Amines Class 12 Important Extra Questions Chemistry Chapter 13

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 13 Amines. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 13 Important Extra Questions Amines

Amines Important Extra Questions Very Short Answer Type

Question 1.
Arrange the following in decreasing order of solubility in water:
(CH3)3N, (CH3)2H, CH3NH2 (CBSE Delhi 2019)
Answer:
CH3NH2 > (CH3)2NH > (CH3)3N

Question 2.
Arrange the following compounds in increasing order of their solubility in water:
C6H5NH2, (C2H5)2NH, C2H5NH2 (CBSE Delhi 2011, CBSE 2013)
Answer:
C6H5NH2 < (C2H5)2NH < C2H5NH2

Question 3.
Write the structure of n-methylhexanamine. (CBSE 2013)
Answer:
CH3 – NH – CH2CH3

Question 4.
Write the structure of 2-amino toluene. (CBSE 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 1
Question 5.
Write the IUPAC name of the given compound: (CBSE Delhi 2016)
Class 12 Chemistry Important Questions Chapter 13 Amines 2
Answer:
2, 4, 6-Tribromoaniline

Question 6.
Arrange the following in increasing order of their basic strength in an aqueous solution:
CH3NH2, (CH3)3N, (CH3)2NH (CBSE Delhi 2013)
Answer:
(CH3)3N < CH3NH2 < (CH3)2NH

Question 7.
Write the IUPAC name of the compound (CBSE Delhi 2014)
Class 12 Chemistry Important Questions Chapter 13 Amines 3
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 4

Question 8.
Arrange the following in increasing order of base strength in the gas phase:
(C2H5)3N, C2H5NH2, (C2H5)2NH (CBSE Delhi 2019)
Answer:
(C2H5)3N > (C2H5)2NH > C2H5NH2

Question 9.
Arrange the following in increasing order of boiling points:
(CH3)3N, C2H5OH, C2H5NH2 (CBSE Delhi 2019)
Answer:
(CH3)3N < C2H5NH2 < C2H5OH

Question 10.
Arrange the following in the increasing order of their pKb values.
C6H5NH2, C2H5NH2, C6H5NHCH3 (CBSE AI 2018)
Answer:
C2H5NH2 < C6H5NHCH3 < C6H5NH2

Question 11.
Write IUPAC name of the following compound:
CH3NHCH(CH3)2 (CBSE Delhi 2017)
Answer:
N-Methylpropan-2-amine

Question 12.
Write IUPAC name of the following compound: (CH3CH2)2NCH3 (CBSE Delhi 2017)
Answer:
N-Ethyl-N-methylhexanamine

Question 13.
Give a chemical test to distinguish between ethylamine and aniline. (CBSE AI 2011)
Answer:
Aniline gives azo dye test while ethylamine does not give azo dye test.

Question 14.
Arrange the following in increasing order of their basic strength:
C6H5NH2, C6H5N(CH3)2, (C6H5)2NH, and CH3NH2 (CBSE AI 2011)
Answer:
(C6H5)2 NH < C6H5NH2 < C6H5N (CH3)2 < CH3NH2

Question 15.
Write the IUPAC name of the compound: (CBSE AI 2016)
Class 12 Chemistry Important Questions Chapter 13 Amines 5
Answer:
N-Methyl-2-methyl propanamine

Amines Important Extra Questions Short Answer Type

Question 1.
Give the structures of A, B, and C in the following reactions: (CBSE Delhi 2013)
Class 12 Chemistry Important Questions Chapter 13 Amines 6
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 7

Class 12 Chemistry Important Questions Chapter 13 Amines 8
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 9
Question 2.
Complete the following reaction equations:
(i) C6H5N2CI + H3PO2 + H20 →
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 10
(ii) C6H5NH2 + Br2(oq) → (CBSE 2012)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 11
Question 3.
Give the structures of products A, B, and C in the following reactions: (CBSE Delhi 2013)
Class 12 Chemistry Important Questions Chapter 13 Amines 12
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 13Class 12 Chemistry Important Questions Chapter 13 Amines 14
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 15
Question 4.
Write the structures of A, B, and C in the following reactions: (CBSE 2016)
Class 12 Chemistry Important Questions Chapter 13 Amines 16
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 17Class 12 Chemistry Important Questions Chapter 13 Amines 18
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 19
Question 5.
Write the structures of A, B, and C In the following: (CBSE Delhi 2016)
Class 12 Chemistry Important Questions Chapter 13 Amines 20
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 21Class 12 Chemistry Important Questions Chapter 13 Amines 22
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 23
Question 6.
Give a chemical test to distinguish between ethylamine and aniline. (CBSE AI 2011)
Answer:
These can be distinguished by the azo dye test. Dissolve the compound in a cone. HCl and add an ice-cold solution of HNO2 (NaNO2 + dil. HCl) and then treat it with an alkaline solution of 2-naphthol. The appearance of brilliant orange or red dye indicates aniline.
Class 12 Chemistry Important Questions Chapter 13 Amines 24
Class 12 Chemistry Important Questions Chapter 13 Amines 25
Ethylamine does not form a dye. It will give brisk effervescence due to the evolution of N2 but the solution remains clear.

Question 7.
How will you distinguish between the following? Give one chemical test:
(i) Aniline and benzylamine
Answer:
These can be distinguished by the azo dye test. Aniline reacts with HNO2 (NaNO2 + dil. HCl) at 273-278K to form stable benzene diazonium chloride which on treatment with an alkaline solution of 2-naphthol gives an orange dye (as given above).
Benzylamine does not give azo dye tests.

(ii) Aniline and N-methylaniline. (CBSE AI 2010)
Answer:
These can be distinguished by carbylamine test. Aniline being primary amines gives carbylamine test, i.e. when heated with an alcoholic solution of KOH and CHCl3, it gives the foul smell of phenyl isocyanide.
Class 12 Chemistry Important Questions Chapter 13 Amines 26
Question 8.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B, and C. (CBSE Sample Paper 2011)
Answer:
(i) Since the compound C of molecular formula C6H7N is formed from B on treatment with Br2 and KOH (Hoffmann bromamide reaction), therefore, the compound ‘B’ must be an amide and ‘C’ must be an amine. The only aromatic amine having molecular formula C6H7N is C6H5NH2 (aniline).

(ii) Since ‘C’ is aniline, the amide from which it is formed must be benzamide (C6H5CONH2).
Class 12 Chemistry Important Questions Chapter 13 Amines 27
Thus, B is benzamide.

(iii) Since B is formed from A with aqueous ammonia and heating, therefore, compound ‘A’ must be benzoic acid.
Class 12 Chemistry Important Questions Chapter 13 Amines 28
Thus, A = C6H5COOH, B = C6H5CONH2, C = C6H5NH2.

Question 9.
Account for the following:
(a) Gabriel phthalimide synthesis is not preferred for preparing aromatic primary amines.
Answer:
Gabriel phthalimide synthesis is not preferred for preparing aryl amines because aryl halides do not undergo nucleophilic substitution reaction with phthalimide.

(b) On reaction with benzene sulphonyl chloride, primary amine yields product soluble in alkali whereas secondary amine yields product insoluble in alkali. (CBSE Al 2019)
Answer:
Answer:
The sulphonamide formed by the reaction of secondary amine and benzene sulphonyl chloride does not contain any hydrogen atom attached to the N atom, so it is not acidic.
Therefore, it is insoluble in alkali.
Class 12 Chemistry Important Questions Chapter 13 Amines 29
Due to the presence of ‘H’ on nitrogen, it is soluble in alkali.
Class 12 Chemistry Important Questions Chapter 13 Amines 30
Since there is no ‘H’ on nitrogen, it is insoluble in alkali.

Amines Important Extra Questions Long Answer Type

Question 1.
Write equations of the following reactions:
(i) Acetylation of Aniline
Answer:
Acetylation of Aniline:
Class 12 Chemistry Important Questions Chapter 13 Amines 31
(ii) Coupling reaction
Answer:
Coupling reaction:
Class 12 Chemistry Important Questions Chapter 13 Amines 32
(iii) Carbylamine reaction (CBSE Delhi 2019)
Answer:
Carbylamine reaction:
Class 12 Chemistry Important Questions Chapter 13 Amines 33

Question 2.
An aromatic compound ‘A’ on heating with Br2 and KOH forms a compound ‘B’ of molecular formula C6H7N which on reacting with CHCl3 and alcoholic KOH produces a foul-smelling compound ‘C’.
Write the structures and IUPAC names of compounds A, B, and C. ((SSE Delhi 2019)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 34
A: C6 H5CONH2: Benzamide B: C6H5NH2: Benzenamine C: C6H5NC: Phenyt isocyanide

Question 3.
Write the structures of main products when benzene diazonium chloride reacts with the following reagents: (CBSE Delhi 2019)
(i) CuCN
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 35
(ii) CH3CH2OH
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 36

(iii) Kl
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 37

Question 4.
(a) Write the product formed when
(i) 2-chioropropane is treated with aic. KOH.
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 38
(ii) Aniline reacts with conc. H2SO4 at 453 – 473 K.
Answer:
Sulphonation: Sulphonation of aniline is carried out by heating aniline with sulphuric acid. The product formed is anilinium hydrogen sulfate which on heating gives sulphanilic acid.
Class 12 Chemistry Important Questions Chapter 13 Amines 39
The sulphanilic acid exists as a dipolar ion (structure II) which has acidic and basic groups in the same molecule. Such ions are called Zwitter ions or inner salts.

(b) When aniline is heated with CHCl3 and aic. KOH, a foul-smelling compound is formed. What is this compound? (CBSE 2019C)
Answer:
Pheriyt isocyanide is formed.
Class 12 Chemistry Important Questions Chapter 13 Amines 40

Question 5.
(a) Identify ‘A’ and ‘B’ In the following reaction:
Class 12 Chemistry Important Questions Chapter 13 Amines 41
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 42
(b) Why does aniline not undergo Friedel-Crafts reaction? (CBSE 2019C)
Answer:
Aniline being a Lewis base reacts with Lewis acid such as AlCl3 to form a salt.
Class 12 Chemistry Important Questions Chapter 13 Amines 43
As a result, N of aniline acquires +ve charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Hence aniline does not undergo Friedel-Crafts reaction.

Question 6.
Complete the following reactions: (CBSE 2013)
(i) CH3CH2NH2 + CHCl3 + ale. KOH →
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 44

(ii) C6H5N2+Cl– →
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 45

Class 12 Chemistry Important Questions Chapter 13 Amines 46
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 47

Question 7.
Write the main products of the following reactions: (CBSE 2013, CBSE AI 2013)
Class 12 Chemistry Important Questions Chapter 13 Amines 48
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 49

Class 12 Chemistry Important Questions Chapter 13 Amines 50
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 51

Class 12 Chemistry Important Questions Chapter 13 Amines 52
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 53

Question 8.
Write the main products of the following reactions: (CBSE 2013)
Class 12 Chemistry Important Questions Chapter 13 Amines 54
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 55

Class 12 Chemistry Important Questions Chapter 13 Amines 56
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 57

Class 12 Chemistry Important Questions Chapter 13 Amines 58
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 59

Question 9.
Account for the following:
(i) Primary amines (R-NH2) have a higher boiling point than tertiary amines (R3N).
(ii) Aniline does not undergo Friedel — Crafts reaction.
(iii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.
OR
Give the structures of A, B, and C in the following reactions: (CBSE 2014)
Class 12 Chemistry Important Questions Chapter 13 Amines 60
Answer:
(i) Primary amines (RNH2) have two hydrogen atoms on the N atom and therefore, form intermolecular hydrogen bonding.
Class 12 Chemistry Important Questions Chapter 13 Amines 61
Tertiary amines (R3N) do not have hydrogen atoms on the N atom and therefore, these do not form hydrogen bonds. As a result of hydrogen bonding in primary amines, they have higher boiling points than tertiary amines of comparable molecular mass. For example, b.p. of n-butylamine is 351 K while that of tert-butylamine is 319 K.

(ii) Aniline being a Lewis base reacts with Lewis acid such as AlCl3 to form a salt.
Class 12 Chemistry Important Questions Chapter 13 Amines 62
As a result, N of aniline acquires +ve charge and hence it acts as a strong deactivating group for electrophilic substitution reactions. Hence aniline does not undergo Friedel Crafts reaction.

(iii) Due to the presence of lone pair of electrons on the N atom, amines are basic in nature. The methyl group is the electron releasing group (+I inductive effect) and therefore, it increases the electron density on the N atom, and therefore, basic character increases, so that (CH3)3N should be more basic than (CH3)2NH. But tertiary ammonium ion formed from tertiary amines is less hydrated than secondary ammonium ion formed from secondary amine. Therefore, (CH3)3N has less tendency to form ammonium ion, and consequently, it is less basic than (CH3)2NH. Thus, (CH3)2NH is more basic than (CH3)3N due to the combined effect of inductive effect and hydration effect.
OR
Class 12 Chemistry Important Questions Chapter 13 Amines 63
Question 10.
Give the structures of A, B, and C in the following reactions:
Class 12 Chemistry Important Questions Chapter 13 Amines 64
OR
How will you convert the following:
(i) Nitrobenzene into aniline
(ii) Ethanoic acid into methanamine
(iii) Aniline into N-phenylethylamine
(Write the chemical equations involved.) (CBSE Delhi 2014)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 65
OR
Class 12 Chemistry Important Questions Chapter 13 Amines 66

Question 11.
Write chemical equations for the following conversions:
(i) Nitrobenzene to benzoic acid.
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 67

(ii) Benzyl chloride to 2-phenytethanamine.
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 68

(iii) Aniline to benzyl alcohol. (CBSE Delhi 2012)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 69

Question 12.
(a) Identify ‘A’ and ‘B’ in the following reaction:
Class 12 Chemistry Important Questions Chapter 13 Amines 70
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 71

(b) Why is ethylamine soluble in water whereas aniline is not?
Answer:
Ethylamine dissolves in water due to intermolecular hydrogen bonding as shown below:
Class 12 Chemistry Important Questions Chapter 13 Amines 72
However, because of the large hydrophobic part (i.e. hydrocarbon part) of aniline, the extent of hydrogen bonding is less and therefore, aniline is insoluble in water.

Question 13.
(a) Identify X and Y in the following:
Class 12 Chemistry Important Questions Chapter 13 Amines 73
Answer:
X =
Class 12 Chemistry Important Questions Chapter 13 Amines 74benzene diazonium ch(orlde,
Y =
Class 12 Chemistry Important Questions Chapter 13 Amines 75 Cyanobenzene

(b) Amino group is o, p-directing for aromatic electrophilic substitution reactions. Why does aniline on nitration give m-nitroaniline? (CBSE 2019C)
Answer:
Under strongly acidic conditions of nitration, most of the aniline is converted into anilinium ion having an NH3+ group. This group is an m-directing group, therefore, m-nitro aniline is also obtained along with o- and p-products.

Question 14.
Give the structures of A, B, and C in the following reactions: (CBSE Delhi 2013)
Class 12 Chemistry Important Questions Chapter 13 Amines 76
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 77

Class 12 Chemistry Important Questions Chapter 13 Amines 78
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 79

Question 15.
Do as directed:
(i) Arrange the following compounds in the increasing order of their basic strength in an aqueous solution:
CH3NH3, (CH3)3N, (CH3)2NH.
Answer:
(CH3)3N < CH3NH2 < (CH3)2 NH

(ii) Identify ‘A’ and ‘B’:
Class 12 Chemistry Important Questions Chapter 13 Amines 80
Answer:
A: C6H5N2+ Cl- B: C6H5OH

(iii) Write the equation of carbylamine reaction. (CBSE 2018C)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 81

Question 16.
(i) Illustrate the following reactions giving a suitable example in each case:
(a) Hoffmann bromamide degradation reaction
(b) Diazotisation
(c) Gabriel phthalimide synthesis
(ii) Distinguish between the following pairs of compounds:
(a) Aniline and N-methylaniline
(b) (CH3)2NH and (CH3)3N
OR
(i) Write the structures of main products when benzene diazonium chloride (C6H5N2+Cl–) reacts with the following reagents:
(a) CuCN/KCN
(b) H20
(c) CH3CH2OH
(ii) Arrange the following:
(a) C2H5NH2, C2H5OH, (CH3)3N – in the increasing order of their boiling point.
(b) Aniline, p-nitroaniline, p-methyl aniline – in the increasing order of their basic strength. (CBSE Delhi 2015)
Answer:
(i) (a) Hoffmann bromamide degradation reaction: Primary amines can be prepared from amides by treatment with Br2 and KOH solution. The amine formed contains one carbon atom less than the parent amide.
Class 12 Chemistry Important Questions Chapter 13 Amines 82

(b) Diazotisation: The reaction of aniline or other aromatic amines, with nitrous acid at 0-5 °C to form diazonium salts is called diazotization. Nitrous acid needed for this reaction is prepared in situ by the action of dil. HCl on NaNO2.
Class 12 Chemistry Important Questions Chapter 13 Amines 83

(c) Gabriel’s phthalimide synthesis. This method is used for preparing only primary amines. In this method, phthalimide is treated with alcoholic KOH to give potassium phthalimide, which is treated with an alkyl halide or benzyl halide to form N-alkyl or aryl phthalimide. The hydrolysis of N-alkyl phthalimide with 20% HCl under pressure or refluxing with NaOH gives primary amine.
Class 12 Chemistry Important Questions Chapter 13 Amines 84
Phthalic acid can again be converted into phthalimide and is used again and again. This method is very useful because it gives pure amines. Aryl halides cannot be converted to arylamines by Gabriel synthesis because they do not undergo nucleophilic substitution with potassium phthalimide.

(ii) (a) Add an alcoholic solution of KOH and CHCl3 to the compounds. Aniline gives the foul smell of isocyanide whereas N-methyl aniline does not give a foul smell.
Class 12 Chemistry Important Questions Chapter 13 Amines 85
(b) When treated with Hinsberg’s reagent (benzene sulphonyl chloride, C6H5SO2CI), dimethylamine, (CH3)2NH gives precipitate which is insoluble in aqueous KOH.
Class 12 Chemistry Important Questions Chapter 13 Amines 86
(CH3)3N does not react with Hinsberg’s reagent.
Or
Class 12 Chemistry Important Questions Chapter 13 Amines 87

(ii) (a) (CH3)2 N < C2H5NH2 < C2H5OH
(b) p-nitroaniline < aniline < p-methylaniline

Question 17.
Give the IUPAC names of the following compounds:
Class 12 Chemistry Important Questions Chapter 13 Amines 88
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 89

Class 12 Chemistry Important Questions Chapter 13 Amines 90
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 91

Class 12 Chemistry Important Questions Chapter 13 Amines 92
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 93

Class 12 Chemistry Important Questions Chapter 13 Amines 94
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 95

Class 12 Chemistry Important Questions Chapter 13 Amines 96
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 97

Class 12 Chemistry Important Questions Chapter 13 Amines 98
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 99

Class 12 Chemistry Important Questions Chapter 13 Amines 100
(CBSE Delhi 2017)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 101

Class 12 Chemistry Important Questions Chapter 13 Amines 102
(CBSE Delhi 2017)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 103

Class 12 Chemistry Important Questions Chapter 13 Amines 104
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 105

Question 18.
Complete the following reactions:
Class 12 Chemistry Important Questions Chapter 13 Amines 106
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 107

Class 12 Chemistry Important Questions Chapter 13 Amines 108
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 109

Class 12 Chemistry Important Questions Chapter 13 Amines 110
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 111

Class 12 Chemistry Important Questions Chapter 13 Amines 112
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 113

Class 12 Chemistry Important Questions Chapter 13 Amines 114
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 115

Question 19.
Write the main products when benzene diazonium chloride (C6H5N2+Cl–) reacts with the following:
(i) CuCN/KCN
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 116

(ii) H20
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 117

(iii) CH3CH2OH (CBSE AI 2015, 2018)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 118

(iv) Copper powder/HCI
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 119

Question 20.
Complete the following chemical equations:
Class 12 Chemistry Important Questions Chapter 13 Amines 120
(CBSE Delhi 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 121

Class 12 Chemistry Important Questions Chapter 13 Amines 122
(CBSE Delhi 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 123

Class 12 Chemistry Important Questions Chapter 13 Amines 124
(CBSE Delhi 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 125

Class 12 Chemistry Important Questions Chapter 13 Amines 126
(CBSE AI 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 127

Class 12 Chemistry Important Questions Chapter 13 Amines 128
(CBSE AI 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 129

Class 12 Chemistry Important Questions Chapter 13 Amines 130
(CBSE AI 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 131

Class 12 Chemistry Important Questions Chapter 13 Amines 132
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 133

Question 21.
Give the structures of A, B, and C In the following reactions:
Class 12 Chemistry Important Questions Chapter 13 Amines 134
(CBSE AI 2017)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 135

Class 12 Chemistry Important Questions Chapter 13 Amines 136
(CBSE AI 2017)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 137

Class 12 Chemistry Important Questions Chapter 13 Amines 138
(CBSE Delhi 2016)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 139

Class 12 Chemistry Important Questions Chapter 13 Amines 140
(CBSE Delhi 2016)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 141

Class 12 Chemistry Important Questions Chapter 13 Amines 142
(CBSE AI 2016)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 143

Class 12 Chemistry Important Questions Chapter 13 Amines 144
(CBSE AI 2016)
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 145

Class 12 Chemistry Important Questions Chapter 13 Amines 146
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 147

Question 22.
An organic compound A’ with molecular formula C7H7NO reacts with Br2/aq KOH to give compound B’, which upon reaction with NaNO2 and HCI at OC gives C’. Compound C’ on heating with CH3CH2OH gives a hydrocarbon D’. Compound B’ on further reaction with Br2 water gives a white precipitate of compound E’. Identify the compounds A, B, C, D, and E; also justify your answer by giving relevant chemical equations. (CBSE Sample Paper 2019)
OR
(a) How will you convert:
(i) Aniline into Fluorobenzene?
(ii) Benzamide into Benzylamine?
(iii) Ethanamine into N, N-Diethylethanamine?
(b) Write the structures of A and B in the following:
Class 12 Chemistry Important Questions Chapter 13 Amines 148
Answer:
Class 12 Chemistry Important Questions Chapter 13 Amines 149
Class 12 Chemistry Important Questions Chapter 13 Amines 150

Important Questions for Class 12 Maths Chapter Wise Pdf with Solutions 2021

Important Questions for Class 12 Maths Chapter Wise Pdf with Solution 2020-2021: Here we are providing CBSE Important Extra Questions for Class 12 Maths with Solutions Pdf download in Hindi and English Medium. Students can get Class 12 Maths NCERT Solutions, Maths Class 12 Important Extra Questions and Answers designed by subject expert teachers.

CBSE Class 12th Maths Important Extra Questions and Answers Chapter Wise Pdf

Maths Important Questions Class 12 | Important Questions of Maths Class 12

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Coordination Compounds Class 12 Important Extra Questions Chemistry Chapter 9

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 9 Coordination Compounds.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 9 Important Extra Questions Coordination Compounds

Coordination Compounds Important Extra Questions Very Short Answer Type

Question 1.
Give an example of linkage isomerism. (CBSE Delhi 2010)
Answer:
[Co (NH3)5 (N02)] Cl2 and [Co (NH3)5 (ONO)]Cl2

Question 2.
Give an example of coordination isomerism. (CBSE Delhi 2010)
Answer:
[CO(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co (CN)6]

Question 3.
Which of the following is more stable complex and why? (CBSE Delhi and Al 2014)
[CO(NH3)6]3+ and [Co(en)3]3+
Answer:
[Co(en)3]3+ because bidentate ligand (ethylenediamine) forms chelate which is more stable.

Question 4.
A coordination compound with molecular formula CrCl3.4H20 precipitates one mole of AgCl with AgN03 solution. Its molar conductivity is found to be equivalent to two ions. What is the structural formula and name of the compound? (CBSE Sample Paper 2017-18)
Answer:
[CrCl2(H20)4]Cl: Tetraaquadichloridochrom ium(III) chloride.

Question 5.
What is coordination isomerism? Give one example. (CBSE Delhi 2010)
Answer:
Coordination isomerism arises when both the cation and anion are complexes and the ligands are exchanged. For example, [CO(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6]

Question 6.
Write IUPAC name of the complex [Co(en)2(H20)(CN)]2+
OR
Using IUPAC norms, write the formula of Ammonium tetrafluoride cobaltates (II). (CBSE AI 2019)
Answer:
Aquacyanidobis (ethane-1,2-diamine) cobalt (III) ion
OR
(NH4)2[Co F4]

Question 7.
Write the IUPAC name of the complex K3[Cr(C204)3].
OR
Using IUPAC norms write the formula of Hexaamminecobalt(III) sulphate. (CBSE AI 2019)
Answer:
Potassium trioxalatochromate(III)
OR
[Co(NH3)6]2 (S04)3

Question 8.
Write IUPAC name of the complex [Co(en)2CI2]
OR
Using IUPAC norms, write the formula of Sodium tetrachloridonickelate(II). (CBSE AI 2019)
Answer:
Dichloridobis (ethylenediamine) cobalt(III) ion.
OR
Na2[NlCl4]

Question 9.
Predict the number of unpaired electrons in the square planar [Pt(CN)4]2- ion. (CBSE 2019C)
Or
Amongst [Fe(C204)3]3- and [Fe(NH3)6]3+ which is more stable and why?
Answer:
The Pt (II) ion has 5d8 electronic configuration. For square planar geometry, dsp2 hybridisation is involved. For this, one empty d-orbital is needed for hybridisation. Therefore, the pairing of electrons takes place in the remaining d-orbitals. Hence, there are no unpaired electrons in [Pt(CN)4]2- ion and it is diamagnetic.
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 1
OR
[Fe(C204)3]3- is more stable because it is a chelate. C2042- is a didentate ligand.

When a didentate or a polydentate ligand uses its two or more donor atoms to bind to the same central metal atom or ion forming a ring structure it is called chelation.

The resulting complex has a ring structure and the ligand coordinating through two or more donor groups is called a chelating ligand. Some common examples of chelating ligands are carbonate ion, oxalate ion (ox2-), ethylenediamine (en), ortho-phenanthroline (ph) and ethylenediaminetetraacetate ion (EDTA).

Chelate ligands form more stable complexes than similar ordinary complexes in which the ligand acts as monodentate. This is called a chelate effect.

Coordination Compounds Important Extra Questions Short Answer Type

Question 1.
Write IUPAC name of the complex [Pt(en)2Cl2]. Draw structures of geometrical isomers for this complex.
OR
Using IUPAC norms write the formulae for the following:
(i) Hexaamminecobalt(III) sulphate
(ii) Potassium trioxalatochromate(III) (CBSE Delhi 2019)
Answer:
Dichlorido bis(ethane-1,2-diamine) platinum (II)
Geometrical isomers.
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 2
OR
(i) [CO(NH3)6]2(S04)3
(ii) K3[Cr(ox)]3

Question 2.
Out of [COF6]3- and [Co(en)3]3+, which one complex is
(i) paramagnetic
Answer:
[CoF6]3- is paramagnetic due to the presence of 4 unpaired electrons.

(ii) more stable
Answer:
[Co(en)3]3+ is more stable because of chelation.

(iii) inner orbital complex and
Answer:
[Co(en)3]3+ forms inner orbital complex involving d2sp3 hybridisation.

(iv) high spin complex (Atomic no. of Co = 27) (CBSE Delhi 2019)
Answer:
[COF6]3- forms a high spin complex (sp3d2 hybridisation).

Question 3.
(i) Write down the IUPAC name of the following complex: (CBSE Delhi 2015)
[Cr(NH3)2Cl2 (en)]Cl
(en = ethylenediamine)
Answer:
Diamminedichlorido(ethane-1,2,- diamine)chromium(III) chloride

(ii) Write the formula for the following complex:
Pentaamminenitrito-O-cobalt (III).
Answer:
[Co(NH3)5 (ONO)]2+

Question 4.
(i) Write down the IUPAC name of the following complex [Co(NH3)5(N02)] (N03)2
Answer:
Pentaamminenitrito-N-cobalt(III) nitrate

(ii) Write the formula for the following complex: Potassium tetracyanonickelate (II) (CBSE 2015)
Answer:
K2[Ni(CN)4]

Question 5.
When a coordination compound CrCl3.6H20 is mixed with AgN03, 2 moles of AgCI are precipitated per mole of the compound. Write (CBSE Delhi 2016)
(i) Structural formula of the complex.
Answer:
For one mole of the compound, two moles of AgCI are precipitated with AgN03, which indicates two ionisable chloride ions in the complex. Hence structural formula is [CrCl(H20)5]Cl2. H20

(ii) IUPAC name of the complex.
Answer:
Pentaaquachloridochromium (III) chloride hydrate

Question 6.
(i) For the complex [Fe(CN)6]3-, write the hybridisation type, magnetic character and spin nature of the complex. (At. number: Fe = 26).
Answer:
[Fe(CN)6]3-:
The oxidation state of Fe = +3
Fe(III)
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 3
Hybridisation: d2sp3
Magnetic character: Paramagnetic
Spin type: Low spin complex

(ii) Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically active. (CBSE Delhi 2016)
Answer:
cis- form is optically active
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 4

Question 7.
When a coordination compound NiCl2.6H20 is mixed with AgNO3, 2 moles of AgCI are precipitated per mole of the compound. Write (CBSE 2016)
(i) Structural formula of the complex
(ii) IUPAC name of the complex
Answer:
For one mole of the compound, 2 moles of AgCI are precipitated with AgNO3, which indicates 2 ionisable Cl” ions in the complex.

  1. Structural formula: [Ni (H20)6]Cl2
  2. IUPAC name: Hexaaquanickel (II) chloride

Question 8.
Write IUPAC name of the complex [Cr(NH3)4Cl2]+. Draw structures of geometrical isomers for this complex.
OR
Using IUPAC norms write the formulae for the following:
(i) Pentaamminenitrito-O-cobalt(III) chloride
(ii) Potassium tetracyanidonickelate(II) (CBSE Delhi 2019)
Answer:
IUPAC name: Tetramminedichloridochromi um(III) ion.
Geometrical isomers:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 5
OR
(i) [CO(NH3)5(ONO)]Cl2
(ii) K2[Ni(CN)4]

Question 9.
Write the hybridisation and magnetic character of the following complexes:
(i) [Fe(H2O)6]2+
(ii) [Fe(CO)5] (Atomic no. of Fe = 26) (CBSE Delhi 2019)
Answer:
[Fe(H20)6]2+
Hybridisation: sp3d2
Magnetic character: Paramagnetic due to 4 unpaired electrons.
Fe(C0)5
Hybridisation: dsp3
Magnetic character: It is diamagnetic.

Question 10.
Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically inactive. Also, write the name of this entity according to the IUPAC nomenclature.
Or
Discuss the bonding in the coordination entity [Co(NH3)6 ]3+ on the basis of valence bond theory. Also, comment on the geometry and spin of the given entity. (Atomic no. of Co= 27) (CBSE Sample Paper 2019)
Answer:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 6
IUPAC Name of the entity:
Dichloridobis(ethane – 1,2-diamine) platinum(IV) ion
Or
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 7

Coordination Compounds Important Extra Questions Long Answer Type

Question 1.
(a) Draw the structures of geometrical isomers of [Fe(NH3)2 (CN)4]–
(b) [NiCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why? [Atomic number of Ni = 28]
OR
Define the following:
(a) Ambidentate ligands
(b) Spectra chemical series
(C) Heteroleptic complexes
Answer:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 8

(b) In the presence of strong field ligand CO, the unpaired d-electrons of Ni pair up so [Ni(CO)4] is diamagnetic but Cl– being a weak ligand is unable to pair up the unpaired electrons, so [Ni(Cl4)]2- is paramagnetic.
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 9
and
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 10
OR
Answer:
(a) Ambidentate ligands: The monodentate ligands which can coordinate with the central atom through more than one site are called ambidentate ligands.
These ligands contain more than one coordinating atoms in their molecules.
For example, NO2 can coordinate to the metal atom through N or 0 as
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 11
Similarly, CN– can coordinate through C or N as
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 12
Thiocyanato (SCN) can coordinate through S or N
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 13

(b) The arrangement of ligands in the increasing order of crystal field splitting is called spectrochemical series. This is shown below:
I– < Br– < SCN– < Cl– < F– < OH– < Ox2- < O2- < H20 < NCS– < py = NH3 <en <NO2– < CN– < CO.

Weak field ligands are those which cause less crystal field splitting. These form high spin complexes. For example, Cl–, F–, etc.

Strong field ligands are those which cause greater crystal field splitting. These form low spin complexes. For example, CN–, NO2–, CO.

(c) The complexes in which the metal is bound to more than one kind of donor groups (ligands) are called heteroleptic complexes. Some common examples of heteroleptic complexes are [NiCl2(H2O)4], [CoCl2(NH3)4]+, etc.

Question 2.
Write the structures and names of all the stereoisomers of the following compounds:
(i) [Co(en)3]Cl3
Answer:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 14

(ii) [Pt(NH3)2Cl2]
Answer:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 15

(iii) [Fe(NH3)4Cl2]Cl (CBSE 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 16

Question 3.
Write the name, stereochemistry and magnetic behaviour of the following:
(i) K4[Mn(CN)6]
(ii) [Co(NH3)2 C1]Cl2
(iii) K2[Ni(CN)4] (CBSE Delhi 2011)
Answer:

Complex Name Stereochemistry Magnetic behaviour
(i) K4[Mn(CN)6] Potassium hexacyanomanganate (I) Octahedral Paramagnetic
(ii) [Co(NH3)5Cl]Cl2 Pentaamminechtorido cobalt (III) chloride Octahedral Diamagnetic
(iii) K2[Ni(CN)4) Potassium tetracyanonicketate (II) Square planar Diamagnetic

Question 4.
For the complex [Fe(en)2Cl2]Cl identify the following:
(i) Oxidation number of iron
Answer:
III.

(ii) Hybrid orbitals and shape of the complex
Answer:
d2sp3 hybridisation, octahedral

(iii) Magnetic behaviour of the complex
Answer:
Paramagnetic due to one unpaired electron

(iv) Number of its geometric isomers
Answer:
Two

(v) Whether there may be optical isomer also
Answer:
One optical, an isomer of cls-geometrical isomer.

