element.

On the other hand, zinc (Z = 30) has the configuration 3d¹⁰ 4s². It does not have partially filled d-subshells in its elementary form or in a commonly occurring oxidation state (Zn²⁺: 3d¹⁰). Therefore, it is not regarded as a transition element.

Question 18.
Name the ox metal anions of the first series of the transition metals in which the metal exhibits an oxidation state equal to its group number. (CBSE Delhi 2017)
Answer:
MnO₄^–: Oxidation state of Mn = +7 (equal to its group number)
CrO₄^2-: Oxidation state of Cr = +6 (equal to its group number)

Question 19.
Give reasons:
(a) Of the d⁴ species, Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidising.
Answer:
Because Cr is more stable in the +3 oxidation state due to the t_2g³ configuration whereas Mn is more stable in the +2 oxidation state due to the half-filled 3d⁵ configuration.

E° value for Cr³⁺/Cr²⁺ is negative (-0.41 V); this means that Cr³⁺ ions are more stable than Cr²⁺. Therefore, Cr²⁺ can readily lose electrons to undergo oxidation to form a Cr³⁺ ion, and hence Cr (II) is strongly reducing. On the other hand, the E° value for Mn³⁺/ Mn²⁺ is positive (+1.57V), which means that Mn³⁺ ions can be readily reduced to Mn²⁺ and hence Mn (III) is strongly oxidizing.

(b) The d1 configuration is very unstable in ions. (CBSE 2019C)
Answer:
After the loss of ns electrons, d1 electron can easily be lost to give a stable configuration. Therefore, the elements having d1 configuration are either reducing or undergo disproportionation.

Question 20.
When chromite ore FeCr₂O₄ is fused, with NaOH in presence of air, a yellow-colored compound (A) is obtained, which on acidification with dilute sulphuric acid gives a compound (B). Compound (B) on reaction with KCI forms an orange colored crystalline compound (C).
(i) Write the formulae of the compounds (A), (B), and (C).
(ii) Write one use of the compound (C). (CBSE Delhi 2016)
OR
Complete the following chemical equations:
(i) 8MnO₄^–+ 3S₂O₃^2-+ H₂O →
(ii) Cr₂O₇^2-+ 3Sn²⁺ + 14H⁺ →
Answer:


(ii) Potassium dichromate (C) is used as a powerful oxidizing agent in redox titrations in the laboratories.
OR
(i) 8MnO₄^–+ 3S₂O₃^2-+ H₂O → 8MnO₂ + 6SO₄^2- + 2OH^–
(ii) Cr₂O₇ ²⁺ + 3Sn²⁺ +14H⁺ → 2Cr3+ + 3Sn⁴⁺ + 7H₂O

Question 21.
Why do the transition elements have higher enthalpies of atomization? In 3d series (Sc to Zn), which element has the lowest enthalpy of atomization and why? (CBSE 2015)
Answer:
The transition elements have high enthalpies of atomization because they have a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interactions and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.

Zinc has the lowest enthalpy of atomization because it has no unpaired electrons and hence weak metallic bonding.

The d-and f-Block Elements Important Extra Questions Long Answer Type

Question 1.
(i) Transition metals and their compounds are generally found to be good catalysts.
(ii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series. (CBSE AI 2011)
Answer:
(i) Some transition metals and their compounds act as good catalysts for various reactions. This is due to their ability to show multiple oxidation states. The common examples are Fe, Co, Ni, V, Cr, Mn, Pt, etc.

The transition metals form reaction intermediates with the substrate by using empty d-orbitals. These intermediates give reaction paths of lower activation energy and therefore, increase the rate of reaction. For example, during the conversion of SO₂ to SO₃, V₂O₅ is used as a catalyst. Solid V₂O; absorbs a molecule of SO₂ on the surface forming V₂O₄ and the oxygen is given to SO₂ to form S0₃. The divanadium tetraoxide is then converted to V₂O₅ by reaction with oxygen:

(ii) The second and third transition series metals show a great tendency to form strong M-M bonds than the first transition series elements. For example, rhenium easily forms ReCl₈^4- which contains a strong Re-Re bond. It does not have a manganese analog.

Similarly, both Niobium and tantalum form M₆X₁₂ⁿ⁺ species and has no vanadium analogue.

Question 2.
Explain the following observations giving an appropriate reason for each.
(i) The enthalpies of atomization of transition elements are quite high.
Answer:
The high enthalpies of atomization are due to a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interaction and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.

(ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition metals (i.e. 3rd series)
Answer:
The heavier transition metals of second and third transition series metals show a great tendency to form strong M-M bonds than first transition series elements. For example, rhenium easily forms ReCl₈^4- which contains a strong Re-Re bond. It does not have a manganese analog. Similarly, both niobium and tantalum form M₆X₁₂ⁿ⁺ species and has no vanadium analog.

(iii) Mn²⁺ is much more resistant than Fe²⁺ towards oxidation. (CBSE Delhi 2012)
Answer:
Mn²⁺ is quite stable because it has a stable half-filled d⁵ electronic configuration and therefore, cannot be easily oxidized. On the other hand, Fe²⁺ has a less stable d⁶ electronic configuration and can be easily oxidized to a stable d⁵ configuration.

Question 3.
How would you account for the following?
(i) Transition metals exhibit variable oxidation states.
(ii) Zr (Z = 40) and Hf (Z = 72) have almost identical radii.
(iii) Transition metals and their compounds act as a catalyst.
OR
Complete the following chemical equations:
(i) Cr₂0₇^2- + 6Fe²⁺ + 14H⁺ →
(ii) 2Cr0₄^2- + 2H⁺ →
(iii) 2Mn0₄^– + 5C₂0₄^2- + 16H⁺ → (CBSE Delhi 2013)
Answer:
(i) The transition elements exhibit variable oxidation states. The variable oxidation states of transition metals are due to the participation of ns and (n – 1) d-electrons. This is because of the very small difference between the energies of (n -1) d and ns orbitals. For the first five elements, the minimum oxidation state is equal to the number of electrons in the 4s orbitals and the other oxidation states are equal to the sum of 4s and some of the 3d-electrons. The highest oxidation state is equal to the sum of 4s and 3d electrons. For the remaining elements, the minimum oxidation state is equal to electrons in 4s-orbitals and the maximum oxidation state is not equal to the sum of 4s and 3d electrons.

In general, the oxidation state increases up to the middle and then decreases.

(ii) Due to lanthanoid contraction, the increase in radii from the second to third transition series vanishes. Therefore, Zr and Hf have almost the same radii.

(iii) Catalytic properties of transition metal ions. Some transition metals and their compounds act as good catalysts for various reactions. The common examples are Fe, Co, Ni, V, Cr, Mn, Pt, etc. For example, iron-molybdenum is used as a catalyst in Haber’s process for the manufacture of NH₃. V₂0₅ is used for the oxidation of S0₂ to S0₃ in the Contact process for the manufacture of H₂S0₄.

The transition metals form reaction intermediates with the substrate by using empty d-orbitals. These intermediates give reaction paths of lower activation energy and therefore, increase the rate of reaction. For example, during the conversion of SO₂ to SO₃, V₂O₅ is used as a catalyst. Solid V₂O₅ absorbs a molecule of SO₂ on the surface forming V₂O₄ and the oxygen is given to SO₂ to form SO₃. The divanadium tetraoxide is then converted to V₂O₅ by reaction with oxygen:

OR
(i) 6Fe²⁺ + Cr₂0₇^2- + 14H+ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O
(ii) 2CrO₄^2- + 2H+ → Cr₂O₇^2-+ H₂O
(iii) 2MnO₄ ^–+ 5C₂O₄^2– + 16H+ → 2Mn²⁺ + 8H₂O + 10CO₂

Question 4.
Give reasons for the following:
(i) Transition elements and their compounds act as catalysts.
Answer:
Transition metals and their compounds act as good catalysts because of their ability to show multiple oxidation states and form complexes, for example, Fe, Co, Ni, V, Cr, V₂O₅, etc.

(ii) E° value for (Mn²⁺|Mn) is negative whereas for (Cu²⁺| Cu) is positive.
Answer:
The negative value of E° (Mn²⁺| Mn) is due to the stability of the half-filled d-subshell in Mn²⁺(3d⁵). The positive E°(Cu²⁺|Cu) is due to the high ionization enthalpy and high enthalpy of atomization of copper. Therefore, the high energy required to convert Cu(s) to Cu²⁺(aq) is not balanced by its hydration enthalpy.

(iii) Actinoids show irregularities in their electronic configuration. (CBSE Delhi 2019)
Answer:
Actinoids have irregularities in the electronic configuration because of almost equal energy of 5f, 6d and 7s orbitals. Therefore, there are some irregularities in the filling of 5f, 6d, and 7s orbitals. The electron may enter either of these orbitals.

Question 5.
(a) Complete the following chemical reactions:
(i) Na₂Cr₂O₇ + KCl →
(ii) 2MnO₄^–+ 5 S0₃^2- + 6 H⁺ →
Answer:
(i) Na₂Cr₂O₇ + KCl → K2Cr2O7 + 2 NaCl
(ii) 5 S0₃^2- + 2MnO₄^– + 6 H⁺ → 2Mn²⁺ + 3H₂O + 5SO₄^2-

(b) How does the colour of Cr₂O₇^2- change when treated with an alkali? (CBSE2019C)
Answer:
The orange color of Cr₂O₇^2- change to yellow due to the formation of chromate ion.
Cr₂O₇^2- + 2OH^– → 2Cr₂O₇^2-  + H₂O (Yellow)

Question 6.
(a) Write chemical equations involved in the preparation of KMn04 from Mn02.
Answer:
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
3MnO₄^2-+ 4H⁺ → 2MnO^–+ MnO₂ + 2H₂O (or any other correct equation of preparation)

(b) Actinoids show wide range of oxidation states. Why? (CBSE 2019C)
Answer:
Lanthanoids show limited number of oxidation states, such as + 2, + 3 and + 4 (+ 3 is the principal oxidation state). This is because of large energy gap between 5d and 4f subshells. On the other hand, actinoids also show principal oxidation state of + 3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of + 3, + 4, + 5 and + 6 and neptunium (Z = 93) shows oxidation states of + 3, + 4, + 5, + 6 and + 7.

This is because of the small energy difference between 5f, 6d, and 7s orbitals.

Question 7.
How would you account for the following?
(i) Many of the transition elements are known to form interstitial compounds.
Answer:
Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard.

These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.

(ii) The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group members of the second (4d) series.
Answer:
The metallic radii of the third (5d) series of transition metals are virtually the same as those of corresponding group members of the (4d) series due to the phenomenon of lanthanoid contraction. The steady decrease in atomic and ionic sizes of lanthanoid elements with increasing atomic numbers is known as lanthanoid contraction.

(iii) Lanthanoids form primarily +3 ions, white the actinoids usually have higher oxidation states in their compounds, +4 or even +6 being typical. (CBSE Delhi 2012)
Answer:
Lanthanoids form primarily +3 ions, while actinoids usually have higher oxidation states, +4 or even +6 in their compounds because in actinoids the 5f, 6d, and 7s energy levels are of comparable energies.

Therefore, all these three subshells can participate.

Question 8.
How would you account for the following:
(i) Among lanthanoids, Ln (III) compounds are predominant. However, occasionally in solutions or in solid compounds, +2 and +4 ions are also obtained.
Answer:
All lanthanoids exhibit a common stable oxidation state of +3. In addition, some lanthanoids show +2 and +4 oxidation states also in solution or in solid compounds. These are shown by these elements which by doing so attain the stable f (empty f-subshell), f (half-filled f-subshell), and f4 (filled f-subshell) configurations. For example,
(a) Ce and Tb exhibit +4 oxidation states. Cerium (Ce) and terbium (Tb) attain f and f configurations respectively when they get +4 oxidation state, as shown below:
Ce4⁺: [Xe] 4f°
Tb4⁺: [Xe] 4 f°

(b) Eu and Yb exhibit + 2 oxidation states.
Europium and ytterbium get f and f⁴ configurations in +2 oxidation state as shown below:
EU²⁺ : [Xe] 4 f⁷
Yb²⁺: [Xe] 4f¹⁴

(ii) The E°M²⁺/M for copper is positive (0.34 V). Copper is the only metal in the first series of transition elements showing this behavior.
Answer:
The E°(M²⁺/M) value for copper is positive and this shows that it is the least reactive metal out of the first transition series. This is because copper has a high enthalpy of atomization and enthalpy of ionization. Therefore, the high energy required to convert Cu(s) to Cu²⁺(aq) is not balanced by its hydration enthalpy.

(iii) The metallic radii of the third (5d) series of transition metals are nearly the same as those of the corresponding members of the second series. (CBSE 2012)
Answer:
The metallic radii of the third transition series elements are virtually the same as those of corresponding members of the second transition series because of the lanthanoid contraction. Due to the presence of lanthanoids between the second and third transition series, the expected increase in radii vanishes. Consequently, the pairs of elements Zr-Hf, Nb-Ta, Mo-W, etc have almost similar sizes.

Question 9.
Explain the following observations:
(i) Many of the transition elements are known to form interstitial compounds.
Answer:
Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon, and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard.

These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.

(ii) There is a general increase in density from titanium (Z = 22) to copper (Z = 29).
Answer:
There is a gradual increase in density from Ti to Cu because as we move along a transition series from left to right, the atomic radii decrease due to an increase in effective nuclear charge. Therefore, the atomic volume decreases but at the same time atomic mass increases. Therefore, density (mass/volume) increases.

(iii) The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members of the lanthanoid series. (CBSE 2012)
Answer:
Lanthanoids show limited a number of oxidation states, such as + 2, + 3, and + 4 (+ 3 is the principal oxidation state). This is because of the large energy gap between 5d and 4/ subshells. On the other hand, actinoids also show a principal oxidation state of + 3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of + 3, + 4, + 5 and + 6 and neptunium (Z = 93) shows oxidation states of + 3, + 4, + 5, + 6 and + 7. This is because of the small energy difference between 5f, 6d, and 7s orbitals.

Question 10.
Following are the transition metal ions of 3d series.
Ti⁴⁺, V²⁺, Mn³⁺, Cr³⁺
(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following:
(i) Which ion is most stable in an aqueous solution and why?
(ii) Which ion is a strong oxidizing agent and why?
(iii) Which ion is colorless and why? (CBSE AI 2017)
Answer:
(i) Cr³⁺ because of the half-filled t_2g³ configuration.
(ii) Mn³⁺ due to stable d⁵ configuration of Mn²⁺
(iii) Ti⁴⁺ because it has no unpaired electrons.

Question 11.
Write chemical equations for the following reactions:
(i) Oxidation of nitrite ion by MnO₄^– in acidic medium.
(ii) Acidification of potassium chromate solution.
(iii) Disproportionation of manganese
(vi) in acidic solution. (CBSE Sample Paper 2011)
Answer:
(i) 5NO₂– + 2MnO₄^– + 6H+ → 2Mn²⁺ + 3H2O + 5NO3^–
(ii) 2K₂CrO4₄ + 2H+ → K₂Cr₂O₇ + 2K+ + H₂O
(iii) 3MnO₄2- + 4H+ → 2MnO₄^– + MnO₂ + 2H₂O

Question 12.
The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series. (CBSE AI 2012, CBSE Delhi 2012)
Answer:
Lanthanoids show a limited number of oxidation states, such as +2, +3, and +4 (+3 is the principal oxidation state). This is because of the large energy gap between 5d and 4f subshells. On the other hand, actinoids also show a principal oxidation state of +3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of +3, +4, +5 and +6 and neptunium (Z = 93) and plutonium (Z = 94) show oxidation states of +3, +4, +5, +6 and +7. This is because of the small energy difference between 5f, 6d, and 7s orbitals.

Question 13.
Explain the following:
(i) Out of Sc³⁺, Co²⁺, and Cr³⁺ ions, only Sc³⁺ is colorless in aqueous solutions.
(Atomic no.: Co = 27; Sc = 21 and Cr = 24)
(iI) The E°Cu²⁺/Cu for copper metal is positive (+0.34), unlike the remaining members of the first transition series
(iiI) La(OH)₃ is more basic than Lu(OH)₃. (CBSE Sample paper 2018)
Answer:
(i) Co²⁺: [Ar]3d⁷, Sc³⁺: [Ar]3d°
Cr³⁺: [Ar]3d³
Co²⁺ and Cr³⁺ have unpaired electrons. Thus, they are colored in an aqueous solution. Sc³⁺ has no unpaired electron. Thus it is colorless.
(ii) Metal copper has a high enthalpy of atomization and enthalpy of ionization. Therefore the high energy required to convert Cu(s) to Cu²⁺(aq) is not balanced by its hydration enthalpy.
(iii) Due to lanthanoid contraction the size of lanthanoid ion decreases regularly with the increase in atomic size. Thus covalent character between lanthanoid ion and OH’ increases from La³⁺ to Lu³⁺. Thus the basic character of hydroxides decreases from La(OH)₃ to Lu(OH)₃

Question 14.
(a) Complete the following chemical reactions:
(i) 2MnO₄^–+ 5 NO₂^–+ 6 H⁺ →
(ii) 3Mn0₄^2- + 4 H⁺ →
Answer:
(i) 5NO₂– + 2MnO4^– + 6H+ → 2Mn²⁺ + 5NO₃^– + 3H₂O
(ii) 3MnO₄^2-+ 4H+ → 2MnO₄^– + MnO₂ + 2H₂O

(b) Name a member of the lanthanoid series which shows a +4 oxidation state. (CBSE 2019C)
Answer:
Cerium / Ce

Question 15.
Give reasons:
(i) E° value for Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺.
Answer:
The comparatively high value for Mn shows that Mn²⁺(d⁵) is particularly stable/Much larger third ionization energy of Mn (where the required change is from d⁵ to d⁴).

(ii) Iron has a higher enthalpy of atomization than copper.
Answer:
Due to the higher number of unpaired electrons.

(iii) Sc³⁺ is colorless in an aqueous solution whereas Ti³⁺ is colored. (CBSE 2018)
Answer:
The absence of unpaired d-electron in Sc³⁺ whereas in Ti³⁺ there is one unpaired electron or Ti³⁺ shows the d-d transition.

Question 16.
(i) Complete the following chemical equations for reactions:
(a) MnO₄^– (aq) + S₂O₃2- (aq) + H₂O (I) →
(b) Cr₂O₇^– (aq) + H₂S(g) + H⁺(aq) →
Answer:
(a) 8MnO₄^– (aq) + 3S₂O₃2-  (aq) + H₂O (l) → 8MnO₂ (s) + 6SO₄2- (aq) + 2OH- (aq)
(b) Cr₂O₇^2- (aq) + 3H₂S(g) + 8H⁺ (aq) → 2Cr³⁺(aq) + 3S(s) + 7H₂O

(ii) Give an explanation for each of the following observations:
(a) The gradual decrease in size (actinoid contraction) from element to element is greater among the actinoids than that among the lanthanoids (lanthanoid contraction).
Answer:
(a) This is because of relatively poor shielding by 5f electrons in actinoids in comparison with shielding of 4f electrons in lanthanoids.

(b) The greatest number of oxidation states are exhibited by the members in the middle of a transition series.
Answer:
This is because, in the middle of the transition series, the maximum number of electrons are available for sharing with others. The small number of oxidation states at the extreme left side is due to a lesser number of electrons to lose or share. On the other hand, at the extreme right-hand side, due to a large number of electrons, only a few orbitals are available in which the electrons can share with others for higher valence.

(c) With the same d-orbital configuration (d⁴) Cr²⁺ ion is a reducing agent but the Mn³⁺ ion is an oxidizing agent. (CBSE Delhi 2009)
Answer:
Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g level. On the other hand, the change from Mn³⁺ to Mn²⁺ results in the half-filled (d⁵) configuration which has extra stability.

Question 17.
(i) Complete the following chemical reaction equations:
(a) Fe²⁺ (aq) + MnO^–4 (aq) + H⁺ (aq) →
(b) Cr₂O₇^2- (aq) + I^– (aq) + H⁺ (aq) →
Answer:
(a) 5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq) → Mn²⁺ (aq) + 5Fe³⁺ (aq) + 4H₂O(l)
(b) Cr₂O₇^2- (aq) + 6I^– (aq) + 14H+ (aq) → 2Cr³⁺(aq) + 3I₂ (s) + 7H₂O(l)

(ii) Explain the following observations:
(a) Transition elements are known to form many interstitial compounds.
Answer:
Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon, and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard.

These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.

(b) With the same d⁴ d-orbital configuration Cr²⁺ ion is reducing while the Mn³⁺ ion is oxidizing.
Answer:
Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g level. On the other hand, the change from Mn³⁺ to Mn²⁺ results in the half-filled (d⁵) configuration which has extra stability.

(c) The enthalpies of atomization of the transition elements are quite high. (CBSE Delhi 2009)
Answer:
The transition metals have strong interactions of electrons in the outermost orbitals and therefore, have high melting and boiling points. These suggest that the atoms of these elements are held together by strong forces and have high enthalpy of atomization.

Question 18.
(i) Complete the following chemical equations:
(a) MnO₄^– (aq) + S₂O₃^2- (aq) + H₂O(l) →
(b) Cr₂O₇^2- (aq) + Fe²⁺ (aq) + H⁺ (aq) →
Answer:
(a) 8MnO₄^– + 3S₂O₃^2- + H₂O → 8MnO₂ + 6SO₄^2-+ 2OH-
(b) Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ +6Fe³⁺ + 7H₂O

(ii) Explain the following observations:
(a) La3+ (Z = 57) and Lu3+ (Z = 71) do not show any colour in solutions.
Answer:
In La3+ there are no f-electrons and in Lu³⁺ the 4f sub-shell is complete (4/14). Therefore, there are no unpaired electrons and consequently, d-d transitions are not possible. Hence, La³⁺ and LU³⁺ do not show any color in solutions.

(b) Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism.
Answer:
Among the divalent cations in the first
transition series, Mn (Z = 25) exhibits the maximum paramagnetism because it has the maximum number of unpaired electrons (3d⁵): five unpaired electrons.

(c) Cu⁺ ion is not known in aqueous solutions. (CBSE 2010)
Answer:
In an aqueous solution, Cu⁺ undergoes disproportionation changing to Cu²⁺ ion.
2Cu⁺ → Cu²⁺ + Cu

Question 19.
(i) Give reasons for the following:
(a) Mn³⁺ is a good oxidizing agent.
(b) E°M²⁺/M values are not regular for first-row transition metals (3d series).
(c) Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF4, whereas the highest oxide is Mn207.
(ii) Complete the following equations:
(a) 2CrO₄^2- + 2H⁺ →
(b) KMnO₄
OR
(i) (a) Why do transition elements show variable oxidation states?
(b) Name the element showing a maximum number of oxidation states among the first series of transition metals from Sc(Z = 21) to Zn (Z = 30).
(c) Name the element which shows only the +3 oxidation state.
(ii) What is lanthanoid contraction? Name an important alloy that contains some of the lanthanoid metals. (CBSE2013)
Answer:
(i) (a) Mn³⁺ (3d⁴) on changing to Mn²⁺ (3d⁵) becomes stable, half-filled configuration has extra stability. Therefore, Mn³⁺ can be easily reduced and acts as a good oxidizing agent.

(b) E°(M²⁺/M) values are not regular in the first transition series metals because of irregular variation of ionization enthalpies (IE₁ + IE₂) and the sublimation energies.

(c) Among transition elements, the bonds formed in +2 and +3 oxidation states are mostly ionic. The compounds formed in higher oxidation states are generally formed by sharing of d-electrons. Therefore, Mn can form MnO4- which has multiple bonds also, while fluorine cannot form multiple bonds.

(ii) (a) 2CrO₄^2- + 2H⁺ → Cr₂O₇^2- + H₂O
(b) 2KMnO₄ K₂MnO₄ + MnO₂ + O₂
OR
(i) (a) The transition elements show variable oxidation states because their atoms can lose a different number of electrons. This is due to the participation of inner (n -1). d-electrons in addition to outer ns electrons because the energies of the ns and (n – 1) d-subshells are almost equal.

(b) Manganese

(c) Scandium

(ii) The steady decrease in atomic and ionic sizes of lanthanide elements with increasing atomic numbers is called lanthanide contraction. In the lanthanoids, there is a regular decrease in the size of atoms and ions with an increase in atomic number. For example, the ionic radii decrease from Ce³⁺ (111 pm) to Lu³⁺ (93 pm).

Cause of lanthanoid contraction. As we move through the lanthanoid series, 4f-electrons are being added, one at each step. The mutual shielding effect of electrons is very little, even smaller than that of d-electrons. This is due to the shape of f-orbitals. The nuclear charge, however, increases by one at each step. Hence, the inward pull experienced by the 4f-electrons increases. This causes a reduction in the size of the entire 4f shell. The sum of the successive reductions gives the total lanthanoid contraction. The important alloy is mischmetal.

