Solid State MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & a = 2\left( {{r_ + } + {r_ - }} \right) \cr & \,\,\,\, = 2\left( {95 + 181} \right) \cr & \,\,\,\, = 552\,pm \cr} $$

$$ccp$$  lattice consists of four atoms per unit cell.
Thus, number of copper atoms per unit cell $$ = 4\left( {ccp} \right)$$
Number of silver atoms per unit cell $$ = \frac{1}{4} \times 12 = 3$$   ( edge centre )
Number of gold atoms per unit cell $$= 1$$   ( body centre )
Thus, formula of the compound $$ = C{u_4}A{g_3}Au.$$

Number of $$O$$ - atoms per unit cell
$$ = \frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 4.$$
Number of octahedral holes per unit cell
$$ = 1 \times 4 = 4$$
Number of $$F{e^{3 + }}\,ions$$   per unit cell
$$ = \frac{{50 \times 4}}{{100}} = 2$$
Number of tetrahedral voids per unit cell
$$ = 2 \times 4 = 8.$$
Number of $$Z{n^{2 + }}\,ions$$   per unit cell $$ = \frac{1}{8} \times 8 = 1$$
Hence, formula $$:ZnF{e_2}{O_4}$$

$$\eqalign{ & d = \frac{{Z \times M}}{{{a^3} \times {N_A}}} \cr & \,\,\,\,\, = \frac{{4 \times 58.5}}{{{{\left( {5.64 \times {{10}^{ - 8}}} \right)}^3} \times 6.023 \times {{10}^{23}}}} \cr & \,\,\,\,\, = 2.16\,g\,c{m^{ - 3}} \cr} $$

$$\eqalign{ & {\text{In }}NaCl{\text{ crystal,}} \cr & {\text{Edge length }} \cr & {\text{ = 2}} \times {\text{distance between }}N{a^ + }{\text{ and }}C{l^ - } \cr & = 2 \times 265 \cr & = 530\,pm \cr} $$

In $$bcc$$ - points are at corners and one in the centre of the unit cell.
Number of atoms per unit cell $$ = 8 \times \frac{1}{8} + 1 = 2.$$
In $$fcc$$ - points are at the corners and also centre of the six faces of each cell.
Number of atoms per unit cell $$ = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4.$$

In solids, the constituent particles have, fixed positions and can only oscillate about their mean positions.

In $$ZnS$$  structure, sulphide ions occupy all $$FCC$$  lattice points while $$Z{n^{2 + }}$$  ions are present in alternate tetrahedral voids.

In $$bcc$$  structure distance between nearest neighbours,
$$\eqalign{ & d = \frac{{AD}}{2} \cr & {\text{In}}\,{\text{right}}\,{\text{angled}}\,\Delta ABC, \cr & A{C^2} = A{B^2} + B{C^2} \cr & A{C^2} = {a^2} + {a^2}\,\,\,{\text{or}}\,\,\,AC = \sqrt 2 a \cr & {\text{Now}}\,{\text{in}}\,{\text{right}}\,{\text{angled}}\,\Delta ADC, \cr & A{D^2} = A{C^2} + D{C^2} \cr & A{D^2} = {\left( {\sqrt 2 a} \right)^2} + {a^2} = 3{a^2} \cr & \Rightarrow AD = \sqrt 3 a \cr & \therefore \,\,d = \frac{{\sqrt 3 a}}{2} \cr & {\text{Radius,}}\,r = \frac{d}{2} = \frac{{\sqrt 3 }}{4}a \cr} $$

$$\eqalign{ & 4R = L\sqrt 2 \cr & {\text{so,}}\,L = 2\sqrt 2 R \cr & {\text{Area of square unit cell}} \cr & = {\left( {2\sqrt 2 R} \right)^2} = 8{R^2} \cr & {\text{Area of atoms present in one unit cell}} \cr & = \pi {R^2} + 4\left( {\frac{{\pi {R^2}}}{4}} \right) \cr & = 2\pi {R^2} \cr & {\text{so, packing efficcirency}} \cr & = \frac{{2\pi {R^2}}}{{8{R^2}}} \times 100 \cr & = \frac{\pi }{4} \times 100 \cr & = 78.5\% \cr} $$