Class 12 Chemistry – Page 2

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines

February 27, 2023

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 13 Amines:

Section Name	Topic Name
13	Amines
13.1	Structure of Amines
13.2	Classification
13.3	Nomenclature
13.4	Preparation of Amines
13.5	Physical Properties
13.6	Chemical Reactions
13.7	Method of Preparation of Diazonium Salts
13.8	Physical Properties
13.9	Chemical Reactions
13.10	Importance of Diazonium Salts in Synthesis of Aromatic Compounds

NCERT INTEXT QUESTIONS

13.1. Classify the following amines as primary, secondary and tertiary:

Ans: (i) 1° (ii) -3° (iii) 1° (iv) 2°

13.2. Write the structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.
(i) Write the IUPAC names of all the isomers
(ii) What type of isomerism is exhibited by different types of amines?
Ans: Eight isomeric amines are possible

Isomerism exhibited by different amines

Chain isomers: (i) and (ii) ; (iii) and (iv) ; (i) and (iv)
Position isomers: (ii) and (iii) ; (ii) and (iv)
Metamers: (v) and (vi) ; (vii) and (viii)
Functional isomers: All the three types of amines are the functional isomers of each other.

13.3. How will you convert:
(i) Benzene into aniline
(ii) Benzene into N,N-dimethylaniline
(iii) Cl-(CH₂)₄-Cl into Hexane -1,6- diamine
Ans:

13.4. Arrange the following in increasing order of their basic strength :
(i) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂, (C₂H₅)₂NH
(ii) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H_sNH₂, C₆H₅CH₂NH₂Ans:
In general, the basic character of ammonia (NH₃) and the amines is linked with the availability of the lone electrons pair on the nitrogen atom. In other words, these are all Lewis bases.
Amines act as Lewis bases due to the presence of lone electron paîr on the nitrogen atom. Since the nitrogen atom is sp³ hybridised, its electron attracting tendency is considerably reduced. It can readily lose its electron pair and acts as a base. For example, amines form hydroxides with water.

Here K_t is called dissociation constant for the base. Greater the K_b value stronger will be the base. The basic strength
of amines can also be expressed as pK_b value which is related to K_b as :

The K_b values are :

13.5. Complete the following acid-base reactions and name the products:
(i) CH₃CH₂CH₂NH₂+HCl ——–>
(ii) (C₂H₅)₃ N+HCl ——–>
Ans:

13.6. Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Ans:

13.7. Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Ans:

13.8. Write structures of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC names of the isomers which will liberated N₂ gas on treatment with nitrons acid.
Ans: In ‘all, four structural isomers are possible. These are:

13.9. Convert:
(i) 3-Methylanilineinto3-nitrotoluene
(ii) Aniline into 1,3,5- Tribromo benzene
Ans:

NCERT EXRECISES

13.1. Write IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines.
(i) (CH₃)₂ CHNH₂ (ii) CH₃(CH₂)₂NH₂ (iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃ CNH₂ (v) C₆H₅NHCH₃(vi) (CH₃CH₂)₂NCH₃
(vii)m-BrC₆H₄NH₂
Ans: (i) Propan-2-amine(1°)
(ii) Propan-1-amine (1°),
(iii) N-Methylpropan-2-amine (2°).
(iv) 2-Methylpropan-2-amine(l°)
(v) N-MethylbenzenamineorN-methylaniline(2°)
(vi) N-Ethyl-N-methylethanamine (3°)
(vii) 3-Bromobenzenamine or 3-bromoaniline (1°)

13.2. Give one chemical test to distinguish between the following pairs of compounds:
(i)Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-Methylaniline.
Ans:

13.3. Account for the following
(i) pKb of aniline is more than that of methylamine
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Ans: (i) In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring.
As a result, electron density on the nitrogen . atom decreases. Whereas in CH₃NH₂,+ I-effect of -CH₃ group increases the electron density on the N-atom. Therefore, aniline is a weaker base than methylamine and hence its pKb value is higher than that of methylamine.
(ii) Ethylamine dissolves in water due to intermolecular H-bonding. However, in case of aniline, due to the large hydrophobic part, i.e., hydrocarbon part, the extent of H-bonding is very less therefore aniline is insoluble in water.

(iii) Methylamine being more basic than water, accepts a proton from water liberating OH^– ions,

(iv) Nitration is usually carried out with a mixture of cone HNO₃ + cone H₂SO₄. In presence of these acids, most of aniline gets protonated to form ahilinium ion. Therefore, in presence of acids, the reaction mixture consist of aniline and anilinium ion. Now, -NH2 group in aniline is activating and o, p-directing while the -+NH₃ group in anilinium ion is deactivating and rw-directing: Nitration of aniline (due to steric hindrance at o-position) mainly gives p-nitroaniline, the nitration of anilinium ion gives m-nitroaniline. In actual practice, approx a 1:1 mixture of p-nitroaniline and m-nitroaniline is obtained. Thus, nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of the amino group.

13.4. Arrange the following:
(i) In decreasing order of pKb values:
C₂H₅NH₂,C₆H₅NHCH₃,(C₂H₅)₂NH and C₆H₅NH₂
(ii) In increasing order of basic strength:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂ NH and CH₃NH₂.
(iii) In increasing order of basic strength:
(а)Aniline,p-nitroaniline andp-toluidine
(b)C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂
(iv) In decreasing order of basic strength in gas phase:
C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃
(v) In increasing order of boiling point:
C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
(vi) In increasing order of solubility in water:
C₆H₅NH₂,(C₂H₅)₂NH,C₂H₅NH₂
Ans: (i) Due to delocalisation of lone pair of electrons of the N-atom over the benzene ring,C₆H₅NH₂ and C₆H₅NHCH₃ are far less basic than C₂H₅NH₂ and (C₂H,)₂NH. Due to +I-effect of the -CH₃ group, C₆H₅NHCH₃ is little more basic that C₆H₅NH₂. Among C₂H₅NH₂ and (C₂H₅)₂NH, (C₂H₅)₂NH is more basic than C₂H₅NH₂due to greater+I-effect of two -C₂H₅ groups. Therefore correct order of decreasing pKb values is:

(ii) Among CH₃NH₂ and (C₂H₅)₂NH, primarily due to the greater +I-effect of the two -C₂H₅ groups over one -CH₃ group, (C₂H₅)₂NH is more basic than CH₃NH₂.In both C₆H₅NH2 and C₆H₅N(CH₃)₂ lone pair of electrons present on N-atom is delocalized over the benzene ring but C₆H₅N(CH₃)₂ is more basic due to +1 effect of two-CH₃ groups.

(iii) (a) The presence of electron donating -CH₃ group increases while the presence of electron withdrawing -NO₂ group decreases the basic strength of amines.

(b) In C₆H₅NH₂ and C₆H₅NHCH₃, N is directly attached to the benzene ring. As a result, the lone pair of electrons on the N-atom is delocalised over the benzene ring. Therefore, both C₆H₅NH₂ and C₆H₅NHCH₃ are weaker base in comparison to C₆H₅CH₂NH₂. Among C₆H₅NH₂ and C₆H₅NHCH₃, due to +1 effect of-CH₃ group C₆H₅NHCH₃ is more basic.

(iv) In gas phase or in non-aqueous solvents such as chlorobenzene etc, the solvation effects i. e., the stabilization of the conjugate acid due to H-bonding are absent. Therefore, basic strength depends only upon the +I-effect of the alkyl groups. The +I-effect increases with increase in number of alkyl groups.Thus correct order of decreasing basic strength in gas phase is,

(v) Since the electronegativity of O is higher than thalof N, therefore, alcohols form stronger H-bonds than amines. Also, the extent of H-bonding depends upon flie number of H-atoms on the N-atom, thus the extent of H-bonding is greater in primary amine than secondary amine.

(vi) Solubility decreases with increase in molecular mass of amines due to increase in the size of the hydrophobic hydrocarbon part and with decrease is the number of H-atoms on the N-atom which undergo H-bonding.

13.5. How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-aminopentane
(iii) Methanol to ethanoic acid.
(iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine
(viii) Propanoic acid into ethanoic acid?
Ans:

13.6. Describe the method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Answer:
The distinction in the three types of amines can be done by the following methods :
(i) Hinsberg’s Test:
This is a very useful test for the distinction of primary, secondary and tertiary amines. An amine is shaken with
Hinsberg’s reagent (benzene suiphonyl chloride) in the presence of excess of aqueous KOH solution. The reactions taking
place are given on the next page.

A primary amine forms N – alkyl benzene suiphonamide which dissolves in aqueous KOH solution to form potassium salt and upon acidification with dilute HCI regenerates the insoluble suiphonamide.
A secondary amine forms N, N – dialkylbenzene suiphonamide which remains insoluble in aqueous KOH and even after acidification with dilute HCl
A tertiary amine does not react with benzene suiphonyl chloride and remains insoluble in aqueous KOH.
However, on acidification with dilute HCI it gives a clear solution due to the formation of the ammonium salt.

(ii) Reaction with nitrous acid:
All the three types of amines, aliphatic as well as aromatic, react with nitrous acid under different conditions to form variety of products. Since nitrous acid is highly unstable, it is prepared in situ by the action of dilute hydrochloric acid on sodium nitrite.

(a) Primary aliphatic amines react with nitrous acid at low temperature (cold conditions) to form primary alcohol and
nitrogen gas accompanied by brisk effervescence. Nitrous acid is unstable in nature and is prepared in situ by reacting sodium
nitrite with dilute hydrochloric acid. For example,

The reaction is used as a tesijôr primary aliphadc amines as no other amine evolves nitrogen with nurous acid.

(b) Primary aromatic amines such as aniline react with nitrous acid under ice cold conditions (273 – 278 K) to form benzen diazonium salt. The reaction is known as diazotisation reaction.

in case, the temperature is allowed to rise above 278 K, benzene diazortium chloride is decomposed by water to form phenol.

Aliphatic primary amines also react with nitrous acid to form alkyl diazonium salts in a similar manner. But these are
quite unstable and decompose to form a mixture of alcohols, alkenes and alkyl halides along with the evolution of N₂ gas.

(c) Secondary amines (both aliphatic and aromatic) react with nitrous acid to form nitrosoamines which separate as
Yellow oily liquids.

(d) Tertiary aliphatic amines dissolve in a cold solution of nitrous acid to form salts which decompose on warming to
give nitrosoamine and alcohol. For example,

(e) Tertiary aromatic amines react with nitrous acid to give a coloured nitrosoderivative. This reaction is called
nitrosation and as a result, a hydrogen atom in the para position gets replaced by a nitroso (-NO) group. For example,

13.7. Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) ‘Hofmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi) Acetylation
(vii) Gabriel phthalimide synthesis
Ans: (i) Carbylamine reaction: Both aliphatic and aromatic primary amines when warmed with chloroform and an alcoholic solution of KOH, produces isocyanides or carbylamines which have very unpleasant odours. This reaction is called carbylamine reaction.

(ii) Diazotisation: The process of conversion of a primary aromatic amino compound into a diazonium salt, is known as diazotisation. This process is carried out by adding an aqueous solution of sodium nitrite to a solution of primary aromatic amine (e.g., aniline) in excess of HCl at a temperature below 5°C.

(iii) Hoffmann’s bromamide reaction: When an amide is treated with bromine in alkali solution, it is converted to a primary amine that has one carbon atom less than the starting amide. This reaction is known as Hoffinann’s bromamide degradation reaction.

(iv) Coupling reaction: In this reaction, arene diazonium salt reacts with aromatic amino compound (in acidic medium) or a phenol (in alkaline medium) to form brightly coloured azo compounds. The reaction generally takes place at para position to the hydroxy or amino group. If para position is blocked, it occurs at ortho position and if both ortho and para positions are occupied, than no coupling takes place.

(v) Ammonolysis: It is a process of replacement of either halogen atom in alkyl halides (or aryl halides) or hydroxyl group in alcohols (or phenols) by amino group. The reagent used for ammonolysis is alcoholic ammonia. Generally, a mixture of primary, secondary and tertiary amine is formed.

(vi) Acetylation: The process of introducing an acetyl (CH₃CO-) group into molecule using acetyl chloride or acetic anhydride is called acetylation.

(vii) Gabriel phthalimide synthesis: It is a method of preparation of pure aliphatic and aralkyl primary amines. Phthalimide on treatment with ethanolic KOH gives potassium phathalimide which on heating with a suitable alkyl Or aralkyl halides gives N-substituted phthalimides, which on hydrolysis with dil HCI or with alkali give primary amines.

13.8. Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2,4,6-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanamine
(vi) Chlorobenzene to p-Chloroaniline
(vii) Aniline to p-bromoaniIine
(viii)Benzamide to toluene
(ix) Aniline to benzyl alcohol.
Ans:

13.9. Give the structures of A,B and C in the following reaction:

Ans:

13.10. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Ans:
From the available information, we find that ‘B’ upon heating with Br₂ and KOH forms a compound ‘C’. The compound ‘B’ is
expected to be an acid amide. Since ‘B’ has been formed upon heating compound ‘A’ with aqueous ammonia, the compound ‘A’ is an aromatic acid.
It is benzoic acid. The reactions involved are given as follows:

13.11. Complete the following reactions:

Ans:

13.12. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Ans: The success of Gabriel phthalimide reaction depends upon the nucleophilic attack by the phthalimide anion on the organic halogen compound.
Since aryl halides do not undergo nucleophilic substitution reactions easily, therefore, arylamines, i.e., aromatic, primary amines cannot be prepared by Gabriel phthalimide reaction.