Question 20.
(i) How do you prepare:
(a) K₂MnO₄ from MnO₂?
(b) Na₂Cr₂O₇ from Na₂CrO₄?
(ii) Account for the following:
(a) Mn²⁺ is more stable than Fe²⁺ towards oxidation to +3 state.
(b) The enthalpy of atomization is lowest for Zn in the 3d series of the transition elements.
(c) Actinoid elements show a wide range of oxidation states.
OR
(a) Name the element of 3d transition series which shows a maximum number of oxidation states. Why does it show so?
(b) Which transition metal of 3d series has a positive E°(M²⁺/M) value and why?
(c) Out of Cr³⁺ and Mn³⁺, which is a stronger oxidizing agent and why?
(d) Name a member of the lanthanoid series which is well known to exhibit a +2 oxidation state.
(e) Complete the following equation:
Mn0₄^– + 8H⁺ +5e^– → (CBSE Delhi 2014)
Answer:
(i) (a) When Mn0₂ is fused with caustic potash (KOH) or potassium carbonate in the presence of air, a green mass of potassium manganate is formed.
2MnO₂ + 4KOH + O₂ 4 2K₂MnO₄ + 2H₂O
Or
2MnO₂ + 2K₂CO₂ + O₂ → 2K₂MnO₄ + 2CO₂.

(b) Na2Cr04 is acidified with dilute sulphuric acid to get Na₂Cr₂O₇.
2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O

(ii) (a) Electronic configuration of Mn²⁺ is [Ar]Bd⁵ which is half-filled and hence is stable. Therefore, the third ionization enthalpy is very high, i.e. the third electron cannot be easily removed. In the case of Fe²⁺, the electronic configuration is 3d⁶. Therefore, Fe²⁺ can easily lose one electron to acquire a 3d⁵ stable configuration. Thus, Mn²⁺ is more stable than Fe²⁺ towards oxidation to + 3 states.

(b) The high enthalpies of atomization of transition elements are due to the participation of electrons (n – 1) d-orbitals in addition to ns electrons in the interatomic metallic bonding. In the case of zinc, no electrons from 3d-orbitals are involved in the formation of metallic bonds. On the other hand, in all other metals of 3d series electrons from d-orbitals are always involved in the formation of metallic bonds.

(c) Lanthanoids show a limited number of oxidation states, such as +2, +3, and +4 (+3 is the principal oxidation state). This is because of the large energy gap between 5d and 4/ subshells. On the other hand, actinoids also show a principal oxidation state of +3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of +3, +4, +5 and +6 and neptunium (Z = 94) shows oxidation states of +3, +4, +5, +6 and +7. This is because of the small energy difference between 5f and 6d orbitals.
OR
(a) Manganese (Z = 25) shows the largest number of oxidation states because it has the maximum number of unpaired electrons. It shows oxidation states from +2 to +7, i.e. +2, +3, +4, +5, +6 and +7.

(b) The E°(M²⁺|M) value for copper is positive and this shows that it is the least reactive metal among the elements of the first transition series. This is because copper has a high enthalpy of atomization and enthalpy of ionization. Therefore, the high energy required to convert Cu(s) to Cu²⁺(aq) is not balanced by its hydration enthalpy.

(c) Mn³⁺ is a stronger oxidizing agent than Cr³⁺ because Mn³⁺(3d⁴) changes to stable half-filled (3d⁵) configuration while in the case of Cr³⁺ (3d⁵) is stable (half-filled) and it cannot be readily changed to Cr²⁺ (3d⁴).

(d) Europium.

(e) MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O.

Question 21.
(a) Account for the following:
(i) Manganese shows the maximum number of oxidation states in 3d series.
(ii) E° value for Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺.
(iii) Ti⁴⁺ is colorless whereas V⁴⁺ is colored in an aqueous solution.
(b) Write the chemical equations for the preparation of KMn04 from Mn02. Why does the purple color of acidified permanganate solution decolorize when it oxidizes Fe²⁺ to Fe³⁺?
OR
(a) Write one difference between transition elements and p-block elements with reference to variability of oxidation states.
(b) Why do transition metals exhibit higher enthalpies of atomization?
(c) Name an element of the lanthanoid series which is well known to show a +4 oxidation state. Is it a strong oxidizing agent or a reducing agent?
(d) What is lanthanoid contraction? Write its one consequence.
(e) Write the ionic equation showing the oxidation of Fe(ll) salt by acidified dichromate solution. (CBSE Al 2019)
Answer:
(a) (i) Manganese has the electronic configuration: 3d⁵ 4s². It has a maximum number of unpaired electrons and hence shows maximum oxidation states.
(ii) Mn²⁺ has 3d⁵ electronic configuration. It is stable because of the half-filled configuration. Therefore, Mn3 easily gets reduced to Mn²⁺. Thus, E°(M³⁺/Mn²⁺) is positive. On the other hand, Cr³⁺ is more stable in the +3 oxidation state due to stable t2g3 configuration.
(iii) Ti⁴⁺ (3d°) does not have any d-electrons. Therefore, there are no d-d transitions whereas V⁴⁺, (3d¹) has one electron in d-subshell and d-d transitions are possible and hence it is colored.

(b) 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
3K₂MnO₄ + 4HCL → 2KMnO + MnO₂ + 2H₂O

When acidified KMnO₄ oxidizes Fe2+ to Fe3+ its purple color gets decolorized due to the formation of Mn2+ from
MnO₄– ion.
MnO₄– + 5Fe²⁺ + 8W → Mn²⁺ + 5Fe³⁺ + 4H₂O
Or
(a) Transition elements, in general, show variable oxidation states which differ by 1 unit, whereas p-block elements show variable oxidation states which differ by 2 units.

(b) Transition eLements have unpaired d-electron and therefore have strong metallic bonding between atoms. Hence, they have high enthalpies of atomization.

(c) Cerium. It is a strong oxidizing agent.

(d) The regular decrease In atomic and ionic radii of the Lanthanides with increasing atomic number is catted Lanthanoid contraction.

Consequence: 5d series elements have almost the same size as the 4d series.

(e) 6Fe²⁺ + Cr₂O₇^2- + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

Question 22.
(i) Account for the following:
(a) Mn shows the highest oxidation state of +7 with oxygen but with fluorine, it shows the highest oxidation state of +4.
(b) Zirconium and Hafnium exhibit similar properties.
(c) Transition metals act as catalysts.
(ii) Complete the following equations:
(a) 2MnO₂ + 4KOH + O₂
(b) Cr₂O₇^2- + 14H⁺ + 6I^–
Or
The elements of 3d transition series are given as:
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Answer the following:
(a) Write the element which is not regarded as a transition element. Give reason.
(b) Which element has the highest m.p.?
(c) Write the element which can show an oxidation state of +1.
(d) Which element is a strong oxidizing agent in the +3 oxidation state and why? (CBSE 2016)
Answer:
(i) (a) Manganese shows the highest oxidation state of +7 with oxygen but +4 with fluorine. This is because oxygen has a tendency to form multiple bonds and hence stabilize the high oxidation state.
(b) Due to lanthanoid contraction, Zr and Hf show similar properties.
(c) The transition metals act as catalysts. This is due to their ability to show multiple oxidation states.

(ii) (a) 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
(b) Cr₂O₇^2- + 14H+ + 6I^– → 2Cr³⁺ + 7H₂O + 3I₂
Or
(a) Zn because of not having partially filled d-orbitals in its ground state or ionic state.
(b) Chromium, Cr
(c) Copper, Cu
(d) Mn, because Mn²⁺ has extra stability due to half-filled d-subshell.
Mn: [Ar] 3d⁵ 4s²; Mn²⁺: [Ar]3d⁵
Mn³⁺ has four electrons (3d⁴) in 3d subshell and requires only one electron to get a half-filled 3d subshell and therefore, acts as a strong oxidizing agent.

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 10 Haloalkanes and Haloarenes.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 10 Important Extra Questions Haloalkanes and Haloarenes

Haloalkanes and Haloarenes Important Extra Questions Very Short Answer Type

Question 1.
Define ambident nucleophile with an example. (CBSE Delhi 2019)
Answer:
The nucleophile which has two sites through which it can attack is called ambident nucleophile, for example, nitrite ion (NO₂^–). It can attack either through 0 (-O^– -N = O) or through N

Question 2.
What happens when bromine attacks CH₂ = CH – CH₂ -C ≡ CH? (CBSE AI2012)
Answer:
The Colour of bromine gets discharged.

Question 3.
What happens when CH₃ – Br is treated with KCN? (CBSE Delhi 2013)
Answer:
Ethane nitrile is formed.
CH₃Br + KCN → CH₃C ≡ N + KBr (Ethane nitrile)

Question 4.
Write the IUPAC name of

Answer:
4-Bromo-4-methylpent-2-ene

Question 5.
What happens when ethyl chloride is treated with aqueous KOH? (CBSE Delhi 2013)
Answer:
Ethyl alcohol is formed.
CH₃CH₂Cl + KOH(aq) → CH₃CH₂OH + KCl Ethyl alcohol

Question 6.
Write the IUPAC name of (CH₃)₂CH.CH(Cl)CH₃. (CBSE Delhi 2013)
Answer:
2-Chloro-3-methyl butane

Question 7.
Which compound in the following pair undergoes a faster S_N1 reaction? (CBSE Delhi 2013)

Answer:

Question 8.
Write the IUPAC name of the following compound: (CBSE AI 2013)

Answer:
3-Chloro-2, 2-dimethylbutane

Question 9.
Which aerosol depletes the ozone layer? (CBSE AI 2013)
Answer:
Chlorofluoro compounds of methane (CCl₂F₂) called freons.

Question 10.
Write the IUPAC name of the following compound: (CBSE AI 2013)

Answer:
2-Bromo-4-chloropentane

Question 11.
Why is CH₂ = CH – CH₂ – Cl more easily hydrolysed than CH₃ – CH₂ – CH₂ – Cl? (CBSE AI 2019)
Answer:
CH₂ = CH – CH₂ – Cl gives allylic carbocation as intermediate, which is resonance stabilised.

Therefore, CH₂ = CH – CH₂ – Cl is more readily hydrolysed.

Question 12.
Write the IUPAC name of the following compound:
CH₂ = CHCH₂Br (CBSE 2011)
Answer:
3-Bromoprop-1-ene

Question 13.
Write the IUPAC name of the following compound: (CH₂ )₃CCH₂Br (CBSE 2011)
Answer:
1 -Bromo 2,2-dimethylpropane

Question 14.
Write the IUPAC name of the following: (CBSE 2012)

Answer:

3-Bromo- 2-methylpropene

Question 15.
Identify the chiral molecule in the following pair: (CBSE 2014)

Answer:

2-Chlorobutane is a chiral molecule.

Question 16.
Which would undergo S_N2 reaction faster in the following pair and why? (CBSE Delhi 2015)

Answer:
CH₃CH₂Br would undergo S_N2 reaction faster because of less steric hindrance.

Question 17.
Out and which is more reactive towards S_N1 reaction and why? (CBSE Delhi 2016)
Answer:
is more reactive.
The S_N1 reaction proceeds through the formation of a carbocation. The compound which forms a more stable carbocation will be more reactive.


Since 2° carbocation is more stable than 1° carbocation, 2-Chlorobutane will be more reactive towards the S_N1 reaction.

Question 18.
Write the structure of the product formed when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether. (CBSE Al 2018)
Answer:

Question 19.
Which of the following two reactions is S_N2 and why? (CBSE AI2016)

Answer:
Reaction (i) is S_N2 because it proceeds by inversion of configuration.

Question 20.
Why is chloroform kept in dark coloured bottles? (CBSE AI 2019)
Answer:
Chloroform reacts with atmospheric oxygen in presence of light to produce an extremely poisonous gas called phosgene.

Chloroform is stored in dark coloured bottles so that it is not exposed to light and the formation of phosgene is prevented.

Question 21.
Write one stereochemical difference between S_N1 and S_N2 reactions. (CBSE Delhi 2019)
Answer:
S_N1 reaction forms a racemic mixture while S_N2 reaction results in inversion of configuration.

Question 22.
A solution of KOH hydrolyses CH₃CHClCH₂CH₃ and CH₃CH₂CH₂CH₂Cl. Which one of these is more easily hydrolysed? (CBSE Delhi 2010)
Answer:
CH₃CHClCH₂CH₃

Question 23.
Write the structure of 2, 4-dinitrochlorobenzene. (CBSE Delhi 2017)
Answer:

Question 24.
Write the structure of 1-Bromo-4- chlorobut-2-ene. (CBSE Delhi 2017)
Answer:

Question 25.
Write the structure of 3-Bromo-2- methylprop-1 -ene. (CBSE Delhi 2017)
Answer:

Question 26.
Write IUPAC name of the given compound: (CBSE Delhi 2019)

Answer:
4-Chlorobenzene sulphonic acid

Question 27.
Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why? (CBSE Sample paper 2018)
Answer:
Benzyl chloride;
Due to resonance, stable benzyl carbocation is formed.

Haloalkanes and Haloarenes Important Extra Questions Short Answer Type

Question 1.
Which one of the following compounds will undergo hydrolysis at a faster rate by S_N1 mechanism? Justify. (CBSE Sample Paper 2019)

Answer:
The following compound will undergo S_N1 faster:

Greater the stability of the carbocation, greater will be its ease of formation from the corresponding halide and faster will be the rate of reaction. The benzylic carbocation formed gets stabilised through resonance.

CH₃CH₂CH₂Cl forms 1° carbocation, which is less stable than benzylic carbocation.

Question 2.
How would you differentiate between S_N1 and S_N2 mechanisms of substitution reactions? Give one example of each. (CBSE 2010)
Answer:
The S_N1 reaction occurs in two steps and the reaction is of the first order.
The S_N2 reaction occurs in one step and the reaction is of second order. These can also be distinguished as:

In S_N1 reaction retention of the configuration takes place while in S_N2 reaction inversion of the configuration takes place.

S_N2 reaction:

S_N1 reaction:

Question 3.
(i) State one use each of DDT and iodoform.
Answer:
DDT is used as an insecticide for killing mosquitoes and other insects. chloroform is used as an antiseptic.

(ii) Which compound in the following couples will react faster in S_N2 displacement and why?
(a) I -Bromopentane or 2-bromopentane
Answer:
1-Bromopentane (1°) will react faster than 2-bromopentane (2°) by S_N2 displacement because It is Less stericaLly hindered in the transition state.

(b) I-Bromo-2-methyl butane or 2-Bromo-2-methyl butane (CBSE Delhi 2010)
Answer:
1 -Bromo-2-methyl butane (1°) reacts faster than 2-bron,o-2-methyl butane (2°) because being primary, it will have Less steric hindrance.

Question 4.
Although chlorine Is an electron-withdrawing group, yet It Is ortho-, para-directing in electrophilic aromatic substitution reactions. Explain why Is It so? (CBSE 2012)
Answer:
Chlorine is an electron-withdrawing group, yet it is ortho-, para-directing in eLectrophiLic aromatic substitution reactions due to eLectron reLeasing resonance effect (+ R-effect).

As can be seen from the above resonating structures that -Ct increases electron density at ortho- and para-positions due to the + R-effect. Therefore, the attack of electrophile takes pLace at ortho- and paro- positions and not at meta- positions.

Question 5.
(i) Which alkyl halide from the following pair Is chiral and undergoes a faster S_N2 reaction? (CBSE Delhi 2014)

Answer:
is chiral

(ii) Out of S_N1 and S_N2 which reaction occurs with
(a) Inversion of configuration
Answer:
S_N2

(b) Racemisation
Answer:
S_N1

Question 6.
Draw the structure of major moon halo product In each of the following reactions: (CBSE Delhi 2014)

Answer:

Answer:

Question 7.
Propose the mechanism of the reaction taking place when
(i) (-) – 2 – Bromooctane reacts with sodium hydroxide to form (+) – octane-2-ol.
Answer:

(ii) 2-Bromopentane is heated with (ale.) KOH to form alkenes. (CBSE Sample Paper 2011)
Answer:

Question 8.
Out of and which is more reactive towards S_N1 reaction
Answer:
is more reactive.
The S_N1 reaction proceeds through the formation of a carbocation. The compound which forms a more stable carbocation will be more reactive.

Since 2° carbocation is more stable than 1° carbocation, 2-Chlorobutane will be more reactive towards the S_N1 reaction.

Haloalkanes and Haloarenes Important Extra Questions Long Answer Type

Question 1.
(i) Out of (CH₃)₃C-Br, and (CH₃)₃C-I, which one is more reactive towards S_N1 and why?
Answer:
(CH₃)₂ C-I, because it can readily form stable carbocation because the C – l bond is weaker than the C – Br bond.

(ii) Write the product formed when p-nitro chlorobenzene is heated with aqueous NaOH at 443 K followed by acidification.
Answer:

(iii) Why are dextro- and laevorotatory isomers of Butan-2-ol difficult to separate by fractional distillation? (CBSE Delhi 2019)
Answer:
Dextro- and laevorotatory isomers of butan-2-ol are enantiomers of each other and both have the same boiling point. Hence, these cannot be separated by fractional distillation.

Question 2.
Give reasons for the following:
(i) Ethyl iodide undergoes S_N2 reaction faster than ethyl bromide.
Answer:
Ethyl iodide undergoes S_N2 reaction faster than ethyl bromide because I- ion is a better leaving group than Br- ion.

(ii) (±) 2 – Butanol is optically inactive.
Answer:
(±) 2 – Butanol represents a racemic mixture of (+) 2-butanol and (-) 2-butanol which rotate the plane polarized light in different directions but to an equal extent. Therefore, the (±) compound is optically inactive.

(iii) C—X bond length in halobenzene is smaller than C—X bond length in CH₃ — X. (CBSE 2013)
Answer:
In halobenzene, there is delocalisation of electrons due to resonance. For example, chlorobenzene is considered to be a resonance hybrid of the following structures:

It is evident that the contribution of structures III, IV and V imparts a partial double bond character to the carbon-chlorine bond. Therefore, the C-X bond length in halobenzene is less than the C-X bond length in CH₃-X, which has only a single C-X bond.

Question 3.
(i) Draw the structures of major moon halo products in each of the following reactions:

Answer:

Answer:

(ii) Which halogen compound in each of the following pairs will react faster in SN2 reaction:
(a) CH₃Br or CH₃I
Answer:
CH₃I

(b) (CH₃)₃C – Cl or CH₃ – Cl (CBSE 2014)
Answer:
CH₃Cl

Question 4.
How would you convert the following:
(i) Prop-1-ene to 1-fluoropropane
(ii) Chlorobenzene to 2-chlorotoluene
(iii) Ethanol to propanenitrile
OR
Write the main products when
(i) n-butyl chloride is treated with alcoholic KOH.
(ii) 2, 4, 6-trinitrochlorobenzene is subjected to hydrolysis.
(iii) methyl chloride is treated with AgCN.
Answer:

OR

Question 5.
Give reasons:
(i) C – Cl bond length in chlorobenzene is shorter than C — Cl bond length in CH₃ — Cl.
Answer:
In chlorobenzene, there is delocalisation of electrons due to resonance.
It is considered to be a resonance hybrid of the following structures.

The contribution of structure III, IV and V imparts a partial double bond character to the C – Cl bond.

This is confirmed by X-ray analysis which shows that the C – Cl bond length in chlorobenzene is 1.69 whereas the C – Cl bond length in ethyl chloride is 1.82 A. There is no resonance in CH₃Cl and hence no partial double bond character.

(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
Answer:
In chlorobenzene, the C of C-Cl bond is sp²-hybridised while the C of C-Cl bond in cyclohexyl chloride is sp³-hybridised.

Therefore, the sp²-hybridised C of chlorobenzene has more s-character and hence more electronegative than the sp³-hybrid C of cyclohexyl chloride. As a result, the sp2 hybrid of the C-Cl bond in chlorobenzene has less tendency to release electrons to Cl than the sp³ hybrid carbon of cyclohexyl chloride. As a result, the C-Cl bond in chlorobenzene is less polar than in cyclohexyl chloride. Thus, chlorobenzene is less polar than cyclohexyl chloride.

(iii) S_N1 reactions are accompanied by racemisation in optically active alkyl halides. (CBSE 2016)
Answer:
The S_N1 reactions proceed through the formation of carbocations. In the case of optically active alkyl halides, the product formed is a racemic mixture. This is because the intermediate carbocation is planar species, therefore, the attack of the nucleophile, OFT ion can take place from both the faces (front and rear) with equal ease. As a result, a 50: 50 mixture of the two enantiomers (laevo and dextro) is formed. Thus, the product formed is a racemic mixture (±) which is optically inactive.

For example, hydrolysis of optically active 2-bromobutane results in the formation of a racemic mixture (±)-butan-2-ol.

Question 6.
The following compounds are given to you:
2-Bromopentane, 2-Bromo-2-methyl butane, 1-Bromopentane
(i) Write the compound which is most reactive towards the S_N2 reaction.
Answer:
1-Bromopentane

(ii) Write the compound which is optically active.
Answer:
2-Bromopentane

(iii) Write the compound which is most reactive towards the S_N2 elimination reaction.
Answer:
2-Bromo-2-methyl butane

Question 7.
Explain why
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. (CBSE Delhi 2016)
Answer:
In chlorobenzene, the C of C-Cl bond is sp²-hybridised while the C of C-Cl bond in cyclohexyl chloride is sp3-hybridised.

Therefore, the sp²-hybridised C of chlorobenzene has more s-character and hence more electronegative than the sp³-hybrid C of cyclohexyl chloride. As a result, the sp² hybrid of the C-Cl bond in chlorobenzene has less tendency to release electrons to Cl than the sp³ hybrid carbon of cyclohexyl chloride. As a result, the C-Cl bond in chlorobenzene is less polar than in cyclohexyl chloride.

Thus, chlorobenzene is less polar than cyclohexyl chloride. In other words, the magnitude of negative charge (δ-) is less on the Cl atom of chlorobenzene than in cyclohexyl chloride. Further, due to the delocalisation of the lone pair of electrons of the Cl atom over the benzene ring due to resonance, the C-Cl bond in chlorobenzene acquires some double-bond character. On the other hand, the C-Cl bond in cyclohexyl chloride is a pure single bond.

Since dipole moment is a product of charge and distance, therefore, chlorobenzene has a lower dipole moment than cyclohexyl chloride due to the lower magnitude of charge (δ-) on Cl atom and small C-Cl distance.

(ii) alkyl halides, though polar, are immiscible with water,
Answer:
Alkyl halides are polar molecules and therefore, their molecules are held together by dipole-dipole forces. On the other hand, the molecules of H₂0 are held together by hydrogen bonds. When alkyl halides are added to water, the new forces of attraction between water and alkyl halide molecules are weaker than the forces of attraction already existing between alkyl halide-alkyl halide molecules and water-water molecules. Hence, alkyl halides are immiscible (not soluble) in water.

(iii) Grignard reagents should be prepared under anhydrous conditions? (CBSE Sample Paper 2011)
Answer:
Grignard reagents are very reactive. They react with the moisture present in the apparatus or the starting materials (RX or Mg).
RMgX + HOH → R-H + Mg(OH)X
Therefore, Grignard reagents must be prepared in anhydrous conditions.

Question 8.
Which one of the following compounds will undergo a faster hydrolysis reaction by the S_N1 mechanism? Justify your answer.

OR
A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in propane. Write the structures of the isomers possible. Give the IUPAC name of the isomer which can exhibit enantiomerism. (CBSE Sample paper 2018-19)
Answer:
C₆ H₅CH₂Cl will undergo S_N1 reaction faster.

The carbocation formed by C₆ H₅CH₂Cl gets stabilised through resonance.
Greater the stability of carbocation, greater will be its ease of formation from the respective halide.

OR

The following isomer will exhibit enantiomerism:

IUPAC name: 1,2-Dichloropropane.

Question 9.
(i) Identify the chiral molecule in the following pair:
and
Answer:

(ii) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.
Answer:

(iii) Write the structure of the alkene formed by dehydrohalogenation of 1-Bromo-1- methylcyclohexane with alcoholic KOH. (CBSE 2018)
Answer:

Question 10.
(a) Out of and which one is more reactive towards S_N2 reaction and why?
Answer:
because it is a primary halide and has less steric hindrance and therefore, undergoes S_N2 action faster.

(b) Out of and which one is more reactive towards nucleophilic substitution reaction and why?
Answer:
is more reactive because of the presence of electron-withdrawing – NO₂ group.

