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Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 15 Polymers.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 15 Polymers

Question 1.
Based on the mode of polymerisation we can classify polymers into addition polymers and condensation polymers. (March – 2010)
a) Classify the polymers given below into addition polymers and condensation polymers. Terylene, polyvinyl chloride, bakelite, polyethene.
b) How will you prepare Nylon 6,6?
Answer:
a) Addition polymers – Polyvinyl chloride, Polythene Condensation polymers-Terylene, Bakelite
b) Nylon 6,6 is prepared by the condensation ‘ polymerisation of hexamethylene diamine with adipic acid under high pressure and at high temperature.

Question 1.
Polymers are high molecular mass compounds having special properties and so used for special purposes. Identify the following polymers X, Y and Z. (Say – 2010)
a) X is a polymer resistant to heat and chemicals. People used it to make non-sticky frying pans.
b) Y is a polymer formed from ethylene glycol and terephthalic acid and used for making heart valves.
c) Z is a polymer used for making unbreakable crockery items.
Answer:
a) Teflon (Polytetrafluoroethylene)
b) Terylene
c) Melamine – formaldehyde polymer

Question 1.
a) LDPE is a homopolymer, while Nylon 6,6 is a co-polymer. Explain. (March – 2011)
b) Classify the following into homopolymer or co-polymer: Nylon-6, HDPE.
Answer:
a) LDPE is Low-Density Poly Ethylene and its monomer is ethylene. The monomers of Nylon 6,6 are hexamethylene diamine and adipic acid. Here two monomers are present and hence it is a copolymer.

b) Nylon – 6 is a homopolymer contains only one type of monomer units, i.e., aminocaproic acid

Question 1.
Monomers polymerise to give polymers. Polymers can be classified in many ways. (Say – 2011)
a) Distinguish between homopolymers and co-polymers.
b) Give the name or formulae of the monomers in the following polymers.
i) Nylon – 6, 6
ii) Dacron
Answer:
a) Homopolymers-Addition polymers formed by the polymerisation of a single monomeric species.
e.g., Polythene.
Co-polymers – Polymers made by addition polymerisation from two different monomers.
e.g., Buna-S, Buna-N
b) i) Nylon 6,6 – hexamethylene diamine with adipic acid
ii) Dacron – Ethylene glycol (HOH₂C-CH₂OH) and Terephthalic acid( HoocCOOH).

Question 1.
a) Rubber is a natural polymer obtained from the bark of rubber trees. (March – 2012)
i) Name the monomer of natural rubber.
ii) Vulcanisation improves the elasticity of rubber. What is vulcanisation?
b) Write two examples for synthetic rubber.
Answer:
a) i) Isoprene, 3-butadiene)
ii) The process of heating a mixture of raw rubber with sulphur and an appropriate additive at a temperature range between 373 K to 415 K.
b) e.g. 1. Neoprene, 2. Buna – N

Question 1.
PVC, bakelite and polythene are plastics. (Say – 2012)
i) Classify the above plastics into thermoplastics and therm osetting plastics.
ii) Name the monomer units of PVC and bakelite.
Answer:
i) Thermo plastics – PVC, Polythene
Thermosetting plastics – Bakelite
ii) The monomer of PVC ¡s ployvinyl chloride (CH₂=CH-Cl)
The monomers of bakelite are phenol (C₆H₅ÇOH) and formaldehyde (HCHO)

Question 1.
a) Synthetic rubber is a vulcanisable rubber-like polymer. (March – 2013)
1) Write one example for synthetic rubber.
ii) Write the method of preparation of the above synthetic rubber.
b) Which are the monomers of Nylon-6 and Nylon-66?
Answer:
a) 1) Neoprene
2) It is obtained by the free radical polymerisation of chloroprene.

b) Polymer Monomer
1) Nylon 6 → Caprolactam
2) Nylon 6,6 → Hexamethylene diamine & Adipic acid

Question 1.
Natural rubber obtained from rubber latex is soft and sticky. (Say – 2013)
a) Suggest a method to improve the stiffness of rubber.
b) Explain the above method.
c) Classify the following into natural and synthetic polymers:
Nylon, Starch, Cellulose, PVC
Answer:
a) Vulcanisation

b) It is the process of heating a mixture of raw rub- ber with sulphur and an appropriate additive at a temperature range between 373 K to 415 K. On vulcanisation, sulphur forms cross-links at the reactive sites of double bonds in the polyisoprene chain and thus rubber gets stiffened. Vulcanisation improves the physical properties of natural rubber-like elasticity, water absorption capacity and solubility. It also increases its resistance to attack by oxidising agents.

c) Natural polymers – Cellulose, Starch Synthetic polymers – Nylon, PVC

Question 1.
a) Write any two differences between step-growth polymerisation and chain-growth polymerisation. (March – 2014)
b) What are the monomers of the following?
i) Neoprene
ii) Nylon-6
Answer:
a)

b) 1) Neoprene: Neoprene (2-Chloro-1 ,3-butadiene)

Question 1.
a) Name two thermoplastics. (Say – 2014)
b) Nylon 6, 6 and Dacron are two synthetic fibres. Suggest the monomers of each.
Answer:
a) 1) Polythene
2) Poly Vinyl Chloride (PVC)

b) Nylon 6,6: Hexamethylenediamine [HOOC(CH₂)₄ COOH] and adipic acid [H₂N(CH₂)₆NH₂],
Dacron : Ethylene glycol (HOH₂C – CH₂OH) and Terephthalic acid (HOOC – C₆H₄ – COOH).

Question 1.
Polymers are macromolecules formed by union of monomers. (March – 2015)
a) Name natural polymer and synthetic polymer.
b) Distinguish between thermoplastic and thermosetting polymers with example.
Answer:
a) Natural polymer – proteins, cellulose, starch, resins, natural rubber (any one)
Synthetic polymer- Polythene, Nylon 6,6, BunaS, Teflon, PVC (any one)
b) Thermoplastic polymers – These are linear or slightly branched long-chain molecules capable of repeatedly softening on heating and hardening on cooling. These polymers possess intermolecular forces of attraction intermediate between elastomers and fibres. These can be reused.

Example – polythene, polystyrene, polyvinyl (any one)

Thermosetting polymers – These polymers are cross linked or heavily branched molecules, which on heating undergo extensive cross-linking in moulds and again become infusible. These cannot be reused.

Example – bakelite, urea-formaldehyde resins (any one)

Question 1.
Polymers are classified into elastomers, fibres, thermoplastic and thermosetting plastics, depending upon the intermolecular forces. Fill in the vacant boxes given below: (Say – 2015)

Answer:

• Bakelite
• Elastomer
• Isoprene (2-Methy-1,3-butadiene)
• Fibres
• Nylon 6
• Thermoplastic

Question 1.
Polymers can be classified based on molecular forces. (March – 2016)
a) Classify the following polymers into elastomers and fibres: Rubber, Nylon 6,6 Buna-S, Terylene
b) What do you mean by thermosetting polymers? Give one example.
Answer:
a) Elastomers – Rubber, Buna-S Fibres – Nylon 6,6, Terylene b) Thermosetting polymers are cross-linked or heavily branched molecules, which on heating undergo extensive cross-linking in moulds and again become infusible.

These cannot be reused.

e.g. Bakelite, urea-formaldehyde resins (any one)

Question 1.
Polymers are of different types (Say – 2016)
a) Identify the thermoplastic polymer from the following:
i) Bakelite
ii) Nylon-6,6
iii) Neoprene
iv) PVC

b) What is biodegradable polymers? Write an example.
Answer:
a) (iv) PVC
b) Biodegradable polymers – Polymers which can be degraded by microorganisms,
eg. Poly β-hydroxybutyrate – co-β-hydroxy- valerate (PHBV), Nylon 2-nylon 6 (any one example)

Question 1.
a) Which of the following is not applicable to Nylon 6,6? (March – 2017)
i) Synthetic polymer
ii) Fibre
iii) Addition polymer
iv) Condensation polymer.

b) Differentiate between thermoplastics and thermosetting plastics. Write one example each to them.
Answer:
a) Addition polymer
b)

Question 1.
a) Distinguish between thermoplastic polymers and thermosetting polymers. (Say – 2017)
b) Name the monomers in the following two polymers.
i) Nylon 6,6
ii) Bu
Answer:
a) March 2017 Question 1 (b)
b) i) Nylon 66 → Adipic acid + Hexamethylene- diamine
ii) Buna-N → 1,3 butadiene + acrylonitrile

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 14 Biomolecules.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 14 Biomolecules

Question 1.
Carbohydrates can be divided into three major classes monosaccharides, oligosaccharides and polysaccharides. (March – 2010)
a) What are polysaccharides?
b) Give two examples for polysaccharides.
c) What is invert sugar?
Answer:
a) Polysaccharides are carbohydrates which on hydrolysis gives large number of monosaccharide units.
b) Starch and cellulose
c) The equimolar mixture of D – (+) – glucose and D – (-) fructose obtained by the hydrolysis of sucrose is called invert sugar.

Question 1.
Glucose (C₆H₁₂O₆) is a monosaccharide, which can be oxidized, reduced and acetylated. What happens when glucose is treated with the following: (Say – 2010)
a) Br₂ water
b) Hl/red P
c) Acetic anhydride
Answer:
a) When glucose is treated with bromine water it is oxidised to gluconic acid.

b) Glucose on prolonged heating with HI forms n-Hexane.

c) Acetylation of glucose with acetic anhydride gives glucose pentaacetate.

Question 1.
a) Names of some carbohydrates, their properties and structural patterns are given below. Match them properly. (March – 2011)

b) Proteins have polypeptide bonds. What are polypeptides?
Answer:
a) Glucose – monosaccharide – Anomers present Sucrose – Disaccharide – Fructoxide Lactose – Reducing -1,4 link Amylopectin – Insoluble in water -1,6 link
b) When the number of amino acid units in a protein is more than ten, then the products are called polypeptides.

Question 1.
Proteins are the polymers of a-amino acids. The structure and shape of proteins can be discussed at four different levels, namely, primary, secondary, tertiary and quaternary. Give an account of structure and shape of proteins considering the above four levels. (Say – 2011)
Answer:
The structure and shape of proteins can be studied at four different levels, i.e., primary, secondary, tertiary and quaternary.
1) Primary structure of proteins – It refers to the sequence of amino acids in a polypeptide chain.
2) They are found to exist in two different types of structures such as α -helix and β-pleated sheet structure. In α -helix the polypeptide chain forms all possible hydrogen bonds by twisting into a right-handed screw with the -NH groups of each amino acid residue hydrogen-bonded to the group of an adjacent tum of the helix. In β -pleated sheet structure all peptide chains are stretched out to nearly maximum extension and laid side by side which are held together by intermolecular hydrogen bonds.
3) Tertiary structure of proteins – It represents overall folding of the polypeptide chains.
4) Quaternary structure of proteins – The spacial arrangement of two or more polypeptide chains.

Question 1.
a) Carbohydrates are classified into monosaccharides, oligosaccharides and polysaccharides. (March – 2012)
i) What is the basis of such classification? Explain.
ii) Give an example for an oligosaccharide.
b) Vitamin ‘C’ is a vitamin found in fruits and vegetables. It cannot be stored in our bodies. Why?
Answer:
a) i) On the basis of their behaviour on hydrolysis, carbohydrates are divided into three major classes.
1) Monosaccharides: These cannot be hydrolysed further into a simpler unit of polyhydroxy aldehyde or ketone. e.g. Glucose, Fructose etc.
2) Oligosaccharides: These carbohydrates which on hydrolysis give 2-10 monosaccharide units. e.g. Sucrose
3) Polysaccharides: These are high molecular mass carbohydrates which give many monosaccharide units on hydrolysis. e.g. Starch, Cellulose, Glycogen etc. ii) Sucrose

b) Vitamin ‘C’ is a water-soluble vitamin and must be supplied regularly in the diet because it are readily excreted in urine and cannot be stored in our body.

Question 1.
Proteins are important polymers of biological systems. (Say – 2012)
i) What is the denaturation of proteins?
ii) Give two examples of denaturation.
Answer:
i) When a protein in its native is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are distrubed, the 2° and 3° structures change and the protein loses its biological activity. This is called denaturation of the protein.
ii) 1) When egg is boiled, it becomes hard because the soluble globular proteins change to in-soluble fibrous proteins.
2) Curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk.

Question 1.
a) Amino acids can be classified into essential amino acids and non-essential amino acids. (March – 2013)
i) What is the basis of such classification?
ii) Write one example each for essential and non-essential amino acids.
b) Write any two differences between DNA and RNA.
Answer:
a) i) The amino acids which can be synthesized in body are known as non-essential amino acids. Those amino acids which can not be synthesized in the body and must be obtained through diet are known as essential aminoacids.
ii) Essential amino acids
e.g. Valine, Lysine
Non -essential amino acids
e.g. Glycine, Alanine
b)

Question 1.
Name the products obtained in the following reactions. (Say – 2013)

c) What is inverted sugar?
d) Name two polysaccharides.
Answer:
a) Gluconic acid.

c) The equimolar mixture of D – (+) – glucose and D – (-) fructose obtained by the hydrolysis of sucrose is called invert sugar.
d) starch and cellulose

Question 1.
Biomolecules are formed by certain specific linkages between simple monomeric units. Write the names of linkages and monomeric units in the following class of biomolecules. (March – 2014)
i) Starch
ii) Protein
iii) Nucleic acid
Answer:
i) Starch – Monomer: α – D – (+) – Glucose
Linkage: Glycosidic linkage
ii) Protein – Monomer: α – amino acids
Linkage: Peptide linkage or Peptide bond
iii) Nucleic acid – Monomer: Nucleotide
Linkage: Phosphodiesterlinkage

Question 1.
a) Name a fat-soluble vitamin. Suggest a disease caused by its deficiency. (Say – 2014)
b) What do you mean by the following:
i) Secondary structure of proteins.
ii) Nucleosides.
Answer:
a) Vitamin A-Xerophthalmia
b) i) It refers to the shape in which a long polypeptide chain can exist. These are found to exist in two different types of structures viz. α-helix and β -pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between and NH- groups of the peptide bond.
ii) Nucleoside is a structural part of nucleic acid. It is a unit formed by the attachment of a base to 1′ position of the sugar.

Question 1.
Carbohydrates are broadly divided into monosaccharides, oligosaccharides, and polysaccharides. (March – 2015)
a) Write one example of each of monosaccharides and oligosaccharides.
b) i) Write any one method for the preparation of glucose.
ii) What is peptide linkage?
Answer:
a) Monosaccharide – Glucose, Fructose, Ribose (anyone) Oligosaccharide – Sucrose, Maltose, Lactose (anyone)
b) i) From Sucrose (cane sugar): Sucrose on hydrolysis gives glucose and fructose.
or
Hydrolysis of starch by boiling it with dilute H₂SO₄ at 393 K under pressure.
ii) Due to peptide linkage an amide farmed between -COOH group and -NH₂ group of amino acids of proteins.

Question 1.
a) Match the following structures of proteins in Column 1 with their characteristic features in column II. (Say – 2015)
b) What is the denaturation of proteins?
Answer:
a)

b) What is the denaturation of proteins?
Answer:
a)

b) When a protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.

Question 1.
Cane Sugar, Glucose and Starch are Carbohydrates. (March – 2016)
a) Represent the structure of Glucose.
b) Write a method to prepare Glucose from Starch. Write the chemical equation of the reaction.
c) Suggest any two uses of Carbohydrates.
Answer:

b) Glucose is obtained by hydrolysis of starch by boiling it with dilute H₂SO₄ at 393 K under pressure.

c)

• Carbohydrates form a major portion of our food.
• Honey, a carbohydrate has been used for a long time as an instant source of energy ¡n ayurvedic system of medicine.
• Carbohydrates are used as storage molecules as starch in plants and glycogen in animals.
• Cell wall of bacteria and plants is made up of cellulose.
• Cellulose in the form of cotton fibre is used for clothing.
• Cellulose in the form of wood is used to build furniture.
• Carbohydrates provide raw materials for many important industries like textiles, paper, lacqures and breweries. (any two)

Question 1.
Proteins are biomolecules (Say – 2016)
a) What is denaturation of protein?
b) Match the following:
Vitamin A – Glucose
Starch – Zymase
Aldohexose – Night blindness
Enzyme – Amylose
– Fructose
Answer:
a) When a protein in its native form is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.
b) VitaminA – Night blindness
Starch – Amylose
Aldohexose – Glucose
Enzyme – Zymase

Question 1.
a) Which of the following is a polysaccharide? (March – 2017)
i) Maltose
ii) Sucrose
iii) Fructose
iv) Cellulose
b) Explain the amphoteric behaviour of amino acid.
Answer:
a) iv) Cellulose
b) This behaviour is due to the presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as a zwitterion. In zwitterionic form, aminoacids show amphoteric behaviour as they react both with acids and bases.

Question 1.
a) a -D-(+) glucose and p -D(+) glucose are (Say – 2017)
i) Metameres
ii) Anomers
iii) Geometrical Isomers
iv) Functional group isomers
b) What is the denaturation of proteins?
c) Differentiate between nucleoside and nucleotide
Answer:
a) ii) Anomers
b) When a protein is treated with acid, alkali or heated or subjected to change in pH, the secondary and primary structure of protein gets ruptured. Denaturation does not change the primary structure of proteins.
c) The repeating structural units of nucleic acids are called nucleotides.

Pentose sugar + Base → nucleoside
Nucleoside + Phosphoric acid → nucleotide

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 13 Amines.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 13 Amines

Question 1.
Aromatic amines are important synthetic intermediates. (March – 2010)
i) What are the products obtained when aniline is treated with bromine water?
ii) How will you convert nitrobenzene to aniline?
iii) Write down the Isocyanide test for the primary amines.
Answer:
i) Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6 tribromoaniline.

ii) Nitrocompounds on reduction gives amines.

iii) Primary amines on heating with chloroform and ethanolic KOH, foul-smelling substances known as isocyanides or carbylamines are formed.
\(\mathrm{R}-\mathrm{NH}_{2}+\mathrm{CHCl}_{3}+3 \mathrm{KOH} \underline{\mathrm{Heat}}, \mathrm{R}-\mathrm{NC}+3 \mathrm{KCl}+3 \mathrm{H}_{2} \mathrm{O}\)

Question 1.
Benzene sulphonyl chloride and aqueous NaOH can be used to distinguish three classes of amines such as primary, secondary and tertiary. (Say – 2010)
a) Name the above test.
b) How will you distinguish the above amines using this test?
c) Give the reactions and justifications,
Answer:
a) Hinsberg test. C₆H₅SO₂Cl
b) 1⁰ Amine + Hinsberg reagent → a compound soluble in alkali
2⁰ Amine + Hinsberg reagent → A compound which is insoluble in alkali
3⁰ Amine +Hinsberg reagent → no reaction.

c) The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali.

Secondary amine react with benzene salphonyl chloride to form N, N-dialkylbenzene- sulphon-amide, which is insoluble in alkali. This is be?cause it does not contain any hydrogen attached to nitrogen atom and is not acidic.

Tertiary amines do not react with benzene salphonyl chloride, because they do not possess any replacable hydrogen.

Question 1.
Amines are versatile functional group useful in the preparation of many organic compounds. How can you convert? (March – 2011)
OR
a) A student tried to prepare p-nitroaniline by nitrating Aniline with Conc. HNO₃ – Coric. H₂SO₄ mixture, but he got only m-nitro aniline. Why?

b) Explain how he should proceed to get pnitroaniline from Aniline.
Answer:

a) In the strongly acidic medium aniline is protonated to form anilium ion which is meta directing.

b) The -NH₂ group in aniline should be protected by acetylation by treating it with acetic anhydride. The acetanilide formed ¡s subjected to nitration to get p-Nitroacetanilide which on hydrolysis gives p-Nitroaniline.

Question 1.
Primary, secondary and teritary amines can be distinguished using Hinsberg’s reagent. (March – 2012)
i) What is Hinsberg’s reagent?
ii) How will you distinguish primary, secondary and tertiary amines using Hinsberg’s reagent?
Answer:
i) Benzene salphonyl chloride (C₆H₅SO₂Cl)
ii) a) The reaction of C₆H₅SO₂Cl with primary amine yields N-alkyl benzenesulphonamide which is soluble in alkali

The hydrogen attached to nitrogen in suiphonamide is strongly acidic due to the presence of strong electron withdrawing suiphonyl group. Hence, it is soluble in alkali.

b) Secondary amine read with benzene salphonyl chloride to form N, N-dialkylbenzene sulphonamide, which is insoluble in alkali. This is because it does not contain any hydrogen attached to nitrogen atom arid is not acidic.

c) Tertiary amines do not react with benzene salphonyl chloride, because they do not pos-sess any replacable hydrogen.

Question 1.
a) Carbyl amines have an offensive smell. (Say – 2012)
i) Write the carbyl amine reaction.
ii) How will you convert aniline into phenol?
b) How will you convert an amide into the following? 0 An amine with one carbon atom less than that
of the amide.
ii) An amine containing the same number of car bon atoms as that in the amide.
Answer:
a) i) Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potas skim hydroxide form foul smelling substances called isocyanides or carbylamines. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines.

ii) Aniline on diazotisation gives benzene diazonium chloride. This on warming with water gives phenol.

b) i) By Hoffman bromamide degradation reaction- when an amide is treated with bromine in an aqueous or ethanolic solution of sodium hy-droxide at about 343 K, an amine with one carbon less than that present in the amide is formed.

ii) By reduction – An amide on reduction with LiAIH₄ or Na and ethanol an amine having the same number of carbon atoms as that in the amide is formed.

Question 1.
a) Amines are basic in nature. (March – 2013)
Arrange the following compounds in the increasing order of their basic strength.
NH₃, C₆H₅NH₂, CH₃-NH₂, (CH₃)₂ NH, (CH₃)₃N.
b) How will you convert aniline (C₆H₅NH₂) to chlorobenzene?
Answer:
a) In aqueous solution when R = CH₃ basic strength increases in the order

Question 1.
Amines can be considered as derivatives of ammonia. (Say – 2013)
a) Arrange the following in the increasing order of their basic strength.
C₆H₅NH₂, C₂H₅-NH₂(C₂H₅)₂NH, NH₃.
b) Represent a reaction of explain the basic character of aniline.
c) Name the reagents used in Hoffmann bromamide reaction.
d) What is the significance of the above reaction?
e) Give one chemical test to distinguish between methyl amine and diethyl amine.
Answer:
a) C₆H₅NH₂ < NH₃ < C₂H₅NH₂, (C₂H₅)₂NH
b) Aniline, being basic reacts with hydrochloric acid to form anilinium chloride salt.

c) An amide, bromine, aqueous or ethanolic solution of NaQH.
d) It is a method for preparation of primary amines. The amine so formed contains one carbon less than that present in the amide.
e) Hinsberg’s test – When methyl amine (1° amine) is treated with Hinsberg’s reagent (benzene suiphonyl chloride), N-Methylbenzene suiphonamide is formed which is soluble in alkali due to the presence of acidic hydrogen.

When dimethyl amine (2° amine) is treated with Hinsberg’s reagent N,N-dimethyl benzene shlphonamide is formed which is insoluble in alkali due to the absence of acidic hydrogen.

Question 1.
a) Write a method of preparation of primary amines. (March – 2014)
b) Describe a chemical reaction given only by primary amines.
c) What is diazotisation?
Answer:
a) Reduction of nitriles – Nitriles on reduction with LiAIH₄ or catalytic hydrogenation produce primary amines.
OR
Reduction of amides with LiAIH₄ produce primary amines.
OR
Pthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.

b) Carbylamine reaction / Isocyanide test – Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines.
R-NH₂ + CHC₃ + 3KH \(\underline{\text { Heat }}\) R-NC + 3KCI + 3H₃O
c) Conversion of primary aromatic amines into dia zonium salts by reaction with nitrous acid is called diazotisation.

Question 1.
a) Amines are basic. Arrange the following amines in the increasing order of basic strength (Say – 2014)
CH₃ NH₂, (CH₃)₂, NH, (CH₃)₃N, C₆H₅NH₂.
b) Two well known reactions are given below: Suggest the main product of each reaction. Also give the name of each reaction.

Answer:
a) In gas phase:

(Hoffmann bromamide degradation reaction)

Question 1.
Amines are classified as primary, secondary and tertiary. (March – 2015)
a) Write the IUPAC name of the following compound: NH₂ – (CH₂) – NH₂
b) Which is stronger base – CH₃NH₂ or C₆H₅NH₂? Why?
Answer:
a) Hexane-1,6-diamine
b) CH₃NH₂ is a stronger base than C₆H₆NH₂.
Alkyl amines are stronger than aniline. This is because the unshared electron pair on nitrogen atom to be in conjugation with the benzene and thus making it less available for protonation.

CH₃NH₂, due to electron releasing nature of the CH₃– group, it pushes electrons towards nitrogen and thus makes the unshared electron pair more available for sharing.