13.13. Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Ans: Both aromatic and aliphatic primary amines react with HNO₂ at 273-278 K to form aromatic and aliphatic diazonium salts respectively. But aliphatic diazonium salts are unstable even at this low temperature and thus decompose readily to form a mixture of compounds. Aromatic and aliphatic primary amines react with

13.14. Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Ans: (i) Loss of proton from an amine gives an amide ion while loss of a proton from alcohol give an alkoxide ion.
R—NH₂—>R—NH^– +H⁺
R—O —H—>R— O^– +H⁺ .
Since O is more electronegative than N, so it wijl attract positive species more strongly in comparison to N. Thus, RO~ is more stable than RNH®. Thus, alcohols are more acidic than amines. Conversely, amines are less acidic than alcohols.
(ii) Due to the presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of H-atom on the N-atom do not undergo H-bonding. As a result, primary amines have higher boiling points than tertiary amines of comparable molecular mass.
(iii) Aromatic amines are far less basic than ammonia and aliphatic amines because of following reasons:
(a) Due to resonance in aniline and other aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalised over the benzene ring and thus it is less easily available for protonation. Therefore, aromatic amines are weaker bases than ammonia and aliphatic amines.
(b) Aromatic amines arS more stable than corresponding protonated ion; Hence, they hag very less tendency to combine with a proton to form corresponding protonated ion, and thus they are less basic.

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NCERT Solutions For Class 12 Chemistry Chapter 1 The Solid State

February 27, 2023

NCERT Solutions For Class 12 Chemistry Chapter 1 The Solid State

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State:

Section Name	Topic Name
1	The Solid State
1.1	General Characteristics of Solid State
1.2	Amorphous and Crystalline Solids
1.3	Classification of Crystalline Solids
1.4	Crystal Lattices and Unit Cells
1.5	Number of Atoms in a Unit Cell
1.6	Close Packed Structures
1.7	Packing Efficiency
1.8	Calculations Involving Unit Cell Dimensions
1.9	Imperfections in Solids
1.10	Electrical Properties
1.11	Magnetic Properties

NCERT TEXTBOOK QUESTIONS SOLVED

1.1. Why are solids rigid?
Ans: The constituent particles in solids have fixed positions and can oscillate about their mean positions. Hence, they are rigid.

1.2. Why do solids have definite volume?
Ans: Solids keep their volume because of rigidity in their structure. The interparticle forces are very strong. Moreover, the interparticle spaces are very few and small as well. As a result, their volumes cannot change by applying pressure.

1.3. Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fibreglass, copper
Ans: Crystalline solids: Benzoic acid, potassium nitrate, copper Amorphous solids: Polyurethane, Teflon, cellophane, polyvinyl chloride, fibreglass

1.4. Why is glass considered as super cooled liquid ? (C.B.S.E. Delhi 2013)
Ans: Glass is considered to be super cooled liquid because it shows some of the characteristics of liquids, though it is an amorphous solid. For example, it is slightly thicker at the bottom. This can be possible only if it has flown like liquid, though very slowly.

1.5. Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?
Ans: As the solid has same value of refractive index along all directions, it is isotropic in nature and hence amorphous. Being amorphous solid, it will not show a clean cleavage and when cut, it will break into pieces with irregular surfaces.

1.6. Classify the following solids in different categories based on the nature of the intermolecular forces: sodium sulphate, copper, benzene, urea, ammonia, water, zinc sulphide, diamond, rubedium, argon, silicon carbide.
Ans: Ionic, metallic, molecular, molecular, molecular (hydrogen-bonded), molecular (hydrogen-bonded), ionic, covalent, metallic, molecular, covalent (network).

1.7. Solid A is a very hard electrical insulator in. solid as well as in molten state and melts at extremely high temperature. What type of solid is it?
Ans: It is a covalent or network solid.

1.8. Why are ionic solids conducting in the molten state and not in the solid-state?
Ans: In the ionic solids, the electrical conductivity is due to the movement of the ions. Since the ionic mobility is negligible in the solid state, these are non-conducting in this state. Upon melting, the ions present acquire some mobility. Therefore, the ionic solids become conducting

1.9. What type of solids are electrical conductors, malleable and ductile?
Ans: Metallic solids

1.10. Give the significance of a lattice point.
Ans: The lattice point denotes the position of a particular constituent in the crystal lattice. It may be atom, ion or a molecule. The arrangement of the lattice points in space is responsible for the shape of a particular crystalline solid.

1.11. Name the parameters that characterise a unit cell.
Ans: A unit cell is characterised by the following parameters:
(i)the dimensions of unit cell along three edges: a, b and c.
(ii)the angles between the edges: α (between b and c); β (between a and c) and γ (between a and b)

1.12. Distinguish between :
(i) Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
Ans:
(i) In a hexagonal unit cell :
a = b # c; α = β = 90° and γ = 120°
In a monoclinic unit cell :
a # b # c and α = γ = 90° and β # 90°
(ii) In a face-centered unit cell, constituent particles are located at all the corners as well as at the centres of all the faces.
In end-centered unit cell, constituent particles are located at all the corners as well as at the centres of two opposite faces. (C.B.S.E Foreign 2015)

1.13. Explain how many portions of an atom located at
(i)corner and (ii)body centre of a cubic unit cell is part of its neighbouring unit cell.
Ans: (i) An atom at the comer is shared by eight adjacent unit cells. Hence, portion of the atom at the comer that belongs to one unit cell=1/8.
(ii)An atom at the body centre is not shared by any other unit cell. Hence, it belongs fully to unit cell.

1.14. What is the two-dimensional coordination number of a molecule in a square close-packed layer?
Ans: In the two-dimensional square close-packed layer, a particular molecule is in contact with four molecules. Hence, the coordination number of the molecule is four.

1.15. A compound forms hexagonal close-packed. structure. What is the total number of voids in 0. 5 mol of it? How many of these are tetrahedral voids?
Ans:
No. of atoms in close packings 0.5 mol =0.5 x 6.022 x 10²³ =3.011 x 10²³
No. of octahedral voids = No. of atoms in packing =3.011 x 10²³
No. of tetrahedral voids = 2 x No. of atoms in packing
= 2 x 3.011 x 10²³= 6.022 x 10²³Total no. of voids = 3.011 x 10²³+ 6.022 x 10²³
= 9.033 x 10²³

1.16. A compound is formed by two elements M and N. The element N forms ccp and atoms of the element M occupy 1/3 of the tetrahedral voids. What is the formula of the compound? (C.B.S.E. Foreign 2015)
Ans: Let us suppose that,
the no. of atoms of N present in ccp = x
Since 1/3rd of the tetrahedral voids are occupied by the atoms of M, therefore,
the no. of tetrahedral voids occupied = 2x/3
The ratio of atoms of N and M in the compound = x : 2x/3 or 3 : 2
∴ The formula of the compound = N₃M₂ or M₂N₃

1.17. Wh ich of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centered cubic and (iii) hexagonal close-packed lattice?
Ans: Packing efficiency of:
Simple cubic = 52.4% bcc = 68% hcp = 74%
hcp lattice has the highest packing efficiency.

1.18. An element with molar mass 2:7 x 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If its density is 2:7 x 103 kg m^-3, what is the nature of the cubic unit cell ? (C.B.S.E. Delhi 2015)
Ans:

Since there are four atoms per unit cell, the cubic unit cell must be face centred (fcc) or cubic close packed (ccp).

1.19. What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
Ans: When a solid is heated, vacancy defect is produced in the crystal. On heating, some atoms or ions leave the lattice site completely, i.e., lattice sites become vacant. As a result-of this defect, density of the substances decreases.

1.20. What types of stoichiometric defects are shown by (C.B.S.E. Delhi 2013)
(i) ZnS
(ii) AgBr?
Ans:
(i) ZnS crystals may show Frenkel defects since the cationic size is smaller as compared to anionic size.
(ii) AgBr crystals may show both Frenkel and Schottky defects.

1.21. Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
Ans: Let us take an example NaCl doped with SrCl, impurity when SrCl2 is added to NaCl solid as an impurity, two Na+ ions will be replaced and one of their sites will be occupied by Sr21- while the other will remain vacant. Thus, we can say that when a cation of higher valence is added as an impurity to an ionic solid, two or more cations of lower valency are replaced by a cation of higher valency to maintain electrical neutrality. Hence, some cationic vacancies are created.

1.22. Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Ans: Let us take an example of NaCl. When NaCl crystal is heated in presence of Na vapour, some Cl^–ions leave their lattice sites to combine with Na to form NaCl. The e^-1 s lost by Na to form Na⁺ (Na⁺ + Cl^–—> NaCl) then diffuse into the crystal to occupy the anion vacancies. These sites are called F-centres. These e-s absorb energy from visible light, get excited to higher energy level and when they fall back to ground state, they impart yellow colour to NaCl crystal.

1.23. A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?
Ans: Impurity from group 15 should be added to get n-type semiconductor.

1.24. What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.
Ans: Ferromagnetic substances make better permanent magnets. This is because when placed in magnetic field, their domains get oriented in the directions of magnetic field and a strong magnetic field is produced. This ordering of domains persists even when external magnetic field is removed. Hence, the ferromagnetic substance becomes a permanent magnet.

NCERT EXERCISES

1.1. Define the term ‘amorphous’. Give a few examples of amorphous solids.
Sol. Amorphous solids are those substances, in which there is no regular arrangement of its constituent particles, (i.e., ions, atoms or molecules). The arrangement of the constituting particles has only short-range order, i.e., a regular and periodically repeating pattern is observed over short distances only, e.g., glass, rubber, and plastics.

1.2. What makes glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Sol. Glass is a supercooled liquid and an amorphous substance. Quartz is the crystalline form of silica (SiO₂) in which tetrahedral units SiO₄ are linked with each other in such a way that the oxygen atom of one tetrahedron is shared with another Si atom. Quartz can be converted into glass by melting it and cooling the melt very rapidly. In the glass, SiO₄ tetrahedra are joined in a random manner.

1.3 Classify each of the following solids as ionic, metallic, modular, network (covalent), or amorphous:
(i) Tetra phosphorus decoxide (P₄O₁₀) (ii) Ammonium phosphate, (NH₄)₃PO₄ (iii) SiC (iv) I₂ (v) P₄(vii) Graphite (viii), Brass (ix) Rb (x) LiBr (xi) Si
Sol.

1.4 (i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atom
(a) in a cubic close-packed structure?
(b) in a body centred cubic structure?
Sol. (i) The number of nearest neighbours of a particle are called its coordination number.
(ii) (a) 12 (b) 8

1.5. How can you determine the atomic mass of an unknown metal if you know its density and dimensions of its unit cell ? Explain your answer. (C.B.S.E. Outside Delhi 2011)
Sol.

1.6 ‘Stability of a crystal is reflected in the magnitude of its melting points’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?
Sol. Higher the melting point, greater are the forces holding the constituent particles together and thus greater is the stability of a crystal. Melting points of given substances are following. Water = 273 K, Ethyl alcohol = 155.7 K, Diethylether = 156.8 K, Methane = 90.5 K.
The intermoleoilar forces present in case of water and ethyl alcohol are mainly due to the hydrogen bonding which is responsible for their high melting points. Hydrogen bonding is stronger in case of water than ethyl alcohol and hence water has higher melting point then ethyl alcohol. Dipole-dipole interactions are present in case of diethylether. The only forces present in case of methane is the weak van der Waal’s forces (or London dispersion forces).

1.7. How will you distinguish between the following pairs of terms :
(a) Hexagonal close packing and cubic close packing
(b) Crystal lattice and unit cell
(c) Tetrahedral void and octahedral void.
Sol.
(a) In hexagonal close packing (hcp), the spheres of the third layer are vertically above the spheres of the first layer
(ABABAB……. type). On the other hand, in cubic close packing (ccp), the spheres of the fourth layer are present above the spheres of the first layer (ABCABC…..type).
(b) Crystal lattice: It deplicts the actual shape as well as size of the constituent particles in the crystal. It is therefore, called space lattice or crystal lattice.
Unit cell: Each bricks represents the unit cell while the block is similar to the space or crystal lattice. Thus, a unit cell is the fundamental building block of the space lattice.
(c) Tetrahedral void: A tetrahedral void is formed when triangular void made by three spheres of a particular layer and touching each other.

Octahedral void: An octahedral void or site is formed when three spheres arranged at the corners of an equilateral triangle are placed over anothet set of spheres.

1.8 How many lattice points are there is one unit cell of each of the following lattices?
(i) Face centred cubic (if) Face centred tetragonal (iii) Body centred cubic
Sol.

1.9 Explain:
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Sol. (i) Metallic and ionic crystals
Similarities:
(a) There is electrostatic force of attraction in both metallic and ionic crystals.
(b) Both have high melting points.
(c) Bonds are non-directional in both the cases.
Differences:
(a) Ionic crystals are bad conductors of electricity in solids state as ions are not free to move. They can conduct electricity only in die molten state or in aqueous solution. Metallic crystals are good conductors of electricity in solid state as electrons are free to move.
(b) Ionic bond is strong due to strong electrostatic forces of attraction.
Metallic bond may be strong or weak depending upon the number of valence electrons and the size of the kernels.
(ii) Ionic solids are hard and brittle.Ionic solids are hard due to the presence of strong electrostatic forces of attraction. The brittleness in ionic crystals is due to the non- directional bonds in them.

1.10 Calculate the efficiency of packing in case of a metal crystal for (i) simple cubic, (ii) body centred cubic, and (iii) face centred cubic (with the assumptions that atoms are touching each other).
Sol. Packing efficiency: It is the percentage of total space filled by the particles.

1.11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 x 10^-8 cm and density is 10.5 g cm^-3, calculate the atomic mass of silver.
Sol.