(c) Out of and which one is optically active and why? (CBSE AI 2019)
Answer:
is optically active because of the presence of chiral carbon. OH

Question 11.
Write the products of the following reactions:
(i) CH₃ CH₂ -CH = CH₂ + HCl →
Answer:

(ii) Me₂CHCH₂OH
Answer:
Me₂CHCH₂OH MeCHCH₂Cl

Answer:

(iv) Me₂C = CMe₂
Answer:

(v) (CBSE AI 2016)
Answer:

Question 12.
(i) In the following pairs of the halogen compounds, which would undergo S_N2 faster?
Answer:
S_N2 reaction proceeds through the formation of a transition state involving the bonding of carbon to five atoms or groups. The reactivity is decided by the stability of the transition state on the basis of steric hindrance. The reactivity follows the order: CH₃ > 1° > 2° > 3° halide.
Therefore,

Answer:
is primary alkyl halide and hence undergoes S_N2 reaction faster.

Answer:
I will undergo S_N2 reaction faster because iodine is a better leaving group due to its large size and hence it will be released at a faster rate in the presence of incoming nucleophile.

Answer:
(1 -Bromo-2, 2-dimethylpentane) reacts faster because it is 1 ° alkyl halide.

(CBSE AI 2009)
Answer:
(2-Methyl-1-bromopropane) reacts faster because it has lesser steric hindrance in the transition state.

Answer:
CH₃ CH₂ Br would undergo S_N2 reaction faster because of less steric hindrance.

(ii) Which one of the following pairs undergoes S_N1 substitution reaction faster and why?
Answer:
S_N1 reaction occurs through the formation of a carbocation. Therefore, the greater the stability of the carbocation, the faster is the rate of an S_N1 reaction. Since the stability of carbocation follows the order 3° > 2° > 1° > CH₃, therefore,

Answer:
Out of or reacts faster because it invoLves the formation of 3° -carbocation while involves the formation of V -carbocation

(CBSE AI 2009)
Answer:
Out of or reacts faster because it forms 2° carbocation while involves 1° carbocation.

(CBSE AI 2015)
Answer:
undergoes SN1 reaction faster. This is because the carbocation derived from CH₃(CH₃)₃ C Br is 3° carbocation and is more stable than carbocation (1°) obtained from CH₃ CH₂ Br.

Question 13.
(i) How would you convert the following:
(a) Prop-1-ene to 1-fluoropropane
Answer:

(b) Chlorobenzene to 2-chlorotoluene
Answer:

(ii) Write the main products when
(a) n-butyl chloride is treated with alcoholic KOH.
Answer:

(b) 2, 4, 6-trinitrochlorobenzene is subjected to hydrolysis.
Answer:

(c) methyl chloride is treated with AgCN. (CBSE AI2015)
Answer:

Question 14.
Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on the concentration of compound and KOH both.
(i) Write down the structural formula of both compounds ‘A’ and ‘B’.
Answer:
The molecular formulae of isomers of C4H9Br are CH3

Since the rate of reaction of the compound ‘A’ (C4H9Br) with aqueous KOH depends upon the concentration of compound ‘A’ only, therefore, the reaction occurs by SN1 mechanism and compound ‘A’ is tertiary bromide, i.e. 2-Bromo-2-methylpropane.
(CH₃)₃CBr + KOH(aq) → (CH₃ )₃COH + KBr rate = k[(CH₃)₃CBr]

(ii) Out of these two compounds, which one will be converted to the product with an inverted configuration. (NCERT Exemplar)
Answer:
Since compound ‘B’ is optically active and is an isomer of compound ‘A’ (C₄H₉Br), therefore, compound ‘B’ must be 2-bromobutane. Since the rate of reaction of compound ‘B’ with aqueous KOH depends upon the concentration of compound ‘B’ and KOH, therefore, the reaction occurs by S_N2 mechanism and the product of hydrolysis will have inverted configuration.

Compound ‘B’ will be converted with an inverted configuration.

Question 15.
(i) Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho, para directing or meta directing.

Answer:
It is clear from resonating structures that there is an increase in electron density at ortho and para positions. Therefore, the functional group ‘X’ is ortho and para directing.

(ii) Classify the following compounds as primary, secondary and tertiary halides.
(a) 1-Bromobut-2-ene
Answer:
Primary

(b) 4-Bromopent-2-ene
Answer:
secondary

(c) 2-Bromo-2-methylpropane (NCERT Exemplar)
Answer:
tertiary

Question 16.
tert-Butylbromide reacts with aq. NaOH by S_N1 mechanism while n-butyl bromide reacts by an S_N2 mechanism. Why?
Answer:
In general, the SN1 reaction proceeds through the formation of a carbocation. The tert-butyl bromide readily loses Br^– ion to form stable 3° carbocation.
Therefore, it reacts with aqueous KOH by S_N1 mechanism as:

On the other hand, n-butyl bromide does not undergo ionisation to form n-butyl carbocation (1°) because it is not stable. Therefore, it prefers to undergo reaction by an S_N2 mechanism, which occurs in one step through a transition state involving the nucleophilic attack of OH^– ion from the backside with simultaneous expulsion of Br^– ion from the front side.

S_N1 mechanism follows the reactivity order as 3° > 2° > 1° while the S_N2 mechanism follows the reactivity order as 1° >2° >3°
Therefore, tert-butyl bromide (3°) reacts by SN1 mechanism while n-butyl bromide (1°) reacts by the S_N2 mechanism.

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 14 Biomolecules. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 14 Important Extra Questions Biomolecules

Biomolecules Important Extra Questions Very Short Answer Type

Question 1.
Write the structure of the product obtained when glucose is oxidised with nitric acid. (CBSE 2012)
Answer:
Saccharic acid (or glucaric acid) is formed.

Question 2.
Write a reaction that shows that all the carbon atoms in glucose are linked in a straight chain. (CBSE 2012)
Answer:
When glucose is heated with HI and red P at 100°C for a long period, it gives n-hexane.

The formation of n-hexane suggests that all six carbon atoms in glucose are arranged in a straight chain.

Question 3.
What are the hydrolysis products of sucrose? (CBSE 2019C)
Answer:
Glucose and fructose

Question 4.
Write the name of the linkage joining two amino acids. (CBSE 2013)
Answer:
Peptide linkage

Question 5.
What are the hydrolysis products of lactose? (CBSE2019C)
Answer:
Lactose on hydrolysis gives β – D – galactose and β – D – glucose.

Question 6.
What are three types of RNA molecules which perform different functions? (CBSE Delhi 2013)
Answer:
(i) messenger RNA: m – RNA
(ii) ribosomal RNA: r – RNA
(iii) transfer RNA: t – RNA

Question 7.
What type of bonding helps in stabilising the a-helix structure of proteins? (CBSE Delhi 2013)
Answer:
Hydrogen bonding between -NH and -C = 0 groups of peptide bonds.

Question 8.
What is a glycosidic linkage? (CBSE Delhi 2013)
Answer:
Two monosaccharide units are linked to each other by a bond called glycosidic linkage.

Question 9.
Which of the two components of starch is water-soluble? (CBSE Delhi 2014)
Answer:
Amylose

Question 10.
Which component of starch is a branched polymer of a-glucose and insoluble in water? (CBSE Delhi 2014)
Answer:
Amylopectin

Question 11.
What are the products of hydrolysis of maltose? (CBSE 2014)
Answer:
Glucose

Question 12.
What is the basic structural difference between glucose and fructose?
OR
Write the products obtained after hydrolysis of lactose. (CBSE Delhi 2019)
Answer:
Glucose contains an aldehyde group (-CHO) and is aldose, which is present at the end of the carbon chain, i.e. C₁.

Fructose contains a keto group (>C = 0) and is called ketose, which is present at a carbon atom next to the terminal, i.e. C₂.
OR
Hydrolysis of lactose gives β – D – glucose and β – D – galactose.

Question 13.
What is the difference between a glycosidic linkage and a peptide linkage?
OR
What is the difference between Nucleotide and Nucleoside? (CBSE AI 2019)
Answer:
Glycosidic linkage is the linkage that joins two monosaccharides through an oxygen atom (-0-). Peptide linkage is the linkage that joins two amino acids through -CONH- bond.
OR
The molecules in which one of the nitrogen bases (purines or pyrimidine) is bonded with a sugar molecule (ribose or deoxyribose) is a nucleoside. The nucleoside linked with the phosphate group is called a nucleotide.

Question 14.
What is meant by the inversion of sugar? (CBSE Sample Paper 2011)
Answer:
The change of specific rotation of sugar from dextrorotatory to laevorotatory is called inversion of sugar.

Question 15.
Glucose does not give 2, 4-DNP test and Schiff’s test. Why? (CBSE Sample Paper 2011)
Answer:
Glucose has a cyclic structure in which the -CHO group is not free because it forms a hemiacetal linkage with the -OH group at C-5. Therefore, it does not give 2, 4-DNP test although it has -CHO group.

Question 16.
Write any two reactions of glucose that could not be explained by an open-chain structure of the glucose molecule.
Answer:
The open-chain structure of a glucose molecule cannot explain the following:

1. Glucose does not react with sodium bisulphite (NaHS0₃) to form an addition product though it has an aldehyde group,
2. Glucose does not give the Schiffs test and 2,4-DNP test like other aldehydes.

Question 17.
Write the product when D-glucose reacts with the cone. HNO₃.
Answer:

Question 18.
(i) Which vitamin deficiency causes rickets?
Answer:
Vitamin D

(ii) Name the base that is found in the nucleotide of RNA only. (CBSE Sample Paper 2017-18)
Answer:
Uracil

Question 19.
Write the Zwitter ion structure of glycine. (CBSE Sample Paper 2011)
Answer:

Question 20.
Name a carbohydrate present in the liver, muscles and brain. (CBSE 2019C)
Answer:
Glycogen

Question 21.
Name the species formed when an aqueous solution of amino acid is dissolved in water. (CBSE Sample Paper 2019)
Answer:
Zwitter ion/dipolar ion

Question 22.
Name the unit formed by the attachment of a base to the 1′ position of sugar in a nucleoside. (CBSE Sample Paper 2019)
Answer:
Nucleotide

Biomolecules Important Extra Questions Short Answer Type

Question 1.
Name the bases present in RNA. Which one of these is not present in DNA? (CBSE Delhi 2011)
Answer:

• The bases present in RNA are guanine, adenine, cytosine and uracil.
• Uracil is not present in DNA. Instead, it contains thymine.

Question 2.
What is meant by biocatalysts? (CBSE Delhi 2012)
Answer:
Biocatalysts. Biocatalysts or enzymes are produced by living cells which catalyse the biochemical reactions occurring in living cells.

Question 3.
Name one fibrous protein and one globular protein.
Answer:
Fibrous: Keratin Globular: Albumin

Question 4.
What is the chemical name of vitamin A and which disease is caused by its deficiency?
Answer:
Retinol, deficiency disease: Xerophthalmia.

Question 5.
What is the chemical name of vitamin C and which disease is caused by its deficiency?
Answer:
Ascorbic acid, deficiency disease: Scurvy.

Question 6.
Which enzyme is present in saliva? What is its function?
Answer:
The enzyme present in the saliva is amylase. It hydrolyses starch into maltose.

Question 7.
What is the difference between DNA and RNA on the basis of the bases they contain?
Answer:
Both DNA and RNA contain two bases derived from purine: guanine and adenine and one base derived from pyrimidine: cytosine. However, they have a fourth different base; DNA contains thymine whereas RNA contains uracil.

Question 8.
Shanti, a domestic helper of Mrs Anuradha, fainted while mopping the floor. Mrs Anuradha immediately took her to the nearby hospital where she was diagnosed to be severely ‘anaemic’. The doctor prescribed an iron-rich diet and multivitamins supplement to her. Mrs Anuradha supported her financially to get the medicines. After a month, Shanti was diagnosed to be normal.

After reading the above passage, answer the following questions:
(i) What values are displayed by Mrs Anuradha?
Answer:
Mrs Anuradha has shown generosity and a caring attitude for the health and life of her domestic helper. Timely help, kindness and financial support saved her domestic helper from a serious health problem.

(ii) Name the vitamin whose deficiency causes ‘pernicious anaemia’.
Answer:
B₁₂

(iii) Give an example of a water-soluble vitamin. (CBSE 2013)
Answer:
Vitamin C

Biomolecules Important Extra Questions Long Answer Type

Question 1.
Differentiate between the following:
(i) Amylose and Amylopectin
(ii) Peptide linkage and Glycosidic linkage
(iii) Fibrous proteins and Globular proteins
OR
Write chemical reactions to show that the open structure of D-glucose contains the following:
(i) Straight chain
(ii) Five alcohol groups
(iii) Aldehyde as carbonyl group (CBSE Delhi 2019)
Answer:
(i) Amylose and Amylopectin:

Amylose	Amylopectin
1. It consists of a long chain of a-D- glucose.	1. It consists of branched polymeric chains of a-D- glucose.
2. It is water-soluble.	2. It is water-insoluble.

(ii) Peptide linkage and Glycosidic linkage:

Peptide linkage	Glycosidic linkage
The linkage joins two amino acids through -CONH- bond.	The linkage joins two monosaccharides through an oxygen atom(-o).

(iii) Fibrous and Globular proteins:

Fibrous	Globular
1. The polypeptide chains run parallel and are held by hydrogen and sulphide bonds.	1. The polypeptide chains coil around to give a spherical shape.
2. These are insoluble in water.	2. These are soluble in water.
3. For example, Keratin in hair.	3. For example, egg albumin.

OR
(i) Glucose on prolonged heating with HI and red P at 100 °C gives n-hexane, which shows the straight-chain structure of glucose.

(ii) Glucose reacts with acetic anhydride in the presence of anhydrous zinc chloride to form glucose pentaacetate. The formation of pentaacetate confirms the presence of five -OH groups in the glucose molecule.

(iii) Glucose reacts with hydrogen cyanide forming glucose cyanohydrin. This shows that glucose contains a carbonyl group (>C=0).

Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on treatment with a mild oxidising agent like Br2 water. This indicates that the carbonyl group present in glucose is an aldehyde (-CHO) group.

Question 2.
Define the following with a suitable example in each:
(i) Oligosaccharides
(ii) Denaturation of protein
(iii) Vitamins
OR
Write the reactions involved when D-glucose is treated with the following reagents:
(i) Br₂ water
(ii) H₂N-OH
(iii) (CH₃CO)₂0 (CBSE Delhi 2019)
Answer:
(i) Oligosaccharides. These are the carbohydrates that give two to ten monosaccharide molecules on hydrolysis, for example, disaccharides such as sucrose, lactose, maltose (C₁₂H₂₂O₁₆), trisaccharides such as raffinose (C₁₈H₃₂O₁₆).

(ii) Deriaturatlon of proteins: A process that changes the physical and biological properties of proteins without affecting the chemical composition of the protein is called denaturation. It is caused by certain physical change like change In temperature or chemical change like change In pH presence of electrolytes, etc., for example, coagulation of albumin present in the white of an egg.

(iii) Vitamins: The organic compounds which cannot be produced by the body and must be supplied in small amounts in the diet to perform specific biological functions for the normal health, growth and maintenance of the body are called vitamins, for example, vitamin A, B, C, D, etc.
OR

Question 3.
Define the following terms with a suitable example of each:
(a) Anomers
Answer:
Anomers: Carbohydrates that differ in configuration at the glycosidic carbon (i.e. C₁ in aldoses and C₂ in ketoses) are called anomers. For example, α-D-glucose and β-D- glucose.

(b) Essential amino acids
Answer:
Essential amino acids: The amino acids which cannot be made by our bodies and must be supplied in our diet are called essential amino acids. For example, valine, leucine, etc.

(c) Denaturation of protein (CBSE AI 2019)
Answer:
A process that changes the native conformation of a protein is called denaturation. The denaturation can be caused by changes in pH, temperature, presence of salts of certain chemical agents. The denatured protein will lose its biological activity. During denaturation, the protein molecule uncoils from an ordered and specific conformation into a more random conformation and protein precipitates from the solution. For example, when an egg is boiled in water, the globular proteins present in it change to a rubber-like insoluble mass.

Question 4.
Define the following terms with a suitable example of each:
(a) Tertiary structure of the protein
Answer:
The tertiary structure of proteins: Tertiary structure of proteins arises because of the folding, coiling or bending of polypeptide chains producing a three-dimensional structure. This structure gives the overall shape of proteins. For example, fibrous proteins such as silk, collagen and α-keratin have large helical structures

(b) Essential amino acids
Answer:
Essential amino acids: The amino acids which cannot be synthesised in our body and must be supplied through our diet are called essential amino acids, for example, valine, leucine, etc.

(c) Disaccharides (CBSE AI 2019)
Answer:
Disaccharides: The carbohydrates which on hydrolysis give two same or different monosaccharides are called disaccharides. Sucrose is an example of a disaccharide, which on hydrolysis gives glucose and fructose.

Question 5.
Describe what do you understand by primary structure and secondary structure of proteins. (CBSE Delhi 2011)
Answer:
Primary structure:
The sequence in which the amino acids are linked in one or more polypeptide chains of a protein is called the primary structure.

The primary structure is usually determined by its successive hydrolysis with enzymes or mineral acids.

Secondary structure: The secondary structure gives the manner in which the polypeptide chains are folded or arranged. Therefore, it gives the shape or conformation of the protein molecule. This arises from the plane geometry of the peptide bond and hydrogen bond between the —C=O and N—H groups of different peptide bonds. Pauling and Corey studied that there are two common types of structures:

(i) α – Helix structure: In this structure, the formation of hydrogen bonding between amide groups within the same chain causes the peptide chains to coil up into a spiral structure (Fig. 1). This is called the a-helix.

(ii) β-pleated sheet structure: In this structure, all polypeptide chains are stretched out to nearly maximum extension and then laid side by side in a zigzag manner to form a flat sheet. Each chain is held to the two neighbouring chains by a hydrogen bond. These sheets are stacked one upon another to form a three-dimensional structure called
β-pleated sheet structure (Fig. 2).

Question 6.
ExplaIn what is meant by the pyranose structure of glucose. (CBSE 2011)
Answer:
The six-membered cyclic structure of glucose is catted pyranose structure similar to the pyran heterocyclic compound. For structure,

Question 7.
Write the main structural difference between RNA and DNA. Of the four bases, name those which are common to both DNA and RNA. (CBSE 2011)
Answer:
Structural differences:

DNA	RNA
(i) It has a double-stranded a-helix structure in which two strands are coiled spirally in opposite directions.	(i) It has a single-stranded a-helix structure.
(ii) The sugar molecule present in DNA is 2-deoxyribose.	(ii) The sugar molecule present in RNA is ribose.
(iii) Nitrogenous base uracil is not present.	(iii) Nitrogenous base thymine is not present.

Common bases present in both: adenine, cytosine and guanine.

Question 8.
(i) Deficiency of which vitamin causes night blindness? (CBSE DelhI 2014)
(ii) Name the base that is found in the nucleotide of RNA only.
(iii) Glucose on reaction with HI gives n-hexane. What does It suggest about the structure of glucose?
Answer:
(i) Deficiency of Vitamin A causes night blindness.
(ii) Uracil
(iii) Glucose reacts with HI to give n-hexane.

This suggests that alt the six carbon atoms in glucose are arranged in a straight chain structure of glucose.

Question 9.
(i) DefIciency of which vitamin causes rickets? (CBSE Delhi 2014)
Answer:
Vitamin D.

(ii) Give an example for each of fibrous protein and globular protein.
Answer:
Fibrous protein: Keratin Globular protein: Albumin

(iii) Write the product formed on reaction of D-glucose with Br2 water.
Answer:

Question 10.
(i) Deficiency of which vitamin causes scurvy? (CBSE DelhI 2014)
Answer:
Vitamin C.

(ii) What type of linkage is responsible for the formation of proteins?
Answer:
Peptide Linkage.

(iii) Write the product formed when glucose is treated with HI.
Answer:

Question 11.
Define the following terms: (CBSE 2014)
(i) Gtycosidic linkage
Answer:
Glycosidic linkage: The condensation of hydroxyl groups of two monosaccharides to form a link between them is called glycosidic linkage.

(ii) Invert sugar
Answer:
Invert sugar: The sugar which on hydrolysis with dilute adds or enzymes gives mixture having specific rotation opposite to the original is called invert sugar. For example, sucrose is inverted sugar.

(iii) Oligosaccharides
Answer:
Oligosaccharides: These are the carbohydrates that give two to ten monosaccharide molecules on hydrolysis. These are further classified as disaccharides, trisaccharides, tetrasaccharides, etc. depending upon the number of monosaccharide units present in their molecules. For example, Disaccharides: Sucrose, lactose, maltose. All these have the molecular

Formula C₁₂H₂₂O₁₁.
Trisaccharide: Raffinose (C₁₈H₃₂O₁₆).
Tetrasaccharides: Stachyose (C₂₄H₄₂O₂₁ ).

Question 12.
Define the following terms: (CBSE 2014)
(i) Nucleotide
Answer:
Nucleotide: A unit formed by the combination of a nitrogen-containing heterocyclic base, a pentose sugar and a phosphoric acid group.

(ii) Anomers
Answer:
Anomers: The anomers are the isomers formed due to the change in the configuration of the -OH group at C-1 of glucose. For example, α-and β-forms of glucose are anomers.

(iii) Essential amino acids.
Answer:
Essential amino acids: The amino acids which cannot be made by our bodies and must be supplied in our diet for the growth of the body are called essential amino acids.

Question 13.
Differentiate between the following:
(a) Fibrous protein and Globular protein
Answer:
Difference between fibrous protein and globútar protein:

Fibrous	Globular
(i) The polypeptide chains run parallel and are held together by hydrogen and disulphide bonds.	1. The polypeptide chains colt around to give a spherical shape.
(ii) These are insoluble in water.	2. These are soluble in water.
(iii) For example, keratin in hair	3. For example, albumin in egg.

(b) Essential amino acids and Non-essential amino acids
Answer:
Difference between essential amino acids and non-essential amino acids:

Essential amino acids	Non-essential amino acids
These are not synthesìsed in our body and must be supplied in the diet.	These are synthesised in our body and not required in our diet.
For example, valine.	For example, alanine

(C) Amylose and Amylopectin (CBSE AI 2019)
Answer:
Difference between amylose and amylopectin:

Amylose	Amylopectin
It consists of branched polymeric chains of α – D – glucose.	It consists of a long straight chain of α – D – glucose.
It is water-insoluble.	It is water-soluble.

Question 14.
(i) Which one of the following is a polysaccharide: (CBSE 2015, 2014)
Starch, Maltose, Fructose, Glucose
Answer:
Starch

(ii) What is the difference between native protein and denatured protein?
Answer:
Native protein is the protein found in a biological system with a unique three-dimensional structure and biological activity.
Denatured protein is the protein which loses its biological activity by certain physical or chemical treatment such as changes in pH, temperature, presence of certain salts or chemical agent, known as denaturation.

(iii) Write the name of the vitamin responsible for the coagulation of blood.
Answer:
Vitamin K.

Question 15.
(i) Write one reaction of D-glucose which cannot be explained by its open-chain structure. (CBSE 2016)
Answer:
Despite having an aldehydic (-CHO) group, glucose does not react with NaHS0₃ to form an addition product.

(ii) What type of linkage is present in nucleic acids?
Answer:
Phosphodiester linkage

(iii) Give one example each for water-soluble vitamins and fat-soluble vitamins.
Answer:
Water-soluble vitamin: vitamin B Fat-soluble vitamin: vitamin A

Question 16.
What is essentially the difference between the α-form of D-glucose and β-form of D-glucose? Explain.
Answer:
α-form and β-form of glucose differ in the orientation of —H and —OH groups around the C₁ atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C₁ are called anomers. The structures of these two may be shown below:

These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|_D = +111° and the β-form of glucose has m.p. 423 K and |α|_D = +19.2°.

Question 17.
(a) What happens when D-glucose is treated with the following reagents:
(i) HI
Answer:
When glucose is treated with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight line.

(ii) cone. HN03
Answer:
Glucose on treatment with nitric acid gives a dicarboxylic acid, saccharic acid.

(b) What is the basic structural difference between starch and cellulose? (CBSE 2019C)
Answer:
Starch consists of two components: amylose and amylopectin. Amylose is a long linear polymer of 200-1000 a-D-(+)-glucose units held by C₁ – C₄ glycosidic linkages. It is soluble in water. Amylopectin is a branched-chain polymer of a-D-(+)-glucose linkages whereas branching occurs by C₁ – C₆ glycosidic linkage. It is insoluble in water.

On the other hand, cellulose is a straight-chain polysaccharide composed only of (β – D – (+)-glucose units which are formed by the glycosidic linkage between C₁ of one glucose unit and C₄ of the next glucose unit.

Question 18.
(1) Write the name of two monosaccharides obtained on hydrolysis of lactose sugar. (CBSE Delhi 2016)
Answer:
Glucose, Galactose

(ii) Why Vitamin C cannot be stored in our body?
Answer:
Vitamin C is water-soluble and hence cannot be stored in our body.

