Plus Two

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)

Chain growth polymerisation	Step growth polymerisation
1) Molecules of the same or different monomers add together on a large scale to form a polymer.	1) There is repetitive con­densation reaction be­tween two bi-functional monomers with elimi­nation of some simple molecules.
2)The monomers are unsaturated com­pounds like alkenes, alkadienes and their derivatives.	2)The monomers are saturated compounds with two functional groups.

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)

Thermoplastic	Thermosetting
1. They can be reused	1. Cannot be reused
2. They can be repeatedly softening on heating and hardening on cooling	2. On heating they undergo extensive cross-linking and become infusible
3. They are linear or lightly branched long-chain polymers	3. They are cross-linked or heavily branched polymers
4. Eg. Polythene, Polystyrene, Polyvinyl	4. Eg. Bakelite, Urea-formaldehyde, Resin.

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)

Glucose	Disaccharide	d-1, 4 link
Sucrose	Reducing	Galactoxide
Lactose	Insoluble (in water)	1, 6-linkage
Amylopectin	Non-reducing	Fructoxide
	Trisaccharide	Anomers present
	Mono saccharide	2-glucose units linked

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)

DNA	RNA
1) Double helix structure 2) Sugar- deoxyribose 3) Bases – A, G, C, T 4) Transmits Traits	1) Single helix 2) Sugar-Ribose 3) Bases A, G, C, U 4) Responsible for protein synthesis

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)

Column I	Column II
(i) Primary structure	(a) Special arrange­ment of polypeptide subunits
(ii) Secondary structure	(b) Structure of amino acids
(iii) Tertiary structure	(c) Folding of peptide chains
Civ) Quaternary structure	(d) Sequence of amino acids
	(e) Fibrous or globular nature

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

Column I	Column II
(i) Primary structure (ii)  Secondary structure (iii) Tertiary structure (iv) Quaternary structure	(d) Sequence of amino acids (c) Folding of peptide chains (e)  Fibrous and globular nature (a) Spacial arrangement of polypeptide subunits

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)

Alcohols		Phenols
1	Do not show any effect on litmus solution.	1	Turn blue litmus red.
2	Do not give any characteristic colour with FeCI3 solution.	2	Give characteristic colours such as violet, red, etc. with FeCI3 solution.
3	Do not react with NaOH solution.	3	React with NaOH solution and form corresponding sodium salt

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)

SN1 reaction	Sn2 reaction
1. Molecularity is 1.	1. Molecularity is 2.
2. Rate of reaction is dependent only on the concentration of the alkyl halide.	2. Rate of reaction is dependent on the concentration of the alkyl halide as well as a nucleophile.
3. Mechanism involves two steps – formation of carbocation followed by the nucleophilic attack.	3. Mechanism involves one step via the formation of a transition state.
4. Starting with an optically active alkyl halide results in partial racemization.	4. Starting with an optically active alkyl halide results in a complete inversion of configuration.

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.

\(4 \mathrm{NH}_{3 \mathrm{gg}}+5 \mathrm{O}_{2(\mathrm{~g})} \frac{\mathrm{Pt} / \mathrm{Rh} \text { gangue catalyst }}{500 \mathrm{~K}, 9 \mathrm{bar}}, 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(9)}\)

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

\(2 \mathrm{NO}_{(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{NO}_{2(g)}\)

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

\(3 \mathrm{NO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}_{(1)} \rightarrow 2 \mathrm{HNO}_{3(\mathrm{aq})}+\mathrm{NO}_{(9)}\)

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.

\(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ;=-46.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

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.

\(\mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(0)} \rightleftharpoons \mathrm{NH}_{4^{+}(\mathrm{aq})}+\mathrm{OH}_{-(\mathrm{aq})}\)

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 English Previous Year Question Papers and Answers.

Kerala Plus Two English Previous Year Question Paper March 2019 with Answers

Board	SCERT
Class	Plus Two
Subject	English
Category	Plus Two Previous Year Question Papers

Time: 2 1/2 Hours
Cool off time: 15 Minutes
Maximum: 80 Score

General Instructions to Candidates:

• There is a ‘cool off-time’ of 15 minutes in addition to the writing time of 2^1/2 hours.
• You are not allowed to write your answers or to discuss anything with others during the cool off time’.
• Read questions carefully before answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Electronic devices except non-programmable calculators are not allowed in the Examina­tion Hall.