1.12. A cubic solid is made of two elements P and Q. Atoms Q are at the corners of the cube and P at the body centre. What is the formula of the compound ? What is the co-ordination number of P and Q?
Sol. Contribution by atoms Q present at the eight corners of the cube = \(\frac { 1 }{ 8 } \)= x 8 = 1
Contribution by atom P present at the body centre = 1
Thus, P and Q are present in the ratio 1:1.
∴ Formula of the compound is PQ.
Co-ordination number of atoms P and Q = 8.

1.13 Niobium crystallises in a body centred cubic structure. If density is 8.55 g cm^-3, calculate atomic radius of niobium, using its atomic mass 93u.
Sol.

1.14 If the radius of the octahedral void is r and radius of the atoms in close-packing is R, derive relation between rand R.
Sol. A sphere is fitted into the octahedral void as shown in the diagram.

1.15 Copper crystallises into a fee lattice with edge length 3.61 x 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 gcm^-3.
Sol.

This calculated value of density is closely in agreement with its measured value of 8.92 g cm³.

Question 16.
Analysis shows that nickel oxide has the formula Ni_0.98 O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?
Solution:
98 Ni-atoms are associated with 100 O – atoms. Out of 98 Ni-atoms, suppose Ni present as Ni²⁺ = x
Then Ni present as Ni³⁺ = 98 – x
Total charge on x Ni²⁺ and (98 – x) Ni³⁺ should
be equal to charge on 100 O^2- ions.
Hence, x × 2 + (98 – x) × 3 = 100 × 2 or 2x + 294 – 3x = 200 or x = 94
∴ Fraction of Ni present as Ni²⁺ = \(\frac { 94 }{ 98 } \) × 100 = 96%
Fraction of Ni present as Ni³⁺ = \(\frac { 4 }{ 98 } \) × 100 = 4%

Question 17.
What are semi-conductors? Describe the two main types of semiconductors and contrast their conduction mechanisms.
Solution:
Semi-conductors are the substances whose conductivity lies in between those of conductors and insulators. The two
main types of semiconductors are n-type and p-type.
(i) n-type semiconductor: When a silicon or germanium crystal is doped with group 15 element like P or As, the dopant atom forms four covalent bonds like Si or Ge atom but the fifth electron, not used in bonding, becomes delocalised and continues its share towards electrical conduction. Thus silicon or germanium doped with P or As is called H-type semiconductor, a-indicative of negative since it is the electron that conducts electricity.

(ii) p-type semiconductor: When a silicon or germanium is doped with group 13 element like B or Al, the dopant is present only with three valence electrons. An electron vacancy or a hole is created at the place of missing fourth electron. Here, this hole moves throughout the crystal like a positive charge giving rise to electrical conductivity. Thus Si or Ge doped with B or Al is called p-type semiconductor, p stands for positive hole, since it is the positive hole that is responsible for conduction.

Question 18.
Non-stoichiometric cuprous oxide, Cu₂O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?
Solution:
The ratio less than 2 : 1 in Cu₂0 shows cuprous (Cu⁺) ions have been replaced by cupric (Cu²⁺) ions. For maintaining electrical neutrality, every two Cu⁺ ions will be replaced by one Cu²⁺ ion thereby creating a hole. As conduction will be due to the presence of these positive holes, hence it is a p-type semiconductor.

Question 19.
Ferric oxide crystallises in a hexagonal dose- packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
Solution:
Suppose the number of oxide ions (O^2-) in the packing = 90
∴ Number of octahedral voids = 90
As 2/3^rd of the octahedral voids are occupied by ferric ions, therefore, number of ferric ions 2 present = \(\frac { 2 }{ 3 } \) × 90 = 60
∴ Ratio of Fe³⁺ : O^2- = 60 : 90 = 2 : 3
Hence, the formula of ferric oxide is Fe₂O₃.

Question 20.
Classify each of the following as being either a p-type or n-type semiconductor :

Ge doped with In
B doped with Si.

Solution:

Ge is group 14 element and In is group 13 element. Hence, an electron deficient hole is created and therefore, it is a p – type semiconductor.
B is group 13 element and Si is group 14 element, there will be a free electron, So, it is an n-type semiconductor.

Question 21.
Gold (atomic radius = 0.144 nm) crystallises in a face centred unit cell. What is the length of the side of the unit cell ?
Solution:
For a face centred cubic unit cell (fcc)
Edge length (a) = \(2\sqrt { 2 } r\) = 2 x 1.4142 x 0.144 mm = 0.407 nm

Question 22.
In terms of band theory, what is the difference

between a conductor and an insulator
between a conductor and a semiconductor?

Solution:
In most of the solids and in many insulating solids conduction takes place due to migration of electrons under the influence of electric field. However, in ionic solids, it is the ions that are responsible for the conducting behaviour due to their movement.

(i) In metals, conductivity strongly depends upon the number of valence electrons available in an atom. The atomic orbitals of metal atoms form molecular orbitals which are so close in energy to each other, as to form a band. If this band is partially filled or it overlaps with the higher energy unoccupied conduction band, then electrons can flow easily under an applied electric field and the metal behaves as a conductor.

If the gap between valence band and next higher unoccupied conduction band is large, electrons cannot jump into it and such a substance behaves as insulator.

(ii) If the gap between the valence band and conduction band is small, some electrons may jump from valence band to the conduction band. Such a substance shows some conductivity and it behaves as a semiconductor. Electrical conductivity of semiconductors increases with increase in temperature, since more electrons can jump to the conduction band. Silicon and germanium show this type of behaviour and are called intrinsic semiconductors. Conductors have no forbidden band.

Question 23.
Explain the following terms with suitable examples :

Schottky defect
Frenkel defect
Interstitial defect
F-centres.

Solution:
(i) Schottky defect : In Schottky defect a pair of vacancies or holes exist in the crystal lattice due to the absence of equal number of cations and anions from their lattice points. It is a common defect in ionic compounds of high coordination number where both cations and anions are of the same size, e.g., KCl, NaCl, KBr, etc. Due to this defect density of crystal decreases and it begins to conduct electricity to a smaller extent.

(ii) Frenkel defect : This defect arises when some of the ions in the lattice occupy interstitial sites leaving lattice sites vacant. This defect is generally found in ionic crystals where anion is much larger in size than the cation, e.g., AgBr, ZnS, etc. Due to this defect density does not change, electrical conductivity increases to a small extent and there is no change in overall chemical composition of the crystal.

(iii) Interstitial defect : When some constituent particles (atoms or molecules) occupy an interstitial site of the crystal, it is said to have interstitial defect. Due to this defect the density of the substance increases.

(iv) F-Centres : These are the anionic sites occupied by unpaired electrons. F-centres impart colour to crystals. The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal.

Question 24.
Aluminium crystallises in a cubic close packed structure. Its metallic radius is 125 pm.

What is the length of the side of the unit cell?
How many unit cells are there in 1.00 cm³ of aluminium?

Solution:

Question 25.
If NaCI is doped with 10^-3 mol % SrCl₂, what is the concentration of cation vacancies?
Solution:
Let moles of NaCI = 100
∴ Moles of SrCl₂ doped = 10^-3
Each Sr²⁺ will replace two Na+ ions. To maintain electrical neutrality it occupies one position and thus creates one cation vacancy.
∴ Moles of cation vacancy in 100 moles NaCI = 10^-3
Moles of cation vacancy in one mole
NaCI = 10^-3 × 10^-2 = 10^-5
∴ Number of cation vacancies
= 10^-5 × 6.022 × 10²³ = 6.022 × 10¹⁸ mol^-1

Question 26.
Explain the following with suitable example:

Ferromagnetism
Paramagnetism
Ferrimagnetism
Antiferromagnetism
12-16 and 13-15 group compounds.

Solution:
(i) Ferromagnetic substances : Substances which are attracted very strongly by a magnetic field are called ferromagnetic substances, e.g., Fe, Ni, Co and CrO₂ show ferromagnetism. Such substances remain permanently magnetised, once they have been magnetised. This type of magnetic moments are due to unpaired electrons in the same direction.

The ferromagnetic material, CrO₂, is used to make magnetic tapes used for audio recording.

(ii) Paramagnetic substances : Substances which are weakly attracted by the external magnetic field are called paramagnetic substances. The property thus exhibited is called paramagnetism. They are magnetised in the same direction as that of the applied field. This property is shown by those substances whose atoms, ions or molecules contain unpaired electrons, e.g., O₂, Cu²⁺, Fe³⁺, etc. These substances, however, lose their magnetism in the absence of the magnetic field.

(iii) Ferrimagnetic substances : Substances which are expected to possess large magnetism on the basis of the unpaired electrons but actually have small net magnetic moment are called ferrimagnetic substances, e.g., Fe₃O₄, ferrites of the formula M²⁺Fe₂O₄ where M = Mg, Cu, Zn, etc. Ferrimagnetism arises due to the unequal number of magnetic moments in opposite direction resulting in some net magnetic moment.

(iv) Antiferromagnetic substances : Substances which are expected to possess paramagnetism or ferromagnetism on the basis of unpaired electrons but actually they possess zero net magnetic moment are called antiferromagnetic substances, e.g., MnO. Antiferromagnetism is due to the presence of equal number of magnetic moments in the opposite directions

(v) 13-15 group compounds : When the solid state materials are produced by combination of elements of groups 13 and 15, the compounds thus obtained are called 13-15 compounds. For example, InSb, AlP, GaAs, etc.

12-16 group compounds : Combination of elements of groups 12 and 16 yield some solid compounds which are referred to as 12-16 compounds. For example, ZnS, CdS, CdSe, HgTe, etc. In these compounds, the bonds have ionic character.

More Resources for CBSE Class 12:

NCERT Solutions Maths Physics Chemistry Biology Science

NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers

December 26, 2022

NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers:

Section Name	Topic Name
15	Polymers
15.1	Classification of Polymers
15.2	Types of Polymerisation Reactions
15.3	Molecular Mass of Polymers
15.4	Biodegradable Polymers
15.5	Polymers of Commercial Importance

NCERT INTEXT QUESTIONS

15.1. What are polymers?
Ans: Polymers are high molecular mass substances (10³ — 10⁷u) consisting of a very large number of simple repeating structural units joined together through covalent bonds in a linear fashion. They are also called macromolecules. Ex: polythene, nylon 6,6, bakelite, rubber, etc.

15.2. How are polymers classified on the basis of structure?
Ans: On the basis of structure, polymers are classified into three types. These are linear chain polymers, branched chain polymers and crossedlinkedpolymers.

1. Linear chain polymers: In this case, the monomer units are linked to one another to form long linear chains. These linear chains are placed one above the other and are closely packed in space. The close packing results in high densities, tensile strength and also high melting and boiling points. High density polyethene is a very common example of this type. Nylon, polyesters and PVC are also linear chain polymers.

2. Branched chain polymers: In this type of polymers, the monomer units are linked to form long chains which have also side chains or branched chains of different Lengths attached to them. As a result of branching, these polymers are not closely packed in space. They have low densities, low tensile strength as well as low melting and boiling points. Some common examples of such polymers are ; low density polyethene, amylopectin, starch, glycogen etc.

3. Cross: linked polymers. In these polymers, also called net—work polymers, the monomer units are linked together to form three dimensionaL net—work as shown in the figure. These are expected to be quite hard, rigid and brittle. Examples of cross linked polymers are bakelite, glyptal. melamine formaldehyde polymer etc.

15.3. Write the names of the monomers of the following polymers:

Ans: (i) Hexamethylene diamine NH₂-(CH₂)6NH₂ and adipic acid HOOC – (CH₂)₄ – COOH
(ii) Caprolactum
(iii) Tetrafluoroethene F₂C = CF₂

15.4. Classify the following as addition and condensation polymers:
Terylene, Bakelite, Polyvinyl chloride,Polythene
Ans: Addition polymers: Polyvinyl chloride, Polythene
Condensation polymers : Terylene, bakelite.

15.5. Explain the difference between Buna-N and Buna-S.
Ans: Both Buna-N and Buna-S are synthetic rubber and are co-polymers in nature. They differ in their constituents.
Buna-N: Constituents are : buta-1, 3-diene and acrylonitrile.
Buna-S: Constituents are : buta-1, 3-diene, and styrene. They condense in the presence of Na.

Buna – S: It is a co—polymer of 1. 3 – butadiene and styrene and is prepared by the polymerisation of these components in the
ratio of 3 : 1 in the presence of sodium.

Buna-N (Nitrile rubber): h is a co-polymer of buta-1. 3-diene and acrylonitrile. It is formed as follows:

15.6. Arrange the following polymers in increasing order of their intermolecuiar forces.
(i) Nylon 6,6, Buna-S, Polythene
(ii) Nylon 6, Neoprene, Polyvinyl chloride
Ans: On the basis of intermolecuiar forces, polymers
are classified as elastomers, fibres and plastics. The increasing order of intermolecuiar forces is: Elastomer < Plastic < fibre.
Thus, we have
(i)Buns-S < Polythene < Nylon 6,6
(ii)Neoprene < Polyvinyl chloride < Nylon 6.

NCERT EXERCISES

15.1. Explain the terms polymer and monomer.
Ans: Polymers are high molecular mass substances consisting of a very large number of simple repeating structural units joined together through covalent bonds in a regular fashion. Polymers are also called macromolecules. Some examples are polythene, nylon-66, bakelite, rubber, etc. Monomers are the. simple and reactive molecules from which the polymers are prepared either by addition or condensation polymerisation. Some examples are ethene, vinyl chloride, acrylonitrile, phenol and formaldehyde etc.