(vi) Name the complex. (CBSE 2011)
Answer:
Dichloridobis(ethylenediamine) iron (III) chloride.

Question 5.
Give the formula of each of the following coordination entities:
(i) CO3+ ion is bound to one Cl–, one NH3 molecule and two bidentate ethylene diamine (en) molecules.
(ii) Ni2+ ¡on is bound to two water molecules and two oxalate ions. Write the name and magnetic behaviour of each of the above coordination entitles. (At. nos. CO = 27, Ni = 28) (CBSE 2012)
Answer:
(i) [Co(NH3) (Cl) (en)2]2+ Amminechtoriðo bls-(ethane -1, 2-diamine) cobalt (III) ion Co(27):
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 17
Since there are no unpaired electrons,
complex is diamagnetic.

(ii) [Ni(H2O)2 (C2O4)2]2- Diaquadioxatatonickelate (II) ion Ni(28):
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 18
The complex has two unpaired electrons, therefore, it will be paramagnetic.

Question 6.
State a reason for each of the following situations:
(i) CO2+ is easily oxidised to CO3+ in presence of a strong ligand.
Answer:
The configuration of CO2+ is t2g6 eg1 and for CO3+, it is t2g6. The crystal field stabilisation energy is more than compensated for the third ionisation enthalpy. Therefore, CO2+ is easily oxidised to CO3+ in the presence of a strong ligand.

(ii) CO is a stronger complexing reagent than NH3.
Answer:
CO is a stronger complexing ligand than NH3 because it contains both σ and π character and can form a back bond (M → CO) also. Therefore, CO forms a stronger bond with the metal. It is also called a strong field ligand.

(iii) The molecular shape of Ni(CO)4 is not the same as that of [Ni(CN)4]2-. (CBSE Delhi 2012)
Answer:
The molecular shape of [Ni(C0)4] is tetrahedral because this complex nickel involves sp3 hybridisation. In [Ni(CN)4]2-, nickel involves dsp2 and its shape is square planer.

Question 7.
Name the following coordination entities and draw the structures of their stereoisomers:
(i) [Co(en)2Cl2]+(en = ethane-1, 2-diamine)
Answer:
(Co(en)2Cl2]+
Name: Dichloridobis (ethane-1, 2-diamine) cobalt(III) ion. It shows two geometrical isomers cis- and trAnswer:c/s-shows optical isomerism.
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 19
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 20
c/s-[Co(en)2Cl2]+ is optically active.

(ii) [Cr(C204)3]3-
Answer:
[Cr(C204)3]3-
Name: trioxalatochromate(III) ion. This complexion shows optical isomerism and their dextro and laevo forms are shown below:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 21

(iii) [CO(NH3)3Cl3]
(Atomic numbers Cr = 24, CO = 27) (CBSE 2012)
Answer:
[Co(NH3)3Cl3]
Name Triamminetrichloridocobalt (III) It shows two geometrical isomers known as facial (fac) and meridional or merisomer as shown below:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 22

Question 8.
Name the following coordination entities and describe their structures:
(i) [Fe(CN)6]4-
(ii) [Cr(NH3)4Cl2]+
(iii) [Ni(CN)4]2-
(Atomic numbers Fe = 26, Cr = 24, Ni = 28) (CBSE 2012)
Answer:
(i) [Fe(CN)6]4-: Hexacyanoferrate (II) ion.
Structure. In this case, iron is in a +2 oxidation state having the outer electronic configuration 3d6. To account for the diamagnetic character, all the electrons get paired leaving two vacant orbitals.

The hybridisation scheme and bonding are as shown below:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 23

(ii) [Cr(NH3)4Cl2]+: Tetraamminedichlorido chromium (III) ion.

The chromium (Z = 24) has the electronic configuration 3d5 4s1. The chromium in this complex is in + 3 oxidation state and the ion is formed by the loss of one 4s and two of the 3d-electrons as shown in Fig. The inner d-orbitals are already vacant and two vacant 3d, one 4s and three 4p-orbitals are hybridised to form six d2sp3-hybrid orbitals. Six pairs of electrons one from each NH3 molecule and Cl- ions (shown by xx) occupy the six vacant hybrid orbitals. The molecule has octahedral geometry.

Since the complex contains three unpaired electrons, it is paramagnetic.
Cr (III):
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 24

(iii) [Ni(CN)4]2-: Tetracyanonickelate (II) ion.
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 8

Question 9.
Write the IUPAC names of the following compounds:
(i) [Cr(NH3)3Cl3]
Answer:
Triamminetrichloridochromium (III)

(ii) K3[Fe(CN)6]
Answer:
Potassium hexacyanoferrate (III)

(iii) [CoBr2(en)2]+, (en = ethylenediamine) (CBSE Delhi 2013)
Answer:
Dibromidobis(ethylenediamine) cobalt (III) ion

Question 10.
Write the types of isomerism exhibited by the following complexes:
(i) [Co(NH3)5Cl]S04
Answer:
ionisation isomerism

(ii) [Co(en)3]3+
Answer:
optical isomerism

(iii) [Co(NH3)6] [Cr(CN)6] (CBSE Delhi 2013)
Answer:
coordination isomerism

Question 11.
For the complex [NiCl4]2-, write
(i) the IUPAC name.
(ii) the hybridisation type.
(iii) the shape of the complex (Atomic no. of Ni = 28)
Or
What is meant by crystal field splitting energy? On the basis of crystal field theory, write the electronic configuration of d4 in terms to f2g and eg in an octahedral field when
(i) Δo > P
(ii) Δo < P (CBSE 2013)
Answer:
(i) Tetrachloridonickelate(II)ion
(ii) sp3 hybridisation
(iii) tetrahedral
OR
The difference between two sets of energy levels when the five degenerate d-orbitals splits in the presence of the electrical field of Ligands is called crystal field splitting energy. In the octahedral field, it is represented as Δo.

Electronic configurations of d4

  • When Δo >P;t2g4
  • When Δo < P; t2g3 eg1

Question 12.
Write the IUPAC name of the complex [Cr(NH3)4Cl2]. What type of Isomerism does it exhibit? (CBSE Delhi 2014)
Answer:
Tetraammlnedlchloridochromium( III) ion. It shows cis and trans isomerism and cis form shows optical isomerism.

Question 13.
(i) Write the IUPAC name of the complex [Cr(NH3)4 Cl2] Cl.
(ii) What type of Isomerism is exhibited by the complex [Co(en)3]3+?
(en = ethane-1, 2-diamine)
(iii) Why Is [NIC(412 paramagnetic but [Ni(CO)4] Is diamagnetic?
(At. nos.: Cr = 24, Co = 27, Ni = 28) (CBSE 2014)
Answer:
(i) TetraamminedichLorìdochromium (III) chloride.
(ii) Optical isomerism
(iii) In [NICl4]2- complexion, nickel is in +2 oxidation state and the configuration is 3d8. Since the molecule is tetrahedral, it involves sp3 hybridisation as shown below:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 25
The molecule is paramagnetic because it contains two unpaired electrons. In [Ni(CO)4], nickel is in O oxidation state and has the configuration 4s2 3d8 or 3d10. The molecule is tetrahedral and involves sp3-hybridisation as given below:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 26

Each CO donates a pair of electrons forming four Ni—CO bond. The compound is diamagnetic since it contains no unpaired electron.

Question 14.
(i) Draw the geometrical isomers of the complex [Pt(NH3)2Cl2].
Answer:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 27
(ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 < P.
Answer:
t2g3 eg1

Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 28

(iii) Write the hybridization and magnetic behaviour of the complex [Ni(CO)4]. (At.no. of Ni = 28) (CBSE Delhi 2015)
Answer:
[Ni(CO)4] involves sp3 hybridization of nickel and the complex is tetrahedral. Magnetic behaviour: Diamagnetic
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 8

Question 15.
(i) Draw the geometrical isomers of complex [Co(en)2Cl2]+.
Answer:
Geometrical isomers of complex [Co(en)2Cl2]+:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 29
(ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 > P.
Answer:
t2g4
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 30

(iii) [NiCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28) (CBSE 2015)
Answer:
In [NiCl4]2- complexion, nickel is in +2 oxidation state and the configuration is 3d8. Since the molecule is tetrahedral, it involves sp3 hybridisation as shown below:
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 31
The molecule is paramagnetic because it contains two unpaired electrons.

In [Ni(CO)4], nickel is in 0 oxidation state and has the configuration 4s2 3d4 or 3d10. The molecule is tetrahedral and involves sp3-hybridisation as given below :
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 32
Each CO donates a pair of electrons forming four Ni-CO bonds. The compound is diamagnetic since it contains no unpaired electron.

Question 16.
What type of isomerism is shown by the complex
(i) [Co (NH3)5 (SCN)]2+ (CBSE AI 2017)
Answer:
Linkage isomerism

(ii) [Co (NH3)6] [Cr (CN)6] (CBSE Delhi 2017)
Answer:
Coordination isomerism

(iii) [Co (en)3] Cl3 (CBSE AI 2014)
Answer:
Optical isomerism

Question 17.
A metal ion Mn+ having d4 valence electronic configuration combines with three bidentate ligands to form a complex compound. Assuming Δo > P:
(i) Write the electronic configuration of the d4 ion.
Answer:
t2g4 eg0

(ii) What type of hybridisation will Mn+ ion have?
Answer:
d2sp3

(iii) Name the type of isomerism exhibited by this complex. (CBSE Sample Paper 2017 – 18)
Answer:
Optical isomerism

Question 18.
Write the state of hybridisation, the shape and the magnetic behaviour of the following complex entities :
(a) [Cr(NH3)4Cl2] Cl
(b) [Co(en)3]Cl3
(c) K3[Ni(CN)4] (CBSE AI 2011)
Answer:

Complex Hybridisation Shape Magnetic behaviour
(a) [Cr(NH3)4Cl2]Cl d2sp3 Octahedral Paramagnetic
(b) [Co(en)3]Cl3 d2sp3 Octahedral Diamagnetic
(c) K2[Ni(CN)4] dsp2 Square planar Diamagnetic

Question 19.
(a) Give one chemical test as evidence to show that [Co(NH3)5Cl] S04 and [Co(NH3)55(S04)]Cl are ionisation isomers.
Answer:
When [Co(NH3)5(S04)]Cl is treated with silver nitrate solution, a white precipitate of AgCl is formed. But [Co(NH3)5Cl]S04 does not give white ppt with AgN03 solution.
[Co(NH3)5S03 ]Cl+AgN03 (Ag) → AgCl White ppt
[Co(NH3)5 Cl]S04 + AgN03 (Ag) → No white ppt

(b) [NiCl4]2- is paramagnetic while [Ni(C0)4] is diamagnetic though both are tetrahedral. Why? (Atomic no. of Ni = 28)
Answer:
In [NiCl4]2-, the oxidation state of nickel is +2 and it has the electronic configuration: 3d8
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 33

Cl– ion is a weak field ligand and it causes the pairing of electrons. Since it has two unpaired electrons, it is paramagnetic. It undergoes sp3 hybridisation forming a tetrahedral structure.

In Ni (CO)4, the oxidation state of nickel is 0 and it has the electronic configuration: 3d8 4s2. CO is a strong field ligand and therefore, it causes the pairing of electrons as
Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds 34
Since it has no unpaired electron, it is diamagnetic. It also undergoes sp3 hybridisation resulting in tetrahedral geometry of the complex,

(c) Write the electronic configuration of Fe(III) on the basis of crystal field theory when it forms an octahedral complex in the presence of (i) strong field ligand, and (ii) weak field ligand. (Atomic no. of Fe = 26) (CBSEAI2019)
Answer:
Fe (III): 3d5
(i) Strong field ligand: t2g5 eg0
(ii) Weak field ligand: t2g3 eg2

Question 20.
A metal complex having the composition Cr (NH3)4Cl2Br has been isolated in two forms A and B. Form A reacts with AgN03 to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.
(i) Write the formulae of isomers A and B.
Answer:
Isomer A: [Cr(NH3)4 BrCl]Cl
Isomer B: [Cr (NH3)4 Cl2]Br

(ii) State the hybridisation of chromium in each of them.
Answer:
The hybridisation of Cr in isomer A and B is d2sp3.

(iii) Calculate the magnetic moment (spin only value) of the isomer A. (CBSE Sample Paper 2018)
Answer:
The number of unpaired electrons in Cr3+(3d3) is 3.
Magnetic moment = \(\sqrt{n(n+2)}\)
= \(\sqrt3(3 + 2)\) = 3.87 BM

Question 21.
(a) Write the formula of the following coordination compound: Iron(III) hexacyanoferrate(II)
Answer:
Fe4[Fe (CN)6]3

(b) What type of isomerism is exhibited by the complex [Co(NH3)5Cl]S04?
Answer:
Ionisation isomerism

(c) Write the hybridisation and number of unpaired electrons in the complex [COF6]3-. (Atomic No. of Co = 27) (CBSE 2018)
Answer:
sp3d2, 4

Question 22.
Write the IUPAC names of the following :
(a) [Ag(NH3)2][Ag(CN)2]
Answer:
diamminesilver(I)
dicyanidoargentate(I)

(b) K3[Fe(C204)3]
Answer:
potassium trioxalatoferrate(III)

(c) [CoCl2(en)2]Cl
Answer:
dichloridobis (ethane-1, 2-diamine) cobalt (III) chloride

(d) K3[Cr(C204)3]
Answer:
potassium trioxalatochromate(III)

(e) K4[Ni(CN)4]
Answer:
potassium tetracyanidonickelate (0)

(f) [Pt(NH3)2Cl(N02)]
Answer:
diamminechloridonitrito-N-platinum(II)

(g) [CoBr2(en)2]+ (CBSE Delhi 2012)
Answer:
dibromidobis (ethylenediamine) cobalt(III) ion

(h) [Co(NH3)5ONO]Cl2
Answer:
pentaamminenitrito-O-cobalt(III) chloride

(i) [Co(NH3)5(N02)](N03)2 (CBSE Al 2015)
Answer:
pentaamminenitrito-N-cobalt(III) nitrate

(j) [Cr(NH3)2Cl2 (en)]Cl (CBSE Delhi 2015)
Answer:
diamminedichlorido(ethane-1, 2-diamine)chromium (III) chloride

Question 23.
Write the formulae of the following coordination compounds:
(a) hexaamminecobalt(III) sulphate
Answer:
[CO(NH3)6]2(S04)3

(b) potassium tetrachloridopalladate(II)
Answer:
K2[PdCl4]

(c) diamminechloridonitrito -N- platinum(II)
Answer:
[Pt(NH3)2Cl(N02)]

(d) pentaamminenitrito -N- cobalt(III)
Answer:
[CO(N02)(NH3)5]2+

(e) pentaamminenitrito – 0- cobalt(III) (CBSE Delhi 2015)
Answer:
[CO(ONO)(NH3)5]2+

(f) sodium dicyanidoaurate(I) (CBSE AI 2017)
Answer:
Na [Au (CN)2]

(g) tetraamminechloridonitrito-N- platinum(IV) sulphate (CBSE AI 2017)
Answer:
[Pt(NH3)4Cl (N02)] S04

(h) potassium tetracyanidonickelate(II) (CBSE AI 2015)
Answer:
K2[Ni(CN)4]

(i) potassium trioxatatachromate(III)
Answer:
K3[Cr(ox)3]

(j) tetracarbonylnickel(O)
Answer:
[Ni(C0)4]

Question 24.
Write IUPAC names of the following :
(i) K3[AI(C204)3]
Answer:
Potassium trioxalatoaluminate (III)

(ii) [Ni(CO)4]
Answer:
tetracarbonylnickel(O)

(iii) Fe4[Fe(CN)6]3
Answer:
iron hexacyanidoferrate(III)

(iv) [CoCI(NH3)5]Cl2
Answer:
pentaamminechloridocobalt (III) chloride

(v) [PtCl2(C5H5N)(NH3)]
Answer:
amminedichlorido (pyridine) platinum(II)

(W) Na[PtBrCI(N02)(NH3)]
Answer:
sodium amminebromidochloridonitri- to-N-platinate(II)

(v/i) [Cr(NH3)3Cl3] (CBSE Delhi 2013)
Answer:
triamminetrichloridochromium (III)

(viii) K3[Fe(CN)6] (CBSE Delhi 2013)
Answer:
potassium hexacyanoferrate (III)

(ix) Na3[AlF6]
Answer:
sodium hexafluoridoaluminate (III).

(x) [Co(en)3]2(S04)3 (CBSE Delhi 2013)
Answer:
tris(ethylenediamine)cobalt(III) sulphate

Question 25.
For the complex [Fe(en)2Cl2]Cl, identify the following:
(i) Oxidation number of iron.
Answer:
III.

(ii) Hybrid orbitals and shape of the complex.
Answer:
d2sp2 hybridisation, octahedral.

(iii) Magnetic behaviour of the complex.
Answer:
Paramagnetic due to one unpaired electron.

(iv) Number of its geometrical isomers.
Answer:
Two.

(v) Whether there may be optical isomer also?
Answer:
An optical isomer of cis-geometrical isomer.

(vi) Name of the complex. (CBSE Delhi 2011)
Answer:
Dichloridobis (ethane-1, 2-diamine) iron (III) chloride.

Electromagnetic Induction Class 12 Important Extra Questions Physics Chapter 6

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 6 Electromagnetic Induction. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 6 Important Extra Questions Electromagnetic Induction

Electromagnetic Induction Important Extra Questions Very Short Answer Type

Question 1.
What is the function of a step-up transformer? (CBSE AI 2011C)
Answer:
The function of a step-up transformer is to step-up the alternating voltage.

Question 2.
State Lenz’s law. (CBSE AI 2012C)
Answer:
It states that the direction of induced emf is such that it opposes the cause of its production.

Question 3.
How can the self-inductance of a given coil having ‘N’ number of turns, area of cross-section of ‘A’ and length T be increased? (CBSE AI 2012C)
Answer:
By inserting a core of high permeability inside the coil.

Question 4.
How does the mutual inductance of a pair of coils change when
(a) the distance between the coils is increased and
(b) the number of turns in the coils is increased? (CBSE AI 2013)
Answer:
(a) decreases
(b) increases.

Question 5.
The motion of the copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping? (CBSE AI 2013)
Answer:
Production of eddy current.

Question 6.
Why is the core of a transformer laminated? (CBSE Delhi 2013C)
Answer:
To reduce the effects of eddy currents.

Question 7.
A metallic piece gets hot when surrounded by a coil carrying a high-frequency alternating current. Why? (CBSE Delhi 2014C)
Answer:
Due to the production of eddy current which generates heat.

Question 8.
Name any two applications where eddy currents are used to advantage. (CBSE Delhi 2016C)
Answer:

  1. Electromagnetic damping
  2. Induction furnace.

Question 9.
A long straight current-carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify. (CBSE Delhi 2017)
Answer:
Yes, as there will be a change in magnetic flux.

Question 10.
Predict the polarity of the capacitor in the situation described below. (CBSE AI 2017)
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 1
Answer:
The upper plate will be positive with respect to the lower plate in the capacitor.

Question 11.
In the figure given, mark the polarity of plates A and B of a capacitor when the magnets are quickly moved towards the coil. (CBSE AI 2017C)
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 2
Answer:
Plate A will be positive with respect to plate B in the capacitor.

Question 12.
A long straight current-carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify. (CBSE Delhi 2017)
Answer:
Yes, as there will be a change in magnetic flux.

Question 13.
An air-cored solenoid has self-inductance 2.8 H. When the core is removed, the self-inductance becomes 2 mH. What is the relative permeability of the core used? (CBSE Delhi 2017C)
Answer:
μr = 2.8 / 2 × 10-3 = 1.4 × 103

Question 14.
A choke and a bulb are in series to a dc source. The bulb shines brightly. How does its brightness change when an iron core is inserted inside the choke coil?
Answer:
There is no change in the brightness as the inductive reactance is zero for dc.

Question 15.
Why does the acceleration of a magnet falling through a long solenoid decrease?
Answer:
It decreases because of the opposing induced emf produced in the soLenoid due to the rate of change of magnetic flux.

Question 16.
Why is the core of a transformer laminated?
Answer:
It is done to reduce the effect of Eddy Currents.

Question 17.
A vertical metallic pole falls down through the plane of the magnetic meridian. Will any emf be Induced between Its ends?
Answer:
No emf will be induced because the pote neither intercepts the vertical component nor the horizontal component of the earth’s magnetic field.

Question 18.
A magnet Is moved towards a coil and an electric charge is induced in it. If the resistance of the coil is increased, how will the induced charge change?
Answer:
On increasing the resistance of the colt, the magnitude of induced charge decreases.

Question 19.
Can a transformer be used In a dc circuit?
Answer:
No, because there is no change in magnetic flux.

Question 20.
Why does a metallic piece become very hot when it is surrounded by a coil carrying high-frequency alternating current?
Answer:
The high-frequency coil induces eddy currents in the metallic piece. These eddy currents produce heat hence the metalLic piece becomes hot.

Question 21.
The figure shows a horizontal solenoid PQ connected to a battery and a switch. A copper ring R is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. What would happen to the ring as the switch S is closed.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 3
Answer:
The ring WilL be repelled due to opposing induced emf produced in it.

Question 22.
An air-core solenoid is connected to an ac source and a bulb. If an iron core Is inserted in the solenoid, how does the brightness of the bulb change? Give reasons for your answer.
Answer:
Insertion of an iron core In the solenoid increases its inductance. This in turn increases the value of inductive reactance. This decreases the current and hence the brightness of the bulb.

Question 23.
A magnet is moved in the direction indicated by an arrow between two coils AB and CD as shown in the figure. Suggest the direction of current in each coil,
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 4
Answer:
In coil AB induced current flows from A to B and in colt CD current flows from C to D.

Question 24.
Predict the directions of induced currents in metal rings 1 and 2 lying in the same place where current I in the wire is increasing steadily. (CBSE Delhi 2012)
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 5
Answer:
1 -clockwise, 2-anticlockwise.

Question 25.
The electric current flowing in a wire In the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept above the wire as shown. (CBSE AI 2014)
Answer:
Clockwise.

Question 26.
Predict the polarity of plate A of the capacitor, when a magnet is moved towards it, as is shown In the figure. (CBSE AI 2014C)
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 6
Answer:
Positive.

Question 27.
The figure below shows two positions of a loop PQR in a perpendicular uniform magnetic field. In which position of the coil is there induced emf?
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 7
Answer:
In position b.

Question 28.
If the self-inductance of an air-core inductor increases from 0.01 mH to 10 mH on introducing an iron core into it, what is the relative permeability of the core used?
Answer:
We know that µ = \(\frac{L}{L_{0}}=\frac{10}{0.01}\) = 1000

Electromagnetic Induction Important Extra Questions Short Answer Type

Question 1.
An induced current has no direction of its own, comment.
Answer:
Yes, it is perfectly correct to say that an induced current has no fixed direction of its own. The direction of induced current depends upon the change in magnetic flux because in accordance with Lenz’s law the induced current always opposes the change in magnetic flux.

Question 2.
How are eddy currents produced? Mention two applications of eddy currents?
Answer:
Eddy currents are the currents induced in the body of a thick conductor when the magnetic flux linked with the conductor changes. When a thick conductor is moved in a magnetic field, magnetic flux linked with it changes. In situations like these, we can have induced currents that circulate throughout the volume of a material.

Because their flow patterns resemble swirling eddies in a river, therefore they are called eddy currents.

  • Electromagnetic braking, and
  • Induction furnace.

Question 3.
Name and define the unit used for measuring the coefficient of mutual inductance. State the relation of this unit with the units of magnetic flux and electric current.
Answer:
In SI the unit of mutual inductance is henry (H). Now from the expression
ε = – \(\frac{d \phi}{d t}\) = – M \(\frac{d l}{d t}\)
we have M = ε l \(\frac{d l}{d t}\).

Let ε = 1 volt and dl/dt = 1 As-1, then
M = 1 volt/1 As-1 = 1 henry.

The mutual-inductance of a coil is said to be 1 henry if a rate of change of current of 1 ampere per sec in the neighbouring coil induces in at an emf of 1 volt.

Question 4.
What are eddy currents? Write any two applications of eddy currents. (CBSE A! 2011)
Answer:
Eddy currents are the currents induced in the body of a thick conductor when the magnetic flux linked with a bulk piece of conductor changes.

  1. Dead Beat Galvanometer, and
  2. Induction furnace.

Question 5.
(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of the magnetic field B, area A and length l of the solenoid.
(b) How is this magnetic energy per unit volume compared with the electrostatic energy per unit volume stored in a parallel plate capacitor? (CBSE Delhi 2011C)
Answer:
The magnetic field stored in a solenoid is given by the expression U = – \(\frac{1}{2}\)Ll2.

But for a solenoid B = μ0nl
or
l = B / μ0 n

Substituting in the above expression we have
U = \(\frac{1}{2}\) × (μ0n2Al)\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) as L = μ0 n2 A l

U = \(\frac{1}{2}\)\(\frac{B^{2} A l}{\mu_{0}}\)

We know that the energy stored per unit volume in a parallel plate capacitor is
UE = \(\frac{1}{2}\)ε0E2

It is clear that in both cases the energy stored per unit volume is proportional to the square of the field intensity.

Question 6.
State Lenz’s Law.
A metallic rod held horizontally along the east-west direction is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer. (CBSE Delhi 2013)
Answer:
Lenz’s law states that the polarity of the induced emf is such that it tends to oppose the cause of its production.

Yes, as it will cut the horizontal component of the earth’s magnetic field.

Question 7.
Starting from the expression for the energy W = 1/2Ll2, stored in a solenoid of self-inductance L to build up the current l, obtain the expression for the magnetic energy in terms of the magnetic field B, area A and length l of the solenoid having n number of turns per unit length. Hence show that the energy density is given by 8z/2m0. (CBSE Delhi 2013C)
(i) The magnetic energy is
U = \(\frac{1}{2}\)LI2 = \(\frac{1}{2}\)L\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) since B = μ0nl

Now L = μ0n2 Al, therefore we have
UB = \(\frac{1}{2}\)(μ0n2 Al)\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) = \(\frac{1}{2 \mu_{0}} B^{2} A l\)

(ii) The magnetic energy per unit volume is
UB = \(\frac{U_{B}}{V}=\frac{U_{B}}{A l}=\frac{B^{2}}{2 \mu_{0}}\)

Question 8.
Define mutual inductance. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? (CBSE Delhi 2016)
Answer:
Mutual inductance is numerically equal to the magnetic flux linked with a coil when the unit current passes through the neighbouring coil.
Given M = 1.5 H, dl = 20 – 0 = 20 A,
dt = 0.5 s, Φ = ?
Φ = – M\(\frac{d l}{d t}\)
or
Φ = – 1.5 × \(\frac{20}{0.5}\) = – 60 Wb

Question 9.
Explain the principle on which the metal detector is used at airports for security reasons.
Answer:
The metal detectors used in airport security checkpoints operate by detecting eddy currents induced in metallic objects. The detector generates an alternating magnetic field. This induces eddy currents in the conduction object carried through the detector. The eddy currents in turn produce an alternating magnetic field. This field induces a current in the detectors receiver coil.

Question 10.
A current is induced in coil C1, due to the motion of current-carrying coil C2.
(i) Write any two ways by which a large deflection can be obtained in the galvanometer G.
(ii) Suggest an alternative device to demonstrate the induced current in place of a galvanometer. (CBSE Delhi 2011)
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 8
Answer:
(ii) The two ways are
(a) Passing a large current through coil C2 and
(b) Moving coil C2 quickly towards the coil.
(ii) A magnetic compass can be placed at the centre of coil C1. Whenever current will be induced it will show a deflection.

Question 11.
Consider a cube EFGHIJKL of side ‘a’ placed in a magnetic field B acting perpendicular to the face FJKG as shown in the figure. Write magnetic flux through the following faces (a) EFGH (b) EFJI (c) EILH (d) FJKG
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 9
Answer:
The magnetic flux depends upon the angle (q) between the magnetic field and area vector.
(a) Here θ = 90°, therefore magnetic flux Φ = BA cos θ = BA cos 90° = 0
(b) Here θ = 90°, therefore magnetic flux Φ = BA cos θ = BA cos 90° = 0
(c) Here θ = 180°, therefore magnetic flux Φ = BA cos 180° = -BA
(d) Here θ = 0°, therefore magnetic flux Φ = BA cos 0° = BA

Question 12.
A metallic rod of ‘L’ length is rotated with an angular frequency of ‘ω’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. (CBSE Delhi 2012, 2013)
Answer:
Let the rod move from OP to OQ through a small sector of angle θ.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 10
The small area covered by the rod is
dA= πL2 × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)L2θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is
ε = \(\frac{d \phi}{d t}=\frac{d B A}{d t}=\frac{B d A}{d t}=B \frac{d}{d t}\left[\frac{1}{2} L^{2} \theta\right]\)

ε = – \(\frac{1}{2}\)L2B\(\frac{d \theta}{d t}=\frac{1}{2}\)BωL2

Question 13.
A metallic rod of length l length is rotated with a frequency ω, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius R = l, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf induced in the rod. If r is the resistance of the rod and the metallic ring has negligible resistance, obtain the expression for the power generated. (CBSE AI 2013C)
Answer:
For the first part refer above question.
The power generated is given by
P = \(\frac{\varepsilon^{2}}{r}=\frac{B^{2} \omega^{2} l^{4}}{2 r}\)

Question 14.
A rectangular loop PQMN with movable ‘ arm PQ of length 10 cm and resistance 4 Ω is placed in a uniform magnetic field of 0.25 T acting perpendicular to the plane of the loop as is shown in the figure. The resistances of the arms MN, NP and MQare negligible.
Calculate the
(i) emf induced in the arm PQand
(ii) current induced in the loop when arm PQ is moved with velocity 20 m s-1. (CBSE Delhi 2014C)
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 11
Answer:
Given L = 10 cm = 0.1 m, B = 0.25 T, v = 20 m s-1
(a) ε = BLv = 0.25 × 0.1 × 20 = 0.5 V
(b) l = ε/R = 0.5/4 = 0.125 A

Question 15.
In the given diagram a coil B is connected to a low voltage bulb L and placed parallel to another coil A as shown. Explain the following observations
(i) Bulb lights, and
(ii) Bulb gets dimmer if coil B is moved upwards.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 12
Answer:
(a) When ac is applied across coil A an induced emf is produced in coil B due to mutual induction between the two coils. This makes the lamp light up.
(b) When coil B is moved upwards the mutual induction and hence induced emf in coil B decreases. This makes the lamp dimmer.

Question 16.
A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.
(a) Write the expression for the instantaneous value of the emf induced in the wire.
Answer:
ε = BLv

(b) What is the direction of the emf?
Answer:
west to east

(c) Which end of the wire is at the higher potential? (CBSEAI2011)
Answer:
East.

Question 17.
Two concentric circular coils one of small radius r1 and the other of large radius r2, such that r1 << r2, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. (NCERT)
Answer:
Let a current l2 flow through the outer circular coil. The field at the centre of the coil is B2 = \(\frac{\mu_{0} I_{2}}{2 r_{2}}\). Since the other coaxially placed coil has a very small radius, B2 may be considered constant over its cross¬sectional area. Hence,
Φ = πr1²B2 = \(\frac{\mu_{0} \pi r_{1}^{2} l_{2}}{2 r_{2}}\) = M12l2

Thus M12 = \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)

But M12 = M21, therefore M12 = M21 = \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)

Note that we calculated M12 from an approximate value of Φ1 assuming the magnetic field B2 to be uniform over the area πr1². However, we can accept this value because r1 << r2.

Question 18.
Consider a magnet surrounded by a wire with an on/off switch S (figure). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain. (NCERT Exemplar)
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 13
Answer:
There is no relative motion between the magnet and the coil. This means that there is no change in magnetic flux, hence no electromotive force is produced and hence no current will flow in the circuit.

Question 19.
A wire in the form of a tightly wound solenoid is connected to a DC source and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will increase. As the wires are pulled apart the flux will leak through the gaps. Lenz’s law demands that induced e.m.f. resist this decrease, which can be done by an increase in current.

Question 20.
A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will decrease. As the iron core is inserted in the solenoid, the magnetic field increases and the flux increases. Lenz’s law implies that induced e.m.f. should resist this increase, which can be achieved by a decrease in current.

Electromagnetic Induction Important Extra Questions Long Answer Type

Question 1.
11 kW of electric power can be transmitted to a distant station at (i) 220 V or (ii) 22,000 V. Which of the two modes of transmission should be preferred and why? Support your answer with possible calculations.
Answer:
1. Consider that 11000 watt of energy has to be transmitted. First at 220 V and then at 22000 V. When the power is transmitted at 220 V then the current flowing through the wires is 11000/220 = 50 A

2. When power is transmitted at 22000 V then the current through the wires is 11000 / 22000 = 0.5 A. If R is the resistance of the line wire then the energy dissipated in the two cases is 2500R joule per sec and 0.25R. joule per sec.

This shows that if energy is transmitted at low voltages there is more loss in energy than when it is transmitted at high voltages. Furthermore, if power is to be transmitted at low voltage then the resistance of the line wire should be low, as such thick wires will be required. To support these thick wires strong poles situated close to each other will be needed. This will increase the cost of transmission. But at high voltages, even thin wires will do.

Question 2.
A coil A is connected to a voltmeter V and the other coil B to an alternating current source D. If a large copper sheet C is placed between the two coils, how does the induced emf in coil A change due to current in coil B. Justify your answer.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 14
Answer:
In the absence of sheet C, an induced emf is set up in coil due to mutual induction phenomenon when an alternating current is passed through coil B.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 15
However, when induced copper sheet C is placed, eddy currents are set up in the sheet due to a change in flux.

Thus, now coil A has a positive effect due to coil B and a negative effect due to eddy currents in C. Consequently, the flux of coil A and hence the induced emf in coil A is decreased, i.e. the reading of voltmeter V is reduced.

Question 3.
A bar magnet is dropped so that it falls vertically through coil C. The graph obtained for the voltage produced across the coil versus time is as shown in figure (b).
(i) Explain the shape of the graph and
(ii) why is the negative peak longer than the positive peak?
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 16
Answer:
(a) As the magnet approaches the coil, an emf is induced in it. As the magnet approaches the coil the magnetic flux linked with the coil increases. As a result the induced emf increases. When the magnet enters the coil, the change in magnetic flux linked with the coil begins to decrease and becomes zero when the magnet is completely inside the coil. This starts decreasing the emf and makes it zero.