(Question Nos. 1 – 6): Answer the questions as directed. (14)

Read the excerpt from ‘When, a Sapling is Planted’ and answer the questions that follow: ‘Later, they became aware of the widespread destruction of the ecosystems, especially through deforestation, Climatic instability and contamination of the soil and waters – all contributed to excruciating poverty and subsequent riots.’

Question 1.
Who does the speaker refer to as ‘they’? (1)
Answer:
The women of Kenya/African women/People of Kenya/Women

Question 2.
What was the main reason for the destruction of the ecosystem? (1)
Answer:
Deforestation/cutting down of trees/destruction of forests

Question 3.
Which word from the options given, can best replace the word ‘excruciating’? (1)
a) Entertaining
b) Progressing
c) Agonising
Answer:
c) Agonising

Question 4.
Here is a diary entry by Sudha Murty.
Fill in the blanks with appropriate similes or adjectives from the options given in brackets. (3)
Village days were memorable. Days were full of fun and frolic and we were all as ……. (a) …….. (dull/cheerful/sharp) as a lark. How I miss those days! In Mumbai, everyone is as busy as a ……… (b) ………. (snail/sloth/bee) and women too, are no different. Girls who were regarded to be as …….. (c) ……… (sour/chill/sweet) as a rose have now transformed themselves into leaders, office and torchbearers.
Answer:
a) cheerful; b) bee; c) sweet

Question 5.
Imagine that you are invited for an interview for the post of the chef at Taj Group of Hotels. Complete the interview suitably. (4)
Interviewer: ……. a ………?
You: I have specialized in Chinese and Continental dishes.
Interviewer: ……. (b) …….?
You: Yes, of course. I am comfortable with all types of delicacies.
Interviewer: ……. (c) …….?
You: I expect a comfortable work atmosphere, fantastic team work, and opportunities to develop.
Interviewer: Where do you see yourselves ten years from now?
You: Ten years from now, I ……. (d) …….
Answer:
a) What have you specialized in?/What are your areas of specialization?
b) Are you comfortable with delicacies?/Do you know how to make delicacies?
c) What are your expectations from this Organization?/ What do you expect from us?
d) I would start a hotel of my own./I may become the Chief Chef here. I would be working in a prestigious hotel abroad.

Question 6.
The passage given below contains a few errors. (4)
Edit the passage appropriately. The main conflict in the story ‘Amigo Brothers’ is that off ambition. Both friends want to be champions. But only one of them can be the champ. So they decide in fight out it. In the end, they realize that they value their bonding more than their ambition.
Answer:
off – of; wants – want; fight out it – fight it out; there bonding – their bonding

(Question Nos. 7 – 12): Answer any 5 questions in not less than 60 words. Each carries 4 scores. (5 × 4 = 20)

Question 7.
Your class is conducting a debate on the topic ‘PUNISHMENTS ARE ESSENTIAL FOR STUDENT DISCIPLINE’.
Write four arguments either favouring or opposing the topic.
Answer:
Punishments are essential for student discipline.
Arguments for:

1. A proverb says, “Spare the rod, spoil the child”.
2. Students can be easily corrected with appropriate punishments.
3. If punishments are not given, students will tend to ignore their studies.
4. Punishments will force the students to be punctual, respectful, obedient and hard-working.

Question 8.
Elaborate the idea conveyed in the lines given below:
‘I am the pillars of the house;
The keystone of the arch am I
Take me away, and roof and wall
Would fall to ruin me utterly.’
Answer:
The main idea in these lines is the importance of a woman in the house. She may be a mother or wife. She says she is the pillars that keep the house in place. She is the keystone of the arch and if she is removed from the house, it will completely collapse. The poem stresses the importance of women for the house and its proper functioning.

Question 9.
The activities of the literary club of your school will be inaugurated by the cine artist and state award winner, Sri. Indrans. The one-act play, ‘Post Early for Christmas’ will be staged on the day. Prepare a notice inviting all for the programme.
Answer:

GOVERNMENT HIGHER SECONDARY SCHOOL,
IRINJALAKUDA
LITERARY CLUB

Date 1 July 2019

NOTICE

The activities of our Literary Club will be inaugurated by the famous Cine Artiste and State Award Winner Sri Indrans. The inaugural meeting will take place at 10.00 a.m. in the Hall of the School. After the inaugural meeting, the one-act play titled “Post Early for Christmas” by R.H. Wood will be staged by our Literary Club Members.

