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Class 12th Solid States

74. Assertion: Graphite is a good conductor of electricity however diamond belongs to the category of insulators.

Reason: Graphite is soft in nature on the other hand diamond is very hard and brittle.

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Payal Gupta

Contributor-Level 10

74. Option (ii) Assertion and reason both are correct statements, but reason is not correct explanation for assertion is the answer since in graphite carbon atoms are arranged in different layers and each atom is covalently bonded to three of its neighbouring atoms in the same layer due to which the fourth valence electron of each atom, present between the different layers is free to move and this makes graphite a good conductor of electricity.

As different layers can slide over each other which makes it soft. while in case of diamond C atom is bonded covalently with their adjacent atoms throughout the crystal. Thus, all the four v

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Class 12th Solid States

Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(i) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(ii) Assertion and reason both are correct statements, but reason is not correct explanation for assertion.

(iii) Assertion is correct statement, but reason is wrong statement.

(iv) Assertion is wrong statement, but reason is correct statement.

73. Assertion: The total number of atoms present in a simple cubic unit cell is one.

Reason: Simple cu

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Payal Gupta

Contributor-Level 10

73. Option (i) Assertion and reason both are correct statements and reason is correct explanation for assertion is the answer since in simple cubic crystal unit cell, the total number of atoms are given as - 8 * 18=1

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Class 12th Solid States

72. Match the type of packing given in Column I with the items given in Column II.

Column I	Column II
(i) Square close packing in two dimensions.	(a) Triangular voids
(ii) Hexagonal close packing two dimensions.	(b) Pattern of spheres is repeated in every fourth layer
(iii) Hexagonal close packing in three dimensions.	(c) Coordination number = 4
(iv) Cubic close packing in three dimensions.	(d) Pattern of sphere is repeated in alternate layers

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Payal Gupta

Contributor-Level 10

72. (i) → (c); (ii) → (a); (iii) → (d); (iv) → (b)

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Class 12th Solid States

71. Match the items given in Column I with the items given in Column II.

Column I	Column II
(i) Mg in solid state	(a) p-Type semiconductor
(ii) MgCl₂in molten state	(b) n-Type semiconductor
(iii) Silicon with phosphorus	(c) Electrolytic conductors
(iv) Germanium with boron	(d) Electronic conductors

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Payal Gupta

Contributor-Level 10

71. (i) → (d); (ii) → (c); (iii) → (b); (iv) → (a)

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Class 12th Solid States

70. Match the types of defects given in Column I with the statement given in Column II.

Column I	Column II
(i) Impurity defect	(a) NaCl with anionic sites called F-centres
(ii) Metal excess defect	(b) FeO with Fe³⁺
(iii) Metal deficiency defect	(c) NaCl with Sr²⁺and some cationic sites vacant

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Payal Gupta

Contributor-Level 10

70. (i) → (c); (ii) → (a); (iii) → (b)

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Class 12th Solid States

69. Match the type of unit cell given in Column I with the features given in Column II.

Column I	Column II
(i) Primitive cubic unit cell	(a) Each of the three perpendicular edges compulsorily have the different edge length i.e., a≠b≠c.
(ii) Body centred cubic unit cell	(b) Number of atoms per unit cell is one.
(iii) Face centred cubic unit cell	(c) Each of the three perpendicular edges compulsorily have the same edge length i.e., a = b = c
(iv) End centred orthorhombic	(d) In addition to the contribution from the unit cell to the corner atoms, the number of atoms present in a unit cell is one.
	(e) In addition to the contribution from the corner atoms the number of atoms present in a unit cell is three.

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Payal Gupta

Contributor-Level 10

69. (i) → (b); (ii) → (c); (iii) → (c); (iv) → (a)

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Class 12th Solid States

In the following questions match the items given in Column I with the items given in Column II.

In some questions more than one item of Column I and Column II may match.

68. Match the defects given in Column I with the statements given.

Column I	Column II
(i) Simple vacancy defect	(a) shown by non-ionic solids and increases density of the solid.
(ii) Simple interstitial defect	(b) shown by ionic solids and decreases density of the solid.
(iii) Frenkel defect	(c) shown by non-ionic solids and density of the solid decreases
(iv) Schottky defect	(d) shown by ionic solids and density of the solid remains the same

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Payal Gupta

Contributor-Level 10

68. (i) → (c); (ii) → (a); (iii) → (d); (iv) → (b)

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Class 12th Solid States

67. Which of the following defects decrease the density?

(i) Interstitial defect

(ii) Vacancy defect

(iii) Frankel defect

(iv) Schottky defect

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Payal Gupta

Contributor-Level 10

67. Options (ii) and (iv) are correct since Vacancy and Schottky, both the defects lead to the decrease in the density while in the Frenkel and Interstitial defect there is no change in the density of the substance.

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Class 12th Solid States

66. Frenkel defect is also known as ________.

(i) Stoichiometric defect

(ii) Dislocation defect

(iii) Impurity defect

(iv) Non-stoichiometric defect

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Payal Gupta

Contributor-Level 10

66. Options (i) and (ii) are correct since in this defect an atom is displaced from its lattice point to an interstitial site and thus creating a vacancy at the lattice point. This usually occurs when the size of anion is quite large as compared to that of the cation due to which it leads to the dislocation of the cation to the interstitial site and hence it is known as dislocation defect and during this defect there is no change in the stoichiometry of the crystal so, it is also known as stoichiometric defect.

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Class 12th Solid States

65. In which of the following arrangements octahedral voids are formed?

(i) Hcp

(ii) Bcc

(iii) Simple cubic

(iv) Fcc

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Payal Gupta

Contributor-Level 10

65. Options (i) and (iv) are correct since when the tetrahedral void of the first layer and the tetrahedral void of the second layer align together, they form an octahedral void, and this is possible in case of hep and fcc arrangements.

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