When the magnet comes out of the coil the direction of induced emf changes direction and begins to increase in the opposite direction. When the magnet moves far away from the coil the induced emf becomes zero.

As the magnet comes out of the coil with a speed greater than the speed at which it approaches the coil, therefore the induced emf is more in the second case. Hence the longer negative peak.

Question 4.
(a) State Lenz’s law: Using this law indicate the direction of the current in a closed loop when a bar magnet with the North Pole is brought close to it. Explain briefly how the direction of the current predicted wrongly results in the violation of the law of conservation of energy.
(b) A rectangular loop and circular loop are moving out of a uniform magnetic field region with a constant velocity v as shown in the figure. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops. (CBSE AI 2011C)
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 17
Answer:
(a) Lenz’s law states that the direction of the induced emf is such that it opposes the cause of its production. When the north pole of the magnet is brought near the coil, the upper end of the coil will acquire north polarity so as to oppose the approaching North Pole (by repelling). This means that the direction of current must be anticlockwise as seen from the side of the magnet.

If the current is wrongly predicted as clockwise, then the upper face will acquire south polarity which will attract the North Pole. This means the current is being produced without doing any work. This leads to the violation of the law of conservation of energy.

(b) It is expected that the induced emf will be constant in the rectangular coil. In the case of the rectangular coil, when pulled out of the magnetic field, the rate of change of magnetic flux will be constant because the rate of change of area is constant. This is not so in the case of the circular coil.

Question 5.
The current through two inductors of self-inductance 15 mH and 25 mH is increasing with time at the same rate. Draw graphs showing the variation of the
(a) emf induced with the rate of change of current
(b) energy stored in each inductor with the current flowing through it. Compare the energy stored in the coils, if the power dissipated in the coils is the same. (CBSE AI 2017C)
Answer:
Given L1 = 15 mH, L2 = 25 mH.
(a) The emf induced across an inductor is given by the expression ε = – L\(\frac{d i}{d t}\) since di/dt is the same for both coils therefore induced emf will depend upon the value of inductance. Hence the graphs are as shown
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 18
(b) The energy stored in an inductor is given by U = \(\frac{1}{2}\)Ll2
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 19
(c) The power dissipated P = l2XL
or
l2 = Pl XL and U = \(\frac{1}{2}\)Ll<sup>1</sup> = \(\frac{1}{2} \frac{L P}{X_{L}}=\frac{1}{2} \frac{P L}{\omega L}=\frac{P}{2 \omega}\)

Since power dissipated is same and co is also same therefore energy stored in the coils will also be the same.

Question 6.
A rectangular frame of wire is placed in a uniform magnetic field directed outwards, normal to the paper. AB is connected to a spring which is stretched to AB and then released at time t = 0. Explain qualitatively how induced e.m.f. in the coil would vary with time. (Neglect damping of oscillations of spring) (CBSE AI 2018C)
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 20
Answer:
As the spring is released AB is pulled out of the field. This increases the area of the loop inside the magnetic field. This increases flux and hence an induced emf is produced. The portion AB does not stop at Ab but moves outwards. Now the spring will push AB inwards. This will decrease the area of the loop thereby decreasing the induced emf. This continues and hence the emf increases and decreases periodically.

Question 7.
Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of the same length wound one over the other. (CBSE AI 2017)
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S1 and S2 with S2 wound over S1.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 21
Let l = length of each solenoid r1, r2 = radii of the two solenoids

A = πr1² = area of cross-section of inner solenoid S1
N1, N2 = number of turns in the two solenoids First we pass a time-varying current l2 through S2. The magnet field set up inside S2 due to l2 is
B2 = μo n2 l2

where n2 = N2/l = the number of turns per unit length of S2.

Total magnetic flux linked with the inner solenoid S1 is
Φ1 = B2AN1 = μ0 n2 l2 × AN1

Mutual inductance of coil 1 with respect to coil 2 is
M12 = \(\frac{\phi_{1}}{l_{2}}\) = μ0n2AN1 = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N1 N2 of their number of turns. This is termed the reciprocity theorem.

Question 8.
Define mutual inductance and write its SI unit.
A square loop of side ‘a’ carrying a current l2 is kept at distance x from an infinitely long straight wire carrying a current l1 as shown in the figure. Obtain the expression for the resultant force acting on the loop. (CBSE Delhi 2019)
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 22
Answer:
(a) Mutual inductance equals the magnetic flux associated with a coil when unit current flows in its neighbouring coil. (b) Force per unit length between two parallel straight conductors
\(\frac{F}{L}=\frac{\mu_{o} I_{1} I_{2}}{2 \pi r}\)

Force on the part of the loop which is parallel to infinite straight wire and at a distance x from it F = \(\frac{\mu_{o} l_{1} l_{2} a}{2 \pi x}\)(away from the infinitely 2KX long wire)

Force on the part of the loop which is at a distance (x + a) from it
F2 = \(\frac{\mu_{0} I_{1} I_{2} a}{2 \pi(x+a)}\) (towards the infinite long wire)

Net force F = F1 – F2
F = \(\frac{\mu_{o} l_{1} l_{2} a^{2}}{2 \pi(x+a)}\) (away from the infinite straight line)

Question 9.
What are the possible causes of energy loss in a transformer? How are these minimised?
Answer:
The possible causes of energy losses in transformers are:
(a) Flux leakage: There is always some flux leakage. It can be reduced by winding the primary and secondary coils one over the other.
(b) Copper: The copper wires used for the windings have some resistance and hence some energy is lost due to heat produced in the wire. It can be minimised by taking thick wire.
(c) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core, which results in loss of electrical energy. To minimise it we use a laminated iron core.
(d) Hysteresis loss: As the magnetisation cycle of the iron core is repeated again and again some loss of energy takes place due to magnetic hysteresis. To minimise it we prefer a soft iron core for which hysteresis loss is less.

Question 10.
Discuss how Faraday’s law of e.m.f induction is applied in an ac generator for converting mechanical energy into electrical energy. Draw graphs to show the ’phase relationship’ between the instantaneous
(a) magnetic flux (Φ) linked with the coil and
(b) induced emf (ε) in the coil. (CBSE Delhi 2016C)
Answer:
In an ac generator, the change in magnetic flux is brought about by rotating the coil in a magnetic field. According to Faraday’s law, induced emf is set up in the coil on changing the magnetic flux linked with it. Hence mechanical energy, which is supplied to rotate the coil, gets converted into electrical energy.

Let the coil be rotated with a constant angular speed, co, the angle 0 between magnetic field 8, and area vector A of the coil at any instant is θ = ωt

Therefore magnetic flux at any instant is Φ = BA cos ωt

From Faraday’s law, induced emf is
ε = – n\(\frac{d \phi}{d t}\) = -nBA\(\frac{d}{d t}\)cos ωt
ε = nBAa sin ωt

Magnetic flux is given by Φ = BA cos ωt, therefore the graph will be a cosine curve as shown
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 23
The induced emf is given by ε = nBA sin ωt, therefore the graph will be a sine curve as shown.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 24
Question 11.
(a) Define self-inductance of a coil. Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current through it grows from zero to l.
(b) A square loop MNOP of the side 20 cm is placed horizontally in a uniform magnetic field acting vertically downwards as shown in the figure. The loop is pulled with a constant velocity of 20 cm s-1 till it goes out of the field.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 25
(c) Depict the direction of the induced current in the loop as it goes out of the field. For how long would the current in the loop persist?
(d) Plot a graph showing the variation of magnetic flux and induced emf as a function of time. (CBSE AI 2015)
Answer:
(a) Let i be the current at some instant through a pure inductor. If di / dt is the rate of change of current through the inductor then the voltage between the terminals of the inductor at this instant is
V=L di / dt

Therefore instantaneous power in the inductor is
P = Vab i = Li \(\frac{d i}{d t}\)

The energy dU supplied to the inductor during an infinitesimal time interval dt is
dU = P dt, so dU = Li di

The total amount of energy supplied while the current increases from zero to a final value l is
U = L ∫o1 i di = \(\frac{1}{2}\)L l2

(b) (i) Direction of induced current – clockwise (MNOP), the duration of induced current is 1s.
(ii) The graph is as shown.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 26
Question 12.
(a) Define mutual inductance and write its SI units.
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil. It is measured in henry.

(b) Derive an expression for the mutual inductance of two long co-axial solenoids of the same length wound one over the other.
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S1 and S2 with S2 wound over S1.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 21
Let l = length of each solenoid r1, r2 = radii of the two solenoids

A = πr1² = area of cross-section of inner solenoid S1
N1, N2 = number of turns in the two solenoids First we pass a time-varying current l2 through S2. The magnet field set up inside S2 due to l2 is
B2 = μo n2 l2

where n2 = N2/l = the number of turns per unit length of S2.

Total magnetic flux linked with the inner solenoid S1 is
Φ1 = B2AN1 = μo n2 l2 × AN1

Mutual inductance of coil 1 with respect to coil 2 is
M12 = \(\frac{\phi_{1}}{l_{2}}\) = μ0n2AN1 = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N1 N2 of their number of turns. This is termed the reciprocity theorem.

(c) In an experiment, two coils c1 and c2 are placed close to each other. Find out the expression for the emf induced in the coil c1 due to a change in the current through the coil c2. (CBSE Delhi 2015)
Answer:
Consider the experimental set up as shown.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 27
When the key K is pressed in coil C1 a magnetic flux linked with C2 changes.

The magnetic flux linked with coil C2 is given by
f2 = M l1 where M is the mutual inductance between the two coils.

But by Faraday’s flux rule we have
ε2 = – \(\frac{d \phi}{d t}\) therefore we have
ε2 = – \(\frac{d M l_{1}}{d t}\) = – M \(\frac{d l_{1}}{d t}\)

Question 13.
(a) When a bar magnet is pushed towards (or away) from the coil connected to a galvanometer, the pointer in the galvanometer deflects. Identify the phenomenon causing this deflection and write the factors on which the amount and direction of the deflection depend. State the laws describing this phenomenon.
(b) Sketch the change In flux, emf and force when a conducting rod PQ of resistance R and length I moves freely to and fro between A and C with speed v on a rectangular conductor placed in the uniform magnetic field as shown in the figure. (CBSE AI 2016)
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 28
Answer:
(a) Phenomenon: electromagnetic induction

Factors:

  • Strength of the magnetic field of the magnet
  • speed of motion of bar magnet.

Direction depends upon

  • the motion of magnet whether inward or outward
  • the direction of the north/south pole.

Law: The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

(b) The sketch is as shown below.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 29
Question 14.
What is induced emf? Write Faraday’s laws of electromagnetic induction. Express it mathematically. A conducting rod of length ‘L’ with one end pivoted is rotated with a uniform angular velocity ‘ω’ in a vertical plane, normal to a uniform magnetic field ‘B’. Deduce an expression for the emf induced in this rod.
Answer:
It is the emf induced when the magnetic flux linked with a coil changes.

Faraday put forward the following laws called Faraday’s laws of electromagnetic induction.

  • Law 1. Whenever the magnetic flux linked with a coil changes an induced emf is produced.
  • Law 2. The induced emf lasts as long as the change in magnetic flux continues.

The expression for induced emf in rotating rod: Let the rod move from OP to OQ through a small sector of angle θ.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 30
The small area covered by the rod is
dA = πL2 × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)L2θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is
ε = \(\frac{d \phi}{d t}=\frac{d B A}{d t}=\frac{B d A}{d t}=B \frac{d}{d t}\left[\frac{1}{2} L^{2} \theta\right]\)
Or
ε = \(\frac{1}{2} L^{2} B \frac{d \theta}{d t}=\frac{1}{2}\)BωL2

Question 15.
How is the mutual inductance of a pair of coils affected when:
(a) the separation between the coils is increased?
Answer:
When the separation between the two coils is increased the mutual inductance decreases. It is because the flux linked with the secondary due to a current in the primary decreases.

(b) the number of turns of each coil is increased?
Answer:
When the number of turns of each coil is increased the mutual inductance increases because M12 (= M21) ∝ N1N2,

i.e. mutual inductance is directly proportional to the number of turns. The linkage of flux increases with an increase in the number of turns in the coils.

(c) a thin iron sheet is placed between the two coils, other factors remaining the same? Explain your answer in each case.
Answer:
When a thin iron sheet is placed between the coils the permeability of the medium between the coil increases. As mutual inductance is directly proportional to the permeability, therefore the mutual inductance increases.

Question 16.
A rectangular coil of area A, having a number of turns N, is rotated at f revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2π f N B A.
Answer:
Consider an armature of the ac generator having n turns and placed in a uniform magnetic field B.

Suppose at any instant t the normal to the plane of the coil makes an angle 0 with the direction of the magnetic field. If ro is the uniform angular velocity with which, the coil rotates then θ = ωt.

The flux through the loop equals its area A multiplied by B⊥ = Bcos θ, the component of magnetic field B perpendicular to the area, hence

Φ = n B A cos Φ = n B A cos ω t …(1)
where is the number of turns in the armature?

By Faraday’s flux rule,
ε = – \(\frac{d \phi}{d t}=-\frac{d}{d t}\) n B A cos ω t
= – n B A \(\frac{d}{d \phi}\)cos ω t = – n B A (- ω sin ω t) ….(2)
Or
ε = n B A ω sin ω t

The induced emf is maximum when sin ωt = maximum = 1, therefore the maximum induced emf is given by
ε0 = n B A ω sin ω t …(3)

Since ω = 2πf, therefore equation (3) becomes ε0 = 2πf nBA

Question 17.
State Lenz’s law. Explain by giving examples that Lenz’s law is a consequence of the law of conservation of energy. (CBSE Delhi 2017C)
Answer:
It states that the direction of induced emf in a coil is such that it opposes the cause of its production. Lenz’s law is a consequence of the law of conservation of energy. To show it, let us consider a bar magnet pushed towards a conducting loop. When N-pole moves towards the loop, the face of the loop facing the North Pole develops north polarity as per Lenz’s law so as to oppose the motion of the magnet.

Again, when N-pole moves away from the loop, the nearby face develops south polarity, thus opposing the motion of the magnet away from the loop. It means that the motion of the magnet is automatically opposed every time. Hence, some work is to be done on the magnet to move it and this mechanical work is transformed into electrical energy. Thus, the conservation law of energy is followed.

Question 18.
Define the term ‘self-inductance’ and write its SI unit.
Obtain the expression for the mutual inductance of two long co-axial solenoids S1 and S2 wound one over the other, each of length L and radii r1 and r2 and n1 and n2 number of turns per unit length when a current I is set up in the outer solenoid S2. (CBSE Delhi 2017)
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S1 and S2 with S2 wound over S1.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 21
Let l = length of each solenoid r1, r2 = radii of the two solenoids

A = πr1² = area of cross-section of inner solenoid S1
N1, N2 = number of turns in the two solenoids First we pass a time-varying current l2 through S2. The magnet field set up inside S2 due to l2 is
B2 = μo n2 l2

where n2 = N2/l = the number of turns per unit length of S2.

Total magnetic flux linked with the inner solenoid S1 is
Φ1 = B2AN1 = μo n2 l2 × AN1

Mutual inductance of coil 1 with respect to coil 2 is
M12 = \(\frac{\phi_{1}}{l_{2}}\) = μon2AN1 = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N1 N2 of their number of turns. This is termed the reciprocity theorem.

Question 19.
Obtain the expression for the magnetic energy stored in an ideal inductor of self-inductance L when a current l passes through it. Hence obtain the expression for the energy density of magnetic field B produced in the inductor. (CBSE Delhi 2016C)
Answer:
Let i be the current at some instant through a pure inductor. If di / dt is the rate of change of current through the inductor then the voltage between the terminals of the inductor at this instant is
V=L di / dt

Therefore instantaneous power in the inductor is
P = Vab i = Li\(\frac{d i}{d t}\)

The energy di supplied to the inductor during an infinitesimal time interval dt is
dU = P dt, so dU = Li di

The total amount of energy supplied while the current increases from zero to a final value l is

U = L ∫01 di = \(\frac{1}{2}\)L l2

This gives the expression for the energy stored in an inductor.

The magnetic energy is
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 31
The magnetic energy per unit volume is
UB = \(\frac{U_{B}}{V}=\frac{U_{B}}{A l}=\frac{B^{2}}{2 \mu_{0}}\)

Question 20.
Derive the expression for the self-inductance of a solenoid.
Answer:
Consider a uniformly wound solenoid with N turns and length l. Let the length of the solenoid be large as compared to its radius and let the core of the solenoid have air. Due to this, we can take the interior field uniform. Therefore the magnetic field in the interior of the solenoid is given by

B = μ0 n l = μ0 \(\frac{N}{l}\) l …(1)

where n is the number of turns per unit length. The flux through each turn is given by
Φm = B A = μ0 \(\frac{N A}{l}\) l …(2)

where A Is the area of cross-section of the solenoid. But
L = \(\frac{N \Phi_{\mathrm{m}}}{I}\) …(3)

Therefore from equations (2) and (3) we have
L = \(\frac{N \Phi_{m}}{l}=\frac{\mu_{0} N^{2} A}{l}\) ….(4)

This shows that L depends on the geometric factors and is proportional to the square of the number of turns. Since N N = n l, we can therefore express the above result as
L = µ0\(\frac{(n l)^{2}}{l}\)A = µ0 n2 A l …(5)

Question 21.
(a) State Faraday’s laws of electromagnetic Induction.
Answer:
Faraday’s Laws of electromagnetic induction
First law: Whenever the magnetic flux (inked with a circuit (or coil) changes, an emf is induced in the circuit. The induced emf Lasts so Long as the change in the magnetic flux continues.

Second law:
The magnitude of induced emf in the circuit (or coil) is directly proportional to the rate of change of magnetic flux linked with the circuit.
i.e. e = – \(\frac{d \phi}{d t}\)

(b) Derive an expression for the emf induced across the ends of a straight conductor of length I moving at right angles to a uniform magnetic field B with a uniform speedy.
Answer:
Let a conductor ab of Length l be placed in a magnetic field \(\vec{B}\) shown by (x) directed towards the reader. When the conductor moves with velocity v perpendicular to B, the force on any free electron of this conductor is given
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 32
F = e vB sin 90° = evß

When this electron moves across the Length of the conductor, the work done by the electron
W = F × l = evBl

After some time when all the free electrons are shifted towards the end b, the end b becomes negative and end a becomes positively charged.

Hence the net induced emf in the conductor
e = \(\frac{W}{q}=\frac{e v B l}{e}=v B l\)
e = Blv

(c) Obtain the expression for the magnetic energy stored in a solenoid in terms of the magnetic field B, area A and length I of the solenoid through which a current j is passed. (CBSE 2019C)
Answer:
Energy stored in an inductor Considers an inductor of inductance L. Let I will be instantaneous current in the inductor and dl/dt be the rate of growth of current.

Induced emf, E = L\(\frac{dl}{dt}\) (in magnitude)

If the source sends a charge dq in time dt, then
dq = ldt

Small amount of work done by the source
dW = edq = eldt = L\(\frac{dl}{dt}\). ldt
or
dW = Ll dl

Total work was done during the growth of current in an Inductor by the external source
W = ∫dW =L∫01 ldl
= L \(\left[\frac{l^{2}}{2}\right]_{0}=\frac{1}{2}\)Ll2
W = \(\frac{1}{2}\) Ll2 …(1)

This work done is stored in the form of magnetic energy.
In a solenoid,
L = μ0n2Al
And B = μ0nl
∴ l = \(\frac{B}{\mu_{0} n}\)

So Eq. (1) becomes
W = \(\frac{1}{2}\)μ0n2Al\(\left(\frac{B}{\mu_{0} n}\right)^{2}=\frac{B^{2} A l}{2 \mu_{0}}\)

Question 22.
(a) A metallic rod of length ‘l’ and resistance ‘R’ is rotated with a frequency ‘v’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius ‘l’, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field ‘B’ parallel to the axis is present everywhere.
(i) Derive the expression for the induced emf and the current in the rod.
(ii) Due to the presence of current in the rod and of the magnetic field, find the expression for the magnitude and direction of the force acting on this rod.
(iii) Hence, obtain an expression for the power required to rotate the rod.
(b) A copper coil l is taken out of a magnetic field with a fixed velocity. Will it be easy to remove it from the same field if its ohmic resistance is increased?
OR
(a) A rectangular coil rotates in a uniform magnetic field. Obtain an expression for induced emf and current at any instant. Also find their peak values. Show the variation of induced emf versus angle of rotation (ωt) on a graph.
(b) An iron bar falling through the hollow region of a thick cylindrical shell made of copper experiences a retarding force. What can you conclude about the nature of the iron bar? Explain. (CBSEAI 2019)
Answer:
(a) (/) Let the rod moves from OP to OQ through a small sector of angle θ.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 33
The small area covered by the rod is
dA = πl2 × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)l2θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 34

Now induced current is given by
i = \(\frac{\varepsilon}{R}=\frac{B \omega l^{2}}{2 R}\)

(ii) \(\vec{F}=i(\vec{l} \times \vec{B})\)
F = \(\frac{l^{3} \omega B}{2 R}\)

Direction of \(\vec{F}\) is perpendicular to both \(\vec{i}\) and \(\vec{B}\),

(iii) P = i2R
= \(\left(\frac{B l^{2} \omega}{2 R}\right)^{2}\)R = \(\frac{B^{2} l^{4} \omega^{2}}{4 R}\)

(b) Yes, since induced current will reduce, it will be a little easier to remove the coil.
Or
(a) Consider an armature of the ac generator having n turns placed in a uniform magnetic field B.

Suppose at any instant t the normal to the plane of the coil makes an angle θ with the direction of the magnetic field. If ω is the uniform angular velocity with which the coil rotates then θ = ωt.

The flux through the loop is Φ = n BA cos Φ = n BA cos ω t
where n is the number of turns in the armature.

By Faraday’s flux rule,
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 35
Now,i0 = \(\frac{\varepsilon_{0}}{R}=\frac{n B A \omega}{R}\)

The graph is as shown.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 36
(b) The bar is magnetic in nature. It experiences retardation in accordance with Lenz’s law.

Question 23.
State Lenz’s law. The energy f required to build up a steady current l, in a given coil, varies with l in the manner as shown. Calculate the self-inductance of the coil.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 37
A circular coil of radius r, is placed coaxially with another coil of radius R (R >> r) with the centre of the two coils coinciding with each other. Obtain an expression for the mutual inductance of the two coils.
Answer:
Lenz’s law states that the direction of induced emf is such that it opposes the cause of its production.
The energy stored in an inductor is given by the relation
E = – \(\frac{1}{2}\)Ll2

From the graph for a current of 200 mA the energy is 4 mJ, therefore self-inductance of the coil.
\(\frac{2 E}{l^{2}}\) = L
or
\(\frac{2 \times 4 \times 10^{-3}}{\left(200 \times 10^{-3}\right)^{2}}\) = 0.2 H
or
L = 200 mH

Consider the two coils of radius r and R placed coaxially, as shown in the figure.
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 38
Let n1 and n2 be the number of turns per unit length in them.

If a current l2 is passed through the outer coil, there will a magnetic field
Bo = μo n1 l1 produced in it.

The flux associated with the inner coil will undergo variation as the field B0 grows.
The effective magnetic flux
Φ1 = Bo × Aeff = \(\frac{\mu_{0} l_{2}}{2 R}\) × πr²

Therefore mutual inductance of the coils is
M = \(\frac{\phi_{1}}{l_{2}}=\frac{\mu_{0} \pi r^{2}}{2 R}\)

Question 24.
(a) Describe a simple experiment (or activity) to show that the polarity of emf induced in a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it.
(b) The current flowing through an inductor of self-inductance L is continuously increasing. Plot a graph showing the variation of
(i) Magnetic flux versus the current
(ii) induced emf versus dl/dt
(iii) Magnetic potential energy stored versus the current. (CBSE Delhi 2014)
Answer:
(a) Consider a coil C connected to a galvanometer. When the North Pole of a bar magnet is pushed towards the coil, the pointer in the galvanometer deflects, indicating the presence of electric current in the coil. When the magnet is pulled away from the coil, the galvanometer shows deflection in the opposite direction, which indicates a reversal of the current’s direction.

We see that the North Pole of a bar magnet is being pushed towards the closed coil. As the North Pole of the bar magnet moves towards the coil, the magnetic flux through the coil increases. Hence current is induced in the coil in such a direction that it opposes the increase in flux.

This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet. Similarly, if the North Pole of the magnet is being withdrawn from the coil, the magnetic flux through the coil will decrease. To counter this decrease in magnetic flux, the induced current in the coil flows in a clockwise direction and the South Pole faces the receding North Pole of the bar magnet. This would result in an attractive force that opposes the motion of the magnet and the corresponding decrease in flux.

(b) The graphs are as shown.
(i) Φ = Ll
Therefore the graph is
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 39
(ii) ε = – L\(\frac{dl}{dt}\)
Therefore the graph is
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 40
(iii) U = \(\frac{1}{2}\)Ll2
Therefore the graph is
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 41
Numerical Problems:

Formulae for solving numerical problems

  • Induced emf ε = –\(\frac{d \phi}{d t}\)
    or
    ε = \(\frac{-\left(\phi_{2}-\phi_{1}\right)}{t_{2}-t_{1}}\)
  • Emf induced in a rod of Length L is ε = Blv.
  • When a conducting rod of length L kept perpendicular to a uniform magnetic field B is rotated about one of its ends with uniform angular velocity, the emf induced between its ends has a magnitude
    \(\frac{1}{2}\)BωL2 = BπvL2.
  • When a current in a coil changes, it induces a back emf in the same coil. The self-induced emf is given by
    ε = – \(\frac{d \phi}{d t}\) = – L \(\frac{d l}{d t}\) inductance of the coil. It is a measure of the inertia of the coil against the change of current through it. Also Φ = L l
  • ε = – M\(\frac{d l}{d t}\) . Also Φ = Ml for mutual induction.
  • \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}=\frac{l_{p}}{l_{s}}\) = k.
  • Average power η = \(\frac{\varepsilon_{s} l_{s}}{\varepsilon_{p} l_{p}}\)

Question 1.
A square loop of side 10 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s-1 in the positive X-direction containing a magnetic field in the positive Z-direction. The field is non-uniform and has a gradient of 10-3 T cm-1 along the negative X-direction (i.e. it increases by 10-3 T cm-1 as one move in the negative X-direction). Calculate the emf induced. (CBSE 2019C)
Answer:
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 42
Question 2.
In a ceiling fan, each blade rotates in a circle of radius 0.5 m. If the fan makes 2 rotations per second and the vertical component of the earth’s magnetic field is 8 × 10-5 T, calculate the emf induced between the inner and outer ends of each blade. (CBSE2019C)
Answer:
Class 12 Physics Important Questions Chapter 6 Electromagnetic Induction 43
Question 3.
A circular coil of radius 10 cm, 500 turns and resistance 2 Ω are placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and the current induced in the coil. The horizontal component of the earth’s magnetic field at the place is 3.0 × 10-5 T. (NCERT)
Answer:
Initial flux through the coil
Φin = BA cos θ = 3.0 × 10-5 × π × 10-2 × cos 0°
= 3π × 10-7 Wb

Final flux after the rotation
Φf = BA cos θ = 3.0 × 10-5 × π × 10-2 × cos 180° = – 3π × 10-7 Wb

Therefore estimated value of induced emf is
ε = n\(\frac{d \phi}{d t}\) = 500 × (6π × 10-7)/ 0.25 = 3.8 × 10-3 V dt

Hence the current induced is
l = ε/R = 3.8 × 10-3 / 2 = 1.9 × 10-3 A

Question 4.
Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area 0.10 m2. The coil rotates at half a revolution per second and is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil? (NCERT)
Answer:
Given n = 100, A = 0.10 m2, B = 0.01 T, f = 0.5 rps
Using the expression ε0 = nBAω

we have ε0 = 100 × 0.01 × 0.10 × 2 × 3.14 × 0.5 = 0.314 V

Question 5.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s-1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? (NCERT)
Answer:
Given L = 8 cm = 8 × 10-2 m, b = 2 cm = 2 × 10-2 m, 8 = 0.3 T, v = 1 cm s-1 = 0.01 m s-1 ε = ?
(a) Using the expression ε = B L v
ε = 0.3 × 8 × 10-2 × 0.01 = 2.4 × 10-4 V

This emf will last till the loop comes out of the magnetic field. Since the shorter side is moving out, therefore it will take 2 s to cover 2 cm with a velocity of 1 cm s-1.

(b) Using the expression e = B L v
ε = 0.3 × 2 × 10-2 × 0.01 = 0.6 × 10-4 V

This emf will last till the loop comes out of the magnetic field. Since the longer side is moving out, therefore it will take 8 s to cover 8 cm with a velocity of 1 cm s-1.

Question 6.
A 1.0 m long metallic rod is rotated with an angular frequency 400 rad s-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. (NCERT)
Answer:
Given L = 1.0 m, ω = 400 rad s-1, B = 0.5 T,

ε = ? Using the relation ε = \(\frac{1}{2}\)BωL2

we have
ε = \(\frac{1}{2}\)0.5 × 400 × 12 =100 V

Question 7.
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s-1 in a uniform horizontal magnetic field of magnitude 3.0 × 10-2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from? (NCERT)
Answer:
Given r = 8.0 cm = 8.0 × 10-2 m, n = 20, ω = 50 rad s-1 B = 3.0 × 10-2 T, ε0 = ?, εa = ? R= 10 Ω l = ?, P = ?
Using the relation
ε0 = nBAω
= 20 × 3 × 10-2 × 3.14 × (8 × 10-2)2 × 50 = 0.603 V

The average induced emf is given by
εa = 0 over one cycle.

Also l0 = ε0 / R
= 0.603 / l0
= 0.0603 A

Now P = 1 /2 ε0 × l0 = 1 /2 × 0.603 × 0.0603
= 0.018 W

The induced current causes a torque opposing the rotation of the coil. An external agent (rotor) must supply torque to counter this torque in order to keep the coil rotating uniformly. Thus the source of the power dissipated as heat in the coil is the external rotor.

Question 8.
Current in a circuit falls from 5.0 A to 0. 0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit (NCERT)
Answer:
Given dl = 5.0 – 0.0 = 5.0 A, dt = 0.1 s, ε = 200 V, L = ?

Using the relation ε = – L \(\frac{dl}{dt}\)
or
L = \(\frac{\varepsilon}{d l / d t}\)

L = \(\frac{200}{5 / 0.1}=\frac{200 \times 0.1}{5}\) = 4 H

Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? (NCERT)
Answer:
Given dl = 20- 0.0 = 20 A, dt = 0.5 s, ε = ?,
M = 1.5 H,
Using the relation ε = – \(\frac{d \phi}{d t}\) = -M\(\frac{dl}{dt}\)
or
dΦ = Mdl = 1.5 × 20 = 30 Wb

Question 10.
A jet plane is travelling towards the west at a speed of 1800 km h-1. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the earth’s magnetic field at the location has a magnitude of 5 × 10-4 T and the dip angle is 30°? (NCERT)
Answer:
Given v= 1800 km h-1, ε = ?, L = 25 m,
B = 5 × 10-4 T, δ = 30°
Now vertical component of earth’s magnetic field is
BV = B × sin 30°
= 5 × 10-4 × 0.5
= 2.5 × 10-4T
Now using the expression ε = B L v
ε = 2.5 × 10-4 × 25 × 500 = 3.125 V

Question 11.
1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of a radius of 0.5 cm for this purpose. Calculate the fraction of ohmic losses to the power transmitted if
(a) power is transmitted at 220 V. Comment on the feasibility of doing this.
(b) a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transformer is used to bring the voltage to 220 V. ρcu = 1.7 × 10-8 SI unit. (NCERT Exemplar)
Answer:
Given P = 1 MW, 2L = 20 km = 2 × 104 m, r = 0.5 cm = 0.5 × 10-2 m
(a) Resistance of the Cu wire used
R = \(\frac{\rho L}{A}=\frac{1.7 \times 10^{-8} \times 2 \times 10^{4}}{\pi \times\left(0.5 \times 10^{-2}\right)^{2}}\) = 4 Ω

Now current at 220 V is

l = \(\frac{P}{V}=\frac{10^{6}}{220}\) = 0.45 × 104 A

Therefore power loss
= l2R = (0.45 × 104)2 × 4 = 106W

This is a huge loss; therefore this method is not feasible.

(b) Now P = Vl’ or
l’ = \(\frac{P}{V^{\prime}}=\frac{10^{6}}{11000}=\frac{1}{1.1}\) × 102 A

Hence power loss
= l’2R = \(\frac{1}{1.21}\) × 4 × 104 = 3.3 × 104 W

Hence fraction of power loss
\(\frac{3.3 \times 10^{4}}{10^{6}}\) = 3.3%

Alcohols, Phenols and Ethers Class 12 Important Extra Questions Chemistry Chapter 11

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 11 Alcohols, Phenols and Ethers. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 11 Important Extra Questions Alcohols, Phenols and Ethers

Alcohols, Phenols and Ethers Important Extra Questions Very Short Answer Type

Question 1.
Arrange the following in increasing order of their boiling point:
CH3CH2OH, CH3CHO, CH3—O—CH3 (CBSE AI 2019)
Answer:
CH3—O—CH3 < CH3CHO < CH3CH2OH

Question 2.
Arrange the following in increasing order of their acidic character:
Ethanol, Phenol, Water (CBSE AI 2019)
Answer:
Ethanol <Water < Phenol

Question 3.
How will you convert ethanol to ethene? (CBSE 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 1

Question 4.
Draw the structural formula of the 2-methylpropan-2-ol molecule. (CBSE Delhi 2012)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 2

Question 5.
Which of the following isomers is more volatile: (CBSE Delhi 2014)
o-nitrophenol or p-nitrophenol?
Answer:
o-nitrophenol

Question 6.
Write the IUPAC name of the given compound: (CBSE 2015)
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 3
Answer:
2-Methylpropane-1, 3-diol

Question 7.
Write the IUPAC name of the given compound: (CBSE Delhi 2015)
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 4
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 5 2, 5- Dinitrophenol

Question 8.
Arrange the following in increasing order of their acidic character: Benzoic acid, Phenol, Cresol (CBSE Al 2019)
Answer:
Cresol < Phenol < Benzoic acid

Question 9.
Rearrange the following compounds in the increasing order of their boiling points: (CBSE Al 2013)
CH3-CHO, CH3-CH2-OH, CH3-CH2-CH3
Answer:
CH3CH2CH3 < CH3CHO < CH3CH2OH

Question 10.
An alkoxide is a stronger base than a hydroxide ion. Justify. (CBSE Sample Paper 2012)
Answer:
Due to the presence of electron-donating alkyl group, there is high electron density in alkoxide ion as compared to hydroxide ion. Therefore, the alkoxide ion is more basic than the hydroxide ion.