(iii) What is the difference between a nucleoside and a nucleotide?
Answer:
A nucleoside contains only two basic components of nucleic acids, namely a pentose sugar and a nitrogenous base.

A nucleotide contains all the three basic components of nucleic acids, namely a phosphoric acid group, a pentose sugar and a nitrogenous base.

Question 19.
Define the following with an example of each: (CBSE 2018)
(i) Polysaccharides
(ii) Denatured protein
(iii) Essential amino acids
OR
(i) Write the product when D-glucose reacts with the cone. HNO₃.
(ii) Amino acids show amphoteric behaviour. Why?
(iii) Write one difference between a-helix and p-pleated structures of proteins.
Answer:
(i) Carbohydrates that give a large number of monosaccharide units on hydrolysis or a large number of monosaccharide units joined together by glycosidic linkage, e.g. starch, glycogen, cellulose.
(ii) Proteins that lose their biological activity or proteins in which secondary and tertiary structures are destroyed, e.g. curdling of milk.
(iii) Amino acids cannot be synthesised in the body. e.g. Valine / Leucine
OR
(i) Saccharic acid / COOH-(CHOH)₄-COOH
(ii) Due to the presence of carboxyl and amino group in the same molecule or due to formation of zwitterion or dipolar ion.
(iii) a-helix has intramolecular hydrogen bonding while p-pleated has intermolecular hydrogen bonding / a-helix results due to regular coiling of polypeptide chains while in p-pleated all polypeptide chains are stretched and arranged side by side.

Question 20.
Explain the following:
(i) Amino acids behave like salts rather than simple amines or carboxylic acids. (CBSE 2018C)
Answer:
Due to the formation of zwitterion.

(ii) The two strands of DNA are complementary to each other.
Answer:
The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases.

(iii) Reaction of glucose indicates that the carbonyl group is present as an aldehydic group in the open structure of glucose.
Answer:

Glucose gets oxidised to gluconic acid on reaction with a mild oxidising agent like Bromine water.

Question 21.
What is essentially the difference between α-glucose and β-glucose? What is meant by the pyranose structure of glucose?
Answer:
α-form and β-form of glucose differ in the orientation of -H and -OH groups around C, atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C₁ are called anomers. The structures of these two may be shown below:

These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|_D = + 111° and the β-form of glucose has m.p. 423 K and |α|_D = + 19.2°.

The α-D-glucose and β-D-glucose can be drawn in a simple six-membered ring form called pyranose structures. These resemble pyran which is a six-membered heterocyclic ring containing five carbon atoms and one oxygen atom.

These are known as pyranose structures and are shown below:

Question 22.
Which sugar is called invert sugar? Why is it called so?
Answer:
Sucrose is called invert sugar. The sugar obtained from sugar beet is a colourless, crystalline and sweet substance. It is very soluble in water and its aqueous solution is dextrorotatory having [α]_D = + 66.5°.

On hydrolysis with dilute acids or enzyme invertase, cane sugar gives an equimolar mixture of D-(+)-glucose and D-(-)-fructose.

So, sucrose is dextrorotatory but after hydrolysis, gives dextrorotatory glucose and laevorotatory fructose. D-(-)-fructose has a greater specific rotation than D-(+)- glucose. Therefore, the resultant solution upon hydrolysis is laevorotatory in nature with a specific rotation of (-39.9°). Since there is a change in the sign of rotation from Dextro before hydrolysis to Laevo after hydrolysis, the reaction is called Inversion reaction and the mixture (glucose and fructose) is called invert sugar.

Question 23.
How do you explain the presence of an aldehydic group in a glucose molecule?
Answer:
Glucose reacts with hydroxylamine to form a monoxime and adds one molecule of hydrogen cyanide to give cyanohydrin.

Therefore, it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a carboxylic acid-containing six carbon atoms.

This indicates that the carbonyl group present in glucose is an aldehydic group.

Question 24.
Describe the term D- and L-configuration used for sugars with examples.
Answer:
The sugars are divided into two families: the D-family and L-family which have definite configurations. These configurations are represented with respect to glyceraldehyde as the standard. The glyceraldehyde may be presented in two forms:

The D-configuration has —OH attached to the carbon adjacent to —CH₂OH on the right while L-configuration has —OH attached to the carbon adjacent to —CH₂OH on left.

The sugars are calLed D- or L- depending upon whether the configuration of the molecule is related to D-glyceraldehyde or L-glyceraldehyde. It has been found that all naturally occurring sugars beLong to D-series, e.g. D-glucose, D-ribose and D-fructose.

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 13 Nuclei. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 13 Important Extra Questions Nuclei

Nuclei Important Extra Questions Very Short Answer Type

Question 1.
What will be the ratio of the radii of two nuclei of mass numbers A₁ and A₂?
Answer:
The ratio is \(\frac{R_{1}}{R_{2}}=\left(\frac{A_{1}}{A_{2}}\right)^{1 / 3}\)

Question 2.
Two nuclei have mass numbers in the ratio 1: 2. What is the ratio of their nuclear densities?
Answer:
The densities of both nuclei are equal as they do not depend upon mass number.

Question 3.
A nucleus of mass number A has a mass defect Δm. Give the formula, for the binding energy per nucleon of this nucleus.
Answer:
The formula is E = \(\frac{\Delta m \times c^{2}}{A}\)

Question 4.
Write the relation between half-life and decay constant of a radioactive sample.
The relation is T_1/2 = \(\frac{0.693}{\lambda}\)

Question 5.
Write the nuclear decay process for β-decay of 15³²P.
Answer:
The process is

Question 6.
State the relation between the mean life (τ) of a radioactive element and its decay constant λ.
Answer:
The two are related as τ = 1 / λ.

Question 7.
Write any two characteristic properties of nuclear force. (CBSE AI 2011)
Answer:

1. Do not obey inverse square law and
2. Spin-dependent.

Question 8.
How is the radius of a nucleus related to its mass number? (CBSE AI 2011C, AI 2013C)
Answer:
The radius R of the nucleus and mass number A is related as R = R_oA^1/3, where R_o is a constant.

Question 9.
A nucleus undergoes β – decay. How does
(i) the mass number,
(ii) atomic number change? (CBSE Delhi 2011C)
Answer:
During β – decay
(i) the mass number remains the same,
(ii) atomic number increases by one.

Question 10.
Define the activity of a given radioactive substance. Write its SI unit. (CBSE AI 2013)
Answer:
The rate of disintegration in a radioactive substance is known as its activity. SI unit is becquerel (Bq).

Question 11.
Why is it found experimentally difficult to detect neutrinos in nuclear β-decay? (CBSE AI 2014)
Answer:
They are very difficult to detect since they can penetrate a large quantity of matter (even earth) without any interaction.

Question 12.
Four nuclei of an element undergo fusion to form a heavier nucleus, with the release of energy. Which of the two — the parent or the daughter nucleus – would have higher binding energy per nucleon? (CBSE AI 2018, Delhi 2018)
Answer:
Daughter nucleus.

Question 13.
Why is nuclear fusion not possible In the laboratory?
Answer:
Because temperature as high as 10⁷ K cannot be sustained in the laboratory.

Question 14.
Why is the penetrating power of gamma rays very large?
Answer:
Because they have high energy and are neutral.

Question 15.
Can it be concluded from beta decay that electrons exist inside the nucleus?
Answer:
No, the beta particle although an electron is actually created at the instant of beta decay and ejected at once. It cannot exist inside the nucleus as its de-Broglie wavelength is much larger than the dimensions of the nucleus.

Question 16.
Why is the ionizing power of α – parties greater than that of γ-rays?
Answer:
Because α – particles are heavy particles and their speed is comparatively small, so they collide more frequently with atoms of the medium and ionize them.

Question 17.
When a nucleus undergoes alpha decay, is the product atom electrically neutral in beta decay?
Answer:
No, in alpha decay, the atomic number decreases by 2 hence the atom is left with 2 extra orbital electrons. It, therefore, has a double negative charge. In beta decay, the atom is left with a net positive charge.

Question 18.
You are given two nuclides ₃⁷X and ₃⁴Y. Are they the isotopes of the same element? Why?
Answer:
Yes, because an atomic number of both nuclides is 3.

Question 19.
The variation of the decay rate of two radioactive samples A and B with time is shown in the figure. Which of the two has a greater decay constant? (NCERT Exemplar)

Answer:
The decay constant of A is greater than that of B but it does not always decay faster than B.

Question 20.
Does the ratio of neutrons to protons in a nucleus increase, decrease, or remain the same after the emission of an alpha particle? (NCERT Exemplar)
Answer:
The ratio of neutrons to protons in a nucleus increases after the emission of an alpha particle.

Question 21.
Which property of nuclear force explains the approximate constancy of binding energy per nucleon with mass number A for nuclei in the range 30 < A < 170? (CBSE2019C)
Answer:
The short-range nature of the nuclear force explains the approximate constancy of binding energy per nucleon with mass number A in the range 30 < A < 170.

Question 22.
Draw a graph showing the variation of decay rate with a number of active nuclei. (NCERT Exemplar)
Answer:
The graph is as shown.

Question 23.
Which sample, A or B, shown in the figure has a shorter mean-life? (NCERT Exemplar)

Answer:
B has shorter mean life as λ is greater forB.

Question 24.
Why do stable nuclei never have more protons than neutrons? (NCERT Exemplar)
Answer:
Protons are positively charged and repel one another electrically. This repulsion becomes so great in nuclei with more than 10 protons or so that an excess of neutrons, which produce only attractive forces, is required for stability.

Question 25.
Why does the process of spontaneous nuclear fission occur in heavy nuclei? (CBSE 2019C)
Answer:
Because heavy nuclei contain a large number of protons that exert strong repulsive forces on one another.

Nuclei Important Extra Questions Short Answer Type

Question 1.
Draw the curve showing the binding energy/nucleon with a mass number of different nuclei. Briefly state, how nuclear fusion and nuclear fission can be explained on the basis of this graph.
Answer:
The diagram is as shown.

Light nuclei have a small value of binding energy per nucleon, therefore to become more stable they fuse to increase their binding energy per nucleon.

A very heavy nucleus, say A 240, has Lower binding energy per nucLeon compared to that of a nucleus with A = 120. Thus if a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightLy bound. This implies energy would be released in the process.

Question 2.
Define decay constant for a radioactive sample. Which of the following radiations α, β, and γ rays
(i) are similar to X-rays,
(ii) are easily absorbed by matter, and
(iii) are similar in nature to cathode rays?
Answer:
The decay constant is defined as the reciprocal of that time duration for which the number of nuclei of the radioactive sample decays to 1 / e or 37 % of its original value.
(i) Gamma
(ii) Alpha
(iii) Beta

Question 3.
State the law of radioactive decay.
Plot a graph showing the number of undecayed nuclei as a function of time (t) for a given radioactive sample having a half-life T_1/2.
Depict In the plot the number of undecayed nuclei at (i) t = 3T_1/2 and (ii) t = 5 T_1/2 (CBSE Delhi 2011)
Answer:
The number of nuclei disintegrating per second is proportional to the number of nuclei present at the time of disintegration and is independent of alt physical conditions like temperature, pressure, humidity, chemical composition, etc.

The plot is as shown.

Question 4.
Draw a plot of the potential energy of a pair of nucleons as a function of their separations. Mark the regions where the nuclear force is (i) attractive and (ii) repulsive. Write any two characteristic features of nuclear forces. (CBSE AI 2012)
Answer:

For r > r₀ (attraction), For r < r_o (repulsion)

1. Strong attractive force (stronger than the repulsive electric force between the protons)
2. Are short-range forces.

Question 5.
(a) Write the relation for binding energy (BE) (in MeV) of a nucleus of mass _Z^AM atomic number (Z) and mass number (A) in terms of the masses of its constituents – neutrons and protons.
Answer:
The required expression is
ΔE = (Zm_p + (A – Z)m_n – M) × 931 MeV

(b) Draw a plot of BE/A versus mass number A for 2 ≤ A ≤ 170. Use this graph to explain the release of energy in the process of nuclear fusion of two light nuclei. (CBSE Delhi 2014C)
Answer:

Since the binding energy of the smaller nuclei like hydrogen is less, therefore they fuse together to form helium in order to increase their binding energy per nucleon and become stable. This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion.

Question 6.
If both the number of neutrons and the number of protons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain. (CBSE AI2016C)
Answer:
We know that the binding energy of a nucleus gives a negative contribution to the mass of the nucleus (mass defect). Now, since proton number and neutron number are conserved in a nuclear reaction the total rest mass of neutrons and protons is the same on either side of a reaction. But the total binding energy of nuclei on the left side need not be the same as that on the right-hand side.

The difference in these binding energies appears as the energy released or absorbed in a nuclear reaction. Since binding energy contributes to mass, we say that the difference in the total mass of nuclei on the two sides gets converted into energy or vice-versa.

Question 7.
State two properties of nuclear forces. Write the relation between half-life and decay constant of a radioactive nucleus. (CBSE AI 2017C)
Answer:

1. They are saturated forces.
2. They are charge-independent.

The required relation is
T = \(\frac{\ln 2}{\lambda}=\frac{2.303 \log 2}{\lambda}=\frac{0.693}{\lambda}\)

Question 8.
(a) Draw a graph showing the variation of binding energy per nucleon (BE/A) vs mass number A for the nuclei in 20 ≤ A ≤ 170.
Answer:

Since the binding energy of the smaller nuclei like hydrogen is less, therefore they fuse together to form helium in order to increase their binding energy per nucleon and become stable. This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion.

(b) A nucleus of mass number 240 and having binding energy/nucleon 7.6 MeV splits into two fragments Y, 1 of mass numbers 110 and 130 respectively. If the binding energy/ nucleon of Y, 1 is equal to 8.5 MeV each, calculate the energy released in the nuclear reaction. (CBSE Al 2017C)
Answer:
Energy released per fission
= (110 + 130) × 8.5 – 240 × 7.6
= 240 × (8.5 – 7.6) MeV
= 240 × 0.9
= 216.0 MeV

Question 9.
Explain with the help of an example, whether the neutron-proton ratio in a nucleus increases or decreases due to beta decay.
Answer:
Consider the following decay

Number of neutrons before beta decay
= 234-90 = 144
Number of neutrons after beta decay
= 234-91 =143
Number of protons before beta decay
= 90
Number of protons after beta decay
= 91
Neutron-proton ratio before beta decay
= \(\frac{144}{90}\) = 1.6

Neutron-proton ratio after beta decay
= \(\frac{143}{91}\) = 1.57

Thus neutron-proton ratio decreases during beta decay.

Question 10.
How is the size of a nucleus experimentally determined? Write the relation between the radius and mass number of the nucleus. Show that the density of the nucleus is independent of its mass number. (CBSE Delhi 2011C)
Answer:
The size of the nucleus can be determined by the Rutherford experiments on alpha particles scattering. The distance of the nearest approach is approximately the size of the nucleus. Here it is assumed that only coulomb repulsive force caused scattering. With alpha rays of 5.5 MeV, the size of the nucleus was found to be less than 4 × 10^-14 m. By doing scattering experiments with fast electrons bombarding targets of different elements, the size of the nuclei of various elements determined accurately.

The required relation is
R = R_oA₁/3, where R_o = 1.2 × 10^-15 m

The density of a nucleus of mass number A and radius R is given by

which is independent of the mass number A.

Question 11.
(a) What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170?
Answer:
The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. This leads to the saturation of forces in a medium or a large-sized nucleus, i.e. nuclei for which A is 30 < A < 170, which is the reason for the constancy of the binding energy per nucleon.

(b) Show that the density of nucleus over a wide range of nuclei is constant- independent of mass number A. (CBSE AI 2012)
Answer:
The size of the nucleus can be determined by the Rutherford experiments on alpha particles scattering. The distance of the nearest approach is approximately the size of the nucleus. Here it is assumed that only coulomb repulsive force caused scattering. With alpha rays of 5.5 MeV, the size of the nucleus was found to be less than 4 x 10^-14 m. By doing scattering experiments with fast electrons bombarding targets of different elements, the size of the nuclei of various elements determined accurately.

The required relation is
R = R_oA₁/3, where R_o = 1.2 × 10^-15 m

The density of a nucleus of mass number A and radius R is given by

which is independent of the mass number A.

Question 12.
A radioactive nucleus ‘A’ decays as given below:
.
If the mass number and atomic number of A₁ are 180 and 73 respectively, find the mass number and an atomic number of A and A₂.
Answer:
For A : Z = 72 and A = 180
For A₂: Z = 71 and A = 176

Question 13.
The sequence of stepwise decay of a radioactive nucleus is. If the nucleon number and atomic number for D₂ are 176 and 71 respectively, what are the corresponding values of D and D₃? Justify your answer in each case.
Answer:
For D: A = 180, Z = 72
For D₃: A = 172, Z = 69

During alpha decay mass number decreases by 4 and the atomic number decreases by 2, while during beta decay the mass number remains the same, and the atomic number increases by 1.

Question 14.
Write symbolically the nuclear β⁺ decay process of ₆¹¹C. Is the decayed product X an isotope or isobar of₆¹¹C?
Given the mass values m (₆¹¹C) = 11.011434 u and m (X) = 11.009305 u. (CBSE AI 2015)
Estimate the Q – value in this process.
Answer:
The required equation is

X is an isobar
Mass defect = m(C) – m(X)
= (11.011434- 11.009305) u = 0.002129 u

Therefore Q = Δm × 931.5 MeV
= 0.002129 × 931.5 = 1.98 MeV

Question 15.
Two radioactive samples, X, Y have the same number of atoms at t = 0. Their half¬lives are 3 h and 4 h respectively. Compare the rates of disintegration of the two nuclei after 12 hours. (CBSE AI 2017C)
Answer:
Let N0 be the nuclei present in X and Y at t = 0. Given T_x = 3 h and T_y = 4 h, t = 12 h.
The number of nuclei present in X and Y after 12 hours is

Question 16.
A radioactive sample has the activity of 10,000 disintegrations per second after 20 hours. After the next 10 hours, its activity reduces to 5,000 dis. sec^-1. Find out its half-life and initial activity. (CBSE Delhi 2017C)
Answer:
Since activity reduces to half in 10 hours from 10000 dis. sec^-1 to 5000 dis. sec^-1, therefore half-life of the sample will be 10 years.

Question 17.
Why is the energy of the beta particles emitted during beta decay continuous?
Answer:
The phenomenon of beta decay arises due to the conversion of a neutron in the nucleus into a proton, electron, and an anti-neutrino. Because the energy available in beta decay is shared by the electron and the anti-neutrino in all possible ratios as they come out of the nucleus, therefore the beta ray energy spectrum is continuous in nature.

Question 18.
Explain, how radioactive nuclei can emit β-particles even though atomic nuclei do not contain these particles. Hence explain why the mass number of a radioactive nuclide does not change during β-decay.
Answer:
Beta-particles (or electrons) as such are not present inside a nucleus. However, in the case of a radioactive nuclide, sometimes a neutron decays into a proton, an electron, and an antineutrino as given by the following equation:

where mass and charge of antineutrino particle is zero. Out of the particles formed, the proton remains within the nucleus itself but electron along with antineutrino comes out of the nucleus. It is this electron that is being emitted as a beta-particle.

As in the process of β-emission, one proton is produced in the nucleus at the expense of a neutron and the mass number of both is the same, hence the mass number of the nuclide remains unchanged during p-decay.

Question 19.
Consider a radioactive nucleus A which decays to a stable nucleus C through the following sequence: A → B → C. Here B is an intermediate nucleus that is also radioactive. Considering that there are N_o atoms of A initially, plot the graph showing the variation of the number of atoms of A and B versus time. (NCERT Exemplar)
Answer:
At t = 0, N_A = N_o while N_B = 0. As time increases, N_A falls off exponentially, while the number of atoms of B increases, becomes maximum, and finally decays to zero  ∞ (following exponential decay law).

Hence the graph is as shown.

Nuclei Important Extra Questions Long Answer Type

Question 1.
Define the terms: half-life period and decay constant of a radioactive sample. Derive the relation between these terms.
Answer:
The half-life is the time required for the number of radioactive nuclei to decrease to one-half the original number.

The decay constant is defined as the reciprocal of that time duration for which the number of nuclei of the radioactive sample decays to 1 / e or 37% of its original value.

To get the relation for half life T and decay constant λ we set N = \(\frac{N_{0}}{2}\) and t = T in the equation N = No e^-λt, obtaining \(\frac{1}{2}\) = e^-λt

Taking the logarithm of both sides and solving for T we have
T = \(\frac{\ln 2}{\lambda}=\frac{2.303 \log 2}{\lambda}=\frac{0.693}{\lambda}\)

Question 2.
(a) Draw a graph showing the variation of the potential energy of a pair of nucleons as a function of their separation. Indicate the regions in which nuclear force is (i) attractive, and (ii) repulsive.
Answer:
Graph showing the variation of potential energy U (in MeV) of a pair of nucleons as a function of their separation r (in fm) is shown here.

1. In the graph region AD (r > r_o) shows the region where nuclear force is strongly attractive.
2. The region DE (r < r_o) shows the region where nuclear force is strongly repulsive.

(b) Write two characteristic features of nuclear force which distinguish it from the Coulomb force.
Answer:
Two characteristics of nuclear forces which distinguish it from Coulomb’s force are

• It is charge Independent.
• It is an extremely short-range force and does not obey the inverse square law.

Question 3.
Prove that the Instantaneous rate of change of the activity of a radioactive substance is Inversely proportional to the square of Its half-life.
Answer:
The activity of a radioactive substance is
A = \(\frac{dN}{dt}\).

We know that the number of nuclei of a radioactive substance Left behind after time t is given by N = N_oe^-λt

Differentiating the above relation with respect to time we have
A = \(\frac{d N}{d t}=\frac{d}{d t}\)N_oe^-λt = – N_oλe^-λt

Differentiating the above equation with respect to time we have

Therefore the Instantaneous rate of change of the activity of a radioactive substance is inversely proportional to the square of Its half Life.

Question 4.
(a) Deduce the expression N = N_oe^-λt the law of radioactive decay.
(b) (i) Write symbolically the process expressing the β⁺ decay of, ₁₁²²Na Also write the basic nuclear process underlying this decay.
(ii) Is the nucleus formed in the decay of the nucleus ₁₁²²Na an Isotope or isobar? (CBSE Delhi 2014)
Answer:
(a) Let N0 be the number of nuclei present in a freshly separated sample of a radioactive substance. Let after time t the number of nuclei left behind be N. Let dN number of nuclei disintegrate in a small time interval dt. Then by the – decay law,

where λ  is a constant of proportionality.

Question 5.
(a) Complete the following nuclear reactions:

Answer:

(b) Write the basic process Involved in nuclei responsible for (i) β^– and (ii) β⁺ decay.
Answer:
The basic nuclear process underlying β^– decay is the conversion of the neutron to proton
n → p + e^– + v^–
while for v+ decay, it is the conversion of a proton into a neutron
p → n + e⁺ + v

(c) Why is it found experimentally difficult to detect neutrinos? (CBSE AI 2015 C)
Answer:
Neutrinos are neutral particles with very small (possibly, even zero) mass compared to electrons. They have only weak interaction with other particles. They are, therefore, very difficult to detect, since they can penetrate a large quantity of matter (even earth) without any interaction.

Question 6.
(a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (B.E./A) versus the mass number A.
Answer:
For the graph

From the plot, we note that

• During nuclear fission: A heavy nucleus in the larger mass region (A > 200) breaks into two middle-level nuclei, resulting in an increase in B.E./ nucleon. This results in the release of energy,
• During nuclear fusion: Light nuclei in the lower mass region (A < 20) fuse to form a nucleus having higher B.E. / nucleon. Hence Energy gets released.

(b) A radioactive Isotope has a half-life of 10 years. How long will It take for the activity to reduce to 3.125%? (CBSE AI2018)
Answer;
3.125% means that the number of nuclei decays to 1/32 of its original value.
Therefore,
\(\frac{N}{N_{0}}=\frac{1}{32}=\left(\frac{1}{2}\right)^{5}\)

Now we know that
N = N0\(\left(\frac{1}{2}\right)^{t / T}\)

Therefore we have
\(\left(\frac{1}{2}\right)^{5}=\left(\frac{1}{2}\right)^{t / T}\)

Therefore
t = 5T = 5 × 10 = 50 years

Question 7.
Group the following six nuclides into three pairs of (?) isotones, (ii) isotopes, and (iii) isobars: ₆¹²C, ₂³He, ₈₀¹⁹⁸Hg, ₁³H, ₇₉¹⁹⁷Au ₆¹⁴C. How does the size of the nucleus depend on its mass number? Hence explain why the density of nuclear matter should be independent of the size of the nucleus.
Answer:
(a) Isotones: ₈₀¹⁹⁸Hg, ₇₉¹⁹⁷Au
(6) Isotopes: ₆¹²C , ₆¹⁴C
(c) Isobars: ₂³He, ₁³H

The size of a nucleus depends upon its mass number as R = R₀ A^1/3

The nuclear density is given by the expression

The calculations show that the nuclear density is independent of the mass number.