15.2. What are natural and synthetic polymers ? Give two examples of each.
Ans:
1. Natural polymers: The polymers which occur in nature mostly in plants and animals are called natural polymers. A few common examples are starch, cellulose, proteins, rubber nucleic acids, etc. Among them, starch and cellulose are the polymers of glucose molecules. Proteins are formed from amino acids which may be linked in different ways. These have been discussed in detail in unit 15 on biomolecules. Natural rubber is yet another useful polymer which is obtained from the latex of the rubber tree. The monomer units are of the unsaturated hydrocarbon 2-methyl-i, 3-butadiene, also called isoprene.
Example of natural polymers: Natural rubber, cellulose, nucleic acids, proteins etc.

2. Synthetic polymers: The polymers which are prepared in the laboraroiy are called synthetic polymers. These are also called man made polymers and have been developed in the present century to meet the ever increasing demand of the modem civilisation.
Example of synthetic polymers: Dacron (or terylene), Bakelite, PVC, Nylon-66, Nylon-6 etc.

15.3. Distinguish between the terms homopolymer and copolymer and give an example of each.
Ans: Polymers whose repeating structural units are derived from only one type of monomer units are called homopolymers, e.g., PVC polyethene, PAN, teflon, polystyrene, nylon- 6 etc.
Polymers whose repeating structural units are derived from two or more types of monomer molecules are copolymers, e.g., Buna-S, Buna-N, nylon-66, polyester, bakelite.

15.4. How do you explain the functionality of a monomer?
Ans: Functionality of a monomer implies the number of bonding sites present in it. For example, monomers like propene, styrene, acrylonitrile have functionality of one which means that have one bonding site.
Monomers such as ethylene glycol, hexamethylenediamine, adipic acid have functionality of two which means that they have two bonding sites.

15.5. Define the term polymerisation?
Ans: It is a process of formation of a high molecular Sol. mass polymer from one or more monomers by linking together a large number of repeating structural units through covalent bonds.

15.6. Is (-NH — CHR—CO-)_n a homopolymer or copolymer?
Ans: It is a homopolymer because the repeating structural unit has only one type of monomer, i.e., NH₂—CHR—COOH.

15.7. In which classes, are the polymers classified on the basis of molecular forces?
Ans: Polymers are classified into four classes on the basis of molecular forces. These are:
elastomers, fibres, thermoplastic polymers and thermosetting polymers.

1. Elastomers: In these polymers, the intermolecular forces are the weakest. As a result, they can be readily stretched by applying small stress and regain their original shape when the stress is removed. The elasticity can be further increased by introducing some cross – links in the polymer chains. Natural rubber is the most popular example of elastomers. A few more examples are of: buna-S, buna-N and neoprene.

2. Fibres: Fibres represent a class of polymers which are thread-like and can be woven into fabrics in a number of ways. These are widely used for making clothes, nets, ropes, gauzes etc. Fibres possess high tensile strength because the chains possess strong intermolecular forces such as hydrogen bonding. These forces are also responsible for close packing of the chains. As a result, the fibres are crystalline in nature and have aJso sharp melting points. A few common polymers belonging to this class are nylon – 66, terylene and polyacrylonitrile etc.

3. Thermoplastics: These are linear polymers and have weak van der Waals forces acting in the various chains and are intermediate of the forces present in the elastomers and in the fibres. When heated, they melt and form a fluid which sets into a hard mass on cooling, Thus, they can be cast into different shapes by using suitable moulds. A few common examples are polyethene and polystyrene polyvinyls etc. These can be used for making toys, buckets, telephone apparatus, television cabinets etc.

4. Thermosetting plastics: These are normally semifluid substances with low molecular masses. When heated, they become hard and infusible due to the cross-linking between the polymer chains. As a result, they also become three dimensional in nature. They do not melt when heated. A few common thermosetting polymers are bakelite, melamine-formaldehyde, urea-formaldehyde and polyurethane etc.

15.8. How can you differentiate between addition and condensation polymerisatiop?
Ans: In addition polymerization, the molecules of the same or different monomers simply add on to one another leading to the formation of a macromolecules without elimination of small molecules like H₂O, NH₃ etc. Addition polymerization generally occurs among molecules containing double and triple bonds. For example, formation of polythene from ethene and neoprene from chloroprene, etc. In condensation polymerisation, two or more bifunctional trifimctional molecules undergo a series of independent condensation reactions usually with the elimination of simple molecules like water, alcohol, ammonia, carbon dioxide and hydrogen chloride to form a macromolecule. For example, nylon-6,6 is a condensation polymer of hexamethylenediamine and adipic acid formed by elimination of water molecules.

15.9. Explain the term copolymerisation and give two examples.
Ans: When two or more different monomers are allowed to polymerise together the product formed is called a copolymer, and the process in called copolymerisation. Example, Buna-S and Buna-N. Buna- S is a copolymer of 1, 3- butadiene and styrene while Buna-N is a copolymer of 1,3-butadiene and acrylonitrile.

15.10. Write the free radical mechanism for the polymerisation of ethene.
Ans:

15.11. Define thermoplastics and thermo setting polymers with two examples of each
Ans: Thermoplastics polymers are linear polymer which can be repeatedly melted and moulded again and again on heating without any change in chemical composition and mechanical strength. Examples are polythene and polypropylene.
Thermosetting polymers, on the other hand, are permanently setting polymers. Once on heating in a mould, they get hardened and set, and then cannot be softened again. This hardening on heating is due to cross- linking between different polymeric chains to give a three dimensional network solid. Examples are bakelite, melamine-foimaldehyde polymer etc.

15.12. Write the monomers used for gettingThe following polymers:
(i) Polyvinylchloride
(ii) Teflon (iii) Bakelite
Ans:

15.13. Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Ans:

15.14. How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Ans: Natural rubber is cis-polyisoprene and is obtained by 1, 4-polymerization of isoprene units. In this polymer, double bonds are located between C₂ and C₃ of each isoprene unit. These cis-double bonds do not allow the polymer chains to come closer for effective interactions and hence intermolecular forces are quite weak. As a result, natural rubber, i.e., cis-polyisoprene has a randomly coiled structure not the linear one and hence show elasticity.

15.15. Discuss the main purpose of vulcanisation of rubber.
Ans: Natural rubber has the following disadvantages:
(a) It is soft and sticky and becomes even more so at high temperatures and brittle at low temperatures. Therefore, rubber is generally used in a narrow temperature range (283-335 K) where its elasticity is maintained.
(b)It has large water absorption capacity, has low tensile strength and low resistance to abrasion.
(c)It is not resistant to the action of organic solvents.
(d)It is easily attacked by oxygen and other oxidising agents. .
To improve all these properties, natural rubber is vulcanised by heating it with about 5% sulphur at 373-415 K. The vulcanized rubber thus obtained has excellent elasticity over a larger range of temperature, has low water absorption tendency and is resistant to the action of organic solvents and oxidising agents.

15.16. What are the monomeric repeating units of Nylon-6 and Nylon 6,6?
Ans:

15.17. Write the names and structures of the monomers of the following polymers:
(i) Buna-S (ii) Buna-N (iii) Dacron (iv) Neoprene
Ans:

15.18. Identify the monomer in the following polymeric structures:

Ans:

15.19. How is dacron obtained from ethylene glycol and terephthalic acid?
Ans: Dacron is obtained by condensation polymerization of ethylene glycol and terephthalic acid with the elimination of water molecules. The reaction is carried out at 420 – 460 K in presence of a catalyst consisting of a mixture of zinc acetate and antimony trioxide.

15.20. What is a biodegradable polymer ? Give an example of a biodegradable aliphatic polyester.
Ans: Polymers which disintegrate by themselves over a period of time due to environment degradation by bacteria, etc., are called biodegradable polymers. Example is PHBV, i. e., Poly-β-Hydroxybutyrate-co-β- Hydroxyvalerate.

More Resources for CBSE Class 12:

NCERT Solutions Maths Physics Chemistry Biology Science

NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life

December 26, 2022

NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 16 Chemistry in Everyday Life:

Section Name	Topic Name
16	Chemistry in Everyday Life
16.1	Drugs and their Classification
16.2	Drug-Target Interaction
16.3	Therapeutic Action of Different Classes of Drugs
16.4	Chemicals in Food
16.5	Cleansing Agents

NCERT INTEXT QUESTIONS

16.1. Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor. Why?
Ans: Most of drugs taken in doses higher than recommended may produce harmful effects and act as poison and cause even death. Therefore, a doctor must always be consulted before taking the drug.

16.2. “Ranitidine is an antacid” With reference to which classification, has this statement been given?
Ans: Ranitidine is labelled as antacid since it is quite effective in neutralising the excess of acidity in the stomach. It is sold in the market under the trade name Zintac.

16.3. Why do we require artificial sweetening agents?
Ans: To reduce calorie intake and to protect teeth from decaying, we need artificial sweeteners.

16.4. Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structures of these compounds are given below:
(i) (C₁₅H₃₁COO)₃C₃H₅-Glyceryl palmitate
(ii) (C₁₇H₃₂COO)₃C₃H₅-Glyceryl oleate
Ans:

16.5. Following type of non-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic part in the molecule. Identify the functional group (s) present in the molecule.

Ans:

Functional groups present in the detergent molecule are:
(i)ether
(ii)1°alcoholic group

NCERT EXERCISES

16.1. Why do we need to classify drugs in different ways?
Ans: Drugs are classified in following different ways:
(a) Based on pharmacological effect.
(b) Based on action on a particular biochemical process.
(c) Based on chemical structure.
(d) Based on molecular targets.
Each classification has its own usefulness.
(а) Classification based on pharmacological effect is useful for doctors because it provides them the whole range of drugs available for the treatment of a particular disease.
(b) Classification based on action on a particular biochemical proc*ess is useful for choosing the correct compound for designing the synthesis of a desired drug.
(c) Classification based on chemical structure helps us to design the synthesis of a number of structurally similar compounds having different substituents and then choosing the drug having least toxicity.
(d) Classification on the basis of molecular targets is useful for medical chemists so that they can design a drug which is most effective for a particular receptor site.

16.2. Explain the following as used in medicinal chemistry
(a) Lead compounds
(b) Target molecules or drug targets.
Ans:
(a) Lead compounds are the compounds which are effective in different drugs. They have specific chemical formulas and may be extracted either from natural sources (plants and animals) or may be synthesised in the laboratory.

(b) Target molecules or drug targets. An enzyme (E) functions by combining with the reactant (called substrate) denoted as ‘S’ to form an activated complex known as enzyme-substrate complex (E-S). The complex dissociates to form product and releases the enzyme for carrying out further activity.

16.3. Name the macro molecules that are chosen as drug targets.
Ans: Proteins, carbohydrates, lipids and nucleic acids are chosen as drug targets.

16.4. Why the medicines should not be taken without consulting doctors?
Ans: No doubt medicines are panacea for most of the body ailments. But their wrong choice and overdose can cause havoc and may even prove to be fatal. Therefore, it is of utmost importance that the medicines should not be given without consulting doctors.

16.5. Define the term chemotherapy.
Ans: It is the branch of chemistry that deals with the treatment of diseases by using chemicals as medicines.

16.6. Which forces are involved in holding the drugs to the active site of enzymes?
Ans: The following forces are involved in holding the drugs to the active site of enzymes:
(a) Hydrogen bonding
(b) Ionic bonding
(c) Dipole-dipole interactions
(d) van der Waals interactions

16.7. Antacids and antiallergic drugs interfere with the function of histamines but do not interfere with the function of each other. Explain.
Ans: They donot interfere with the functioning of each other because they work on different receptors in the body.Histamine stimulates the secretion of pepsin and hydrochloric acid in the stomach. The drug cimetidine (antacid) was designed to prevent the interaction of histamine with the receptors present in the stomach wall. This resulted in release of lesser amount of acid. Antacid and antiallergic drugs work on different receptors.

16.8. Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Name two drugs.
Ans: In case of low level of neurotransmitter, . noradrenaline, tranquilizer (antidepressant) drugs are required because low levels of noradrenaline leads to depression. These drugs inhibit the enzymes which catalyse the degradation of noradrenaline. If the enzyme is inhibited, noradrenaline is slowly metabolized and can activate its receptor for longer periods of time thereby reducing depression. Two important drugs are iproniazid and phenylzine.

16.9. What is meant by the term broad spectrum antibiotics? Explain.
Ans: Broad spectrum antibiotics are effective against several different types or wide range of harmful bacteria. For example, tetracycline, chloramphenicol and of loxacin. Chloramphenicol can be used in case of typhoid, dysentry, acute fever, urinary infections, meningitis and pneumonia.

16.10. How do antiseptics differ from disinfectants ? Give one example of each.
Ans: Many times, the same substance can act as an antiseptic as well as disinfectant by changing the concentration of the solution used. For example, a 0.2 per cent solution of phenol acts as an antiseptic and its 1 percent solution is a disinfectant. Chlorine is used in India for making water fit for drinking at a concentration of 0.2 to 0.4 ppm (parts per million). Low concentration of sulphur dioxide is used for sterilizing squashes for preservation. A few points of distinction between antiseptics and disinfectants are listed.

Antiseptics	Disinfectants
1. Can kill or prevent the growth of micro-organisms.	1.Can kill micro-organisms.
2. Do not harm the living tissues. Therefore, these can be applied to the skin.	2. Toxic to the living tissues. Therefore, these cannot be applied to the skin.
3. These are used for the dressing of wounds, ulcers and in the treatment of diseased skin.	3. These are used for disinfecting floors, toilets, drains, instruments etc.