Question 1.
a) Aromatic and aliphatic amines are basic in nature like ammonia. Arrange the following compounds in the increasing order of their basic strength: (Say – 2015)
CH₃NH₂,(CH₃),NH,NH₃,C₆H₅ -NH₂
b) How will you carry out the following reactions?
i) Hoffmann bromamide reaction
ii) Carbylamine reaction (Chemical equations not required)
Answer:
a) C₆H₅-NH₂<NH₃<CH₃NH₂<(CH₃)₂NH
b) i) Hoffmann bromamide reaction – When an amide is treated with bromine in an aqueous or ethanolic solution of NaOH an amine with one carbon atom less than that present in the amide is formed. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom.

ii) Carbylamine reaction – Aliphatic and aromatic primary amines on heating with chloroform and ethanolic KOH form isocyanides or carbylamines which are foul smelling substances.
R – NH₂ + CHCI₃ + 3KOH → R – NC + 3KCI + 3H₂O

Question 1.
Amines are classified as primary, secondary and tertiary amine. (March – 2016)
a) Represent the structure of secondary and tertiary amine.
b) How will you convert nitrobenzene to aniline?
c) Aniline does not undergo Friedel-Crafts reaction. Why?
Answer:

b) Nitrobenzene on reduction by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum and also by reduction with metals(Sn or Fe) in acidic medium gives aniline.

Or, the chemical equation:

c) Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation) due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst. Due to this, nitrogen of aniline acquires positive charge and hence acts as a strong deactivating group forfurther reaction.

Question 1.
Amines are basic in nature. (Say – 2016)
a) Arrange the following compounds in the increasing order of their basic strength
NH₃,C₂H₆NH₂, C₆H₅NH₂, (C₂H₅)₂ NH
b) How will you convert aniline to chlorobenzene?
Answer:
a) C₆H₅ – NH₂ < NH₃ < C₂H₅NH₂ < (C₂H₅)₂NH

Question 1.
a) Classify the following amines as primary, secondary and tertiary (March – 2017)

Identity the products B and C and write their formulae.
Answer:
a) Primary amines:

Secondary amine: (C₂H₅)₂NH
Tertiary amine:

Product B is aniline and product C ¡s 2,4,6- tribromoaniline

Question 1.
a) The most basic compound among the following is (Say – 2017)
i) C₂H₅NH₂
ii) C₆H₅NH₂
iii) NH₃
iv) (C₂H₅)₂NH

b) Compound A is treated with Ethanolic NaCN to give the compound C₂H₅CN(B). Compound B on reduction gives compound C. Identify compounds A and C.
Answer:
a) iv) (C₂H₅)₂NH
b) A-C₂H₅-X ethyl halide
C → C₂H₅-CH₂NH₂ propanamide

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 12 Aldehydes, Ketones and Carboxylic Acids.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 1.
a) Aldehydes and ketones are organic compounds containing carbonyl group. (March – 2010)
i) Write a chemical reaction to distinguish between aldehydes and ketones.
ii) Aldehydes and ketones can be subjected to Clemmensen reduction and Wolf-Kishner reduction. Name the reagents used in both cases.

b) How will you make the following conversions?
i) Ethanoic acid to ethanol.
ii) Propanoic acid to 2-chloropropanoic acid.
iii) Toluene to benzoicacid.
Answer:
a) i) Tollens’ Test: on warming an aldehyde with freshly prepared ammonical silver nitrate solution (Tollens’ reagent), a bright silver mirror is produced due to the formation of silver metal.
\(\begin{array}{r}
\mathrm{RCHO}+2\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}+3 \mathrm{OH}^{-} \rightarrow \\
\quad \mathrm{RCOO}^{-}+2 \mathrm{Ag}+2 \mathrm{H}_{2} \mathrm{O}+4 \mathrm{NH}_{3}
\end{array}\)
Ketones, will not anwserTollens’ test.
ii) Clemensons reduction – Zinc amalgam and concentrated hydrochloric acid
Woif-Kishner reduction – Hydrazine, KOHI Ethylene glycol

b) i) By reduction – When ethanoic acid is treated with lithium aluminium hydride it is reduced to ethanol.
\(\mathrm{CH}_{3} \mathrm{COOH} \frac{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{OH}\)
ii) By Hell-Volhartl-Zelinsky HVZ) reaction – When propanoic acid is treated with chlorine in presence of small amount of red phosphorus followed by hydrolysis 2-chloropropanoic acid is formed.

iii) When toluene is heated with alkaline solution of potassium permanganate the methyl side chain is oxidised to form benzoic acid.

Question 2.
Following are a group of compounds showing acidic behavior: (Say – 2010)

a) Give the IUPAC names of these compounds.
b) does not contain a carboxylic group, still it is acidic. Why?
c) Phenols are less acidic than carboxylic acids Why?
d) Formic acid is stronger than acetic acid. Why?
Answer:
a) HCOOH → Methanoic acid

b) Thus, phenol always remains ionised in solution giving H⁺ ions and is acidic in nature.

Resonance in phenol:

Resonance in phenoxide ion:

c) Carboxylic acids are more acidic than simple phenols because the carboxylate anion is more resonance stabilised by two equivalent resonance structures than phenoxide ion.

d) In acetic acid due to the I effect of the CH₃ – group attached to the – COOH group, the resonance stabilisation of the corresponding carboxylate anion is decreased while there is not I effect for H in formic acid. Hence, formic acid is stronger than acetic acid.

Question 3.

a) What is its IUPAC name? (March – 2011)
b) Explain the conversion of the above acid to the following:

Answer:
a) 3,4 – Dinitrobenzoic acid.

Question 4.
Aldehydes resemble ketones in many respects. (Say – 2011)
a) Give the reason fortheir resemblance.
b) Give a reaction in which aldehydes resemble ketones.
c) Write two tests to distinguish between aldehydes and ketones.
d) What is Cannizaro reaction?
Answer:
a) Both aldehydes and ketones contain the carbonyl functional group . The general formulae of aldehydes and ketones are given below:

(R and R’ ar same or different alkyl or aryl groups)

b) Both aldehydes and ketones undergo nucleophilic addition reactions. For example, both aldehydes and ketones react with hydrogen cyanide (HCN) to yield cyanohydrins.

c) 1) Tollens’ Test – On warming an aldehyde with freshly prepared ammonical silver nitrate solution (Tollens’ reagent), a bright silver mirror is produced due to the formation of silver metal. Here the aldehydes reduce Ag+ to metallic silver and are oxidised to the corre sponding carboxylate anion.

2) Fehlina’s Test – On heating an aldehyde with Fehling’s reagent (FehlingssolutionA- aqueous copper sulphate and Fehling’s solution B – alkaline sodium potassium tartarate), a red dish brown precipitate is obtained. Here the aldehydes reduce Cu² to Cu²O and are oxidised to corresponding carboxylate anion.

Ketones, being less reactive than aldehydes will not anwserTollens’ test and Fehling’s test.

d) Cannizzaro reaction – Aldehydes which do not have α-hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction on treatment with concentrated alkali. This reaction is called Cannizzaro reaction. In this reaction, one molecule of the aldehyde is reduced to corresponding alcohol while another molecule is oxidised to corresponding carboxylic acid salt.

Question 5.
Aniline is an aromatic pnmary amine. Starting from aniline a number of organic compounds can be prepared.
a) How is aniline converted to benzene diazonium chlonde?
b) How are the following obtained from benzene diazonium chloride?
i) Ch loro benzene
ii) Phenol
Answer:
a) Aniline is treated with nitrous acid. Nitrous acid is produced in the reaction mixture by the reaction of sodium nitrite with hydrochloric acid.
b) i) Benzenediazonium chloride when treated with cuprous chloride and HCI, the diazonium group is replaced by Cl ion to form chioroben zene.

ii) When the aqueous solution of benzene-dia zonium chloride is warmed upto 283 K it is hydrolysed to phenol.

Question 6.
a) Which named reaction is used to reduce CH₃COCI to CH₃CHO? (March – 2012)
b) Aldehydes and Ketones undergo reactions due to the presence of α – hydrogen atom.
i) Write the name of the reaction of aldehyde which takes place only because of the presence of α – hydrogen atom.
ii) How will you bring about the above reaction?
c) i) CH₂CICOOH is a stronger acid than CH₃COOH Why?
ii) How will you convert CH₃COOH to CH₂CICOOH
Answer:
a) Rosenmund Reduction.

b) i) Aldol condensation.
ii) Aldehyde or ketone containing at least one x-hydrogen undergo self-condensation reaction with dil. alkali (e.g. dii. NaOH) as catalyst to form β – hydroxy aldehydes (atdol) or β – hydroxy ketones (ketol) respec travel.

c) i) Electron with drawing group or-leffect group ‘Cl’ stabilises the conjugate acid of the carboxylate anion through delocalisation of negative charge and strengthens the carboxylic acid.
CH₂CI COOH > CH₃ COOH

ii) When a carboxylic acid that contains α – hydrogen is treated with Cl₂ or Br₂ in the presence of small amount of red phosphorus the α – hydrogen atoms are replaced by chlonne or bromine atoms to give α – halo carboxylic acids. This reaction is known as the Hell-Volhard-Zelinsky (HVZ) Reaction.

Question 7.
a) Complete the following. Write down the structure of A, B and C. (Say – 2012)

b) Write down the IUPAC names of A, B and C.
c) Explain the following reactions.
i) Cannizzaro reaction
ii) Esterification
Answer:

b) A – Propanoic acid
B – n-Butance
C – 2-Bromobutanoic acid

c) i) Canizzaro reaction -Aldehydes which do not have an α-hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction on treatment with concentrated alkali. In this reaction, one molecule of the aldehyde is reduced to alcohot while another is oxidised to carboxylic acid salt. e.g.

ii) Carboxylic acids react with alcohols or phenols in the presence of mineral acids such as concentrated H₂SO₄ or HCI gas as catalysts to form esters. This reaction is known as esterification.

Question 8.
a) Suggest a method of preparation of benzaldehyde from toluene. (March – 2013)
b) Aldehydes and ketones differ in their chemical reactions. How do they react with the following?
i) Tollen’s reagent
ii) Alcohol.
c) How will you convert propanoic acid into the following compounds?
i) Ethane
ii) Butane.
Answer:
a) Toluene on treating with chromyl chloride give benzaldehyde.This reaction is called Etard reaction.

b) i) Tollens’ Test: on warming an aldehyde with freshly prepared ammonical silver nitrate solution (Tollens’ reagent), a bright silver mirror is produced due to the formation of silver metal.
\(\begin{array}{r}
\mathrm{RCHO}+2\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}+3 \mathrm{OH}^{-} \rightarrow \\
\quad \mathrm{RCOO}^{-}+2 \mathrm{Ag}+2 \mathrm{H}_{2} \mathrm{O}+4 \mathrm{NH}_{3}
\end{array}\)
Ketones, will not anwserTollens’ test.

ii) Aldehydes react with one equivalent of monohydric alcohol in the presence of dry hydrogen chloirde to yield alkoxy alcohol known as hemiacetalswhich further react with one more molecule of alcohol to give a gem-dialkoxy compound known as acetal.

Ketones react with ethylene glycol in presence of dry hydrogen chloride to form cyclic products known as ethylene glycol ketals.

c) Propanoic acid → Ethane

Question 9.
a) Among formaldehyde, acetaldehyde, benzalde hyde and formic acid, which compounds undergo Cannizzaro reaction? Give reason. (Say – 2013)
b) What is esterification?
c) Write the chemical reaction to effect the transformation of sodium acetate to ethane.
d) Write the IUPAC names of the compounds given below,
i) CH₃-CH₂-CO-CH₃
ii) HOOC-CH₂-COOH
Answer:
a) i) Canizzaro reaction -Aldehydes which do not have an α-hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction on treatment with concentrated alkali. In this reaction, one molecule of the aldehyde is reduced to alcohot while another is oxidised to carboxylic acid salt. e.g.

ii) Carboxylic acids react with alcohols or phenols in the presence of mineral acids such as concentrated H₂SO₄ or HCI gas as catalysts to form esters. This reaction is known as esterification.

b) Formation of ester is known as esterification. Carboxylic acid react with alcohols or phenols in presence of acids like HCI to give ester.
c) By Kolbe’s electrolytic method – An aqueous solution of sodium acetate on electrolysis gives ethane at the anode.

Question 10.
a) Aldol condensation reaction is a special reaction of aldehydes. (March – 2014)
i) What is a Idol condensation reaction?
ii) Write the structural formula of aldol formed from ethana I.
b) Write simple chemical tests and observations used to distinguish between the following compounds:
i) Propanal and propanone
ii) Phenol and benzoic acid
c) Write the names of the reagents used to bring about the following transformations:

Answer:
a) i) Aldehydes and ketones having at least one α-hydrogen undergo a condensation reaction in the presence of dilute alkali to form β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol) respectively. This reaction is known as Akiol condensation.

b) i) Propanal and propanone – They can be distin guished by Tollens’ Test – When propanal is warmed with freshly prepared ammonical silver nitrate solution (Tollens’ reagent), a bright silver mirror is produced due to the formation of silver metal. Here propanal is oxidised to propanoate anion while it reduces Ag⁺ to metallic silver. Since ketones are less reactive, propanone will not answer this test.

[Other tests: (1) Fehling’s test – Propanal gives a red precipitate on heating with Fehling’s reagent while propanone does not answer this test. (2) iodoform test – Propanone being a methyl ketone, when heated with NaOH and 12 sOlUtiOn an yellow precipitate of lodoform is formed. But propanal does not answer the jodo-form test.]

ii) Phenol and benzoic acid – When benzoic acid is treated with NaHCO₃ solution there is bnsk effervescence of CO₂. But phenol being less acidic than benzoic acid will not react with NaH CO₃ solution.

c)

This reaction is caNed Rosenmund reduction.

This reaction is called Hell-Volhard-Zelinsky (HVZ) reaction.

Question 11.
a) Methanal (HCHO) is an aldehyde having no α – hydrogen atom. What are the products formed when methanal is treated with strong KOH solution? (Say – 2014)
b) How are the following conversions achieved?
i) Benzoyl chloride (C₆H₅COCI) to benzalde hyde (C₆H₅ – CHO)
ii) Acetic acid (CH₂COOH) to chioro acetic acid (CH₂CI – COOH)
iii) Ethanal (CH₃ – CHO) to Ethane (CH₃ – CH₃)
Answer:
a) Aldehydes which do not have α -hydrogen atom, undergo self oxidation and reduction (dispropor tionation) reaction on treatment with concentrated alkali. This reaction is called Cannizzaro reaction. ie, one molecule of the aldehyde is reduced to corresponding alcohol while another molecule is oxidised to the carboxylic acid salt. eg when methanal is treated with strong KQH solution it under goes self oxidation and reduction to give a mixture of potassium formate and methanol.

b) i) Benzoyl chloride is hydrogenated over catalyst, palladium on barium sulphate to get benzaldehyde (Rosenmund reduction).

ii) Acetic acid on treatment with chlorine in presence of small amount of red phosphorus is chlorinated at the a position to get α-chioroacetic acid (HeIl-Volhard-Zelinsky reaction).

iii) When ethanal is treated with zinc amalgam and concentrated hydrochloric acid the carbonyl group is reduced to -CH₂ group to get ethane.

Question 12.
Aldehydes, Ketones and Acids contain . (March – 2016)
a) Name the product obtained by the reaction between Acetic acid and Ethanol.
b) a) Give any Iwo tests to distinguish between aldehydes and ketones.
ii) Two chemical reactions are given below:
1) Identify the products of each reaction.
2) Give the name of each reaction.

Answer:
a) Acetic acid reacts with ethanol in presence of mineral acids such as concentrated H₂SO₄ as catalyst to form the ester ethyl acetate or ethyl ethanoate (CH₃COOC₂H. This reaion is known as esterification.

b) i) 1) Tollens test (Silver mirror test) – on warming an aldehyde with freshly prepared am moniacal silver nitrate solution (Tollens’ reagent) a bright silver mirror is produced due to the formation of silver metal. Here the aldehydes are oxidised to the corresponding carboxylate anion while they reduœ Ag to metallic silver.

2) Fehling’s Test – on heating an aldehyde with Fehlings reagent (mbdure of aqueous copper sulphate and alkaline sodium potassium tartarate), a reddish brown precipi tate is obtained. Here aldehydes are oxidised to the corresponding carboxylate anion while they reduce Cu²⁺ to Cu₂O.

Since ketones are less reactive than aldehydes they will not answer these two tests.

Question 13.
a) Explain aldol condensation taking CH₂-CHO example. (Say – 2015)
b) Write the named reactions involved in the following conversions:

c) How are the following conversions achieved?

Answer:
a) Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form β-hydroxy aIde hydes (aldol) or β-hydroxy ketones (ketol), respectively. This is known asAldol reaction. The aldol and ketol readily lose water to give α, β unsaturated carbonyl compounds which are al-dol condensation products and the reaction is called Aldol condensation.

e.g. CH₃-CHO undergo Aldol reaction in presence of dil NaOH to form 3-Hydroxybutanal which on heating lose water to form the Aldol condensation product But-2-enal.

b) i) Rosenmund reduction
ii) Cannizzaro reaction

Question 14.
Aldehydes, Ketones and Carboxylic acids are Car bonyl compounds. (March – 2016)
a) Aldehydes differ from Ketones in their oxidation reactions. Illustrate with one example.
b) How will you prepare benzaldehyde by Gatterman Koch reaction?
c) Write the reactions of carboxylic acid with the following reagents. (Write the chemical equations)
i) Thinoyl chloride (SOCl₂)
ii) Chlorine in presence of small amount of red phosphorous.
iii) Lithium Aluminium hydride (LiAlH₄) I ether.
a) Write a test to distinguish between aldehydes and ketones.
b) How will you prepare benzaldehyde by Etard’s rea dio n?
c) Howwillyou bring about the followng conversions? (Write the chemical equations)
i) Ethanol → Ethanoic acid
ii) Benzamide → benzoic acid
iii) Benzaldehyde → meta nitro benzaldehyde
Answer:
a) Aldehydes are easily oxidised to carboxylic acids containing same number of carbon atoms on treatment with mild or strong oxidising agents. Ketones are oxidised under vigorous conditions Le., with strong oxidising agents and at elevated temperatures to give carboxylic acids containing lesser number of carbon atoms.

OR

On warming an aldehyde with freshly prepared ammoniacal silver nitrate solution (Tollens’ reagent), a bright silver mirror is produced due to formation of silver metal. The aldehydes are oxidised to corresponding carboxylate anion.

Ketones will not give this reaction because Tollens’ reagent being a mild oxidising agent cannot oxidise ketones.

[Or, any other suitable example – Reaction with Fehling’s reagent, Reaction with Benedict’s reagent etc.]

b) When benzene is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride, it gives benzaldehyde.

Or, the chemical equation:

a) Fehling’s Test – on heating an aliphatic aldehyde with Fehling’s reagent (aqueous copper sulphate + alkaline sodium-potassium tolerate), a reddish-brown precipitate is obtained. Aldehydes are oxidised to corresponding carboxylate anion.

Aromatic aldehydes and ketones do not answer this test. This is because Fehling’s reagent being a mild oxidising agent cannot oxidise them.
[Or, any other suitable example – Fehling’s test, Benedict’s test etc.]

b) Toluene on treating with chromyl chloride (CrO₂Cl₂) in CS₂ forms a chromium complex WNCII on hydrolysis gives benzaldehyde.

Or, the chemical equation:

Question 15.
Aldehydes and ketones are the compounds having >C = O group (Say – 2016)
a) Choose the IUPAC name of the compound CH₃ – CH = CH – CHO
i) propen-1 -al
ii) But-2-en-1 -al
iii) Butanal
iv) But-2-en-2-al

b) Complete the following reaction:

OR
Aldehydes, Ketones and acids contain > C= O group.
a) Choose the IUPAC name of the compound (CH₃)₂CHCOOH
i) Butanoic acid
ii) Ethanoic acid
iii) 2-methyl propanoic acid
iv) Propanoic acid

b) Complete the following reaction:

Answer:

Question 16.
a) The product obtained when benzene is treated with carbon monoxide and hydrogen chloride in presence of anhydrous AICI₃ is (March – 2017)
i) Chlorobenzene
ii) Phenol
iii) Benzaldehyde
iv) Benzoic acid

b) How will you carry out the following conversions?

OR
Explain the following:
i) Esterification
ii) Tollen’s test
iii) HVZ reaction
iv) Decarboxylation of Carboxylic acid.
Answer:
a) iii) Benzaldehyde
b) i) Toluene on heating with alkaline KMnO₄ followed by acid hydrolysis give benzoic acid.

ii) Benzoic acid reacts with ammonia to give ammonium benzoate which on further heating at high temperature gives benzamide.

iii) Propanoic acid on reduction using LiAIH₄ gives Propan-1 -al.

iv) Ethanoic acid on heating with mineral acids such as H₂SO₄ or with P₂O₅ gives ethanoic anhydride.

i) Estenfication: Carboxylic acids are estenfied with alcohols or phenols in the presence of a mineral acid such as concentrated H₂SO₄ or HCI gas as a catalyst.

ii) Tollens’s Test: On warming an aldehyde with freshly prepared ammoniacal silver nitrate solution (Tollens’ reagent), a bright silver minor is produced due to formation of silver metal. The aldehydes are oxidised to corresponding carboxylate anion.

Ketones will not give this reaction because Tollens’ reagent being a mild oxidising agent cannot oxidise ketones.

iii) HVZ reaction: Carboxylic acids having an α – hydrogen are halogenated at the α – position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give halocarboxylic acids. The reaction is known as Hell-Volharti-Zelinsky (HVZ) reaction.

iv) Decarboxylation: Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with sodalime (NaOH and CaÇ. in the ratio of 3: 1). The reaction is kriownas decar carboxylation.

Question 17.
a) Which among the following reduces Tollen’s reagent? (Say – 2017)
i) Methanal
ii) Propanone
iii) Benzophenone
iv) Acetophenone

b) Since both aldehydes and ketones possess carbonyl functional group, they undergo similar chemical reactions.
i) Explain the structure of carbonyl group.
ii) Explain Aldol condensation with an example.
OR

a) Which among the following does not give red precipitate with Fehling’s solution?
i) Ethanal
ii) Propanal
iii) Butanal
iv) Benzaldehyde

b) How will you bring about the following conversions?
i) Toluene into Benzaldehyde
ii) Benzoic Acid to Benzamide

c) Explain Cannizaro reaction with an example.
Answer:
a) i)Methanal
b) i) The carbonyl C atom is sp² hybridized. Carbon forms 3 α bonds and one π bond. 

The C = O bond is polarised due to higher electronegativity of oxygen relative to carbon.

ii) Aldehyde or ketone containing at least one x-hydrogen undergo self-condensation reaction with dil. alkali (e.g. dil. NaOH) as catalyst to form β – hydroxy aldehydes (atdol) or β – hydroxy ketones (ketol) respec travel.

OR

a) iv) Benzaldehyde
b) i) Etard’s reaction
Toluene on treatment with CrO₃ and acetic unhydride gives benzaldehyde

c) i) Canizzaro reaction -Aldehydes which do not have an α-hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction on treatment with concentrated alkali. In this reaction, one molecule of the aldehyde is reduced to alcohot while another is oxidised to carboxylic acid salt. e.g.

ii) Carboxylic acids react with alcohols or phenols in the presence of mineral acids such as concentrated H₂SO₄ or HCI gas as catalysts to form esters. This reaction is known as esterification.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 11 Alcohols, Phenols and Ethers.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 11 Alcohols, Phenols and Ethers

Question 1.
Phenols are more acidic than alcohols. (March – 2010)
i) Name the product obtained when phenol is treated with chloroform in the presence of NaOH.
ii) Name the above reaction.
iii) What is the product obtained when phenol is treated with con. HNO₃?
iv) Write the structure and IUPAC name of the above product.
v) Ethanol and propane have comparable molecular masses but their boiling points differ widely. Which of them has a higher boiling point? Substantiate your answer.
Answer:
i) Salicylaldehyde or 2-Hydroxy benzaldehyde.

ii) Reimer-Tiemann Reaction
iii) When phenol is treated with concentrated nitric acid, it is converted to 2,4,6 trinitrophenol, commonly known as picric acid.


v) Ethanol : Ethanol molecules can associate through intermolecular hydrogen bonding. But propane has no hydrogen bonding. Therefore, ethanol has high boiling point (351 K) than propane (231 K).

Question 2.
A compound A reacts with thionyl chloride to give compound B. B reacts with magnesium in ether medium to form a Grignard reagent which is treated with acetone and the product on hydrolysis gives. (Say – 2010)

Identify (A) and (B). Write down the chemical equations for the reactions involved.
Answer:

Question 3.
Ethers are generally non-reactive compounds. One of the important reactions of Ethers is the action of HI. (March – 2011)

ldentifyA& B. Explain the reaction.
Answer:
Alkyl aryl ethers on reaction with Hl are cleaved at the alkyl – oxygen bond. Cleavage does not occur at the aryl – oxygen bond. This is due to the fact that the aryl – oxygen bond has high stability caused by resonance. This reaction yields phenol and the corresponding alkyl iodide.