Question 8.
Define the term ‘decay constant’ of a radioactive sample. The rate of disintegration of a given radioactive nucleus is 10,000 disintegrations/s and 5,000 disintegration/s after 20 hr and 30 hr respectively from start. Calculate the half-life and an initial number of nuclei at t = 0. (CBSE Delhi 2019)
Answer:
The decay constant of a radioactive element is the reciprocal of the time in which the number of its nuclei reduces to 1 /e of its original number.

We have R = λN
R(20hrs) = 100o0 = λN₂₀
R(30hrs) = 5000 = λN₃₀
\(\frac{N_{20}}{N_{30}}\) = 2

This means that the number of nuclei, of the given radioactive nucleus, gets halved in a time of (30 – 20) hours = 10 hours
Half-life = 10 hours

This means that in 20 hours (= 2 half-Lives), the original number of nuclei must have gone down by a factor of 4.
Hence rate of decay at t = 0
λ N₀ = 4 λ N₂₀
R₀ = 4 × 10,000 = 40,000 disintegration per second

Question 9.
(a) Write the relation between half-life and an average life of a radioactive nucleus.
Answer:
The relation is τ = 1 .44T^1/2

(b) In a given sample two isotopes A and B are initially present in the ratio of 1:2. Their half-lives are 60 years and 30 years respectively. How long will it take so that the sample has these isotopes in the ratio of 2:1? (CBSE Delhi 2019)
Answer:


Question 10.
Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy release in MeV in the deuterium-tritium fusion reaction:

Using the data m(21H) = 2.014102 u,
m(₁³H) = 3.016949u, m(₂⁴He) = 4.002603 u,
m_n = 1.008665 u,1 u = 931.5 MeV/c² (CBSE Delhi 2015)
Answer:
The distinction is shown in the table below.

Nuclear Fission	Nuclear Fusion
1. It is the splitting of a heavy nucleus into two or tighter unstable nuclei.	1. It is the combining of two light nuclei into a heavier nucleus.
2. It may or may not be a chain reaction.	2. It is always a chain reaction.
3. It is independent of temperature.	3. It is temperature-dependent.
4. It can be controlled.	4. It can’t be controlled.
5. Tremendous amount of energy is released.	5. Energy released per unit mass is seven times the energy released during fission.
6. By-products are harmful.	6. By-products are not harmful.
7. Example of reaction – The atom bomb.	7. Example of reaction – Reaction in stars, hydrogen born

In both reactions, there is a mass defect that is converted into energy.
Now energy released in the reaction

Question 11.
(a) Draw a plot showing the variation of the potential energy of a pair of nucleons as a function of their separation. Mark the regions where the nuclear force is (a) attractive and (b) repulsive.
Answer:

For r > r_o, the force is attractive For r < r_o, the force is repulsive

(b) In the nuclear reaction

determine the values of a and b. (CBSE Delhi 2018 C)
Answer:
We have,
1 + 235 = a + 94 + 2 × 1
∴ a = 236 – 96 = 140

Also
0 + 92 = 54+ 6 + 2 × 0
∴ b = 92 – 54 = 38

Question 12.
Binding energy per nucleon versus mass number curve is as shown. _Z^AS, _Z1^A1w, _Z2^A2X, and _Z3^A3Y, are four nuclei indicated on the curve.

Based on the graph:
(а) Arrange X, W, and S in the increasing order of stability.
Answer:
(a) S, W, and X

(b) Write the relation between the relevant A and Z values for the following nuclear reaction. S → X + W
Answer:
The equation is
Z = Z₁ + Z₂ and A = A₁ + A₂

(c) Explain why binding energy for heavy nuclei is low. (CBSE Sample Paper 2018-19)
Answer:
Reason for low binding energy: For heavier nuclei, the Coulomb repulsive force between protons increases considerably and offsets the attractive effects of the nuclear forces. This can result in such nuclei being unstable.

Question 13.
(a) Derive the law of radioactive decay,
viz. N = Noe-λt
Answer:
Let N0 be the number of nuclei present in a freshly separated sample of a radioactive substance. Let after time t the number of nuclei left behind be N. Let dN number of nuclei disintegrate in a small time interval dt. Then by the – decay law,

where λ is a constant of proportionality.

(b) Explain, giving necessary reactions, how energy is released during
(i) fission
Answer:
Nuclear Fission: It is a process in which a heavy nucleus splits up into two Lighter nucLei of nearly equal masses. It is found that the sum of the masses of the product nuclei and particles is less than the sum of the masses of the reactants, i.e. there is some mass defect. This mass defect appears as energy. One such fission reaction is given below:

The Q value of the above reaction is about 200 MeV. The sum of the masses of Ba, Kr, and 3 neutrons is less than the sum of the masses of U and one neutron.

(ii) fusion
Nuclear Fusion: It is the process in which two light nuclei combine together to form a heavy nucleus. For fusion very high temperature of l is required. One such fusion reaction is given below:

The Q value of this nuclear reaction is 24 MeV. It is the energy equivalent of the mass defect in the above reaction. The energy released in fusion is much less than in fission but the energy released per unit mass infusion is much greater than that released in fission.

Question 14.
(a) Distinguish between isotopes and isobars, giving one example for each.
(b) Why is the mass of a nucleus always less than the sum of the masses of its constituents? Write one example to justify your answer.
Or
(a) Classify the following six nuclides into (i) isotones, (ii) Isotopes, and (iii) isobars: (CBSEAI2019)

(b) How does the size of a nucleus depend on its mass number? Hence explain why the density of nuclear matter should be independent of the size of the nucleus.
Answer:
(a) Isotopes have the same atomic number while isobars have the same mass number
Examples of isotopes ₆¹²C, ₆¹⁴C
Examples of isobars ₂³He, ₁³H

(b) Mass of a nucleus is less than its constituents because in the bound state some mass is converted into binding energy which is energy equivalent of mass defect e.g., the mass of 1860 nucleus is less than the sum of masses of 8 protons and 8 neutrons
Or
(a) Isotones: ₈₀¹⁹⁸Hg, ₇₉¹⁹⁷Au
(6) Isotopes: ₆¹²C , ₆¹⁴C
(c) Isobars: ₂³He, ₁³H

(b) The radius of the nucleus is given by
R = R_oA^1/3

Volume of the nucleus \(\frac{4}{3}\)πR³ = \(\frac{4}{3}\)πR_o³ A

If m is the average mass of the nucleon then the mass of the nucleus M = mA
Hence nuclear density

Which is independent of the A i.e., the size of the nucleus.

Numerical Problems:

• Radius of the nucleus R = R_oA^1/3
• Mass defect, Δm = Z m_p + (A – Z) m_n– M
• Energy released ΔE = Δm × 931 MeV
  or
  ΔE = [Z m_p + (A – Z) m_n – M] × 931 MeV
• Binding energy per nucleon BE/N = ΔE/A
• Relation between original nuclei (N) and nuclei left (N_o) after time t N = N_oe^-λt
• Relation between decay constant (λ) and half-life (T) = \(\frac{\ln 2}{\lambda}=\frac{2.303 \log 2}{\lambda}=\frac{0.693}{\lambda}\)
• Half-life is also given by the expression N = N_o\(\left(\frac{1}{2}\right)^{n}\) where n = t/T
• The average life is given by
  τ = \(\frac{1}{\lambda}=\frac{T}{\ln 2}=\frac{T}{0.693}\) = 1.44 T
• Activity is given by A = -λN = \(\frac{0.693N}{T}\)

Question 1.
Calculate the binding energy per nucleon of Fe⁵⁶₂₆ Given m_Fe = 55.934939 u, m_n = 1.008665 u and m_p = 1.007825 u
Answer:
Number of protons Z = 26
Number of neutrons (A – Z) = 30
Now mass defect is given by
Δm = Z m_p + (A – Z)m_n – M
Δm = 26 × 1.007825 + 30 × 1.008665 – 55.934939
= 0.528461 u

Therefore binding energy
BE = Δm × 931 MeV = 0.528461 × 931
= 491.99 MeV

BE/nucleon = 491.99/56 = 8.785 MeV

Question 2.
The activity of a radioactive element drops to one-sixteenth of its initial value in 32 years. Find the mean life of the sample.
Answer:

Or
32/T = 4 or 7 = 32 / 4 = 8 years.
Therefore mean life of the sample is τ = 1.44 7 = 1.44 × 8 = 11.52 years.

Question 3.
A radioactive sample contains 2.2 mg of pure 116C which has a half-life period of 1224 seconds. Calculate (i) the number of atoms present initially and (ii) the activity when 5 pg of the sample will be left.
Answer:
Mass of sample = 2.2 pg
Now 11 g of the sample contains 6.023 × 10²³ nuclei, therefore the number of nuclei in 2.2 mg = 2.2 × 10^-3 g are

Question 4.
The half-life of 238 92U is 4.5 × 10⁹ years. Calculate the activity of 1 g sample of ₉₂²³⁸U.
Answer:
Given T = 4.5 × 10⁹ years.
Number of nuclei of U in 1 g
= N = \(\frac{6.023 \times 10^{23}}{238}\) = 2.5 × 10²¹

Therefore activity

Question 5.
The decay constant for a given radioactive sample is 0.3456 per day. What percentage of this sample will get decayed in a period of 4 days?
Answer:
Given λ = 0.3456 day^-1
or
T_1/2 = 0.693/λ = 0. 693/ 0.3456 = 2.894 days, t = 4 days.

Let N be the mass left behind, then N = N_oe^-λt
or
N = N_o e^{-0 3456 × 4}
or
N = N₀ e^{-1 3824} = N_o × 0.25

Therefore the percentage of undecayed is

Question 6.
It is observed that only 6.25 % of a given radioactive sample is left undecayed after a period of 16 days. What is the decay constant of this sample per day?
Answer:
Given N/N_o = 6.25 %, t = 16 days, λ = ?

Or
16/ T = 4 or T = 4 days.

Therefore λ = 1/T = 1/4 = 0.25 day^-1

Question 7.
A radioactive substance decays to 1/32^th of its initial value in 25 days. Calculate its half-life.
Answer:
Given t = 25 days, N = N_o / 32, using

Or
25/7= 5 or T= 25 / 5 = 5 days.

Question 8.
The half-life of a radioactive sample is 30 s.
Calculate
(i) the decay constant, and
Answer:
Given T₁/2 = 30 s, N = 3N_o / 4, λ = ?, t = ?
(i) Decay constant
λ = \(\frac{0.693}{T_{1 / 2}}=\frac{0.693}{30}\) = 0.0231 s^-1

(ii) time taken for the sample to decay to 3/4 th of its initial value.
Answer:
Using N = N_oe^-λt we have

Question 9.
The half-life of 14 6C is 5700 years. What does it mean?
Two radioactive nuclei X and Y initially contain an equal number of atoms. Their half-lives are 1 hour and 2 hours respectively. Calculate the ratio of their rates of disintegration after 2 hours.
Answer:
It means that in 5700 years the number of nuclei of carbon decay to half their original value.
Given N_ox = N_oY, T_X = 1 h, T_Y = 2 h, therefore
\(\frac{\lambda_{X}}{\lambda_{Y}}=\frac{2}{1}\) = 2

Now after 2 hours X will reduce to one- fourth and Y will reduce to half their original value.
If activities at t = 2 h are R_x and R_y respectively, then

Thus their rate of disintegration after 2 hours is the same.

Question 10.
A star converts all its hydrogen to helium achieving 100% helium composition. It then converts helium to carbon via the reaction.

The mass of the star is 5 × 10³² kg and it generates energy at the rate of 5 × 10³⁰ watt. How long will it take to convert all the helium to carbon at this rate?

As 4 × 10^-3 kg of He consists of 6.023 × 10²³ He nuclei so 5 × 10³² kg He will contain
\(\frac{6.023 \times 10^{23} \times 5 \times 10^{32}}{4 \times 10^{-3}}\) = 7.5 × 10⁵⁸ nuclei

Now three nuclei of helium produce 7.27 × 1.6 × 10^-13 J of energy
So all nuclei in the star will produce
E = \(\frac{7.27 \times 1.6 \times 10^{-13}}{3}\) × 7.5 × 10⁵⁸
= 2.9 × 10⁴⁶ J

As power generated is P = 5 × 10³⁰ W, therefore time taken to convert all He nuclei into carbon is
t = \(\frac{E}{P}=\frac{2.9 \times 10^{46}}{5 \times 10^{30}}\) = 5.84 × 10¹⁵ s
or
1.85 × 10⁸ years

Question 11.
Radioactive material is reduced to (1/16)^th of its original amount in 4 days. How much material should one begin with so that 4 × 10^-3 kg of the material is left after 6 days?
Answer:
N = N_o / 16, t = 4 days,
N = 4 × 10^-3 kg,
t = 6 days

To calculate half-life of the material we have

Now using the expression 4 × 10^-3 = No\(\left(\frac{1}{2}\right)^{6 / 1}\)
Solving we have N_o = 0.256 kg

Question 12.
Two different radioactive elements with half-lives T₁ and T₂ have N₁ and N₂ (undecayed) atoms respectively present at a given instant. Determine the ratio of their activities at this instant.
Answer:
The activity of a radioactive sample is given by the relation
A = – λN

Therefore the ratio of activity of these two radioactive elements is
\(\frac{A_{1}}{A_{2}}=\frac{-\lambda_{1} N_{1}}{-\lambda_{2} N_{2}}=\frac{T_{2} N_{1}}{T_{1} N_{2}}\)

Question 13.
Given the mass of the iron nucleus as 55.85 u and A = 56. Find the nuclear density? (NCERT)
Answer:
Given m_Fe = 55.85 u = 9.27 × 10^-26kg

The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these Neutron stars has been compressed to such an extent that they resemble a big nucleus.

Question 14.
We are given the following atomic masses: ₉₂²³⁸U = 238.05079 u, ₂⁴He = 4.00260 u, ₉₀²³⁴Th = 234.04363 u ₁¹H = 1.00783 u, ₉₁²³⁷Pa =237.05121 u Here the symbol Pa is for the element protactinium (Z = 91). (a) Calculate the energy released during the alpha decay of ₉₂²³⁸U. (b) Show that cannot spontaneously emit a proton. (NCERT)
Answer:
(i) The alpha decay of ₉₂²³⁸Uis given by

The energy released in this process is given by
Q= (M_u – M_Th – M_He) × 931.5 MeV

Substituting the atomic masses as given in the data we find that
Q = (238.05079 – 234.04363 – 4.00260) × 931.5 MeV ⇒ Q = 4.25 MeV.

(ii) If 29®U spontaneously emits a proton, the decay process would be

The Q for this process to happen is Q = (M_u – M_pa – M_H) × 931.5 MeV
Q = (238.05079 – 237.05121 – 1.00783) × 931.5 MeV ⇒ Q = – 7.68 MeV

Thus the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply energy of 7.68 MeV to the ₉₂²³⁸U nucleus to make it emit a proton.

Question 15.
The half-life of 90Sr is 28 years. What is the disintegration rate of 15 mg of this isotope? (NCERT)
Answer:
Given T_1/2 = 28 years, m = 15 mg
Now the rate of disintegration is given by

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 9 Ray Optics and Optical Instruments. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 9 Important Extra Questions Ray Optics and Optical Instruments

Ray Optics and Optical Instruments Important Extra Questions Very Short Answer Type

Question 1.
When light undergoes refraction at the surface of separation of two media, what happens to its frequency/wavelength?
Answer:
There is no change in its frequency, but its wavelength changes.

Question 2.
Define the refractive index.
Answer:
The Refractive index of a medium is defined as the ratio of the speed of light in a vacuum to the speed of light in the given medium.

Question 3.
What is the distance between the objective and eyepiece of an astronomical telescope in its normal adjustment?
Answer:
Distance between objective and eyepiece of telescope = f_o + f_e

Question 4.
Name the phenomenon responsible for the reddish appearance of the sun at sunrise and sunset.
Answer:
Atmospheric refraction.

Question 5.
What are the two main considerations that have to be kept in mind while designing the ‘objective’ of an astronomical telescope?
Answer:
Two main considerations are

1. Large light gathering power
2. Higher resolution (or resolving power)

Question 6.
Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid? (CBSE Delhi 2012)
Answer:
When the refractive index of the liquid is equal to the refractive index of a glass of which the lens is made.

Question 7.
Write the relationship between the angle of incidence ‘i’, angle of prism ‘A’ and angle of minimum deviation for a triangular prism. (CBSE Delhi 2013)
Answer:
2i = A + δ_m

Question 8.
Why can’t we see clearly through the fog? Name the phenomenon responsible for it. (CBSE Al 2016)
Answer:
Because it scatters light. Scattering of light.

Question 9.
How does the angle of minimum deviation of a glass prism vary if the incident violet light is replaced by red light? Give reason. (CBSE AI 2017)
Answer:
It decreases as δ_m ∝ \(\frac{1}{λ}\)

Question 10.
The objective lenses of two telescopes have the same apertures but their focal lengths are in the ratio 1: 2. Compare the resolving powers of the two telescopes. (CBSE AI 2017C)
Answer:
Same as resolving power does not depend upon the focal length of lenses.

Question 11.
An object is kept in front of a concave lens. What is the nature of the image formed? (CBSE Delhi 2017C)
Answer:
Virtual and erect.

Question 12.
The refractive index of the material of a concave lens is n₁. It is immersed in a medium of refractive index n₂. A parallel beam of tight is incident on the lens. Trace the path of the emergent rays when n₂ > n₁.
Answer:
The path of rays is as shown

Question 13.
A convex lens made of glass of refractive index μ_L is immersed in a medium of refractive index μ_m. How will the lens behave when μ_L < μ_m?
Answer:
The lens will continue to behave as a convex lens.

Question 14.
The image of an object formed by a lens on the screen is not in sharp focus. Suggest a method to get clear focusing of the image on the screen without disturbing the position of the object, the lens or the screen.
Answer:
Limit the field of view of the lens by using a blackened glass having a small circular hole in the middle.

Question 15.
In the figure given below, the path of a parallel beam of light passing through a convex lens of refractive index ng kept in a medium of refractive index nm is shown. Is (i) n_g = n_m or (ii) n_g > n_m, or (iii) n_g < n_m?

Answer:
As the rays of light do not suffer any deviation, therefore n_g = n_m.

Question 16.
In the figure path of a parallel beam of light passing through a convex lens of refractive index ng kept in a medium of refractive index, nm is shown. Is (i) n_g = n_m or (ii) n_g > n_m, or (iii) n_g < n_m?

Answer:
As the rays of light diverge, therefore n_g < n_m.

Question 17.
The refractive index of the material of a concave lens is μ₁. It is immersed in a medium of refractive index μ₂. A parallel beam of light is incident on the lens. Trace the path of emergent rays when μ₂ < μ₁.
Answer:
The path of rays is as shown below.

Question 18.
Suppose that the lower half of the concave mirror’s reflecting surface is covered with an opaque (non-reflective) material. What effect will this have on the image of an object placed in front of the mirror?
Answer:
The image of the whole object will be formed. However, as the area of the reflecting surface has been reduced the intensity of the image will below (in this case, half).

Question 19.
For the same value of angle of incidence, the angles of refraction in three media A, B and C are 15°, 25° and 35° respectively. In which medium would the velocity of light be minimum? (CBSE Al 2012)
Answer:
Medium A.

Question 20.
When red light passing through a convex lens is replaced by the light of blue colour, how will the focal length of the lens change? (CBSE AI 2013C)
Answer:
It will decrease.

Question 21.
A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason. (CBSE AI 2014)
Answer:
The diverging lens as its focal length will become negative.

Question 22.
A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens? (CBSE Delhi 2015)
Answer:
It will behave as a convex lens.

Question 23.
Why can’t we see clearly through a fog? Name the phenomenon responsible for it. (CBSE AI 2016)
Answer:
Because it scatters light. Scattering of light.

Question 24.
How does the angle of minimum deviation of a glass prism vary if the incident violet light is replaced by red light? Give reason. (CBSE AI 2017)
Answer:
It decreases as δ_m ∝ \(\frac{1}{λ}\)

Question 25.
The objective lenses of two telescopes have the same apertures but their focal lengths are in the ratio 1: 2. Compare the resolving powers of the two telescopes. (CBSE AI 2017C)
Answer:
Same as resolving power does not depend upon the focal length of lenses.

Question 26.
An object is kept in front of a concave lens. What is the nature of the image formed? (CBSE Delhi 2017C)
Answer:
Virtual and erect.

Question 27.
A convex lens is held in water. What change, if any, do you expect In its focal length?
Answer:
Focal Length of the given Lens increases in accordance with tens maker’s formula
\(\frac{1}{f}\) = (μ – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

This is because _wμ_{g a}μ_g

Question 28.
Violet light is incident on a converging lens of focal length f. State with reason, how the focal length of the lens will change 1f the violet light is replaced by a red light.
Answer:
As μ_r < μ_v hence in accordance with the relation
\(\frac{1}{f}\) = (μ – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) focal Length of Lens for red colour will be more.

Question 29.
What is the minimum value of the refractive index of the prism shown in the figure below?

Answer:
μ = \(\frac{1}{\sin i_{c}}=\frac{1}{\sin 45^{\circ}}=\frac{1}{\sqrt{2}}\)

Question 30.
How does the resolving power of a telescope change on decreasing the aperture of its objective lens? Justify your answer.
Answer:
As the resolving power of a telescope is \(\frac{D}{1.22λ}\) hence on decreasing the aperture of its objective lens, the resolving power of the telescope decreases in the same ratio.

Question 31.
What will happen to a ray of light incident normally on the interface of air and glass?
Answer:
It will pass un-deviated into the glass.

Question 32.
What is the speed of light in glass having a refractive index of 1.5?
Answer:
Speed of light in glass v = c/n = 3 × 10⁸ / 1.5 = 2 × 10⁸ m s^-1

Question 32.
Light of wavelength 600 nm in air enters a medium of refractive index 1.5. What will be its frequency in the medium?
Answer:
Frequency does not change when light moves from one medium into another. Therefore frequency of light
v = \(\frac{c}{\lambda}=\frac{3 \times 10^{8}}{600 \times 10^{-9}}\) = 5 × 10¹⁴ Hz

Question 34.
How does the focal length of a convex lens change if monochromatic red light Is used instead of monochromatic blue
light?
Answer:
Focal Length increases, i.e. f_r > f_v.

Question 35.
Two thin lenses of power + 5 D and – 3 D are in contact. What Is the focal length of the combination?
Answer:
Here P = P₁ + P₂ = + 5 – 3 = + 2 D
Hence f = 1/P = ½ m = + 50 cm.

Question 36.
Use the maker’s formula to write an expression for the (relative) refractive index, μ, of the material in terms of its focal length, f and the radii of curvature, r₁ and r₂, of Its two surfaces.
Answer:
The formula is \(\frac{1}{f}\) = (μ – 1)\(\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)\)

Question 37.
Two thin lenses of power – 4 D and 2 D are placed in contact coaxially. Find the focal length of the combination. (CBSE Ai 2012C)
Answer:
Total power P = – 4 + 2 = – 2 D
Now f = 1/P = 1/ -2 = – 0.5 m

Question 38.
A convex lens is placed In contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens? (CBSE Delhi 2014)
Answer:
20 cm

Question 39.
The focal length of a biconvex lens is equal to the radius of curvature of either face. What is the refractive index of the material of the lens? (CBSEAI 2015)
Answer:
1.5
Using \(\frac{1}{f}\) = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

Given R₁ = R

Question 40.
An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the ens are reversed? (NCERT Exemplar)
Answer:
No, the reversibility of the Lens does not change the lens formula.

Ray Optics and Optical Instruments Important Extra Questions Short Answer Type

Question 1.
The aperture of the objective lens of an astronomical telescope is doubled. How does it affect
(i) the resolving power of the telescope and
(ii) the intensity of the image? (CBSE Sample Paper 2018-19)
Answer:
The resolving power of a telescope is given by the expression \(\frac{D}{1.22λ}\).

(i) When the aperture of the objective lens is increased, the resolving power of the telescope increases in the same ratio.
(ii) The intensity of the image is given by the expression β ∝ D₂, thus when the aperture is doubled, the intensity of the image becomes four times.

Question 2.
How does the resolving power of a compound microscope change on (a) decreasing the wavelength of light used, and (b) decreasing the diameter of the objective lens?
Answer:
The resolving power of a microscope is given by the expression RP = \(\frac{2 n \sin \theta}{\lambda}\)

(a) If the wavelength of the incident tight is decreased, the resolving power of the microscope increases.
(b)There is no effect of the decrease in the diameter of the objective on the resolving power of the microscope.