16.11. Why are cimetidine and ranitidine better antacids than sodium hydrogencarbonate or magnesium or aluminium hydroxide?
Ans: If excess of NaHCO₃ or Mg(OH)₂ or Al(OH)₃ is used, it makes the stomach alkaline and thus triggers the release of even more HCl which may cause ulcer in the stomach. In contrast, cimetidine and ranitidine prevent the interaction of histamine with the receptor cells in the stomach wall and thus release of HCl will be less as histamine stimulates the secretion of acid.

16.12. Name a substance which can be used as an antiseptic as well as disinfectant.
Ans: 0.2% solution of phenol acts as antiseptic while 1% solution acts as a disinfectant.

16.13. What are the main constituents of dettol?
Ans: Chloroxylenol .and α-terpineol in a suitable solvent.

16.14. What is tincture of iodine? What is its use?
Ans: 2-3% solution of iodine in alcohol and water is called tincture of iodine. It is a powerful antiseptic. It is applied on wounds.

16.15. What are food preservatives?
Ans: Preservation has a major role in the food products. Chemically preserved squashes and crushes can be kept for a fairly long time even after opening the seal of bottle.
A preservative may be defined as the substance which is capable of inhibiting or arresting the process of fermentation, acidification or any other decomposition of food. Salting i.e. addition of table salt is a well known method for food preservation and was applied in ancient times for preserving raw mangoes, tamarind, meat, fish etc. Sugar syrup can also act as a preservative. Vinegar is a useful preservative for pickles. Apart from these, sulphur dioxide and benzoic acid can be employed for the preservation of food. The major source of sulphur dioxide is potassium metabisulphite (K₂S₂O₅). It is fairly stable in neutral and alkaline medium but gets decomposed by weak acids such as carbonic, citric, tartane and malic acids. Benzoic acid is used either as such or in the form of sodium benzoate. However, sulphur dioxide has a better preservative action than sodium benzoate against bacteria and moulds. It also retards the development of yeast in juice but fails to arrest their multiplication once the number has reached a high value. Sorne salts of sorbic acid and propionic acid are also being used these days for the preservation of the food.
The use of preservatives must be properly controlled as their indiscriminate use is likely to be harmful. The preservative should not be injurious to health and should be also non-irritant.

16.16. Why is the use of aspartame limited to cold foods and drinks?
Ans: This is because it decomposes at baking or cooking temperatures and hence can be used only in cold foods and drinks as an artificial sweetener

16.17. What are artificial sweetening agents? Give two examples.
Ans: Artificial sweeteners are chemical substances which are sweet in taste but do not add any calories to our body. They are excreted as such through urine. For example, saccharin, aspartame, alitame etc.

16.18. Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Ans: Saccharine, aspartame or alitame may be used in the preparation of sweets for a diabetic patient.

16.19. What problem arises in using alitame as artificial sweetener?
Ans: Alitame is a high potency artificial sweetener.Therefore, it is difficult to control the sweetness of the food to which it is added.

16.20. How are synthetic detergents better than soaps?
Ans: They can be used in hard water as well as in acidic solution. The reason being that sulphonic acids and their calcium and magnesium salts are soluble in water thus they do not form curdy white precipitate with hard water but the fatty acids and their calcium and magnesium salts of soaps are insoluble. Detergents also works in slightly acidic solution due to formation of soluble alkyl hydrogen sulphates. Soaps react with acidic solution to form insoluble fatty acids.

16.21. Explain the following terms with suitable examples:
(i) cationic detergents (ii) anionic detergents and (iii) non-ionic detergents
Ans: (i) Cationic detergents: These are quaternary ammonium salts, chlorides, acetates, bromides etc containing one or more long chain alkyl groups. For example, cetyltrimethyl ammonium chloride.
(ii) Anionic detergents are called so because a large part of their molecules are anions. ‘These are of two types:
(a) Sodium alkyl sulphates: For example, sodium lauryl sulphate, C₁₁H₂₃CH₂OSO₃ Na⁺.
(b) Sodium alkylbenzenesulphonates.Vor example, sodium 4-(l-dodecyl) benzenesu Iphphonate (SDS).

(iii) Neutral or non-ionic detergents: These are esters of high molecular mass alcohols with fatty acids. These can also be obtained by treatment of long chain alcohols by with excess of ethylene oxide in presence of a base. For example, polyethylene glycol stearate,CH₃(CH₂)₁₆COO (CH₂CH₂O)₁₁ CH₂CH₂OH Polyethylene glycol stearate.

16.22. What are biodegradable and non-biodegradable detergents? Give one example of each.
Ans: Detergents having straight chain hydrocarbons are easily degraded (or decomposed) by microorganisms and hence are called biodegradable detergents while detergents containing branched hydrocarbon chains are not easily degraded by the microorganisms find hence are called non-biodegradable detergents. Consequently, non-biodegradable detergents accumulate in rivers and water ways thereby causing severe water pollution. Examples of biodegradable detergents are sodium lauryl sulphate, sodium 4-(-l-dodecyl) benzenesulphonate and sodium 4-(2-dodecyl) benzenesulphonate.
Examples of non-biodegradable detergents is sodium 4-(1, 3,5,7 – tetramethyloctyl) benzene sulphonate.

16.23. Why do soaps not work in hard water? (C.B.S.E. Outside Delhi 2009, 2011)
Ans: Soaps are water soluble sodium or potassium salts of higher fatty acids like palmitic acid (C₁₅H₃₁COOH), oleic acid (C₁₇H₃₃COOH) and stearic acid (C₁₇H₃₅COOH). Hard water contains certain calcium and magnesium salts which combine with soaps to form corresponding magnesium compounds. These being insoluble, get separated as curdy white precipitates resulting in wastage of soap.

16.24. Can you use soaps and synthetic detergents to check the hardness of water?
Ans: Soaps get precipitated as insoluble calcium and magnesium soaps in hard water but detergents do not. Therefore, soaps but not synthetic detergents can be used to check the hardness of water.

16.25. Explain the cleansing action of soaps.
Ans: Cleansing action of soaps : Soaps contain two parts, a large hydrocarbon which is a hydrophobic (water repelling) and a negative charged head, which is hydrophillic (water attracting). In solution water molecules being polar in nature, surround the ions & not the organic part of the molecule. When a soap is dissolved in water the molecules gather together as clusters, called micelles. The tails stick inwards & the head outwards. The hydrocarbon tail attaches itself to oily dirt. When water is agitated, the oily dirt tends to lift off from the dirty surface & dissociates into fragments. The solution now contains small globules of oil surrounded detergent molecules. The negatively charged heads present in water prevent the small globules from coming together and form aggregates. Thus the oily dirt is removed from the object.

16.26. If water contains dissolved calcium hydrogencarbonate, out of soaps and synthetic detergents, which one will you use for cleaning clothes?
Ans: Calcium hydrogencarbonate makes water hard. Therefore, soap cannot be used because it gets precipitated in hard water. On the other hand, a synthetic detergent does not precipitate in hard water because its calcium salt is also soluble in water. Therefore, synthetic detergents can be used for cleaning clothes in hard water.

16.27. Label the hydrophilic and hydrophobic parts in the following compounds.
(i)cCH₃(CH₂)₁₀CH₂OSO₃ ^–Na⁺
(ii) CH₃(CH₂)₁₅ -N+(CH₃)₃Br^–
(iii) CH₃(CH₂)₁₆C00(CH₂CH₂O)₁₁CH₂CH₂OH
Ans:

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NCERT Solutions For Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

December 26, 2022

NCERT Solutions For Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements:

Section Name	Topic Name
6	General Principles and Processes of Isolation of Elements
6.1	Occurrence of Metals
6.2	Concentration of Ores
6.3	Extraction of Crude Metal from Concentrated Ore
6.4	Thermodynamic Principles of Metallurgy
6.5	Electrochemical Principles of Metallurgy
6.6	Oxidation Reduction
6.7	Refining
6.8	Uses of Aluminium, Copper, Zinc and Iron

NCERT IN TEXT QUESTIONS

6.1. Which of the ores mentioned can be concentrated by magnetic separation method?
Ans: Ores Which are magnetic in nature can be separated from non-magnetic gangue particles by magnetic separation method. For ex: ores of iron such as haemetite (Fe₂O₃), magnetite (Fe₃O₄), siderite (FeCO₃) and iron pyrites (FeS₂) being magnetic can be separated from non-magnetic silica and other impurities by magnetic separation method.

6.2. What is the significance of leaching in the extraction of aluminium?
Ans: Leaching or chemical separation is quite effective to purify bauxite an ore of aluminium associated with the impurities of iron oxide. The ore is leached with concentrated solution of NaOH to form a soluble complex leaving behind the impurities.

6.3.

Ans: This is explained on the basis of K_eq, the equilibrium constant. In the given redox reaction, all reactants and products are solids at room temperature, so, there is no equilibrium between the reactants and products and hence the reactions does not occur at RT. At high temperature, Cr melts and values of TAS increases. As a result, the value of

6.4. Is it true that under certain conditions, Mg can reduce Al203 and Al can reduce MgO? What are those conditions?
Ans:

NCERT EXRECISES

6.1. Copper can be extracted by hydrometallurgy but not zinc. Explain.
Ans:

But with water, these metals (Al, Mg, Ca and K) forms their corresponding ions with the evolution of H₂ gas.
Thus, Al, Mg, Ca, K, etc., cannot be used to displace zinc from zinc solution, and only copper can be extracted by hydrometallurgy but not the zinc.

6.2.What is the role of depressant in froth-floatation process?
Ans: The role of depressant is to prevent one type of sulphide ore particles from forming the froth with air bubbles. NaCN is used as a depressant to separate lead sulphide (PbS) ore from zinc sulphide (ZnS) ore. NaCN forms a zinc complex, Na₂[Zn(CN)₄] on the surface of ZnS thereby preventing it from the formation of the froth.

In this condition, only lead sulphide forms froth and thus can be separated from zinc sulphide ore.

6.3. Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Ans:

6.4. Explain:
(i)Zone refining
(ii)Column chromatography.
Ans: (i) Zone refining: This method is used for production of semiconductors and other metals of very high purity, e.g., Ge, Si, B, Ca and In.
It is based on the principle that the impurities
are more soluble in the molten state (melt) than in the solid state of the metal.
The impure metal in the form of bar is heated at one end with a moving circular heater. As – the heater is slowly moved along the length of the rod, .the pure metal crystallises out of the melt whereas the impurities pass into the adjacent molten zone. Thi,s process is repeated several times till the impurities are completely driven to one end of the rod which is then cut off and discarded.
(ii) Chromatography: It is based on the principle that the different components of a mixture are adsorbed to different extents on an adsorbent.
In column chromatography, an adsorbent, such as alumina (Al₂O₃) or silica gel is packed in a column. This fonns the stationary phase. The mixture to be separated is dissolved in a suitable solvent (mobile phase) and applied to the top of the column. The adsorbed components are extracted (eluted) from the column with a suitable . solvent (eluent). The component which is more strongly adsorbed on the column takes longer time to travel through the column than a component which is weakly adsorbed. Thus, the various components of the mixture are seperated as they travel through absorbent (stationary phase).

6.5. Out of C and CO which is a better reducing agent at 673 K?
Ans: This can be explained thermodynamically, taking entropy and free energy changes into account.

As can be seen from ΔG° Vs T plot (Ellingham diagram), lines for the reactions, C ——–> C0₂ and C ——–> CO cross at 983 K. Below 983 K, the reaction (a) is energetically more favourable but above 673 K, reaction (b) is favourable and preferred. Thus, below 673 K both C and CO can act as a reducing agent but since CO can be more easily oxidised to C0₂ than C to C0₂ , therefore, below 673 K, CO is more effective reducing agent than carbon.

6.6. Name the common elements present in anode mud in the electro-refining of copper. Why are they so present?
Ans: Anode mud contains metals like Ag, Au, Pt etc. which are less reactive than Cu. Actually, they are not in a position
to lose electrons though they constitute the electrode which acts as anode. All these metals are left as residue under anode (known as anode mud) while the entire copper present participates in the oxidation half reaction.
Cu(s) → Cu²⁺(aq) + 2e^–

6.7. Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Ans: In the blast furnace reduction of iron oxides take place at different temperature ranges as shown below.

6.8. Write chemical reactions taking place in the extraction of zinc from zinc blende.
Ans: The following processes are involved in the extraction of zinc from zinc blende:
(i) Concentration: Zinc blende ore is crushed and the concentration done by froth- floatation process.
(ii) Roasting: The concentrated ore is then roasted in presence of excess of air at about 1200 K as a result zinc oxide is formed.

(iii) Reduction : Zinc oxide obtained above is mixed with powdered coke and heated to 1673 K in a fire clay retort where it is reduced ‘ to zinc metal.

At 1673 K, zinc metal being volatile (boiling point 1180 K), distills over and is condensed.
(iv) Electrolytic refining: Impure zinc is made the anode while pure zinc strip is made the cathode. ZnSO₄ solution acidified with dil. H₂SO₄ is the electrolyte used. On passing electric current, pure zinc gets deposited on the cathode.

6.9. State the role of silica in the metallurgy of copper.
Ans: Silica (SiO₂) acts as an acidic flux in the metallurgy of copper and combines with FeO (the main impurity) to form FeSiO₃ which is a slag.

6.10. What is meant by the term “chromatography”?
Ans: Chromatography is a technique used for separation, purification, identification and characterization of the components of a mixture whether coloured or colourless. The term chromatography was originally derived from the Greek word ‘chroma’ meaning colour and ‘graphy for writing because the method was first used for the separation of coloured substances (plant pigments) into individual components.

