Solid States 2023-2019

March 2023

Q. The relation between radius of sphere and edge length in body centered cubic lattice is given by formula :
(a) \sqrt{3}r=4a
b) r=\frac{\sqrt{3}}{a}\times 4
(c) r=\frac{\sqrt{3}}{4}a
(d) r=\frac{\sqrt{2}}{4}\times a

Ans :

(c) r=\frac{\sqrt{3}}{4}a

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Q. Explain the following terms :
(i) Substitutional impurity defect
(ii) Interstitial impurity defect

Ans :

(i) Substitutional impurity defect :

In this defect, the foreign atoms are found at the lattice sites in place of host atoms. The regular atoms are displaced from their lattice sites by impurity atoms.

e.g. : Brass

(ii) Interstitial Impurity defect :

In this defect, the impurity atoms occupy interstitial spaces of lattice structure.

e.g. : Steel

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Q. Silver crystallizes in fcc structure. If edge length of unit cell is 400 pm, calculate density or silver (Atomic mass of Ag = 108).

Ans :

Given data :

Edge length (a) = 400 pm = 4.00 x 10^-8 cm

Atomic mass of Ag = M = 108 g mol^-1

To find :

Density (ρ)

Formula :

Density (ρ) = \frac{Mn}{a^3N_A}

Calculation :

For an fcc lattice, number of atoms per unit cell is 4.

∴ n = 4

Using formula,

Density (ρ) = \frac{Mn}{a^3N_A}

\therefore \rho =\frac{108\;g\;mol^{-1}\times 4\;atom}{\left( 4.00\times 10^{-3}\right)cm^3\times 6.022\times 10^{23}\;atom\;mol^{-1}}

= 11.2 g/cm³ (The density of Silver)

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March 2022

Q. The co-ordination number of in body centred cubic Structure (bcc) is ______
(a) 4
(b) 6
(c) 8
(d) 12

Ans :

(b) 6

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Q. Write the consequences of Schottky defect with reasons.

Ans :

Consequences of Schottky defect :

(i) As the number of ions decreases, mass decreases. However, volume remains unchanged. Hence, the density of a substance decreases.

(ii) The number of missing cations and anions is equal. Hence, the electrical neutrality of the compound is preserved.

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Q. Explain metal deficiency defect with example.

Ans :

Metal deficiency defect :

This defect is possible only in compounds of metals that show variable oxidation States.

In some crystals, positive metal ions are missing from their original lattice sites. The extra negative charge is balanced by the presence of cation of the same metal with higher oxidation state than that of missing cation.

e.g. In the compound NiO, one Ni²⁺ ion is missing creating a vacancy at its latticc site. The deficiency of two positive charges is made up by the presence of two Ni³⁺ ions at the other lattice sites of Ni2+ ions. The composition of NiO then becomes Ni_0.97 O_1.0.

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Q. Gold crystallises into face-centred cubic cells. The edge length of unit cell is 4.08 x 10^-8 cm. Calculate the density of gold. [Molar mass of gold = 197 g mol^-1 ]

Ans :

Given :

Edge length (a) = 4.08 x 1^-8 cm

Molar mass of gold = M = 197 g mol^-1

N_A = 6.022 x 10²³ atoms mol^-1

To find :

Density (ρ)

Formula :

Density (ρ) = \frac{Mn}{a^3N_A}

Calculation :

For an fcc lattice, number of atoms per unit cell is 4.

∴ n = 4

Using formula,

Density (ρ) = \frac{Mn}{a^3N_A}

\therefore \rho =\frac{197\;g\;mol^{-1}\times 4\;atom}{\left( 4.08\times 10^{-3}\right)cm^3\times 6.022\times 10^{23}\;atom\;mol^{-1}}

= 19.27 g/cm³ (The density of Gold) or 19.27 x 10³ kg/m³

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March 2020

Q. The number of atoms per unit cell of body centred cube is :
(a) 1
(b) 2
(c) 4
(d) 6

Ans :

(b) 2

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Q. Classify the following solids into different types :
(i) Silver
(ii) P₄
(iii) Diamond
(iv) NaCl

Ans :

(i) Silver – Metallic Solid

(ii) P₄ – Molecular Solid

(iii) Diamond – Covalent network solid

(iv) NaCl – Lonic sodun

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Q. Unit cell of a metal has edge length of 288 pm and density of 7.86 g cm^-3. Determine the type of crystal lattice. [Atomic mass of metal = 56 g mol ^-1]

Ans :

Given :

Edge length 9a) = 288 pm = 2.88 x 10^-8 cm,

Atomic mass = 56 g mol^-1,

Density (ρ) = 7.86 g cm^-3

To find :

Type of crystal lattice

Formula :

Density (ρ) = \frac{Mn}{a^3N_A}

Calculations :

By using above formula,

7.86\;cm^{-3}=\frac{56\;g\;mol^{-1}\times\;n}{\left(2.88\;\times\;10^{-8}\right)^3cm^3\;\times\;6.022\;\times\;10^{23}\;atom\;mol^{-1}} n=\frac{7.86\;cm^{-3}\;\times\;\left(2.88\;\times\;10^{-8}\right)^3cm^3\;\times\;6.022\;\times\;10^{23}\;atom\;mol^{-1}}{56\;g\;mol^{-1}} n=\frac{7.86\;\times\;\left(2.88\right)^3\;\times\;6.022\;\times\;10^{-1}\;}{56} ]

Calculations using log table :

= Anti\log_{10}\left[\log_{10}\left(7.86\right)+\left(3\times\log_{10}\left(2.88\right)\right)+\log_{10}\left(6.022\right)-\log_{10}\left(56\right)\right]

= Anti\log_{10}\left[0.8954+\left(3\times0.4594\right)+0.7797-1.7482\right]

= Anti\log_{10}\left[1.6751+1.39728-1.7482\right]

= Anti\log_{10}\left[3.0533-1.7482\right]

= Anti\log_{10}\left[1.3051\right]

= 20.18 x 10^-1

= 2.018

∴ Number of atoms in unit cell = 2

Since unit cell contains 2 atoms, it has body-centered cubic (bcc) structure.

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Mar 19

Q. Distinguish between Crystaline solids and amorphous solids. [Mar 19, 17, 14, 13]

Ans :

Sr. No.	Crystalline Solids	Amorphous Solids
1.	The constituent particles are arranged in a regular and periodic manner.	The constituent particles are arranged randomly.
2.	They have sharp and characteristic melting point.	They do not have sharp melting point. They gradually soften over a range of temperature.
3.	They are anisotropic, i.e. have different physical properties in different direction.	They are isotropic, i.e. have same physical properties in all directions.
4.	They have long range order.	They have only short range order.
5.	Ice, NaCl, etc.	Glass, rubber, plastics, etc.

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