Question 3.
The layered lens shown in the figure is made of two kinds of glass. How many and what kinds of images will be produced by this lens with a point source placed on the optic axis? Neglect the reflection of light at the boundaries between the layers.

Answer:
Two images will be formed as the lens may be thought of, as two separate lenses of different focal lengths. The images will be surrounded by bright halos.

Question 4.
Monochromatic light is refracted from air into a glass of refractive index n. Find the ratio of wavelengths of the incident and refracted light.
Answer:
Using the relation λ₁n₁ = λ₂n₂ we have
\(\frac{\lambda_{1}}{\lambda_{2}}\) = n

Question 5.
Draw a labelled ray diagram to show the image formation in a compound microscope.
Answer:
The labelled diagram is as shown.

Question 6.
A ray of light while travelling from a denser to a rarer medium undergoes total internal reflection. Derive the expression for the critical angle in terms of the speed of light in the two media.
Answer:
Snell’s law can be used to find the critical angle. Now Snell’s law, when the ray moves from denser medium ‘b’ to rarer medium ‘a’, is given by


Now we know that n = \(\frac{c}{v}\) , substituting in the above relation we have
\(\frac{c}{v}=\frac{1}{\sin i_{c}}\) or sin i_c = \(\frac{v}{c}\)

Question 7.
Draw a labelled diagram for a refracting type astronomical telescope. How will its magnifying power be affected by increasing for its eyepiece (a) the focal length and (,b) the aperture? Justify your answer. Write two drawbacks of refracting type telescopes. (CBSE Sample Paper 2018-19)
Answer:
The labelled diagram of the telescope is as shown in the figure.

(a) The magnifying power of a telescope is given by M = \(\frac{f_{0}}{f_{\mathrm{e}}}\). If the focal length of the eyepiece is increased, it will decrease the magnifying power of the telescope.

(b) Magnifying power does not depend upon the aperture of the eyepiece. Therefore there is no change in the magnifying power if the aperture of the eyepiece is increased.

Drawbacks:

• Large-sized lenses are heavy and difficult to support.
• Large-sized lenses suffer from chromatic and spherical aberration.

Question 8.
Draw a labelled ray diagram of a Newtonian type reflecting telescope. Write any one advantage over refracting type telescope.
Answer:
The labelled diagram is shown below.

Due to the large aperture of the mirror as compared to a lens the image formed is much brighter than that formed by a refracting type telescope.

Question 9.
A right-angle crown glass prism with a critical angle of 41° is placed between the object PQ in two positions as shown in figures (a) and (b). Trace the path of rays from P and Q passing through the prisms in the two cases.

Answer:
The path of rays is as shown.


Question 10.
Write two conditions necessary for total internal reflection to take place.
Answer:
(a) The incident ray should travel from the denser to the rarer medium.
(b) The angle of incidence, in the denser medium, should be greater than the critical angle for the given pair of media.

Question 11.
(a) Explain the working of a compound microscope with the help of a labelled diagram.
Answer:
A compound microscope is an instrument used to see highly magnified images of tiny objects.

Working:
Let a tiny object AB be placed in front of the objective lens at a distance more than F₀. Its real and enlarged image is formed at A ‘ B ‘. The image A’ B’ acts as an object for the eyepiece and forms the final image at A” B “,

i.e. at a distance D, the least distance of distinct vision.

(b) Write the considerations that you keep in mind while choosing lenses to be used as eyepiece and objective in a compound microscope. (CBSE 2019C)
Answer:
The objective and eyepiece should have a short focal length for large magnification.

Question 12.
Explain the working of a refracting telescope with the help of a labelled diagram. What are the main limitations of this type of telescope and how are these overcome in a reflecting telescope? (CBSE 2019C)
Answer:
Refracting telescope

Working:
When the rays of light are made to incident on the objective from a distant object, the objective forms the real and inverted image at its focal plane. The lens is so adjusted that the final image is formed at least distance of distinct vision or at infinity.

Limitations:

• Large-sized lenses are needed which are expensive.
• Large-sized lenses suffer from spherical aberration and distortions.

Reflecting telescope (To overcome limitations):

• Reflecting telescopes are free from chromatic aberration and spherical aberration is very small.
• They are less heavy and easier to support.

Question 13.
Explain why the colour of the sky is blue.
Answer:
As sunlight travels through the earth’s atmosphere, it gets scattered (changes its direction) by the atmospheric particles. Light of shorter wavelengths is scattered much more than light of longer wavelengths. (The amount of scattering is inversely proportional to the fourth power of the wavelength. This is known as Rayleigh scattering.)

Hence, the bluish colour predominates in a clear sky, since blue has a shorter wavelength than red and is scattered much more strongly. In fact, violet gets scattered even more than blue, having a shorter wavelength. But since our eyes are more sensitive to blue than violet, we see the sky blue.

Question 14.
Why does the sun look reddish during sunrise and sunset?
Answer:
At sunset or sunrise, the sun’s rays have to pass through a larger distance in the atmosphere (Figure). Most of the blue and other shorter wavelengths are removed by scattering. The least scattered light reaching our eyes, therefore, the sun looks reddish. This explains the reddish appearance of the sun near the horizon.

Question 15.
What are the two ways of adjusting the position of the eyepiece while observing the final image in a compound microscope? Which of these is usually preferred and why?
Answer:
The two ways
(a) Final image formed at least distance of distinct vision.
(b) Final image formed at infinity.

The second one is usually preferred as it helps the observer to observe the final image with his/her eye in a relaxed position.

Question 16.
Write the relation between the angle of incidence (i), the angle of emergence (e), the angle of the prism (A) and the angle of deviation (δ) for rays undergoing refraction through a prism. What is the relation between ‘i’ and ‘e’ for rays undergoing minimum deviation? Using this relation obtain an expression for the refractive index (μ) of the material of the prism in terms of ‘A’ and angle of minimum deviation.
Answer:
The relation is i + e = A + δ_m
In the minimum deviation position ∠i = ∠e
In the minimum deviation position we have A = r₁ + r₂ = r + r = 2r
or
r = A / 2
and i + i = A + δ_m
or
2i = A + δ_m
or
i = \(\frac{A+\delta_{m}}{2}\)

substituting for i and r in the expression for Snell’s law we have
μ = \(\frac{\sin i}{\sin r}=\frac{\sin \left[\frac{A+\delta_{m}}{2}\right]}{\sin \left(\frac{A}{2}\right)}\)

Question 17.
Draw a ray diagram showing the formation of the image by a concave mirror of an object placed beyond its centre of curvature. If the lower half of the mirror’s reflecting surface is covered, what effect will it have on the image? (CBSE AI 2011C)
Answer:
The required ray diagram is shown below.

When the lower half of the mirror’s reflecting surface is covered, the intensity of the image will be reduced.

Question 18.
An object AB is kept in front of a concave mirror as shown in the figure.

(a) Complete the ray diagram showing the image formation of the object.
(b) How will the position and intensity of the image be affected if the lower half of the mirror’s reflecting surface is painted black? (CBSE AI 2012)
Answer:
The completed ray diagram is as shown.

No change in the position of the image, but the intensity will decrease.

Question 19.
Draw a ray diagram to show the formation of the image by an astronomical telescope when the final image is formed at the near point. Answer the following, giving reasons
(a) Why the objective has a larger focal length and a larger aperture than the eyepiece?
(b) What would be the effect on the resolving power of the telescope if its objective lens is immersed in a transparent medium of the higher refractive index? (CBSE AI 2012C)
Answer:
The ray diagram is as shown.

(a) The objective lens has a large focal length and large aperture for large magnification and resolving power.
(b) The resolving power will increase as the wavelength of light used will decrease.

Question 20.
Two convex lenses of the same focal length but of aperture A₁ and A, (A, < A₁), are used as the objective lenses in two astronomical telescopes having identical eyepieces. What is the ratio of their resolving power? Which telescope will you prefer and why? Give reason. (CBSE Delhi 2011)
Answer:
The resolving power is directly proportional to the aperture. Therefore the ratio of their resolving power is RP = \(\frac{A_{1}}{A}\).

Since A₁ > A therefore we will prefer the telescope with aperture A1 as it will gather a larger amount of light than the telescope of aperture A.

Question 21.
Which two of the following lens L₁ L₂ and L₃ will you select as objective and eyepiece for constructing the best possible (a) telescope and (b) microscope. Give a reason to support your answer. (CBSE Delhi 2015C)

Lens	Power	Aperture
L1	6 D	1 cm
L2	3 D	8 cm
L3	10 D	1 cm

Answer:
(a) For telescope

• Objective lens L₂: Because it has a large focal length and large aperture
• Eye lens L₃: Because it has a small focal length and small aperture

(b) For Microscope

• Objective lens L₃: Because it has a small aperture and small focal length
• Eye lens L₃: because it has a large aperture and large focal length

Question 22.
Draw a ray diagram to show how a right-angled isosceles prism may be used to bend the path of light by 90°.
Write the necessary condition in terms of the refractive index of the material of this prism for the ray to bend by 90°. (CBSE Delhi 2016C)
Answer:
The ray diagram is as shown.

The angle of incidence in the denser medium should be greater than the critical angle for the given pair of media, i.e.

Question 23.
The image of a candle is formed by a convex lens on a screen. The lower half of the lens is painted black to make it completely opaque. Draw the ray diagram to show the image formation. How will this image be different from the one obtained when the lens is not painted black?
Answer:
The ray diagram depicting the image is shown

The image formed will be less bright as compared to that when half of the lens is not painted black. This is because every part of the lens forms the image. When the lower half is blackened, light from this portion will be blocked, hence the intensity of light in the image will be less.

Question 24.
A magician during a show makes a glass lens with n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water?
Answer:
The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear. This means n₁ = n₂. This gives 1 /f = 0 or f = ∞. The lens in the liquid will act as a plain sheet of glass. No, the liquid is not water, it could be glycerine.

Question 25.
Explain with reason, how the resolving power of an astronomical telescope will change, when
(a) frequency of the incident light on the objective lens is increased,
(b) the focal length of the objective lens is increased, and
(c) the aperture of the objective lens is halved?
Answer:
Resolving power of a telescope = \(\frac{D}{1.22 \lambda}\) hence
(a) on increasing the frequency of incident light, wavelength X decreases and consequently resolving power is increased.
(b) increase in focal length of the objective lens will have no effect on the resolving power, and
(c) if the aperture of the objective lens is halved, then resolving power is also halved.

Question 26.
Explain with reason, how the resolving power of a compound microscope will change when
(a) frequency of the incident light on the objective lens is increased,
(b) the focal length of the objective lens is increased, and
(c) the aperture of the objective lens is increased.
Answer:
Resolving power of a compound microscope RP = \(\frac{2 \mu \sin \theta}{\lambda}\), hence
(a) on increasing the frequency of incident light, its wavelength X decreases and consequently resolving power increases.
(b) on increasing the focal length of the objective lens, the value of sine and hence resolving power decreases.
(c) on increasing the aperture of the objective lens, the value of sin 0 and hence resolving power increases.

Question 27.
A convex lens of the focal length is kept in contact with a concave lens of focal length f₂. Find the focal length of the combination. (CBSE AI 2013)
Answer:
For a convex lens +f₁ and for a concave lens – f₂, using the expression
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)
or

Question 28.
(a) The relation between the angle of incidence ‘i’ and the corresponding angle of deviation (δ), for a certain optical device, is represented by the graph shown in the figure. Identify this device. Draw a ray diagram for this device and use it for obtaining an expression for the refractive index of the material of this device in terms of an angle characteristic of the device and the angle, marked as δ_m, in the graph. (CBSE Al 2016C)

Answer:
(a) The device is a prism.
The relation is i + e = A + δ_m
In the minimum deviation position ∠i = ∠e
In the minimum deviation position we have A = r₁ + r₂ = r + r = 2r
or
r = A / 2
and i + i = A + δ_m
or
2i = A + δ_m
or
i = \(\frac{A+\delta_{m}}{2}\)

substituting for i and r in the expression for Snell’s law we have
μ = \(\frac{\sin i}{\sin r}=\frac{\sin \left[\frac{A+\delta_{m}}{2}\right]}{\sin \left(\frac{A}{2}\right)}\)

(b) The diagram is as shown.

Ray Optics and Optical Instruments Important Extra Questions Long Answer Type

Question 1.
Draw a labelled ray diagram to show the image formation in a refracting type astronomical telescope. Obtain an expression for the angular magnifying power and the length of the tube of an astronomical telescope in its ‘normal adjustment’ position. Why should the diameter of the objective of a telescope be large?
Answer:
A labelled diagram of the telescope is shown in the figure.

The object subtends an angle at the objective and would subtend essentially the same angle at the unaided eye. Also, since the observers’ eye is placed just to the right of the focal point f’₂, the angle subtended at the eye by the final image is very nearly equal to the angle β.

Therefore, M = \(\frac{\beta}{\alpha}=\frac{\tan \beta}{\tan \alpha}\) …(1)

From right triangles ABC and ABC’ as shown in figure, we have
tan α = \(\frac{\mathrm{AB}}{\mathrm{CB}}=\frac{-h}{f_{0}}\) and
tan β = \(\frac{\mathrm{AB}}{\mathrm{C}^{\prime} \mathrm{A}}=\frac{-h}{f_{\mathrm{e}}}\)

substituting the above two equations in equation (1), we have
M = \(\frac{\beta}{\alpha}=\frac{-h^{\prime}}{f_{e}} \times \frac{f_{0}}{-h^{\prime}}=\frac{f_{0}}{f_{e}}\)

The length of the telescope is the distance between the two lenses which is L = f_o + f_e The diameter of the objective of a telescope should be large so that it can collect more light and image of distant objects is formed clear.

Question 2.
Draw a ray diagram to show the formation of an erect image of an object kept in front of a concave mirror. Hence deduce the mirror formula. (CBSE 2019C)
Answer:
An object AB is placed between P and F. The course of rays for obtaining erect image A₁B₁ of object AB is shown in the figure.

Draw DG ⊥ on the principal axis.
Triangles DGF and A₁ B₁C are similar

Since Point G is close to P, so GF = PF

Multiplying and dividing both sides by uvf, we get
\(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)

Question 3.
(a) Define the term power of a lens. Write its SI units.
Answer:
The power of a lens is the reciprocal of its focal length in the metre.
i.e. P = \(\frac{1}{f(\text { in } m)}\)

S.I. unit of power of a lens is dioptre (D).

(b) Derive the expression for the power of two thin lenses placed coaxially in contact with each other. (CBSE 2019C)
Answer:
Power of two thin lenses in contact. Consider an object O placed at a distance u on the principal axis of the lens A. Rays of light starting from O forms, the image at I₁
∴ \(\frac{1}{v_{1}}-\frac{1}{u}=\frac{1}{f_{1}}\) …(1)

Place lens B in contact with A. The image I₁ will serve as a virtual object and forms a real image I.
∴ Again apply lens formula
\(-\frac{1}{v_{1}}+\frac{1}{v}=\frac{1}{f_{2}}\) …(2)

Adding (1) and (2), we have

where f is the focal length of the combination

From (3), \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

The total power of the lens combination is given by
P = P₁ + P₂

Question 4.
A biconvex lens of refractive index μ1 focal length ‘f and radius of curvature R is immersed in a liquid of refractive index μ₂. For (i) μ₂ > μ₁, and (ii) μ₂ < μ₁, draw the ray diagrams in the two cases when a beam of light coming parallel to the principal axis is incident on the lens. Also, find the focal length of the lens in terms of the original focal length and the refractive index of the glass of the lens and that of the medium. (CBSE AI 2013C)
Answer:

(b) Given: As the lens is equiconvex therefore R₁ = R₂ = R, _aμ_g = μ₁; _aμ₁ = μ₂ f_a = f, f_L = ?

Using the expression

Question 5.
Write the conditions for observing a rainbow. Show, by drawing suitable diagrams, how one understands the formation of a rainbow. (CBSE AI 2014C)
Answer:
The conditions for observing a rainbow are that the sun should be shining in one part of the sky (say near the western horizon) while it is raining in the opposite part of the sky (say the eastern horizon). An observer can therefore see a rainbow only when his back is towards the sun.

Sunlight is first refracted as it enters a raindrop, which causes the different wavelengths (colours) of white light to separate. The longer wavelength of light (red) is bent the least while the shorter wavelength (violet) are bent the most. Next, these component rays strike the inner surface of the water drop and get internally reflected if the angle between the refracted ray and normal to the drop surface is greater than the critical angle (48°, in this case).

The reflected light is refracted again as it comes out of the drop as shown in the figure. It is found that the violet light, emerges at an angle of 40° related to the incoming sunlight and red light emerges at an angle of 42°. For other colours, angles lie in between these two values.

Question 6.
Explain the basic differences between the construction and working of a telescope and a microscope. (CBSE AI 2015)
Answer:
(a) In a telescope the objective has a large aperture and a large focal length, while in a microscope both the aperture and focal length of the objects are small (f_e > f_o). In a telescope, the eyepiece has a small aperture as compared to the objective, while in a microscope the eyepiece has a bigger aperture than the objective (f_e < f_o).

(b) A telescope increases the angle the object subtends at the eye thereby increasing its clarity by bringing it closer, while a microscope actually magnifies the object or a telescope magnifies distant objects, while a microscope magnifies nearby objects.

Question 7.
Why does white light disperse when passed through a glass prism?
Using the lens maker’s formula show how the focal length of a given lens depends upon the wavelength of light incident on it. (CBSE Delhi 2015C)
Answer:
It is because a glass prism offers a different refractive index to different wavelengths of light.
Lens maker’s formula is \(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\).

The refractive index depends upon wavelength as n ∝ \(\frac{1}{λ}\), therefore f ∝ λ.

Hence the focal length of a lens increases with the increase in the wavelength of light.

Question 8.
A thin converging lens has a focal length f in air. If it is completely immersed in a liquid, briefly explain, how the focal length of the lens will vary?
Answer:
The focal length of a converging lens is given by
\(\frac{1}{f}\) = (_an_g – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
where ang = refractive index of Lens with respect to air.

And In a given liquid medium the focal Length of the Lens is given by
\(\frac{1}{f_{m}}\) = (_mn_g – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

Therefore the ratio of the totaL focal Length is

(a) If _mn_g < _an_g then f is +ve and has a value greater than f, i.e. focal length of the lens increases when immersed in a given liquid medium.

(b) If _mn_g >_an_g, then f is -ve, i.e. the lens will begin to behave as a diverging lens.

Question 9.
When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why?
Answer:
Reflection and refraction arise through the interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators that take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus the frequency of scattered light equals the frequency of incident light.

Question 10.
The image of an object formed by the combination of a convex lens (of focal length f) and a convex mirror (of the radius of curvature R), set up as shown is observed to coincide with the object. Redraw this diagram to mark on it the position of the centre of curvature of the mirror. Obtain the expression for R in terms of the distances marked as ‘a’ and ‘d’ and focal length f of the convex lens. (CBSE Delhi 2016C)

Answer:
The final image, formed by the combination, is coinciding with the object itself. This implies that the rays, from the object, are retracing their path, after refraction from the tens and reflection from the mirror. The (refracted) rays are, therefore, fatling normally on the mirror. It follows that the rays A B, and A’ B’, when produced, are meeting at the centre of curvature C of the mirror. Hence C is the centre of curvature of the mirror.

From the figure, we then see that for the convex lens we have

u = – a and v = + (d + R). If f is the focal length of the lens, we have


Question 11.
Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope have short focal lengths? Explain. (CBSE Delhi 2017)
Answer:
It is defined as the ratio of the angle subtended on the eye by the final image when it lies at infinity to the angle subtended on the eye by the object when it lies at the least distance of distinct vision from the eye or the near point of the eye. The magnifying power of a microscope is given by
M = M_e × M_o = \(\frac{L}{f_{0}} \times\left(1+\frac{D}{f_{e}}\right)\)

Since it depends inversely on the focal lengths of the objective and the eyepiece, therefore both have to be small to get a large magnification.

Question 12.
(a) Draw a ray diagram for the formation of the image by a compound microscope.
Answer:
The ray diagram is as shown.

(b) You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct a compound microscope?

Lenses	Power(D)	Aperture(cm)
L1	3	8
L2	6	1
L3	10	1

Answer:
The objective should have a short focal length and large aperture while the eyepiece should have a short focal length and small aperture. Thus for objective lens L₃ should be used and for eyepiece lens, L₂ should be used.

(c) Define the resolving power of a microscope and write one factor on which it depends. (CBSE AI 2017)
Answer:
The resolving power of a microscope is given by the expression
RP = \(\frac{2 \mu \sin \theta}{1.22 \lambda}=\frac{2 \mu v \sin \theta}{1.22 c}\)

It depends upon the wavelength/ frequency of the incident light.

Question 13.
Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. (CBSE Delhi 2019)
A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is 3.48 × 10⁶ m, and the radius of the lunar orbit is 3.8 × 10⁸ m.
Answer:
A Labelled ray diagram is as shown.

Given f_o = 15 m, f_e = 1.0 cm = 0.01 m, M = ? D_m = 3.48 × 10⁶ m, r = 3.8 × 10⁸ m,
Using M = \(\frac{f_{0}}{f_{e}}=\frac{15}{0.01}\) = 1500

The angle subtended by the moon at the objective of the telescope

Question 14.
(a) Amobilephoneliesalongtheprincipal axis of a concave mirror. Show with the help of a suitable diagram the formation of its image. Explain why magnification is not uniform.
Answer:
The image of the mobile phone formed by the concave mirror is shown in the below figure.

The part of the mobile phone that is at C will form an image of the same size only at C. In the figure, we can see that B’C = BC. The part of the mobile phone that lies between C and F will form an enlarged image beyond C as shown in the figure. It can be observed that the magnification of each part of the mobile phone cannot be uniform on account of different locations. That is why the image formed is not uniform.

(b) Suppose the lower half of the concave mirrors reflecting surface is covered with an opaque material. What effect this will have on the image of the object. Explain. (CBSE Delhi 2014)
Answer:
The intensity of the image will decrease while the whole image will be for.

Question 15.
(a) Draw a labelled ray diagram of a compound microscope, when the final image forms at the least distance of distinct vision.
(b) Why is its objective of short focal length and of short aperture compared to its eyepiece? Explain.
(c) The focal length of the objective is 4 cm while that of the eyepiece is 10 cm. The object is placed at a distance of 6 cm from the objective lens.
(i) Calculate the magnifying power of the compound microscope, if its final image is formed at the near point.
(ii) Also calculate the length of the compound microscope. (CBSE AI 2019)
Or
(a) With the help of a labelled ray diagram, explain the construction and working of a Cassegrain reflecting telescope.
(b) An amateur astronomer wishes to estimate roughly the size of the sun using his crude telescope consisting of an objective lens of focal length 200 cm and an eyepiece of focal length 10 cm. By adjusting the distance of the eyepiece from the objective, he obtains an image of the sun on a screen 40 cm behind the eyepiece. The diameter of the sun’s image is measured to be 6-0 cm. Estimate the sun’s size, given that the average earth-sun distance is 1.5 × 10¹¹ m.
Answer:
(a) The diagram is as shown.

(b) The magnifying power of compound microscope m = m₀ × m_e = \(\frac{L}{f_{o}}\left(1+\frac{D}{f_{e}}\right)\)

To have high magnifying power and high resolution, the focal length of the objective and its aperture should be short.

The focal length of the eyepiece is comparatively greater than the objective so an image formed by the objective lens may form within the focal length of the eyepiece and the final magnified image may be formed.

Aperture is short for higher resolution,

(c) Given u_o = – 6 cm
For objective lens we have

(i) For magnifying power of compound microscope we have
m = \(\frac{v_{o}}{u_{o}}\left(1+\frac{D}{f_{e}}\right)=\frac{12}{6}\left(1+\frac{25}{10}\right)\) = 7

(ii) Length of the compound microscope L = v₀ + u_e = 12 + 7.14 = 19.14 cm
Or
(a) Cassegrain reflecting telescope consists of a large concave (primary) parabolic mirror having a hole in its centre. There is a small convex (secondary) mirror near the focus of the concave mirror. The eyepiece is placed near the hole of the concave mirror. The parallel rays from a distant object are reflected by the large concave mirror. These rays fall on the convex mirror which reflects these rays outside the hole. The final magnified image is formed at infinity.

The diagram is as shown.