Question 11.
Why is (±) butan-2-ol optically inactive? (CBSE Delhi 2013, CBSE AI 2013)
Answer:
(±)-Butan-2-ol represents a racemic mixture of (+)-butan-2-ol and (-)-2-butanol which rotate the plane polarized light in different directions but to an equal extent. Therefore, (±) compound is optically inactive.

Question 12.
Write the IUPAC name of the following compound: (CBSE Al 2017)
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 6
Answer:
2-Methoxy-2-methylpropane

Question 13.
Write the IUPAC name of the following compound: (CBSE AI 2017)
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 7
Answer:
3-Phenylprop-2-en-1-ol

Question 14.
Write the IUPAC name of the following: (CBSEAI 2018)
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 8
Answer:
3, 3-Dimethylpentan-2-ol

Alcohols, Phenols and Ethers Important Extra Questions Short Answer Type

Question 1.
Predict the major product obtained when t-butyl bromide reacts with sodium methoxide. Also, give its IUPAC name.
Or
(a) Show the chemical reaction with bond movements and arrows for the nucleophilic attack of water on carbocation in acid catalysed hydration of alkenes.
(b) Give IUPAC name for the following: (CBSE 2019C)
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 9
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 10
OR
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 11
Mechanism: hydration of alkenes
The mechanism of the reaction involves the following steps:
Step 1. Alkene gets protonated to form a carbocation by the electrophilic attack of H3O+:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 12
Step 2. Nucleophile (H20) attacks the carbocation forming protonated alcohol.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 13
Step 3. Loss of H+ from oxygen (deprotonation) to form alcohol
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 14
(b) 2,6-dimethylphenol

Question 2.
How are the following conversions carried out?
(i) Propene to propan-2-ol
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 15

(ii) Ethyl magnesium chloride to propan-1-ol (CBSE Delhi 2010)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 16

Question 3.
How would you obtain:
(i) Picric add (2, 4, 6-trlnitrophenol) from phenol.
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 17
(ii) 2-Methyl propene from 2-methyl propanol? (CBSE Delhi 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 18

Question 4.
Write the chemical equations involved in the following reactions:
(i) Kolbe’s reaction
(ii) Friedel-Crafts acetylation of anisole (CBSE Delhi 2016)
OR
How do you convert:
(i) Phenol to toluene
(ii) Formaldehyde to ethanol?
Answer:
(i) Kolbe’s reaction. Sodium phenoxide reacts with CO2 under pressure (6-7 atm) at 400 K to form sodium salicylate which upon acidification with HCL gives salicylic acid.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 19
(ii) Friedel-Crafts acetylation of anisole. On reacting anisole with acetyl chloride in the presence of anhyd. AlCl3, 2-methoxy acetophenone and 4-methoxy acetophenone are formed.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 20

(i) Phenol to toluene
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 21

(ii) Formaldehyde to ethanol
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 22

Question 5.
Explain the mechanism of the following reaction: (CBSE Delhi 2013, 2016)
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 23
Answer:
The mechanism for the formation of diethyl ether from ethanol at 413 K is given below:
(i) Ethyl alcohol gets protonated in the presence of H+.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 25
(ii) Nucleophilic attack by another alcohol (unprotonated) molecule occurs on the protonated alcohol with the elimination of a molecule of water.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 26

(iii) Oxonium ion [oses a proton to form an ether.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 27

Question 6.
How will you convert:
(i) Propene to Propan-2-ol?
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 28
(ii) Phenol to 2, 4, 6 – trinitrophenol? (CBSE Delhi 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 29
Question 7.
How will you convert:
(i) Propene to propan-1 -ol?
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 30

(ii)Ethanal to propan-2-ol?
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 31
Question 8.
Give chemical tests to distinguish between (CBSE Sample Paper 2011)
(i) Isopropyl alcohol and n-propyl alcohol
Answer:
Isopropyl alcohol gives the iodoform test. On heating with NaOH/l2 or NaOl, isopropyl alcohol forms a yellow precipitate of iodoform (CHI3).
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 32
(ii) Phenol and alcohol.
Answer:
Phenol reacts with neutral FeCl3 solution to give red-violet complex, whereas alcohol does not give this test.

Question 9.
(i) Arrange the following compounds in the increasing order of their acid strength:
p-cresol, p-nitrophenol, phenol
Answer:
p-cresol < phenol < p-nitrophenol

(ii) Write the mechanism (using curved arrow notation) of the following reaction:
(a) CH2 = CH2 Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 33 CH3-CH2 + H2O
OR
Write the structures of the products when butan-2-ol reacts with the following: (CBSE AI 2017)
(i) CrO3
(ii) SOCl2
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 34
Or
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 35
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 36

Question 10.
Alcohols are easily protonated in comparison to phenols. Explain. (CBSE AI 2016)
Answer:
In alcohols, the electron-releasing inductive effect (+ I effect) of the alkyl group attached to the carbon having the -OH group increases the electron density on the oxygen atom. Therefore, alcohols are easily protonated.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 37
On the other hand, in the case of phenol, the oxygen atom acquires a partial positive charge due to resonance.
Thus, it is not protonated.

Question 11.
Explain why ortho nitrophenol is more acidic than ortho methoxy phenol. (CBSE AI 2015)
Answer:
This is because -NO2 (nitro group) is an electron-withdrawing group and will increase the +ve charge on oxygen to make it more acidic. On the other hand, the -OCH3 group is an electron-releasing group and will decrease +ve charge on oxygen making it less acidic as the O-H bond will not break easily.

Question 12.
Anisole on reaction with HI gives phenol and CH3I as main products and not iodobenzene and CH3OH. Why? (CBSE AI 2016)
Answer:
It this reaction protonated anisole, i.e. methyl phenyl oxonium ion, is first formed and then the halide ion attacks this protonated anisole. Due to steric hindrance of the bulky phenyl group, the attack preferably occurs to the alkyl group forming methyl iodide and phenol.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 38

Question 13.
The boiling points of ethers are lower than their corresponding isomeric alcohols. Explain. (CBSE AI 2012)
Answer:
Ethers have low polarity and therefore, do not show any association by intermolecular hydrogen bonding. On the other hand, their isomeric alcohols have strong intermolecular hydrogen bonding and therefore, their boiling points are high.

Question 14.
Butan-1-ol has a higher boiling point than diethyl ether. Give reason.
Answer:
Butan-1-ol has intermolecular hydrogen bonding between their molecules. Therefore, it exists as associated molecules and a large amount of energy is required to break these bonds and hence its boiling point is high. But diethyl ether does not show any association by intermolecular hydrogen bonding. Hence, its boiling point is low.

Question 15.
Write the mechanism of acid dehydration of ethanol to yield ethene. (CBSE Sample Paper 2018)
Answer:
The reaction is believed to occur as follows:
(i) Formation of protonated alcohol: Alcohol combines with a proton to form a protonated alcohol.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 39
(ii) Formation of the carbocation: The protonated alcohol loses a water molecule to form a carbocation.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 40
(iii) Elimination of a proton to form alkene: The carbocation then eliminates a proton and undergoes rearrangement of electrons to form the alkene.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 41
The acid used in step (i) is released in step (iii). To drive the equilibrium reaction in the forward direction, ethene is removed as soon as it is formed.

Alcohols, Phenols and Ethers Important Extra Questions Long Answer Type

Question 1.
What happens when
(a) Sodium phenoxide is treated with CH3Cl?
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 42
(b) CH2 = CH – CH2 – OH is oxidised by PCC?
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 43
(c) Phenol is treated with CH3COCI/anhydrous AlCl3?
Write chemical equations in support of your answer.
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 44
Question 2.
(a) How will you convert the following:
(i) Phenol to benzoquinone
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 45

(ii) Propanone to 2-methyl propane-2-ol
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 46
(b) Why does propanol have a higher boiling point than that butane? (CBSE 2019C)
Answer:
The molecules of propanol are held together by intermolecular hydrogen bonding while butane molecules have only weak van der Waals forces of attraction. Since hydrogen bonds are stronger than van der Waals forces, therefore, propanol has a higher boiling point than butane.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 47

Question 3.
Identify the product formed when propan-1 -ol is treated with a cone. H2S04 at 413 K. Write the mechanism involved for the above reaction. (CBSE Sample Paper 2019)
Answer:
(a) 1-Propoxypropane is formed.
Mechanism involved:
Step 1: Formation of protonated alcohol. Propanol gets protonated in the presence of H+.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 48
Step 2: Nucleophilic attack. Due to the presence of a +ve charge on the oxygen atom, the carbon of the CH2 part becomes electron deficient. As a result, a nucleophilic attack by another alcohol molecule (unprotonated) occurs on the protonated alcohol with the elimination of a molecule of water.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 49
Step 3: Deprotonation. Oxonium ion loses a proton to form an ether.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 50

Question 4.
Write the equations involved in the following reactions: (CBSE 2013, 2014)
(i) Reimer-Tiemann reaction
Answer:
Reimer-Tiemann reaction: When phenol is refluxed with chloroform in the presence of aqueous caustic alkali at 340 K, an aldehydic group (CHO) gets introduced in the ring at a position ortho to the phenolic group. Ortho hydroxy benzaldehyde or salicylaldehyde is formed as the product of the reaction.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 51

(ii) Williamson’s ether synthesis
Answer:
Williamson’s ether synthesis. This is used to prepare symmetrical and unsymmetrical ethers by treating alkyl halide with either sodium alkoxide or sodium phenoxide.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 52
Aryl halides cannot be used for the preparation of alkyl-aryl ethers because of their low reactivity.

Question 5.
Draw the structure and name the product formed if the following alcohols are oxidised. Assume that an excess of the oxidising agent is used. (CBSE Delhi 2012)
(i) CH3CH2CH2CH2OH
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 53
(ii) 2-butenol
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 54
(iii) 2-methyl-1-propanol
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 55

Question 6.
Write the main product(s) in each of the following reactions: (CBSE Delhi 2016)
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 56
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 57

Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 58
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 59

Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 60
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 61

Question 7.
Give reasons for the following: (CBSE Delhi 2016, Outside Delhi)
(i) Protonation of phenols Is difficult whereas ethanol easily undergoes protonation.
Answer:
In phenols, the oxygen atom acquires a partial positive charge due to resonance and therefore, it is not easily protonated. On the other hand, in ethanol, the alkyl group is an electron-releasing group and increases the electron density on 0 atoms. Therefore, ethanol is easily protonated.

(ii) Bolting point of ethanol is higher than that of dimethyl ether.
Answer:
The molecules of ethanol are held together by intermolecular hydrogen bonding while dimethyl molecules have only weak van der Waals forces of attractions. Since hydrogen bonds are stronger than van der Waals forces, ethanol has a higher boiling point than dimethyl ether.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 62

(iii) Anisole on reaction with Hl gives phenol and CH3-I as main products and not iodobenzene and CH3OH.
Answer:
This is because, during the reaction, the attack of halide ion occurs to the protonated anisole, i.e. methyl phenyl oxonium ion, which is formed during the protonation of anisole. Due to steric hindrance of the bulky phenyl group, the attack preferably occurs to the alkyl group forming methyl iodide and phenol.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 63

Question 8.
How do you convert the following: (CBSE Delhi 2015, Delhi)
(i) Phenol to anisole
(ii) Propan-2-ol to 2-methylpropan-2-ol
(iii) Aniline to phenol?
OR
(i) Write the mechanism of the following reaction:
2CH3CH2OH Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 64CH3CH2—O—CH2CH3
(ii) Write the equation involved In the acetylation of salicylic acid.
Answer:
(i) Phenol to anisole
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 65

Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 66

(iii) Aniline to phenol
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 67
Or
(i) The mechanism for the formation of ether from ethanol at 413 K is a nucleophilic bimolecular reaction as given below:
(a) Ethyl alcohol gets protonated in the presence of H+
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 68

(b) Due to the presence of a +ve charge on the oxygen atom, the carbon of CH2 part of CH3CH2 becomes electron deficient. As a result, nucleophilic attack by another alcohol molecule (unprotonated) occurs on the protonated alcohol with the elimination of a
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 69

(c) Oxonium ion loses a proton to form an ether.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 70
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 71

Question 9.
Give reasons for the following: (CBSE 2015, Outside Delhi)
(i) o-nitrophenol is more acidic than o-methoxyphenyl.
Answer:
This is because —NO2 (nitro group) is an electron-withdrawing group and will increase the +ve charge on oxygen to make it more acidic. On the other hand, the -OCH3 group is an electron¬releasing group and will decrease +ve charge on oxygen making it less acidic as O-H bond will not break easily.

(ii) Butan-1-oi has a higher boiling point than diethyl ether.
Answer:
Butan-1 -ol has intermolecular hydrogen bonding between their molecules. Therefore, it exists as associated molecules and large amount of energy is required to break these bonds and hence, its boiling point is high. But diethyl ether does not show any association by intermolecular hydrogen bonding. Hence, its boiling point is low.

(iii) (CH3)3C – O – CH3 on reaction with HI gives (CH3)3C – I and CH3 – OH as the main products and not (CH3)3C – OH and CH3 – I.
Answer:
The reaction:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 72
gives (CH3)3C – I and CH3OH as the main products and not (CH3)3COH and CH3I. This is because the reaction occurs by the SN1 mechanism and the formation of products is governed by the stability of carbocation formed from the cleavage of the C-0 bond in the protonated ether. Since tert. butyl carbocation, (CH3)3C+ is more stable than methyl carbocation, CH3, the cleavage of C-0 gives a more stable carbocation, [(CH3)3C]+ and methanol. Then, iodide ion, I– attacks this tert. butyl carbocation to form tert. butyl iodide.

Question 10.
(i) Write the mechanism of the following reaction: (CBSE 2015, Outside Delhi)
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 73
Answer:
HBr → H+ + Br–
(a) H+ attacks oxygen of O-H to form protonated alcohol
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 74
(b) Protonated alcohol Loses a molecule of water to form a carbocation.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 75
(C) Br attacks the carbocation to form bromoatkane
CH3CH2+ + Br– → CH3CH2Br (Bromoethane)

(ii) Write the equation involved in the Reimer-Tiemann reaction.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 76

Question 11.
(A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C4H80. Isomers (A) and (C) give positive Tollens’ test whereas isomer (B) does not give Tollens’ test but gives positive Iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCl, give the same product (D).
(i) Write the structures of (A), (B), (C) and (D).
Answer:
A = CH3CH2CH2CHO
B = CH3COCH2CH3
C = (CH3)2CHCHO
D = CH3CH2CH2CH3

(ii) Out of (A), (B) and (C) isomers, which one is least reactive towards the addition of HCN? (CBSE 2018 C)
Answer:
B

Question 12.
(a) How will you convert the following:
(i) Phenol to benzene
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 77

(ii) Propene to propanol
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 78

(b) Why is ortho-nitrophenol more acidic than ortho-methoxy phenol? (CBSE 2018C) OH
Answer:
Due to the strong -R and -I effect of the -NO2 group, the electron density in the O-H bond decreases and hence the loss of a proton becomes easy.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 79
Moreover, the o-nitrophenoxide formed after the Loss of a proton is stabilised by resonance.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 80

o-nitrophenoxide ion is stabilised by resonance and hence o-nitrophenol is a stronger acid. On the other hand, due to the +R effect of the -OCH3 group, the electron density in the O-H bond increases and this makes the loss of proton difficult.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 81
Furthermore, after the Loss of proton o-methoxyphenoxide ion left is destabilised by resonance.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 82
The two negative charges repel each other and therefore, destabilise the o-methoxyphenoxide ion. Thus, o-nitrophenol is more acidic than o-methoxyphenyl.

Question 13.
(a) How will you convert the following:
(i) Ethanal to propan-2-ol
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 83
(ii) Phenol to 2-hydroxy acetophenone
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 84

(b) Why is phenol more acidic than ethanol? (CBSE 2019C)
Answer:
It is because phenoxide ion is more stable than ethoxide ion. Phenols are more acidic than alcohols. The greater acidic character of phenoLs as compared to alcohols can be explained on the basis of resonance. Phenol is a resonance hybrid of the following structures:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 85
It is clear that three structures of phenol (III, IV and V) have a +ve charge on the oxygen of the —OH group. This oxygen attracts the electron pair of O—H bond strongly towards itself, weakens the O—H bond and, therefore, facilitates the release of W. Similarly, the phenoxide ion is resonance stabilised as follows:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 86
Thus, we observe that both phenol and phenoxide ion are stabilised by resonance. Now if we carefully observe the resonance structures, we observe that phenoxide ion is more resonance stabilised than phenol. In phenol three contributing structures (III, IV and V) have both positive and negative charges and therefore, these will be unstable. These structures will require energy to separate the charge and therefore, will be unstable.

On the other hand, there is no structure in phenoxide ion which requires charge separation. Thus, the resonance hybrid of phenol is less stable than phenoxide ion and the reaction is very much in favour of the phenoxide ion. Therefore, the phenol is more acidic.

On the other hand, in the case of alcohols, neither the alcohol nor the alkoxide ion is stabilised by resonance.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 87
Thus, phenols are more acidic than alcohols.

Question 13.
(a) How do you convert the following?
(i) Phenol to Anisole
(ii) Ethanol to Propan-2-ol
(b) Write the mechanism of the following reaction:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 90
(c) Why phenol undergoes electrophilic substitution more easily than benzene?
OR
(a) Account for the following:
(i) o-Nitrophenol is more steam volatile than p-nitrophenol.
(ii) Tert-butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of tert-butyl methyl ether.

(b) Write the reaction involved in the following:
(i) Reimer-Tiemann reaction
(ii) Friedel-Crafts Alkylation of Phenol

(c) Give a simple chemical test to distinguish between Ethanol and Phenol. (CBSE Delhi 2019)
Answer:
(i) Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 88
(ii) Ethanol to Propan-2-ol
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 89

(b) Mechanism CH3
(i) Ethanol combines with a proton to form protonated alcohol.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 91

(ii) Protonated alcohol loses a water molecule to form a carbocation.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 92

(iii) Carbocation eliminates a proton and undergoes rearrangement of electrons to form an alkene.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 93

(c) In phenol, the -OH group is an activating group and therefore, electrophilic substitution reaction undergoes more easily in phenol than benzene.
OR
(a) (i) o-Nitrophenol is more steam volatile than p-nitrophenol because of chelation due to intramolecular hydrogen bonding.

(ii) Tert-Butyl chloride reacts with sodium methoxide to give 2-methylpropene.
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 94
This is because alkoxides are not only nucleophiles but strong bases as well. They react with alkyl halides leading to an elimination reaction.

(b) Reimer-liemann reaction
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 95

(C) Phenols react with neutral Fecl3 to give violet colour, while ethanol does not give any colour.

Question 14.
(i) Write steps to carry out the conversion of phenol to aspirin.
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 96

(ii) What happens when benzene diazonium chloride Is heated with water?
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 97

(iii) Write the structures of the Isomers of alcohols with molecular formula C4H10O. Which of these exhibits optical activity?
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 98
(ii) is optically active.

Question 15.
How are the following conversions carried out?
(i) Benzyl chloride to benzyl alcohol
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 99

(ii) Methyl magnesium bromide to 2-methyl propane-2-ol
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 100

(iii) Propene to propan-2-ol
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 101

(iv) Ethylmagnesium chloride to propan-1-ol. (CBSE Delhi 2010, 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 102

Question 16.
How will you convert:
(i) Phenol to 2, 4, 6-trlnftrophenot
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 103

(ii) Propan-2-ol to propanone
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 104

(iii) Phenol to 2, 4, 6-trlbromophenol
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 105

(iv) Propene to propan-1-ol
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 106

(v) Ethanal to propan-2-ol? (CBSE DelhI 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 107

Question 17.
How will you convert
(i) Propan-2-ol to 2-methylpropan-2-ol (CBSE Delhi 2015)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 108

(ii) Aniline to phenol (CBSE Delhi 2015)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 109

(iii) Ethanol to propane nitrile (CBSE AI 2015)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 110

(iv) Phenol to toluene (CBSE AI 2016)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 111

(v) Formaldehyde to ethanol? (CBSE AI 2016)
Answer:
Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers 112

Electromagnetic Waves Class 12 Important Extra Questions Physics Chapter 8

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 8 Electromagnetic Waves. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 8 Important Extra Questions Electromagnetic Waves

Electromagnetic Waves Important Extra Questions Very Short Answer Type

Question 1.
Name the part of the electromagnetic spectrum which has the longest wavelength and write its one use. (CBSE 2019C)
Answer:

  • In the electromagnetic spectrum, long radio waves have the longest wavelength.
  • Radio waves are used in communication systems.

Question 2.
The small ozone layer on the top of the stratosphere is crucial for human survival. Why?
Answer:
The ozone layer absorbs the ultraviolet rays, emitted by the sun, which are harmful to the living tissues of human beings.

Question 3.
Name the part of the electromagnetic spectrum which is used in the “greenhouse” to keep plants warm.
Answer:
Infrared rays.

Question 4.
How are radio waves produced? (CBSE AI 2011)
Answer:
They are produced by rapid acceleration and decelerations of electrons in aerials.

Question 5.
How are X-rays produced? (CBSE Al 2011)
Answer:
By the transition of inner-shell electrons.

Question 6.
How are microwaves produced? (CBSE AI 2011)
Answer:
By using a magnetron.

Question 7.
A plane electromagnetic wave travels in a vacuum along the z-direction. What can you say about the direction of electric and magnetic field vectors? (CBSE Delhi 2011)
Answer:
The electric and magnetic field vectors will be along the x and y directions.

Question 8.
What is the frequency of electromagnetic waves produced by the oscillating charge of frequency v? (CBSE Delhi 2011C)
Answer:
The frequency of electromagnetic waves produced by the oscillating charge of frequency v is also v.

Question 9.
What are the directions of electric and magnetic field vectors relative to each other and relative to the direction of propagation of electromagnetic waves? (CBSE AI 2012)
Answer:
The three are mutually perpendicular to one other.

Question 10.
Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiation. Name the radiations and write the range of their frequency. (CBSE Al 2013)
Answer:
UV radiations, 1015 to 1017 Hz

Question 11.
To which part of the electromagnetic spectrum does a wave of frequency 5 × 1019 Hz belong? (CBSEAI 2014)
Answer:
X – rays

Question 12.
The available frequency AC source is connected to a capacitor. Will the displacement current change if the frequency of the AC source is decreased? (CBSE Al 2015C)
Answer:
No

Question 13.
Why microwaves are considered suitable for radar systems used in aircraft navigation? (CBSE Delhi 2016)
Answer:
They have a small wavelength and travel along a straight line with deflecting.

Question 14.
Do electromagnetic waves carry energy and momentum? (CBSE AI 2017)
Answer:
Yes.

Question 15.
How is the speed of em-waves in vacuum – determined by the electric and magnetic field? (CBSE Delhi 2017)
Answer:
c = \(\frac{E_{0}}{B_{0}}\)

Question 16.
Why is skywave propagation of signals restricted to a frequency of 30 MHz? (CBSE Al 2017 C)
Answer:
The atmosphere is transparent to frequencies higher than 30 MHz.

Question 17.
Name the electromagnetic radiations used for
(a) water purification
Answer:
UV radiation

(b) eye surgery. (CBSEAI 2018, Delhi 2018)
Answer:
Visible light

Question 18.
Write the range of frequencies of electromagnetic waves which propagate through sky wave mode. (CBSE Al 2018 C)
Answer:
A few MHz up to 30 to 40 MHz.

Question 19.
Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?
Answer:
This is because of the fact that X-rays are absorbed by the atmosphere, whereas light and radio waves penetrate through it.

Question 20.
A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic wave produced by the oscillator?
Answer:
Same as that of the oscillating charged particle, i. e. 109 Hz.

Question 21.
Why are infrared radiations referred to as heat waves also? Name the radiations which are next to these radiations in the electromagnetic spectrum having
(a) Shorter wavelength and
(b) Longer wavelength.
Answer:
This is because they produce a heating effect.
(a) Visible light and
(b) Microwaves.

Question 22.
What physical quantity are the same for X-rays of wavelength 10-10 m, the red light of wavelength 680 nm, and radio waves of wavelength 500 m?
Answer:
Since all of them are electromagnetic waves, their speed in a vacuum will be the same, i.e. 3 × 108 m s-1.

Question 23.
A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:
The frequency of the electromagnetic waves produced by the oscillator is the same as its frequency of vibration, i.e. 109 Hz.

Question 24.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in a vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave?
Answer:
Given B0 = 510 nT = 510 × 10-9 T, E0 = ?

Using the relation E0 = c × B0, we have
E0 = 3 × 108 × 510 × 10-9 = 153 N C-1

Question 25.
Why does a microwave oven heat up a food item containing water molecules most efficiently?
Answer:
It is because the frequency of the microwave matches the resonant frequency of water molecules. This makes the molecules vibrate with maximum amplitude thereby producing heat.

Question 26.
Name the most energetic electromagnetic radiation and write its frequency range. (CBSE AI 2019 C)
Answer:
The most energetic radiations are Gamma Rays Frequency range of gamma rays is: 1018 Hz to 1023 Hz
Or
Name the electromagnetic radiations used in eye surgery or to kill germs in water purifiers. Write its frequency, range.
Answer:

  1. Radiations used for eye surgery or to kill germs are Ultraviolet Rays.
  2. The frequency range of ultraviolet rays: 105 Hz to 1017 Hz.

Electromagnetic Waves Important Extra Questions Short Answer Type

Question 1.
Radio waves and gamma rays both are transverse in nature and electromagnetic in character and have the same speed in a vacuum. In what respect are they different?
Answer:
The radio waves have an atomic origin, while gamma rays have a nuclear origin. Further owing to their very small wavelength, gamma rays are highly penetrating in comparison to radio waves.

Question 2.
Show that the average energy density of the electric field equals the average density of the magnetic field.
Answer:
The average density of the electric field is given by
Ue = \(\frac{1}{2}\)ε0E2 and the average energy density of the magnetic field is given by UB = \(\frac{B^{2}}{2 \mu_{0}}\).

But B = \(\frac{E}{c}\) and c = \(\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\) , hence the above equation becomes UB = \(\frac{B^{2}}{2 \mu_{0}}=\frac{E^{2}}{2 \mu_{0} c^{2}}\)

UB = \(\frac{E^{2}}{2 \mu_{0} \times \frac{1}{\mu_{0} \varepsilon_{0}}}=\frac{1}{2} \varepsilon_{0} E^{2}\). Hence the result.

Question 3.
State four properties of electromagnetic waves.
Answer:
(a) They do not require any material medium to travel.
(b) They are transverse in nature, i.e. electric and magnetic fields are perpendicular to each other and also to the direction of the propagation of the wave.
(c) The energy of the wave is divided equally amongst the electric and the magnetic field.
(d) They travel, in free space, with a velocity of 3 × 108 m s-1.

Question 4.
Electromagnetic radiations with wavelength
(a) λ1 are used to kill germs in water purifiers.
(b) λ2 are used in TV communication systems.
(c) λ3 plays an important role in maintaining the earth’s warmth.
Name the part of the electromagnetic spectrum to which these radiations belong. Arrange these wavelengths in decreasing order of their magnitude.
Answer:
(a) λ1 – Ultraviolet radiations.
(b) λ2 – Microwaves
(c) λ3 – Infrared rays
Their order is λ1 < λ3 < λ2.

Question 5.
Name the constituent radiation of the electromagnetic spectrum which
(a) is used in satellite communication.
Answer:
Microwaves.

(b) is used for studying crystal structure.
Answer:
X-rays

(c) is similar to the radiations emitted during the decay of a radioactive nucleus.
Answer:
Gamma rays

(d) is absorbed from sunlight by the ozone layer.
Answer:
UV rays

(e) produces an intense heating effect.
Answer:
Infrared rays

(f) has its wavelength range between 390 nm and 770 nm.
Answer:
Visible light.

Question 6.
Name the radiations of the electromagnetic spectrum which are used in
(a) warfare to look through the haze.
Answer:
Infrared rays

(b) radar and geostationary satellites
Answer:
Microwaves.

(c) studying the structure and properties of atoms and molecules.
Answer:
Gamma rays.

Question 7.
Why are microwaves used in RADAR?
Answer:
Microwaves are electromagnetic waves of very short wavelength. Such waves are used in RADAR due to the reason that they can travel in a particular direction in the form of a beam without being deflected.

Question 8.
Electromagnetic waves with wavelength
(a) λ1 are used to treat muscular strain.
(b) λ2 are used by an FM radio station for broadcasting.
(c) λ3 are used to detect fractures in bones.
(d) λ4 are absorbed by the ozone layer of the atmosphere.
Identify and name the part of the electromagnetic spectrum to which these radiations belong. Arrange
these wavelengths in decreasing order of magnitude.
Answer:
(a) Infrared radiations are used to treat muscular strain.
(b) Radio and microwave radiations are used for FM transmission.
(c) X-rays are used to detect fractures in bones.
(d) Ultraviolet radiation is absorbed by the ozone layer of the atmosphere.
The decreasing order of their wavelength is
λ2 > λ1 > λ4 > λ3.

Question 9.
(a) Draw a graph of a linearly polarised em wave propagating in the Z-direction showing the directions of the oscillating electric and magnetic fields.
Answer:
Graph of linearly polarised em wave propagating in the Z-axis.
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 1
(b) Write the relations (i) between the speed of light and the amplitudes of electric and magnetic fields, (ii) for the speed of em wave in terms of a permittivity e0, and magnetic permeability p0, of the medium. (CBSE 2019C)
Answer:
(i) Relation between speed of light and amplitudes of electric and magnetic field c = \(\frac{E_{0}}{B_{0}}\)

(ii) Speed of light in terms of ε<sub>0</sub> and μ0,
c = \(\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\)

Question 10.
Arrange the following electromagnetic waves in the order of their increasing wavelength:
(a) Gamma rays
(b) Microwaves
(c) X-rays
(d) Radio waves
How are infrared waves produced? What role does infrared radiation play in
(a) maintaining the Earth’s warmth and
(b) physical therapy? (CBSE Al 2015)
Answer:
Gamma(γ) rays, X-rays, Microwaves, Radio waves:
Infrared rays are produced by hot bodies/ vibration of atoms and molecules Infrared rays: (a) Maintain the earth’s warmth through the greenhouse effect, (b) produce heat

Question 11.
Name the parts of the electromagnetic spectrum which is
(a) suitable for radar systems used in aircraft navigation.
Answer:
Microwave: They are produced by oscillating circuits.

(b) used to treat muscular strain.
Answer:
Infrared rays: They are produced by the vibration of atoms and molecules.

(c) used as a diagnostic tool in medicine. Write in brief, how these waves can be produced. (CBSE Delhi 2015)
Answer:
X-rays: They are produced by bombarding high atomic number targets with electrons.

Question 12.
An e.m wave Y1 has a wavelength of 1 cm while another e.m wave, Y2 has a frequency of 1015 Hz. Name these two types of waves and write one useful application for each. (CBSE AI 2016 C)
Answer:
Y1 – Microwaves and
Y2 – Ultraviolet waves.

  • Microwaves: used for communication.
  • Ultraviolet waves: used for sterilization.

Question 13.
Identify the electromagnetic waves whose wavelengths vary as
(a) 10-12 m < λ < 10-8 m
Answer:
X – rays: a study of crystal structure

(b) 10-3 m < λ < 10-1 m. Write one of their uses. (CBSE Al 2017)
Answer:
Microwaves: radar and communication

Question 14.
Name the type of e.m waves having a wavelength range of 0.1 m to 1 mm. How are these waves generated? Write their two uses. (CBSE Al 2017 C)
Answer:

  • Microwaves: These are generated with the help of special vacuum tubes (called klystrons, magnetrons, and Gunn diodes).
  • Uses: Cooking, radar, and communication

Question 15.
(a) Give one use of electromagnetic radiations obtained in nuclear disintegrations.
Answer:
Treating cancer.

(b) Give one example each to illustrate the situation where there is
(i) displacement current but no conduction current and
Answer:
Between the plates of a capacitor

(ii) only conduction current but no displacement current.(CBSE Al 2018 C)
Answer:
Outside the plates of a capacitor

Question 16.
Scientists predict that a global nuclear war on the earth will be followed by a severe nuclear winter, with devastating effects on the earth. What is the basis of this prediction?
Answer:
The explosions will produce so much dust, which will cover the whole atmosphere, thereby blocking the sun’s rays from reaching the earth. This will cause the setting in of a long winter, which is called nuclear winter.

Question 17.
If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
Answer:
Due to the Greenhouse effect, the temperature of the earth’s surface is raised in the presence of the atmosphere. In the absence of the atmosphere, the heat received by the earth during the day is completely lost during the night. Hence the average surface temperature will be lower than the preset temperature.

Question 18.
A capacitor of capacitance ‘C is being charged by connecting it across a dc source along with an ammeter. Will the ammeter show a momentary deflection during the process of charging? If so, how would you explain this momentary deflection and the resulting continuity of current in the circuit? Write the expression for the current inside the capacitor. (CBSE AI 2012)
Answer:
Yes, this is due to the rate of change of electric flux inside the capacitor due to the production of displacement current. The expression for the current inside the capacitor is lD = εo \(\frac{d \phi_{E}}{d t}\)

Question 19.
Define displacement current. What role does it play while charging a capacitor by dc source? Is the value of displacement current the same as that of the conduction current? Explain. (CBSE AI 2019)
Answer:

  • Displacement current is the current due to the change of electric flux.
  • It provides continuity of current in circuits containing capacitors.
  • Yes, the value of displacement current is equal to the conduction current.
  • ld = εo \(\frac{d \phi_{e}}{d t}\)

Question 20.
Why is the orientation of the portable radio with respect to the broadcasting station important? (NCERT Exemplar)
Answer:
This is because the electromagnetic waves are plane polarised; hence the receiving antenna should be parallel to the electric/magnetic part of the wave.