“Post Early for Christmas” is a hilarious comedy in which you find different kinds of people coming to the Post Office and the havoc created by a parcel that was ticking. Some people think it is a bomb. There are many frantic activities going on. See what happens on stage!

All of you are invited!

Eva Saifu
Secretary

Question 10.
You have come to know that your close friend is having a severe toothache. Frame four sets of suggestions you would make before him/her.
You may use expressions like:

I think you should ……/ If I were you, I ……/ Would you like ……/ Perhaps you could ……/ I’d like to suggest ……
Answer:
I think you should immediately keep some salty, warm water in your mouth as it will reduce your pain. If I were you, I would then go to the dentist. Would you like to have a warm drink or take some pain killers? Perhaps you could take a Panadol to suppress your pain for the time being till you meet the doctor.

Question 11.
Prepare a profile of Shaheen Mistri using the hints given below:
Birth: 16th March 1971
Place of birth: Mumbai
Academic Qualifications: B.A. degree (Sociology), M.A. in Sociology
Almamater: the University of Mumbai, University of Manchester
Occupation: CEO, Teach for India
Famous as: Indian social activist and educator
Founder: Akansha Foundation, Teach for India
Awards won: Ashoka Fellow (2001), Global Leader for tomorrow (2002), Asia Society 21 Leader (2006)
Authored: Redrawing India – The Teach for India Story (2014)
Answer:

PROFILE OF SHAHEEN MISTRI

Shaheen Mistri was born in Mumbai on 16 March 1971. After her +2, she joined the University of Mumbai and got her B.A. in Sociology. Later she went to England and passed her M.A. in Sociology with a First Class, from the University of Manchester. Currently, she is the CEO of the Organization called “Teach India” which is doing a laudable job in educating slum children.

She is famous as a social activist and educator. She is the Founder of two famous Organizations named “Akanksha” and “Teach India”. She has received numerous awards which include Ashoka Fellow (2001), Global Leader for Tomorrow (2002), and Asia Society 21 Leader (2006). She has authored a well-known Book – “Redrawing India – The Teach for India Story” which was published in 2014.

Question 12.
Wangari Maathai says, ‘Women are often the first to become aware of environmental damage ………..’
Do you agree with her opinion? Substantiate your answer.
Answer:
I quite agree with the opinion of Wangari Maathai that “Women are often the first to become aware of environmental damage”. Wangari Maathai is a Kenyan. She is an environmentalist and political activist. She got the Nobel Prize in 2004 for her contribution to sustainable development, democracy, and peace. She made the remark quoted above in her Nobel Prize Acceptance speech.

Being a Kenyan she was especially speaking of the women of Africa. She spoke from her childhood experiences and observations of nature in rural Kenya. As she was growing up, she noticed that forests were cleared and were replaced by commercial plantations. This destroys the local biodiversity and the ability of the forests to conserve water.

Because of large scale deforestation, the African women lacked firewood, clean drinking water, balanced diets, shelter, and income. In Africa, women are the primary caretakers. They till the land and feed their families. As a result, they are the first to notice the environmental damage as resources become scarce, making it difficult for them to maintain their families.

(Question Nos. 13 – 19): Answer any 5 questions in about 80 words. Each carries 6 scores. (5 × 6 = 30)

Question 13.
Read the following lines from the poem ‘Rice’.
‘I’ll reach home in good time, at last
just as my mother drains the well-cooked rice.’
This is better money – what good times!’
How is the expression ‘good times’ used in the above two contexts? Substantiate your answer.
Answer:
The expression “good time/s” is used with two different meanings in the poem. I’ll reach home in good time means I will reach home soon or promptly or without much delay. The son, who is in North India, studying 15. for his doctorate, is tired of eating chappatis all the time and now he is dreaming of eating the well-cooked rice which his mother has drained. So he wants to reach home as quickly as he can. He hopes he will reach home in good time, that is very soon, to eat the rice. But in the second use ‘what good times’ the phrase good times refers to prosperity, the time when one has plenty of money. People of his village have abandoned rice cultivation and have started cultivating cash crops for good times and better money.