Question 4.
Mixture of conc. HCI and anhydrous ZnCl₂ is an important reagent which helps to distinguish between 10, 20 and 3° alcohols. (Say – 2011)
a) Give the name of the above reagent.
b) Give one example each for 10, 20 & 30 alcohols.
c) Explain how the above reagent helps to distinguish above three types of alcohols.
Answer:
a) Lucas reagent

c) Lucas test 10, 20 and 3° alcohols can be distin guished by Lucas test. Alcohols are soluble in Lucas reagent while their halides are immiscible and produce turbidity in solution. The difference in reactivity of three classes of alcohols with HCI distinguishes from one another. The order of reactivity of alcohols with Lucas reagent is in the order 3> 20> 1°. Thus, in the case of 30 alcohols turbidity is produced immediately as they form the halides easily. In the case of 2° alcohols turbidity is produced within 5 minutes at room tem perature. In the case of 1° alcohols no turbidity is produced at room temperature since they are least reactive.

Question 5.
a) Write the name or structure of the compounds A and B in the following reactions: (March – 2012)

b) Vapours of an alcohol ‘C’ on passing over heated copper produce compound ‘D’. ‘D’ on reaction with CH₃MgCI followed by hydrolysis produce 2-Methyl butan-2-ol. Write the name or structure of compounds ‘C’ and ‘D’.
Answer:

Question 6.
Methanol and ethanol are two commercially important alcohols. (Say – 2012)
i) Write one method of preparation of methanol and ethanol.
ii) Name the products obtained when ethanol is treated with CrO₃ in an anhydrous medium.
iii) The boiling point of ethanol is higher than that of methoxy methane. Give reason.
Answer:
i) Methanol is produced by catalytic hydrogenation of carbon monoxide at high pressure (200 atm – 300 atm) and temperature (573 K -673 K) and in presence of ZnO – Cr₂O₃ catalyst.

Ethanol is obtained commercially by fermentation of molasses. The enzyme invertase present in yeasf converts sugar (sucrose) glucose and fructose invertase converts sugar to glucose and fructose Zymase converts glucose and fructose to ethanol and CO₂.

ii) When ethanol is treated with CrO₃ in anhhdrous medium it is oxidised to ernanal (acetaldehyde).
\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \stackrel{\mathrm{CrO}_{3}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CHO}\)

iii) In ethanol there is a presence of hydrogen bond but in methoxy methane no hydrogen bond. Therefore boiling point of ethanol is higher than methoxy methane.

Question 7.
a) Write the IUPAC names of all the possible isomers with molecular formula C₃H₈O. (March – 2013)
b) Phenol is usually manufactured from cumene. Write the structure of cumene.
c) Primary, secondary and tertiary alcohols can be distinguished by Lucas test.
i) What is Lucas reagent?
ii) Write the observation for primary, secondary and tertiary alcohols in Lucas test.
Answer:
a) C₃H₈O

b) Cumene is isopropyl benzene. Its structure is

c) i) Mixture of concentrated HCI and ZnCl₂
ii) Refer SAY 2011, Question 1.c

Question 8.
How are the following conversions carried out? Represent the chemical reactions. (Say – 2013)
a) Ethanol to Ethanal
b) Phenol to Picric acid
c) Phenol to Benzene
d) Phenol to Tribromophenol
Answer:
a) Ethanol to Ethanal – When the vapors of ethanol are passed over heated copper at 573 K, dehydrogenation takes place to form ethanal.

b) Phenol to Picric acid – When phenol is treated with concentrated nitric acid it is converted to 2,4,6-Tnnitrophenol commonly known as picric acid.

c) Phenol to benzene – When phenol is heated with zinc dust it is converted to benzene.

d) Phenol to Tnbromophenol – When phenol is treated with bromine water 2,46-Tribromophenol is formed.

Question 9.
a) How will you prepare the following compounds using Grignard reagent? (March – 2014)
i) Primary alcohol
ii) Secondary alcohol
b) How will you distinguish primary and secondary alcohols using the Lucas test?
c) Write the correct pair of reactants for the preparation of t-butyl ethyl ether by Williamson synthesis.
Answer:
a) i) When a suitable Grignard reagent is treated with formaldehyde gives which on hydrolysis primary alcohol.

ii) When a suitable Grignard reagent is treated with any suitable aldehyde other than formaldehyde an adduct is formed which on hydrolysis yields secondary alcohol.

b) Refer SAY 2011, Question f.c
c) When sodium tert-butoxide is treated with ethyl bromide, t-butyl ethyl ether is formed.

Question 10.
a) Write the name or formula of the following: (Say – 2014)
i) A simple ether
ii) A mixed ether
iii) A dihydnc alcohol
iv) A trihydric alcohol.

b) Phenol on treatment with Br₂ in CS₂ at low temperature gives two isomenc monobromo phenols ‘X’ and ‘Y’. But phenol on treatment with bromine water gives a white precipitate ‘Z’. Identify the products ‘X’, ‘Y’. and ‘Z’.
Answer:
a) I) Dimethyl ether(Methoxymethane) – CH₃ – O – CH₃
ii) Ethylmethyl ether (Methoxyethane) – CH₃– O – C₂H₅
iii) Ethylene glycol (Ethane-1 ,2-diol) – HO – CH₂– CH₂ – OH
iv) Glycerol (Propane-1, 2, 3-trio I) – HO – CH₂ – CH(OH) – CH₂ – OH

Question 11.
Alcohols are compounds with general formula R-OH. (March – 2015)
a) Alcohols are soluble in water. What is the reason?
b) i) Explain a method formanufadure of Ethanol.
ii) How will you convert phenol to benzene?
Answer:
a) Solubility of alcohols in water is due to their ability to form hydrogen bonds with water molecules.
b) i) Ethanol (C₂H₅OH) is obtained commercially by fermentation. The sugar in molasses, sugarcane or fruits such as grapes is converted to glucose and fructose in the presence of an enzyme invertase. Glucose and fructose un dergo fermentation in the presence of an other enzyme zymase, which is found in yeast.

OR
By hydration of ethene – Ethene reacts with water in the presence of acid as catalyst to form ethanol.
OR
By hydration of ethene – Ethene reacts with water in the presence of acid as catalyst to form ethanol.
CH₂ = CH₂ + H₂O CH₃CH₂OH
ii) Phenol is converted to benzene by heating with zinc dust.

Question 12.
a) Write suitable reagent or reagents used for the following conversions: (Say – 2015)

b) Wnte a test to distinguish between phenol and alcohol.
Answer:
a) i) aq.NaOHoraq,KOH
ii) Conc. H₂SO₄ at 413 K
iii) CHCI₃ + aq. NaOH

b)

Question 13.
a) Complete the following: (March – 2016)

b) Explain the following:
i) Estenfication
ii) Williamson Synthesis
Answer:

b) i) Esterification – Alcohols and phenols react with carboxylic acids, and acid anhydrides in presence of small amount of concentrated sulphuric acid and with acid chlorides in presence of pyridine to form esters. This reaction is known as estenfi cation.

Or, any of the following chemical equations:

ii) Wihiamson synthesis – When and alkyl halide is treated with sodium alkoxide ether is formed. It is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers.

Or, the chemical equation:
R-X + R’ – ONa → R-O-R’ + NaX

Or, any other specific example.

Question 14.
a) Phenol when treated with Con. HNO3 gives, (Say – 2016)
i) o-Nitrophenol
ii) p-Nitropbenol
iii) 2,4,6 -Thnitro phenol
iv) a mixture of o-nitrophenol and p-nitrophenol.

b) Methanol and ethanol are two commercially important alcohols. Write one method each for the preparation of methanol and ethanol.
Answer:
a) iii) 2,4,6-Tnnitrophenol
b) Methods of preparatIon of methanol :
1. Destructive distillation of wood.
2. Catalytic hydrogenation of carbon monoxide at high pressure and temperature in the presence of ZnO – Cr₂O₃ catalyst.

Methods of preparation of ethanol:
Fermentation: The sugar in sugarcane or grapes is converted to glucose and fructose in the presence of an enzyme invertase. Glucose and fructose are converted to ethanol and CO₂ by the enzyme zymase, which is found in yeast.

OR

Question 15.
a) Arrange the following compounds in the order of increasing boiling points: (March – 2017)
Ethanol, Propan-1-ol, Butan-1 -01, Butan-2-ol

b) In the lab students were asked to carry out the reaction between phenol arid conc. HNO₃. But one student, ‘A’ carried out the reaction between phenol and dil. HNO₃. Do you think that the student ‘A’ got the same result as others. Substantiate with suitable explanations. (Also write the chemical equations wherever necessary.)
Answer:
a) Ethanol < Propan-1-ol < Butan-2-ol < Butan-1-ol
b) Student ‘A’ will get a mixture of o-Nitropehnol and p-Nitrophenol.

Other students will get 2,4,6-tnnirophenol (picncacid) as the product.

Question 16.
a) Identify the product (Say – 2017)

b) Complete the following:

Answer:
a) ii) CH₃CH₂OH
b) i) Picric acid or 2, 4, 6, Trinitro phenol

ii) Salicyl aldehyde or2-hydroxy benzaldehyde or

iii) C₆H₅OH + CH₃l
phenol + iodomethane

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 10 Haloalkanes and Haloarenes.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 10 Haloalkanes and Haloarenes

Question 1.
Most of the organic chlorides, bromides and iodides react with certain metals to give compounds containing carbon-metal bonds. (March – 2010)
i) Give one example for such a compound.
ii) How will you prepare the above compound?
b) Write any two electrophilic substitution reactions of chlorobenzene.
Answer:
i) Grignard reagent; CH₃MgCI (Methyl magnesium chloride)
ii) Grignard reagents are prepared by treating haloalkanes with magnesium metal in dry ether.
\(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{Mg}_{\frac{\text { dry ether }}{\longrightarrow}} \mathrm{CH}_{3} \mathrm{MgCl}\)

b) Halogenation : When chlorobenzene is treated with chlorine in presence of anhydrous FeCl₃ a mixture of 1,4-Dichlorobenzene (major product) and 1.2-Dichlorobenzene is formed.

Nitration : When chlorobenzene is treated with nitrating mixture (mixture of conc.HNO₃ and cone. H₂SO₄) a mixture of 1-Chloro-4-nitrobenzene (major product) and 1-Chloro-2-nitrobenzene (minor product) is formed.

Question 2.
An organic compound A reacts with metallic sodium in an ether medium to form ethane. A reacts with Magnesium in ether medium to give B, which on hydrolysis gives methane. Identify A and B. Write down the chemical equations for the reactions involved. (Say – 2010)
OR
Bromoethane, when treated with alcoholic KOH, gives ethene, KBr, and H₂O.
a) Identify the type of reaction.
b) Instead of bromoethane, if you take 2- bromobutane, what is the major product obtained? Write down the chemical equation for the reaction.
c) Explain the rule behind the above reaction.
Answer:
A → A CH₃ Cl (Methyl chloride)
B → CH₃Mg Cl (Methyl magnesium chloride)

OR

a) β – elimination (Dehydrohalogenation)

c) Saytzeffs rule – It states that in dehydro-halogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

Question 3.
Haloalkanes and Haloarenes react with metals to give Hydrocarbons or products from which hydrocarbons are obtained easily.
(March – 2011)

Identify the product and name the reaction.

Identify the product and name the reaction.

Identify A & B.
Answer:

This reaction is called the Wurtz-Fitting reaction.

Question 4.
Alkyl halides are the starting materials for the synthesis of a number of organic compounds. How are the following compounds obtained from the alkyl halide CH₃ – CH₂ – Br? (Say – 2011)
a) Ethene
b) Ethanol
c) Butane
d) Ethoxy Ethane
Answer:
a) Conversion of bromoethane to ethene – By β – elimination reaction.
OR

b) Conversion of bromoethane to ethanol – by alkaline hydrolysis.
When bromoethane is treated with aqueous KOH or moist Ag20 ethanol is formed.
OR
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{KOH}_{(\mathrm{aq})} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}+\mathrm{KBr}\)
c) Conversion of bromoethane to n-butane – by Wurtz reaction.
When bromoethane is treated with sodium in dry ether n-Butane is formed.
OR

d) Conversion of bromoethane to ethoxyethane – by Williamson’s synthesis.
When bromoethane is treated with sodium ethoxide ethoxyethane is formed.
OR

Question 5.
Nucleophilic substitution reactions are of two types – S_N1 reactions and S_N2 reactions. (March – 2012)
i) Write any two differences between S_N1 and S_N2 reactions.
ii) Write any two reasons for the less reactivity of aryl halides towards nucleophilic substitution reactions.
Answer:
i)

ii) Aryl halides are much less reactive than haloalkane or alkyl halides towards nucleophilic substitution reaction due to

1) Resonance effect – in haloalkanes the electron pairs on halogen atom are in conjugation with -IT electrons of the ring and thus the C – X bond acquires a partial double bond character. Asa result the bond cleavage in haloarenes is difficult than in haloalkanes.

2) Difference in hybridization of carbon atom in C – X bond – in haloalkanes, the carbon atom attached to halogen is sp³ hybridized while in case of haloarenes the carbon atom attached to halogen is sp² hybridized which is more electronegative. Hence, the C – X bond length in haloarenes (169 pm) is less than that in haloalkanes (177 pm). It is difficult to break a shorter bond than a longer bond. Therefore, Haloarenes are less reactive towards nucleophilic substitution reaction.

3) InstabilIty of phenyl cation – in case of haloarenes, the phenyl cation formed as a result of self-ionization will not be stabilized by resonance and therefore, S_N1 mechanism Is ruled out.

4) Because of the possible repulsion, it is less likely for the electron-rich nucleophile to approach electron-rich arenes,

Question 6.
Haloarenes undergo electrophilic substitution race fans. Explain the Important electrophilic substitution reactions of chlorobenzene. (Write down the chemical equation) (Say – 2012)
Answer:
Halogenation: When chlorobenzene Is treated with chlorine In presence of anhydrous FeCl₃ a mixture of 1. 4-Dichlorobenzene (major product) and I 2-Dichlorobenzene b formed.

Nitration: When chlorobenzene is treated with nitrating mixture (mixture of conc.HNO₃ and conc. H₂SO₄) a mixture of 1-Chloro-4-nitrobenzene (major product) and 1-Chloro-2-nitrobenzene (minor product) is formed.

Sulphonation: When chlorobenzene is heated with conc.H₂SO₄ a mixture of 4-Chiorobenzene suiphonic acid (major product) and 2-Chiorobenzene sulphonic acid (minor product) is formed.

Question 7.
a) For the preparation of alkyl chlorides from alcohols, thionyl chloride (SOCl₂) is preferred. Give reason. (March – 2013)
b) Haloalkanes undergo b – elimination reaction in presence of alcoholic potassium hydroxide.
i) Which is the major product obtained by the b- elimination of 2-Bromo pentane?
ii) Name the rule, which leads to the product in the above elimination reaction.
c) Write the chemical equation for the preparation of toluene by Wurtz-Fitting reaction.
Answer:
a) S02 and HCI being escapable gases, the reaction of alcohols with thionyl chloride gives pure alkyl chlorides.
R-OH + SOCI₂ → R-CI + SO₂ + HCI
b) i) pent-2-ene is the major product since it has a greater number of alkyl groups attached to the doubly bonded carbon atoms.

Question 8.
Haloarenes undergo nucleophilic and electrophilic substitution reactions. (Say – 2013)
a) Write two examples for ambident nucleophiles.
b) Write one example for the nucleophilic substitution reaction of chlorobenzene.
c) Write any two examples of electrophilic substitution reaction of chlorobenzene.
Answer:
a) Cyanide ion (CN^–) and nitrite ion (NO₂).
b) Chlorobenzene when heated with aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atmosphere gets converted to phenol by a nucleophilic substitution reaction.

c) Nitration – When chloroform is treated with the nitrating mixture (a mixture of conc, HNO₃ and cone. H₂SO₄), a mixture of 1-Chloro-2-nitrobenzene (minor product) and 1-Chloro-4-nitrobenzene (major product) is obtained.

Friedel-Crafts alkylation – When chloroform is treated with methyl chloride in presence of anhydrous AICI₃ a mixture of 1 – Chloro – 2 – methylbenzene (minor product) and 1-Chloro-4-methylbenzene (major product) are obtained.

Question 9.
a) Most important chemical reactions of haloalkanes are their substitution reactions. (March – 2014)
i) What is S_N1 reaction?
ii) Arrange the four isomeric bromo butanes in the increasing order of their reactivity towards SN1 reaction.
b) How will you prepare chlorobenzene from benzene diazonium chloride?
Answer:
a) i) S_N1 reaction is a unimolecular nucleophilic substitution reaction. It involves the substitution of a weaker nucleophile by a stronger one and follows first-order kinetics, i.e., the rate of reaction depends upon the concentration of only one reactant. It occurs in two steps. In step I, the polarised C – X bond undergoes slow cleavage to produce a carbocation.
ii) S_N1 reaction follows the order 10 alkyl halides < 2° alkyl halides < 3° alkyl halides.
CH₃CH₂CH₂CH₂Br < (CH₃)₂CHCH₂Br < CH₃CH₂CH(Br)CH₃ < (CH₃)₃CBr

b) By Sandmeyer reaction – When freshly prepared benzene diazonium chloride solution is mixed with cuprous chloride the diazonium group is replaced by -Cl to form chlorobenzene.

OR
By Gatterman reaction – When freshly prepared benzene diazonium chloride solution is treated with hydrochloric acid in presence of copper pow- der the diazonium group is replaced by -Cl to form chlorobenzene.

Question 10.
a) i) Write ‘Saytzeff rule’. (Say – 2014)
ii) The products A and B of the following reaction are two isomeric alkenes. Identify Aand B.

b) Identify the main product of the following reactions. Suggest whether the reaction is S_N1 or S_N2.

Answer:
a) i) The Saytzeff rule states that in dehydrohalogenation reactions, the preferred product is that alkene which has a greater number of alkyl groups attached to the doubly bonded carbon atoms.

b) i) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br} \stackrel{\mathrm{aqNaOH}}{\longrightarrow}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}\) This reaction follows S_N1 mechanism. Tertiary alkyl halides undergo S_N1 reaction very fast because of the high stability of 3° carbocations.

ii) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\)
This reaction follows S_N1 mechanism. Ben- zylic halides show high reactivity towards the S_N1 reaction because the benzylic carbocation formed is stabilised through resonance.

Question 11.
a) Among the following which one is chlorine-containing insecticide? (March – 2015)
i) DOT
ii) Freon
iii) Phosgene
iv) lodoform

b) Halo arenes undergo Wurtz-Fittig reaction.
i) What is Wurtz-Fittig reaction?

Write the formulae of A and B in the above reaction.
Answer:
a) i) DDT
b) i) A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in dry ether.

Question 12.
i) State Saytzeff Rule. (Say – 2015)
ii) Identify the major and minor products obtained by the reaction between 2-bromo butane and alcoholic KOH.
iii) Write the product obtained by the reaction between 2-bromo butane and aqueous KOH.
iv) 2-bromo butane exhibit optical isomerism. What is optical isomerism?
Answer:
i) The Saytzeff rule states that in dehydro-halogenation reactions, the preferred product is that alkene which has a greater number of alkyl groups attached to the doubly bonded carbon atoms.

iv) Optical isomerism is the phenomenon in which molecules having the same molecular, as well as structural formulae, differ in the direction of rotation of the plane of plane polarised light.

Question 13.
a) Aryl halides are less reactive in nucleophilic substitution reactions. (March – 2016)
i) Write any two reasons for less reactivity,
ii) Give one example for nucleophilic substitution reactions of aryl halides.
b) Write a method for the preparation of alkyl halides.
c) Which of the following is not a polyhalogen compound?
a) Chloroform
b) Freon
c) Carbon tetrachloride
d) Chlorobenzene
Answer:
i) 1. Resonance Effect- In aryl halides the electron pairs on halogen atom are in conjunction with π-electrons of the ring and different resonating structures are possible, e.g. resonance in chlorobenzene:

The C – X bond acquires a partial double bond character due to resonance. As a result, the bond cleavage in aryl halides is difficult than in alkyl halides and therefore, they are less reactive towards nucleophilic substitution reaction.

2. Difference in hybridisation of carbon atom in C-Xbond: In alkyl halides, the carbon atom attached to halogen is sp³ hybridised while in the case of aryl halides, the carbon atom attached to halogen is sp² hybridised.

The sp² hybridised carbon with a greater s-character is more electronegative and can hold the electron pair of C – X bond more tightly than sp³ hybridised carbon in alkyl halides with less s-character. Thus, the C – X bond length in aryl halides is less than that in alkyl halides. Since it is difficult to break a shorter bond than a longer bond, therefore, aryl halides are less reactive than alkyl halides towards nucleophilic substitution reaction.

3. Instability of phenyl cation – In the case of aryl halides, the phenyl cation formed as a result of self-ionization will not be stabilised by resonance and therefore, S_N1 mechanism is ruled out.

4. Because of the possible repulsion, it is less likely for the electron-rich nucleophile to approach electron-rich arenes. (any two reasons)

ii) Chlorobenzene when heated in aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atmospheres followed by acidification gets converted to phenol.

b) Alkyl halides can be prepared by treating alcohols with concentrated halogen acid in presence of anhydrous zinc chloride as a catalyst.

c) d) Chlorobenzene

Question 14.
Haloalkanes and haloarenes are compounds containing halogen atom. They undergo many types of reactions. (Say – 2016)
a) Identify the product formed in the following reaction:

b) i) Chloroform is stored in closed, dark coloured bottles completely filled up to the neck. Give reason.
ii) Write any two differences between SN1 and SN² reactions.
Answer:
a) iii) CH₃ – CH₂ = CH₂
b) i) Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carbonyl chloride, also known as phosgene. It is therefore stored in closed dark coloured bottles completely filled so that air is kept out.
\(2 \mathrm{CHCl}_{3}+\mathrm{O}_{2} \stackrel{\text { light }}{\longrightarrow} 2 \mathrm{COCl}_{2}+2 \mathrm{HCl}\)
ii) Refer March 2012, Question 1 .(i)

Question 15.
a) An ambident nucleophile is …………. (March – 2017)
i) Ammonia
ii) Ammonium ion
iii) Chloride ion
iv) Nitrite ion

b) Haloalkanes and Haloarenes are organohalogen compounds.
i) Suggest a method for the preparation of alkyl chloride.
ii) Aryl halides are less reactive towards Nucleophilic substitution reactions.
Give reasons.
Answer:
a) iv) Nitrite ion
b) i) Alkyl chlorides can be prepared by passing dry hydrogen chloride gas through a solution of alcohol or by heating a solution of alcohol in concentrated aqueous HCI in presence of anhydrous zinc chloride as a catalyst.
\(\mathrm{R}-\mathrm{OH}+\mathrm{HCl} \quad \stackrel{\mathrm{ZnCl}_{2}}{\longrightarrow} \mathrm{R}-\mathrm{Cl}+\mathrm{H}_{2} \mathrm{O}\)

Or, by the action of alcohols with PCl₃, PCI₅ or SOCI₂.

3R-OH + PCl₃ → 3R-CI + H₃PO₃
R-OH + PCl₅ → R-CI + POCl₃ + HCl
R-OH + SOCl₂ → R-CI + SO₂ + HCl

Or, chlorination of hydrocarbons in presence of light or heat.
\(\begin{array}{l}
\text { eg. } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \frac{\mathrm{Cl}_{2} \text { NV ligtt }}{\text { or heat }} \longrightarrow \\
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_{3}
\end{array}\)
Or, by the addition of hydrogen chloride to alkenes.
\(\text { eg. } \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{HCl} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}_{3}\) (Any one method)
ii) Refer March 2012 Question 1 (ii)

Question 16.
On kinetic consideration, nucleophilic substitution in aryl/alkyl halides may be S_N1 or S_N2 mechanisms. (Say – 2017)
a) Briefly explain S_N2 mechanism with an example.
b) In dehydrohalogenation of 2-Bromopentane why Pent-2-ene is major product and Pent-ene is minor product.
Answer:
a) S_N2 mechanism (bimolecular nucleophilic substitution)
1. Takes place in a single step
2. Through the formation of the intermediate transition state
3. Inversion of configuration occurs

b) Saytzeff Rule: In dehydrohalogenation reaction, the more substituted alkene is the major product.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 9 Coordination Compounds.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 9 Coordination Compounds

Question 1.
[Cr(NH₃)₄CI₂] Br is a co-ordination compound. (March – 2010)
a) Identify the central metal ion of the above compound.
b) Name the ligands present in it.
c) What is its coordination number?
d) Write its IUPAC name.
e) Write the Ionization isomer of the above compound.
Answer:
a) Chromium (Cr)
b) Ammine (NH₃), Chloride (Cl- )
c) 6
d) Tetraamminedichloridochromium (lll)bromide
e) [Cr(NH₃)₄CIBr] Cl. By exchanging the ions inside and outside the coordination sphere.