6.11. What criterion is followed for the selection of the stationary phase in chromatography?
Ans: In chromatography, particularly in adsorption chromatography, the stationary phase is the adsorbent. It should fulfil certain criteria for better results.
(i) It should have high but selective adsorption power.
(ii) The particles should be spherical in shape and of uniform size.
(iii) The adsorbent should not react chemically with the solvents used for elution or with the components of the mixture under investigation.
(iv) The adsorbent should contain as small amount of the soluble components as possible.
(v) The adsorbent should be catalytically inactive and must have a neutral surface.
(vi) The adsorbent should be easily available.
(vi) The adsorbent should be perfectly white.

6.12. Describe a method for refining nickel.
Ans: When impure nickel is heated in presence of CO at 330-350 K, it forms volatile nickel tetracarbonyl leaving behind the impurities. The nickel tetracarbonyl thus obtained is then heated to higher temperature (450-470K), then it undergoes thermal decomposition to give pure nickel.

6.13. How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.
Ans: Pure alumina can be separated from silica in bauxite by Baeyer’s process. The bauxite ore associated with silica is heated with a concentrated solution of NaOH at 473-523 K and 35-36 bar pressure. Under these conditions, alumina dissolves as sodium meta-aluminate and silica as sodium silicate leaving behind the impurities.

The resulting solution is filtered to remove the undissolved impurities, sodium meta-aluminate can be precipitated as hydrated aluminium oxide by passing CO₂ vapours. The sodium silicate formed cannot be precipitated and can be filtered off.

6.14. Giving examples, differentiate between ‘roasting’ and ‘calcination’.
Ans: Calcination is a process of converting carbonates and hydroxide ores of metals to their respective oxides by heating them, strongly below their melting points either in absence or limited supply of air.

Roasting is a process of converting sulphide ores into its metallic oxides by heating strongly below its melting point in excess of air.

6.15. How is ‘cast-iron’ different from ‘pig iron’?
Ans: Cast iron differs from pig iron with respect to the carbon contents. Whereas carbon contents in pig iron are nearly four percent (4%), cast iron contains carbon to the extent of nearly three percent (3%).

6.16. Differentiate between “minerals” and “ores’.
Ans: Minerals: The natural substances in which the metals or their compounds occur in the earth is called minerais.
Ores: The minerals from which the metals can be coaveniently and economically extracted are called ores.
Note : All ores are minerals but all minerals are not ores.

6.17. Why copper matte is put in silica lined converter?
Ans: Copper matte consists of Cu₂S along with some unchanged FeS. When a blast of hot air is passed through molten matte placed in silica lined converter, FeS present in matte is oxidised to FeO which combines with silica (SiO₂) to form FeSiO₃slag.
₂S undergoes oxidation to form Cu₂0 which then reacts with more Cu₂S to form copper metal.

Thus, copper matte is heated in silica lined converter to remove FeS present in matte as FeSiO₃ slag.

6.18. What is the role of cryolite in the metallurgy of aluminium?
Ans: (a) It lowers the fusion (melting) point of the bath from 2323 K to about 1140 K.
(b) It makes alumina a good conductor of electricity.

6.19. How is leaching carried out in case of low grade copper ores?
Ans: Leaching in case of low grade copper ores is carried out with acids in presence of air. In this process, copper is oxidised to Cu²⁺ ions which pass into the solution.

6.20. Why is zinc not extracted from zinc oxide through reduction using CO?
Ans: The chemical reaction involving the reduction of ZnO by CO is :
ZnO(s) + CO(g) → Zn(s) + CO₂(g)
The process is thermodynamically not feasible because there is hardly any change in entropy as a result of the reaction. This is quite evident from the physical states of the reactants and products involved in the reaction

6.21. The value of Δ_fG° for formation of Cr₂O₃ is – 540 kJ mol^-1 and that of Al₂0₃ is – 827 kJ mol^-1 . Is the reduction of Cr₂O₃ possible with Al?
Ans: Chemical equation for the formation of Cr₂O₃ and Al₂0₃ are as follows :

6.22. Out of C and CO, which is a better reducing agent for ZnO?
Ans: The two reduction reactions are :

In the first case, there is increase in the magnitude of ΔS° while in the second case, it almost remains the same. In other words ΔG° will have more negative value in the first case when C(s) is used as the reducing agent than in the second case when CO(g) acts as the reducing agent. Therefore, C(s) is a better reducing agent.

6.23. The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Ans: We can study the choice of a reducing agent in a particular case using Ellingham diagram.
It is evident from the diagram that metals for which the standard free energy of formation oftheir oxides is more negative can reduce those metal oxides for which the standard free energy of formation of their respective oxides is less negative. It means that any metal will reduce the oxides of other metals which lie above it in the Ellingham diagram. This is because the standard free energy change (Δ_rG°) of the combined redox reaction will be negative by an amount equal to the difference in Δ_f G° of the two metal oxides. Thus both Al and Zn can reduce FeO to Fe but Fe cannot reduce Al₂0₃ to A1 and ZnO to Zn. In the same way, G can reduce ZnO to Zn but not CO.
Note : Only that reagent will be preferred as reducing agent which will lead to decrease in free energy value (ΔG°) at a certain specific temperature.

6.24. Name the processes from which chlorine is obtained as a by-product What will happen if an aqueous solution of NaCl is subjected to electrolysis?
Ans: Down process is used for the preparation of sodium metal, where chlorine is obtained as a by- product. This -process involves the electrolysis of a fused mixture of NaCl and CaCl₂ at 873 K.Sodium is discharged at the cathode while Cl₂ is obtained at the anode as a by-product.

If, an aqueous solution of NaCl is electrolysed, H₂ is evolved at the cathode while Cl₂ is obtained at the anode.

6.25. What is the role of graphite rod in the electrometallurgy of aluminium?
Ans: In the electrometallurgy of aluminium, oxygen gas is evolved at anode. O₂ reacts with graphite or carbon (graphite electrodes) to form carbon monoxide and carbon dioxide. In case if some other metal electrodes is used as anode, then oxygen will react with aluminium formed during the process to form aluminium oxide(Al₂O₃) which will pass into the reaction mixture resulting into wastage of Al. Since graphite is cheaper than aluminium, its wastage or can be tolerated.

6.26. Outline the principles of refining of metals by the following methods:
(i)Zone refining
(ii)Electrolytic refining
(iii)Vapour phase refining
Ans: (i) Zone refining: This method is used for production of semiconductors and other metals of very high purity, e.g., Ge, Si, B, Ca and In.
It is, based on the principle that the impurities are more soluble in the molten state (melt) than in the solid state of the metal.
The impure metal in the form of bar is heated at one end with a moving circular heater. As the heater is slowly moved along the length of the rod, the pure metal crystallises out of the melt whereas the impurities pass into the adjacent molten zone. This process is repeated several times till the impurities are completely driven to one end of the rod which is then cut off and discarded.
(ii)Electrolytic refining: Many metals, such as Cu, Ag, Au, Al, Pb, etc., are purified by this method. The impure metals is made the anode while a thin sheet of pure metal acts as a cathode. The electrolytic solution consists of a salt or a complex salt solution of the metal. On passing the current, the pure metal is deposited on the cathode while the impurities fall down as anode mud.
(iii)Vapour-phase refining: The crude metal is freed from impurities by first converting it into a suitable volatile compound by heating it with a specific reagent at a lower temperature and then decomposing the volatile compound at some higher temperature to give the pure metal.
(a)Mond’s process: When impure nickel is heated is a current of CO at 330-350 K, it forms volatile nickel tetracarbonyl complex leaving behind the impurities. The complex then heated to a higher temperature (450-470K) when it undergoes thermal decomposition giving pure nickel.

(b)Van Arkel method: This method is Used for preparing ultra-pure metals by removing all the oxygen and nitrogen present as impurities in metals like zirconium and titanium (which are used in space technology).Crude Zr is heated in an evacuated vessel with iodine at 870 K. Zirconium tetraiodide thus formed is separated. It is then decomposed by heating over a tungsten filament at 1800 – 2075 K to give pure Zr.

6.27. Predict conditions under which Al might be expected to reduce MgO.
Ans: The equations for the formation of the two oxides are

If we look at the plots for the formation of the two oxides of the Ellingham diagram, we find that they intersect at certain point. The corresponding value of ΔG° becomes zero for the reduction of MgO by Al metal.

This means that the reduction of MgO by Al metal can occur below this temperature. Aluminium (Al) metal can reduce MgO to Mg above this temperature because Δ°G for Al₂O₃ is less as compared to that of MgO.

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NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements

December 26, 2022

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 7 The p Block Elements:

Section Name	Topic Name	Section Name	Topic Name
7	The p-Block Elements	7.12	Simple Oxides
7.1	Group 15 Elements	7.13	Ozone
7.2	Dinitrogen	7.14	Sulphur – Allotropic Forms
7.3	Ammonia	7.15	Sulphur Dioxide
7.4	Oxides of Nitrogen	7.16	Oxoacids of Sulphur
7.5	Nitric Acid	7.17	Sulphuric Acid
7.6	Phosphorus – Allotropic Forms	7.18	Group 17 Elements
7.7	Phosphine	7.19	Chlorine
7.8	Phosphorus Halides	7.20	Hydrogen Chloride
7.9	Oxoacids of Phosphorus	7.21	Oxoacids of Halogens
7.10	Group 16 Elements	7.22	Interhalogen Compounds
7.11	Dioxygen	7.23	Group 18 Elements

NCERT Solutions CBSE Sample Papers Chemistry Class 12 Chemistry

NCERT IN TEXT QUESTIONS

7.1. Why are pentahalides more covalent than trihalidcs?
Ans: The group 15 elements have 5 e^-1 s in their valence shell. It is difficult to lose 3e^-1s to form E³⁺ and even more difficult to lose 5e^-1s to form E⁵⁺. Thus, they have very little tendency to form ionic compounds. Further, since the elements in +5 state have less tendency to lose e^-1s than in the +3 state, elements in +5 state have more tendency to share e^-1s and hence pentahalides are more covalent than trihalides.

7.2. Why is BiH₃ the strongest reducing agent amongst all the hydrides of group 15 elements? (C.B.S.E. 2013)
Ans: Down the group, the atomic size of the element (E) increases and the bond length of the corresponding E—H bond also increases. This adversely affects the bond dissociation enthalpy. This means that amongst the trihydrides of the members of nitrogen family, the bond dissociation enthalpy of Bi—H bond is the least. Therefore, BiH₃ is the strongest reducing agent among the hydrides of group 15 elements.

7.3. Why is N₂ less reactive at room temperature?
Ans: Due to presence of triple bond between two N-atoms (N = N), the bond dissociation energy of N₂ is very high. As a result, N₂ becomes less reactive at room temperature.

7.4. Mention the conditions required to maximise the yield of ammonia.
Ans: Ammonia is prepared by Haber’s process as given below:

7.5. How does ammonia react with a solution of Cu²⁺?
Ans:

7.6. What is the covalence of nitrogen in N₂O₅ ?
Ans: In N₂O₅ , each N-atom has four shared pairs of e-1 s as shown:

7.7. Why is bond angle in \({ PH }_{ 4 }^{ + }\) ion higher than in PH₃ ? (Pb. Board 2009)
Ans: In both PH₃ and \({ PH }_{ 4 }^{ + }\) ion, the phosphorus atom is sp³ hybridised. However, in PH₃ the central atom has apyramidal structure due to the presence of lone electron pair on the phosphorus atom.

Because of lone pair : shared pair repulsion which is more than that of shared pair : shared pair repulsion, the bond angle in PH₃ is nearly 93-6°. In \({ PH }_{ 4 }^{ + }\) ion, there is no lone electron pair on the phosphorus atom. It has a tetrahedral structure with bond angle of 109°-28′. Thus, the bond angle in \({ PH }_{ 4 }^{ + }\) ion is higher than in PH₃.

7.8. What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂?
Ans:

7.9. What happens when PCl₅ is heated?
Ans:

7.10. Write a balanced equation for the hydrolytic reaction of PC is in heavy water.
Ans:

7.11. What is the basicity of H₃PO₄?
Ans:

7.12. What happens when H₃PO₄ is heated?
Ans: On heating, H₃PO₃disproportionates to form PH₃ and H₃PO₄ with O.S. of-3and + 5.

7.13. List the important sources of sulphur.
Ans: Sulphur mainly occurs in the combined states in earth’s crust in the form of sulphates and sulphides.
Sulphates : gypsum (CaSO₄.2H₂O); epsom (MgSO₄.7H₂O); baryte (BaSO₄), etc.
Sulphides : Galena (PbS); zinc blende (ZnS); copper pyrites (CuFeS2); iron pyrites (FeS₂), etc. Traces of sulphur occur’as H₂S and in organic materials such as eggs, proteins, garlic, onion, mustard, hair and wool.

7.14. Write the order of thermal stability of the – hydrides of Group 16 elements.
Ans: The thermal stability of hydrides of group 16 elements decreases down the group. This is because down the group, size of the element (M) increases, M-H bond length increases and thus, stability of M-H bond decreases so that it can be broken down easily. Hence, we have order of thermal stability as H₂O > H₂S > H₂Se > H₂Te > H₂PQ

7.15. Why is H₂O a liquid and H₂S a gas?
Ans: Due to high electronegativity of O than S, H₂O undergoes extensive intermolecular H-bonding. As a result, H₂O exists as an associated molecule in which each O is tetrahedrally surrounded by four H₂O molecules. Therefore, H₂O is a liquid at room temperature.
On the other hand,H₂S does not undergo H- bonding. It exists as discrete molecules which are held together by weak van der waals forces of attraction. A small amount of energy is required to break these forces of attraction. Therefore, H₂S is a gas at room temperature.

7.16. Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe
Ans: Platinum (Pt) is a noble metal and does not react with oxygen directly.