(b)

Magnification produced by the eye piece is
me = \(\frac{v_{e}}{u_{e}}=\frac{40}{40 / 3}\) = 3

Diameter of the image formed by the objective is
D = 6/3 = 2 cm

If D is the diameter of the sun then the angle subtended by it on the objective will be
α = \(\frac{D}{1.5 \times 10^{11}}\) rad

Now, angle subtended by the image at the objective = angle subtended by the sun
α = \(\frac{\text { size of image }}{f_{0}}=\frac{2}{200}=\frac{1}{100}\) rad

Therefore,
\(\frac{D}{1.5 \times 10^{11}}=\frac{1}{100}\)
D = 1.5 × 10⁹

Question 16.
A convex lens is placed in contact with a plane mirror. An axial point object, at a distance of 20 cm from this combination, has its image coinciding with itself. What is the focal length of the convex lens?
Answer:
Figure (a) shows a convex lens L in contact with a plane mirror M. P is the point object, kept in front of this combination at a distance of 20 cm, from it. Since the image of the object is coinciding with the object itself, the rays from the object, after refraction from the lens, should fall normally on the mirror M, so that they retrace their path and form an image coinciding with the object itself.

This will be so, if the incident rays from P form a parallel beam perpendicular to M, after refraction from the lens. For clarity, M has been shown at a finite distance from L, in figure (b). For lens L, since the rays from P form a parallel beam after refraction, P must be at the focus of the lens. Hence the focal length of the lens is 20 cm.


Question 17.
A convex lens and a convex mirror (of the radius of curvature 20 cm) are placed co-axially with the convex mirror placed at a distance of 30 cm from the lens. For a point object, at a distance of 25 cm from the lens, the final image, due to this combination, coincides with the object itself. What is the focal length of the convex lens?
Answer:
The ray diagram is as shown.

The final image, formed by the combination, is coinciding with the object itself. This implies that the rays, from the object, are retracing their path, after refraction from the lens and reflection from the mirror. The (refracted) rays are, therefore, falling normally on the mirror. It follows that the rays A B, and A’ B’, when produced, are meeting at the centre of curvature C of the mirror. Hence O₂C = 20 cm, the radius of curvature of the mirror. From the figure, we then see that for the convex lens u = – 25 cm and v = + (30 + 20) cm = + 50 cm. If f is the focal length of the lens, we have
\(\frac{1}{50}-\frac{1}{(-25)}=\frac{1}{f}\)
or
f = 16.67 cm

Question 18.
Derive a relation between focal length, image distance and object distance for a
concave mirror.
Answer:
Consider a concave mirror of small aperture, image distance and object distance for a Figure below shows two rays of light leaving the tip of the object. One of these rays passes through the centre of curvature C of the mirror, hitting the mirror head-on and reflecting back on itself. The second ray BP strikes the mirror at its pole P and reflects as shown obeying the laws of reflection. The image of the tip of the arrow is located at the point where these two rays intersect. Thus the image is formed between the focus and centre of curvature.

Consider the two right-angled triangles BAP and B’A’P

Now from triangle BAP using trigonometry, we have
tan θ = \(\frac{AB}{PA}\) …(1)

Therefore equation (7) becomes
\(\frac{\mathrm{PA}}{\mathrm{PA}^{\prime}}=\frac{\mathrm{PA}-\mathrm{PC}}{\mathrm{PC}-\mathrm{PA}^{\prime}}\)

By Cartesian sign conventions we have PA = -u, PA’ = -v and PC = – R, substituting in equation (8), we have

simplifying we have 2uv = vR + uR dividing both sides by uvR we have
\(\frac{1}{u}+\frac{1}{v}=\frac{2}{R}\) …(11)

but R = 2 f, where f is the focal length of the mirror. Therefore the above equation
reduces to \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)
which is the required mirror formula.

Question 19.
With the help of a suitable ray diagram, derive a relation between the object distance (u) image distance v and radius of curvature (R) for a convex spherical surface, when a ray of light travels from rarer to denser medium. (CBSE Delhi 2011C)
Answer:
Let a convex spherical surface XPY separate a rarer medium of refractive index (n₁) from a denser medium of refractive (n2). The real image of object 0 is formed by refraction from the convex spherical surface of radius of curvature R. The angle α β γ are shown in the figure.

Let ∠AOP = α, ∠AIP = β and ∠ACP = γ.

From point A, drop AN perpendicular to the principal axis of the spherical refracting surface. From triangle AOC, we have
i = α + γ

Since the aperture of the spherical refracting surface is small, point A will be close to point P and hence angles α, β and γ will be small. As such, these angles may be replaced by their tangents.

Therefore, equation (1) may be written as
i = tanα + tanγ …(2)

From right angled triangles ΔALO and ΔALC, we have

tan α = \(\frac{AL}{LO}\) and tan γ = \(\frac{AL}{LC}\)

Substituting for tan α and tan γ in equation (2), we have

i = \(\frac{AL}{LO}\) + \(\frac{AL}{LC}\) …(3)

Again, as aperture of the refracting surface is small, point L will be close to point P, the pole of the refracting surface.
Therefore,
LO ≈ PO and LC ≈ PC

Therefore, equation (3) becomes
i = \(\frac{AL}{PO}\) + \(\frac{AL}{PC}\) …(4)

Now, from triangle ΔACI, γ = r + β
or
r = γ— β

Since angles γ and β are small, we have
r = tan γ— tan β …(5)

From right angled triangles ΔALC and ΔALI,
we have
tan γ = \(\frac{A L}{L C} \approx \frac{A L}{P C}\) and tan β = \(\frac{A L}{L I} \approx \frac{A L}{P I}\)

Substituting for tan β and tan γ in equation (5) we have
r = \(\frac{\mathrm{AL}}{\mathrm{PC}}-\frac{\mathrm{AL}}{\mathrm{PI}}\) …(6)

Now by Snell’s Law at point A we have
n₁ sin i = n₂ sin r

Since angles are small, therefore the above relation becomes
n₁ i = n₂ r …(7)

Substituting the values of i and r from equations (4) and (6) we have

Applying new Cartesian sign conventions:
PO = – u (a distance of the object is against incident Light)

PI = + v (distance of image Is along incident Light)
PC = + R (a distance of the centre of curvature is along with incident Light) we have

The above equation connects u, V and R to the absolute refractive indices of the material of the refracting surface and that of the rarer medium.

Question 20.
A ray of light Is an Incident on one face of a glass prism and emerges out from the other face. Trace the path of the ray and derive an expression for the refractive index of the glass prism. Also, plot a graph between the Angle of incidence and angle of deviation. (CBSE Delhi 201 IC)
Answer:
The graph is as shown

Consider a cross-section XYZ of a prism as shown in the figure. Let A be the angle of the prism. A ray PQ of monochromatic light is incident on face XY of the prism at an angle i. The ray is called the incident ray and the angle is called the angle of incidence. This ray is refracted towards the normal NQE and travels in the prism along QR. This ray is called the refracted ray at the face XY.

Let r₁ be the angle of refraction at this surface. The refracted ray QR is incident at an angle r₂ on the surface XZ. The ray QR again suffers refraction and emerges out of face XZ at an angle e along with RS. The ray is called the emergent ray and the angle e is called the angle of emergence. When the ray SR is extended backwards it meets the extended ray PQ at point D such that δ is the angle of deviation of the ray.

As seen from figure we have in triangle QDR that ∠DQR = i – r₁ and
∠DRQ = e – r₂. Therefore from triangle QDR we have
δ = ∠DQR + ∠DRQ = (i – r₁) + (e – r₂)
or
δ = (i + e) – (r₁ + r₂) …(1)

Now from the quadrilateral XQER, we have
A + E = 180° …(2)

In triangle QER we have r₁ + r₂ + e – 180° …(3)

From equations (2) and (3) we have
A = r₁ + r₂ …(4)

Substituting in equation (1) we have
δ = i + e – A
or
i + e = A + δ … (5)

The deviation produced by a prism depends upon (i) the angle of incidence (ii) the angle of prism and (Hi) the refractive index of the material of the prism. It is found that as the angle of incidence changes, the angle of deviation also changes.

A graph between the angle of incidence and the angle of deviation is shown in the figure above. As the angle of incidence increases, the angle of deviation first decreases becomes a minimum for a particular angle of incidence and then increases. The minimum value of the angle of deviation is called the angle of minimum deviation. It is denoted by δ_m. In this position the ray of light passes symmetrically through the prism, i.e. the refracted ray QR is parallel to the base of the prism. In this position, the angle of incidence is equal to the angle of emergence, i.e. i = e. Also in this position, the angle of refraction at the faces of the prism are equal, i.e. r₁ = r₂.

Substituting these values in equations (4) and (5) we have
A = r₁ + r₂ = r + r = 2r
or
r = A/2 …(6)
and i + i = A + δm
or
2i = A + δm
or
i = \(\frac{A+\delta_{m}}{2}\) …(7)

substituting for i and r in the expression for Snell’s law we have
μ = \(\frac{\sin i}{\sin r}=\frac{\sin \left[\frac{A+\delta_{m}}{2}\right]}{\sin \left(\frac{A}{2}\right)}\)

Question 21.
(a) Draw a ray diagram showing the image formation by a compound microscope. Obtain an expression for total magnification when the image is formed at infinity.
Answer:
The ray diagram is as shown.


(b) How does the resolving power of a compound microscope get affected, when
(i) the focal length of the objective is decreased.
(ii) the wavelength of light is increased? Give reasons to justify your answer. (CBSE AI 2015C)
Answer:
The resolving power of a microscope is given by the expression RP = \(\frac{2 n \sin \theta}{\lambda}\)
(i) There is no effect of the increase in the focal length of the objective on the resolving power of the microscope.
(ii) If the wavelength of the incident light is increased, the resolving power of the microscope also decreases.

Question 22.
An optical instrument uses eye-lens of power 12.5 0 and an object lens of power 50 D and has a tube length of 20 cm. Name the optical instrument and calculate its magnifying power, if it forms the final image at infinity. (CBSE Delhi 2017C)
Answer:
It is a compound microscope.
The ray diagram is as shown.

Let m_o and me be the magnifications produced by the objective lens and eye lens respectively, the total magnifying power of the microscope
M = m_o x m_e

Now m_o = – \(\frac{v_{0}}{u_{0}}\)

Also, the magnifying power of the eyepiece when the final image is formed at infinity
m_e = \(\frac{D}{f_{e}}\)

Thus magnifying power of the microscope is
M = \(\frac{-v_{0}}{u_{0}} \times \frac{D}{f_{e}}\)

As a first approximation, u_e ≈ f_e and v_e ≈ L = distance between objective and eyepiece (or length of microscope tube), then we have
M = \(\frac{-L}{f_{0}} \times \frac{D}{f_{e}}\)

Question 23.
Use the mirror equation to show that
(i) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
Answer:
By mirror formula we have \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
Therefore \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\) and for concave mirror u and f both are having negative sign, i.e. f < 0 and u < 0.

It means that v > 2f, having a negative sign. Thus, the image is real and lies beyond 2f.

(ii) a convex mirror always produces a virtual image independent of the location of the object.
Answer:
(Ii) For a convex mirror \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\) but f is positive and u is negative, i.e. f >0 but u < 0.

Therefore, it is self-evident from the above relation that irrespective of the value of u, the value of v is aLways +ve. It means that the Imaged formed by the convex mirror is always virtual independent of the location of the object.

(iii) an object placed between the pole and the focus of a concave mirror produces a virtual and enlarged image. (CBSE AI 2011)
Answer:
As for a concave mirror f and u, both are negative, hence

It means that for an object placed between the pole and principal focus of a concave mirror the image formed (as v is +ve) is virtual and magnification

m = \(\frac{1}{O}=\frac{|v|}{|u|}\) > 1, i.e. the image is an enlarged image.

Question 24.
(i) Obtain Lens makers formula using the expression \(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{\left(n_{2}-n_{1}\right)}{R}\), here the ray of light propagating from a rarer medium of refractive index (n₁) to a denser medium of refractive index (n₂) is incident on the convex side of spherical refracting surface of radius of curvature R.
(ii) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. Using this diagram, derive the magnification formula for the Image formed. (CBSE DelhI 2011)
Answer:
(i) The course of rays through the Lens Is as shown. For refraction at the spherical surface XP₁Y, we have


Consider the second surface XP₂Y. Actually, the materiaL of the Lens does not extend beyond XP₁Y. Therefore, before the refracted ray from A₁ could meet the principal axis, it will suffer refraction at point A₂ on the second face XP₂Y and the Lightray will finally meet the principal axial. Such that l is the final image. Thus point lies the real Image of the virtual object I. Hence for refraction at the surface XP₂Y we have

The above equation becomes

But n₂/n₁ = n, the absolute refractive index of the material of the lens, therefore the above equation takes the form
\(\frac{1}{v}-\frac{1}{u}\) = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

But by lens formula we have \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
Therefore, from the above two equations we have
\(\frac{1}{f}\) = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) …(7)

This is the Lens maker’s equation or formula.

(ii) The ray diagram is as shown

Now from triangle BAP we have
tan θ = \(\frac{AB}{PA}\) …(1)

Also from triangle B’A’P, we have A’B’
tan θ = \(\frac{A^{\prime} B^{\prime}}{P A^{\prime}}\) …(2)

Comparing equations (1) and (2), we have

By Cartesian sign convention AB = h, A’B’ = h’, PA = – u and PA’= v . Substituting in equations (3) we have

But \(\frac{h^{\prime}}{h}\) = m ( Linear magnification),
therefore
m = – \(\frac{\text { image distance }}{\text { object distance }}\)

Question 25.
(a) Two thin convex lenses L₁ and L₂ of focal lengths f₁ and f₂ respectively, are placed coaxially in contact. An object is placed at a point beyond the focus of lens L₁. Draw a ray diagram to show the image formation by the combination and hence derive the expression for the focal length of the combined system.
Answer:
Consider two lenses A and B of focal length f₁ and f₂ placed in contact with each other. Let the object be placed at a point O beyond the focus of the first lens L₁ as shown.

The first lens L₁ produces an image at l₁ Since image l₂ is real, it serves as a virtual object for the second lens L₂, producing the final image at l. Since the lenses are thin, we assume the optical centres of the lenses to be coincident. Let this central point be denoted by P.

For the image formed by the first lens L₁, we get
\(\frac{1}{v_{1}}-\frac{1}{u}=\frac{1}{f_{1}}\) …(i)

For the image formed by the second lens L₂, we get
\(\frac{1}{v}-\frac{1}{v_{1}}=\frac{1}{f_{2}}\) …(ii)

Adding equations (i) and (ii) we have
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) …(iii)

If the two lens-system is regarded as equivalent to a single lens of focal length f, we have
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) …(iv)

Therefore from equations (iii) and (iv) we get
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

(b) A ray PQ incident on the face AB of a prism ABC, as shown in the figure, emerges from the face AC such that AQ = AR.

Draw the ray diagram showing the passage of the ray through the prism. If the angle of the prism is 60° and the refractive index of the material of the prism is \(\sqrt{3}\), determine the values of angle of incidence and angle of deviation. (CBSE AI 2015)
Answer:
The ray diagram is as shown

The prism, in this situation, is in the minimum deviation position, therefore we have
r = \(\frac{A}{2}=\frac{60}{2}\) = 30°

Hence n = \(\frac{\sin i}{\sin r}=\frac{\sin i}{\sin 30^{\circ}}\) = \(\sqrt{3}\)

This gives, i = 60°
Hence from
i = \(\frac{A+\delta_{m}}{2}\) we have
δ_m = 2i – A = 2 × 60° – 60° = 60°

Question 26.
(a) Plot a graph to show the variation of the angle of deviation as a function of the angle of incidence for light passing through a prism. Derive an expression for the refractive index of the prism in terms of angle of minimum deviation and angle of prism.
(b) What is the dispersion of light? What is its cause?
(c) A ray of light incident normally on one face of a right isosceles prism is totally reflected as shown in the figure. What must be the minimum value of the refractive index of glass? Give relevant calculations. (CBSE Delhi 2016)

Answer:

For a prism we have
i + e = A + δ …(1)

From the graph between the angle of incidence and the angle of deviation it follows that as the angle of incidence increases, the angle of deviation first decreases becomes a minimum for a particular angle of incidence and then again increases. The minimum value of the angle of deviation is called the angle of minimum deviation. It is denoted by 5m. In this position the ray of light passes symmetrically through the prism, i.e. the refracted ray QR is parallel to the base of. the prism. In this position, the angle of incidence is equal to the angle of emergence, i.e. i = e. Also in this position, the angle of refraction at the faces of the prism are equal, i.e. r₁ = r₂. Substituting these values in equations (4) and (5) we have
A = r₁ + r₂ = r + r = 2r
or
r = A/2 …(2)
and
i + i = A + δ_m
or
2i = A + δ_m
or
i = \(\frac{A+\delta_{m}}{2}\) …(3)

Substituting for i and r in the expression for Snell’s law we have a speed of light in a vacuum

(b) The phenomenon of splitting a ray of white tight into its constituent colours (wavelengths) is catted dispersion.
Cause: It is because different wavelengths travel at different speeds In a medium other than vacuum.

(C) The diagram of the path of rays through the prism is as shown.
Here θ = 45° > i_c

Now n = \(\frac{1}{\sin i_{c}}=\frac{1}{\sin 45^{\circ}}=\frac{1}{1 / \sqrt{2}}=\sqrt{2}\)
Therefore n > \(\sqrt{2}\)

Numerical Problems :

Formulae for solving numerical problems

• When Light travels from an optically denser medium to an optically rarer medium It bends away from the normal.
• When light travels from an optically rarer medium to an optically denser medium It bends towards from the normal. speed of Light in vacuum
• μ = \(\frac{\text { speed of light in vacuum }}{\text { speed of light in the medium }}\)
• μ = \(\frac{1}{\sin \theta_{c}}\)
• In minimum deviation position A = 2r or
  
• Lens maker’s formula
  \(\frac{1}{f}\) = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
• MagnifyIng power of a compound microscope
  M = M_e × M_o = \(\frac{L}{f_{0}} \times\left(1+\frac{D}{f_{e}}\right)\)

Question 1.
A convex lens made up of a glass of refractive index 1.5 is dipped, In turn, In
(a) a medium of refractive index 1.65,
Answer:
When dipped in the medium of refractive index 1.65, it will behave as a concave lens and when dipped in the medium of refractive index 1.33, it will behave as a convex lens.

(b) a medium of refractive index 1.33.
(i) Will it behave as a converging or a diverging lens in the two cases?
Answer:
Its focal length in another medium is given by

Thus f_m = -5.5 f_a, i.e. focal length increases and becomes negative.

(ii) How will Its focal length change In the two media? (CBSE AI 2011)
Answer:
Similarly

Thus f_m = 3.3 f_a, i.e. focal length increases.

Question 2.
A compound microscope uses an objective lens of focal length 4 cm and an eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also, calculate the length of the microscope. (CBSE Al 2011)
Answer:
f_o = 4 cm, f_e = 10 cm, u_o = – 6 cm, M = ?, L = ?
Using

Hence angular magnification


Question 3.
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1 cm is used, find the angular magnification of the telescope.
If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 × 10⁶ m and the radius of the lunar orbit is 3.8 × 10⁸ m. (CBSE AI 2011, Delhi 2015)
Answer:
Given f_o = 15 m, f_e = 1.0 cm = 0.01 m, M = ?
Dm = 3.48 × 10⁶ m, r = 3.8 × 10⁸ m,

Using M = \(\frac{f_{0}}{f_{\mathrm{e}}}=\frac{15}{0.01}\) = 1500

The angle subtended by the moon at the objective of the telescope

Question 4.
A beam of light converges at a point P. A concave lens of focal length 16 cm is placed in the path of this beam 12 cm from P. Draw a ray diagram and find the location of the point at which the beam would now converge. (CBSE Delhi 2011C)
Answer:
The ray diagram is shown in the figure. In the absence of the concave lens the beam converges at point P. When the concave lens is introduced, the incident beam of light is diverged and now comes to focus at point Q. Thus for the concave lens P serves as a virtual object giving rise to a real Image at Q.

Here u = + 12 cm, f = – 16 cm, v = ?, Now for a lens

Hence v = 48 cm
i. e. the point at which the beam is focused is 48 cm from the lens.

Question 5.
Two convex lenses of focal length 20 cm and 1 cm constitute a telescope. The telescope is focused on a point that is 1 m away from the objective. Calculate the magnification produced and the length of the tube, if the final image Is formed at a distance of 25 cm from the eyepiece. (CBSE Delhi 2011 C)
Answer:
Given f_o = 2o cm, f_e = 1 cm, u = – 100 cm, M =?, y =?

Fora lens \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
or
\(\frac{1}{v}-\frac{1}{-100}=\frac{1}{20}\)
or
v = 25cm

Since the eye lens forms the image of the virtual object at the distance of distinct vision for the eye lens
v = – 25 cm, f_e = 1 cm,

Now \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
or
\(\frac{1}{-25}-\frac{1}{u}\) = 1
or
u = – \(\frac{25}{26}\)cm

Now magnification produced by the object lens
m_o = \(\frac{v}{u}=-\frac{25}{100}=-\frac{1}{4}\)

Magnification produced by the eye Lens
m_e = \(\frac{v}{u}=\frac{-25}{-25}\) × 26 = 26

Hence total magnification
M = m_o × m_e = -1 /4 × 26 = – 6.5

Question 6.
(a) Under what conditions are the phenomenon of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium.
(b) Three lenses of focal lengths +10 cm, -10 cm and +30 cm are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination. (CBSE Delhi 2019 C)

Answer:
(a) (i) Light travels from a denser medium to a rarer medium.
(ii) Angle of Incidence in the denser medium is more than the critical angle for a given pair of media.
For the grazing incidence n sin i_C = l sin 90°
n = \(\frac{1}{\sin i_{c}}\)

(b) For convex lens f = + 10 cm

Object distance for concave lens u₂ = 15 – 5 = 10 cm

For third lens
\(\frac{1}{f_{3}}=\frac{1}{v_{3}}-\frac{1}{\infty}\) ⇒ v₃ = 30 cm

Question 7.
A ray of light incident on an equilateral glass prism (μ_g = \(\sqrt{3}\)) moves parallel to the baseline of the prism inside it. Find the angle of Incidence for this ray. (CBSE Delhi 2012)
Answer:
Given A = 60°, μ_g = \(\sqrt{3}\), i = ?

Using the expression μ = \(\frac{\sin i}{\sin A / 2}\)
or

Question 8.
Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism. (CBSE AI 2014)

Answer:
The critical angle for the two rays is

This shows that the angle of Incidence for ray ‘2’ Is greater than the critical angle. Hence it suffers total internal reflection, white ray ‘1’ does not. Hence the path of rays is as shown.

Question 9.
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed. (CBSE AI 2014)
Answer:
The ray diagram is as shown.

For the convex lens, we have
u₁ = – 60 cm, f = + 20 cm, v = ?

Using lens formula we have

Had there been only the Lens, the image would have been formed at Q₁, which acts as a virtual object for the convex mirror.
Therefore u₂ = OQ₁ – OO’ = 30 – 15 = 15 cm

Using mirror formuLa we have
\(\frac{1}{v_{2}}+\frac{1}{u_{2}}=\frac{2}{R}\)
or
\(\frac{1}{v_{2}}+\frac{1}{u_{15}}=\frac{2}{20}\)

Solving for v₂ we have
v₂ = 30cm

Hence the final image is formed at (Point Q) a distance of 30 cm behind the mirror.

Question 10.
A ray PQ is an incident normally on the face AB of a triangular prism refracting angle of 60°, made of a transparent material of refractive index 2 / \(\sqrt{3}\), as shown in the figure. Trace the path of the ray as it passes through the prism. Also, calculate the angle of emergence and angle deviation. (CBSE Delhi 2014C)

Answer:
Critical angle for glass
µ = \(\frac{1}{\sin i_{c}}\)
or
sin i_c = \(\frac{1}{\mu}=\frac{\sqrt{3}}{2}\)= 0.866
or
i_c = 60°

Now the ray is incident at an angle of 60° which is equal to the critical angle, therefore the ray graces the other edge of the prism

Therefore the angle of emergence is = 90°
Hence δ = 30°

This is as shown.

Question 11.
An object Is placed 15 cm In front of a convex lens of focal length 10 cm. Find the nature and position of the image formed. Where should a concave mirror of radius of curvature 20 cm be placed so that the final image is formed at the position of the object itself? (CBSE AI 2015)
Answer:
Given u = -15 cm, f = + 10 cm, v = ?
For lens we have

Nature of image-real, magnified Final image formed will be at the object itself only if the image formed by the lens is at the position of the centre of curvature of the mirror
∴ Distance of mirror from lens, D = 30 + R = 30 + 20 cm = 50 cm

Question 12.
Define the critical angle for a pair of media. A point source of monochromatic light ‘S’ is kept at the centre of the bottom of a cylinder with the radius of 15 cm. The cylinder contains water (refractive index 4/3) to a height of 7.0 cm. Draw the ray diagram and calculate the area of the water surface through which the light emerges in the air. (CBSE Delhi 2015C)
Answer:
It the angle of incidence in the denser medium for which the angle of refraction in the rare medium is 90°.
Given h = 7 cm, A = πr² = ?, μ = 4/3 = 1.33

Due to total internal reflection light from the bulb will not come out of the entire surface of the water as shown.