Question 2.
When CuSO₄ is mixed with an excess of NH₃, a deep blue coloured solution is obtained. (Say – 2010)
a) Write the formula of the compound formed.
b) What is the IUPAC name of the compound?
c) What do you understand by the term coordination number and ligand in a coordination compound?
d) Give the oxidation number and coordination number of the central metal atom of the deep blue coloured compound.
Answer:
a) [Cu (NH₃)₄]SO₄
b) Tetraammine copper (ll) sulphate
c) It is the number of ligand atoms to which the metal is directly bonded. Ligands are neutral molecules or ions bounded to the central atom/ion in the coordination entity.
d) Oxidation number of Cu in [Cu(NH₃)₄]SO₄ is +2 Coordination number of Cu in [Cu(NH₃)₄]SO₄ is 4.

Question 3.
The geometry and magnetic properties of complexes can be explained by V.B. Theory. (March – 2011)

The octahedral complex [Co(NH₃]₆]³⁺ is diamagnetic while the octahedral complex [C0F₆]^3- is diamagnetic. Explain using VB. Theory,
Answer:
In the complex, [Co(NH₃)₆]³⁺ the cobalt ion is in +3 oxidation state and has the electronic configuration 3a⁶. It undergoes d²sp³ hybridisation as shown below:

Here, the six pairs of electrons, one from each NH₃ molecule, occupy the six hybrid orbitals. Thus, the complex has octahedral geometry and is diamagnetic be cause of the absence of an unpaired electron.

In [CoF₆]^3- also he cobalt ion is in +3 oxidation state and has the electronic configuration 3d⁶. Since F ion provides a weak ligand field one 4s, three 4p and two 4d orbitaIs hybridise to yield six sp³d² hybrid or bitaIs pointing towards the six ends of an octahedron. The six Fions then donate a pair of electrons to each of these vacant orbitals to have an octahedral geometry. The presence of unpaired electrons in 3d orbit ais makes the complex paramagnetic.

Question 4.
The central ion Ag⁺ with coordination number 2 forms a positive complexion with NH₃ ligand. Also Ag forms a negative complex with CN- ligand.
a) Writetheformula of above positive and negative complexions. Give the IUPAC name of each.
b) Give the denticity of NH₃ and CN- ligands.
c) Write the formula and name of a hexadentate ligand.
Answer:
a) [Ag(NH₃)₂]⁺ – Diamminesilver (l) ion [Ag(CN)₂] – Dicya noargentate(l) ion
b) Denhcìty of NH₃ is 1, since N isthe only donor atom. Densicity of CN is 2, since both C and N atoms can act as donor atoms.
c) (EDTA⁴) ion – Ethylenediaminetetraacetate ion.

Question 5.
Consider the coordination compound [Co(NH₃)₅SO₄] Br. (March – 2012)
a) Write the IUPAC name of the above coordination compound.
b) What is the primary valence and secondary valence of the central metal, cobalt, in the above coordination compound?
c) Which type of structural isomerism is exhibited by the above coordination compound?
Answer:
a) Pentaamminesulphatocobalt (lll) bromide
b) (Co (NH₃)₅ SO₄] Br Primary valency = I Secondary valency =6
c) Ionisation isomerism.

Question 6.
[Cr(NH₃)₅CO₃] Cl is a coordination compound. (Say – 2012)
i) Name the central metal ion of the above compound.
ii) What is its IUPAC name?
iii) Name the ligands present in the above compound.
iv) Whether the ligands present in the above compound are ambdentate ligands? Why?
Write the ionisation isomer of the above compound.
Answer:
i) Chromium (Cr)
ii) Pentamminecarbonatoch romium(lll) chloride
iii) Ammonia, Carbonate ion
iv) No. Both ammonia and carbonate ion are not ambidentate ligands because they have only one donor site to bind with the metal.
v) [Cr(NH₃)₅Cl] CO₃ Pentamminechloridochromium (lll) carbonate

Question 7.
The magnetic behaviour of a complex can be explained on the basis of the Valence Bond (VB) theory. (March – 2013)
a) [CO(NH₃)₆]³⁺ is a diamagnetic complex and [COF₆]^3- is a paramagnetic complex. Substantiate the above statement using VB theory.
b) Classify the above-mentioned complexes into inner orbital and outer orbital complexes.
Answer:
a) Refer (March 2011 Question No.1
b) Inner orbital complex → [Co (NH₃)₆]³⁺ since the innerd orbital (3d) is used for hybridisation in this complex.
Outer orbital complex → [CoF₆]^3- since, the outer d orbital (4d) is used for hybridisation in this complex.

Question 8.
Many theories have been put forth to explain the nature of bonding in coordination compounds. (Say – 2013)
a) On the basis of valence bond theory account for the diamagnetic behaviour of [Ni(CN)₄]^2-.
b) What is the shape of the above complex?
c) Arrange the following ligands in the increasing order of their field strengths, (as per the spectrochemical series) CI-, CO, H₂0, OH
Answer:
a) In the complex ion [Ni(CN)₄]^2- the central metal atom Ni is in +2 oxidation state and has the electronic configuration 3d8. Since CN – is a strong field ligand two unpaired electrons are forced to pair up against and give fourdsp2 hybrid orbitals. Each of the hybrid orbitals receive a pair of electrons from a cyanide ion to form [Ni(CN)₄]². Since there is no unpaired electron, the complexion is diamagnetic.

b) Since the hybridisation involved is dsp2 this complex has a square planar geometry.

c) Cl < OH < H₂O < CO

Question 9.
[CO(NH₃)₅SO₂]CI is an octahedral coordination compound. (March – 2014)
a) Write the IUPAC name of the above coordination compound.
b) Write the formula of the ionisation isomer of the above compound.
c) How do ‘d’ orbitals split in an octahedral crystal field?
d) Draw the diagram which indicates the splitting of ‘d’ orbitals in the tetrahedral field.
Answer:
a) Pentaamminesulphatocobalt(lli) chloride
b) [CO(NH₃)₅CI]SO₄

c) In an octahedral crystal field the ligands approach the central metal atom/ion along the coordinate axes. Thus, the d_{x² – y²} and d_z² orbitais which point towards the axes along the direction of the ligand will experience more repulsion and will be raised in energy; and the d_xy, d_yz and d_xz orbitais which are directed between the axes will be lowered in energy relative to the average energy in the spherical crystal filed. The crystal filed splitting in an octahedral crystal field yield three orbitaIs of lower energy (t_2g set) and two orbita Is of higher energy (e_g set). The energy separation is denoted by Δ₀. The energy of the two e_g orbitais will increase by (3/5)Δ₀, and that of the three t_2g will decrease by (2/5)Δ₀. The d-orbitai splitting in an octahedral crystal field is shown below:

d)

Question 10.
a) Valence Bond Theory (VBT) can explain the magnetic behaviour and shape of complexes. Using VBT explain the diamagnetism and square planar shape of [NKCN)₄]^2- (Say – 2014)
b) i) Suggest the shape of the following complexes Ni(CO)₄ and [CoF₆]^3-.
ii) The central ion Co³⁺ with coordination number 6 is bonded to the ligands NH₃ and Br to form a dipositive complexion. Write the formula or IUPAC name of the complexion.
Answer:
a) In the complex ion [Ni(CN)₄]² the central metal atom Ni is in +2 oxidation state and has the electronic configuration 3d⁸. Since CN – is a strong field ligand two unpaired electrons are forced to pair up against Hund’s rule and makes available a vacant 3d, one 4s and two 4p orbitals, which hybridise to give four dsp2 hybrid orbitals. Each of the hybrid orbitals receives a pair of electrons from a cyanide ion to form [Ni(CN)₄]^2-. Since there is no unpaired electron, the complexion is diamagnetic.

Since the hybridisation involved is dsp² this complex has a square planar geometry.

b) i) In [Ni(CO)₄] the central atom Ni undergoes sp³ hybridisation. Hence, it has a tetrahedral geometry. In [CoF₆]³ the central metal ion Co³⁺ is in the sp³d² hybridised state. Hence, it has an octahedral geometry,
ii) [Co(NH₃)₅Br]²⁺
OR
Pentaamminebromidocobalt(lll) ion

Question 11.
Coordination compounds contains central metal atom/ion and ligands.
a) Primary valency of central metal atom/ion in [Co(NH₃)₆] Cl₃ is
i) 3
ii) 6
iii) 4
iv) 9
b) i) What are the postulates of Werner’s theory?
ii) Write the IUPAC names of K₃[Fe(CN)₆], [Co(NH₃)₆] Cl₃.
a) i) 3
b) i)
1) In coordination compounds metals show two types of linkages (valencies) – primary and secondary.
2) The primary valences are normally ionisable and are satisfied by negative ions. ie. oxidation state.
3) The secondary valences are non – ionisable. These are satisfied by neutral molecules or negative ions. The secondary valence is equal to the coordination number and is fixed for a metal.
4) The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination numbers.
ii) K₃[Fe(CN)₆] – Potassium hexacyanoferrate(lll) [CO(NH₃)₆]CI₃ – Hexaamminecobalt(lll) chloride

Question 12.
a) Write the IUPAC name of the complex K₃[Cr(C₂O₄)₃]. (Say – 2015)
b) Draw the figure to show the splitting of ‘d’ orbitals in octahedral crystal field.
c) [Fe(H₂O)₆]^3- is strongly paramagnetic, whereas [Fe(CN₆)]^3- is weakly paramagnetic. Write the reason.
Answer:
a) Potassium trioxalatochromate(lll)

c) H₂O is a weaker ligand compared to CN. There is no pairing of electrons in the d-orbital of Fe in [Fe(H₂O)₆]³⁺. But there is the pairing of electrons in the d-orbital of Fe in [Fe(CN₆)]^3-.

Question. 13.
a) Write down the ionization isomer of [Co(NH₃), Cl] SO₄. (March – 2016)
b) Write the IUPAC name of the above compound.
c) [Ni(CO)₄] is diamagnetic while [Nid4]2 is paramagnetic though both are tetrahedral. Why?
Answer:
a) [Co(NH₃)₅SO₄]Cl
b) Pentaamminechlondocobalt(lll) sulphate
c) In [NICI₄]^2- nickel is in +2 oxidation state and the Ni² ion has the electronic configuration 3d⁶. It undergoes sp³ hybridisation.

Each CI ion donates a pair of electrons. The compound ¡s paramagnetic since it contains two unpaired electrons.

[Ni(CO₄)] has tetrahedral geometry but is diamagnetic since nickel is in zero oxidation state and contains no unpaired electron.

Question 14.
Consider the co-ordination compound [CoNH₃)₅Cl]Cl₂ (Say – 2016)
a) Write the IUPAC name of the above coordination compound.
b) i) What is the primary valency and secondary valency of the central metal ion in the above co-ordination compound?
ii) Write the name of isomerism exhibited by the complex [Pt(NH₃)₂Cl₂] Represent the possible isomers.
Answer:
a) Pentaamminechloiidocobalt(lll) chloride
b) i) Primary valency: +3 Secondary valency: 6
ii) Geometric isomerism or cis-trans isomerism

Question 15.
(Co(NH₃)₅SO₄]CI and [Co(NH)₅Cl]SO₄ are coordination compounds. (March – 2017)
a) Identify the isomerism shown by the above compounds.
b) Write the IUPAC names of the above compounds.
c) Identify the ligands ¡ri each of the above compounds.
Answer:
a) Ionisation isomerism
b) [Co(NH₃)₅SO₄]CI – Pentaamm in sulphate cobalt(III) chloride
[Co(NH₃)₅Cl]SO₄ – Pentaamminechloridocobalt(III) sulphate
c) In [Co(NH₃)₅SO₄]CI the ligands are NH₃ and SO₄.
In [Co(NH₃)₅Cl]SO₄ the ligands are NH₃ and Cl.

Question 16.
a) In which of the following, the central atom/ion is in zero oxidation state. (Say – 2017)
i) [Ni(CN)₄]^2-
ii) [NiCl₄]^2-
iii) [Ni(CO)₄]
iv) [Ni(NH₃)₅]²⁺

b) [Ni(CN)₄]^2- has square planar structure and it is diamagnetic.
i) On the basis of valence bond theory explain why [Ni(CN)₄)^2- exhibit these properties.
ii) Identify the ligand in the above-mentioned complex.
Answer:
a) iii) or [NiCo)₄]
b) i) dsp² hybridisation or No unpaired electrons or CN is a strong field ligand.
ii) CN- or cyanide ion

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 8 The d and f Block Elements.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 8 The d and f Block Elements

Question 1.
a) Transition elements are ‘d’ block elements. (March – 2010)
i) Write any four characteristic properties of transition elements.
b) Lanthanoids and actinoids are f – block elements.
i) What is the common oxidation state of Lanthanoids?
ii) Name the Lanthanoid with common oxidation state +4.
iii) It is difficult to separate Lanthanoids in the pure state. Explain.
Answer:
a)

• They form coloured compounds.
• They exhibit variable oxidation state.
• They form complex.
• They are good catalysts,

b)

• + 3
• Cerium (Ce)
• Due to lanthanide contraction size is approxi mately equal. So separation is difficult.

Question 2.
Transition metals are widely used as catalysts in industrial processes. (Say – 2010)
a) Name any two industrial processes in which transition elements are used as catalysts.
b) Transition metals exhibit catalytic properties. Why?
c) Why do the transition elements exhibit greater similarity in properties compared to main group elements along the period as well as down the group?
Answer:
a) Fe – Haber’s process forthe manufacture of NH₃ Ni – Hydrogenation of Oil for the manufacture of vanaspati ghee
b) Because of

• variable oxidation state of metals
• ability to form complexes

c) The outer electronic configuration remains almost same and hence they show horizontal similarity.

Question 3.
a) Atomic sizes increase as we come down a group, but in 4th group of the Periodic Table Zr, Hf have almost the same atomic sizes. Why? (March – 2011)
b) E° (standard electrode potential) values generally become less negative as we move across a transition series, but E° values of Ni²⁺/Ni and Zn²⁺/Zn values are exceptions. Justify.
Answer:
a) This is due to Lanthanide contraction. It is the phenomenon of regular decrease in atomic size across the lanthanoid series.
b) Due to the stability of completely filled d¹⁰ configuration of Zn²⁺ its Δ_iHΘ₂ is less. This is responsible for the high negative value of \(E_{z_{n}^{2+} / z_{n}}^{0}\) ( – 0.76 V).

Question 4.
Transition elements are d – block elements, with some exceptions. Usually they are paramagnetic. They show variable oxidation states. They and their compounds show the catalytic property. (Say – 2011)
a) Zn (Atomic number 30) is not a transition element, though it is a d – block element. Why?
b) Which is more paramagnetic Fe²⁺ or Fe³⁺? Why?
c) Why do transition elements show variable oxidation states?.
d) What is the reason for their catalytic property?
Answer:
a) A true transition element is one which has incompletely filled d – orbitals in its ground or in their common oxidation states. The zinc atom has completely filled d – orbitals.
b) Fe³⁺ is more paramagnetic.
The paramagnetic character of a transition metal ion depends on the number of unpaired d – electrons present in it. Fe³⁺ wich 3d⁵ configuration has 5 unpaired d – electrons while Fe²⁺ with 3d⁶ configuration has only 4 unpaired d – electrons.
c) In transition elements the energy difference between (n – 1)d and ns orbitals is very less. Hence, along with ns electrons (n – 1)d electrons can also take part in chemical reactions.
d) Due to their ability to adopt multiple oxidation states and to form complexes.

Question 1.
a) Potassium dichromate (K₂ Cr₂ O₇) is an important compound of chromium. Describe the method of preparation of potassium dichromate from chromite ore. (March – 2012)
b) The gradual decrease in the size of lanthanoid elements from lanthanum to lutetium is known as lanthanoid contraction. Write anyone consequence of lanthanoid contraction.
Answer:
a) K₂ Cr₂ O₇ is prepared from chromate one Fe Cr₂ O₄.
Step I: The powdered ore is heated with molten alkali in free access of air to form soluble sodium chromate.
4 Fe Cr₂ O₄ + 16 NaOH + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8H₂O

Step II: Sodium chromate (Na₂Cr O₄) is filtered and acidified with dil. H₂SO₄ to form sodium dichromate.
2Na₂CnO₄ + H₂SO₄ → Na₂Cn₂O₇ + Na₂SO₄ + H₂O

Step III: Na₂Cn₂O₇ solution is treated with KCI to form K₂Cr₂O₇.
Na₂Cn₂O₇ + 2 KCI → K₂Cn₂O₇ + 2NaCI

b) Consequences of lanthanoid contraction ane

• Difficulty in separation of lanthanoids due to similanity in chemical properties.
• The similarity in size of elements belonging to same group of second & third transition series.

Question 1.
Assume that you are going to present a seminar on transition elements. Prepare a seminar paper by stressing any four important properties of transition elements. (Say – 2012)
Answer:
The transition elements are the elements in groups 3 – 12 of the periodic table in which the d – orbitals are progressively filled.
1) Magnetic properties : Most of transition elements show paramagnetism due to the presence of unpaired electrons. The magnetic moment \((\mu) \mu=\sqrt{n(n+2)}\)
2) Formation of coloured ions : Transition elements form coloured compounds due to the presence of unpaired d – electrons, which can take part in d – d transition.
3) Formation of complex compounds: Transition metals form a large number of complex compounds. This is due to the comparatively smaller sizes of the metal ions, their high ionic charges and the availability of d – orbitals for bond formation, e.g., K₄[Fe(CN)₆], [Co(NH₃)₆]CI₃
4) Catalytic properties : Transition elements and their compounds act as good catalysts. This is attributed to their ability to adopt show multiple oxidation states because of and to form complexes due to the presence of partially filled d – orbitals, e.g., Finely divided Fe is used as a catalyst in Haber’s process.

Question 1.
Account for the following trends in atomic and ionic radii of transition metals. (March – 2013)
i) Ions of the same charge in a given series (3d, 4d or 5d) show progressive decrease in radii with icreasing atomic number.
ii) The atomic radii of elements in 4d series are more than that of corresponding elements in 3d series.
iii) The atomic radii of the corresponding elements in ‘4d’ series and ‘5d’ series are virtually the same.
Answer:
i) This is because each time a new electron enters a ‘d’ orbital the nuclear change increases by unity. The shielding effect of a ‘d’ electron is not that effective. Hence, the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases.

ii) The effect of addition of new shells in 4d series overtakes the effect of increase in nuclear charge.
Thus, electrostatic attraction between nucleus and valence electron decreases and hence atomic size increases.

iii) This phenomenon is due to the intervention of the 4f – orbitals which must be filled before the 5d series of elements begins. The filling of 4f before 5d – orbitals results in a regular decrease in atomic radii called Lanthanoid Contraction due to imperfect shielding of intervening 4f – orbital electrons which compensate for the expected increase in atomic size on moving down the group, with increasing atomic number. The net result of Lanthanoid Contraction is that the second and the third d series exhibit similar radii.

Question 1.
i) d – Block elements belong to group 3 – 12 in the periodic table, in which the d orbitals are progressively filled. (Say – 2013)
a) What is their common oxidation state?
b) Name two important compounds of transition elements.
c) Transition elements form a large number of complex compounds, why?
ii) What is mischmetal?
Answer:

1. a) (n – 1)d^1-10 ns^1-2
   b) Potassium dichromate, Potassium permanganate. KMnO₂
   c) This is due to the following two factors: Cations of transition metals are very
   • small in size
   • high effective nuclear
   • have vacant d – orbitals
2. ‘Misch metal’ is an alloy which consists of a lanthanoid metal (- 95%) and iron (- 5%) and traces of S, C, Ca and Al. It is used in Mg-based alloy to produce bullets, shell and lighter flint.

Question 1.
Potassium dichromate is an orange coloured crystal and is an important compound used as an oxidant in many reactions. (March – 2014)
a) How do you prepare K₂Cr₂O₇ from chromite ore?
b) How will you account for the colour of potassium dichromate crystals?
Answer:
a) The chromite ore is fused with sodium carbonate in free access of airto get sodium chromate.
\(\begin{aligned}
4 \mathrm{FeCr}_{2} \mathrm{O}_{4}+8 \mathrm{Na}_{2} \mathrm{CO}_{3}+& 7 \mathrm{O}_{2} \longrightarrow \\
& 8 \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{Fe}_{2} \mathrm{O}_{3}+8 \mathrm{CO}_{2}
\end{aligned}\)
The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to get sodium dichromate.
\(2 \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{H}^{+} \rightarrow \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{4}+2 \mathrm{Na}^{+}+\mathrm{H}_{2} \mathrm{O}\)
The sodium dichromate solution is treated with potassium chloride to get potassium dichromate.
\(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{KCl} \rightarrow \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{NaCl}\)
Orange crystals of potassium dichromate crystallise out.

b) This is due to charge transfer spectra i.e., Chromium being a transition element has vacant d – orbitals.

Question 1.
Potassium permanganate and potassium dichromate are two transition metal compounds. (Say – 2014)
a) Write any four characteristics of transition metals.
b) i) Write any two uses of potassium permanganate.
ii) Draw the structure dichromate ion.
Answer:
a)

• Variable oxidation states.
• Formation of coloured ions in aqueous solution.
• Formation of complex compounds.
• Formation of interstitial compounds,

b) i)

• Lab reagent
• For bleaching of wool, cotton, silk and other textile fibres.

ii) The dichromate ion consists of 2 tetrahedra sharing one comer with Cr – O – Crbond angle of 126°.

Question 1.
Fourteen elements following Lanthanum are called Lanthanoids: (March – 2015)
a) What is Lanthanoid contraction? Give reason for it.
b) KMnO₄ is a purple coloured crystal and it acts as an oxidant. How will you prepare KMnO₄ from MnO₂?
Answer:
a) The overall decrease in atomic and ionic radii from lanthanum to lutetium is called lanthanoid contraction.

Lanthanoid contraction is caused by the imperfect shielding of one 4f electron by another in the same set of orbitals. The shielding of one 4f electron by another is less than that of one d electron by another. Hence, as the nuclear charge increases along the lanthanoid series, there is fairly regular derease in the size of the entire 4fn orbitals.

b) MnO₂ is fused with an alkali metal hydroxide and an oxidising agent like KNO₃ to get dark green potassium manganate, K₂MnO₄.
2MnO₂ + 4KOH + O₂ → K₂MnO₄ + 2H₂O

Potassium manganate disproportionates in a neutral or acidic solution to give potassium permanganate.
3MnO^2-₄ + 4H⁺ → 2MnO^4- + MnO₂ + 2H₂O

Question 1.
a) Which of the following oxidation state is common for lanthanids? (Say – 2015)
i) +2
ii) +3
iii) +4
iv) +5
b) Drawthe structures of chromate and dichromate ions.
c) Zirconium (Zr) belongs to ‘4d’ and Hafnium (Hf) belongs to ‘5d’ transition series. It is difficult to separate them. Explain.
Answer:
a) ii) +3


c) It is a consequence of lanthanoid contraction, a cumulative effect of the contraction of atomic radii of the lanthanoid series caused by the imperfect shielding of one electron by another in the 4f sub-shell.

Question 1.
a) Which of the following oxidation state is not shown by Maganese? (March – 2016)
a) +1
b) +2
c) +4
d ) +7

b) Represent the structure of dichromate ion.
c) Potassium permanganate (KMnO₄) is a strong oxidizing agent. Write any two oxidizing reactions of KMnO₄.
Answer:
a) +1

c) 1. Permanganate ion oxidises iodide to iodine in acid medium:
\(\left.10\right|^{+}+2 \mathrm{MnO}_{4}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{I}_{2}\)

2. Permanganate ion oxidises Fe2+ ion (green) to Fe³⁺ ion (yellow) in acid medium:
\(5 \mathrm{Fe}^{2+}+\mathrm{MnO}_{4}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{Fe}^{3+}\)

Question 1.
Transition elements are d-block elements and inner transition elements are f-block elements. (Say – 2016)
i) Write any two properties of transition elements.
ii) Name a transition metal compound and write one use of it.
iii) What is Lanthanoid Contraction?
iv) Write any two consequences of Lanthanoid Contraction.
Answer:
i) Transition elements are metals, have high melting points and high enthalpy of atomisation, exhibit variable oxidation states, show paramagnetism, form coloured compounds, form complex compounds, show catalytic properties, form interstitial compounds, form alloys etc. (any two properties)
ii) Fe to makes steal
iii) The overall decrease in atomic and ionic radii from lanthanum to lutetium, caused by the poor shielding of one 4f electron by another is called lanthanoid contraction.
iv)
1. The atomic radii of second row of transition elements are almost similar to those of third row of transition elements.
2. The almost identical radii of Zr (160 pm) and Hf (159 pm).
3. All the lanthanoids have quite similar properties and due to this they are difficult to be separated.
4. The basic strength of hydroxides decreases from La(OH)₃ to Lu(OH)₃ due to decrease in size of M³⁺ ions and consequent increase in the covalent character of M – OH bond.