Question 14.
You have dreams of setting up a business of your own after your studies. You are greatly inspired by Irfan Alam’s views and ideas. Draft a letter of inquiry to Sammaan foundation asking for clarification of the doubts regarding the financial investment of the organization, mode of operation, the security of the members, etc.
Answer:
Jerry John
16/IV, Azad Road
Irinjalakuda North P.O.
Kerala
PIN 680125
3 July 2019

Mr. Irfan Alam
CEO, SammaaN
Kalina
Mumbai-21

Dear Sir,
Sub : Information about SammaaN

First of all let me congratulate you for founding such a useful organization like SammaaN which is helping thousands of auto-rickshaw workers to live better lives without getting into any kind of debt traps.

Being inspired by your model, I also want to embark upon some such entrepreneurship after my +2 studies. The financial condition of my family does not permit me to go for higher education as my father is sick and my mother does not go for any outside work.

I want to get some matters clarified by you as I also want to bean entrepreneur like you. What was your initial investment in the Organization? Did the banks or any government agency advance you some funds to start the operations?

I also would want to know the modes of your operation. Do you give financial help to the members who join you? What are the security measures you have in place if a member runs into a loss or is disabled due to some disease or accident?

Please find some time to send a reply to my letter. You can contact me on the phone. My phone number is 282 228450. My email address is jerry@amail.com.
Thank you in advance,

Yours sincerely,
Jerry John.

Question 15.
After the competition at Tompkin’s Square Park, the Amigo Brothers are interviewed.
Reporter: Has the bout affected your friendship?
Felix: No, we both never take fighting into our hearts.
Reporter: So then, who among you must have won the fight?
Antonio: We both are always winners. The question itself is unimportant.
How would you report the above conversation?
Answer:
The reporter asked the Amigo brothers if/whether the bout had affected their friendship. Felix replied in the negative saying that they both never took fighting into their hearts. Then the reporter asked them who among them must have won the fight. Antonio replied saying that they both were always winners and that the question itself was unimportant.

Question 16.
As the secretary of the Anti-Narcotics Club in your school, you decide to write an article on the dangers of drug abuse to make aware of this growing menace among schoolmates. Prepare the article to be published in your school magazine.
Answer:

DANGERS OF DRUG ABUSE

The problem of drug abuse has become a serious menace threatening the future and even the life of the youth, especially the school and college-going youth. This age group is more vulnerable to drug abuse because most of them are teenagers. Teenage is a time of stress and strain. To overcome their stress and strain, many youngsters abuse drugs. These drugs give them temporary pleasure, relieving them of their tensions for some time.

There are many reasons for drug abuse among youth. They get a lot of money from their parents. Since most families have only one child or two, the parents pamper them. Many parents do not find time to spend with their children. Another reason for drug abuse is peer influence. A third reason is the blind imitation of the so-called glamour boys and girls in films and sports who use drugs. A 4th reason is the easy availability of drugs. Another reason is the moral laxity of the times. Law-enforcing agencies do not do their work sincerely. There is corruption everywhere and one can do anything if one has money.

Sensual drugs play havoc with the body and mind of their users. The drug addict experiences sensory deprivation. He has a general feeling of physical discomfort and there are personality changes in him. The addict feels depressed. His mental disturbance can be like paranoia. The addict knows he has a problem. But he does not know its source. He looks at external objects with suspicion. Anything outside scares him and he withdraws further and further into himself.

Drugs also affect the body adversely. Dirty needles and solutions can cause liver diseases, venereal diseases and infection of the kidney and brain. Sniffing cocaine and amphetamines can damage the tissue of the nose. Marijuana and tobacco smoking can cause lung diseases. Babies of women who are addicted to drugs are likely to be born with a lot of problems. A drug addict can easily get pneumonia, tuberculosis and have problems of malnutrition and weight loss. An overdose of any drug can cause respiratory or cardiac failure and death.

The drug problems can be solved only through the concerted efforts of the parents, teachers, community leaders and the law-enforcing agencies. Awareness programmes should be conducted about the dangerous effects of drug abuse. The community leaders should help the youth to channel their energies in the right directions. The police must ensure that sensual drugs are not easily available.

Drug abuse is a demon that should be exterminated from our midst with all the might we can muster.