Question 21.
(a) Give one use of electromagnetic radiations obtained in nuclear disintegrations.
Answer:
used to destroy cancer cells

(b) Give one example each to illustrate the situation where there is
(i) displacement current but no conduction current and
Answer:
The region, between the plates of a capacitor, connected to a time-varying voltage source, has a displacement current but no conduction current.

(ii) only conduction current but no displacement current. (CBSE Delhi 2018C)
Answer:
The wires, connected to the plates of a capacitor, joined to a time-varying or steady voltage source, carry a conduction current but no displacement current. (Alternatively, A circuit, having no capacitor in it, and carrying a current has conduction current but no displacement current.)

Electromagnetic Waves Important Extra Questions Long Answer Type

Question 1.
Answer the following:
(a) Name the em waves which are used for the treatment of certain forms of cancer. Write their frequency range.
Answer:
Gamma rays.
Frequency range > 3 × 1020 Hz

(b) Thin ozone layer on top of the stratosphere is crucial for human survival. Why?
Answer:
The thin ozone layer on top of the stratosphere is crucial for human survival because it absorbs most of the ultraviolet rays coming from the sun. If the ozone layer had not been there, then ultraviolet rays would have entered the earth and caused danger to the survival of the human race.

(c) An em wave exerts pressure on the surface on which it is incident. Justify. (CBSE Delhi 2014)
Answer:
An em wave carries a linear momentum with it. The linear momentum carried by a portion of a wave having energy U is given by p = U/c.

Thus, if the wave incident on a material surface is completely absorbed, it delivers energy U and momentum p = U/c to the surface. If the wave is totally reflected, the momentum delivered is p = 2U/c because the momentum of the wave changes from p to – p. Therefore, it follows that an em wave incident on a surface exerts a force and hence a pressure on the surface.

Question 2.
Answer the following questions:
(a) Why is the thin ozone layer at the top of the stratosphere crucial for human survival? Identify to which part of the electromagnetic spectrum does this radiation belongs and write one important application of the radiation.
Answer:
The thin ozone layer on top of the stratosphere is crucial for human survival because it absorbs most of the ultraviolet rays coming from the sun. If the ozone layer had not been there, then ultraviolet rays would have entered the earth and caused danger to the survival of the human race. This radiation is UV radiation. It is used in sterilization.

(b) Why are infrared waves referred to as heat rays? How are they produced? What role do they play in maintaining the earth’s warmth through the greenhouse effect? (CBSE Delhi 2015C)
Answer:
Infrared radiations heat up the material on which they fall, hence they are also called heat rays. They are produced by the vibration of atoms and molecules. After falling on the earth, they are reflected back into the earth’s atmosphere. The earth’s atmosphere does not allow these radiations to pass through as such they heat up the earth’s atmosphere.

Question 3.
How are electromagnetic waves produced? What is the source of energy of these waves? Write mathematical expressions for electric and magnetic fields of an electromagnetic wave propagating along the z-axis. Write any two important properties of electromagnetic waves. (CBSE AI 2016)
Answer:
Electromagnetic waves are produced by accelerated charges which produce an oscillating electric field and magnetic field (which regenerate each other).

  • Source of the Energy: Energy of the accelerated charge or the source that accelerates the charges.
  • Expression: Ex = Eo sin (kz – ωt) and By = Bo sin (kz – ωt)
    (a) They are transverse in nature.
    (b) They don’t require a medium to propagate.

Question 4.
How are em waves produced by oscillating charges?
Draw a sketch of linearly polarised em waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields. (CBSE Delhi 2016)
Answer:
(a) An oscillating charge produces an oscillating electric field in space, which produces an oscillating magnetic field. The oscillating electric and magnetic fields regenerate each other, and this results in the production of em waves in space.
(b) See Figure.
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 2
Question 5.
Write Maxwell’s generalization of Ampere’s Circuital Law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is i = εo\(\frac{d \phi_{E}}{d t}\) where ΦE is the electric flux produced during charging of the capacitor plates. (CBSE Delhi 2016)
Answer:
The generalized form of Maxwell ampere law is
\(\oint \vec{B} \cdot \overrightarrow{d l}\)= μo(l + lD) where lD = εo\(\frac{d \phi_{E}}{d t}=\frac{d q}{d t}\)

The electric flux Φ between the plates of the parallel plate capacitor through which a time-dependent current flow is given by:
ΦE = E A, but E = σ/εo
Therefore we have
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 3
Question 6.
(a) Why are Infrared waves often called heatwaves? Explain.
Answer:
Infrared waves have frequencies lower than those of visibLe Light; they have the ability to vibrate not only the electrons but the entire atoms or molecules of a body. This vibration increases the internal energy and temperature of the body. That is why infrared waves are often called heat waves.

(b) What do you understand by the statement, “Electromagnetic waves transport momentum”? (CBSE AI, Delhi 2018)
Answer:
If we consider a plane perpendicular to the direction of propagation of the electromagnetic wave, then electric charges present on the plane will be set and sustained in motion by the electric and magnetic fields of the electromagnetic wave. The charges present on the surface thus acquire energy and momentum from the waves. This just illustrates the fact that an electromagnetic wave (like other waves) transfers energy and momentum.

Question 7.
(a) When the oscillating electric and magnetic fields are along the x- and y-direction respectively
(i) point out the direction of propagation of the electromagnetic wave,
Answer:
Z-axis

(ii) express the velocity of propagation in terms of the amplitudes of the oscillating electric and magnetic fields.
Answer:
c = Eo / Bo

(b) How do you show that the em wave carries energy and momentum? (CBSE A! 2013C)
Answer:
Consider a plane perpendicular to the direction of propagation of the electromagnetic wave. If there are, on this plane, electric charges, they will be set and sustained in motion by the electric and magnetic fields of the electromagnetic wave. The charges thus acquire energy and momentum from the waves. This illustrates the fact that an electromagnetic wave carries energy and momentum.

Question 8.
Answer the following questions:
(a) Show, by giving a simple example, how em waves carry energy and momentum.
Answer:
Consider a plane perpendicular to the direction of propagation of the electromagnetic wave. If there are, on this plane, electric charges, they will be set and sustained in motion by the electric and magnetic fields of the electromagnetic wave. The charges thus acquire energy and momentum from the waves. This just illustrates the fact that an electromagnetic wave (like other waves) carries energy and momentum.

(b) How are microwaves produced? Why is it necessary in microwave ovens to select the frequency of microwaves to match the resonance frequency of water molecules?
Answer:
Microwaves (short-wavelength radio waves), with frequencies in the gigahertz (GHz) range, are produced by special vacuum tubes (called klystrons, magnetrons, and Gunn diodes). In such ovens, the frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves is transferred efficiently to the kinetic energy of the molecules. This raises the temperature of any food containing water.

(c) Write two important uses of infrared waves. (CBSE Delhi 2014C)
Answer:
Remote control of electronic devices, heating

Question 9.
A parallel plate capacitor is being charged by a time-varying current. Explain briefly how Ampere’s circuital law is generalized to incorporate the effect due to the displacement current. (CBSE AI 2011)
Answer:
Consider the curve C bounding two surfaces S1 and S2. For the bound surface S1; the current passes through it, as such, we can write Ampere’s circuital law as
\(\oint_{s_{1}} \vec{\phi} \cdot \vec{d}\) = μol …(1)

If however, we choose the bound surface S2 that passes through the plates of the capacitor and is not pierced by a current-carrying conductor, then Ampere’s circuital law is written as
\(\oint_{s_{2}} \vec{\phi} \cdot \vec{d}\) = 0 ….(2)
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 4
The above two equations contradict each other. To resolve this contradiction, Maxwell showed that this inconsistency is due to the assumed discontinuity of the current. According to Maxwell a current called displacement current “flows” between the plates of the capacitor, where there is no conduction current.

Therefore Ampere’s circuital law takes the generalised expression \(\oint \vec{B} \cdot d \vec{l}\) = μo (l + lD),

here l is the conduction current and lD is the displacement current given by
lD = εo \(\frac{d \phi_{E}}{d t}=\frac{d q}{d t}\).

Outside the plates of the capacitor, conduction current flows and displacement current is zero, whereas inside the plates of the capacitor, displacement current exists and there is no conduction current.

Question 10.
(a) Identify the part of the electromagnetic spectrum used in (i) radar and (ii) eye surgery. Write their frequency range.
Answer:
Microwaves:
Frequency range (1010 to 1012 Hz) Ultraviolet rays:
Frequency range (1015 to 1017 Hz)

(b) Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field. (CBSE Delhi 2019)
Answer:
The average density of the electric field is
uE = \(\frac{1}{2}\)εo E2
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 5
Question 11.
Answer the following questions:
(a) Long-distance radio broadcasts use short-wave bands. Why?
Answer:
The long-distance radio broadcast is not possible using long or medium wave bands because these waves, traveling as ground waves, can cover a maximum distance of 200 km. When used as sky waves, the short waves pass through the lower portion of the atmosphere but are reflected back from the ionosphere. In this way, short waves can travel very large distances and can even travel around the earth.

(b) It is necessary to use satellites for long-distance TV transmission. Why?
Answer:
TV waves have a frequency range of 47 MHz to 940 MHz. These frequencies are not reflected by the ionosphere. As space waves, they can cover a distance of 50-60 km only. Therefore, for long-distance TV transmission, we make use of satellites that reflect the TV signal wave back towards the earth.

(c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?
Answer:
In optical and radio-telescopes we use visible light and radio waves, respectively, which can pass through the atmosphere. Hence, such telescopes are built on the ground. However, X-rays have extremely small wavelengths and are absorbed by the atmosphere. Hence, X-ray astronomy is not possible from ground stations. X-ray astronomy is possible only from satellites orbiting the earth at a height of 500 km or more.

(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
Answer:
The thin ozone layer on top of the stratosphere is crucial for human survival because it absorbs most of the ultraviolet rays coming from the sun. If the ozone layer had not been there, then ultraviolet rays would have entered the earth and caused danger to the survival of the human race.

(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
Answer:
If the earth did not have an atmosphere, then its average surface temperature would be lesser than what it is now because in that case greenhouse effect will be absent.

(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe nuclear winter’ with a devastating effect on life on the earth. What might be the basis of this prediction?
Answer:
The prediction is based on the assumption that the large dust clouds produced by global nuclear war would perhaps cover a substantial part of the sky and solar radiations will not be able to reach the earth. It may cause a severe winter on the earth with a devastating effect on life on the earth.

Numerical Problems

Question 1.
The frequency values v1 and v2, for two spectral lines of the e.m. spectrum are found to be 5 × 1020 Hz and 2.5 × 1011 Hz respectively. Find the ratio, λ1/λ2 of their wavelengths.
Answer:
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 6
Question 2.
In a plane electromagnetic wave, the electric field oscillates with a frequency of 2 × 1010 Hz and an amplitude of 40 VM.
(i) What is the wavelength of the wave and
(ii) what is the energy density due to the field?
Answer:
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 7
Question 3.
A plane electromagnetic wave of frequency 25 MHz travels In free space along the x-direction. At a particular point In space and time, the electric vector is \(\vec{E}\) = 6.3 Vm-1 ĵ. Calculate \(\vec{B}\) at this point. (NCERT)
Answer:
The magnitude of B and E are related as c = \(\frac{E}{B}\), therefore B = \(\frac{E}{c}=\frac{6.3}{3 \times 10^{8}}\) = 2.1 × 10-8. This field is along the z-axis, i.e. perpendicular to both the propagation of the wave and the electric field.

Question 4.
The magnetic field in a plane electromagnetic wave Is given by B = 2 × sin (0.5 × 10-3 x + 1.5 × 1011 t) T
(a) What are the wavelength and frequency of the wave?
Answer:
(a) Comparing the given equation with the equation
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 8
(b) Write an expression for the electric field. (NCERT)
Answer:
Eo = cBo = 2 × 107 × 3 × 108 = 60 Vm-1

The electric field component is perpendicular to the direction of propagation and the direction of the magnetic field. Therefore, the electric field component aLong the z-axis is obtained as
Ez = 60 sin(0.5 × 103 × + 1.5 × 1011 t)Vm-1

Question 5.
A radio can tune In to any station in the 7.5 MHz to 12 MHz bands. What is the corresponding wavelength band? (NCERT)
Answer:
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 9
Thus the wavelength band is 40 m to 25m.

Question 6.
Suppose that the electric field amplitude of an electromagnetic wave E0 = 120 N C-1 and that Its frequency is v = 50.0 MHz. (a) Determine, B0, ω k, and λ. (b) Find expressions for E and B. (NCERT)
Answer:
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 10
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 11
Question 7.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m-1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the 8 fields. [c = 3 × 108ms-1.] (NCERT)
Answer:
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 12
(c) The average density of electric field is
given by Ue = \(\frac{1}{2}\)ε0E2 and the average energy density of the magnetic field is given by UB = \(\frac{B^{2}}{2 \mu_{0}}\). But B = \(\frac{E}{c}\) and C = \(\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\) , hence the above equation becomes UB = \(\frac{B^{2}}{2 \mu_{0}}=\frac{E^{2}}{2 \mu_{0} c^{2}}\),
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 13
Question 8.
Suppose that the electric field part of an electromagnetic wave in a vacuum is
E = {(3.1 NC-1>)cos(1.8 rad m-1)y + (5.4 × 106 rad s-1)t}î
(a) What is the direction of propagation?
(b) What is the wavelength?
(c) What is the frequency v?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave. (NCERT)
Answer:
The electric field is of the form
Class 12 Physics Important Questions Chapter 8 Electromagnetic Waves 14
(e) Expression for magnetic field part of the wave
\(\vec{B}\) = Bo cos (ky + ωt) k̂
or
\(\vec{E}\) = {(10.3 n T)cos[(1.8 rad m-1)y + (5.4 × 106 rad s-1)t} î

Atoms Class 12 Important Extra Questions Physics Chapter 12

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 12 Atoms. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 12 Important Extra Questions Atoms

Atoms Important Extra Questions Very Short Answer Type

Question 1.
Name the spectral series which lies in the visible region.
Answer:
Balmer series.

Question 2.
What is the maximum number of spectral lines emitted by a hydrogen atom when it is in the third excited state? (CBSE AI 2013C)
Answer:
Six.

Question 3.
When is Hα line of the Balmer series in the emission spectrum of hydrogen atom obtained? (CBSE Delhi 2013C)
Answer:
It is obtained when an electron jumps from n =3 to n = 2 level.

Question 4.
A mass of lead is embedded in a block of wood. Radiations from a radioactive source incident on the side of the block produce a shadow on a fluorescent screen placed beyond the block. The shadow of the wood is faint but the shadow of lead is dark. Give a reason for this difference.
Answer:
The shadow of the wood is faint because only the a-radiations are stopped by the wood (since a-radiations are least penetrating). The shadow of lead is dark because p and y-radiations are also stopped by lead.

Question 5.
What was the source of alpha particles in Rutherford’s alpha scattering experiment?
Answer:
The source was 21483Bi.

Question 6.
If the radius of the ground level of a hydrogen atom is 5.3 nm, what is the radius of the first excited state?
Answer:
It is 4 × 5.3 = 21.2 nm ( ∵ r = n²ro)

Question 7.
Calculate the ratio of energies of photons produced due to the transition of electron of a hydrogen atom from its
(a) Second permitted energy level to the first level, and
Answer:
energy of photon E1 = – 3.4 – (-13.6) = 10.2 eV

(b) Highest permitted energy level to the second permitted level.
Answer:
energy of photon E2 = 0 – (-3.4) = 3.4 eV
Ratio \(\frac{E_{1}}{E_{2}}=\frac{10.2}{3.4}\) = 3

Question 8.
The mass of an H-atom is less than the sum of the masses of a proton and electron. Why is this? (NCERT Exemplar) Answer:
Einstein’s mass-energy equivalence gives E = mc². Thus the mass of an H-atom is mp + me – B/c² where B ≈ 13.6 eV

Atoms Important Extra Questions Short Answer Type

Question 1.
Define electron-volt and atomic mass unit. Calculate the energy in joule equivalent to the mass of one proton.
Answer:
Electron volt: It is defined as the energy gained by an electron when accelerated through a potential difference of 1 volt. Atomic mass unit: It is defined as one-twelfth of the mass of one atom of carbon 12.

The mass of a proton is 1.67 × 10-27 kg. Therefore, energy equivalent of this mass is E = mc² = 1.67 × 10-27 × (3 × 108)2 = 1.5 × 10-10 J

Question 2.
State Bohr’s quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong. (CBSE Delhi 2019)
Or
Calculate the orbital period of the electron in the first excited state of the hydrogen atom.
Answer:
Bohr’s Quantisation condition: Only those orbits are permitted in which the angular momentum of the electron is an integral multiple of h/2π.

For Brackett Series,
The shortest wavelength is for the transition of electrons from ni = ∞ to nf = 4

Using the equation
Class 12 Physics Important Questions Chapter 12 Atoms 1
Question 3.
Write two important limitations of the Rutherford nuclear model of the atom. (CBSE AI2018, Delhi 2018)
Answer:

  1. Rutherford’s model fails to explain the line spectra of the atom.
  2. Rutherford’s model cannot explain the stability of the nucleus.

Question 4.
Find out the wavelength of the electron orbiting in the ground state of the hydrogen atom. (CBSEAI 2018, Delhi 2018)
Answer:
The wavelength of an electron in the ground state of hydrogen atom is given by
E = \(\frac{hc}{λ}\)
or
λ = \(\frac{hc}{E}\)

For ground state
E = – 13.6 eV = 13.6 × 1.6 × 10-19 J

Hence wavelength of electron in the first orbit
λ = \(\frac{h c}{E}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{13.6 \times 1.6 \times 10^{-19}}\) = 0.9 × 10-7 J

Question 5.
(a) State Bohr’s postulate to define stable orbits in a hydrogen atom. How does de Broglie’s hypothesis explain the stability of these orbits?
Answer:
Bohr’s postulate for stable orbits states the electron in an atom revolves around the nucleus only in those orbits for which its angular momentum is an integral multiple of h/2π (h = Planck’s constant), (n = 1, 2, 3 …)

As per de Broglie’s hypothesis λ = h/p = h/mv
For a stable orbit, we must have a circumference of the orbit = nλ (n = 1, 2, 3,…)
∴ 2πr = nλ
or
mvr = nh/2π

Thus de-Broglie showed that the formation . of stationary patterns for integral “n” gives rise to the stability of the atom.
This is nothing but Bohr’s postulate.

(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon. (CBSE AI 2018, Delhi 2018)
Answer:
Energy in the n = 4 level n1 = 1 and n2 = 4
Class 12 Physics Important Questions Chapter 12 Atoms 2
Class 12 Physics Important Questions Chapter 12 Atoms 3
Question 6.
An alpha particle moving with initial kinetic energy K towards a nucleus of atomic number Z approaches a distance ‘d’ at which it reverses its direction. Obtain an expression for the distance of closest approach ‘d’ in terms of the kinetic energy of the alpha particle, K. (CBSEAI2016C)
Answer:
At the distance of the closest approach, the kinetic energy of the alpha particle is converted into the electrostatic potential energy of the alpha particle-nucleus system. Therefore, at the distance of the closest approach

we have
Kinetic energy = Potential energy
Therefore,
Class 12 Physics Important Questions Chapter 12 Atoms 4
where K is the kinetic energy.

Question 7.
The figure shows the energy level diagram of the hydrogen atom.
Class 12 Physics Important Questions Chapter 12 Atoms 5
(a) Find out the transition which results in the emission of a photon of wavelength 496 nm.
Answer:
The wavelength of photon emitted is given by \(\frac{1}{λ}\) = RH\(\left(\frac{1}{n_{\mathrm{f}}^{2}}-\frac{1}{n_{\mathrm{i}}^{2}}\right)\)

None of these transitions correspond to a wavelength of 496 nm. The closest is 4 to 2 of 489 nm

(b) Which transition corresponds to the emission of radiation of maximum wavelength? Justify your answer. (CBSE AI 2015 C)
Answer:
Transition 4 to 3 as the frequency of this radiation is maximum.

Question 8.
A nucleus makes a transition from one permitted energy level to another level of lower energy. Name the region of the electromagnetic spectrum to which the emitted photon belongs. What is the order of its energy in electron-volts? Write four characteristics of nuclear forces.
Answer:
(a) Emitted photon belongs to gamma-rays part of the electromagnetic spectrum.
(b) the energy is of the order of MeV.
(c) Four characteristics of nuclear forces are:

  1. Nuclear forces are independent of charges.
  2. Nuclear forces are short-range forces.
  3. Nuclear forces are the strongest forces in nature, in their own small range of few fermis.
  4. Nuclear forces are saturated forces.

Atoms Important Extra Questions Long Answer Type

Question 1.
Explain Rutherford’s experiment on the scattering of alpha particles and state the significance of the results.
Answer:
The schematic arrangement in the Geiger Marsden experiment is shown in the figure.
Class 12 Physics Important Questions Chapter 12 Atoms 6
Alpha-particles emitted by a Bismuth (21483Bi) radioactive source were collimated into a narrow beam by their passage through lead bricks. The beam was allowed to fall on a thin foil of gold of thickness 2.1 × 10-7 m. The scattered alpha-particles were observed through a rotatable detector consisting of a zinc sulfide screen and a microscope. The scattered alpha-particles on striking the screen produced bright light flashes or scintillations. These scintillations could be viewed through the microscope and counted at different angles from the direction of the incident beam.

Significance: The experiment established the existence of a nucleus that contained the entire positive charge and about 99.95% of the mass.

Question 2.
Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to the Balmer series occur due to the transition between energy levels. (CBSE Delhi 2013)
Answer:
The electron revolving around the nucleus has two types of energy:

Kinetic energy due to its motion.
Potential energy due to it lying in the electric field of the nucleus.

Thus the total energy of the electron is given by
E = K. E. + P. E. …(1)

An electron of mass m moving around the nucleus with an orbital velocity v has kinetic energy given by
K.E. = \(\frac{1}{2}\)mv² = \(\frac{1}{2} \frac{k e^{2}}{r}\) …(2)

Now the potential energy of the electron at a distance r from the nucleus is given by
PE = potential due to the nucleus at a distance r × charge on the electron = V × – e …(3)

Now the potential at a distance r from the nucleus having a charge e is given by
V = k \(\frac{e}{r}\) …..(4)

Substituting in equation (3) we have
P.E. = V × – e = -k\(\frac{e^{2}}{r}\) …(5)

Substituting equations (2) and (3) in equation 1 we have
Class 12 Physics Important Questions Chapter 12 Atoms 7
But the radius of the nth orbit is given by
rn = \(\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2} k}\)

Substituting in equation (6) we have
E = – \(\frac{2 \pi^{2} m e^{4} \mathrm{k}^{2}}{n^{2} \mathrm{~h}^{2}}\) …(7)

This gives the expression for the energy possessed by the electron in the nth orbit of the hydrogen atom.

Question 3.
Hydrogen atoms are excited with an electron beam of energy of 12.5 eV. Find
(a) The highest energy level up to which the hydrogen atoms will be excited.
Answer:
The maximum energy that the excited hydrogen atom can have is
Class 12 Physics Important Questions Chapter 12 Atoms 8
For n=4, E4 = \(\frac{-13.6}{16}\) = – 0.85eV
(> – 1.1 eV)

∴ The eLectron can only be excited up to n = 3 states.

(b) The longest wavelengths in the (i) Lyman series, (ii) Balmer series of the spectrum of these hydrogen atoms. (CBSE 2019C)
Answer:
From energy tevet of hydrogen atom,
we have
\(\frac{1}{λ}\) = R\(\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\)

Longest wavelength of Lyman senes
Class 12 Physics Important Questions Chapter 12 Atoms 9
Question 4.
Using Bohr’s postulates of the atomic model derive the expression for the radius of the 11th electron orbit. Hence obtain the expression for Bohr’s radius. (CBSE AI 2014)
Answer:
Let us consider a mechanical model of the hydrogen atom as shown in the figure that incorporates this quantization assumption.
Class 12 Physics Important Questions Chapter 12 Atoms 10
This atom consists of a single electron with mass m and charge – e revolving around a single proton of charge + e. The proton is nearly 2000 times as massive as the electron, so we can assume that the proton does not move. As the electron revolves around the nucleus the electrostatic force of attraction between the electron and the proton provides the necessary centripetal force. Therefore, we have
Class 12 Physics Important Questions Chapter 12 Atoms 11
This gives the radius of the nth orbit of the hydrogen atom.
If n = 1 we have r = ao which is called Bohr’s radius.
ao = \(\frac{h^{2}}{4 \pi^{2} m e^{2} k}\)

Question 5.
State Bohr’s postulate of the hydrogen atom successfully explains the emission lines in the spectrum of the hydrogen atoms.
Use the Rydberg formula to determine the wavelength of Ha line. [Given Rydberg constant R = 1.03 × 107 m-1] (CBSE AI 2015)
Answer:
It states that an electron might make a transition from one of its specified non¬radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by

hv = Ei – Ef where Ei and Ef are the energies of the initial and final states
Using the formula \(\frac{1}{λ}\) = RH\(\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)\) we have for Ha Line ni = 3 and nf = 2
Class 12 Physics Important Questions Chapter 12 Atoms 12

Question 6.
Using Bohr’s postulates derive the expression for the frequency of radiation emitted when an electron In a hydrogen atom undergoes a transition from a higher energy state (quantum number n-) to the towering state (n,). When an electron in a hydrogen atom jumps from the energy state ni =4 to n = 3, 2, 1, identify the spectral series to which the emission lines belong. (CBSE Delhi 201 1C)
Answer:
According to Bohr’s frequency condition, if an electron jumps from an energy Level E to E1, then the frequency of the emitted radiation is given by
hv = E – E1 …(1)

Let ni and nf be the corresponding orbits then
Class 12 Physics Important Questions Chapter 12 Atoms 13
This gives the frequency of the emitted radiation.
When ni =4 and nf = 3, Paschen series
When ni = 4 and nf = 2, Balmer series
When ni = 4 and nf = 1, Lyman senes

Question 7.
Calculate the ratio of the frequencies of the radiation emitted due to the transition of the electron In a hydrogen atom from Its (i) second permitted energy level to the first level and (ii) highest permitted energy level to the second permitted level. (CBSE Delhi 2018C)
Answer:
We have
Class 12 Physics Important Questions Chapter 12 Atoms 14
Question 8.
Monochromatic radiation of wavelength 975 A excites the hydrogen atom from its ground state to a higher state. How many different spectral lines are possible In the resulting spectrum? Which transition corresponds to the longest wavelength amongst them? (CBSE Sample Paper 201819)
Answer:
The energy corresponding to the given wavelength:
Class 12 Physics Important Questions Chapter 12 Atoms 15
The longest wavelength Will correspond to the transition n = 4 to n = 3

Question 9.
(a) Using postulates of Bohr’s theory of hydrogen atom, show that
(i) the radii of orbits increases as n², and
Answer:
Let us consider a mechanical. model of the hydrogen atom as shown in the figure.
Class 12 Physics Important Questions Chapter 12 Atoms 16
This atom consists of a single electron with mass m and charge – e revolving around a single proton of charge + e. As the electron revolves around the nucleus the electrostatic force of attraction between the electron and the proton provides the necessary centripetal force. Therefore we have,
Class 12 Physics Important Questions Chapter 12 Atoms 17
Substituting equation 3 in equation 2 we have
Class 12 Physics Important Questions Chapter 12 Atoms 18
This gives the radius of the nth orbit of the hydrogen atom which shows that E ∝ \(\frac{1}{n^{2}}\)

(ii) the total energy of the electron increases as 1/n², where n is the principal quantum number of the atom.
Answer:
the total energy possessed by an electron in the nth orbit of the hydrogen atom is given by
E = T + U …(1)
i.e. the sum of its kinetic and electrostatic potential energies.

An electron of mass m moving around the nucleus with an orbital velocity v has kinetic energy given by
K.E. = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\frac{k e^{2}}{r}\) ….(2)

Now the potential energy of the electron at a distance r from the nucleus is given by
PE = potential due to the nucleus at a distance r × charge on the electron
= V × – e …(3)

Now the potential at a distance r from the nucleus having a charge e is given by
V = k\(\frac{e}{r}\) ….(4)

Substituting in equation 2 we have
P.E. = V × – e = – k \(\frac{e^{2}}{r}\) …(5)

Substituting equations 2 and 5 in equation 1 we have
Class 12 Physics Important Questions Chapter 12 Atoms 19

But the radius of the nth orbit is given by rn = \(\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2} k}\)

Substituting in equation 6 we have
E = – \(\frac{2 \pi^{2} m e^{4} k^{2}}{n^{2} h^{2}}\) …(7)

This gives the expression for the energy possessed by the eLectron in the nth orbit of the hydrogen atom which shows that E ∝ \(\frac{1}{n^{2}}\)

(b) Calculate the wavelength of H2 line In Balmer series of hydrogen atom, given Rydberg constant R = 1.097 × 107 m-1. (CBSE AI 2011C)
Answer:
For H2 Line in Balmer series n1 = 2 and n2 = 3
Class 12 Physics Important Questions Chapter 12 Atoms 20
Question 10.
State Bohr’s quantization condition for defining stationary orbits. How does de Brogue hypothesis explain the stationary orbits?
Find the relation between the three wavelengths λ1 λ2 and λ3 from the energy level diagram shown below. (CBSE Delhi 2016)
Class 12 Physics Important Questions Chapter 12 Atoms 21
Answer:
It states that only those orbits are permitted in which the angular momentum of the electron about the nucleus is an integral multiple of \(\frac{h}{2π}\), where his Planck’s constant.

According to de Broglie, an electron of mass m moving with speed v would have a wavelength λ given by
λ = h/mv.

Now according to Bohr’s postulate,
mvrn = \(\frac{n h}{2 \pi}\)
or
2πrn = \(\frac{nh}{mv}\)

But h / mv = A is the de BrogUe wavelength of the electron, therefore, the above equation becomes 2πrn = nλ where 2πrn is the circumference of the permitted orbit. If the wavelength of a wave does not close upon itself, destructive interference takes place as the wave travels around the loop and quickly dies out. Thus only waves that persist are those for which the circumference of the circular orbit contains a whole number of wavelengths.
Class 12 Physics Important Questions Chapter 12 Atoms 22
Numerical Problem :
Formulae for solving numerical problems

  • Distance of closest approach ro = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 z E^{2}}{E_{k}}\)
  • Radius of the nth orbit of hydrogen atom rn = \(\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2} k}\)
  • Velocity of eLectron in the ntt orbit v = \(v=\frac{c}{137 n}\)
  • Wavelength of radiation emitted when electron jumps form ni to nf \(\frac{1}{\lambda}=R_{\mathrm{H}}\left(\frac{1}{n_{\mathrm{f}}^{2}}-\frac{1}{n_{\mathrm{i}}^{2}}\right)\)
  • Energy of electron in the nth orbit of hydrogen atom
    E = – \(\frac{2 \pi^{2} m e^{4} k^{2}}{n^{2} h^{2}}\)
    or
    E = – \(\frac{13.6}{n^{2}}\) eV

Question 1.
A hydrogen atom in its excited state emits radiations of wavelengths 1218 Å and 974.3 Å when it finally comes to the ground state. Identify the energy levels from where transitions occur. Given Rydberg constant R = 1.1 × 107 m-1. Also, specify the spectral series to which these lines belong. (CBSE AI 2019)
Answer:
We know that
Class 12 Physics Important Questions Chapter 12 Atoms 23
Class 12 Physics Important Questions Chapter 12 Atoms 24
On solving n = 4
Lyman series.

Question 2.
Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom, i.e. an atom where the electron is replaced by a negatively charged muon (μ–) of mass about 207 me that orbits around a proton. (Given for hydrogen atom, the radius of first orbit and ground state energy are 0.53 × 10-10m and – 13.6 eV respectively) (CBSE AI 2019)
Answer:
In Bohr’s model of hydrogen atom the radius of nth orbit is given by
rn = \(\frac{n^{2} h^{2}}{4 \pi^{2} e^{2} m_{e} k}\)

As n = 1
Therefore, we have 1
Class 12 Physics Important Questions Chapter 12 Atoms 25
Class 12 Physics Important Questions Chapter 12 Atoms 26

Question 3.
The electron in a given Bohr orbit has a total energy of – 1.5 eV. Calculate Its
(a) kinetic energy.
(b) potential energy.
(C) the wavelength of radiation emitted, when this electron makes a transition to the ground state.
(Given Energy in the ground state = – 13.6 eV and Rydberg’s constant = 1.09 × 1o7 m-1) (CBSE Delhi 2011C)
Answer:
Total energy of the electron In a Bohr’s orbit is – 1.5 eV

We know that kinetic energy of the electron in any orbt is half of the potential energy in magnitude and potential energy is negative
(a) Total energy = kinetic energy + potential energy
– 1.5 = Ek – 2Ek
1.5 = Ek

(b) Ep = – 2 × 1.5 = – 3 eV

(c) Energy released when the transition of this electron takes place from this orbit to the ground state
= – 1.5 – (- 13.6)
= 12.1 eV
= 12.1 × 1.6 × 1019 = 1.936 × 10-18 J

Let be the wavelength of the Light emitted then,
Class 12 Physics Important Questions Chapter 12 Atoms 27

Question 4.
In a Geiger—Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when an a-particle of 8 MeV energy Impinges on it before It comes momentarily to rest and reverses its direction. How will the distance of the closest approach be affected when the kinetic energy of the a-particle is doubled? (CBSEAI 2012)
Answer:
Given Z = 80, E = 8 MeV = 8 × 106 × 1.6 × 1019 J, ro =?
Using the expression
Class 12 Physics Important Questions Chapter 12 Atoms 28
When the kinetic energy of a-particle has doubled the distance of the closest approach becomes half its previous value, i.e. 1.44 × 1014 m

Question 5.
The ground state energy of the hydrogen atom is – 13.6 eV. If an electron makes a transition from an energy level – 0.85 eV to – 3.4 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong? (CBSE AI 2012)
Answer:
Energy released = – 0.85 – (- 3.4) = 2.55 eV = 2.55 × 1.6 × 10-19 J
Using E = hc/λ we have
λ = \(\frac{h c}{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{2.55 \times 1.6 \times 10^{-19}}\) = 4.87 × 10-7 m

It belongs to the Balmer series.