(any two consequences required)

Question 1.
a) Transition elements are ‘d’ block elements. (March – 2017)
i) Write any four characteristic properties of transition elements.
ii) Cr²⁺ and Mn³⁺ have d⁴ configuration. But Cr²⁺ is reducing and Mn³⁺ is oxidising. Why?
b) Which of the following is not a lanthanoid element?
i) Cerium
ii) Europium
iii) Lutetium
iv) Thorium
Answer:
a) i) Transition elements are metals, have high melting points and high enthalpy of atomisation, exhibit variable oxidation states, show paramagnetism, form coloured compounds, form complex compounds, show catalytic properties, form interstitial compounds, form alloys etc. (any four properties)

ii) For Cr²⁺ to Cr³⁺ configuration changes from d⁴ to d³ For Mn³⁺ to Mn²⁺ d⁵ configuration results in extra stability due to half-filled configuration,

b) iv) Thorium

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 7 The p Block Elements.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 7 The p Block Elements

Question 1.
Elements in the groups 13 to 18 in the Periodic table constitute the ‘p’ block elements. (March – 2010)
i) Name the most important oxo acid of Nitrogen.
ii) How will you prepare the above oxo acid on large scale?
iii) In general, noble gases are least reactive. Why?
Answer:
i) Nitric acid (HNO₃)
ii) HNO₃ can be prepared on a large scale by Gstwald process. It involves three steps.
a) NH₃ is oxidised catalytically by atmospheric oxygen.

b) Nitric oxide thus formed combines with oxy gen giving NO₂.

C) Nitrogen dioxide so formed dissolves in water to give HNO₃.

iii)

• The noble gases except helium (1s²) have completely filled ns2np6 electronic configuration in their valence shell.
• They are octet completed and stable so they are inert.

Question 2.
After a discussion about the structures of hydrides of the group – 15 elements, Neethu wrote the order of bond angles as NH₃ < PH₃ < AsH₃ (Say – 2010)
a) is this the correct order?
b) Justify your answer.
c) Give the hybridization and shape of these hydrides.
d) Also arrange the above hydrides in the increasing order of their thermal stability. Justify your answer.
Answer:
a) No.
b) From top to bottom in the group the bond angle of group 15 hydrides decreases. As the electronegativity of the central atom decreases on moving down the group, the bond pair-bond pair repulsion decreases. Hence the bond angle decreases ¡n the order NH₃ > PH₃ > AsH₃.

C) In all hydrides the central atom is sp³ hybridised. The molecules assume trigonal pyramidal geometry with a lone pair on the central atom.

d) Thermal stability of the group 15 hydrides increases BiH₃ < AsH₃ < PH₃ < NH₃ This is due to the fact that moving up the group the EH bond dissociation enthalpy (‘E’ is a group 15 element) increases due to a decrease in size of the central atom and the molecules will decompose only at higher temperatures.

Question 3.
The Discovery of Haber’s process for the manufacture of Ammonia is considered to be one of the principal discoveries of the twentieth century. (March – 2011)
a) Which is the promoter used ¡ri the earlier process when Iron was used as a catalyst?
b) What is the temperature condition for the maximum yield of Ammonia? Justify.
c) Explain how can you convert NH₃ to HNO₃, on a large scale commercially.
Answer:
a) Molybdenum (Mo)

b) By Le – Chaltiers principle, the rate of an exothermic reaction increases with a decrease in temperature of 500°C to get a good yield of the product.

C) NH₃ is converted to HNO₃ commercially by Ostwald’s process. It involves three steps.
i) NH₃ is oxidised catalytically by atmospheric oxygen.

ii) Nitric oxide thus formed combines with oxygen giving NO₂.
\(2 \mathrm{NO}_{(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{NO}_{2(g)}\)

iii) Nitrogen dioxide so formed dissolves in water to give HNO₃.
\(3 \mathrm{NO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}_{(1)} \rightarrow 2 \mathrm{HNO}_{3(\mathrm{aq})}+\mathrm{NO}_{(g)}\)

Question 4.
The phosphorus of group 15 and sulphur of group 16 are two industrially important ‘p’ block elements. Their compounds are also industrially important. (Say – 2011)
a) \(4 \mathrm{H}_{3} \mathrm{PO}_{3} \stackrel{\text { heat }}{\longrightarrow} 3 \mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{PH}_{3}\) show that this is a disproportionation reaction.
b) PCl₃ fumes in moisture. Give reason.
c) Sulphuric acid can be manufactured from sulphur using V₂O₅ as a catalyst.
i) Give the name of the method.
ii) Outline the principle.
Answer:
a) Disproportionation reactions are a special type of redox reaction in which an element in one oxidation state is simultaneously oxidised and reduced. In phosphorous acid (H₃PO₃) phosphorus is in the intermediate oxidation state of +3. It is increased to +4 (oxidation) in phosphoric acid (H₃PO₄) and decreased to – 3 (reduction) in phosphine (PH₃).

b) PCI₃ undergoes hydrolysis in presence of moisture giving fumes of HCI. PCI₃ + 3H₂O → H₃PO₃ + 3HCI

c) i) Contact Process

Question 5.
a) Important allotropic forms of phosphorus are white phosphorus, red phosphorus and black phosphorus. Among these which allotropic form is more reactive? Why? (March – 2012)
b) In the manufacture of sulphuric acid (H₂SO₄) the final product obtained is oleum.
i) What is Oleum?
ii) Write chemical equation for the conversion of oleum to sulphuric acid.
c) Name the halogen which forms only one oxoacid and also write the formula of the oxo acid of that halogen.
d) Which element among inert gases form a maximum number of compounds? Write the formula of one of the compounds formed by the element.
Answer:
a) White Qhosfhorus It consists of discrete tetrahedral P4 molecules.
b) i) Pyrosuiphuncacid
ii) H₂S₂O₇ + H₂O → 2H₂SO₄
c) Flounne or HOF or hypoflourous acid
d) Compounds – Xe F₂, XeF₄ Xe O₃, Xe OF₂ etc.,

Question 6.
i) What are the products obtained when copper reacts with concentrated nitric acid? (Say – 2012)
ii) Name two important xenon fluorides.
iii) Interhalogen compounds are compounds formed by the combination of different halogen atoms. Which are more reactive, halogens or interhalogen compounds? Give reason.
Answer:
i) Copper reacts with concentrated nitric acid to give copper nitrate and nitrogen dioxide.
Cu + 4HNO_3(conc) + Cu(NO₃)₂ + 2NO₂ + 2H₂O
ii) Two important xenon fluorides are XeF₂ and XeF₄.
iii) Interhalogen compounds are more reactive than halogens (except fluorine). Because, the bond between different halogen atoms (X – X’) in interhalogen compounds is weakerthan the bond between similar halogen atoms. Due to the difference in size and electronegativity. The F – F bond is the weakest due to interelectronic repulsion.

Question 7.
a) Nitrogen forms a number of oxides in the different oxidation stats. Write the names and structural formulae of any four oxides of nitrogen. (March – 2013)
b) Boiling point of H₂O (373 K) is very much higher than that of H₂S (213k). Give reason.
c) Suggest a method for the quantitative estimation of ozone (O₃).
Answer:

b) Molecules of water are highly associated through hydrogen bonding resulting in its high boiling point. Hydrogen bonding is not possible ¡n H₂S.
c) When O₃ reacts with an excess of Kl solution buffered with a borate buffer (pH = 9.2) 12 is liberated, which can be titrated against a standard solution of sodium thiosulphate. 2Kl+ H₂O + O₃ → O₂ + KOH + l₂

Question 8.
a) Name the products obtained when the copper reacts with concentrated nitric acid. (Say – 2013)
b) Write down the chemical reaction between concentrated nitric acid and aluminium.
c) What is the basicity of H₃PO₃?
d) How do you account for the basicity of H₃PO₃?
e) Write down the three steps involved in the manufacture of sulphuric acid by the Contact Process.
f) Write any two important uses of noble gas elements.
Answer:
a) Copper nitrate, Nitrogen dioxide and Water. [Cu + 4HNO_3(conc) → Cu(NO₃) + 2NO₂ + 2H₂O]
b) Aluminium does not dissolve in concentrated nitric acid because it is rendered passive due to the formation of a thin protective layer of metal oxide on the surface of the metal which cuts off the further action.
c) T’ do.
d) The basicity of oxo acids of phosphors is deter mined by the number of P – OH bonds, because only those H atoms which are attached with oxygen ¡n P – OH form are ionisable and cause the basicity. H₃PO₃ has two P – OH bonds.

e) i) Preparation of suiphurdioxide by burning sulphur. S(s) + O₂(g) → SO₂(g)
ii) Oxidation of sulphur dioxide to suphurtrioxide catalytically with atmospheric oxygen.

iii) Preparation of oleum by absorbing sulphur trioxide in sulphuric acid. It is diluted with enough water to get sulphuric acid of desired concentration.
\(\begin{array}{l}
\mathrm{SO}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{I}) \rightarrow \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}(\mathrm{I}) \\
\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}(\mathrm{I})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})
\end{array}\)

Orthophosphorous acid Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes. Argon is used to provide an inert atmosphere in high-temperature metallurgical processes.

Question 9.
Compounds of nitrogen, phosphorus and sulphur such as ammonia, phosphoric acid and sulphuric acid are used in the fertilizer industry. (March – 2014)
a) Describe Haber process for the manufacture of ammonia.
b) Write the chemical equation forthe preparation of phosphoric acid (H₃PO₄) from ortho phosphorous acid (H₃PO₃).
c) Describe contact process for the manufacture of sulphuric acid.
Answer:
a) On a large scale, ammonia is manufactured by Haber’s process. In this process nitrogen gas reacts with hydrogen gas to form ammonia gas as per the reaction:
\(\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)}=2 \mathrm{NH}_{3(g)} ; \Delta_{f} \mathrm{H}^{\circ}=-46.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

According to Le Chatelier’s principle, high pressure favours the formation of ammonia. The optimum conditions for the production of ammonia are a pressure of 200 x 10⁵ Pa (about 200 atm), a temperature of 700 K and the use of catalysts such as iron oxide with small amounts of K₂O and Al₂O₃ to increase the rate of attaintment of equilibrium. The flow chart for the production of ammonia is shown below:

b) Orthophosphorous acid on heating disproportionates to give orthophosphoric acid and phosphine.
4H₃PO₃ → 3H₃PO₄ + PH₃
c) i) Bunning of sulphur or sulphide ores in air to generate SO₂. S_(s) + O_2(g) → SO_2(g))
ii) Conversion of SO₂ to SO₃ by the reaction with oxygen in the presence of V₂O₅ catalyst.
\(\begin{array}{l}
2 \mathrm{SO}_{2(\mathrm{~g}}+\mathrm{O}_{2(\mathrm{~g})} \stackrel{\mathrm{V}_{2} \mathrm{O}_{3}}{\longrightarrow} 2 \mathrm{SO}_{3(\mathrm{~g})} \Delta_{\mathrm{r}} \mathrm{H}^{\circ} \\
=-196.6 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{array}\)
A pressure of 2 bar and a temperature of 720 K are applied.

iii) Absorption of SO₃ gas in H₂SO₄ to give oleum (H₂S₂O₇) SO₃ + H₂SO₄ → H₂S₂O₇ Dilution of oleum with water gives H2S04 of the desired concentration. H₂S₂O₈ + H₂O → 2H₂SO₄

Question 10.
Ammonia and Nitric acid are two industrially mportant compounds. (Say – 2014)
a) Write any two uses of ammonia.
b) Complete the following equations. (Balancing is not required)
i) NH₃ + O₂ > \(\frac{\mathrm{Pt}}{500 \mathrm{~K}, 9 \mathrm{ber}}]\)
ii) Cu + Con. HNO₃ →
iii) Zn + dil. HNO₃ →
iv) NH₃ + excess Cl₂ →

OR

a) Phosphorus forms a number of oxoacids. Write the name or formulae of any two dibasic oxoacids of phosphorus.
b) Account for the following:
i) PCl₃ fumes in moist air.
ii) Nitrogen does not form a pentahalide.
iii) Boiling point of PH₃ is less than that of NH₃
iv) NO₂ undergone dimerisation.
Answer:
a)

• To produce various nitro geneous fertilisers.
• In the manufacture of some inorganic nitrogen compounds like nitric acid.

OR

a) i) Orthophosphorous acid (H₃ PO₃)
ii) Pyrophosphorous acid (H₄P₂O₅)
b) i) PCI₃ hydrolyses in the presence of moisture giving fumes of HCI. PCl₃ + 3H₂O → H₃PO₃ + 3HCI
ii) It does not have ‘d’ orbitais to expand its covalence beyond four. That is why it does not form pentahalide.
iii) Unlike NH₃, PH₃ molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of PH₃ s lower than that of NH₃.
iv) NO₂ contains odd number of valence electrons. It behaves as a typical odd electron molecule. On dimerisation, it is converted to stable N₂O₄ molecule with even number of electrons.

Question 11.
Some elements in p – block shows allotropy. (March – 2015)
a) What are the allotropic forms of sulphur?
b) i) How will you manufacture Sulphuric Acid by contact process?
ii) What are interhalogen compounds?
Answer:
a) Rhombic sulphur (α – sulphur) and Monoclinic sulphur (β – sulphur)
b) i) The manufacture of sulphuric acid by contact process involves three steps:
1) Burining of sulphur or sulphide ores in air to generate SO₂. S(s) + O_2(g) → SO_2(g)
2) Conversion of SO₂ to SO₃ by the reaction with ozygen in the presence of V₂O₅ catalyst.
\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \frac{\mathrm{V}_{2} \mathrm{O}_{5}}{2 \mathrm{~S} \mathrm{O}_{3}(\mathrm{~g}) \mathrm{\Delta}_{\mathrm{r}} \mathrm{H}^{\circ}}=-196.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

This reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are the favaourable conditions for maximum yield. In practice, a pressure of 2 bar and a temperature of 720 K are applied.

3) Absorption of SO₃ gas in H₂SO₄ to give oleum (H₂S₂O₇) SO₃ + H₂SO₄ → H₂S₂O₇ Dilution of oleum with water gives H2S04 of the desired concentration. H₂S₂O₈ + H₂O → 2H₂SO₄

The flow diagram for the manufacture of sulphuric acid by Contact Process is

ii) Compounds formed by the reaction between two different halogens are called interhalogen compounds. They can be assigned general compositions as XX’, XX₃’, XX₅ and XX₇’ where X is halogen of larger size and X’ of smaller size and X is more electropositive than X’.

Question 12.
a) Name two oxoacids of Sulphur.
b) i) How will you manufacture ammonia by Haber process?
ii) Write any two uses of inert gases.
Answer:
a) Sulphurous acid (H₂S₂O₅), Sulphuric acid (H₂SO₄), Peroxodisuiphuric acid (H₂S₂O₅), Pyrosulphunc acid or Oleum (H₂S₂O₇) – Any two.
b) i) Ammonia is manufactured by Haber’s process. In this process nitrogen gas reacts with hydrogen in the ratio 1:3 to form ammonia: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta_{f} \mathrm{H}^{\odot}=-46.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
High pressure favours the formation of ammonia 200 atm, a temperature of 700 K and the use of catalysts such as iron oxide.

ii) Helium is used for filling balloons for meteorological observations, Neon is used in discharge tubes and fluorescent bulbs. Argon is used to provide an inert atmosphere in high-temperature metallurgical processes and for filling electric bulbs, Xenon and Krypton are used in light bulbs designed for special purposes. (Any two)

Question 13.
a) What are interhalogen compounds? Write any two examples. (Say – 2015)
b) Write a method of preparation of phosphine from white phosphorus.
c) Write the name or formula of oxo acid of chlorine, in which chlorine possess oxidation number +7. Draw the structure of XeO₃ and XeF₂.
Answer:
a)These are compounds formed by the reaction of two different halogens. They can be assigned general compositions as XX’, XX, XX and XX where X is halogen of larger size and X’ of smaller size and X is more electropositive than X’. e.g. dF, dF₃, BrF₅, IF₇ (any two)

b) Phosphifle is prepared by heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO₂.
P₄ + 3NaOH + 3H₂O → 4 PH₃ + 3NaH₂PO₂

c) Perchloric acid or Chionc (VII) acid (HOCIO₃)

d)

Question 14.
a) Account for the following: (March – 2O16)
i) NH₃ acts as a Lewis base.
ii) PCI₃ fumes ¡n moist air.
iii) Fluorine shows only – 1 oxidation state.

b) i) SuggestanytwofluoridesofXenon.
ii) Write a method to prepare any one of the above mentioned Xenon fluorides.
OR
a) Account for the following:
i) H₂O is a liquid while H₂S is a gas.
ii) Noble gases have very low boiling points.
iii) NO₂ dimerises to N₂O₄.
b) i) What are interhalogen compounds?
ii) Suggest any two examples of interhalogen compounds.
Answer:
a) i) Nitrogen atom in NH₃ has one lone pair of electrons which is available for donation. There fore, it acts as a Lewis base.
ii) PCI₃ hydrolyses in the presence of moisture giving fumes of HCI. PCI₃ + 3H₂O → H₃PO₃ + 3HCI
iii) Fluorine is the most electronegative element and cannot exhibit any positive oxidation state. Fluorine atom has no d orbitals in its valence shell and therefore cannot expand its octet.

b) i) Xenon ditluoride, XeF₂
Xenon tetrafluoride, XeF₄
Xenon hexafluonde, XeF₆ (any two)
ii) XeF₂ is prepared by treating Xe with excess fluorine at 673 Kandl bar.
\(\mathrm{Xe}(\mathrm{g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 673 \mathrm{~K}, 1 \text { bar } \quad>\mathrm{XeF}_{2}(\mathrm{~s})\)
Or, XeF₄ is prepared by treating Xe with excess fluorine in 1: 5 ratio at 873 K and 7 bar.
\(\mathrm{Xe}(\mathrm{g})+2 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{B} 73 \mathrm{~K}, 7 \mathrm{bar} \quad \rightarrow \mathrm{XeF}_{4}(\mathrm{~s})\)
Or, XeF₆ is prepared by treating Xe with excess fluorine in 1 :20 ratio at 573 K and 60 – 70 bar.
\(\mathrm{Xe}(\mathrm{g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow 573 \mathrm{~K}, 60-70 \mathrm{bar} \quad \longrightarrow \mathrm{XeF}_{6}(\mathrm{~s})\)

OR

a) i) Due to small size and high electronegativity of oxygen it is capable of forming hydrogen bond. Thus, water molecules can associate through intermolecular hydrogen bonds and hence ¡t exists as a liquid.

Due to big large and low electronegativity of sulphur t is not capable of forming hydrogen bond. So hydrogen sulphide molecules cannot associate through intermolecular bonds and hence it exists as a gas.

ii) Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.

iii) NO₂ contains odd numberof valence elecrons. It behaves as a typical odd electron molecule. On dimensation, it is converted to stable N₂O₄ molecule with even number of electrons.
b) i) These are compounds formed by the reaction between two different halogens.
ii) dF, BrF, IF, BrCI, ICI, dF₃, BrF₃, IF₃, ICI₃, IF₅,
BrF₅, dF₅, IF₇ (any two)

Question 15.
Nitrogen shows different oxidation states in different oxides. (Say – 2016)
a) In which of the fof lowing oxides, nitrogen is in + 4 oxidation state?
a) NO
ii) N₂O
iii) N₂O₃
iv) NO₂
b) Prepare a short write upon Nftric acid highlighting its structure, manufacture and any two properties.
OR
Phosphorous forms oxoacids
a) In which of the following phosphorous is in + 1 oxidation state?
i) H₃PO₂
ii) H₃PO₃
iii) H₄P₂O₇
iv) H₃PO₄
b) Prepare a short write up on Ammonia highlighting its structure, manufacture and properties.
Answer:
a) iv) NO₂
b) Nitric acid is the most important oxoacid of nitrogen. HNO₃ exists as planar molecule.

Manufacture of nitric acid: On a large scale, nitric acid is prepared mainly by Ostwald’s process. This method is based upon catalytic oxidation of NH₃ by atmospheric oxygen.

4NH₃(g) + \(5 \mathrm{O}_{2}(\mathrm{~g}) \frac{\text { PUR } \text { guage catalyst }}{500 \mathrm{~K}, \text { bar }}\) 4NO_(g) + 6H₂O_(g)

Nitric oxide thus formed combines with oxygen giving NO₂.

2NO_(g) + O_2(g) \(\rightleftharpoons\) 2NO_2(g)

Nitrogen dioxide so formed, dissolves in water to give HNO₂.

3NO_2(g) + H₂O_(l) → 2HNO_3(aq) + NO_(g)

Properties of nitric acid: It is a colourless liquid. In aqueous solution nitric acid behaves as a strong acid giving hydronium and nitrate ions.

HNO_2(g) + H₂O₍₁₎ → H₃O_+(aq) + NO_3(aq)

Concentrated nitric acid is a strong oxidising agent and attacks most metals except noble metals such as gold and platinum.

OR

a) i) H₃PO₂
b) Structure of ammonia: The ammonia molecule is trigonal pyramidal with the nitrogen atom at the apex. It has three bond pairs and one lone pair of electrons.

Manufacture of ammonia: On a large scale ammonia is manufactured by Haber’s process.

High pressure would favour the formation of ammonia. (about 200 atm), a temperature of —700 K and the use of catalyst such as iron oxide with small amounts of K₂O and Al₂O₃.

Ammonia gas is highly soluble in water. Its aqueous solution is weakly basic due to the formation of OH ions.

It forms salts with acids. It precipitates the hydroxides of many metals from their salt solutions. The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule makes it a Lewis base.

Question 16.
Nitrogen forms a number of oxides and oxoacids. (March – 2017)
a) Which of the following is a neutral oxide of Nitrogen.
i) N₂O
ii) N₂O₅
iii) NO₂
iv) N₂O₄

b) Prepare a short write – up on Nitric acid high lighting its laboratory preparation, chemical properties and uses.

OR

Phosphorous forms a number of compounds.

a) The gas liberated when calcium phosphide is treated with die. HCl is
i) Cl
ii) H₂
iii) PH₃
iv) All the above

b) Prepare a short write up on PCl₃ and PCI₅ highlighting the preparation and chemical properties of PCl₃ and structure of PCl₅.
Answer:
a) i) N₂O

b) Laboratory preparation: In the laboratory, nitric acid is prepared by heating KNO₃ or NaNO₃ and concentrated H₂SO₄ in a glass retort.

NaNO₃ + H₂SO₄ → NaHSO₄ + HNO₃

Uses: in the manufacture of ammonium nitrate for fertilisers and other nitrates for use in explosives and pyrotechnics; for the preparation of nitroglycerin, trinitrotoluene and other organic nitro compounds; in the pickling of stainless steel etching of metals and oxidiser in rocket fuels.

OR

a) iii) PH₃
b) Preparation of PCI₃: By passing dry chlorine over heated shite phosphorus.

P₄ + 6Cl₂ → 4PCl₃

Or, by the action of thionyl chloride with white phosphorus.

P₄ + 8SOCl₂ → 4PCI₃ + 4SO₂ + 2S₂Cl₂

Properties of PCI₃: It is a colourless oily liquid and hydrolyses in the presence of moisture. Hence, it fumes in moist air.

PCI₃ + 3H₂O → H₃PO₃ + 3HCI

Structure of PCI₅: In gaseous and liquid phases, PCI₅ has a trigonal bipyramidal structure. The three equational P – CI bonds are equivalent, while the two axial bonds are longer than equatorial bonds. This is due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond pairs.

Question 17.
a) Identify the most acidic compound from the following (Say – 2017)
i) H₂O
ii) H₂S
iii) H₂Se
iv) H₂Te
b)
i) Explain step P and Q.
ii) Give a reaction which indicates dehydration property of conc. H₂SO₄.
iii) Write any two uses of sulphuric acid.
OR
a) Identify the least basic compound among the following:
i) NH₃
ii) PH₃
iii) AsH₃
iv) SbH₃

b) i) Halogens have maximum negative electron gain enthalpy in the respective periods. Give reason.
ii) Draw the structure of Perch bric acid (HClO₄)
iii) Write the formulae of any two interhalogen compounds.
Answer:
a) iv or H₂Te
b) i)

ii) Charring action of cane sugar to carbon
C₁₂H₂₂O₁₁ + Con H₂SO₄ → 12 C+ 11 H₂O

iii) Dehydrating agent, laboratory reagent
OR
a) iv) SbH₃
b) i) by getting one electron octet ¡s completed. So electronegative ¡s very high

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 6 General Principle and Processes of Isolation of Elements.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 6 General Principle and Processes of Isolation of Elements

Question 1.
Analyse the table given below: (March – 2010)

a) Which of the ores mentioned in the above table can be concentrated by magnetic separation method? Justify your answer.
b) Identify the ores that can be concentrated by leaching.
c) What do you mean by leaching?
Answer:
a) Haematite, Magnetite, Iron pyrites Magnetic separation is based on difference in magnetic properties of the ore components. If either the ore or the gangue (one of these two) is capable of being attraced by a magnetic field, then such separations are carried out. In the case of iron ores mentioned above the ore particles are magnetic while the impurities are non-magnetic. Thus, when the ground ore is carried on a conveyer belt which passes over a magnetic roller, the ore particles are attracted towards the magnetic roller while the non-magnetic particles are collected away from the magnetic roller,

b) Bauxite

c) It is a chemical method used for the concentration of ore. For example, the ore of Al, bauxite is concentrated by leaching (Baeyer’s process). Bauxite is heated with NaOH solution. As a result of this reaction, sodium meta aluminate is formed. The aluminate solution is neutralised by passing CO₂ solution and hydrated Al₂O₃ is precipitated by seeding with freshly prepared samples of hydrated Al2O₃. Hydrated alumina is filtered, dried and heated to obtain pure Al2O₃.

