$$\eqalign{ & {\text{For }}bcc,Z = 2 \cr & d = \frac{{Z \times M}}{{{N_A} \times {a^3}}} \Rightarrow M = \frac{{d \times {N_A} \times {a^3}}}{Z} \cr & \therefore M = \frac{{4 \times 6 \times {{10}^{23}} \times {{\left( {500 \times {{10}^{ - 10}}} \right)}^3}}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4 \times 6 \times {{10}^{23}} \times 125 \times {{10}^{ - 24}}}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 150\,g\,mo{l^{ - 1}} \cr} $$

Let the number of $$Z$$ atoms in the $$ccp$$  arrangement = 100
Thus the number of tetrahedral sites = 200
Since all the tetrahedral sites are occupied by $$X$$ atoms, the number of $$X$$ atoms = 200
Hence ratio of $$X : Z = 2 : 1$$
Thus the formula is $${X_2}Z$$

$$\eqalign{ & {\text{Effective number of}}\,\,\,'A'\,\,{\text{atoms}} \cr & = \left( {8 \times \frac{1}{8}} \right) + \left( {4 \times \frac{1}{2}} \right) = 3 \cr & {\text{Effective number of}}\,\,\,'B'\,\,{\text{atoms}} \cr & = \left( {12 \times \frac{1}{4}} \right) + 1 = 4 \cr & \therefore \,{\text{ Formula of the solid}} = {A_3}{B_4}. \cr} $$

Molecular crystals are bad conductors of electricity as the electrons are localised in the bonds.

Crystalline solids are anisotropic in nature.

Body diagonal $$(d)$$  of a cubic crystal of edge length $$(a)$$  is given by,
$$\eqalign{ & d = a\sqrt 3 \cr & {\text{putting}}\,\,a = 400\,pm,{\text{we get}} \cr & d = \sqrt 3 \times 400\,pm = 692.8\,pm \approx 693\,pm. \cr} $$

$$\eqalign{ & M = \frac{{\rho \times {a^3} \times {N_A} \times {{10}^{ - 30}}}}{Z} \cr & = \frac{{10 \times {{\left( {100} \right)}^3} \times 6.02 \times {{10}^{23}} \times {{10}^{ - 30}}}}{4} \cr & = 15.05 \cr & \therefore \,\,{\text{Number of atoms in }}100{\text{ }}g \cr & = \frac{{6.02 \times {{10}^{23}}}}{{15.05}} \times 100 \cr & = 4 \times {10^{25}} \cr} $$

Low co-ordination compound can be changed into high co-ordination compound by increasing pressure and decreasing temperature.
$$\underset{\left(6,6\right)}{NaCl}\underset{Temp.}{\overset{\mathrm{Pr}.}{\rightleftharpoons }}\underset{\left(8,8\right)}{CsCl}$$

Number of A ions in the unit cell $$ = \frac{1}{8} \times 8 = 1$$
Number of B ions in the unit cell $$ = \frac{1}{2} \times 6 = 3$$
Hence empirical formula of the compound = $$A{B_3}$$

$$\eqalign{ & \rho = \frac{{Z \times M}}{{{N_A} \times {a^3}}} \cr & = \frac{{2 \times 100}}{{6.023 \times {{10}^{23}} \times {{\left( {400 \times {{10}^{ - 10}}} \right)}^3}}} \cr & = 5.188\,g/c{m^3} \cr} $$