7.17. Complete the following reactions:
(i)C₂H₂ + O₂ -> (ii) 4Al + 3 O₂ ->
Ans:

7.18. Why does O₃ act as a powerful oxidising agent?
Ans: On heating, O₃ readily decomposes to give O₂ and nascent oxygen.

Since nascent oxygen is very reactive, therefore, O₃ acts as a powerful oxidising agent.

7.19. How is O₃ estimated quantitatively?
Ans: When O₃ is treated with excess of KI solution buffered with borate buffer (pH = 9.2), I2 is liberated quantitatively.

The I₂ thus liberated is titrated against a standard solution of sodium thiosulphate using starch as an indicator.

7.20. What happens when sulp’hur dioxide is passed through an aqueous solution of Fe(III) salt?
Ans: SO₂ acts as a reducing agent and reduces aqueous solution of Fe (III)salt to Fe (II) salt.

7.21. Comment on the nature of two S-O bonds formed in S02 molecule. Are the two S-O bonds in this molecule equal ?
Ans: SO₂ exists as an angular molecule with OSO bond angle of 119.5°. It a resonance hybrid of two canonical-forms:

7.22. How is the presence of SO₂ detected?
Ans: SO₂is a pungent smelling gas. It can be detected by two test:

7.23. Mention three areas in which H₂SO₄ plays an important role.
Ans: (i) Sulphuric acid is used for the manufacture of a number of chemicals like hydrochloric acid, phosphoric acid, nitric acid along with a large number of organic compounds.
(ii) A mixture of concentrated nitric acid and concentrated sulphuric acid is used in the manufacture of explosives like picric acid, T.N.T, dynamite etc.
(iii) Dilute solution of acid is employed in petroleum refining in order to remove the unwanted impurities of sulphur.

Question 24.
Write the conditions to maximise the yield of H₂SO₄ by Contact process.
Solution:
The key step in the manufacture of sulphuric acid is oxidation of SO₂ to SO₃ in presence of V₂O₅ catalyst.

The reaction is exothermic and reversible. Hence, low temperature and high pressure are the favourable conditions for maximum yield of SO₃. In practice a pressure of 2 bar and temperature of 720 K is maintained.

Question 25.
Why is K_a2 « K_a1 for H₂SO₄ in water?
Solution:
H₂SO₄ is a very strong acid in water largely because of its first ionisation to H₃O⁺ and HSO₄^–The ionisation of HSO₄^– to H₃O⁺ and SO₄^2-is very very small. That is why, K_a2« K_a1.

Question 26.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising powers of F₂ and Cl₂.
Solution:
The oxidising powers of both the members of halogen family are expressed in terms of their electron accepting tendency and can be compared as their standard reduction potential values.
F₂ + 2e^– → 2F^–; E° = 2-87 V, Cl₂ + 2e^– → 2Cl^– ; E° = 1-36 V
Since the E° of fluorine is more than that of chlorine, it is a stronger oxidising agent.
Explanation : Three factors contribute towards the oxidation potentials of both the halogens. These are :
(i) Bond dissociation enthalpy: Bond dissociation enthalpy of F2 (158 kJ mol^-1) is less compared to that of Cl₂ (242·6 kJ mol^-1).
(ii) Electron gain enthalpy: The negative electron gain enthalpy of F (- 332·6 kJ mol^-1) is slightly less than of Cl (-348·5 kJ mol^-1).
(iii) Hydration enthalpy: The hydration enthalpy of F- ion (515 kJ mol^-1) is much higher than that of Cl- ion (381 kJ mol^-1) due to its smaller size.
From the available data, we may conclude that lesser bond dissociation enthalpy and higher hydration enthalpy compensate lower negative electron gain enthalpy of fluorine as compared to chlorine. Consequently, F₂ is a more powerful oxidising agent than Cl₂.

Question 27.
Give two examples to show the anomalous behaviour of fluorine.
Solution:

Ionisation enthalpy, electro-negativity and electrode potential are higher for fluorine than the expected trends of other halogen.
Fluorine does not show any positive oxidation state except in HOF.

Question 28.
Sea is the greatest source of some halogens. Comment.
Solution:
Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium but sodium chloride being the maximum makes sea water saline. Various sea weeds contain upto 0.5% iodine.

Question 29.
Give the reason for bleaching action of Cl₂.
Solution:
Chlorine bleaches by oxidation Cl₂ + H₂O → HCl + HOCl → HCl + [O]
The nascent oxygen reacts with dye to make it colourless.

Question 30.
Name two poisonous gases which can be prepared from chlorine gas.
Solution:
COCl₂ (phosgene), CCl₃NO₂ (tear gas)

Question 31.
Why is ICI more reactive than l2?
Solution:
In general, interhalogen compounds are more reactive than halogens due to weaker X-X’ bonding than X-X bond. Thus, ICI is more reactive than I₂.

Question 32.
Why is helium used in diving apparatus?
Answer:
Helium along with oxygen is used in the diving apparatus by the sea divers. Since it is very little soluble in blood, it reduces decompression and causes less discomfort to the diver in breathing. A mixture of helium and oxygen does not cause pain due to very low solubility of helium in blood as compared to nitrogen.

Question 33.
Balance the following equation :
XeF₆ + H₂O → XeO₂F₂ + 4HF
Solution:

Question 34.
Why has it been difficult to study the chemistry of radon?
Solution:
Radon is radioactive with very short half-life which makes the study of chemistry of radon difficult.

NCERT EXERCISES

7.1. Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
Sol: In group 15 of the Periodic Table, the elements, nitrogen (₇N), phosphorus (₁₅P), arsenic (₃₃As), antimony (₅₁Sb) and bismuth (₈₃Bi) are present. The elements of this group can exhibit various oxidation states ranging between -3 to + 5. Negative oxidation state will be exhibited when they combine with less electronegative element andpositive oxidation state will be exhibited with more electronegative element. Positive oxidation state becomes more favourable as we more down the group due to increasing metallic character & electropositivity. Although due to inert pair effect the stability of +5 state will also decrease. The only stable compound of Bi (V) is BiF₅.
The atomic (covalent) and ionic radii (in a particular oxidation state) of the elements of nitrogen family (group 15) are smaller than the corresponding elements of carbon family (group 14). On moving down the group, the covalent and ionic radii (in a particular oxidation state) increase with increase in atomic number. There is a considerable increase in covalent radius from N to P. However, from As to Bi, only a small increase is observed.
As the size increases on moving down the group, the ionisation enthalpy increases. The ionisation enthalpy of nitrogen group elements is more than the corresponding elements of oxygen group. This is because of more stable half-filled outermost p- subshell of nitrogen group elements. Electronegativity decreases down the group with increase in atomic size.

7.2. Why is the reactivity of nitrogen different from that of phosphorus?
Sol: Molecular nitrogen exists as a diatomic molecule (N₂) in which the two nitrogen atoms are linked to each other by triple bond (N≡N). It is a gas at room temperature. Multiple bonding is not possible in case of phosphorus due to its large size. It exists as P₄ molecule (solid) in which P atoms are linked to one another by single covalent bonds. Because of greater bond dissociation enthalpy (946 kJ mol^-1) of N≡N bond, molecular nitrogen is very less reactive as compared to molecular phosphorus.

7.3. Discuss the trends in chemical reactivity of group 15 elements.
Sol: Hydrides: All elements of group 15 form gaseous hydrides of the type MH₃.
In all the hydrides the central atom is sp³ hybridized and their shape is pyramidal due to presence of lone pair of electrons.
(a)The basic strength of the hydrides decreases as we move down the group.
Thus, NH₃ is the strongest base.
NH₃ > PH₃ > AsH₃ > SbH₃
(b)The thermal stability of the hydrides decreases as the atomic size increases, i.e., the M – H bond strength decreases which means reducing character increases.
(c)In the liquid state, the molecules of NH₃are associated due to hydrogen bonding. The molecules of other hydrides are not associated.
(d)NH₃ is soluble in water whereas other hydrides are insoluble.
(e)All the hydrides, except NH₃, are strong reducing agents and react with metal ions (Ag⁺, Cu²⁺, etc.) to form phosphides, arsenides or antimonides.
Halides: The elements of group 15 form two series of halides MX₃ and MX₅.
(a)All the elements of the group form trihalides. The ionic character of trihalides increases as we move down the group. Except NCl₃ all the trihalides are hydrolysed by water. This is due to the absence of d-orbitals in nitrogen.
(b)PF₃ is not hydrolysed because fluorine being more electronegative than oxygen forms more stable bonds with phosphorus than P – O bonds.
(c)N cannot form NX₅ because of non-availability of rforbitals. Bi cannot form BiX₃ because of reluctance of 6s electrons of Bi to participate in bond formation.
(d)The hybridisation of M in MX₃ is sp³ and shape is pyramidal. M in MX₅ is sp³ as hybridised and shape is trigonal pyramidal. The axial bonds in MX₅ are weaker and longer, So MX₅ are less stable and decompose on heating eg:

Oxides:
(a)Nitrogen forms a number of oxides. The rest of the members (P, As, Sb and Bi) of the group form two types of oxides : E₂0₃ and E₂O₅.
(b)The reluctance of P, As, Sb and Bi to enter into pπ -pπ multiple bonding leads to cage structures of their oxides and they exist as dimers, E₄O₆ and E₅O₁₀.
(c)The basic nature of die oxides increases with increase in atomic number of the element. Thus, the oxides of nitrogen (except N₂0 and NO), P (III) and As (III) are acidic, Sb (III) oxide is amphoteric and Bi (III) oxide is basic.

7.4. Why does NH₃ form hydrogen bond but PH₃ does not?
Sol: Nitrogen has an electronegativity value 3.0, which is much higher than that of H (2.1). As a result, N – H bond is quite polar and hence NH₃ undergoes intermolecular H – bonding.

Phosphorus have an electronegativity value 2-1. Thus, P – H bond is not polar and hence PH₃ does not undergo H – bonding.

7.5. How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions . involved.
Sol: In laboratory, nitrogen is prepared by heating an equimolar aqueous solution of ammonium chloride and sodium nitrite. As a result of double decomposition reaction, ammonium nitrite is formed. Ammonium nitrite is unstable and decompose to form nitrogen gas.

7.6. How is ammonia manufactured industrially?
Sol: Commercially, by Haber’s process.

iron oxide, K₂O, Al₂0₃The optimum conditions for the production of NH₃ are pressure of 200 atm and temperature of 100K.

7.7. Illustrate how copper metal can give different products on reaction with HN0₃.
Sol: On heating with dil HN0₃, copper gives copper nitrate and nitric oxide.

7.8. Give the resonating structures of N0₂ and N₂O₅.
Sol: Resonating structures of N0₂ are:

7.9. The HNH angle value is higher than HPH, H AsH and HSbH angles. Why?
(Hint: Can be explained on the basis of sp³ hybridisation in NH₃ and only s-p bonding , between hydrogen and other elements of the group).
Sol: In all these cases, the central atom is sp³ hybridized. Three of the four sp³ orbitals form three σ-bonds, while the fourth contains the lone pair of electrons. On moving down from N to Sb, the electronegativity of the central atom goes on decreasing. As a result of this, bond pairs of electrons lie away and away from the central atom. This is because of the force of repulsion between the adjacent bond pairs goes on decreasing and the bond angles keep on decreasing from NH₃ to SbH₃. Thus, bond angles are in the order:

7.10. Why does R₃P=0 exist but R₃N=0 does not (R is an alkyl group) ?
Sol: Nitrogen does not have vacant d-orbitals on its valence shell. Therefore, it cannot extend its dπ-pπ bonding is not possible. As a result, the molecules of R₃N = 0 does not exist. However, phosphorus and rest of the members of the group 15 have vacant d-orbitals in the valence shell which can be involved in dπ-pπ bonding. Under the circumstances, R₃P=0 molecule can exist.

7.11. Explain why NH₃ is basic while BiH₃ is only feebly basic.
Sol: In both NH₃ and BiH₃, N and Bi have a lone paif of electrons on the central atom and hence should behave as Lewis bases. But NH₃ is much more basic than BiH₃. Since the atomic size of N is much smaller than that of Bi, therefore, electron density on N-atom is much higher than that on Bi-atom. Thus, the tendency of N in NH₃ to donate its lone pair of electrons is much more in comparison to tendency of Bi in BiH₃. Hence, NH₃ is more basic than BiH₃.

7.12. Nitrogen exists as diatomic molecule and phosphorus as P₄. Why?
Sol: Nitrogen exists as a diatomic molecule having a triple bond between the two N-atoms, This is due its small size that it forms pπ-pπ multiple bonds with itself and with carbon /oxygen as well. On the other hand, phosphorus due to its larger size does not form multiple pπ-pπ bonds with itself. It prefers to form P – P single bonds and hence it exists as tetrahedral P₄ molecule.

7.13. Write main differences between the properties of white phosphorus and red phosphorus.
Sol:

Structure of white and red phosphorus are given below:

7.14. Why does nitrogen show catenation properties less than phosphorus ? (C.B.S.E. Foreign 2009)
Sol: The valence shell electronic configuration of N is 2s²2p³. In order to complete the octet, the two nitrogen atoms share three electron pairs in the valence p-sub-shell and get linked by triple bond (N=N). Thus molecular nitrogen exists as discrete diatomic species and there is no scope of any self linking or catenation involving a number of nitrogen atoms. However, in case of phosphorus, multiple bonding is not feasible due to comparatively large atomic size of the element. Molecular phosphorus exists as tetra-atomic molecule (P4) in white phosphorus. These tetrahedrons are further linked by covalent bonds to form red variety which is in polymeric form. Thus, catenation in nitrogen is less than in phosphorus.