The angle of the cone through which tight will spread out is twice the critical angle.

Therefore the radius of the circular patch will be
Using the relation r= \(\frac{h}{\sqrt{\mu^{2}-1}}\) we have
r = \(\frac{7}{\sqrt{(1.33)^{2}-1}}\) = 7.98 cm

Therefore area of the surface of water through which light comes out
A = πr² = 3.14 × (7.98)² = 199.95 cm²

Question 13.
A ray PQ incident on the refracting face BA Is refracted in the prism BAC as shown In the figure and emerges from the other refracting face AC as RS such that AQ = AR. If the angle of prism A = 60° and refractive Index of the material of prism is \(\sqrt{3}\), calculate angle θ. (CBSE AI 2016)

Answer:
Here As AQ = AR, therefore QR is parallel to BC, hence prism is in minimum deviation position
A = 60°, θ = δ_m = ?, n = \(\sqrt{3}\)

Solving θ = 60°

Question 14.
(i) A ray of light incident on face AB of an equilateral glass prism shows a minimum deviation of 30°. Calculate the speed of light through the prism.

(ii) Find the angle of incidence at face AB so that the emergent ray grazes along with the face AC. (CBSE Delhi 2017)
Answer:
(i) Given


Question 15.
A small illuminated bulb is at the bottom of a tank, containing a liquid of refractive index p up to a height h. Find the expression for the diameter of an opaque disc, floating symmetrically on the liquid surface in order to cut off light from the bulb. (CBSE Sample Paper 2018-19)
Answer:
Let d be the diameter of the disc. The bulb shall be invisible if the incident rays from the bulb at 0 to the surface at d/2 are at the critical angle.
Let l be the angle of incidence
Then sin i = \(\frac{1}{μ}\) = tan i.
Now

Question 16.
A ray of light is incident on a glass prism of refractive index μ and refracting angle A. If it just suffers total internal reflection at the other face, obtain an expression relating to the angle of incidence, angle of prism and critical angle.
Answer:
For a ray to just suffer total internal reflection at the second face of the prism r₂ = i_c
Now A = r₁ + i_c, therefore r₁ = A – i_c

Now by Snells Law, we have

or
μ = \(\frac{\sin i}{\sin \left(A-1_{c}\right)}=\frac{1}{\sin i_{c}}\)

Question 17.
(a) Define, the refractive index of a medium.
(b) In the following ray diagram, calculate the speed of light In the liquid of unknown refractive index. (CBSE AI 2017C)

Answer:
(a) It is defined as the ratio of the speed of light in a vacuum to the speed of light in the given medium.
(b) The speed of Light can be found by using the formula
\(\frac{v}{c}=\frac{\sin 1}{\sin r}\)
or
v = \(\frac{\sin i}{\sin r}\) × c

From the diagram, we find that

Question 18.
A concave lens made of material of refractive index n₁ is kept In a medium of refractive index n2. A parallel beam of light is incident on the lens. Complete the path of the rays of light emerging from the concave lens If (i) n₁ > n₂ (ii) n₁ < n₂ and (iii) n₁ n₂
Answer:
The path of rays in the three cases is as shown.

Question 19.
A convex ten made of a material of refractive index n₁ is kept in a medium of refractive index n₂. A parallel beam of light is incident on the lens. Complete the path of rays of light emerging from the convex lens if (i) n₁ > n₂, (ii) n₁ = n₂, and (iii) n₁ < n₂.
Answer:
The path of the rays in the three cases is as shown.

Question 20.
Write the Lens Maker’s formula and use it to obtain the range of values of μ (the refractive index of the material of the lens) for which the focal length of an equi-convex lens, kept in air, would have a greater magnitude than that of the radius of curvature of its two surface. (CBSE Delhi 2016C)
Answer:
Lens maker’s formula is given by
\(\frac{1}{f}\) = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

For equiconvex Lens we have R₁ = + R and
R₂ = – R
Therefore we have
\(\frac{1}{f}\) = (n – 1)\(\frac{2}{R}\)

For f to be greater than R
2(n – 1) < 1
2n – 2 < 1
2n < 3
n < 1.5
Hence range is 1.0< n < 1.5

Question 21.
A ray of light passing from the air through an equilateral glass prism undergoes minimum deviation when the angle of Incidence is 3/4th of the angle of the prism. Calculate the sp..d of light In the prism. (CBSE AI 2017)
Answer:
Given A = 60°, i = 3/4 × 60 = 45°,
c = 3 × 10⁸ m s^-1,

Now δ_m = 2i – A
= 90 – 60 = 30°

Using the formula

Question 22.
Calculate the value of the angle of Incidence when a ray of light incident on one face of an equilateral glass prism produces the emergent ray, which just grazes along the adjacent face. Refractive index of the prism is \(\sqrt{2}\). (CBSE 2017C)
Answer:
The diagram is as shown.


Question 23.
How is the working of a telescope different from that of a microscope?
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment. (CBSE Delhi 2012)
Answer:
A microscope increases the size of the object whereas a telescope brings the object closer for better vision.
Given f_o = 1.25 cm, f_e = 5 cm, M = 30 cm, u = l
In normal adjustment final image is formed at infinity, therefore we have

Question 24.
(a) Draw a labelled ray diagram showing the image formation of a distant object by a refracting telescope. Deduce the expression for its magnifying power when the final image is formed at infinity.
(b) The sum of focal lengths of the two lenses of a refracting telescope is 105 cm. The focal length of one lens is 20 times that of the other. Determine the total magnification of the telescope when the final image is formed at infinity. (CBSE AI 2014C)
Answer:
(a) For the diagram

From right triangles ABC and ABC’ as shown in figure, we have
tan α = \(\frac{\mathrm{AB}}{\mathrm{CB}}=\frac{-h}{f_{0}}\) and
tan β = \(\frac{\mathrm{AB}}{\mathrm{C}^{\prime} \mathrm{A}}=\frac{-h}{f_{\mathrm{e}}}\)

From the above we have we have
M = \(\frac{\beta}{\alpha}=\frac{-h}{f_{e}} \times \frac{f_{0}}{-h}=\frac{f_{0}}{f_{e}}\)

(b) L = 105 cm, f_o = 20f_e,

Now L = f_o + f_e = 21f_e
Or
fe = 105/21 = 5 cm
Hence f_o = 20 × f_e 20 × 5 = 100 cm
Hence M = f_o/f_e = 100/ 5 = 20

Question 25.
(a) Explain with reason, how the power of a diverging lens changes when
(i) it is kept in a medium of refractive index greater than that of the lens.
(ii) Incident red light is replaced by violet light.
(b) Three lenses L₁, L₂, and L₃ each of focal length 30 cm are placed coaxially as shown in the figure. An object is held at 60 cm from the optic centre of L₁. The final image is formed at the focus of L₃. Calculate the separation between (i) L₁ and L₂ and (ii) L₂ and L₃ (CBSE AI 2017C)

Answer:
(a) (i) The power of is given by the expression P = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) medium changes the power of the lens decreases as its focal length increases.
(ii) We know that f ∝ X, therefore a decrease in the wavelength from red to violet decreases the focal length and increases the power of the lens.

(b) Given f₁ = f₂ = f₃ = 30 cm
For lens L1, u₁ = 60 cm = 2f₁, therefore the image will be formed at If on the other side of the lens L₁.

Since the final image for lens L₃ is formed at the focus, therefore the rays of light falling on lens L₃ should come from infinity. This is possible if the image of L₁ lies at the focus of L₂.

Thus distance L₁L₂ = 60 + 30 = 90 cm

Also, distance L₂L₃ can have any value as the rays between L₂ and L₃ will be parallel.

Question 26.
Calculate the radius of curvature of an equi-concave lens of refractive Index 1.5, when It Is kept In a medium of refractive index 1.4, to have a power of -5 D?
OR
An equilateral glass prism has a refractive Index of 1.6 In the air. Calculate the angle of minimum deviation of the prism, when kept In a medium of refractive Index \(\frac{4 \sqrt{2}}{5}\). (CBSE Delhi 2019)
Answer:
Given n_L = 1.5, n_M = 1.4, P = -5 D
Focal length f \(\frac{1}{P}=\frac{1}{-5}=\frac{-100}{5}\) = 20 cm

Using the lens makers formula and putting
R₁ = – R and
R₂ = +R

Solving for R, we get
R = 20/7 = 2.86cm
or
Given A = 60, n = 1.6, δm =?,n_M = \(\frac{4 \sqrt{2}}{5}\)

Using the prism formula

Question 27.
A ray of light passes through an equilateral glass prism, such that the angle of Incidence Is equal to the angle of emergence. If the angle of emergence is v times the angle of the prism, calculate the refractive index of the glass prism.
Answer:
Given the angle of prism A = 60°,
the angle of incidence i = angle of emergence ‘e’ and under this condition angle of deviation is minimum.

Question 28.
The magnifying power of an astronomical telescope In the normal adjustment position is 100. The distance between the objective and the eyepiece is 101 cm. Calculate the focal length of the objective and the eyepiece.
Answer:

Question 29.
A double convex lens made of glass of refractive Index 1.5 has both radii of curvature of magnItude 20 cm. An object 2 cm high is placed at 10 cm from the lens. Find the position, nature and size of the image.
Answer:
Given n = 1.5 and for a double convex Lens R₁ = + 20 cm and R₂ = – 20 cm

Now size of the object O = + 2 cm and u = – 1o cm

The image ¡s virtual and erect.

Question 30.
Three rays of light-red (R), green (G) and blue (B) are Incident on the face AB of a right-angled prism ABC. The refractive indices of the material of the prism for red, green and blue wavelengths are 1.39, 1.44 and 1.47, respectively. Trace the path of the rays through the prism. How will the situation change if these rays were incident normally on one of the faces of an equilateral prism?

Answer:
The course of rays through the prism is as shown below.
Critical angle for a light ray is given by
sin i_c = \(\frac{1}{n}\)
or
i_c = sin^-1\(\left(\frac{1}{n}\right)\)

Now critical angle for red ray (R)
i_R = sin^-1\(\left(\frac{1}{n}\right)\) = sin-1\(\left(\frac{1}{1.39}\right)\) = 46°

Critical angle for green ray (G)
i_G = sin^-1\(\left(\frac{1}{n}\right)\) = sin-1\(\left(\frac{1}{1.44}\right)\) = 44°

and Cnticat angLe for bLue ray (B)
i_B = sin^-1\(\left(\frac{1}{n}\right)\) = sin-1\(\left(\frac{1}{1.47}\right)\) = 42°52′

Thus the angle of incidence for blue and green rays inside the prism will be greater than the critical angle for these two colours. Therefore blue and green rays undergo total internal reflection. The angle of incidence for red colour is smaller than its critical angle; therefore it does not undergo total internal reflection. This is depicted in the figure below.

For an equilateral prism, the light rays pass through the prism suffering deviation. The angle of deviation δ = (n – 1) A. Hence, the deviation for red coloured ray will be the least and that for blue ray maximum. Paths of rays are shown in the figure below.

Question 31.
You are given three lenses of powers 0.5 D, 4 D, 10 D. State, with reason, which two lenses will you select for constructing a good astronomical telescope.
Calculate the resolving power of this telescope, assuming the diameter of the objective lens to be 6 cm and the wavelength of light used to be 540 nm.
Answer:
The focal lengths of the three lenses are
f₁ = 1 /P₁ = 1 / 0.5 = 2 m = 200 cm,
f₂ = 1 /P₂ = 1 / 4 = 0.25 m = 25 cm,
f₃ = 1 /P₃ = 1 /10 = 0. 1 m = 10 cm,

For an astronomical telescope, the objective should have a large focal length and the eyepiece should have a small focal length. Therefore for the objective, we will use the lenses of power 0.5 D and for the eyepiece the lens of power 10 D.

Given D = 6 cm = 0.06 m, λ = 540 nm = 540 × 10^-9 m

Now resolving power
R.P. = \(\frac{D}{1.22 \lambda}=\frac{0.06}{1.22 \times 540 \times 10^{-9}}\) = 91074.6

Question 32.
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different positions separated by 20 cm. Calculate the focal length of the lens.
OR
A convex lens of focal length 20 cm and a concave lens of focal length 15 cm are kept 30 cm apart with their principal axes coincident. When an object is placed 30 cm in front of the convex lens, calculate the position of the final image formed by the combination. Would this result change if the object were placed 30 cm In front of the concave tens? Give reason. (CBSE AI 2019)
Answer:
The image of an object can be obtained at two different positions of a convex lens for a fixed value of the distance between the object and the screen if values of u and v are interchanged in these two positions. This situation is as shown below.

Here x + 20 + x = 90 cm or x = 35cm
Therefore, u = -35 cm, v = + 55 cm, f = ?

or

For image formed by a convex Lens
f₁ = + 20 cm, u = – 30 cm

Therefore, u for concave Lens = 60 – 30 = + 30 cm and f₂ = – 15 cm

Now for concave lens

No, the result will not change as per the principle of reversibility.

Question 33.
For an equilateral glass prism, the minimum angle of deviation, for a parallel beam of monochromatic Incident light, is measured to be 30°. What is the refractive index of the glass used for making this prism?
If this prism were to be placed in a medium of refractive Index 1.414, how would a ray of light incident on one of its refracting faces pass through it? Draw the ray diagram for the same.
Answer:
Given A = 60°, δ_m = 30°, μ = ?
Using the relation

When this prism is placed in a medium of refractive index 1.414, it will behave like a plane glass piece as both the prism and the medium have the same refractive index.

The ray diagram is as shown.

Question 34.
Find the position of the image formed by the lens shown in the figure.

Another lens is placed in contact with this lens to shift the image further away from the lens. What is the nature of the second lens?
Answer:
f = + 10 cm, u = – 30 cm, v =?
Using the lens equation we have

The second lens Is concave.

Question 35.
A beam of light of wavelength 400 nm is incident normally on a right-angled prism as shown. It is observed that the light just grazes along with the surface AC after falling on it. Given that the refractive index of the material of the prism varies with the wavelength as per the relation
μ = 1.2 + \(\frac{b}{\lambda^{2}}\) calculate the value of b and the refractive index of the prism material for a wavelength λ = 500 nm. [(Given θ = Sin^-1 (0.625)]

Answer:
The figure can be redrawn as shown below.

By symmetry angle, θ will be the critical angle.
Therefore
μ = \(\frac{1}{\sin \theta}=\frac{1}{0.625}\) = 1.6

Now the value of the constant b is obtained as follows:
μ_A = 1.2 + \(\frac{b}{\lambda^{2}}\)
or
1.6 = 1.2 + \(\frac{b}{(400)^{2}}\)

Solving for b we have b = 64000 nm²
Now for a wavelength λ = 500 nm the refractive index of the material of the prism is
μ = 1.2 + \(\frac{(64000)^{2}}{(500)^{2}}\) = 1.456

Question 36.
A convex lens, of focal length 20 cm, has a point object placed on its principal axis at a distance of 40 cm from It. A plane mirror Is placed 30 cm behind the convex lens. Locate the position of the image formed by this combination.
Answer:
We first consider the effect of the lens. For the Lens, we have

u = – 40 cm and f = + 20 cm
Using the lens formula, we get
\(\frac{1}{v}-\frac{1}{(-40)}=\frac{1}{20}\)
or
v = + 40 cm

Had there been the lens only the image would have been formed at Q₁. The plane mirror M is at a distance of 30 cm from the lens L. We can, therefore, think of a Q₁ as a virtual object, located at a distance of 10 cm, behind the plane mirror M. The plane mirror, therefore, forms a real image (of this virtual object Q₁) at Q₁, 10 cm in front of it. This is shown in the figure.

Question 37.
(i) Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of the image also.
(ii) Using the mirror formula, explain why does a convex mirror always produce a virtual image. (CBSE Delhi 2016) Answer:
Given R = – 20 cm or f = – 10 cm, m = + 2 (real image), u = ?, v = ?
Using the relation m = \(\frac{f}{f-u}\) we have

(b) For a convex mirror \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\) but f is positive and u is negative i.e., f > 0 but u < 0.
Therefore, it is self-evident from the above relation that irrespective of the value of u, the value of v is always +ve. It means that the image formed by a convex mirror is always virtual Independent of the location of the object.

Question 38.
A convex lens of focal length 20 cm and a concave mirror of focal length 10 cm are placed co-axially 50 cm apart from each other. An incident beam parallel to its principal axis is incident on the convex lens. Locate the position of the (final) image formed due to this combination.
Answer:
The incident beam, on lens L_, is parallel to its principal axis. Hence the lens forms an image Q₁ at its focal point, i.e. at a distance OQ₁ (= 20 cm) from the lens. This image, Q₁, now acts as a real object for the concave mirror. For the mirror, we then have: u = – 30 cm, and f = – 10 cm,

Hence using the mirror formula, we get
\(\frac{1}{v}+\frac{1}{(-30)}=\frac{1}{(-10)}\)
or
v = – 15 cm

The lens-mirror combination, therefore, forms a real image Q at a distance of 15 cm from M. The ray diagram is as shown in

Question 39.
A convex lens of focal length 25 cm and a concave mirror of radius of curvature 20 cm are placed coaxially 40 cm apart from each other. An incident beam parallel to the principal axis is incident on the convex lens. Find the position and nature of the image formed by the combination. (CBSE Al 2016)
Answer:
The image of the object is formed at the focus of the convex lens, i.e. 25 cm. This image acts as an object for the concave mirror and ties at a distance (40 – 25) = 15 cm from the mirror.

Hence for the mirror
u = – 15 cm, v =? f = R/2 = 20/2 = 10 cm

Using mirror formula we have

The image is formed in front of the mirror.

Question 40.
(i) A screen is placed at a distance of 100 cm from an object. The image of the object is formed on the screen by a convex lens for two different locations of the lens separated by 20 cm. Calculate the focal length of the lens used.
(ii) A converging lens is kept coaxially in contact with a diverging lens both the lenses being of equal focal length. What is the focal length of the combination? (CBSE AI 2016)
Answer:
(i) Given u + v= 100 or v= 100 – u
Using lens formula From lens first \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) we have

Or
f = 24 cm

(ii) f = infinity. \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{-f_{2}}\) = 0

Question 41.
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on the face of the prism. By rotating the prism, the angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light. The refracting angle of the prism is 60°. (NCERT)
Answer:
Given δ_m = 40°, aμg = ?, _aμ_w = 1.33, A = 60°

Solving for δ_m we have δ_m = 100°

Question 42.
A beam of light converges to a point P. A lens is placed in the path of the convergent beams 12 cm from P. At what point does the beam converge if the lens is
(i) a convex lens of focal length 20 cm,
Answer:
(i) Here u = 12 cm, f = + 20 cm and v = ?
Using the formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) we have
\(\frac{1}{v}=\frac{1}{12}+\frac{1}{20}=\frac{3+5}{60}=\frac{8}{60}\)
or
v = 7.5 cm

(ii) a concave lens of focal length 16 cm? (NCERT)
Answer:
Here u = 12 cm, f = – 16 cm and v =?
Using the formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)we have
\(\frac{1}{v}=\frac{1}{12}+\frac{1}{-16}=\frac{-3+4}{48}=\frac{1}{48}\)
or
v = 48 cm

Question 43.
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (i) the least distance of distinct vision (25 cm), (ii) infinity? What is the magnifying power of the microscope in each case? (NCERT)
Answer:
Given f0 = 2.0 cm, fe = 6.25 cm, L = 15 cm, u0 = ?,

Now applying Lens formula for objective Lens we have \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \text { or } \frac{1}{u_{0}}=\frac{1}{v_{0}}-\frac{1}{f_{0}}\)

Question 44.
A small telescope has an objective lens of a focal length of 144 cm and an eyepiece of a focal length of 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece? (NCERT)
Answer:
Given f_o = 144 cm, f_e = 6.0 cm, M = ?,
L = f_o + f_e = ?

Using the formula M = \(\frac{f_{0}}{f_{e}}=\frac{144}{6}\) = 24 and
L = f_o + f_e = 144 + 6 = 150 cm

Question 45.
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope Is used to view the moon, what is the diameter of the Image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 10⁶ m, and the radius of lunar orbit Is 3.8 × 10⁸ m. (NCERT)
Answer:
Given f_o = 15 m, f_e = 1.0 cm = 0.01 m, M = ?
D_m = 3.48 × 10⁶ m, r = 3.8 × 10⁸m,
(a) Using M = \(\frac{f_{0}}{f_{e}}=\frac{15}{0.01}\) = 1500
(b) AngLe subtended by moon at the objective of the teLescope
α = \(\frac{D_{m}}{r}=\frac{3.48 \times 10^{6}}{3.8 \times 10^{8}}\)

Therefore angLe subtended by the image
β = M × α = 1500 × \(\frac{3.48 \times 10^{6}}{3.8 \times 10^{8}}\)

If I be the Linear size of the image then

Question 46.
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524. (NCERT)
Answer:
Given A = 60°, i = ?, μ = 1.524
For a ray to just suffer total internal reflection at the second face of the prism r₂ = i_c
Now sin ic = \(\frac{1}{\mu}=\frac{1}{1.524}\) = 0.6562
or
i_c = 41°

Now A = r₁ + r₂, therefore r₁ = A – r2
r₁ = 60 – 41 = 19°

Now by Snell’s law we have
μ = \(\frac{\sin i}{\sin r_{1}}\)
or
sin i = μ × sinr₁ = 1.524 × sin 19°
or
i = 29.75° = 30°

Question 47.
A small telescope has an objective lens of a focal length of 140 cm and an eyepiece of a focal length of 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e. when the final image is at infinity.
Answer:
Given f_o = 140 cm, f_e = 5.0 cm, M = ?
For normal adjustment we have
M = \(\frac{f_{0}}{f_{e}}=\frac{140}{5}\) = 28

(b) the final image is formed at the least distance of distinct vision (25 cm) (NCERT)
Answer:
When the final image is formed at the least distance of distinct vision we have
M = \(\frac{f_{0}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)=\frac{140}{5}\left(1+\frac{5}{25}\right)\) = 33.6

Question 48.
(a) The refractive index of glass is 1.5. What is the speed of light In a glass? (Speed of light in a vacuum is 3.0 × 10⁸ ms^-11 )
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?
Answer:
(a) Given μ = 1.5, c = 3.0 × 10⁸ m s^-1
v = \(\frac{c}{\mu}=\frac{3 \times 10^{8}}{1.5}\) = 2.0 × 10⁸ ms^-1

(b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour or mean light. Now we know violet colour deviates more than red in a glass prism, i.e. μ_v > μ_R. Therefore, the violet component of white light travels slower than the red component.

Question 49.
Three immiscible liquids of densities d₁ > d₂ > d₃ and refractive indices μ₁ > μ₂ > μ₃ are put in a beaker. The height of each liquid column is d₃. A dot is made at the bottom of the beaker. For near-normal vision, find the apparent depth of the dot. (NCERT Exemplar)
Answer:
The situation is as shown.

Let the apparent depth be O₁ for the object as seen from μ2, then
Apparent depth = real depth/μ
O₁ = \(\frac{h / 3}{\mu_{1} / \mu_{2}}=\frac{\mu_{2} h}{3 \mu_{1}}\)

If seen from O₃ the apparent depth O₂ is
Apparent depth = real depth/μ
O₂ = \(\frac{\left(h / 3+O_{1}\right)}{\mu_{2} / \mu_{3}}=\frac{\mu_{3}}{\mu_{2}}\left(\frac{h}{3}+\frac{\mu_{2}}{\mu_{1}} \frac{h}{3}\right)\)

Seen from outside the apparent height is
Apparent depth = real depth/μ

Question 50.
For a glass prism (μ = \(\sqrt{3}\)) the angle of minimum deviation Is equal to the angle of the prism. Find the angle of the prism. (NCERT Exemplar)
Answer:

Question 51.
In many experimental set-ups, the source and screen are fixed at a distance say D and the lens Is movable. Show that there are two positions for the lens for which an image Is formed on the screen. Find the distance between these points and the ratio of the Image sizes for these two points. (NCERT Exemplar)
Answer:
The situation is shown.

If there was no cut then the object would have been at a height of 0.5 cm from the principal axis OO’. Consider the image for this case by lens formuLa we have
\(\frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50}\)
Or
v = 50 cm
Hence magnification of the image is.
m = \(\frac{v}{u}=\frac{-50}{50}\) = -1

Thus the image would have been formed at 50 cm from the pole and 0.5 cm below the principal axis. Hence with respect to the X-axis passing through the edge of the cut lens, the co-ordinates of the image are (50 cm, -1 cm)