Question 2.
You are provided with samples of some impure metals such as Titanium and Nickel. (Say – 2010)
a) Which method would you recommend for the purification of each of these metals?
b) Briefly explain each method.
Answer:
a) Ti – van Arkel method
Ni – Monds process

b) The crude metal is heated in an evacuated vessel with iodine. The metal iodide is formed. The metal iodide is decomposed on a tungsten filament, at about 1800 K. The pure metal is ’ deposited on the filament.

Mond process: In this method impure Ni is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl.

The carbonyl is subjected to higher temperature so that it is decomposed to give pure metal.

Question 3.
The concept of AG° of coupled reactions are used to explain reductions in metallurgy. (March – 2011)
a) Explain the above statement.
b) In the blast furnace for manufacturing iron, most of the reduction is carried out by CO rather than C(Coke). How can you account for this?
Answer:
a) The reduction of a metal oxide which is not feasible (\(\Delta G^{\ominus}\) positive) is coupled with the oxidation of a suitable reducing agent, which is usually a highly feasible reaction (\(\Delta G^{\ominus}\) highly negative) so that the \(\Delta G^{\ominus}\) of the overall reaction (coupled reaction) becomes negative and the reduction process occurs spontaneously.This is in accordance with Ellingham diagram. The \(\Delta G^{\ominus}\) value should be negative to make the reaction feasible.

b) CO is a good reducing agent at low temperature than coke. This is because, in the Ellingham diagram, at low temperature, the CO → CO₂ line is below the C → CO line. Thus, oxidation of CO is more feasible than that of coke. Hence, in the blast furnace, CO reduces Fe₂O₃ even at the lower tem perature range. At high-temperature coke reacts with CO₂ to form carbon monoxide. Thus CO is the actual reducing agent.

Question 4.
Bauxite, Al₂O₃, xH₂O, is an important ore of aluminium. It is concentrated by leaching. Explain the method. (Say – 2011)
Answer:
Leaching of alumina from bauxite (Baeyer’s process) The powered bauxite ore is diagested with a concentrated solution of NaOH at 473 – 523 K and 35 – 36 bar pressure. AI₂O₃ is leached out as sodium aluminate along with SiO₂ as sodium silicate. Other impurities like iron oxides and TiO₂ are left behind.

AI₂O_3(s) + 2NaOH_(aq) + 3H₂O_(l) → 2Na[AI(OH)₄]_(aq)

The solidum aluminate in solution is neutralised by passing CO₂ gas and hydrated Al₂O₃ is precipitated. At this stage, the solution is seeded with freshly prepared samples of hydrated Al₂O₃ which induces the precipitation of hydrated Al₂O₃.

2Na [AI(OH)₄]_(aq) + CO_2(g) → AI₂O₃. XH₂O_(s) + 2NaHCO_3(aq)

The sodium silicate remains in the solution and hy-drated alumina is filtered, dried and heated to give back pure Al₂O₃.

Question 5.
a) All ores are minerals, but all minerals are not ores. Why? (March – 2012)
b) Carbonate ores are usually subjected to calcination, while sulphide ores are subjected to roasting. Comment on the statement.
Answer:
a) The naturally occuring materials in which the metals are present either in the native or in the combined state are called minerals.

The minerals from which the metals can be extracted economically are called ores.

Hence all ores are minerals but all minerals are not ores.

b) Calcination is the proœss of heating the ore in a limited supply of air below its metting point. This removes the volatile impurities and moisture from the ore. Oxygen is not used up during calcination.

Roasting is the process of heating the concentrated ore in a regular supply of air in a furnace below the melting point of the metal. It is usually employed in the concentration of sulphide ores.

2 Zns + 3O₂ → 2ZnO + 2SO₂
2 PbS + 3O₂ → 2PbO + 2SO₂
2 Cu2S + 3O₂ → 2Cu₂O + 2SO₂

Question 6.
Concentrated ore of iron, coke and limestone are fed into a blast furnace from the top. (Say – 2012)

i) Write down the reason for adding limestone along with the concentrated ore of iron.
ii) Write down the reactions taking place at the higher temperature range in the blast furnace.
OR
Metals are extracted from their chief ore.
i) Name the pencil pal ore of aluminium.
ii) Write the equations for the reactions taking place at the anode and at the cathode during the extraction of aluminium by the electrolytic process.
Answer:
i) Lime stone is added to blast furnace morder to remove acidic impurities (gangue) like silica (SiO₂). At high temperature, lime stone decomposed to form calcium oxide, which acts as a basic flux and removes acidic silica gaunge as calcium silicate slag.

ii) The following reaction take place at the higher temperature range (900 K – 1500 K) in the blast furnace:

OR

i) The principal ore of Al is Bauxite (Al₂O₃ 2H₂O). In Hall-Heroult process for the electrolytic extration of Al, purified Al₂O₃ mixed with Na₃ALF₆ or CaF₂ acts as the electrolyte, steel cathode and graphite anode. The following reactions take place during electrolysis:

At cathode: Al₃(melt) + 3e → 4 Al(l)

At anode : The oxygen liberated at anode reacts with the carbon of anode producing CO and CO₂.

The overall reaction is

Question 7.
a) Match the items of Column I with hems of Column II. (March – 2013)

b) The reduction of the metal oxide is easier if the metal formed is in liquid state, at the temperature of reduction. Give reason.
Answer:
a)

b) The entropy ¡s higher if the metal is in liquid state than when it is in solid state. So the value of entropy change or the reduction process will be more positive. Thus, the value of AGe becomes more negative and the reduction becomes easier.

Question 8.
The scientific and technological processes used for isolation of the metal from its ore is known as metallurgy. (Say – 2013)
a) Name the method used for removing gangue from suiphide ores.
b) Explain the above method.
c) Give two examples for alloy steel.
Answer:
a) Froth floatation method

b) It is used for removing gangue from suiphide ores. The mineral particles become wet by oils while the gangue particles by water. Finely powdered ore is agitated with water containing little frothing agent (e.g. pine oil) and froth stabilizers (e.g. cresols) by passing a forceful current of air. Heavier gangue particles are left to the bottom. The froth is skimmed off and then dried.

c)

• Stainless steel (Fe-74%, Cr-18%, Ni-8%)
• Nickel steel (Fe-96%, Ni-4%)

Question 9.
a) Calcination and roasting are pre-treatments in metallurgy before metal extraction. Differentiate between calcination and roasting. (March – 2014)
b) Match the items of Column I with items of Column II.

a) Calcination is the process of heating the ore in the absence or limited supply of air when the volatile matter escapes leaving behind the metal oxide. Here oxygen ¡s not consumed, It is applied to hydrated oxides, hydroxides and carbonates.

Roasting is the process of heating the ore in a regular supply of air in a furnace at a temperature below the melting point of the metal. Here oxygen is consumed. it is applied to suiphide ores.

b)

Question 10.
Sulphide ores are concentrated by froth floatation process. (Say – 2014)
a) Write the name or formula of any two sulphide ores of copper.
b) Explain froth floatation process.
Answer:
a)

• Copper pyrites (CuFeS₂)
• Copper glance (Cu₂S)

b) This method is used for removing gangue from sulphide ores and is based on the principle of preferential wetting of solid surface by vanousliq ulds. i.e., the mineral particles become wet by oils while the gangue particles by water. Finely powered ore is agitated with water containing collectors (e.g. pine oils, fatty acids, xanthates, etc.) and froth stabilizers (e.g. cresols, aniline) by passing a forceful current of air. The collectors enhance non-wettability of the mineral particles and froth stabilisers stabilise the froth. The froth which is formed at the surface of water carries up lighter ore particles and heavier gangue particles are left to the bottom. The froth is skimmed off and then dried for recovery of the ore particles.

Question 1.
a) Name two metals which can be refined by van Arkel Method. (March – 2015)
b) Match the items of Column I with items of Column II.

Answer:
a) Zirconium (Zr) or Titanium (Ti)
b) i – d;
ii – c;
iii – a;
iv – b

Question 11.
The process involved in metallurgy are the concentration of the ore, isolation of the metal from its concentrated ore and purification of the metal. (Say – 2015)
a) Froth floatation method is an ore concentration method. What is the principle behind the process?
b) What is the role of limestone (CaCO₃) in the extraction of iron?
c) Monds process is used for refining of Ni and Van Arkel method is used for refining Zr (Zirconium). Write one similarity between these processes.
Answer:
a) The principle behind froth floatation is adsorption. The ore particles are preferentially wetted by the oil and are carried to the surface by the froth. The gangue material wetted by water sinks to of the tank. The froth is light and is skimmed off.

b) Inside the blastfumace lime stone is decomposed to CaO which acts as a basic flux and removes the silicate impurity (acidic gangue) of the ore as calcium silicate slag. The slag is in molten state and separates our from iron.

c) Vapour phase refining techniques. The metal is converted into its volatile compound and collected elsewhere. It is then deèomposed to give pure metal.

Question 12.
a) Which of the following is the ore of zinc? (March – 2016)
a) Bauxite
b) Magnetite
c) Malachite
d) Calamine

b) There are several methods for refining metals. Explain a method for refining Zirconium.
Answer:
a) d) Calamine
b) Van Arkel method. The crude metal is heated in an evacuated vessel with iodine. The metal iodide is formed.

Zr+ 2l₂ → Zrl₄

The metal iodide is decomposed on a tungsten filament, at 1800 K. Pure metal is deposited on the filament.

Zrl₄ → Zr + 2l₂

Question 13.
Metals are extracted from their ores (Say – 2016)
a) Among the following which metal is extracted from bauxite:
i) Zinc
ii) Iron
iii) Aluminium
iv) Copper

b) Suiphide ores are subjected to roasting while carbonate ores are subjected to calcination. Comment on the statement.
Answer:
a) iii) Aluminium

b) In roasting, the ore is heated in a regular supply of air in a furnace at a temperature below the melting point of the metal. Here, oxygen is consumed. The sulphide ores need to be converted to oxides. Hence these are subjected to roasting.

e.g. 2ZnS + 3O₂ → 2ZnO + 2SO₂

In calcination, the ore ¡s heated in a limited supply fair below its melting point morder to remove he volatile matter. Here, oxygen is not consumed. Hence, the carbonate ores are subjected to calcination.

e.g. ZnCO₃(s) → ZnO(s) + CO₂(g)

Question 14.
Leaching is a process of concentration of ores. Explain the leaching of alumina from bauxite. (March – 2017)
Answer:
Bauxite usually contains silica (SiO₂), iron oxides and titanium oxide (TiO₂) as impurities. The powdered ore is digested with a concentrated solution of NaOH at 473 – 523 K and 35 – 36 bar pressure. Al₂O₃ is leached out as sodium aluminate leaving the impurities behind. (SiO₂ is also leached as sodium silicate).

Al₂O₃(s) + 2NaOH_(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq)

The aluminate in solution is neutralised by passing CO₂ gas and hydrated Al₂O₃ is precipitated. At this stage, the solution is seeded with freshly prepared samples of hydrated Al₂O₃ to induce the precipitation.

2Na[Al(OH)₄](aq) + CO₂(g) → Al₂O₃ xH₂O + 2NaHCO₃(aq)

The sodium silicate remains in the solution and hydrated alumina is filtered, dried and heated to give back pure Al₂O₃.

Question 15.
a) Which of the following is not an Ore of Iron? (Say – 2017)
i) Haematite
ii) Magnetite
iii) Malachite
iv) Sidenote

b) Explain froth floatation process for the concentration of Ore.
Answer:
a) i) Haematite

b) It is used for removing gangue from sulphide ores. The mineral particles become wet by oils while the gangue particles by water. The finely powdered ore is agitated with water containing little frothing agent (e.g. pine oil) and froth stabilizers (e.g. cresols) by passing a forceful current of air. Heavier gangue particles are left to the bottom. The froth is skimmed off and then dried.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 5 Surface Chemistry.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 5 Surface Chemistry

Question 1.
Colloids exhibit certain special properties. (March – 2010)
a) Name the property of colloid involved in the construction of ultramicroscope.
b) Explain the above property.
c) What are the conditions to be satisfied to exhibit Tyndall Effect?
Answer:
a) Tyndall effect

b) When a colloid is viewed at right angles to the passage of light, the path of the beam is illuminated by a bluish light. This effect is called the Tyndall effect and the bright cone of light observed is called the Tyndall cone. Tyndall effect is caused by the scattering of light by colloidal particles in all directions in space.

c) There are two conditions for observing the Tyndall effect:

• The diameter of the dispersed particles is not much smaller than the wavelength of the light used.
• The refractive indices .of the dispersed phase and the dispersion medium differ greatly in magnitude.

Question 1.
In an attempt to prepare ferric hydroxide sol by adding small amount of ferric chloride to water, one person got a precipitate of ferric hydroxide. (Say – 2010)
a) How can you help him to convert Fe(OH)₃ precipitate to Fe(OH)₃ sol?
b) Name the phenomenon behind this.
c) What happens when BaCI₂ is added to Fe(OH)₃ sol?
d) Give reason forthe above.
Answer:
a) By adding a solution of FeCI₃ to the fresh precipitate of ferric hydroxide.
b) Peptization
c) Precipitation of Fe(OH)₃ sol will take place.
d) BaCI₂ being an electrolyte ionises to Ba²⁺ + and Cl ions. The particles of Fe(OH)₃ precipitate adsorb Ba²⁺ ions to their surface and get positively charged. These positively charged particles of the precipitate repel each other and ultimately break up into smaller particles of the size of a colloid. Here BaCI₂ is acting as a peptizing agent.

Question 1.
Physisorption and Chemisorption are 2 types of ad-sorption. (March – 2011)
a) What is the effect of temperature on physisorption and chemisorption?
b) In certain cases physisorption transfers into chemisorption as temperature is increased. Explain with an example.
c) Explain how colloids get coagulated on addition of salts.
Answer:
a) Physical adsorption or physisorption decreases with increase of temperature where as chemical adsorption increases with increase of temperature reaches a maximum value at an optimum temperature and then decreases with increase in temperature.

b) Physisorption of a gas adsorbed at low temperature may transform into chemisorption at a high temperature. This is due to the fact that no activation energy is required for physisorption while chemisorption requires activation energy. For example, dihydrogen is first adsorbed on nickel by van der Waals’ forces. This is physisorption. Molecules of dihydrogen then dissociate to form hydrogen atoms which are held on the surface by chemisorption.

c) When excess of a salt (electrolyte) is added, the colloidal particles interact with ions carrying charge opposite to that present on themselves. (According to Hardy-Schultz rule).

Question 1.
Ferric hydroxide sol can be prepared from freshly prepared ferric hydroxide precipitate. It can also be prepared by adding ferric chloride solution to boiling water. In both cases the sol particles are positively charged. (Say – 2011)
a) Name the above two methods of preparation of ferric hydroxide sol.
b) What happens when an electric potential is applied across two platinum electrodes dipping in ferric hydroxide sol? Explain.
Answer:
a) The method of preparation of ferric hydroxide sol from freshly prepared ferric hydroxide precipitate is called peptization. The method of preparation of ferric hydroxide sol by the addition of ferric chloride solution to boiling water is called hydrolysis,
b) When an electric potential is applied across two platinum electrodes dipped in ferric hydroxide sol electrophoresis will occur. Since ferric hydroxide sol is positively changed the colloidal particles will move towards the platinum electrode which acts as the cathode (negative electrode).

Question 1.
Colloids have many characteristic properties. Among this Tyndall effect is an optical property and coagulation is the process of settling of colloidal particles. (March – 2012)
i) What is the Tyndall effect?
ii) State Hardy Schulze rule which deals with the coagulation of colloids by the addition of an electrolyte.
iii) What is a protective colloid?
Answer:
i) When light passes through colloids the path of light becomes visible. This effect is called Tyndall effect. This is due to scattering of light by colloidal particles.

ii) It state that ‘Thegreaterthe valency of flocculating ion, the greater will be its coagulating, flocculating or precipitating power.

Note:
i) In the coagulation of a positive sol, the flocculating power increases in the order Na⁺ < Ba²⁺ + <AI³⁺
ii) Inthe coagulation of a negative sol, the flocculating power increases in the order Cl < S0²₄ < PO³₄ < [Fe(CN)₆]⁴ Na⁺ < Ba²⁺ < AI³⁺
iii) The lyophilic particles form a layer around the lyophobic particles and thus protect the latter from electrolytes. Such colloids are called protective colloids.

Eg : Gold sol can be protected by adding a little gelatin. Here, gelatin is the protective colloid.

Question 1.
Colloids are widely used in industry and in daily life. (Say – 2012)
i) What are colloids?
ii) Write any four applications of colloids.
Answer:
i) A colloid is a heterogeneous system in which one substance called a dispersed phase is dispersed as very fine particles in another substance called a dispersion medium. The particles in a colloid are larger than simple molecules but small enough to remain suspended. The diameter of colloidal particles ranges between 1nm and 1000 nm.

ii)
1) Electrical precipitation of smoke using Cottrell smoke precipitator-Smoke is a colloid of solid particles such as carbon, arsenic compounds, dust etc., in air. These particles are precipitated using high voltage electrodes.

2) Purification of drinking water – The suspended impurities present in water obtained from natural sources is coagulated by adding alum and is made fit for drinking purposes.

3) Medicines – Most of the medicines are colloidal in nature. Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated.

4) In the rubber industry – Rubber latex is a colloid of negatively charged rubber particles which is coagulated to rubber by adding formic acid.

Question 1.
a) The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is termed adsorption. (March – 2013)
i) What is adsorption isotherm?
ii) Write the mathematical expression of Freundlich adsorption isotherm.
b) Enzymes are known as biochemical catalysts. Write any two important characteristics of enzyme catalysis.
Answer:
a) i) A plot between the amount of gas adsorbed pergram of adsorbent (x/m) and the pressure of the adsorbate at constant temperature is called Adsorption isotherm.

\(\frac{x}{m}=k p^{1 / n}\)
OR
\(\log \frac{x}{m}=\frac{1}{n} \log P+\log k[latex]

x → Amount of gas adsorbed by ‘m’ gram of the adsorbent at a pressure P. ‘k’ and ‘n’ are constants.

b)

• Enzymes are highly specific in nature.
• Enzymes are highly efficient.
• They are highly active under an optimum temperature.

Question 1.
There are mainly two types of adsorption of gases on solids (Say – 2013)
a) What are the two types of adsorption of gases on solids?
b) Write any two characteristics of each of the above two types of adsorption.
Answer:
a) Physical adsorption or physisorption and chemical adsorption or chemisorption,
b)

Question 1.
Sols are colloidal systems in which dispersion medium is liquid and dispersed phase is solid. (March – 2014)
a) Write any four differences between lyophilic sols and lyophobic sols.
b) Peptisation is a method .of preparation of sols. Write a general procedure for peptisation.
Answer:
a)

b) Peptization is the process of converting a precipitate into colloidal sol by shaking it with a dispersion medium in the presence of small amount of electrolyte. The electrolyte used for this purpose is called peptizing agent. This method is applied to convert a freshly prepared precipitate into a colloidal sol. During peptization, the precipitate adsorbs one of the ions of the electrolyte on its surface. This causes the development of positive or negative charges on precipitates, which ultimately break up into smaller particles of the size of a colloid.

Question 1.
a) ‘Adsorption’ has many applications. Write any two applications of adsorption. (Say – 2014)
b) Physisorption and chemisorption are the two types of adsorption. Write any four differences between them.
Answer:
a) 1) In production of high vacuum
2) In gas masks
b)

Question 1.
a) Which of the following is Lyophobic colloid? (March – 2015)
1) Starch in water
ii) Gum in water
iii) Soap in water
iv) Gold sol

b) Write four applications of colloids.
Answer:
a) iv) Gold sol
b)

• Electrical precipitation of smoke
• Purification of drinking water
• Medicines
• Tanning
• The cleansing action of soaps and detergents
• Photographic plates and films
• Coagulation of rubber latex etc. (Any four)

Question 1.
a) Which of the following is an example of absorption? (Say – 2015)
i) Water on silica gel
ii) Water on CaCI₂
iii) Hydrogen on finely divided Nickel
iv) Oxygen on the metal surface

b) Write any two differences between absorption and adsorption.
OR
Based on particles of the dispersed phase, colloidal systems are classified into multimolecular, macromolecular, and associated colloids.
a) Which of the following colloidal system is an example of the multimolecular system?
i) Starch water
ii) Soap solution
iii) Ferric hydroxide in water
iv) Polyvinyl alcohol in water

b) Associated colloids are also known as micelles. How are they formed?
Answer:
a) ii) Water on CaCl₂
b)

OR
a) iii) Ferric hydroxide in water.

b) Associated colloids or micelles are formed by the aggregation of ions of an electrolyte above a particular concentration and temperature. Soap is an example of associated colloid. It is sodium or potassium salt of higher fatty acids and can be represented as RCOO-Na⁺ or RCOO K⁺. When dissolved in water, it dissociates into RCOO and Na⁺ or K³ ions. The RCOO ions consist of two parts – a long hydrocarbon chain R (non-polar ‘tail’) which is hydrophobic (water-repelling), and a polar group COO (polar ionic ‘head’) whic is hydrophobic (water-loving). Therefore, the RCOO ions are present on the surface with their COO groups in water and the hydrocarbon chains (R) staying away from it and remain at the surface. At critical micelle concentraion, the anions are pulled into the bulk of the solution and aggregate to form a spherical shape with their hydrocarbon cains pointing towards the centre of the sphere with COO part remaining outward on the surface of the sphere. An aggregate thus formed is known as ionic micelle.

Question 1.
i) Catalysis can be classified into two groups homogenous and heterogeneous. (March – 2016)
a) What do you mean by homogenous catalysis?
b) Write one example for heterogeneous catalysis.
ii) Which of the following is an emulsifying agent?
a) Milk
b) Butter
c) Gum.
d) Lampblack
Answer:
i) a) When the reactants and the catalyst are in the same phase (i.e., liquid or gas), the process is said to be homogenous catalysis.
b) e.g. Oxidation of sulphur dioxide into sulphur trioxide in the presence of Pt or V₂O₅.
[latex]2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \stackrel{\mathrm{Pt}(\mathrm{s})}{\longrightarrow} 2 \mathrm{SO}_{3}(\mathrm{~g})\)
In this process the reactants, sulphur dioxide and dioxygen are in the gaseous state while the catalyst Pt or V₂O₅ is in the solid state.
ii) c) gum (for O/W emulsions)
d) lamp black (for W/O emulsions)

Question 1.
Dispersed phase and dispersion medium are two phases of the colloidal system (Say – 2016)
a) Name the colloid in which dispersed phase is liquid and dispersion medium is solid.
i) Sol
ii) foam
iii) Emulsion
iv) Gel

b) Physisorption and Chemisorption are two types of adsorption. Write any four differences between them.
Answer:
a) iv) Gel
b) Refer March 2017 Question No 1 (a)

Question 1.
There are mainly two types of adsorption. They are physisorption and chemisorption. (March – 2017)
a) Differentiate between physisorption and chemisorption.
b) Write any two applications of adsorption.
Answer:
a) Ant two

b) Production of high vacuum, gas masks, control of humidity, removal of colouring matter from solutions, heterogeneous catalysis, separation of inert gases, in curing diseases, Froth floatation process, adsorption indicators, chromatographic analysis etc. (Any two applications required).

Question 1.
a) Which among the following Is not an electrical property of colloids? (Say – 2017)
i) Electrophoresis
ii) Electro osmosis
iii) Coagulation
iv) Tyndal effect

b) Freundlich adsorption Isotherm is
x/m = a kp^1/n where n > 1
Answer the following questions based on Freundlich adsorption isotherm:
i) What Is adsorption isotherm?
ii) Explain the terms In the above equation.
Answer:
iv) Tyndaleffed
b) i) lt is a curve obtained by plotting extent of adsorption against pressure at a constant temperature.

ii) x = mass of gas adsorbed
m = mass of adsorbent
p = pressure
k, n are constant n > 1

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 4 Chemical Kinetics.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 4 Chemical Kinetics

Question 1.
The order of a chemical reaction can be zero and even a fraction but molecularity cannot be zero or a non-integer. (March – 2010)
i) What do you mean by the order of a reaction?
ii) What is the molecularity of a reaction?
iii) The conversion of molecules ‘A’ to ‘B’ follows second order kinetics. If the concentration of A is increased to three times, how will it affect the rate of formation of ‘B’?
Answer:
i) It is sum of powers of the concentration – terms of the reactants in the rate law expression.
ii) The number of reacting species in an elementary reaction.
iii) Increases by 9 times.
Rate = k[R]²
Rate’= k[3R]² = 9 [R]²; i.e., Rate = 9 Rate

Question 2.
In a class room discussion about order and molecularity of a chemical reaction, Ramu argued that “there are reactions which appear to be of higher order but actually follow first order kinetics”. (Say – 2010)
a) How far is his statement true? Give your opinion in this regard. Justify your answer using suitable example.
b) List out any three important differences between order and molecularity.
Answer:
a) The reaction appears to be of higher order but actually follows a lower order kinetics. Such reactions are called pseudo order reactions. For example, hydrolysis of ethyl acetate. This reaction appears to be of second order but actually follows first order kinetics.