Question 6.
A hydrogen atom in the third excited state de-excites to the first excited state. Obtain the expressions for the frequency of radiation emitted in this process.
Also, determine the ratio of the wavelengths of the emitted radiations when the atom de-excites from the third excited state to the second excited state and from the third excited state to the first excited state. (CBSEAI2019)
Answer:
We know that
Class 12 Physics Important Questions Chapter 12 Atoms 29
Class 12 Physics Important Questions Chapter 12 Atoms 30
Question 7.
Obtain the expression for the ratio of the de Broglie wavelengths associated with the electron orbiting in the second and third excited states of the hydrogen atom. (CBSE Delhi 2019)
Answer:
We know that
2πr = nλ ….(i)
For the second excited state (n = 3)
r = 0.529(n)² A = 0.529(3)²

Putting in (i) we get 2π(0.529)(3)² = 3 λ2

For third excited state n = 4
r = 0.529 (4)²
Putting in (i) we get 2π (0.529)(4)² = 4 λ3
Class 12 Physics Important Questions Chapter 12 Atoms 31

Question 8.
(a) The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10-10 m. Calculate its radius in n = 3 orbit.
(b) The total energy of an electron in the first excited state of the hydrogen atom is 3.4 eV. Find out its (i) kinetic energy and (ii) potential energy in this state. (CBSE Delhi 2014C)
Answer:
(a) The radius is given by
rn = \(\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2} k}\) = n²ao

where n is the number of orbit, hence r3 = 5.3 × 10-10 × 32 = 4.77 × 10-9 m

(b) Total kinetic energy = + 3.4 eV
Total potential energy = – 6.8 eV

Question 9.
Given the ground state energy E0 = – 13.6 eV and Bohr radius a0 = 0.53 Å. Find out how the de Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state.
Answer:
The de-Broglie wavelength is given by 2πrn = nλ.
In ground state, n = 1 and ro = 0.53 Å, therefore, λo = 2 × 3.14 × 0.53 = 3.33 Å

In first excited state, n = 2 and r1 = 4 × 0.53 Å = 2.12 Å, therefore,
r1 = (2 × 3.14 × 2.12)/2 = 6.66Å

Therefore, λ1 – λo = 6.66 – 3.33 = 3.33 Å.
In other words, the de-Broglie wavelength becomes double.

Question 10.
When is the Hα line in the emission spectrum of hydrogen atom obtained? Calculate the frequency of the photon emitted during this transition. (CBSE AI 2016)
Answer:
Hα is obtained when ni = 3 and nf = 2
Class 12 Physics Important Questions Chapter 12 Atoms 32

Question 11.
A 12.5 eV beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelength and the corresponding series of lines emitted. (CBSE AI 2017)
Answer:
Given the energy of electron beam E = 12.5 eV
This will make the energy of the electrons 12.5 – 13.6 = – 1.1 eV.
Thus, the electrons of hydrogen will be raised to the third orbit.
Lines emitted will be ni = 3, nf = 2 Balmer series.

The wavelength is given by
Class 12 Physics Important Questions Chapter 12 Atoms 33
Class 12 Physics Important Questions Chapter 12 Atoms 34

For first member ni = 2, therefore, the wavelength of the first member of the Lyman series.
Class 12 Physics Important Questions Chapter 12 Atoms 35

Question 12.
Calculate the longest wavelength of the photons emitted in the Balmer series of the hydrogen spectrum. Which part of the e.m. spectrum does it belong to? [Given Rydberg constant, R = 1.1 × 107 m-1 J] (CBSE Delhi 2017C)
Answer:
The longest wavelength is obtained when the electron jumps from ni = 3 to nf = 2 in the Balmer series. Therefore we have,
Class 12 Physics Important Questions Chapter 12 Atoms 36
It lies in the visible region.

Question 13.
10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. ‘ Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite. (NCERT Exemplar)
Answer:
From Bohr’s postulate, we have mvn rn = nh/2π
Here m = 10 kg and rn = 8 × 106 m

We have the time period T of the circling satellite as 2 h.
That is T = 7200 s
Thus the velocity vn = 2πrn/T

The quantum number of the orbit of the satellite
N = (2πr)² × m/(T × h).

Substituting the values,
n = (2π × 8 × 106 )² × 10/(7200 × 6.64 × 10-34)
= 5.3 × 1045

Note that the quantum number for the satellite motion is extremely large. In fact, for such large quantum numbers, the results of quantization conditions tend to those of classical physics.

Question 14.
Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?
Answer:
No, because according to the Bohr model, E = – 13.6 /n² eV and electrons having different energies belong to different levels having different values of n.

So, their angular momenta will be different, as nh/2π = mvr

Question 15.
Using the Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state. (NCEFU Exemplar)
Answer:
Let v be the velocity of the electron in the ground state. Let a0 be the Bohr radius.
Now time is taken to complete one revolution
Class 12 Physics Important Questions Chapter 12 Atoms 37
Question 16.
What is the minimum energy that must be given to an H atom in the ground state so that it can emit a Hy line in the Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such Hy photon? (NCERJ Exemplar)
Answer:
The Hy in Balmer series corresponds to transition n = 5 to n = 2. So, the electron in ground state n = 1 must first be put in state n = 5.

Hence energy required for the same.
Energy required = E1 – E5 = 13.6 – 0.54 = 13.06 eV.

If angular momentum is conserved, angular momentum of photon = change in angular momentum of electron = L5 – L2 = 5h – 2h = 3h = 3 × 1.06 × 10-34 = 3.18 × 10-34 kg m² s-1

Communication Systems Class 12 Important Extra Questions Physics Chapter 15

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

Class 12 Physics Chapter 15 Important Extra Questions Communication Systems

Question 1.
flow will you classifi.’ communication systems?
Answer:
Communication systems can be classified based on the nature of source. mode of communication. type of modulation and nature of channel used.

Question 2.
What are the lips of channels used for transmission?
Answer:
i. Space communication (Broadcasting, microwave mobile etc.)
ii. Line communication (Two wire, co-axial cables, fiber optical etc.)

Question 3.
What is the length of antenna required to transmit wave of frequency 40 Hz and 40 MHz?
Answer:
The minimum length of antenna required is \(\frac {λ}{4} \)
Velocity,c = υλ
λ = \(\frac {c}{υ} \)
Class 12 Physics Important Questions Chapter 15 Communication Systems 1

Question 4.
Identify the sound that can travel a longer distance – siren from a factory or horn of a car. Why?
Answer:
Siren from a factory. High intensity.

Question 5.
Mention the factors on which power of electromagnetic wave transmitted depends.
Answer:
The power of electromagnetic wave is related to the length of antenna and wavelength of the wave.
Power α \(\left(\frac{l}{\lambda}\right)^{2}\), where l is the length and λ-the wavelength.

Question 6.
Which range of wave is more reliable of intermixing – shorter or longer wavelength?
Answer:
Longer wavelength

Question 7.
is there any change in the frequency or phase due to amplitude modulation?
Answer:
No change of either frequency or phase.

Question 8.
Which physical quantity of wave is varied in AM. FM and PM?
Answer:
In AM. the physical quantity of carrier that changes is amplitude.
In FM, the physical quantity of carriers that change is frequency.
In PM, the physical quantity of carrier that changes is a phase.

Question 9.
What are the advantages and limitations of AM and FM?
answer:

Advantages Limitations
AM Wireless transmission possible, simple circuit, two-sided bands Low efficiency, small operating range. noisy receptions, interference-effect.
FM More resistant to noise, a large number of sidebands, carrier frequency high (television broadcast), more economical space wave propagation Bandwidth wide, circuit more complex, a smaller area of reception

Question 10.
What is ground wave propagation?
Answer:
Ground wave follows curvature of the earth and has carrier frequencies up to 2MHz. e.g. AM radio.
Ground waves progress along the surface of the earth and must be vertically polarized to prevent short circuiting the electric equipments. A wave induces currents in the ground over which it passes and thus loses some energy by absorption. This is made up by energy diffracted downwards from the upper portion of the wavefront.
Class 12 Physics Important Questions Chapter 15 Communication Systems 2
There is another way also by which the ground waves get attenuated. Because of diffraction, the wavefront gradually tilts over, as shown in the figure. As the wave propagates over the earth, the tilt increases and this tilt causes greater short-circuiting of the electric component of the wave. Hence there is a reduction in the field strength. Eventually, at some distance from the antenna, the wave gets weakened and dies off. The maximum range of such a transmitter depends on its frequency and power. The ground wave propagation is effective only at VLF.

Question 11.
Ground waves are not sustained for long-range communication. Why?
Answer:
Because of damping by earth surface.

Question 12.
What is the range of frequencies used in ground wave propagation? Why?
Answer:
VLF. The attenuation of surface waves increases very rapidly with an increase in frequency.

Question 13.
How can we overcome this limitation?
Answer:
By changing to space wave communication.

Question 14.
What is the basic requirement of space wave communication?
Answer:
There should be a transmitter and receiving antenna.

Question 15.
Why is space communication also known as the line of sight communication?
Answer:
The transmitting and receiving antenna are on sight.

Question 16.
Why are repeaters needed in the line of sight communication?
Answer:
To compensate for the loss of energy during propagation.

Question 17.
What are the limitations of space wave propagation?
Answer:
This method needs repeaters and a suitable antenna length (height).

Question 18.
What is skywave communication?
Answer:
Skywave communication is otherwise called ionospheric communication. In this the electromagnetic wave of high frequency is directed towards the ionosphere which reflects the wave back to earth.

Question 19.
Which space transmission technology makes use of total internal reflection?
Answer:
ionospheric transmission.

Question 20.
is sky wave propagation possible on moon? Why?
Answer:
No. Moon has no ionosphere.

Question 21.
Can all frequencies be transmitted using sky wave propagation?
Answer:
No. Only frequencies below the critical value.

Question 22.
How does sky wave propagation depend on refractive index of atmosphere?
Answer:
The refractive index of ionosphere decreases below that of free space by the change of velocity of electrons in the ionosphere and the electromagnetic saves undergo total internal reflection.

Question 23.
Through which atmospheric layer. does the propagation take place in ground. space and sky communications?
Answer:
Ground wave – Troposphere
Space wave – Troposphere
Sky wave – Ionosphere

Question 24.
Which type of space communication has maximum range of transmission?
Answer:
Satellite

Question 25.
Compare the principle applied for each type of communication.
Answer:
Ground wave – Wireless
Space wave – Line of sight
Skywave – Total internet reflection by the ionosphere.

Question 26.
What is the range of frequency used ¡n each case?
Answer:
Ground wave – <2 MHz
Space wave -> 30 MHz
Skywave -< 10 MHz

Question 27.
Point out the limitations and uses in each case.
Answer:
Ground wave – Damping effect, wireless communication.
Space wave – Finite curvature of the earth, line of sight
Skywave – Critical frequency, long-distance coverage

Question 28.
Name the type of channel used in telephone, cable TV, and high-speed internet connections.
Answer:
Telephone – Two-wire
Cable TV – Coaxial cable
Internet – Space (satellite) .

Question 29.
Which ¡s the cheapest mode of line communication?
Answer:
Two-wire system

Question 30.
What are the merits and demerits of two wire communication?
Answer:
Signals can travel kilometres without amplification, digital and analogue signals can be sent cheap.
Attenuation of signal, interference etc.

Question 31.
Why twisted wires are preferred?
Answer:
To reduce interference of electromagnetic radiations.

Question 32.
Under which condition, does maximum power transmission occur through two wire lines?
Answer:
When the impedance of the detecting device at the receiver (load) is matched (i.e., equal) to the characteristic impedance of the two wire system.

Question 33.
What kind of cable is used to connect VC’R to TV?
Answer:
Coaxial

Question 34.
Draw the figure of coaxial cable.
Answer:
Class 12 Physics Important Questions Chapter 15 Communication Systems 3

Question 35.
What is the structure of coaxial cable?
Answer:
Coaxial cables are shielded i.e., outer conductor surrounds the insulated inner wire and the outer is always grounded.

Question 36.
Which type of material  is suitable to use as spacer in coaxial cable? Why?
Answer:
Solid dielectric material, for insulation.

Question 37.
What are the merits of coaxial cables?
Answer:
These cables do not suffer from radiation problem and can he used for microwaves.

Question 38.
For establishing a communication between a transmitting and receiving station, a physical medium is used.
Answer:
(a) Name the two principal classes of communication based on the physical medium used for propagation.
(b) Construct a table showing advantages and one practical application each for the two wire, coaxial cable and optic fiber communication.
(c) in cable TV transmission usually channel in UHF band carries relatively more noise, compared to VHF band. Justify
Answer:
(a) Line communication and space communication.
Class 12 Physics Important Questions Chapter 15 Communication Systems 4
(c) At higher frequency, radiation loss is high.

Question 39.
Schematic diagram for three types of satellite orbits are shown below and named as A.B.C.
Answer:
Class 12 Physics Important Questions Chapter 15 Communication Systems 5
(a) Identify the polar orbit and give its approximate height from earth.
(b) Give the criteria for selecting frequency of em wave to be used in photographs from satellites.
(c) A satellite T V company attempts to use 25,000 kHz for up linking signal to a sat ellite. Say whether they have selected apt frequency. Justify.
Answer:
(a) Orbit C. Its height is about 1000 km.
(b) i. Nature of the atmosphere.
ii. Reluctance of the object.
(c) No. Because frequency below 20 MHz will undergo total internal reflection at the ionosphere.

Question 40.
The following diagrams represent some of ihe modulated signals.
GSEB Solutions Class 12 Physics Chapter 15 Communication Systems image - 9
GSEB Solutions Class 12 Physics Chapter 15 Communication Systems image - 10
GSEB Solutions Class 12 Physics Chapter 15 Communication Systems image - 11
Which among is following correct
a. i only
b. ii only
c. iii only
d. both i and ii
Answer:
d. both i and ii

Question 41.
in communication the concept of information is the central thing. Instead of information we deal with messages, since there is no precise definition for the word information.
(a) Name the two distinct message categories.
(b) Explain them with examples.
Answer:
(a) In electronic communication systems, we use
i. analog
ii. digital signals

(b) i. Analog signal is continuous in amplitude and time variables. e.g. Speech converted microphone signal, the ECO etc.
ii Digital – This signal is discrete in amplitude and time. Here, the analog signal is subjected to time-sampling and amplitude quantization. e.g. Digital video stream, Data files etc.

Question 42.
List the various types of communications according to
(a) nature of information
(b) mode of transmission
(c) transmission channel
(d) types of modulation
Answer:
(a) Speech. picture, fax, data transmission
(b) Analog and digital communication
(c) i. Space communication
ii. Line communication
(d) i. Sinusoidal waves – AM, FM, PM
ii. Pulsed carrier waves – PAM, PTM, PPM, PNM, PCM

Question 43.
Explain the necessity of modulation.
Answer:
The unmodified signal from the source will be usually weak to be transmitted to long distance through channel. The Long-wave signal is then suitably combined with a high frequency (short wave) wave called carrier. During combining some property of the carrier is allowed to vary in proportion to that of the signal. This process is called modulation.

Question 44.
(a) What is meant by demodulation?
(b) What is its necessity?
(c) What are the different types of demodulation?
Answer:
(a) The process of extracting the information from a modulated wave is called demodulation or detection.
(b) Demodulation is an essential process for realization at the receiving end.
(c) There are different types of detectors depending on the type of modulation. e.g. optical detector, diode detector, etc.

Question 45.
(a) What does the figure represent?
(b) What is the function of ‘C’?
(c) What is the function of ‘R’?
Class 12 Physics Important Questions Chapter 15 Communication Systems 9
Answer:
(a) This is an AM demodulate (or detector)
(b) The capacitor charges to the peak voltage and then discharges through R. It serves as a capacitor filter. The diode rectifies the incoming AM signal. The rectified wave, while passing through the network, gets the RF carrier component removed thus producing the original signal.
(c) The voltage across R is the envelope of modulated wave (the signal)

Question 46.
Space wave communication is called troposphere wave propagation or LOS.
Answer:
It takes place at line of sight condition.

Question 47.
Name the four areas in which space technology finds application.
Answer:
Meteorology, climatology, oceanography, and coastal studies

Question 48.
“For long TV transmission, we need satellites “. Give reason.
Answer:
The bandwidth of picture from TV camera is about 64 Hz. Because of various factors that alternate the wave, a signal directly transmitted may not reach the destination.

Question 49.
Flow Kepler’s III law plays an important role in satellite communication?
Answer:
The stable orbit is designed by the condition that square of time period is x to the cube of mean distance of the satellite from earth.

Question 50.
(a) What do you understand by synchronous satellite?
(b) Why are such satellites used for world wide communications?
Answer:
(a) A satellite with period of revolution 24 hrs.
(b) High availability, reliability and wide coverage area.

Chemistry in Everyday Life Class 12 Important Extra Questions Chemistry Chapter 16

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 16 Chemistry in Everyday Life. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 16 Important Extra Questions Chemistry in Everyday Life

Chemistry in Everyday Life Important Extra Questions Very Short Answer Type

Question 1.
Name a substance that can be used as an antiseptic as well as a disinfectant. (CBSE 2019C)
Answer:
Phenol / C6H65OH

Question 2.
Differentiate between disinfectants and antiseptics. (CBSE 2012)
Answer:
Both antiseptics and disinfectants kill or prevent the growth of microorganisms. Antiseptics are applied to living tissues (cuts, wounds). But disinfectants are applied to inanimate objects only (floor, instruments).

The same substance can act as an antiseptic as well as a disinfectant by varying the concentration of the solution used. For example, 0.2% solution of phenol acts as an antiseptic and its 1% solution is a disinfectant.

Question 3.
What is the cause of a feeling of depression in human beings? Name a drug that can be useful in treating this depression. (CBSE 2012)
Answer:
The cause of depression in human beings is a low level of noradrenaline. Because of the low level of noradrenaline, the signal-sending activity of the hormones becomes low and the person suffers from depression. Phenelzine is useful in treating this depression.

Question 4.
Name one substance that can act as both:
(i) Analgesic and antipyretic. (CBSE Sample Paper 2012)
Answer:
Aspirin

(ii) Antiseptic and disinfectant. (CBSE Sample Paper 2011)
Answer:
Phenol

Question 5.
What is the tincture of iodine? What is its use? (CBSE 2019C)
Answer:
2 – 3% iodine solution of alcohol-water is called tincture of iodine. It is a powerful antiseptic and is applied on wounds.

Question 6.
(a) Which one of the following is a food preservative?
Equanil, Morphine, Sodium benzoate. (CBSE Delhi 2013)
Answer:
Sodium benzoate

(b) If the water contains dissolved Ca2+ ions, out of soaps and synthetic detergent, which will you use for cleaning clothes? (CBSE AI 2013)
Answer:
Synthetic detergent.

Question 7.
Among the following which one acts as a food preservative?
Aspartame, Aspirin, Sodium benzoate, Paracetamol
Answer:
Sodium benzoate.

Chemistry in Everyday Life Important Extra Questions Short Answer Type

Question 1.
What are food preservatives? Name two such substances. (CBSE 2012)
Answer:
Food preservatives are chemical substances that are added to food materials to prevent spoilage and to retain their nutritive value for long periods. For example, sodium benzoate, potassium metabisulphite.

Question 2.
(i) Why is bithional added to soap?
Answer:
Bithional acts as an antiseptic agent and reduces the odors produced by the bacterial decomposition of organic matter on the skin.

(ii) Which class of drugs is used in sleeping pills? (CBSE AI 2018)
Answer:
Tranquilizers.

Question 3.
(i) What class of drug is Ranitidine?
Answer:
Antacid

(ii) Which of the following is an antiseptic?
0. 2% phenol, 1% phenol(CBSE AI 2013)
Answer:
0.2% phenol: antiseptic; 1% phenol: disinfectant

Question 4.
Give one example for each of the following:
(i) An artificial sweetener whose use is limited to cold drinks.
Answer:
Aspartame.

(ii) A non-ionic detergent. (CBSE Sample Paper 2011)
Answer:
Ester of stearic acid and polyethylene glycol
CH3(CH2)16COO(CH2CH2O)nCH2CH2OH

Question 5.
Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take their doses without consultation with the doctor. Why? (CBSE Sample Paper 2011)
Answer:
Most of the drugs taken in doses higher than recommended may cause harmful effects and act as poison leading to death. Therefore, a doctor must always be consulted before taking any medicine, who will advise the patient for proper and safe doses of the drug.

Question 6.
Why do soaps not work in hard water? (CBSE Delhi 2011)
Answer:
Hard water contains calcium and magnesium salts. Therefore, in hard water soap gets precipitated as insoluble calcium and magnesium soaps which being insoluble stick to the cloth as gummy mass and blocks the ability of soap to remove oil or grease from the cloth.

Question 7.
Explain the following term with one suitable example:
Antifertility drugs (CBSE 2010)
Answer:
Antifertility drugs. These are the chemical substances used to control the pregnancy. These are also called oral contraceptives birth control pills. The common drugs used as antifertility drugs are norethindrone, ethynylestradiol (nostril), mifepristone, ormeloxifene, etc.

Chemistry in Everyday Life Important Extra Questions Long Answer Type

Question 1.
(i) What type of drug is used in sleeping pills?
(ii) What type of detergents is used in toothpaste?
(iii) Why is the use of alitame as an artificial sweetener not recommended?
OR
Define the following terms with a suitable example in each:
(i) Broad-spectrum antibiotics
(ii) Disinfectants
(iii) Cationic detergents (CBSE Delhi 2019)
Answer:
(i) Tranquilizers
(ii) Anionic detergents
(iii) With alitame, it is difficult to control the sweetness of food to which they are added.

OR
(i) Broad-spectrum antibiotics: These are the antibiotics that are effective against several types of harmful microorganisms and are used for curing a variety of diseases, for example, chloramphenicol.

(ii) Disinfectants: The chemical substances which are used to kill microorganisms but cannot be applied on living tissues are called disinfectants, for example, phenol (1% solution).

(iii) Cationic detergents: The substances which have a long hydrocarbon chain with a positive charge on the nitrogen atom (cationic part) which is involved in cleaning action are called cationic detergents. These are quaternary ammonium salts of amines with acetates, chlorides, or bromides as anions, for example, cetyltrimethylammonium bromide.

Question 2.
What is the following substance? Give one example of it:
Antacids (CBSE2011, CBSE Delhi 2012)
Answer:
Antacids: These are the chemical substances that neutralize excess acid in the gastric juices and give relief from acid indigestion, acidity, heartburns, and gastric ulcers. Until 1970, the antacids such as sodium bicarbonate or a mixture of aluminum and magnesium hydroxide have been commonly used for the treatment of acidity.

However, excessive bicarbonate can make the stomach alkaline and trigger the production of even more acid. Nowadays, acidity is cured by drugs such as cimetidine (Tagamet), ranitidine (Zantac), omeprazole, lansoprazole, etc.

Question 3.
(i) Which one of the following is a food preservative?
Equanil, Morphine, Sodium benzoate
Answer:
Sodium benzoate

(ii) Why is bithional added to soap?
Answer:
Bithional acts as an antiseptic agent and reduces the odors produced by the bacterial decomposition of organic matter on the skin.

(iii) Which class of drugs is used in sleeping pills? (CBSE Delhi 2013)
Answer:
Tranquilizers.

Question 4.
(i) What class of drug is Ranitidine?
Answer:
Antacid

(ii) If the water contains dissolved Ca2+ ions, out of soaps and synthetic detergent, which will you use for cleaning clothes?
Answer:
Synthetic detergent

(iii) Which of the following is an antiseptic:
0. 2% phenol, 1% phenol? (CBSE 2013)
Answer:
0.2% phenol: antiseptic; 1% phenol: disinfectant

Question 5.
(i) Give two examples of macromolecules that are chosen as drug targets.
Answer:
Carbohydrates, proteins.

(ii) What are antiseptics? Give an example.
Answer:
The chemical substances which are used to either kill or prevent the growth of micro-organisms are called antiseptics. These are not harmful to live tissues and can be safely applied on wounds, cuts, ulcers, diseased skin surfaces, etc.
For example, Dettol.

(iii) Why is the use of aspartame limited to cold foods and soft drinks? (CBSE Delhi 2014)
Answer:
Aspartame is unstable at cooking temperatures and therefore, it is used as sugar substitute for cold foods and
soft drinks.

Question 6.
(i) Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Answer:
Saccharin

(ii) What are antibiotics? Give an example.
Answer:
Antibiotics are chemical substances that are produced by micro-organisms (bacteria, fungi, moulds) and can inhibit the growth or even destroy other micro-organisms. For example, Penicillin.

(iii) Give two examples of macromolecules that are chosen as drug targets. (CBSE Delhi 2014)
Answer:
Proteins, lipids.

Question 7.
(a) Differentiate between antiseptic and disinfectant. Give one example of each.
(b) Why do we require artificial sweetening agents?
OR
Define the following terms:
(a) Tranquilizers
(b) Antacids
(c) Analgesics (CBSE2019C)
Answer:
(a) Antiseptics: Antiseptics are the chemical substances that are used to either kill or prevent the growth of micro-organisms. These are not harmful to live tissues and can be safely applied on wounds, cuts, diseased skin surfaces. For example, Dettol, Savlon, furnace, tobramycin, etc.

Disinfectants: Disinfectants are chemical substances that kill micro-organisms but cannot be applied to living tissues. In other words, they also kill micro-organisms like antiseptics but are not safe for living tissues. These are commonly applied to inanimate objects such as the floor, drainage systems, instruments, etc. Some common examples of disinfectants are phenol (1% solution), chlorine (0.2 to 0.4 ppm), etc.

(b) Artificial sweetening agents are used to reduce calorie intake. These also protect teeth from decay.
OR
(a) The chemical substances used for the treatment of stress, fatigue, mild and severe mental diseases are called tranquilizers.

Tranquilizers are neurologically active drugs that affect the message transfer mechanism from nerve to receptor. These are used to relieve or reduce mental tension, irritability, excitement, and anxiety leading to calmness. These form an essential component of sleeping pills.

(b) Acidic stomach is necessary for good health, but excessive acidity in the stomach can cause discomforts such as acid indigestion, heartburn, irritation, or pain of gastric ulcers. The chemical substances which neutralize excess acid in the gastric juices and give relief from add indigestion, acidity, heartburns, and gastric ulcers are called antacids.

(c) The chemical substances which are used to relieve pains without causing impairment of consciousness, mental confusion, incoordination or paralysis, or some other disturbances of the nervous system are called analgesics.

These are of two types:

  1. Non-narcotic (non-addictive) drugs
  2. Narcotic drugs

Question 8.
Explain the following terms with one suitable example for each:
(i) A sweetening agent for diabetic patients
Answer:
A sweetening agent for diabetic patients: The chemical substances which give a sweetening effect to food but do not add any calorie to our body are called artificial sweetening agents. The sweetening agent for diabetic patients is saccharin.

(ii) Enzymes
Answer:
Enzymes: Enzymes are biological catalysts produced by living cells that catalyze the biochemical reaction in living organisms. Chemically enzymes are naturally occurring simple or conjugated proteins and some enzymes are non-proteins also. These increase the rates of biochemical reactions by providing alternative paths of lower energy. For example, maltase, which catalyzes the hydrolysis of maltose to glucose.
Class 12 Chemistry Important Questions Chapter 16 Chemistry in Everyday Life 1

(iii) Analgesics (CBSE AI 2011)
Answer:
Analgesics: These are medicines used to relieve pains. Aspirin and some other antipyretics are also used as analgesics. Certain narcotics (which produce sleep and unconsciousness) are also used as analgesics. For example, morphine, codeine, heroin, marijuana. These are known to be habit-forming.

Question 9.
Define and write an example for the following:
(a) Broad-spectrum antibiotics.
Answer:
Broad-spectrum antibiotics: These are the antibiotics that are effective against several types of harmful micro-organisms. Therefore, these are used for curing a variety of diseases. The common examples are tetracycline, Chloromycetin, and chloramphenicol which are effective against a variety of diseases.

Other important broad-spectrum antibiotics used are vancomycin and ofloxacin. The antibiotic dysidazirine is found to be toxic to certain strains of cancer cells.

(b) Analgesics (CBSE Sample Paper 2019)
Answer:
Analgesics: The chemical substances which are used to relieve pains without causing impairment of consciousness, mental confusion, incoordination or paralysis, or some other disturbances of the nervous system are called analgesics.

These are of two types:

  1. Non-narcotic (non-addictive) drugs
  2. Narcotic drugs

1. Non-narcotic drugs: The common non-addictive analgesics are aspirin and paracetamol. Aspirin (2-acetoxy benzoic acid) is the most familiar example. It inhibits the synthesis of compounds known as prostaglandins which stimulate inflammation in the tissues and cause pain. These drugs are effective in relieving skeletal pain such as that due to arthritis.

Aspirin has also been very popular because it has antipyretic (temperature lowering) properties. Now, aspirin also finds use in the prevention of heart attack because it has anti-blood-clotting action. In addition, many other potential applications of aspirin, presently under investigation, include pregnancy-related complications, viral inflammation in AIDS patients, Alzheimer’s disease, dementia, cancer, etc.

Because of the shortcomings of aspirin, other analgesics like naproxen, ibuprofen, and diclofenac sodium or potassium find use as alternatives.

2. Narcotic drugs: Certain narcotics (which produce sleep and unconsciousness) are also used as analgesics. For example, morphine and its derivatives codeine, heroin, and marijuana are used in severe pain as analgesics. These are known to be habit-forming. When used in medicinal doses, these relieve pain and produce sleep. However, in excessive (poisonous) doses these produce stupor coma, convulsions and ultimately leading to death.

These analgesics are mainly used for relief in postoperative pains, cardiac pain, and pains related to childbirth and terminal cancer.

Question 10.
(a) Why are metal hydroxides better alternatives than sodium hydrogen carbonate in antacids?
(b) Why is aspirin used in the prevention of heart attacks?
(c) Why antihistamines do not affect the secretion of acid in the stomach?
OR
Define the following terms with a suitable example of each:
(a) Tranquilizers
(b) Antibiotics
(c) Non-ionic detergents (CBSE AI 2019)
Answer:
(a) Metal hydroxides are better alternatives than sodium hydrogen carbonates because of being insoluble. These do not increase the pH above neutrality. However, excessive hydrogen carbonates can make the stomach alkaline and may trigger the production of even more acid.

(b) Aspirin is a blood thinner and has anti-blood-clotting action and therefore, used in the prevention of heart attacks.

(c) Antihistamines do not affect the secretion of acid in the stomach because anti-allergic and antacid drugs work on different receptors. The receptors present in the stomach do not react with antihistamines.
OR
(a) Tranquilizers: These are substances used to relieve mental diseases. They reduce tension and anxiety. They act on higher centers of the nervous system. These are the constituents of sleeping pills. The common examples are Derivatives of barbituric acid, Equanil, luminal, diazepam, methedrine, etc.

(b) Antibiotics: Antibiotics are chemical substances that are produced by micro-organisms (bacteria, fungi, molds) and can inhibit the growth or even destroy other micro-organisms. For example, penicillin.

(c) Non-ionic detergents: These detergents do not contain any iron in their constitution and therefore, are non-ionic like the esters of high molecular mass. However, these contain polar groups which can form hydrogen bonds with water. For example, polyethylene glycol stearate.

Question 11.
(i) Why bithional is added in soap?
(ii) Why magnesium hydroxide is a better antacid than sodium bicarbonate?
(iii) Why soaps are biodegradable whereas detergents are non-biodegradable?
OR
Define the following terms with a suitable example in each:
(i) Antibiotics
(ii) Artificial sweeteners
(iii) Analgesics (CBSE Delhi 2019)
Answer:
(i) Bithional is added to soap because it acts as an antiseptic agent and reduces the odors produced by the bacterial decomposition of organic matter on the skin.

(ii) Magnesium hydroxide is insoluble and does not increase the pH above neutrality. But excess sodium bicarbonate can make the stomach alkaline and may trigger the production of even more acid.

(iii) The synthetic detergents have hydrocarbon chains that are highly branched. Bacteria cannot degrade them easily. Therefore, detergents are non-biodegradable. On the other hand, soaps have unbranched chains which can be biodegraded more easily. Hence, soaps are biodegradable.
OR
(i) Antibiotics: Antibiotics are chemical substances that are produced by microorganisms (bacteria, fungi, and molds) and can inhibit the growth or even destroy other micro-organisms, for example, penicillin, ampicillin, Amoxycillin, etc.

(ii) Artificial sweeteners: These are the chemical compounds that give a sweetening effect to the food and enhance its odor and flavor, for example, saccharin, aspartame, alitame, etc.

(iii) Analgesics: These are neurologically active pain-killing drugs that reduce or abolish pain without causing impairment of consciousness, incoordination, mental confusion, paralysis, or some other disturbances of the nervous system, for example, aspirin, paracetamol, naproxen, ibuprofen, etc.

Question 12.
Give reasons for the following:
(i) Use of aspartame as an artificial sweetener is limited to cold foods.
Answer:
It is unstable at cooking temperature.

(ii) Metal hydroxides are better alternatives than sodium hydrogen carbonate for the treatment of acidity.
Answer:
Excessive hydrogen carbonate can make the stomach alkaline and trigger the production of even more acid. Metal hydroxides being insoluble do not increase the pH above neutrality,

(iii) Aspirin is used in the prevention of heart attacks. (CBSE Sample Paper 2018, 2019)
Answer:
Aspirin has anti-blood-clotting action.

Question 13.
(i) Why is bithional added to soap?
Answer:
To impart antiseptic properties

(ii) What is the tincture of iodine? Write its one use.
Answer:
2-3% solution of iodine in the alcohol-water mixture is the tincture of iodine. Iodine dissolved in alcohol, used as an antiseptic, and applied on wounds.