Question 17.
Ahoregallu is essential in any journey. It is more so when people around us are too busy in their own worlds. You feel that it is necessary to post counsellors in schools so that students can reveal their fears and joys to them. Write a letter to the Minister of Education requesting him to take necessary steps in this regard.
Answer:

15/XV Azad Road
Irinjalakuda, North P.O.
10, July 2019

Prof. C. Ravindranath
Education Minister of Kerala
The Secretariat
Trivandrum – 001

Honourable Education Minister,

Sub.: Posting Counsellors in schools

Every day we watch on the TV and read in the newspapers about cases of child abuse. It is strange that even parents, guardians, teachers and religious . leaders abuse children who are entrusted to their care. These people are supposed to work for the welfare of the children and lead them into the right paths to succeed in life and make their valuable contributions to humanity at large. But unfortunately the fact is that many children are abused in many ways, even sexually, by the very people who are supposed to protect them.

The main reason for child abuse is the lack of ‘horegallus’. A horegallu is a stone bench where the weary traveller sits and talks to a sympathetic listener. Thus a horegallu is a patient and sympathetic listener. In the journey of life we all need horegallus. Children need them all the more. A child is afraid to -talk to others about the abuse he/she is facing from one of his relatives, teachers, neighbours or religious leaders. So he/she silently suffers the abuse which can even change his/her personality and attitude towards life and other people.

If there are counsellors in schools, they can listen to the problems of the children patiently and suggest appropriate solutions. By telling their problems to a sympathetic counsellor, the abused children feel relieved. The counsellors can initiate actions against the offenders and-th*is free the abused children from further abuses.

I humbly request your honour to post Counsellors to all schools so that the children can have horegallus to whom they can reveal their hearts.

Yours sincerely,
Mehboob Saithu

Question 18.
Robert Baldwin in ‘The Hour of Truth’ stands as an epitome of honesty throughout the play. His decisions are never influenced by any financial offers. Prepare a character sketch of Robert Baldwin.
Answer:
Robert Baldwin is the hero of the play “The Hour of Truth” by Percival Wilde. Baldwin lives in a trim cottage with his wife Martha and their son John and daughter Evie. He is working as the governor of a bank owned by John Gresham. His salary is very low, 60 dollars a week. His son John earns only 30 dollars a week. With this limited income they live reasonably happily.

Suddenly there is a problem in their lives. John Gresham has misappropriated bank money and he is in jail. The bank is closed and Baldwin will have no job. Only Baldwin is the witness for the misappropriation. If he gives his true testimony, John Gresham will definitely go to jail for a long period. Baldwin is honest and he has taught his family the importance of honesty.

When Baldwin returns home after the arrest of John Gresham, his wife and children rush to him to know the latest news. Initially his wife and the children tell him that he should say the truth during the trial even if it means jail term for Gresham. But when they come to know that Gresham has offered him a bribe of 100,000 dollars just to say ‘I don’t remember’ when some incriminating questions are asked in the court, they change their stance. Theytrytofind out all kinds of loopholes to make Baldwin accept the bribe. Martha, John and Evie do their best to make Baldwin change his mind. But he asserts that he wants to go to his grave clean.

He does love Gresham. He even named his son after him. Even now he remembers how he worked with him for so long. They were boys together. They worked side by side. All this is true, but he is not willing to tell a lie to save his friend and employer even when he is offered a colossal sum of 100,000 dollars.

Baldwin’s honesty is repaid abundantly. Even Gresham is proud of him and he recommends him to Mr. Marshall, the President of the Third International. At the end of the play we see Marshall coming to the house of Baldwin and offering him a job at the Third International. We see Baldwin crying in the end. He must have been crying for two reasons – his friend Gresham has confessed his guilt and he will surely be punished. Secondly he is shedding tears of joy in gratitude to God who has amply repaid his honesty. Honesty, Baldwin proves, beyond an iota of doubt, is the best policy.

Question 19.
The story of Amigo Brothers is always heart, warming. You want to share your appreciation of the story with your friend in Bengaluru. Draft an e-mail conveying the essence of the story so as to inspire your friend to read it at the earliest.
Answer:
ashalatha@gmail.com
Hi Asha, I got your mail. Good to know that you are planning to visit the Taj Mahal during the Onam Vacation.