Here the concentration of water does not get altered much during the course of the reaction.
Hence, [H₂O] can be taken as a constant. The equation thus becomes
Rate = k[CH₃CQOH] where k = k [H₂O]
Thus, the reaction behaves as a first-order reaction.

(b)

Question 3.
The hydrolysis of an ester in acid medium is a first-order reaction. (March – 2011)
a) What do you mean by a first-order reaction?
b) What is the relation between Rate Constant and Half-Life Period of a reaction?
c) Half-Life Period of a first-order reaction is 20 seconds. How much time will it take to complete 90% of the reaction?
Answer:
a) When the sum of the powers of the concentra¬tion terms in the rate expression is one, that reaction is called a first-order reaction.
Rate = k[A]¹

b) In the case of the first-order reaction,

Question 4.
The value of rate constant K of a reaction depends on temperature. From the values of K at two different temperatures, the Arrhenius parameters E_a, and A can be calculated. (Say – 2011)
a) The rate constant of a reaction at 600K and 900K are 0.02 s^-1 and 0.06 s^-1 respectively. Find the values of E_a and A.
b) Write the unit of rate constant ‘K’ of a reaction if the concentration is in mol L^-1 and time in s. (order of the reaction is two)
Answer:

Question 5.
Rate of a reaction is the change in concentration of any one of the reactants or any one of the products in unit time. (March – 2012)
i) Express the rate of the following reaction in terms of reactants and products:
2H I → H₂ + l₂
ii) If rate expression for the above reaction is, rate = k[H I]², What is the order of the reaction?
iii) Define order of a reaction.
iv) Whether the molecularity and the order of the above reaction are the same? Give reason.
Answer:
i) In terms of reactants

iii) Order is the sum of the powers of the concentration terms in the rate law.
rate = K [A]^x [B]^y
∴ Order = x + y

iv) Yes.
2H I = H₂ + l₂
rate = K [H I]² ∴ Order = 2
Molecularity is the number of reacting species taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction.

∴ Here molecularity = 2

Question 6.
For a first-order reaction, the half-life period (t_1/2) is independent of initial concentration of its reacting species. (Say – 2012)
i) What is meant by half-life period of a reaction?
ii) By deriving the equation for t_1/2 of first-order reaction, prove that it is independent of initial concentration of its reacting species.
[Hint: Fora first ortler reaction, \(\left.\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}\right]\)
Answer:
i) The half-life period of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration,
ii) For a first order reaction,

Thus, for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species.

Question 7.
a) Zero-order reaction means that the rate of a reaction is independent of the concentration of reactants. (March – 2013)
i) Write an example for a zero-order reaction.
ii) Write the integral rate expression for the zero-order reaction, R→ P.

b) The temperature dependence of the rate of a chemical reaction can be accurately explained by the Arrhenius equation. With the help of the Arrhenius equation calculate the rate constant for the first-order reaction C₂H₅l(g) → C₂H₄(g) + Hl(g) at 700K. Energy of activation (E_a) for the reaction is 209 kJ moh1 and rate constant at 600 K is 1.60 x 10^-5 S^-1 [Universal gas constant R = 8.314 JK^-1 mol^-1]
Answer:
a) i) Zero-order reaction

where ‘I is the constant of integration.
At t = 0 1 = [R]₀
[R]₀ → InitiaI concentration of the reactant
[R] → concentration at time
∴ Equation (1) becomes. [R] = ^–kt + [R]₀

Question 8.
The conversion of molecule A to B follows second order kinetics. (Say – 2013)
a) If the concentration of A is increased to 4 times, how will it affect the formation of B?
b) Indicate the order and molecularityofthe reaction given below.
\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)
Answer:
a) Rate r,= k[A]²
If the concentration of A is increased by four times, the new rate, r2 = k[4A]²
or r₂ = 16 k [A]²
or r₂ = 16 r₁
i.e., rate is increased by 16 times.

b) This is a pseudo first order reaction.
Order = 1, Molecularity = 2

Question 9.
a) Consider a general reaction aA + bB → cC + dD. The rate expression for the reaction is Rate = K[A]*[B]^y (March – 2014)
i) Establish the significance of ‘(a+b)’ and ‘(x+y)’ in terms of order and molecularity.
ii) Write any two differences between order and molecularity.
b) “Reactions with zero order is possible, but zero molecularity is not”. Justify the statement.
Answer:
a) i) (a + b) – Molecularity of the reaction
(X + y) – Order of the reaction

ii)

b) The order of a reaction can be zero which means that the rate of the reaction is independent of the concentration of the reactants. But, molecularity of a reaction cannot be zero which means that there is no reacting species and hence no reaction is possible.

Question 10.
a) Unit of rate constant (K) of a reaction depends on the order of the reactions. (Say – 2014)
Values of ‘K’ of two reactions are given below. Find the order of each reaction.
i) K = 3 x 10^-2 molL^-1 s^-1
ii) K = 5 x 10^-3 mol^-1 Ls^-1
b) i) Write integrated rate equation for a first order reaction.
ii) Write the relation between half life (t^1/2) and rate constant (K) of a first order reaction.
iii) Rate constant (K) of a reaction is 5 x 10.2 s^-1.
Find the half life (t_1/2) of the reaction.
Answer:
a) i) Zero-order reaction by analysing the unit.
ii) Second-order reaction

Question 11.
The terms order and molecularity are common in chemical kinetics. (March – 2015)
a) What do you mean by order and molecularity?
b) i) Write two factors influencing rate of a reaction.
ii) WnteArrhenius equation.
Answer:
a) Order of a chemical reaction is the sum of powers of the concentration of the reactants in the rate law expression.
Molecularity of a reaction is the number of reating species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction.

b) i) Temperature, Nature of the reactant, Concentration of the reactant (Pressure in the case of gases). Presence of catalyst, Presence of radiation/light, Surface area etc. (any one)

Question 12.
Integrated rate expression for rate constant of first-order reaction is given by \(\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}\), for a general reaction R → R (Say – 2015)
i) Derive an expression for half life period of first order reaction.
ii) A first order reaction has a rate constant 1.15 x 10^-3s^-1. How long will 5g of the reactant take to reduce 3 g?
Answer:

Question 13.
i) The molecularity of the reaction 2NO + O₂ → 2NO₂ is, (March – 2016)
a) 5
c) 2
c) 3
d) O

ii) a) What do you mean by rate oía reaction?
b) What will be the effect of temperature on rate of a reaction?
iii) A first order reaction is found to have a rate constant, k = 5.5 x 10^-14 s^-1. Find out the half-life of the reaction.
Answer:
c) 3

ii) a) The rate of a reaction is defined as the change in concentration of any one of the reactants or products ¡n unit time. i.e.,

b) The rate of most of the chemical reactions (endothermic reactions) increase with increase in temperature. For a chemical reaçtion with rise ¡n temperature by 100, the rate constant is nearly doubled. The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation, \(k=A e^{\frac{-E_{0}}{R T}}\)

iii) For a first order reaction, half-life period,

Question 14.
Rate of a reaction is the change in concentration of any one of the reactants or any one of the products in unit time (Say – 2016)
a) Express the rate of the following reaction in terms of reactants and products \(2 \mathrm{NO}_{(g)}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{NO}_{2(g)}\).
b) i) \(\mathrm{N}_{2} \mathrm{O}_{5(g)} \rightarrow 2 \mathrm{NO}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)}\) is a first order reaction. Find the unit of K.
ii) Calculate the time required for the completion of 90% of a first order reaction. (K = 0.2303s^-1)
Answer:

Question 15.
a) Plot a graph showing variation in the concentration of reactants against time for a zero-order reaction. (March – 2017)
b) What do you mean by zero-order reaction?
C) The initial concentration of the first-order reaction, \(\mathrm{N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+1 / 2 \mathrm{O}_{2(g)}\) was 1 24 x 10^-2 mol L^-1 at 300 K. The concentration of N₂O₅ after ‘1’ hour was 0.20 x 10^-2 mol L^-1. Calculate the rate constant of the reaction at 300 K.
Answer:

b) It is a reaction for which the rate of the reaction is proportional to zero power of the concentration of reactants, i.e, order is zero.
Or, this is a reaction for which the rate of the reaction is independent of the concentration of the reactants.

Or, this is a reaction for which rate of the reaction,

Question 16.
The effect of temperature on rate of reaction is given by Arrhenius equation. (Say – 2017)
i) Write Arrtenius equation.
ii) Define activation energy (Ea).
iii) Rate constant K₂ of a reaction at 310 K is two times of its rate constant K., at 300 K. Calculate activation energy of the reaction. (1og2 0.3010, log 1=0)
Answer:
\(\begin{array}{l}
\text { I) } \mathrm{K}=\mathrm{A} \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}} \text { or } \\
\log \mathrm{K}=\log \mathrm{A} \frac{-\mathrm{Ea}}{2.303 \mathrm{RT}}
\end{array}\)

ii) Activation energy is the energy required to form an activated complex or It is the energy difference between the activated complex and the reactant molecules.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 3 Electrochemistry.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 3 Electrochemistry

Question 1.
From the position of elements in the electrochemical series, copper (Cu) can displace silver (Ag) from silver nitrate solution. (March – 2010)
a) Represent the cell constructed with silver and copper electrodes.
b) Write down the reaction taking place at the anode.
c) Write down the reaction taking place at the cathode.
d) Write the Nernst equation for the above cell reaction.
Answer:

Question 2.
In a class room, the teacher has explained the quantitative aspects of electrolysis by stating the Faraday’s laws of electrolysis. (May – 2013)

a) State the Faraday’s laws of electrolysis.
b) Explain the term electrochemical equivalent.
c) Calculate the quantity of electricity required to deposit 0.09 g of Aluminium during the following electrode reaction:
Al³⁺ + 3e → Al (Atomic mass of Al = 27)
Answer:
a) First law : The amount of a substance which is deposited or liberated at any electrode during ectrolysis is directly proportional to the quantity of electricity flowing through the electrolyte.

Second law: If the same quantity of electricity is passed through different electrolytes the amount of substances formed is directly proportional to their chemical equivalent weights.

b) It is the quantity of a substance formed when one-ampere current is passed through an electrolyte for one second.

c) Quantity of electricity required to deposit 27 g of
Al = 3F = 3 x 96500 C = 289500 C
∴ the quantity of electricity required to deposit 0.09
\(g \text { of } A l=\frac{289500 \times .09}{27}=965 C\)

Question 3.
The limiting molar conductivity of an electrolyte is to Aained by adding the limiting molar conductivities of cation and anion of the electrolyte. (March – 2011)
a) Name the above law.
b) What is meant by limiting molar conductivity?
c) Explain how conductivity measurements help to determine the ionization constant of a weak electrolyte like Acetic Acid.
d)Explain the change of conductivity and molar conductivity of a solution with dilution.
Answer:
a) Kohlrauschs law.
b) It is the conductivity of an electrolyte when the concentration of the solution approaches zero (or at infinite dilution).
c) The limiting molar conductivity of acetic acid \(\left(\Lambda_{\mathrm{Gh}_{3} \mathrm{COOH}}\right)\) isdeterrnined by app’ying Kohlrauschts law \(\left(\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{0}=\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}+\Lambda_{\mathrm{HCl}}^{0}-\Lambda_{\mathrm{NaCl}}^{0}\right)\). Then, degree of dissociaijon,OE is determined using the re lation, \(a=\frac{\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\mathrm{C}}}{\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\mathrm{O}}}\)

From α, the diissociation constant can be deter mined using the relation, \(K_{a}=\frac{c a^{2}}{(1-\alpha)}\)

Substituting for CL we get,

d. Conductivity decreases with dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution. The vanation of molar conductivity with dilution is different for strong and weak electrolytes. For strong electrolytes molar conductivity in creases steadily with increase in dilution due to decrease in interionic attraction. Thus, a straight line is obtained when ∧_m is plotted against C^1/2. For weak electrolytes molar conductivity increases with dilution steadily intially and shows a steep increase, especially at lower concentration due to increase in degree of dissociation.

Thus, a plot of ∧_m against C^1/2 gives a curve,

Question 4.
The standard electrode potentials of some electrodes are given below: (May – 2011)
E°_(zn2+,zn) = – 0.76V
E°_{(cu2+, Cn)} = + 0.34V
E°_{(Ag+, Ag)} = + 080V
E°_{(H+, H2)} = 0V

a) Can CuSO₄ solution be kept in silver vessel?
b) Zinc or Copper which can displace hydrogen from dii. H₂SO₄?
c) What is the reaction taking place at SHE when its connected to Ag/Ag electrode to form a galvanic cell?
d) Find the value of K_C (equillibnum constant) in the Daniel cell at 298k.
Answer:
a) Yes.
Since the standard reduction potential of silver is more than that of copper it is less reactive than copper and hence cannot react with CuSO₄.

b) Zinc with negative standard electrode potential is more active than hydrogen and hence can displace hydrogen from dil.H₂SO₄. But copper with a positive standard electrode potential is less active than hydrogen and hence cannot displace hydrogen from dil. H₂SO₄.

c) Since silver is less active than hydrogen, when silver electrode is connected with S.H.E, silver will act as cathode and S.H.E will act as anode. At the anode H is oxidiseci to H⁺. Hence, the reaction taking place at S.H.E is

Question 5.
Daniell cell is a galvanic cell made of zinc and copper electrodes. (March – 2012)
i) Write anode and cathode reactions in Daniell cell.
ii) Nernst equation for the electrode reaction
Mⁿ⁺ + + ne- 2 M is
\(\mathrm{E}_{\left(\mathrm{M}^{n+} / \mathrm{M}\right)}=\mathrm{E}_{\left(\mathrm{M}^{n+} / \mathrm{M}\right)}^{0}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \frac{1}{\left[\mathrm{M}^{\mathrm{n}^{+}}\right]}\)
Derive Nernst equation for Daniel cell.
OR
Leclariche cell, Lead storage cell and Fuel cell are galvanic cells having different uses.

i) Among these, the Leclanche cell is a primary cell and Lead storage cell is a secondary cell. Write any two differences between primary cells and secondary cells.
ii) What is a Fuel cell?
iii) Write the overall cell reaction in H₂ – O₂ Fuel cell.
Answer:
i) Anode : Zn → Zⁿ²⁺ + 2e- (oxidation half reaction)
Cathode : Cu²⁺ + 2e^– → Cu (reduction half reaction)
Overall cell reaction is Zn + Cu²⁺ → Zn²⁺ + Cu Theceilcanberepresentedas: Znl Zn²⁺llCu²⁺lCu
Anode : Zn I ZnSO₄
Cathode : Cu I CuSO₄

which is the Nernst equation for Daniell cell.
OR

ii) Fuel cells are galvanic cells in which chemical energy from fuels like H₂, CO, CH₄ etc are converted to electrical energy.

Question 6.
Innumerable number of galvanic cells can be constructed on the pattern of Daniell cell by taking combination of different half cells. (May – 2012)
i) What is galvanic cell?
ii) Name the anode and cathode of the Daniell cell.
iii) Write the name of the half-cell represented by Pt_(S)/H_2(g)/H⁺_(aq).
iv) What is the potential of the above half-cell at all temperate res?
Answer:
i) It is a device for converting chemical energy released into electrical energy.
ii) Anode – Zn rod dipped in ZnSO₄
Cathode – Cu rod dipped on CauSO₄
iii) Standard Hydrogen Electrode (S.H.E)
iv) S.H.E is assigned a zero potential at all temperatures.

Question 7.
With decrease in concentration of an electrolytic solution, conductivity (K)decreases and molar conductivity (∧_m) increases. (March – 2013)
i) Write the equation showing the relationship between conductivity and molar conductivity.
ii) How will you account for the increase in molar conductivity with decrease in concentration?
iii) Limiting molar conductivity (L°_m) of a strong electrolyte can be determined by graphical extrapolation method. Suggest a method for the determination of limiting molar conducivity of a weak electrolyte, taking acetic acid (CH₃COOH) as example.
Answer:
i) Molar conductivity \(\left(\Lambda_{m}\right)=\frac{\kappa \times 1000}{M}\)
where M → Molanty and K → Conductivity
OR
\(\lambda_{\mathrm{m}}=\frac{\mathrm{K}}{\mathrm{C}}\) Concentration of where solution in mol/litre.

ii) Molar conductivity increase with decrease in con centration or increase in dilution as number of ions as well as mobility of ions increases with dilution.

a) For strong electrolytes, the number of ions do not increase appreciably on dilution and only mobility of ions increases due to decrease in interionic attraction.

∴ increases a little as shown in the above graph.

b) For weak electrolytes, the number of ions, as well as mobility of ions, increases on dilution. Hence, increases steeply on dilution, especially near lower concentrations as shown in graph.

iii) Using Kohlrauschs law

Thus the molar conductivity of CH3COQH at infinite dilution can be determined from the knowledge Of \(\lambda_{\mathrm{m}\left(\mathrm{CH}_{3} \mathrm{COONa}\right)}^{0}, \lambda_{\mathrm{m}(\mathrm{HCl})}^{0}, \lambda_{\mathrm{m}(\mathrm{NaCl})}^{0}\)

Question 8.
We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combi nation of different half cells. (May – 2013)
a) What is a galvanic cell?
b) Name the cathode and anode used in the Daniell cell.
c) Name the cell represented by Pt_(S), H_2(g)/H⁺_(aq).
d) According to convention what is the potential of the above cell at all temperatures?
e) Write the use of the above cell.
Answer:
a) It is a device for converting chemical energy into electrical energy. The decrease in free energy in a spontaneous chemical process appears as elec trical energy. e.g., Daniell cell,
b) A zinc rod dipped in 1 M solution of ZnSO₄ acts as the anode. Here oxidation takes place. A copper rod dipped in 1 M solution of CuSO₄ acts as the cathode. Here reduction takes place.
c) This represents the andard Hydrogen Electrode (S.H.E), when it acts as the anode.
d) According to convention, S.H.E is assigned a zero potential at all temperatures
e) It is used as a primary reference electrode for determining the standard electrode potential of an unknown electrode. The electrode whose standard potential is to be determined is coupled with a reference electrode of known potential i.e., S.H.E to get a galvanic cell. The potential of the resulting galvanic cell is determined experimentally. E = E – E Knowing the potential of one electrode that of the other can be calculated.

Question 9.
a) The cell reactìon in Daniell cell is Zn_(s) + CU²⁺_(aq) → Zn²⁺_(aq) + CU_(s) and Nernst equation for single electrode potential for general electrode reaction \(\mathrm{M}^{\mathrm{n}+}{ }_{(\mathrm{aq})}+\mathrm{ne}^{-} \longrightarrow \mathrm{M}_{(\mathrm{s})}\) is \(E_{M^{n+} M}=E_{M^{n+} / M}^{0}-\frac{2.303 R T}{n F} \log \frac{[M]}{\left[M^{n+}\right]}\) Derive Nernst equation for Daniell cell. (March – 2014)
b) Daniell cell is a primary cell while lead storage cell is a secondary cell. Write any one difference between primary cells and secondary cells.
Answer:
a) In Daniell cell, the electrode potential for any given concentration of Cu²⁺ and Zn²⁺ ions, we can write, For Cathode:

b) In primary cells the reaction occurs only once and after use over a period of time cell becomes dead and cannot be reused again. Here the cell reaction is irreversible. e.g., dry cell Secondary cells after use can be recharged by passing current through them in the opposite direction so that they can be used again. Here the cell reaction is reversible. e.g., Lead storage cell.

Question 10.
Fuel cells are special types of galvanic cells. (May – 2014)
a) i) Whataregalvaniccells?
ii) Write any two advantages of fuel cells.
b) Write the electrode reactions is H₂ – O₂ fuel cells.
a) i) These are devices for converting chemical energy into electrical energy. The decrease in free energy in a spontaneous chemical process appears as electrical energy. e.g., Daniell oeil.
ii) 1) Fuel cells are pollution free.
2) Fuel cells are highly efficient (about 70%) compared to thermal plants (about 40%)
3) Fuel cells run continuously as long as the reactants are supplied. (any two)

b) At cathode:
O₂(g) + 2H₂O(l) + 4e → 4OH(aq)
At anode:
2H₂(g) + 4OH(aq) → 4H₂O(l)

Question 11.
You are supplied with the following substances: Copper rod, Zinc rod, Salt bridge, two glass bea kers, a piece of wire, 1 M CuSO₄ solution, 1 M ZnSO₄ solution. (March – 2015)
a) Represent the cell made using the above materials.
b) i) Write the Nemst equation for the above cell.
ii) Calculate the standard EMF of the cell if
\(\begin{array}{l}
E_{\left(\mathrm{Zn}^{2+} \mid \mathrm{Zn}\right)}=-0.76 \mathrm{~V} \\
\mathrm{E}_{\left(\mathrm{Cu}^{2+} \mid \mathrm{Cu}\right)}=+0.34 \mathrm{~V}
\end{array}\)
Answer:
a) Zn(s)|Zn²⁺(1 M)||Cu²⁺(1 M)|Cu(s)
OR

Question 12.
a) Conductance (G), conductivity (K) and molar conductivity (∧_m) are terms used in electrolytic conduction. (May – 2015)
i) Write any two factors on which conductivity depends on.
ii) How do conductivity and molar conductivity vary with concentration of electrolytic solution?
b) Write any one difference between primary cell and secondary cell.
Answer:
a) i) 1. the nature of the electrolyte added
2. size of the ions produced and their solvation
3. the nature of the solvent and its viscosity
4. concentration of the electrolyte
5. temperature (any two factors)

ii) Conductivity always decreases with decrease in concentration both for weak and strong electrolytes. This can be explained by the fact that the number of ions per unit volume that carry the current in a solution decreases on dilution.

Molar conductivity, ∧_m = _kV

∧_m increases with decrease in concentration. This is because the total volume (V) of the solution containing one mole of electrolyte also increases. The decrease in K on dilution is more than compensated by increase in its volume.

In the case of strong electrolytes Am increases slowly with dilution and can be represented by the equation:
\(\Lambda_{m}=\Lambda_{m}^{0}-A c^{1 / 2}\)

where ‘c’ is the molar concentration, ‘A’ is a constant (equato to -ve of slope) and ∧_m° is the limiting molar conductivity. Here, the plot of ∧_m against ‘C^1/2‘ will be a straight line. In the case of weak electrolytes ∧_m increases steeply on dilution, especially near lower concentrations due to increase in degree of dissociation.

b) Pnmary cell – Cell in which the reaction occurs only once and after use over a period of time the cell becomes dead and cannot be reused again. Secondary cell – Cell which can be recharged after use by passing current through it in the opposite direction so that it can be used again.

Question 13.
a) Which of the following is a secondary cell? (March – 2016)
a) Dry cell
b) Leclanche cell
C) Mercury cell
d) None of these

b) What is the relationship between resistance and conductance?
c) One of the fuel cells uses the reaction of hydrogen and oxygen to form water. Write down the cell re action taking place in the anode and cathode of that fuel cell.
Answer:
d) None of these
b) Resistance is inversely proportional to conductance. Or, Conductance is the inverse of resistance.

Question 14.
Galvanic cells are classified into primary and secondary cells (May – 2016)
a) Write any two differences between primary cell and secondary cell.
b) i) What is a fuel cell?
ii) Write the overall cell reaction in H₂ – O₂ fuel cell.
Answer:
a) Primary cell
Cell reaction cannot be reversed and hence can not be recharged, cannot be reused again. e.g. Dry cell, Mercury cell

Secondary cell
Cell reaction can be reversed and hence can be recharged, can be resued again. e.g. Lead storage battery, nickel-cadmium cell

b) i) Fuel cell is a galvanic cell that is designed to convert the energy of combustion of fuels directly into electncal energy.
ii) 2H₂(g) + O₂(g) → 2H₂O(l)

Question 15.
a) Represent the galvanic cell based on the cell reaction given below (March – 2017)
\(\mathrm{Cu}_{(\mathrm{s})}+2 \mathrm{Ag}_{(\mathrm{aq})} \rightarrow \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{Ag}_{(\mathrm{s})}\)
b) Write the half cell reaction of the above cell.
c) ∧_m⁰ for NaCI. HCI and NaAc are 126.4, 425.9 and 91.0 S cm² mol^-1 respectively. Calculate ∧_m⁰ for HAc.
Answer:

Question 16.
a) Identify the weak electrolyte from the following: (May – 2017)
i) KCl
ii) NaCl
iii) KBr
iv) CH₃COOH

b) Kohlrausch’s law helps to determine the degree of dissociation of a weak electrolyte at a given concentration.
i) State Kohlrausch’s law.
ii) The molar conductivity ∧_m of .001 M acetic acid is 4.95 x 10^-5 S cm² mol^-1. Calculate the degree of dissociation (α) at this concentra tion if limiting molar conductivity \(\wedge_{m}^{0}\) for H+ is 340 x 10^-5 S cm² mol^-1 and for CH₃COO is 50.5 x 10^-5 S cm² mol^-1.
Answer:
a) iv) CN₃COOH
b) i) Kohlrausch’s law: Limiting molar conductivity of an electrolyte is the sum of individual contnbutions of anion and cation respectively.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 2 Solutions.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 2 Solutions

Question 1.
Colligative properties are properties of solutions which depend on the number of solute particles irrespective of their nature. (March – 2010)
a) Name the four important colligative properties.
b) What happens to the colligative properties when ethanoic acid is treated with benzene? Give reason.
Answer:
a) 1) Relating lowering of vapour pressure of the solvent
2) Depression of freezing point of the solvent
3) Elevation of Boiling point of the solvent
4) Osmotic pressure of the solution

b) Molecules of ethanoic add (acetic acid) dimenses in benzene due to hydrogen bonding. As a result of dimerisation, the actual number of solute particles in the solution is decreased. As colligative property decreases molecular mass increases.