7.15. Give the disproportionation reaction of H₃ P0₃.
Sol: On heating, H₃ P0₄ undergoes self – oxidation-reduction, i.e: disproportionation to form PH₃.

7.16. Can PCl₅ act as an oxidising as well as a reducing agent Justify.
Sol: The oxidation state of P in PCl₅ is+5. Since P has five electrons in its valence shell, therefore, it cannot donate electron and cannot increase its oxidation state beyond + 5, Thus, PCl₅ cannot act as a reducing agent. It can act as oxidizing agent by itself undergoing reduction.

7.17. Justify the placement of O, S, Se, Te and Po in the same group’of the periodic table in terms of electronic configuration, oxidation state and hydride formation.
Sol: (1)Electronic configuration:
O (At. no. = 8) = [He] 2s² 2p⁴
S (At. no. = 16) = [Ne] 3s² 3p⁴
Se (At. no. = 34) = [Ar] 3d¹⁰ 4s² 4p⁴
Te (At. no. = 52) = [Kr] 4d¹⁰ 5s² 5p⁴ ,
Po (At. no. = 84) = [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁴ ,
Thus, all these elements have the same ns² np⁴ (n = 2 to 6) valence shell electronic configuration, hence are justified to be placed in group 16 of the Periodic Table.
(2)Oxidation state : Two more electrons are needed to acquire the nearest noble gas configuration. Thus, the minimum oxidation state of these elements should be – 2. O and to some extent S show – 2 oxidation state. Other element being more electropositive than O and S, do not show negative oxidation state. As these contain six electrons, thus, maximum oxidation state shown by them is+ 6. Other oxidation state shown by them are + 2 and + 4. O do not show+4 and + 6 oxidation state, due to the absence of d-orbitals. Thus, on the basis of maximum and minimum oxidation states, these elements are justified to be placed in the same group 16 of the periodic table.
(3)Hydride formation: All these elements share two of their valence electrons with 1 s- orbital of hydrogen to form hydrides of the general formula EH₂, i.e., H₂0, H₂S, H₂Se, H₂Te and H₂Po. Thus, on the basis of hydride formation, these elements are justified to be placed in the same group 16 of the Periodic Table.

7.18. Why is dioxygen a gas but sulphur a solid?
Sol: Due to the small size and high electronegativity, oxygen forms pπ- pπ multiple bonds. As a result, oxygen exists as diatomic (O₂) molecules. These molecules are held together by weak van der Waal’s forces of attraction which can be overcome by collisions of the molecules at room temperature. Therefore, O₂ is a gas at room temperature. Due to its bigger size and lower electronegativity, sulphur does not form pn-pn multiple bonds. It prefers to form S – S single bonds. S – S single bond is stronger then O-O single bond. Thus, sulphur has higher tendency for catenation than oxygen. Due to higher tendency for catenation and lower tendency for pπ – pπ multiple bonds sulphur exits as octa-atomic (S_g) molecule. Due to bigger size, the force of attraction holding the S_g molecules together are much stronger which cannot be overcome by collisions of molecules at room temperature. Therefore, sulphur is solid at room temperature.

7.19. Knowing the electron gain enthalpy values of O—>O^– and O—>O^2- as -141 and 702 kJ mol^-1 respectively, how can you account for the formation of a large number of oxides having O^2- species and not O^–?
Sol: Let us consider the reaction of oxygen with monopositve metal, we can have two compounds. MO(O in -1 state) and M₂O (O in -2 state). The energy required for formation of O^-2 is compensated by increased coulombic attraction between M⁺ and O^-2. Coulombic force of attraction, F_A is proportional to product of charges on ions i.e.

where q₁ and q₂ are charges on ions and r is distance between ions. Same logic can be applied if metal is dispositive.

7.20. Which aerosols deplete ozone?
Sol: Aerosols like chlorofluorocarbons (CFC’s), i.e., freon (CCl₂F₂), depletes the ozone layer by supplying Cl* free radicals which convert O₃ to O₂

7.21. Describe the manufacture of H₂SO₄ by contact process?
Sol: Preparation of sulphuric acid:By Contact Process: Burning of sulphur or sulphide ores in presence of oxygen to produce SO₂. Catalytic oxidation of SO₂ with O₂to give SO₃ in the presence of V₂O₅.

Then SO₃ made to react with sulphuric acid of suitable normality to obtain a thick oily liquid called oleum.

Then oleum is diluted to obtain sulphuric acid of desired concentration.

The sulphuric acid obtained by contact process is 96-98% pure.

7.22. How is SO₂ an air pollutant?
Sol: (1) SO₂dissolves in moisture present in air to form H₂SO₄ which damages building materials especially marble (acid – rain).- CaCO₃ + H₂SO₃ ——->CaSO₃ + H₂0 + CO₂
(2)It corrodes metals like Fe and steel. It also brings about fading and deterioration of fabrics, leather, paper, etc., and affecting the colour of paints.
(3)Even in low concentration (= 0.03 ppm), it has damaging effect on the plants. If exposed for a long time, i.e., a few days or weeks, it slows down the formation of chlorophyll i. e., loss of green colour. This is called chlorosis.
(4)It is strongly irritating to the respiratory track. It cause throat and eye irritation, resulting into cough, tears and redness in eyes. It also cause breathlessness and effects larynx i. e. „ voice box.

7.23. Why are halogens strong oxidising agents?
Sol: Members of the halogen family act as strong oxidising agents on account of their electron accepting tendency both in the molecular as well as atomic form.

This is attributed to their high electronegativity, negative electron gain enthalpy values and alsp low bond dissociation enthalpies sinve they contain single covalent bonds(X — X) in their molecules. Fluorine is most reactive among the halogens and the reactivity down the group.

7.24. Explain why fluorine forms only one oxoacid, HOF.
Sol: Cl, Br and I form four series of oxo acids of general formula HOX, HOXO, HOXO₂, and H0XO₃. In these oxo-adds, the oxidation states of halogens are + 1, + 3, + 5, and + 7 respectively. However, due to high electronegativity, small size and absence of d-orbitals, F does not form oxo-acids with + 3, + 5 and + 7, oxidation states. It just forms one oxo-acid (HOF).

7.25. Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.
Sol: Both .nitrogen (N) and chlorine (Cl) have electronegativity of 3.0. However, only nitrogen is involved in the hydrogen bonds (e.g., NH3) and not chlorine. This is due to smaller atomic size of nitrogen (atomic radius =70 pm) as compared to chlorine (atomic radius = 99) pm), therefore, N can cause greater polarisation of N-H bond than Cl in case of Cl—H bond.Consequently, N atom is involved in hydrogen bonding and not chlorine.

7.26. Write two uses of ClO₂
Sol: (1) ClO₂ is an excellent bleaching agent. It is 30 times stronger bleaching agent then the Cl₂. It is used as a bleaching agerit for paper pulp in paper industry and in textile industry. (2) ClO₂ is also a powerful oxidising agent and chlorinating agent. It acts as a germicide for disinfecting water. It is used for purifying drinking water.

7.27. Why are halogens coloured?
Sol: The halogens are coloured because their molecules absorb light in the visible region. As a result of which their electrons get excited to higher energy levels while the remaining light is transmitted. The color of halogens is the color of this transmitted light.

7.28. Write the reactions of F₂ and Cl₂ with water.
Sol:

7.29. How can-you prepare Cl₂ from HCl and HCl from CI₂? Write reactions only.
Sol:

7.30. What inspired N. Bartlett for carrying out reaction between Xe and PtF₆?
Sol: N. Bartlett observed that PtF₆ reacts with O₂to give an compound O₂+ [PtF₆]^–.
PtF₆ (g) + O₂ (g) ——–>O₂+[PtF₆]^–
Since the first ionization enthalpy of Xe (1170 kJ mol^-1 )is fairly close to that of 02 molecule (1175 kJ mol^-1 ), he thought that PtF₆should also oxidise Xe to Xe⁺. This inspired Bartlett to carryout the reaction between Xe and PtF₆. When PtF₆ and Xe were made to react, a rapid reaction took place and a red solid, Xe+[PtF₆]^– was obtained.

7.31. What are the oxidation states of phosphorus in the following: –
(i) H₃PO₃ (ii)PCl₃
(iii) Ca₃P₂(iv)Na₃PO₄
(v) POF₃
Sol:

7.32. Write balanced equations for the following:
(i) NaCl is heated witlrsulphuric acid in the presence of MnO₂
(ii) Chlorine gas is passed into a solution of Nal in water.
Sol:

7.33. How are xenon fluorides XeF₂, XeF₄ and XeF₆obtained?
Sol: XeF₂, XeF₄ and XeF₆ are obtained by direct reaction between Xe and F₂ as follows:

7.34. With which neutral molecule is ClO^– isoelectronic? Is this molecule Lewis acid or base ? (Pb. Board 2009)
Sol: ClO^– has (17 + 8 + 1) = 26 electrons. It is iso-electronic with two neutral molecules.
Oxygen difluoride (OF₂) : 8 + 18 = 26 electrons
Chlorine fluoride (ClF) : 17 + 9 = 26 electrons
Out of these, ClF can act as Lewis base. The atom chlorine has three lone electron pairs which it donates to form compounds like ClF₃, ClF₅ and ClF₇.

7.35. How are XeO₃and XeOF₄prepared?
Sol:

7.36. Arrange the following in the order of property indicated for each set: –
(i) F₂ , Cl₂ , Br₂ , I₂ – increasing bond dissociation enthalpy.
(ii) HF, HCI, HBr, HI – increasing acid . strength.
(iii) NH₃, PH₃, AsH₃, SbH₃, BiH₃ – increasing Sol. base strength.
Sol: (i) Bond dissociation enthalpy decreases as the bond distance increases from F₂ to I₂ due to increase in the size of the atom, on moving from F to I.
F – F bond dissociation enthalpy is smaller then the Cl – Cl and even smaller than Br – Br. This is because F atom is very small and have large electron-electron repulsion among the lone pairs of electrons in F₂ molecule where they are much closer to each other than in case of Cl₂. The increasing order of bond dissociation enthalphy is I, < F₂ < Br₂ < Cl₂
(ii) Acid strength of HF, HCI, HBr and HI depends upon their bond dissociation enthalpies. Since the bond dissociation enthalpy of H – X bond decreases from H – F to H-l as the size of atom increases from F to I.
Thus, the acid strength order is HF < HCI < HBr < HI
The weak acidic strength of HF is also due to H-bonding due to which release of H⁺ becomes difficult.
(iii) NH₃, PH₃, ASH₃, SbH₃ and BiH₃ behaves as Lewis bases due to the presence of lone pair of electrons on the central atom. As we move from N to Bi, size of atom increases. Electron density on central atom decreases and hence the basic strength decreases from NH₃ to BiH₃. Thus basic strength order is BiH₃<SbH₃<AsH₃<PH₃<NH₃

7.37. Which one of the following does not exist ?
(i)XeOF₄ (ii)NeF₂
(iii)XeF₄ (iv)XeF₆
Sol: NeF₂ does not exist. This is because the sum of first and second ionization enthalpies of Ne are much higher than those of Xe. Consequently, F₂ can oxidise Xe to Xe²⁺ but cannot oxidise Ne to Ne²⁺.

7.38. Give the formula and describe the structure of a noble gas species which is isostructural with: (i) ICI₄^– (ii) IBr₂^– (iii) Br0₃^–
Ans: (i) ICI₄^–: In ICI₄^–, central atom I has seven valence electrons and one due to negative charge. Four out of these 8 electrons are utilized in forming four single bonds with four Cl atoms. Four remaining electrons constitutes the two lone pairs. It is arranged in square planar structure. ICI₄^– has 36 valence electrons. A noble gas species having 36 valence electrons is XeF₄ (8 + 4 x 7 = 36). XeF₄ is also square planar.

(ii) IBr₂^–: In IBr₂^–, central atom I has eight electrons. Two of these are utilized in forming two single bonds with two Br atom. Six remaining electrons constitutes three lone pairs. It is arranged in linear structure.

IBr₂^– has 22 valence electrons. A noble gas species having 22 valence electrons is XeF₂ (8+2 x 7=22).
XeF₂ is also linear.
(iii) In Br0₃^– ion the central Br atom has 8 valence electrons (7 +1). Out of these, it shares 4 with two atoms of O forming Br = O bonds. Out of the remaining four .electrons, 2 are donated to the third O atom which accounts for its negative charge. The remaining 2 electrons constitute one lone pair. In order to minimise the force of repulsion, the structure of Br0₃^– ion must be pyramidal. Br0₃^– ion has (7 + 3 x 6 + 1) = 26 valence electrons and is isoelectronic as well as iso-structural with noble gas species Xe0₃ which has also 26(8 + 3 x 6) electrons.

7.39. Why do noble gases have comparatively large atomic size?
Sol: The members of the noble gas family have comparatively large atomic size as compared to rest of the members present in the same period. Actually, for these elements, van der Waals’ radii are considered while for rest of the elements either covalent radii or metallic radii are taken into account. Since van der Waals’ radii arise simply due to van der Waals’ forces of attraction, these are expected to have comparatively large magnitude.

7.40. List the uses of neoirand argon gases.
Sol: Uses of Neon
Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes. Glow’of different colours ‘neon signs’ can be produced by mixing neon with other gases. Neon bulbs and used in botanical gardens and in green’ houses.
Uses of Argon
Argon is used mainly to provide an inert atmosphere in high temperature metallurgical processes such as arc welding of metals and alloys. In the laboratory, it is used for handling substance which are air sensitive.
It is used in filling incandescent and fluorescent lamps where its presence retards the sublimation of the filament and thus increases the life of the lamp.It is also used in “neon signs” for obtaining lights of different colours.

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