I am writing this email with a special purpose. I read a story titled “Amigo Brothers” written by Piri Thomas. The story impressed me greatly. It is about friendship. As you are my best friend, I want you to read this story. I’m sure you will enjoy it.

Antonio Cruz and Felix Vargas were both 17. They were so together in friendship that they felt like brothers. They had known each other from childhood. They grew up in the same building on the Lower East Side of Manhattan. They both had a dream – becoming a light-weight champion of the world.

Whenever they got a chance they exercised. They would run everyday morning. They had a collection of fight magazines. They also kept the torn tickets of all the boxing matches they had gone to see. They also had some clippings of their own.

After a series of elimination bouts, they were told that they were to meet each other in the division finals which would be 2 weeks away. The winner would represent the Boys’ Club in the Golden Gloves Championship Tournament.

They were in a big fix. They both wanted to win. But neither of them wanted to defeat the other. They decided to go into the ring as if they had never met. They have to fight it out.

The fight was at Tompkins Square Park. The Park was full of people as the fight was well publicised. Antonio and Felix enter the ring. The crowd explodes into a roar. The referee whistles to start the fight. The fight is on. Punches fly back and forth but none is floored. Each punch is applauded thunderously by the crowd.

It is now the 3rd round. Each contestant wants to win. They punch each other very hard. Nobody falls to the canvas. The final bell rings. But the fight continues. The bell rings again and again. But there is no stopping to the fighting. People start getting worried. It looks as if they are witnessing a do or die fight and not a mere contest. There is utter silence. Then the referee and the trainers separate the contestants.

Now the contestants embrace. They have forgotten they were fighting like bulls up to a moment ago. The announcer makes an announcement. He is trying to name the winner. But Antonio and Felix had left like good old friends, arm in arm, declaring their bond of friendship.

In my mind both are winners, aren’t they Asha?
I am coming to Bengaluru next week. We will talk more about the story then! Bye for now!

Love,
Johny

(Question Nos. 20 – 22): Answer any 2 questions, each in about 140 words. Each carries 8 scores. (2 × 8 = 16)

Question 20.
Given below is a poster displayed at the Assembly Hall in connection with the International Women’s Day celebrations conducted by the Souhrida Club of your school. As the convenor of the programme, you are asked to give a detailed talk on ‘EMPOWERING WOMEN – CHALLENGES AHEAD’.Prepare the likely speech taking hints from the poster.

Answer:
Respected Principal and teachers and my dear friends,

Today we are celebrating International Women’s Day and I am asked to talk about “Empowering Women – Challenges Ahead”. It is a great pity that Indians speak very highly of women, but they do not show much respect to them in their actions. They will say woman is the mother, Devi guru and the light of the home. They even name their daughters with names of goddesses and saintly persons. Even in our own school, we have so many Lakshmis and Saraswatis, Aishas and Khadeejas and Marys and Theresas. But do they receive the respect they deserve from the society? Even in the buses and roads they are troubled and teased. We hear of so many crimes against women every day. Rapes and murder of women have ceased to shock people as they have become everyday affairs.

I say we are doing only lip-service to women. Even in Kerala which has more women than men, women are not given their rights. Look at the number of women candidates in the Lok Sabha elections for 2019. We have three coalitions in Kerala, UDF, LDF and NDA. Each of these groups is fielding 20 candidates for the 20 Lok Sabha seats. What percentage of women candidates do they field? Just 10%. No political party is free from this fault.

Women don’t get equal pay with men in non¬governmental services. A man gets Rs. 800 a day but a woman is given only Rs. 400 although she is doing similar or even harder work than the man. Women are refused opportunities for employment because employers feel that married women will take maternity leave.

I strongly feel that the Parliament and State Assemblies should have equal number of men and women representatives. Men have no business keeping women under a patriarchal system. Let the women also feel they are persons with the dignity and freedom that men enjoy in our society.

We have to change our mindset. Man and woman are created as equals and they should be treated equally by the society in all walks of life.

As I urge all the women here to fight for their rights,
I wish them success in their endeavors.

Jai Hind!

Question 21.
Read the poem given below. Compare and contrast it with ‘Mending Wall’ and attempt a critical appreciation.

NEIGHBOUR

Iain Crichton Smith

Build me a bridge over the stream
to my neighbour’s house
where he is standing in dungarees*
in the fresh morning.