Question 2.
a) Mr. Raju has determined the molecular masses of different solutes in different solvents by osmotic pressure measurements and presented them in the following table. Please help him to complete the table. (May – 2010)

b) The extent of deviation from ideal behaviour of a solution is explained by van’t Hoff factor, i. What is meant by van’t Hoff factor?
Answer:
a) 2B, C, 2D, E/3, F, G/5.
b) 2B and 2 D are due to association
\(\mathrm{i}=\frac{\text { Normal molar mass }}{\text { Abnormal molar mass }}\)
OR
\(\mathrm{i}=\frac{\text { Observed colligative property }}{\text { Calculated colligative property }}\)

Question 3.
Colligative properties can be used to determine the molecular mass of solutes in solutions. (March – 2011)
a) What do you mean by ‘Colligative Property’?
b) Fordeterminingthe molecular mass of polymers, osmotic pressure is preferred to other properties. Why?
c) For intravenous injections only solutions with freezing point depression equal to that of 0.9% NaCI solution is used. Why?
Answer:
a) Colligative properties are those properties which depend upon no. of particles in the solution.

b) i) Pressure measurement can be done around the room temperature.
ii) Molarity of the solution is used instead of molality.
iii) Its magnitude is large compared to other colligative properties.
iv) Polymers have poor solubility.

c) This is because the osmotic pressure associated with the fulid inside the blood cell is equivalent to that of 0.9 (m/w) sodium chloride solution, i.e., blood is isotonic with this solution.

Question 4.
Relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure are important colligative properties of dilute solutions. (May – 2011)
a) Relative lowering of vapour pressure of an aqueous dilute solution of glucose is 0.018. What is the mole fraction of glucose in the solution?
b) An aqueous dilute solution of a non-volatile solute boils at 373.052 K. Find the freezing point of the solution.
For water K_b = 0.52K Kg mol^-1
For water K_f = 1.86 K Kgmol^-1
Normal boiling point of water = 373K
Normal freezing point of water = 273K
Answer:
a) 0.018
Because according to Raoult’s law for solutions containing non-volatile solutes, the relaive lower-ing of vapour pressure is equal to the mole fraction of the non-volatile solute.

Question 5.
Vapour pressure of a solution is different from that of pure solvent. (March – 2012)
i) Name the law which helps us to determine partial vapour pressure of a volatile component in solution.
ii) State the above law.
iii) Vapour pressure of chloroform (CHCI₃) and dichloro methane (CH₂CI₂) at 298 K are 200 mm and 415 mm of Hg respectively. Calculate the vapour pressure of solution prepared by mixing 24 g of chloroform and 17 g of dichloro methane at 298 K. [At. Mass : H -1, C – 12, Cl – 35.5]
Answer:
i) Raoult’s law.

ii) It states that at a given temperature for a solution of volatile liquids, the partial v.p. of each component in the solution is directly proportional to its mole fraction.

Question 6.
Colligative properties are properties of solutions which depend on the number of solute particles in the solution. (May – 2012)
i) Write the names of four important colligative properties.
ii) The value of van’t Hoff factor, ‘i’ for aqueous KCI solution is close to 2, while the value of ‘i’ for ethanoic acid in benzene is nearly 0.5. Give reason.
i) The Colligative Properties are:
1) Relative lowering of vapour pressure
2) Elevation of boiling point
3) Depression of freesing point
4) Osmotic pressure

ii) This is caused by dissociation in the case of KCI and association in the case of acetic acid.

KCI in aqueous solution undergoes dissociation as KCI → K⁺ + Cl^–

Thus, if complete ionisation occurs the number of particles in solution becomes double and hence van’t Hoff factor (i) for aqueous KCI solution is close to 2.

In the case of ethanoic acid (acetic acid) association (dimerisation) occurs in benzene through in- termolecular hydrogen bonding. Thus, if complete association occurs the number of particles in solution becomes half and hence van’t Hoff factor (i) for ethanoic acid in benzene is nearly 0.5.

Question 7.
Elevation of boiling point is a colligative property, (March – 2013)
i) What are colligative properties?
ii) Elevation of boiling point (D T_b) is directly proportional to molality (m) of solution.
Thus, DT_b = K_bm, K_b is called the molal elevation constant.
From the above relation derive an expression to obtain molar mass of the solute,
iii) The boiling point of benzene is 353.23K. When 1.80 g of a non-volatile solute was dissolved in 90g of benzene, the boiling point is raised to 354.11K. Calculate the molar mass of the solute. K_b for benzene is 2.53K kg mol^-1.
Answer:
i) Colligative properties are those properties which depend upon the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.

Question 8.
Liquid solutions can be classified into ideal and non-ideal solutions on the basis of Raoult’s law. (May – 2013)
a) State Raoult’s law.
b) What are ideal solutions?
c) Write two important properties of ideal solutions.
d) What type of deviation is shown by a mixture of chloroform and acetone? Give reason.
Answer:
a) Raoult’s law states that for a solution of volatile liquids, the partial vapour of each component in the solution is directly proportional to its mole fraction. Or The partial vapour pressure of any volatile component of a solution is equal to the product of the vapour pressure of pure component and mole fraction of that component in the solution.

b) Ideal solutions are solutions which obey Raoult’s law over the entire range of concentration.

c) i) Enthalpy of mixing of the pure components to form the solution is zero, i.e.; Δ_mixH = 0
ii) Volume of mixing is zero, i.e.; Δ_mixV = 0

d) Negative deviation. This is because chloroform molecule is able to form hydrogen bond with acetone molecule.

Question 9.
Osmotic pressure is a colligative property and it is proportional to the molarity of the solution. (March – 2014)
a) What is osmotic pressure?
b) Molecular mass of NaCI determined by osmotic pressure measurement is found to be half of the actual value. Account for it.
c) Calculate the osmotic pressure exerted by a solution prepared by dissolving 1.5 g of a polymer of molar mass 185000 in 500 mL of water at 37°C. [R = 0.0821 L atm K^-1 mol^-1].
Answer:
a) Osmotic pressure is the extra pressure applied on the solution to stop osmosis
b) NaCI, as a strong electrolyte dissociates to form Na+ and Ch ions. The number of particles doubles colligative property also doubles. Observed molecular mass will be half.

Question 10.
Molarity (M), molality (m) and mole fraction (x) are some methods for expressing concentrations of solutions. (May – 2014)
a) Which of these are temperature independent?
b) i) Define ‘molefraction’.
ii) A mixture contains 3.2 g methanol (molecular mass = 32 u) and 4.6 g ethanol, (molecular mass = 46 u) Find the molefraction of each ’ component)
Answer:
a) Molality and Mole fraction are temperature independent because these are mass to mass relationships. Mass is independent of temperature,
b) i) Mole fraction is defined as the ratio of the number of moles of that component to the total number of moles of all the components in solution.

Question 11.
a) Among the following which is not a colligative property? (March – 2015)
i) Osmotic pressure
ii) Elevation of boiling point
iii) Vapour pressure
iv) Depression of freezing point

b) i) 200 cm³ of an aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure of solution at 300 K is found to be 8.3 x 10^-2 bar. Calculate the molar mass of protein. (R = 0.083 L bar K^-1 mol^-1)
ii) What is the significance of van’t Hoff factor?
Answer:
a) iii) Vapour pressure
b) i) Osmotic pressure, n = 8.3 x 10^-2 bar
Volume of the solution, V = 200 cm³ = 0.200 L
Temperature, T = 300 K
R = 0.083 L bar mol^-1 K^-1

Thus, in the case of association, the value of ‘i’ is less than unity while for dissociation ‘i’ is greater than unity. If i =1 it means that there is no association or dissociation of the solute particles in solution.

Question 12.
a) Draw a vapour pressure curve, by plotting vapour pressure against mole fraction of an ideal solution of two volatile components A and B (not to scale). Indicate partial vapour pressure of A and B (P_A and P_B) and total vapour pressure (P_total) (May – 2015)
b) What is an ideal solution?
c) Modify the above plot for non-ideal solution showing positive deviation. (Draw the above plot once again and modify).
Answer:

b) A solution which obeys Raoult’s law over the entire range of concentration is known as an ideal solution.

Question 13.
a) Number of moles of the solute per kilogram of the solvent is (March – 2016)
a) Mole fraction
b) Molality
c) Molarity
d) Molar mass

b) The extent to which a solute is dissociated or associated can be expressed by Van’t Hoff factor.’ Substantiate the statement.
c) The vapour pressure of pure benzene at a certain temperature is0.850 bar. Anon volatile, non-electrolyte solid weighing 0.5 g when added to 39 g of benzene (molar mass 78 g mol^-1), the vapour pressure becomes 0.845 bar. What is the molar mass of the solid substance?
Answer:
a) b) Molality
b) b) The van’t Holf factor,

When i = 1 ⇒ there is no association or dissociation of solute particles.
When i < 1 ⇒ there is association of solute particles. When i > 1 ⇒ there is dissociation of solute particles.

Question 15.
Osmotic pressure is a colligative property. (May – 2016)
a) What is osmotic pressure?
b) 1.00 g of a non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40 k. The freezing point depression constant of benzene is 5.12 K kg/mol. Find the molar mass of solute.
Answer:
a) Osmotic pressure is the extra pressure applied on the solution side to just stop osmosis i.e., the flow of the solvent from its side to solution side through the semipermeable membrane,

Question 16.
a) Henry’s law is related to solubility of a gas in liquid. (March – 2017)
i) State Henry’s law.
ii) Write any two applications of Henry’s law.
b) 1000cm³ of an aqueous solution of a protein contains 1.26 gm of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 x 10^-3 bar. Calculate molar mass of the protein. (R = 0.083 L bar mol^-1 K^-1)
Answer:
a) i) Henry’s law states that at constant tempera-ture, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. p = K_H x where, p is the partial pressure of the gas in vapour phase, K_H is the Henry’s law constant and x is the mole fraction of the gas in the solution.

ii) 1. To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.

2. To avoid bends and the toxic effects of high concenration of nitrogen in blood, the tanks used by scuba divers are filled with air diluted with helium.

3. Low partial pressure of oxygen at high altitudes leads to low concentration of oxygen in the blood and tissues of people living at high altitudes or climbers and causes anoxia. (Any two applications)

Question 17.
a) The mole fraction of water in a mixture containing equal number of moles of water and ethanol is (May – 2017)
i) 1
ii) 0.5
iii) 2
iv) 0.25

b) The following are the vapour pressure curves of a pure solvent and a solution of a non-volatile solute in it.

Based on the above curves answer the following questions:
i) What do the curves A and B indicate?
ii) Explain why the value of TB is greater than that of Tb⁰.
Answer:
a) ii) 0.5
b) i) A-Vapour pressure curve of solvent Vapour pressure curve of solution
ii) Due to the presence of a non-volatile solute vapour pressure of solution is less than solvent and the boiling point is increased.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 1 The Solid State.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 1 The Solid State

Plus Two Chemistry The Solid State 4 Marks Important Questions

Question 1
a) Schottky defects and Frenkel defects are two stoichiometric defects shown by crystals. (March – 2010)
i) Classify the following crystals into those showing Schottky defects and Frenkel defects:
NaCI, AgCI, CsCI, CdCI₂
ii) Name a crystal showing both Schottky defect and Frenkel defect.
b) Schematic alignment of magnetic moments of ferromagnetic, antiferromagnetic and ferrimagnetic substances are given below. Identify each of them.
i) ↑↓↑↓↑↓↑↓
ii) ↑↑↓↑↓↑↑↑
iii) ↑↑↑↑↑↑↑↑
Answer:
a) i) Schottchydefect – NaCI, CsCI Frenkel defect – AgCI, CdCI₂
ii) AgBr

b) i) Antifero magnetism
ii) Ferrimagnetism
iii) Ferromagnetism

Question 2.
Based on the nature of order present in the arrangement of the constituent particles, solids are classified into two, crystalline and amorphous. (May – 2010)
a) List out any four points of difference between crystalline and amorphous solids.
b) A list of solids are given below:
Quartz, glass, iodine, ice.
From this, identify crystal (s)
i) having sharp melting point.
ii) which is/are isotropic
Answer:

b) i) Quartz, Iodine, Ice
ii) Glass

Question 3.
Cristal defects give rise to certain special properties in the solids. (March – 2011)
a) What is meant by Frenkel Defect?
b) Why does LiCI not exhibit Frenkel Defect?
c) Explain the pink colour of LiCI when heated in . the vapours of Li.
Answer:
a) The dislocation of a cation from its original site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location.
b) The size of the cation is bigger than the void.
c) Due to F – centre. It is an electron trapped anion vacancy.

Question 4.
A cubic unit cell is characterized by a = b = c and α = β = γ = 90° (May – 2011)
a) Name three important types of cubic unit cells and calculate the number of atoms in one unit cell in the above three cases.
b) A metal forms cubic crystals. The mass of one unit cell of it is M/NA gram, where M is the atomic mass of the metal and NA is Avogardo Number. What is the type of cubic unit cell possessed by the metal?
Answer:
a) Simple cubic unit cell or Primitive unit cell, Body – centred cubic unit cell (bcc) and Face – centred cubic unit cell (fee).
b) Primitive cubic unit cell: This unit cell has atoms only at its comers. There are 8 corners for a cube.
Contribution by atom at the corner = 1/8
Total number of atoms in one unit cell \(=8 \times \frac{1}{8}=1\) atom

Body – centred cubic unit cell:
This unit cell has an atom at each of its corners and also one atom at its body centre.
8 corners \(\times \frac{1}{8}\) per corner atom \(=8 \times \frac{1}{8}=1\) atom
1 body centre atom = 1 x 1 = 1 atom
∴ Total number of atoms per unit cell = 2 atoms

Face – centred cubic unit cell:
This unit cell contains atoms at all the corners and at the centre of all the faces of the cube.
8 corners \(\times \frac{1}{8}\) per corner atom \(=8 \times \frac{1}{8}=1\) atom Contribution by an atom at the face centre = \(\frac{1}{2}\)

6 face – centred atoms x \(\frac{1}{2}\) atom per unit cell \(=6 \times \frac{1}{2}=3\) atoms

∴ Total number of atoms per unit cell = 4 atoms

c) Mass of one unit cell = \(\frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}} \mathrm{g}\)

Mass of 1 mole of unit cells \(\mathrm{N}_{\mathrm{A}} \times \frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}\) = M gram = Gram atomic mass It means that 1 unit cell contains one atom of the metal. Hence, the type of unit cell is primitive cubic or simple cubic.

Question 5.
Solids can be classified into three types on the basis of their electrical conductivities. (March – 2012)
i) Name three types of solids classified on the basis of electrical conductivities.
ii) How will you explain such classification based on Band theory?
Answer:
i) Conductors, insulators & semiconductors
ii) In conductors, the valance band overlaps with Ir e conduction band or no energy gap exists between the valance band and conduction band.

∴ The electrons can easily go into the conduction band and hence metals are good conductors. In insulators, the energy gap between valance band and conduction band is very large. Hence electrons from valance band cannot move into the conduction band.

Semi conductors have energy gap between conductors and insulators. At room temperature, these are not good conductors. But with an in crease in temperature electrons acquire sufficient energy to move from valance band into conduction band resulting in an increase in conductivity.

Question 6.
Schottky and Erenkel defects are stoichiometric defects. (May – 2012)
i) Write any two differences between Schottky defect and Frenkel defect.
ii) When pure NaCI (Sodium Chloride) crystal is heated in an atmosphere of sodium vapours, it turns yellow. Give reason.
Answer:
i)

ii) It is caused by metal excess defect due to anion vacancies. When crystals of NaCI are heated in a atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl- ions diffuse to the surface of the crystal and combine with Na atoms to give NaCI. The electrons released from Na atoms diffuse into the crystal and occupy anionic sites to form Fcentres, which impart yellow colourto the crystal. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

Question 7.
a) NaCI has fcc structure. Calculate the number of NaCI units in a unit cell of NaCI. (March – 2013)
b) Calculate the density of NaCl, if edge length
of NaCI unit cells is 564pm. [Molar mass of
NaCI =58.5g/mol].
Answer:
a) Number of Na ions = 12 (at edge centres) x \(\frac{1}{4}\) +
= 1(at body centre) x 1
= 3 + 1 = 4
Number of Cl- ions = 8 (at the corners) x \(\frac{1}{8}\) +
= 6 (at face centres) x \(\frac{1}{2}\)
= 1 + 3 = 4
∴ Number of NaCI units per unit cell (z) = 4

Question 8.
Unit cells can be broadly classified into 2 categories primitive and centred unit cells. (May – 2013)
a) What is a unit cell?
b) Name the three types of centred unit cells.
c) The unit cell dimension of a particular crystal system is a = b = c, α = β = γ = 90°. ldentify the crystal system.
d) Give one example for the above crystal system.
Answer:
a) The smallest repeating unit of a crystal.
b) Body centred unit cell, Face centred unit cell and End centred unit cell.
c) Cubic crystal system.
d) NaCI (Rock salt struãture)

Question 9.
a) Every substance has some magnetic properties associated with it. How will you account for the following magnetic properties? (March – 2014)
i) Paramagnetic property
ii) Ferromagnetic property

b) A compound is formed by two elements P and Q. Atoms of Q (as anions) make hep lattice and those of the element P (as cations) occupy all the tetrahedral voids. What is the formula of the compound?
Answer:
a) i) Paramagnetic substances are weakly attracted by a magnetic field. They are magnetised in a magnetic field in the same direction. They lose their magnetism in the absence of magnetic field. Paramagnetism is due to presence of one or more unpaired . electrons, e g., O₂, Cu²⁺

ii) Ferromagnetic substances are strongly attracted by a magnetic field. They are permanantly magnetised. When a ferromagnetic substance is placed in a magnetic field all the magnetic domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced, e.g,, Fe, Co

For hep lattice, No. of particles per unit cell = 6

∴ Number of anions of Q in the unit cell = 6
Number of tetrahedral voids = 2 x N = 2 x 6 = 12
∴ Number of cations of P in the unit cell = 12
Hence, formula of the compound = P₁₂Q₂ = P₂Q

Question 10.
a) Crystalline solids are ‘anisotropic’. What is ‘anisotropy’? (May – 2014)
b) Copper crystals have fee unit cells.
i) Compute the number of atoms per unit cell of copper crystals.
ii) Calculate the mass of a unit cell of copper crystals. (Atomic mass of copper = 63.54 u)
Answer:
a) Anisotropy means physical properties shows different values along different directions, eg. refrative index electrical resistance,
b) i) Face centred cubic unit cell (fee) – It contains atoms at all the corners and at the centre of all the faces of the cube.
8 corners x \(\frac{1}{8}\) per corner atom

Question 11.
Unit cells can be divided into two categories, primitive and centred unit cells. (March – 2015)
a) Differentiate between Unit Cell and Crystal Lattice.
b) Calculate the number of atoms per unit cell in the following:
i) Body centred cubic unit cell (bcc)
ii) Face centred cubic unit cell (fee)
Answer:
a) Unit Cell – It is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.

Crystal Lattice – It is the regularthree dimensional arrangement of points in space.

b) i) Body centred cubic unit cell (bcc) – It has atoms at each of its corners and one atom at its body centre.
8 corners x \(\frac{1}{8}\) per corner atom = 8 x \(\frac{1}{8}\) = 1 atom
1 body centre atom = 1 x 1 = 1 atom
∴ Total number of atoms per unit cell = 2 atoms

ii) Face centred cubic unit cell (fee) – It contains atoms at all the corners and at the centre of all the faces of the cube.

8 corners x \(\frac{1}{8}\) per corner atom
\(=8 \times \frac{1}{8}=1\) atom

6 face – centred atoms x \(\frac{1}{2}\) atom per unit cell \(=6 \times \frac{1}{2}=3\) atoms
∴ Total number of atoms per unit cell = 4 atoms

Question 12.
a) Which of the following is not a characteristic of a crystalline solid? (May – 2015)
i) Definite heat of fusion
ii) Isotropic nature
iii) A regular ordered arrangement of constituent particles
iv) A true solid

b) Frenkel defect and Shottky defects are two stoichiometric defects found in crystalline solids.
i) What are stoichiometric defects?
ii) Write any two differences between Frenkel defect and Schottky defect.
Answer:
a) ii) Isotropic nature
b) i) Stoichiometric defects are those point defects which do not disturb the stoichiometry of the solid.

ii)

Question 13.
a) Which of the following is a molecular solid? (March – 2016)
a) Diamond
b) Graphite
c) Ice
d) Quartz

b) Unit cells can be classified into primitive and centered unit cells. Differentiate between primitive and centered unit cells.
c) Presence of excess Sodium makes NaCI crystal coloured. Explain on the basis of crystal defects.
Answer:
a) Ice
b) In primitive unit cell constituent particles are present only at the corner positions. Unit cells in which one constituent particles are present at the centres of a faces in addition to those at corners.

c) Such anionic sites occupied by unpaired electrons are called F – centres (colour centres). They impart yellow colourto the crystals of NaCI. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

Question 14.
A unit cell is a term related to crystal structure. (May – 2016)
a) What do you mean by unit cell?
b) Name any two types of cubic unit cells.
c) Calculate the number of atoms in each of the above – mentioned cubic unit cells.
d) Identify the substance which shows Frenkel defect:
i) NaCI
ii) KCI
iii) ZnS
iv) AgBr
Answer:
a) Unit cell is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.
b) Simple cubic unit cell, body centred cubic (bcc) unit cell, face centred cubic (fee) unit cell (any two)
c) Number of atoms per unit cell
i) Simple cubic unit cell:
Total number of atoms in one unit cell \(=8 \times \frac{1}{8}=1\) atom

ii) Body centred cubic (bcc) unit cell: Total number of atoms in one unit cell
\(=\left(8 \times \frac{1}{8}\right)+(1 \times 1)=1+1=2\) atoms

iii) Face centred cubic (fee) unit cell: Total number of atoms in one unit cell \(=\left(8 \times \frac{1}{8}\right)+\left(6 \times \frac{1}{2}\right)=1+3=4\)
(any two required)

d) ZnS or AgBr
(AgBr shows both Schottky and Frenkel defects)

Question 15.
a) Identify the non – stoichiometric defect (March – 2017)
i) Schottky defect
ii) Frenkel defect
iii) Interstitial defect
iv) Metal deficiency defect
b) What type of substance could make better permanent magnets – ferromagnetic or ferrimagnetic? Justify your answer.
c) In terms of Band theory write the differences between conductor and insulator.
Answer:
a) iv) Metal deficiency defect

b) Ferromagnetic substances could make better permanent magnets because when these substances are placed in a magnetic field all the magnetic moments (domains) get oriented in the direction of the magnetic field and a strong magnetic effect is produced. This ordering of domains persist even when the magnetic field is removed and the ferromagentic substance becomes a permanent magnet.

The schematic alignment of magnetic moments in a ferromagnetic substance is as shown below:

c) In conductors the valence band is either partially filled or it is overlaped with a higher energy unoccupied conduction band so that the electrons can flow easily under an applied electric field. Whereas in insulators the energy gap between the filled valence band and the next higher unoccupied conduction band is large so that electrons cannot jump to it.

Question 16.
a) From the following choose the incorrect statement about crystalline solids. (May – 2017)
i) Melt at sharp temperature.
ii) They have definite heat of fusion.
iii) They are isotropic
iv) They have long range order.

b) Cubic unit cells are divided into primitive, bcc and fee.
i) Calculate the number of atoms in a unit cell of each of the following:
* bcc
* fcc

ii) Write two examples for covalent solids.
a) iii) They are isotropic
b) i) \(\begin{array}{l}
\text { bcc }-2\left(8 \times \frac{1}{8}+1=2\right) \\
\text { fCc }-4\left(8 \times \frac{1}{8}+6 \times \frac{1}{2}=4\right)
\end{array}\)
ii) Graphite, Diamond