(iii) Among the following, which one acts as a food preservative?
Sodium benzoate, Aspartame (CBSE 2016)
Answer:
Sodium benzoate

Question 14.
(a) Which one of the following is a disinfectant?
0-2% solution of phenol or 1% solution of phenol
(b) What is the difference between agonists and antagonists?
(c) Write one example of each of
(i) Artificial sweetener
(ii) Antacids
OR
Define the following terms with a suitable example of each:
(a) Antiseptics
(b) Bactericidal antibiotics
(c) Cationic detergents (CBSE Al 2019)
Answer:
(a) 1% solution of phenol

(b) Agonists are substances that bind the receptor and produce a biological response.
Antagonists are the substances that bind to the receptor but inhibit its natural biological response.

(c) (i) Saccharin
(ii) Ranitidine
OR
(a) Antiseptics: The chemicals which either kill or prevent the growth of microorganisms when applied on living tissues are called antiseptics, for example, tobramycin.

(b) Bactericidal antibiotics: The chemicals which have a killing effect on microbes are called bactericidal antibiotics, for example, penicillin.

(c) Cationic detergents: The molecules which have long hydrocarbon chains with a positive charge on the nitrogen atom (cationic part) which is involved in cleansing action are called cationic detergents. These are quaternary ammonium salts of amines with acetates, chlorides, or bromides as anions. The cationic part of the molecule is involved in the cleansing action, for example, cetyltrimethylammonium bromide.

Question 15.
(i) How are synthetic detergents better than soaps?
Answer:
The detergents are better than soaps because of the following reasons:
(a) Detergents can be used for washing even in hard water. On the other hand, soaps cannot be used in hard water.
(b) Detergents can be used in acidic solutions because they are not readily decomposed in an acidic medium. On the other hand, soaps cannot be used in an acidic medium because they are decomposed into carboxylic acids in an acidic medium.
(c) Detergents have a stronger cleansing action than soap.

(ii) Can you use soaps and synthetic detergents to check the hardness of water? (CBSE 2015)
Answer:
Soaps give an insoluble precipitate of calcium and magnesium in hard water whereas detergents do not give a precipitate. Therefore, soaps but not detergents can be used to check the hardness of the water.

Question 16.
Explain the role of the allosteric site in enzyme inhibition?
Answer:
Some drugs do not bind to the enzyme’s active site. These bind to a different site of enzyme which is called the allosteric site. This bonding of inhibitor at the allosteric site changes the shape of the active site in such a way that the substrate cannot recognize it. As a result, the affinity of the substrate for the active site is reduced.
Class 12 Chemistry Important Questions Chapter 16 Chemistry in Everyday Life 2
It may be noted that if the bond formed between enzyme and inhibitor is a strong covalent bond and therefore cannot be broken easily, then the enzyme gets blocked permanently. The body then degrades the enzyme-inhibitor complex and synthesizes a new enzyme.

Question 17.
What are food additives? Describe the following with suitable examples:
(i) Preservatives
(ii) Artificial sweetening agents
Answer:
Food additives are the chemicals that are added to food for their preservation and enhancing their appeal. Some common food additives are:
(a) Flavours and sweeteners
(b) Food colors (dyes)
(c) Stabilising agents
(d) Antioxidants
(e) Preservatives
(f) Nutritional supplements such as vitamins, minerals, and amino acids.

(i) Preservatives: These are the chemical substances that are added to the food materials to prevent their spoilage and to retain their nutritive value for long periods. These preservatives prevent the rancidity of food and inhibit the growth or kill the micro-organisms.

The common salt is generally added to resist the activity of micro-organisms in food. The preservation of food by adding a sufficient amount of salt to it is called salting. It is used for the preservation of raw mango, amla, beans, tamarind, fish, meat, etc. The salt prevents the water from being available for microbial growth.

Sugar syrup is also used for preserving many fruits such as apples, mango, strawberry, carrot, etc. Besides these vinegar, oils, spices, citric acid is also used as food preservatives, which are used for pickles, ketchup, jams, squashes, etc.

The growth of microbes in food material can also be prevented by adding certain chemical substances. The most common preservative used is sodium benzoate (C6H5COONa), salts of propanoic acid, sorbic acid, and potassium metabisulphite (source of sulfur dioxide). Certain food preservatives such as butylated hydroxyanisole (BHA) and butylated hydroxytoluene (BHT) for edible oils also act as antioxidants.

(ii) Artificial sweetening agents. These are the chemical compounds that give a sweetening effect to the food and enhance its odor and flavor. Ortho- sulphobenzimide known as saccharin is the most popular sweetening agent and has been used for many articles of food. It has a very sweet taste and is about 550 times sweeter than sucrose.

Other artificial sweeteners commercially used in food articles are aspartame (methyl ester), alitame, dulcin (urea sweetener), dihydrochalcones (DHC), sucralose, etc.

General Principles and Processes of Isolation of Elements Class 12 Important Extra Questions Chemistry Chapter 6

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 6 General Principles and Processes of Isolation of Elements. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 6 Important Extra Questions General Principles and Processes of Isolation of Elements

General Principles and Processes of Isolation of Elements Important Extra Questions Very Short Answer Type

Question 1.
What is the composition of copper matte? (CBSE Delhi 2013)
Answer:
Copper matte contains cuprous sulphide (Cu2S) and iron sulphide (FeS).

Question 2.
Name the method used for refining of
(i) Nickel
(ii) Zirconium (CBSE Sample Paper 2007, 2014)
Answer:
(i) Mond’s process
(ii) Van Arkel method

Question 3.
Write the overall reaction taking place in the process used for the electrolysis of alumina by Hall-Heroult process. (CBSE Sample Paper 2011)
Answer:
2Al2O3 + 3C → 4Al + 3CO2

Question 4.
Write a non-exothermic reaction taking place in the blast furnace during extraction of iron. (CBSE Sample Paper 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 1

Question 5.
Name the method of refining of metals such as germanium. (CBSE Delhi 2016)
Answer:
Zone refining.

Question 6.
In the extraction of Al, impure Al2O3 is dissolved in cone. NaOH to form sodium aluminate and leaving impurities behind. What is the name of the process? (CBSE Delhi 2016)
Answer:
Al2O3 + 2NaOH + 3H2O → 2Na [Al(OH)4]
This process is called leaching.

Question 7.
Name the method of refining which is based on the principle of adsorption.
(CBSE AI 2016)
Answer:
Chromatography.

Question 8.
Out of PbS and PbCO3 (ores of lead), which one is concentrated by froth floatation process preferably? (CBSE Delhi 2017)
Answer:
PbS

Question 9.
What is the role of depressants in the froth floatation process? (CBSE Al 2017)
Answer:
The depressants in the froth floatation process selectively prevent one of the sulphide ores from forming the froth with air bubbles.

Question 10.
What is the role of collector and froth stabilizer in froth floatation process?
(CBSE AI 2012)
Answer:

  • Collector enhances non-wettability of the mineral particles.
  • Froth stabilisers stabilise the froth.

Question 11.
Name the method used for refining of copper. (CBSE AI 2013)
Answer:
Electrolytic refining

Question 12.
What is the role of graphite in the electrometallurgy of aluminium? (CBSE Delhi 2012)
Answer:
The graphite rod is useful in the electrometallurgy of aluminium for reduction of alumina to aluminium.
2Al2O3 + 3C → 4Al + 3CO2

Question 13.
Why is it that only sulphide ores are concentrated by ‘froth floatation process’? (CBSE Delhi 2011)
Answer:
This is because the sulphide ore particles are preferentially wetted by oil and the gangue particles by water.

Question 14.
What is the role of collectors in Froth- Floatation process? (CBSE AI 2012, 2017)
Answer:
The collectors such as pine oil, eucalyptus oil and fatty acids enhance the non – wettability of the mineral particles in froth – floatation process.

Question 15.
Name the substance used as depressant in the separation of two sulphide ores in froth floatation method. (CBSE Sample Paper 2019)
Answer:
Sodium cyanide.

Question 16.
Which reducing agent is employed to get copper from the leached low grade copper ore? (CBSE Delhi 2014)
Answer:
Iron scraps.

Question 17.
What is the role of zinc metal in the extraction of silver? (CBSE 2014)
Answer:
Zinc acts as a reducing agent.

General Principles and Processes of Isolation of Elements Important Extra Questions Short Answer Type

Question 1.
Out of C and CO which is a better reducing agent for FeO?
(i) In the lower part of blast furnace (Higher temperature)
(ii) In the upper part of blast furnace (Lower temperature) (CBSE Sample Paper 2011)
Answer:
(i) C is better reducing agent at higher temperature (lower part of blast furnace).
(ii) CO is better reducing agent at lower temperature (higher part of blast furnace).

Question 2.
What is the role of limestone in the extraction of iron from its oxides? (CBSE Al 2016)
Answer:
Limestone decomposes to form lime (CaO) and carbon dioxide (CO2). The lime thus produced acts as a flux and combines with silica (present as impurity in oxides of iron) to produce slag.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 2

Question 3.
How is copper extracted from a low grade ore of it? (CBSE Al 2012)
Answer:
Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria. The solution containing copper ions (Cu2+) is treated with scrap iron or H2 as:
Cu2+(aq) + H2(g) → Cu(s) + 2H+(aq)
In this way, copper is obtained.

Question 4.
(i) Which solution is used for the leaching of silver metal in the presence of air in the metallurgy of silver?
(ii) Out of C and CO, which is a better reducing agent at the lower temperature range in the blast furnace to extract iron from the oxide ore? (CBSE Delhi 2013)
Answer:
(i) Dilute solution (0.5%) of NaCN or KCN in the presence of air.
(ii) CO

Question 5.
(i) Which of the following ores can be concentrated by froth floatation method and why?
Fe2O3, ZnS, Al2O3
(ii) What is the role of silica in the metallurgy of copper? (CBSE Delhi 2013)
Answer:
(i) ZnS, because in sulphide ores, the sulphide ore particles are preferentially wetted by oil and gangue particles by water.
(ii) During roasting, the copper pyrites are converted into a mixture of FeO and Cu2O.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 3
To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with ferrous oxide present to form fusible slag of iron silicate.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 4
The slag being lighter floats and forms the upper layer and is removed through slag hole. Therefore, silica helps to remove FeO in the metallurgy of copper.

Question 6.
(i) Name the method used for removing gangue from sulphide ores.
(ii) How is wrought iron different from steel? (CBSE Al 2013)
Answer:
(i) Froth floatation method.
(ii) Wrought iron is the pure form of iron and contains carbon and other impurities not more than 0.5%.
Steel contains 0.5 to 1.5% carbon along with small amounts of other elements such as Mn, Cr, Ni, etc. and other impurities.

Question 7.
Outline the principles behind the refining of metals by the following methods: (CBSE Delhi 2014)
(i) Zone refining method
(ii) Chromatographic method
Answer:
(i) Zone refining method. It is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
(ii) Chromatographic method. This is based on the principle that different components of a mixture are differently adsorbed on an adsorbent.

Question 8.
Write the principle behind the froth floatation process. What is the role of collectors in this process? (CBSE 2014)
Answer:
The froth floatation process is based on the principle of difference in the wetting properties of the ore and gangue particles with water and oil. The collectors such as pine oil, eucalyptus oil, fatty acids, etc. are used to enhance the non-wettability of the mineral particles.

Question 9.
What is meant by vapour phase refining? Write any one example of the process which illustrates this technique, giving the chemical equations involved.
OR
Write and explain the reactions involved in the extraction of gold. (CBSE Sample Paper 2019)
Answer:
Vapour phase refining: This method is based on the fact that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal. Thus, the two requirements are:

  • The metal should form a suitable compound with a suitable reagent.
  • The volatile compound should be easily decomposable so that the metal can be easily recovered.

For example, nickel is refined by this technique and the method is known as Mond process. In this method, nickel is heated in a stream of carbon monoxide to form volatile nickel tetracarbonyl, Ni(CO)4, complex.
The carbonyl vapours when subjected to higher temperature (450-470 K) undergo thermal decomposition giving pure nickel.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 5
OR
The extraction of gold involves leaching of metal present in the ore with CN– ions. This is an oxidation-reaction because during the leaching process, Au is oxidised to Au+ which then combines with CN– ions to form their respective soluble complexes.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 6
The metal is then recovered from these complexes by reduction or displacement method using a more electropositive zinc metal.
2 [Au (CN)2]– (aq) + Zn(s) → 2 Au (s) + [Zn (CN)4]2- (aq)
Zinc acts as a reducing agent.
This process is called hydro metallurgy.

General Principles and Processes of Isolation of Elements Important Extra Questions Long Answer Type

Question 1.
(i) Give one example of each the following:
(a) Acidic flux
(b) Basic flux
(ii) What happens when
(a) Cu2O undergoes self reduction in a silica line converter
(b) Haematite oxidises carbon to carbon monoxide? (CBSE Sample Paper 2012)
Answer:
(i) (a) Acidic flux: SiO2
(b) Basic flux: CaO

(ii) (a) Cu2O undergoes self reduction to form blister copper as:
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 7
(b) Haematite oxidises carbon to carbon monoxide forming iron.
Fe2O3 + 3C → 3CO + 2Fe

Question 2.
What is a flux? What is the role of flux in the metallurgy of iron and copper? (CBSE Sample Paper 2011)
Answer:
Flux is a substance which combines with gangue which may still be present in the calcined or roasted ore to form an easily fusible material called the slag.
Flux + Gangue → Slag (fusible)
In the metallurgy of copper, most of the ferrous sulphide present as impurity gets oxidised to ferrous oxide which combines with silica (flux) to form fusible slag.
FeO + SiO2 → FeSiO3
The slag being lighter floats and forms the upper layer which is removed from the slag hole from time to time.
In the blast furnace for metallurgy of iron, lime acts as a flux and combines with silica present as impurity to form slag.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 8

Question 3.
Describe the principle involved in each of the following processes.
(i) Mond process for refining of Nickel.
(ii) Column chromatography for purification of rare elements. (CBSE 2012)
Answer:
(i) Mond’s process for refining of nickel. Impure nickel is heated in a stream of carbon monoxide forming volatile complex, tetra carbonylnickel (0).
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 9
The volatile complex is decomposed at high temperature to give pure metal
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 10

(ii) Column chromatography for purification of rare elements. This technique is based on the differences in the adsorbing capacities of the metal and its impurities. The metal and the impurities are differently adsorbed. Impure metal is put in a liquid or gaseous medium (called moving phase) which is moved through an adsorbent (stationary phase).
Metal and impurities are adsorbed at different levels in column.

Question 4.
Which methods are usually employed for purifying the following metals?
(i) Nickel
(ii) Germanium
Mention the principle behind each one of them. (CBSE 2012)
Answer:
(i) Nickel. By vapour phase refining. This method is based on the principle that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal.

For example, nickel is refined by this technique and the method is known as Mond process. In this method, nickel is heated in a steam of carbon monoxide to form volatile nickel carbonyl Ni(CO)4.
The carbonyl vapours when subjected to higher temperature (450-470 K) undergo thermal decomposition giving pure nickel.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 11

(ii) Germanium. By Zone refining method. This method is based on the principle that the impurities are more soluble in melt than in the solid state of the metal. Therefore, an impure metal on solidification will deposit pure metal and the impurities will remain behind in the molten part of the metal.

Question 5.
Explain the principle of the method of electrolytic refining of metals. Give one example. (CBSE 2014)
Answer:
In electrolytic refining method, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. Both anode and cathode are placed in a suitable electrolytic bath containing soluble salt of the same metal. On passing the electrical current, metal ions from the electrolyte are deposited at the cathode in the form of pure metal, while equivalent amount of metal dissolves from the anode into the elctrolyte in the form of metal ions. The impurities fall down below the anode as anode mud. The reactions occurring at the electrodes are:
At cathode: Mn+ + ne– → M
At anode: M → Mn+ + ne–
Copper is refined using electrolytic refining method.

Question 6.
(i) Name the method used for the refining of zirconium. (CBSE 2015)
(ii) What is the role of CO in the extraction of iron?
(iii) Reduction of metal oxide to metal becomes easier if the metal obtained is in liquid state. Why?
Answer:
(i) Van Arkel method
(ii) CO acts as reducing agent. It reduces oxides of iron to iron.
FeO + CO → Fe + CO2
3Fe2O3 + CO → 2Fe3O4 + CO2
Fe3O4 + 4CO → 3Fe + 4CO2
Fe2O3 + CO → 2FeO + CO2

(iii) This is because if the metal is in liquid state, its entropy is higher than when it is in solid state. Therefore, the value of entropy change (ΔS) of the reduction process is more on +ve side when the metal formed is in the liquid state and the metal oxide being reduced is in solid state. As a result, the value of ΔG° becomes more on negative side and hence the reduction becomes easier.

Question 7.
(i) Name the method of refining which is based on the principle of adsorption. (CBSE 2008, 2016)
(ii) What is the role of depressant in froth floatation process?
(iii) What is the role of limestone in the extraction of iron from its oxides?
Answer:
(i) Chromatography

(ii) The depressants are used to prevent certain types of particles from forming the froth with bubbles in froth floatation process. This helps to separate two sulphide ores. For example, in case of an ore containing zinc sulphide (ZnS) and lead sulphide (PbS), sodium cyanide (NaCN) is used as a depressant. It forms a layer of zinc complex Na2[Zn(CN)4] with ZnS on the surface of ZnS and therefore, prevents it from forming the froth. Therefore, it acts as a depressant.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 12
However, NaCN does not prevent PbS from forming the froth and allows it to come with the froth.

(iii) Limestone decomposes to form lime (CaO) and carbon dioxide (CO2). The lime thus produced acts as a flux and combines with the silica (present as impurity in oxides of iron) to produce slag.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 13

Question 8.
(i) Name the method of refining of metals such as germanium. (CBSE Delhi 2016)
(ii) In the extraction of Al, impure Al2O3 is dissolved in cone. NaOH to form sodium aluminate and leaving impurities behind. What is the name of this process?
(ii) What is the role of coke in the extraction of iron from its oxides?
Answer:
(i) Zone refining
(ii) Al2O3 + 2NaOH + 3H2O → 2Na[Al(OH)4]
This process is called leaching.
(iii) The coke serves as a fuel as well as a reducing agent. It burns to produce CO2 and heat which raises the temperature to about 2200 K (in combustion zone).
C + O2 → CO2, ΔH = -393.4 kJ
It also reduces Fe2O3 to iron.
Fe2O3 + 3C → 2Fe + 3CO + Heat

Question 9.
(i) Indicate the principle behind the method used for the refining of zinc.
(ii) What is the role of silica in the extraction of copper?
(iii) Which form of the iron is the purest form of commercial iron? (CBSE Delhi 2015)
Answer:
(i) Zinc is refined by electrolytic refining:
In this method, the impure zinc is made anode, while a plate of pure zinc is made the cathode. These are placed in a suitable electrolytic bath containing suitable salt of the same metal (e.g. zinc sulphate) containing a small amount of dilute sulphuric acid. On passing current, zinc from the solution is deposited at the cathode, while an equivalent amount of zinc from anode goes into the electrolytic solution. Therefore, pure zinc is obtained at cathode.

(ii) During roasting, the copper pyrites are converted into a mixture of FeO and Cu2O.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 14
To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with ferrous oxide present to form fusible slag of iron silicate.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 15
The slag being lighter floats and forms the upper layer and is removed through slag hole. Therefore, silica helps to remove FeO in the metallurgy of copper.

(iii) Wrought iron is the purest form of commercial iron.

Question 10.
(a) What is the difference between calamine and malachite?
(b) Why is zinc used instead of Cu for recovery of Ag from [Ag(CN)2]?
(c) What is the role of cryolite in metallurgy of Al? (CBSE 2019C)
OR
(a) Give two points of differences between pig iron and cast iron.
(b) Outline the principle of zone refining.
Answer:
(a) Calamine is an ore of Zn while malachite is an ore of copper. Calamine is ZnCO3 while malachite is CuCO3.Cu(OH)2

(b) Zinc is more electropositive than silver and therefore, zinc displaces silver from its solution.
2[Ag(CN)2]– + Zn → [Zn(CN)4]2- + 2Ag
On the other hand, copper is less electropositive than silver and therefore, cannot displace silver from its solution. Zn is more reactive than Cu, so reduction will be faster in case of Zn.

(c) Fused alumina is a bad conductor of electricity. Purified alumina is mixed with molten cryolite (Na3AlF6) and is electrolysed in an iron tank lined inside with carbon. The molten cryolite decreases the melting point and increases the electrical conductivity. It acts as a solvent.

OR

(a)

Cast Iron Pig Iron
1. It is moulded pig iron. 1. It is obtained directly from the blast furnace.
2. It is made by melting pig iron with scrap iron and coke using hot air blast. 2. It dissolves some sulphur, silicon, phosphorus and manganese as impurities.
3. It contains about 3% carbon. 3. It contains about 4% carbon.
4. It is extremely hard and brittle. 4. It is more brittle.

(b) Zone refining – It is based on the principle that impurities are more soluble in the melt than the solid state of the metal. Therefore, an impure metal on solidification will deposit crystals of pure metal and the impurities will remain behind in the molten part of the metal.

Question 11.
How will you convert the following:
(i) Impure Nickel to pure Nickel
(ii) Zinc blende to Zinc metal
(iii) [Ag(CN)2]– to Ag (CBSE Delhi 2019)
Answer:
(i) Impure nickel is heated in a stream of carbon monoxide forming volatile complex tetracarbonylnickel (0). The volatile complex is decomposed at high tempreature to give pure nickel.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 16
This is called Mond’s process.

(ii) The zinc blende ore is concentrated by froth floatation process and then roasted in the presence of excess air at about 1200 K.
2ZnS + 3O2 → 2ZnO + 2SO2
Zinc oxide formed is reduced to zinc by heating with crushed coke at 1673 K in vertical fire clay retorts.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 17

(iii) [Ag(CN)2]– complex is heated with zinc to get silver.
2[Ag(CN)2]– + Zn → [Zn(CN)4]2- + 2Ag

Question 12.
(a) Name the method of refining which is
(i) used to obtain semiconductor of high purity,
(ii) used to obtain low boiling metal.
(b) Write chemical reactions taking place in the extraction of copper from Cu2S. (CBSE Delhi 2019)
Answer:
(a) (i) Zone refining
(ii) Distillation

(b) Cu2S is first roasted and converted to oxide
2Cu2S + 3O2 → 2Cu2O + 2SO2
The oxide is then easily reduced to metallic copper with coke.
Cu2O + C → 2Cu + CO

Question 13.
Describe how the following steps can be carried out.
(i) Recovery of Gold from leached gold metal complex.
(ii) Conversion of Zirconium iodide to pure Zirconium.
(iii) Formation of slag in the extraction of copper.
(Write the chemical equations also for the reactions involved)
OR
Explain the use of the following:
(i) NaCN in Froth Floatation Method.
(ii) Carbon monoxide in Mond process.
(iii) Coke in the extraction of Zinc from Zinc Oxide.
Answer:
(i) Leached gold complex is treated with Zinc and gold is recovered by displacement method.
2Au[(CN)2]r(aq) + Zn(s) → 2Au(s) + [Zn(CN)4]2-(aq)

(ii) Zirconium iodide is decomposed on a tungsten filament, electrically heated to 1800 K. Pure Zr metal is deposited on the filament.
ZrI4 → Zr + I2

(iii) Silica is added to the ore and heated. It helps to slag off iron oxide as iron silicate.
FeO + SiO2 → FeSiO3 (slag)

OR

(i) NaCN is used as depressants to separate two sulphide ores (ZnS and PbS) in Froth Floatation Method.
(ii) Carbon monoxide forms a volatile complex of nickel, nickel tetracarbonyl.
(iii) Coke is used as a reducing agent to reduce zinc oxide to zinc.

Question 14.
Write the role of
(i) NaCN in the extraction of gold from its ore.
(ii) Cryolite in the extraction of aluminium from pure alumina.
(iii) CO in the purification of Nickel. (CBSE 2018)
Answer:
(i) Gold is leached out in the form of a complex with dil. solution of NaCN in the presence of air. Here NaCN acts as leaching agent.
(ii) It lowers the melting point of alumina and makes it a good conductor of electricity.
(iii) CO forms a volatile complex with nickel which is further decomposed to give pure Ni metal.

Question 15.
(i) Write the role of ‘CO’ in the purification of nickel.
(ii) What is the role of silica in the extraction of copper?
(iii) What type of metals are generally extracted by electrolytic method? (CBSE Delhi 2019)
Answer:
(i) Carbon monoxide forms a volatile complex, Ni(CO)4, on heating nickel in a stream of CO while the impurities remain unaffected. The compound on further heating decomposes to give pure nickel.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 18
Thus, CO helps to purify nickel.

(ii) During roasting, the copper pyrites are converted into a mixture of Cu2O and FeO.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 19
To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with FeO to form fusible slag of iron silicate.
FeO + SiO2 → FeSiO3
The slag being lighter floats and forms the upper layer and is removed. Thus, silica helps to remove FeO.

(iii) Very reactive metals such as sodium, potassium, calcium, aluminium, etc. are extracted by electrolytic methods. Certain less reactive metals such as copper are also purified by using electro-refining method.

Question 16.
Write down the reactions taking place in blast furnace related to the metallurgy of iron in the temperature range 500 K to 800 K. What is the role of limestone in the metallurgy of iron?
OR
What happens when
(a) Silver is leached with NaCN in the presence of air?
(b) Copper matte is charged into silica-lined converter and hot air blast is blown?
(c) NaCN is added in an ore containing PbS and ZnS during concentration by froth floatation method? (CBSE Delhi Al 2019)
Answer:
Reactions taking place in blast furnace at 500K to 800K: The oxides of iron are reduced by CO.
FeO + CO → Fe + CO2
3Fe2O3 + CO → 2Fe3O4 + CO2
Fe3O4 + 4CO → 3Fe + 4CO2
Fe2O3 + CO → 2FeO + CO2
Role of Limestone: It acts as a flux which decomposes to calcium oxide.
CaCO3 → CaO + CO2
Limestone
CaO combines with impurity (e.g. Si02) to form slag which is then removed.
CaO + SiO2 → CaSiO3

OR

(a) Silver is leached with dilute solution (0.5%) of NaCN in the presence of atmospheric oxygen. The metal dissolves forming the complex.
4Ag + 8CN– + 2H2O + O2 → 4[Ag(CN)2]– + 4OH–
The metal is extracted from water-soluble complex by zinc metal.
2[Ag(CN)2]– + Zn → [Zn(CN)4]2- + 2Ag

(b) The copper matte containing Cu2S and FeS is put in silica-lined convertor and hot air blast is blown to convert remaining FeS to FeO, which is removed as slag with silica.
2FeS + 3O2 → 2FeO + 2SO2
FeO + SiO2 → FeSiO3
Cu2S or CuO gets converted to copper
2Cu2S + 3O2 → 2Cu2O + 2SO2
2Cu2O + Cu2S → 6Cu + SO2

(c) NaCN acts as a depressant in preventing ZnS from forming the froth. If forms a layer of zinc complex Na2[Zn(CN)4] on the surface of ZnS and thereby prevents it from forming the froth.
4NaCN + ZnS → Na2[Zn(CN)4] + Na2S

Long Answer Type Questions (LA-II)

Question 1.
Explain the following:
(i) Although thermodynamically feasible, in practice magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why?
(ii) Why is zinc and not copper used for the recovery of silver from the complex [Ag(CN)2]–?
(iii) The extraction of Au by leaching with NaCN involves both oxidation and reduction. Justify giving equations. (CBSE Sample Paper 2011)
(iv) Lime-stone is used in the manufacture of pig iron from haematite. Why?
Answer:
(i) Inspection of Ellingham diagram shows that ∆G vs T curves for Al2O3 and MgO intersect at a point corresponding to very high temperature of the order of 2000 K. This means above this temperature, ∆G for the reaction:
Al2O3 + 3Mg → 2Al + 3Mgo
would become negative and hence reduction will be feasible. However, this temperature is very high so that the process is uneconomical and technologically difficult.

(ii) Zinc is a stronger reducing agent (E° = – 0.76 V) and more electropositive than copper (E° = + 0.34 V). Therefore, zinc is used for recovery of Ag from [Ag(CN2)]– complex.

(iii) During leaching process, gold (Au) is first oxidised by O2 of the air to Au+ which then combines with CN– ions to form the soluble complex, dicyanoaurate (I).
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 20
Gold is then extracted from this complex by displacement method by using a more electropositive zinc metal. In this method, zinc acts as a reducing agent and it reduces Au+ to Au. Zinc itself gets oxidised to Zn2+ ions which combine with CN– ions to form soluble complex, sodium tetracyanozincate (II).
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 21
Thus, extraction of gold by leaching with NaCN involves both oxidation and reduction.

(iv) Haematite is an ore of iron and contains silica (SiO2) as the main impurity. The purpose of limestone is to remove SiO2 as calcium silicate (CaSiO3) slag.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 22

Question 2.
Name the principal ore of aluminium. Explain the significance of leaching in the extraction of aluminium. (CBSE AI 2013)
Answer:
The principal ore of aluminium is bauxite, Al2O3.2H2O. Leaching process is used to concentrate the ore of aluminium, bauxite, which is contaminated with impurities of silica (SiO2), iron oxide (Fe2O3), titanium oxide (TiO2), etc. Leaching is done by treating the powdered ore with hot cone. (45%) solution of NaOH at about 473-523 K and 35-36 bar pressure.
Al2O3.2H2O(s) + 2NaOH(aq) + H2O(l) → 2Na [Al(OH)4](aq)
The impurities of ferric oxide and silica are insoluble and are removed by filtration.
The solution containing sodium aluminate is neutralised by passing C02 gas and hydrated alumina is precipitated.
2Na [Al(OH)4] (aq) + CO2 (g) → Al2O3. x H2O (s) + 2Na HCO3(s)
Hydrated alumina is heated to 1473 K to get pure alumina.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 23

Question 3.
Describe the principle controlling each of the following processes:
(i) Vapour phase refining of titanium metal.
(ii) Froth floatation method of concentration of sulphide ore.
(iii) Recovery of silver after silver ore was leached with NaCN. (CBSE AI 2011, CBSE Delhi 2011)
Answer:
(i) The principle of vapour phase refining is that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal. During refining of titanium metal, the titanium metal is heated with l2 to form a volatile compound Til4, which on further heating at higher temperature decomposes to give pure titanium metal.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 24

(ii) This is based upon the different wetting properties of the ore and the gangue particles with oil and water. The mineral particles become wet by oil and the gangue particles by water. When the ore is mixed with water containing small quantities of pine oil and then by agitating the water by blowing air violently, the froth is formed. The froth carries the lighter ore particles to the surface and the heavier impurities sink to the bottom.

(iii) This is a chemical method and is useful for the ores which are soluble in certain suitable solvents but the impurities are not soluble. The impurities left undissolved are removed by filtration. The ore of silver, Ag2S (argentite), reacts with NaCN as
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 25

Sodium sulphide thus formed is oxidised to sodium sulphate by blowing air into the solution. This helps the reaction to occur in the forward direction.
4Na2S + 5O2 + 2H2O → 2Na2SO4 + 4NaOH + 2S

The above solution after filtration and removing insoluble impurities is heated with zinc to get silver.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 26

Question 4.
Describe the role of the following:
(i) NaCN in the extraction of silver from a silver ore
(ii) Iodine in the refining of titanium
(iii) Cryolite in the metallurgy of aluminium (CBSE 2010)
Answer:
(i) The role of NaCN in the extraction of silver is to do the leaching of silver ore in the presence of atmospheric air from which silver is obtained later by replacement with zinc as:
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 27

(ii) Iodine is used to form a volatile unstable compound of the metal, which is decomposed to get the pure metal. For example, titanium is heated with iodine to 523 K forming unstable titanium iodide which is decomposed to Ti as
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 28

(iii) Cryolite is added to bauxite ore before electrolysis because of the following reasons:
(a) It acts as a solvent.
(b) It lowers the melting point of alumina to about 1173 K.
(c) Addition of cryolite to alumina increases the electrical conductivity.

Question 5.
Describe the role of the following:
(i) SiO2 in the extraction of copper from copper matte
(ii) NaCN in froth floatation process
Answer:
(i) The copper matte containing Cu2S and FeS is put in silica lined converter. Some silica is also added and hot air blast is blown to convert remaining FeS to FeO, which is removed as slag with silica.
2FeS + O2 → 2FeO + 2SO2
FeO + Si02 → FeSiO3
Cu2S or CuO gets converted to copper.
2Cu2S + 3O2 → 2Cu2O + 2SO2
2Cu2O + Cu2S → 6Cu + SO2

(ii) NaCN in froth floatation process is used to separate one sulphide ore from another and is known as depressant.
For example, sodium cyanide can be used as a depressant in the separation of zinc sulphide ore (ZnS) and lead sulphide ore (PbS). Sodium cyanide forms a layer of zinc complex, Na2[Zn(CN)4], on the surface of ZnS and therefore, prevents it from forming the froth. Therefore, it acts as a depressant.
Class 12 Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements 29
However, NaCN does not prevent PbS from forming the froth. Thus, it selectively prevents ZnS from coming to the froth but allows PbS to come with the froth. Thus, the two ores can be separated by the use of a depressant.

Surface Chemistry Class 12 Important Extra Questions Chemistry Chapter 5

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 5 Surface. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 5 Important Extra Questions Surface Chemistry

Surface Chemistry Important Extra Questions Very Short Answer Type

Question 1.
Give an example of a shape selective catalyst. (CBSE Delhi 2010, 2011)
Answer:
Zeolite catalyst known as ZSM-5.

Question 2.
Why are medicines more effective in colloidal state? (CBSE Delhi 2019)
Answer:
Medicines are most effective in colloidal state because colloids have large surface area and hence the medicines can be easily adsorbed and assimilated.

Question 3.
What is difference between an emulsion and a gel? (CBSE Delhi 2019)
Answer:
Emulsions are colloidal solutions in which both the dispersed phase and the dispersion medium are liquids. Gel is a colloidal system in which a liquid is dispersed in a solid.

Question 4.
Define ‘peptization’. (CBSE2012)
Answer:
The process of converting a freshly prepared precipitate into colloidal form by addition of a suitable electrolyte is called peptization.