O ring of snowdrops
spread wherever you want
and you also blackbird
sing across the fences.

My neighbour, if the rain fails on you,
let it fall on me also
from the same black cloud
that does not recognize gates.

*dungarees: a garment consisting of trousers held up by straps over the shoulders.
Answer:
“Mending Wall” by Robert Frost is a delightful poem. He once said: “A poem begins in delight and ends in wisdom.” He starts the poem in a delightful way saying that there is something that does not like a wall. That something makes the ground under the wa|l swell which results in cracks in the wall. Gradually the stones that make the wall fall to either side.

The fallen stones have lost their shapes. It is not easy to keep them back in their place. The gaps are so big that even two people can walk abreast through them. The poet strongly feels there is no need for a wall between him and his neighbour because he grows apple trees and the neighbour grows pine. Apple trees won’t go and eat the pine and pines won’t come to eat the apples. A wall was fine if they had cows as they could get mixed up without a wall. When the poet says there is no need for a wall between them, the neighbour tells him “Good fences make good neighbours.” Thus the poem ends in wisdom.

In “Mending Wall”, the poet has used many poetic devices such a metaphor, simile, personification and repetition. The language is simple and the imagery is exquisite. It gives a fine message, a priceless truth. “Neighbour” by lain Crichton Smith is a beautiful poem of just 12 lines. It gives a fine message. We see two neighbours here separated by a stream, ring of snowdrops, fences and gates. The poet can see his neighbour standing in his house wearing his trousers with shoulder straps. He wants a bridge to be built across the stream to connect the two neighbours. The blackbird sings across the fences and the rain falls on both the neighbour and himself. The rain which comes from the black cloud does not differentiate his neighbour from him. Nature is treating them equally and so the poet thinks that the neighbours should be linked by a bridge.

The poem is highly melodious. The imagery is superb as we can visualize the stream, the ring of snowdrops, the blackbird singing, and the rain falling. We can see the neighbour standing in his house in his trousers held by shoulder straps. The message is loud and clear – There should be love between neighbours.

Between the two poems, I prefer “Mending Wall”, as it is more dramatic and action packed.

Question 22.
In the light of reading excerpts from the lives of Irfan Alam and Shaheen Mistri, you arrive at the conclusion that ‘Entrepreneurs are not made, they are born.’ Our nation benefits from such enterprises too. Write an essay on the topic highlighting the merits of promoting entrepreneurship among the youth.
(HINTS : youth made responsible – dignity of labour – self-confident – self-reliance – hard work and commitment – job satisfaction – financial stability – creating employment opportunities – initiatives like kudumbasree, mango cabs, ubereats, flipkart, etc.)
Answer:

ENTREPRENEURS ARE NOT MADE, THEY ARE BORN

There is a wrong tendency in our society which makes the youth to look for government jobs. After getting their degrees, they go on writing PSC tests in the
hope of getting clerical jobs in government services. This is really bad for our country because the government can’t provide jobs for all the people. Here comes the importance of entrepreneurship.

Entrepreneurship makes us responsible and the success of the enterprise depends on our planning, vision and hard work. It also adds to the concept of dignity of labour’. An entrepreneur is prepared to do any kind of work that is needed of him for the success of his enterprise. In the Government service employees are graded as Class I, II, III and IV. Class IV employees are usually sweepers, cleaners and peons. Class III consists of clerks. Class II is of the Supervisory staff and Class I is that of High Officers. But in a private enterprise, there is no such division.

Entrepreneurship enhances self-confidence, self reliance and hard work. There is a lot of job satisfaction when we see our enterprises are succeeding. Entrepreneurs give jobs to others instead of being job hunters. It brings financial stability to ourselves and also the people whom we employ.

There are so many private enterprises which are very popular today. We have the Kudumbasree, Mango Cabs, Uber Eats and Flipkart. They are all successful enterprises and they originated in the minds of ordinary people. With some thinking, willingness to work hard, determination and self-confidence we too can embark upon some project that will prove successful.

Rome was not built in a day. Let us remember Dhirubhai Ambani, the father of the richest man in India today, Mukesh Ambani. Dhirubhai Ambani worked as a clerk in Yemen and returned to India and started a textile trading company in 1958. His initial capital was a mere Rs. 15,000. Let’s remember, “Where there is a will there is a way.”