Solid State MCQ Questions & Answers in Physical Chemistry | Chemistry

Monoclinic crystals have two Bravais lattices - primitive and end centred.

$$\eqalign{ & {\text{Molar mass of the element}} \cr & = \frac{{200}}{{24 \times {{10}^{23}}}} \times 6.023 \times {10^{23}} \cr & = 50.19\,g\,mo{l^{ - 1}} \cr} $$
$${\text{For}}\,\,fcc,Z = 4,$$    $$V = {a^3} = {\left( {200 \times {{10}^{ - 10}}} \right)^3}$$
$$\eqalign{ & d = \frac{{Z \times M}}{{{N_A} \times V}} \cr & \,\,\,\, = \frac{{4 \times 50.19}}{{6.023 \times {{10}^{23}} \times {{\left( {200 \times {{10}^{ - 10}}} \right)}^3}}} \cr & \,\,\,\, = 41.66\,g\,c{m^{ - 3}} \cr} $$

Packing efficiency ( i.e., space occupied ) for $$bcc$$  is 68%
∴ % of empty space = 100 - 68 = 32%

$$\eqalign{ & {\text{For }}MgO,\frac{{{r_c}}}{{{r_a}}} = \frac{{0.65}}{{1.4}} = 0.47 \cr & M{g^{2 + }}\,{\text{is in octahedral void so}}\,\,C.N = 6 \cr & {\text{For}}\,MgS,\frac{{{r_c}}}{{{r_a}}} = \frac{{0.65}}{{1.84}} = 0.353 \cr & M{g^{2 + }}\,{\text{is in tetrahedral void so}}\,C.N. = 4 \cr} $$

Silicon Carbide from covalent solid, $$Ca{F_2}$$  and $$NaCl$$  form ionic crystal. $$C{H_4}$$  form molecular solid.

Packing efficiency for $$hcp$$  and $$ccp$$  is 74% and that of $$bcc$$  is 68% and of simple cubic cell is 52.4%.

$$\eqalign{ & {\text{For a }}fcc{\text{ unit cell}} \cr & r = \frac{{\sqrt 2 a}}{4} \cr & a = \frac{{4r}}{{\sqrt 2 }} \cr & = 2\sqrt 2 \times 0.14 \cr & = 0.39 \approx 0.4\,nm. \cr} $$

In Frenkel defect, the smaller cation is dislocated from its normal position to an interstitial site.

$$\eqalign{ & {\text{For}}\,\,{\text{simple}}\,\,{\text{cubic}}\,\,{\text{structure,}} \cr & V = {a^3}\left( {{\text{volume}}\,\,{\text{of}}\,\,{\text{the}}\,\,{\text{unit}}\,\,{\text{cell}}} \right) \cr & V' = \frac{4}{3}\pi {r^3}\left( {{\text{volume}}\,\,{\text{of}}\,\,{\text{one}}\,\,{\text{atom}}} \right) \cr & = \frac{4}{3}\pi {\left( {\frac{a}{2}} \right)^3} \cr & = \frac{{\pi {a^3}}}{6} \cr & {\text{Packing}}\,\,{\text{efficiency}} \cr & = \frac{{V'}}{V} \cr & = \frac{{\frac{{\pi {a^3}}}{6}}}{{{a^3}}} \cr & = \frac{\pi }{6} \cr & = 0.52\,\,{\text{or}}\,\,52\% \cr & {\text{For}}\,\,fcc\,\,{\text{structure,}} \cr & V' = 4 \times \frac{4}{3}\pi {r^3}\left( {{\text{4}}\,\,{\text{atoms}}\,\,{\text{per}}\,\,{\text{unit}}\,\,{\text{cell}}} \right) \cr & r = \frac{a}{{2\sqrt 2 }} \cr & V' = \frac{{16}}{3}\pi {\left( {\frac{a}{{2\sqrt 2 }}} \right)^3} \cr & \,\,\,\,\,\,\,\,\, = \frac{\pi }{{3\sqrt 2 }}{a^3} \cr & {\text{Volume}}\,\,{\text{of}}\,\,{\text{unit}}\,\,{\text{cell}} = V = {a^3} \cr & {\text{Packing}}\,\,{\text{efficiency}} \cr & = \frac{{V'}}{V} \cr & = \frac{{\pi {a^3}}}{{3\sqrt 2 {a^3}}} \cr & = \frac{\pi }{{3\sqrt 2 }} \cr & = 0.74\,\,{\text{or}}\,\,{\text{74% }} \cr & {\text{For}}\,\,bcc\,\,{\text{structure,}} \cr & V' = 2 \times \frac{4}{3}\pi {r^3}\left( {{\text{2}}\,\,{\text{atoms}}\,\,{\text{per}}\,\,{\text{unit}}\,\,{\text{cell}}} \right) \cr & r = \frac{{\sqrt 3 }}{4}a \cr & V' = 2 \times \frac{4}{3}\pi {\left( {\frac{{\sqrt 3 }}{4}a} \right)^3} \cr & \,\,\,\,\,\,\,\, = \frac{{\sqrt 3 \pi {a^3}}}{8} \cr & V = {a^3} \cr & {\text{Packing}}\,\,{\text{efficiency}} \cr & = \frac{{V'}}{V} \cr & = \frac{{\sqrt 3 \pi {a^3}}}{{8{a^3}}} \cr & = \frac{{\sqrt 3 }}{8}\pi \cr & = 0.68\,\,{\text{or}}\,\,{\text{68% }} \cr} $$

$$CsCl$$  has structure with $$C{l^ - }\,ions$$   at the corners and $$C{s^ + }$$ at the centre of the cube.