Question 5.
What is meant by ‘shape-selective catalysis? (CBSE 2012)
Answer:
The catalytic reaction which depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysis.

Question 6.
What happens when a freshly precipitated
Fe(OH)3 is shaken with little amount of dilute solution of FeCl3? (CBSE 2013)
Answer:
A reddish brown colloidal solution of Fe(OH)3 is obtained. This process is called peptization.

Question 7.
What is especially observed when a beam of light is passed through a colloidal solution? (CBSE 2013)
Answer:
Tyndall effect

Question 8.
Give one example each of sol and gel. (CBSE Delhi 2014)
Answer:
Sol: As2S3 sol.
Gel: Cheese.

Question 9.
Give one example each of ‘oil in water’ and ‘water in oil’ emulsion. (CBSE 2014)
Answer:
Oil in water emulsion:
Milk Water in oil emulsion: Butter

Question 10.
Write a method by which lyophobic colloids can be coagulated. (CBSE 2015)
Answer:
By electrophoresis lyophobic colloids can be coagulated. Other methods are by mixing two oppositely charged sols or by boiling or by addition of electrolyte.

Question 11.
Write the main reason for the stability of colloidal sols. (CBSE Delhi 2016)
Answer:
The stability of colloidal sols is because of the presence of electrical charge on the particles which prevents the particles to come close together.

Question 12.
What are lyophobic colloids? Give one example for them. (CBSE Delhi 2011)
Answer:
The colloidal solutions in which the particles of the dispersed phase have no affinity for the dispersion medium are called lyophobic colloids. For example, metal sulphides like As2S3.

Question 13.
Write one similarity between physisorption and chemisorption. (CBSE Delhi 2017, CBSE AI 2017)
Answer:
Both are surface phenomena and increase with increase in surface area.

Question 14.
Why are lyophilic colloidal sols more stable than lyophobic colloidal sols? (CBSE Delhi 2015)
Answer:
The lyophilic colloidal sols are more stable because they are highly hydrated in solution.

Question 15.
What is the difference between a sol and a gel? (CBSE Delhi 2017)
Answer:
In a sol, dispersion medium is liquid and dispersed phase is solid. On the other hand, in a gel, dispersion medium is solid and dispersed phase is liquid.

Question 16.
What type of colloid is formed when a liquid is dispersed in a solid? Give an example. (CBSE AI 2017)
Answer:
Gel
For example: Cheese

Question 17.
Which of the following is most effective electrolyte in the coagulation of Fe2O3.x H2O/Fe3+ sol?
KCl, AlCl3, MgCl2, K4[Fe(CN)6] (CBSE Sample Paper 2011)
Answer:
Since Fe(OH)3 sol is positively charged, the anion having highest charge will be most effective, i.e. [Fe(CN)6]4.

Question 18.
Why is colloidal gold used for intramuscular injection? (CBSE Sample Paper 2011)
Answer:
Colloidal gold is more effective because of larger surface area and therefore, is easily assimilated with blood which is colloidal.

Question 19.
What is colloidion? (CBSE Sample Paper 2011)
Answer:
4% solution of nitrocellulose in a mixture of alcohol and ether.

Question 20.
Leather gets hardened after tanning. Why? (CBSE Delhi 2015)
Answer:
Animal hide is colloidal in nature and has positively charged particles. When it is soaked in tannin which has negatively charged colloidal particles, it results in mutual coagulation. This results in the hardening of leather.

Question 21.
It is necessary to remove CO when ammonia is prepared by Haber’s process. Explain. (CBSE Delhi 2015)
Answer:
Carbon monoxide acts as a poison for the catalyst in Haber’s process and therefore, it will lower the activity of the catalyst. Thus, CO must be removed when ammonia is obtained by Haber’s process.

Question 22.
Addition of alum purifies water. Why? (CBSE AI 2015)
Answer:
Alum coagulates the impurities present in water by neutralising the charge.

Question 23.
Out of sulphur sol and proteins, which one forms multimolecular colloids? (CBSE Delhi 2016)
Answer:
Proteins are macromolecules which cannot form multimolecular colloids while sulphur sol has smaller S8 molecules which can coagulate to form multimolecular colloids.

Question 24.
Write the chemical method by which Fe(OH)3 sol is prepared from FeCl3. (CBSE AI 2017)
Answer:
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 1

Question 25.
Out of MgCl2 and AlCl3 which one is more effective in causing coagulation of negatively charged sol and why? (CBSE Delhi 2016)
Answer:
According to Hardy-Schulze rule, for negatively charged sol, greater the valency of the positive ion of the electrolyte added, greater is its coagulating power. Thus, AlCl3 (Al3+ ions) is more effective in causing coagulation of negatively charged sol than MgCl2 (Mg2+ ions).

Question 26.
Give one example each of lyophobic sol and lyophilic sol. (CBSE Delhi 2014)
Answer:
Lyophobic : As2S3, Lyophilic : Gelatin

Question 27.
Out of BaCl2 and KCl which one is more effective in causing coagulation of negatively charged colloidal sol. Give reason. (CBSE Delhi 2015)
Answer:
BaCl2 because greater the valency of the coagulating ion (positive ion), greater is its tendency to coagulate.

Question 28.
What are the dispersed phase and . dispersion medium in milk? (CBSE AI 2014)
Answer:
Liquid (dispersed phase), Liquid (dispersion medium)

Question 29.
What type of colloid is formed when a solid is dispersed in liquid? Give an example. (CBSE AI 2017)
Answer:
Sol, paints

Question 30.
CO (g) and H2(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions? (CBSE AI 2018)
Answer:
Selectivity of a catalyst. It is the ability of a catalyst to selectively form a particular product.

Surface Chemistry Important Extra Questions Short Answer Type

Question 1.
Describe a conspicuous change observed when
(i) a solution of NaCl is added to a sol of hydrated ferric oxide.
(ii) a beam of light is passed through a solution of NaCl and then through a sol. (C.B.S.E. Delhi 2012)
Answer:
1. Hydrated ferric oxide is positively charged. When an electrolyte such as NaCl is added, the excess Cl– ions neutralise positive charge and cause coagulation of the sol.

2. When a beam of light is passed through a colloidal solution, the path of the light becomes visible when viewed from the direction at right angle to that of incident beam. This phenomenon is called Tyndall effect. This is because of scattering of light by colloidal particles. The particles first absorb the incident light and then a part of it gets scattered by them and therefore the part becomes visible when seen at right angle to the direction of incident beam.

Question 2.
Name the two groups into which phenomenon of catalysis can be divided. Give an example of each group with the chemical equation involved. (CBSE Delhi 2012)
Answer:
Catalysis can be divided into two groups
(i) Homogeneous catalysis
(ii) Heterogeneous catalysis

(i) Homogenous catalysis: When the catalyst is present in the same phase as the reactants and products, it is called homogeneous catalysis. For example,
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 2
SO2(g) is oxidised to SO3(g) in the presence of gaseous nitric oxide as catalyst.

(ii) Heterogeneous catalysis: When the catalyst is in different phase than the
reactants, it is called heterogeneous catalysis. For example,
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 3

Question 3.
Write the dispersed phase and dispersion medium of the following colloidal systems: (CBSE Delhi 2013)
(i) Smoke
(ii) Milk
Answer:
Colloidal system: Dispersed phase/Dispersion medium
(i) Smoke: Solid/Gas
(ii) Milk: Liquid/Liquid

Question 4.
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the addition of small amounts of electrolytes? (CBSE Delhi 2013)
Answer:
The colloidal solutions in which the particles of the dispersed phase have great affinity for the dispersion medium are called lyophilic sols. For example, glue, gelatin, starch, proteins, etc.

The colloidal solutions in which there is no affinity between the particles of the dispersed phase and the dispersion medium are called lyophobic sols. For example, solutions of . metals like Ag, Au, Al(OH)3, Fe(OH)3, etc.

Lyophobic sols can be easily coagulated on the addition of small amounts of electrolytes.

Question 5.
What is the difference between oil/water (O/W) type and water/oil (W/O) type emulsions? Give an example of each type. (CBSE Delhi 2013)
Answer:
1. Oil-in-water emulsion (o/w). In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

2. Water-in-oil emulsion (w/o). In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

Question 6.
Peptizing agent is added to convert precipitate into colloidal solution. Explain. (CBSE Sample Paper 2011)
Answer:
Peptization is a process of converting a freshly prepared precipitate into colloidal form by the addition of an electrolyte called peptizing agent. The suitable ions from the peptizing agent (electrolyte) are adsorbed by the particles of the precipitate giving it positive or negative charge. The charged particles repel one another and break up the precipitate into smaller particles of the size of the colloid. Therefore, it results into the formation of colloid.

For example, on treating a precipitate of iron (III) oxide with a small amount of FeCl3 solution gives a reddish brown coloured colloidal solution.
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 4

Question 7.
Cottrell’s smoke precipitator is fitted at the mouth of chimney used in factories. Give reasons. (CBSE Sample Paper 2011)
Answer:
Smoke coming out of chimney of a factory is a colloidal solution of solid carbon particles which are charged in nature.

The mouth of the chimneys used in factories is fitted with Cottrell smoke precipitator. In this method, the smoke is allowed to pass through a chamber having a series of plates charged to very high potential (20,000 to 70,000 V).

Charged particles of smoke get attracted by charged plates, get precipitated and the gases coming out of chimney become free of charged carbon and dust particles.

Question 8.
Differentiate between peptization and coagulation.
(CBSE Sample Paper 2011, CBSE AI 2017)
Answer:
Peptization is the process of converting a freshly prepared precipitate into collodial form by the the addition of a suitable electrolyte. The electrolytes used for the purpose are called peptizing agents. On the other hand, coagulation is the phenomenon of precipitation of a collodial solution by the addition of excess of an electrolyte.

Question 9.
Explain:
(i) Sky appears blue in colour.
(ii) A freshly formed precipitate of ferric hydroxide can be converted to a colloidal sol by shaking it with a small quantity of ferric chloride. (CBSE Sample Paper 2011)
Answer:
1. Dust particles along with water suspended in air scatter blue light which reaches our eyes and therefore, sky looks blue to us.

2. When we add FeCl3 to a freshly formed precipitate of Fe(OH)3, peptization occurs. The Fe3+ ions are adsorbed on the surface of the precipitate which ‘ ultimately breaks down into smaller particles of colloidal size.

Question 10.
What happens when
(i) a freshly prepared precipitate of Fe(OH)3 is shaken with a small amount of FeCl3 solution?
(ii) persistent dialysis of a colloidal solution is carried out?
(iii) an emulsion is centrifuged? (CBSE 2018)
Answer:
(i) Peptization occurs / Colloidal solution of Fe(OH)3 is formed.
(ii) Coagulation occurs
(iii) Demulsification or breaks into constituent liquids

Surface Chemistry Important Extra Questions Long Answer Type

Question 1.
Classify colloids where the dispersion medium is water. State their characteristics and write one example of each of these classes.
OR
Explain what is observed when
(i) an electric current is passed through a sol.
(ii) a beam of light is passed through a sol.
(iii) an electrolyte (say NaCl) is added to ferric hydroxide sol. (CBSE 2011)
Answer:
These colloids are of two types:
(i) Hydrophilic
(ii) Hydrophobic

Characteristics: In lyophilic colloids, there is affinity between dispersed phase and the dispersion medium while in the lyophobic colloids, there is no affinity rather hatred between the phases.

The main points of distinction between lyophilic and lyophobic sols are summarized below:

Lyophilic colloids (Hydrophilic) Lyophobic collodis (Hydrophobic)
1. These are easily formed by direct mixing. 1. These are formed only by special methods.
2. These are reversible in nature. 2. These are irreversible in nature.
3. The particles of colloids are true molecules and are big in size. 3. The particles are aggregates of many molecules.
4. The particles are not easily visible even under ultramicroscope. 4. The particles are easily detected under ultramicroscope.
5. These are very stable. 5. These are unstable and require traces of stabilizers.
6. The addition of small amount of electrolytes causes precipitation (called coagulation) of colloidal solution. 6. The addition of small amount of electrolytes has less effect. Larger quantities of electrolytes are required to cause coagulation.
7. The particles do not carry any charge. The particles may migrate in any direction or even not under the influence of an electric field. 7. The particles move in a specific direction i.e., either towards anode or cathode depending upon their charge.
8. The particles of colloids are heavily hydrated due to the attraction for the solvent. 8. The particles of colloids are not appreciably hydrated due to the hatred for the solvent.
9. The viscosity and surface tension of the sols are much higher than that of the dispersion medium. 9. The viscosity and surface tension are nearly the same as that of the dispersion medium.
10. They do not show Tyndall effect. 10. They show Tyndall effect.

Example: Hydrophilic: Starch,
Hydrophobic : Metal sulphide

OR

(i) On passing electric current through colloidal solution, the colloidal particles move towards the oppositely charged electrodes, where they lose their charge and get coagulated. This is electrophoresis process.

(ii) When a beam of light is passed through a colloidal solution, the path of the light becomes visible when viewed from the direction at right angle to that of incident beam. This phenomenon is called Tyndall effect. This is because of scattering of tight by colloidal particles. The particles first absorb the incident light and then a part of it gets scattered by them and therefore the part becomes visible when seen at right angle to the direction of incident beam.

(iii) Hydrated ferric oxide is positively charged. When an electrolyte such as NaCl is added, the excess Cl– ions neutralise positive charge and cause coagulation of the sol.

Question 2.
Explain how the phenomenon of adsorption finds application in each of the following processes:
(i) Production of vacuum
(ii) Heterogeneous catalysis
(iii) Froth floatation process (CBSE Delhi 2011)
Answer:
(i) Production of vacuum: The adsorption of air in liquid air helps to create a high vacuum in a vessel. This process is used in high vacuum instruments such as Dewar flask for storage of liquid air or liquid hydrogen. The remaining traces of air can be adsorbed by charcoal from the vessel evacuated by a vacuum pump to give a very high vacuum.

(ii) Heterogeneous catalysis: The phenomenon of adsorption is useful in the heterogeneous catalysis. Adsorption of reactants on the solid surface of catalysts increases the rate of reaction. The metals such as Fe, Ni, Pt, Pd, etc. are used in the manufacturing processes. Manufacture of ammonia using iron as catalyst (Haber process), manufacture of sulphuric acid by Contact process and use of finely divided nickel in the hydrogenation of oils are excellent examples of heterogeneous catalysis. Its use is based upon the phenomenon of adsorption.

(iii) In froth floatation process: A low grade sulphide ore is concentrated by separating it from silica and other earthly matter by adsorption using pine oil and frothing agent.

Question 3.
What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out. (CBSE 2012)
Answer:
The phenomenon of precipitation of a colloidal sol by the addition of excess of an electrolyte is called coagulation of colloidal sol. Coagulation of lyophobic sols can be carried out by the following methods:

1. By addition of an electrolyte: When an electrolyte is added to the sol, colloidal particles take up ions carrying opposite charge from the electrolyte and get neutralised resulting in coagulation.

2. By mutual precipitation: When two oppositely charged sols are mixed in equimolar proportions, they mutually neutralise their charge and both get coagulated.

3. By electrophoresis: When electrophoresis is carried out for a long time, the particles of dispersed phase move towards oppositely charged electrodes, lose their charge and get coagulated.

Question 4.
Explain the following terms giving a suitable example for each:
(i) Aerosol
(ii) Emulsion
(iii) Micelle (CBSE 2012)
Answer:
(i) Aerosol: It is a colloidal dispersion of a liquid in a gas. For example, fog, mist, cloud.

(ii) Emulsion. Emulsions are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium. Types of emulsion. These are of two types:

(a) Oil-in-water emulsion. In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

(b) Water-in-oil emulsion. In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

(iii) Micelle: The particles of colloidal size formed due to aggregation of several ions or molecules with lyophobic as well as lyophilic particles are called micelles. The common micelles system is soap. The micelles behave as normal electrolytes at low concentrations but as colloids at higher concentrations.

Question 5.
Write three distinct features of chemisorption which are not found in physisorption. (CBSE 2012)
Answer:
Characteristics of Chemisorption:

  1. The forces between the adsorbate molecules and the adsorbent are strong chemical forces similar to chemical bonds. But no chemical bonds exist in physisorption.
  2. Chemisorption is irreversible in nature but physisorption is reversible.
  3. Chemisorption forms mono – molecular layers, but in physisorption multimolecular layers are present.

Question 6.
What are the characteristics of the following colloids? Give one example of each.
(i) Multimolecular colloids
(ii) Lyophobic sols
(iii) Emulsions (CBSE 2013)
Answer:
(i) The colloidal solutions in which atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimensions. The particles thus formed are called multimolecular colloids. For example, sols of gold atoms and sulphur (S8) molecules.

(ii) The colloidal solutions in which there in no affinity (or love rather they have hatred) between the particles of the dispersed phase and the dispersion medium are called lyophobic sols.
For example, metal sulphides like As2S3.

(iii) The colloidal solutions in which both the dispersed phase and the dispersion medium are liquids are called emulsions. For example, milk.

Question 7.
Write the dispersed phase and dispersion medium of the following colloidal systems:
(i) Smoke
(ii) Milk
OR
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the addition of small amounts of electrolytes? (CBSE Delhi 2013)
Answer:

Colloidol System Dispersed phase Dispersed medium
(i) Smoke Solid Gas
(ii) Milk Liquid Liquid

OR
The colloidal solutions in which the particles of the dispersed phase have great affinity for the dispersion medium are called lyophilic sols. For example, glue, gelatin, starch, proteins, etc.

The colloidal solutions in which there is no affinity between the particles of the dispersed phase and the dispersion medium are called lyophobic sols. For example, ‘ solutions of metals like Ag, Au, Al(OH)3, Fe(OH)3, etc.
Lyophobic sols can be easily coagulated on the addition of small amounts of electrolytes.

Question 8.
What are emulsions? What are their different types? Give one example of each type. (CBSE 2014)
Answer:
Emulsions are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium.

Types of emulsions. These are of two types:
1. Oil-in-water emulsion. In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

(ii) Water-in-oil emulsion. In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

Question 9.
(i) In reference to Freundlich adsorption isotherm, write the expression for adsorption of gases on solids in the form of an equation.
(ii) Write an important characteristic of lyophilic sols.
(iff) Based on type of particles of dispersed phase, give one example each of associated colloid and multimolecular colloid. (CBSE Delhi 2014)
Answer:
(i) \(\frac{x}{m}\) = kp1/n
where \(\frac{x}{m}\) is extent of adsorption, x is the mass of adsorbate, m is the mass of adsorbent, p is pressure at a particular temperature, k is a constant and n takes any whole number value.

(ii) in lyophilic sols, the particles of dispersed phase have a great affinity for the dispersion medium and are reversible in nature.

(iii) Associated colloids: Soap (e.g. sodium stearate).
Multimolecular colloids: Sol of gold (atoms).

Question 10.
Give reasons for the following observations:
(i) Leather gets hardened after tanning.
(ii) Lyophilic sol is more stable than lyophobic sol.
(iii) It is necessary to remove CO when ammonia is prepared by Haber’s process. (CBSE Delhi 2015)
Answer:
(i) Animal hide is colloidal in nature and has positively charged particles. When it is soaked in tannin which has negatively charged colloidal particles, it results in mutual coagulation. This results in the hardening of leather.

(ii) Lyophilic sol is more stable than lyophobic sol because it is highly hydrated or solvated in solution.

(iii) Carbon monoxide acts as a poison for the catalyst in Haber’s process and therefore, it will lower the activity of the catalyst.
Thus, CO must be removed when ammonia is obtained by Haber’s process.

Question 11.
Give reasons for the following observations:
(i) Physisorption decreases with increase in temperature.
(ii) Addition of alum purifies the water.
(iii) Brownian movement provides stability to the colloidal solution. (CBSE 2015)
Answer:
(i) In physisorption, the attractive forces between adsorbent and adsorbate molecules are weak vander Waals forces. When temperature is increased, the kinetic energy of the molecules of the gas increases and they can easily leave the surface of adsorbent.
Therefore, physisorption decreases with increase in temperature.

(ii) Alum coagulates the impurities present in water by neutralising the charge.

(iii) The Brownian movement is due to the unbalanced bombardment of the particles by the molecules of the dispersion medium. This results in zigzag motion. Because of brownian movement, the particles do not settle down and hence is responsible for the stability of the sols.

Question 12.
Define the following terms:
(i) O/W Emulsion
(ii) Zeta potential
(iii) Multimolecular colloids (CBSE 2016)
Answer:
(i) O/W Emulsion: It is the emulsion in which oil is the dispersed phase and water is the dispersion medium. For example, milk is an emulsion of soluble fats in water.

(ii) Zeta potential: The potential difference between the fixed layer and the diffused layer of opposite charges is called Zeta potential.

(iii) Multimolecular colloids: When a dissolution, atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimension, the particles thus formed are called multimolecular colloids.

Question 13.
(i) Differentiate between adsorption and absorption.
(ii) Out of MgCl2 and AlCl2, which one is more effective in causing coagulation of negatively charged sol and why?
(iii) Out of sulphur sol and proteins, which one forms multimolecular colloids? (CBSE Delhi 2016)
Answer:
(i)

Absorption Adsorption
1. It is the phenomenon in which the particles of gas or liquid get uniformly distributed throughout the body of the solid. 1. It is the phenomenon of higher concentration of particles of gas or liquid on the surface than in the bulk of the solid.
2. The concentration is the same throughout the material. Therefore, it is a bulk phenomenon. 2. The concentration on the surface of the adsorbent is different from that in the bulk.  Therefore, it is a surface phenomenon.
3. Absorption occurs at uniform rate. 3. Adsorption is rapid in the beginning and its rate slowly decreases.

(ii) According to Hardy-Schulze rule, for negatively charged sol, greater the valency of the positive ion of the electrolyte added, greater is its coagulating power. Thus AlCl3 (Al3+ ion) is more effective in causing coagulation of negatively charged sol than MgCl2 (Mg2+ ions).

(iii) Proteins are macromolecules which cannot form multimolecular colloids while sulphur sol has smaller S8 molecules which can coagulate to form multimolecular colloids.

Question 14.
What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out.
(CBSE Delhi 2013)
Answer:
The phenomenon of precipitation of a colloidal sol by the addition of excess of an electrolyte is called coagulation of colloidal sol.
Coagulation of lyophobic sols can be carried out by the following methods:
(i) By addition of an electrolyte: When an electrolyte is added to the sol, colloidal particles take up ions carrying opposite charge from the electrolyte and get neutralised resulting in coagulation.

(ii) By mutual precipitation: When two oppositely charged sols are mixed in equimolar proportions, they mutually neutralise their charge and both get coagulated.

(iii) By electrophoresis: When electrophoresis is carried out for a long time, the particles of dispersed phase move towards oppositely charged electrodes, lose their charge and get coagulated.

Question 15.
(i) What is the role of activated charcoal in gas mask?
(ii) A colloidal sol is prepared by the given method in figure. What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented?
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 5
(iii) How does chemisorption vary with temperature? (CBSE Delhi 2019) Answer:
(i) Activated charcoalin gas mask adsorbs the poisonous or toxic gases from air. Therefore, these masks help to purify the air for breathing.

(ii) Negatively charged sol
Representation: Fe2O3. xH2O/OH–

(iii) In chemisorption, adsorption initially increases with rise in temperature and then decreases.

Question 16.
Define the following terms with an example in each case:
(i) Macromolecular sol
(ii) Peptization
(iii) Emulsion (CBSE Al 2013)
Answer:
(i) Macromolecular sol: In this type, the particles of the dispersed phase are sufficiently big in size (macro) to be of colloidal dimensions. In this case, a large number of small molecules are joined together through their primary valencies to form giant molecules. These molecules are called macromolecules and each macromolecule may consist of hundreds or thousands of simple molecules. The solutions of such molecules are called macromolecular solutions. For example, colloidal solution of starch, cellulose, etc.

(ii) Peptization: The process of converting a freshly prepared precipitate into colloidal form by addition of a suitable electrolyte is called peptization. The electrolytes used for the purpose are called peptizing agents.

(iii) Emulsions: They are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium. Normally, these are obtained by mixing an oil with water. Since the two do not mix well, the emulsion is generally unstable and is stabilised by adding a suitable reagent called emulsifier or emulsifying agent. The substances that are commonly employed for the purpose are gum, soap, glass powder, etc.

Question 17.
Define the following with a suitable example of each:
(a) Coagulation
(b) Multimolecular colloid
(c) Gel
Answer:
(a) Coagulation. The phenomenon of precipitation of a colloidal solution by the addition of excess of an . electrolyte is called coagulation. For
example, Fe(OH)3 colloidal solution is coagulated by adding HCl.

(b) Multimolecular colloid. When on dilution, atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimensions, then the particles thus formed are called multimolecular colloids, for example, sols of gold atoms.

(c) Gel. A gel is a colloidal system in which a liquid is dispersed in a solid. For example, jellies, cheese.

OR

(a) Out of starch and ferric hydroxide sol, which one can easily be coagulated and why?
(b) What is observed when an emulsion is centrifuged?
(c) What is the role of promoters and poisons in catalysis? (CBSE Al 2019)
Answer:
(a) Ferric hydroxide sol is easier to coagulate because it is a lyophobic sol.
(b) De-emulsification occurs.
(c) Promoters increase the efficiency of catalyst whereas poisons inhibit the efficiency of catalyst.

Question 18.
Give reason for the following observations:
(i) When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed.
(ii) Finely divided substance is more effective as an adsorbent.
(iii) Lyophilic colloids are also called reversible sols.
Answer:
(i) The precipitated silver iodide adsorbs iodide ions from the dispersion medium resulting in the negatively charged colloidal solution.
(ii) Due to large surface area
(iii) If the dispersion medium is separated from the dispersed phase, the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols.

Question 19.
Answer the following questions:
(a) Which of the following electrolytes is most effective for the coagulation of Agl/Ag+ sol?
MgCl2, K2SO4, K4[Fe(CN)6]
(b) What happens when a freshly precipitated Fe(OH)3 is shaken with a little amount of dilute solution of FeCl3?
(c) Out of sulphur sol and proteins, which one forms macromolecular colloids? (CBSE Sample Paper 2019)
Answer:
(a) K4[Fe(CN)6]
Hardy Schulze rule: The coagulation tendency of different electrolytes is different. It depends upon the valency of the active ion called flocculatins ion, which is the ion-carrying charge opposite to the charge on the colloidal particles. According to Hardy Schulze rule, greater the valency of the active ion or flocculating ion, greater will be its coagulating power. For example, to coagulate negative sol of AS2S3, the coagulating power of different cations has been found to decrease in the order as:
Al3+ > Mg2+ > Na+
Similarly, to coagulate a positive sol such as Fe(OH)3, the coagulating power of different anions has been found to decrease in the order:
[Fe(CN)6]4– > PO43- > SO42- > Cl–

(b) Due to the phenomenon of peptization, a reddish brown positively charged sol of ferric hydroxide is obtained when a freshly precipitated Fe(OH)3 is shaken with a little amount of dilute solution of FeCl3. It involves the adsorption of suitable ions from the electrolyte by . the particles of the precipitate. The charged particles repel one another and form a colloidal sol. Ferric ions from ferric chloride are adsorbed by Fe(OH)3, precipitate and get converted into colloidal sol.
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 6

(c) Proteins are macromolecules which cannot form multi-molecular colloids while sulphur sol has smaller S8 molecules which can coagulate to form multi-molecular colloids.

Question 20.
Answer the following:
(a) Why is ester hydrolysis slow in the beginning and then becomes faster after some time?
(b) In a solution of methylene blue, animal charcoal is added, the solution is then well shaken. What will be observed and why?
(c) Give an example of oil in water emulsion.
Answer:
(a) This is because of the process of autocatalysis. Ester on hydrolysis gives an acid which starts acting as a catalyst after sometime and therefore, the reaction becomes fast.

(b) Adsorption of a dye by charcoal. When animal charcoal is shaken with a solution of an organic dye such as methylene blue it is observed that the solution turns colourless. The discharge of the colour is due to the fact that the coloured component (generally an organic dye) gets adsorbed on the surface of animal charcoal. Therefore, animal charcoal is used for decolourising a number of organic substances in the form of their solutions.
(c) Milk, vanishing cream.

OR

Define the following:
(a) Associated colloids
(b) Electrophoresis
(c) Zeta potential (CBSE 2019C)
Answer:
(a) Associated colloids: These are the substances which when dissolved
in a medium behave as normal electrolytes at low concentration but behave as colloidal particles at higher concentration due to the formation of aggregated particles. The aggregate particles thus formed are called micelles. For example, in aqueous solution, soap (sodium stearate) ionises as:
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 7
In concentrated solution, these ions get associated to form an aggregate of colloidal size.

(b) The phenomenon of movement of colloidal particles under an applied electric field is called electrophoresis. If the particles accumulate near the negative electrode, the charge on the particles is positive. On the other hand, if the sol particles accumulate near the positive electrode, the charge on the particles is negative.

(c) The combination of two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer. The first layer of ions is firmly held and is termed fixed layer while the second layer is mobile which is termed diffused layer. The charges of opposite signs on the fixed and diffused parts of the double layer result in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charges is called the electrokinetic potential or zeta potential.

Question 21.
Write the differences between physisorption and chemisorption With respect to the following:
(i) Specificity
(ii) Temperature dependence
(iii) Reversibility and
(iv) Enthalpy change (CBSE Delhi 2013)
Answer:

Property Physisorption Chemisorption
(i) Specificity: not specific highly specific
(ii) Temperature dependence: Occurs at low temperature and decreases with increasing temperature Occurs at high temperature and increases with increasing temperature
(iii) Reversibility: Reversible in nature Irreversible in nature
(iv) Enthalpy change: Low enthalpy of adsorption (20 – 40 kJ mol-1) High enthalpy of adsorption (80 – 240 kJ mol-1)

Question 22.
Describe some features of catalysis by zeolites.
Answer:
Zeolites are microporous aluminosilicates of the general formula Mx/n [(AlO2)x (SiO2)y] mH2O. These are most important oxide catalysts. These are used in petrochemical industries for cracking of hydrocarbons and isomerisation. The reactions in zeolites depend upon the size of the cavities (cages) or pores (apertures) present in them.

The most remarkable feature of zeolite catalysis is the shape selectivity. Therefore, the selectivity of catalyst depends on the pores structure. It has been observed that the pore size in zeolites generally varies between 260 pm and 740 pm. Depending upon the size of the reactants and products compared to the size of the cages or pores of zeolite, reactions proceed in specific manner.

A zeolite catalyst called ZSM-5 converts alcohols to gasoline, by first dehydrating the alcohol by loss of water.

Question 23.
What do you understand by adsorption isotherm? Discuss briefly Freundlich adsorption isotherm.
Answer:
Adsorption isotherm. A graph between amount of adsorption and gas pressure keeping temperature constant is called an adsorption isotherm.

Freundlich adsorption isotherm. The behaviour of adsorption with pressure can be expressed by adsorption isotherm as shown in figure.
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 8
The inspection of the graph reveals the following facts:
(i) At low pressure, the graph is almost straight line which indicates that x/m is directly proportional to the pressure. This may be expressed as:
\(\frac{x}{m}\) ∝ P or \(\frac{x}{m}\) = kP ……. (i)
where k is constant.

(ii) At high pressure. The graph becomes almost constant which means that x/m becomes independent of pressure. This may be expressed as:
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 9

(iii) Thus, in the intermediate range of pressure, x/m will depend upon the power of pressure which lies between 0 and 1, i.e. fractional power of pressure. This may be expressed as:
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 10
where n can take any value between 1 and a larger integer depending upon the pressure. The above relationship is also called Freundlich’s adsorption isotherm.

Question 24.
Explain the following in connection with colloids:
(i) Hardy-Schulze rule
(ii) Dialysis.
Answer:
(i) Hardy-Schulze rule: This rule states that greater the valency of the active ion or flocculating ion, greater will be its coagulating power. According to this rule:

(a) The ions carrying the charge opposite to that of sol particles are effective in causing coagulation of the sol.

(b) Coagulation power of an electrolyte is directly proportional to the valency of the active ions.
For example, for negative As2S3 sol, the coagulating power of cations decreases as Al3+ > Mg2+ > K+
Similarly, for positively charged sol such as Fe (OH)3, the coagulating power decreases as
[Fe(CN)3]4- > PO43- > SO42- > Cr–

(ii) Dialysis. The method used to separate the impurities from the colloidal solution is called dialysis. Its principle is based upon the fact that colloidal particles cannot pass through a parchment or cellophane membrane while the ions of the electrolyte can pass through it. The colloidal solution is taken in a bag made of cellophane or parchment. The bag is suspended in fresh water. The impurities slowly diffuse out of the bag leaving behind pure colloidal solution. Dialysis can be used for removing HCl from the ferric hydroxide sol. The method is shown in figure.
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 11

The ordinary process of dialysis is slow. To increase the process of purification, the dialysis is carried out by applying electric field. This process is called electrodialysis.

Question 25.
Explain the following terms:
(i) Electrodialysis
(ii) Tyndall effect.
Answer:
(i) Electrodialysis. It is the process of separating the impurities from the colloidal solution by applying electric field. It is based on the fact that colloidal particles cannot pass through a parchment or cellophane membrane while the ions of the electrolyte can pass through it. The colloidal solution is taken in a bag made of cellophane or parchment and is suspended in a fresh water and electric current is applied. The impurities slowly diffuse out of the bag leaving behind pure colloidal solution.
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 12

(ii) Tyndall effect. When a strong beam of light is passed through a true solution placed in a beaker, in a dark room, the path of the light does not become visible. However, if the light is passed through a sol, placed in the same room, the path of the light becomes visible when viewed from a direction at right angle to that of the incident beam.
Class 12 Chemistry Important Questions Chapter 5 Surface Chemistry 13
This phenomenon was studied for the first time by Tyndall and therefore it is called Tyndall effect. The cause to Tyndall effect is the scattering of light by colloidal particles, i.e. these particles first absorb the incident light and then a part of it gets scattered by them. Since the intensity of the scattered light is at right angle to the plane of the incident light, the path becomes visible only when seen in that direction. The particles in true solution are too small in size to cause any scattering, i.e. the Tyndall effect is not noticed in true solutions.