$$\eqalign{ & {\text{Volume of unit cell}}\, = a \times b \times c \cr & = 5 \times {10^{ - 8}} \times 8 \times {10^{ - 8}} \times 4 \times {10^{ - 8}} \cr & = 1.6 \times {10^{ - 22}}\,c{m^3} \cr & {\text{Mass of unit cell}} = 1.6 \times {10^{ - 22}} \times 5.2 \cr & = 8.32 \times {10^{ - 22}}g \cr & {\text{Number of molecules in one unit cell}} \cr & = \frac{{8.32 \times {{10}^{ - 22}}g}}{{166.4\,g\,mo{l^{ - 1}}}} = 3 \cr} $$

For an $$Fcc$$  crystal
$$\eqalign{ & {r_{cation}} + {r_{anion}} = \frac{{{\text{edge}}\,{\text{length}}}}{2};\,110 + {r_{anion}} = \frac{{508}}{2} \cr & {r_{anion}} = 254 - 110 = 144pm \cr} $$

$$\eqalign{ & {\text{For }}bcc{\text{ structure}} \cr & {\text{Interionic distance}} \cr & = {r^ + } + {r^ - } = \frac{{\sqrt 3 }}{2}a\,\,\left( {a = {\text{edge length}}} \right) \cr & r_{NH_4^ + }^ + + r_{C{l^ - }}^ - = \frac{{\sqrt 3 }}{2}a \cr & r_{NH_4^ + }^ + + 180 = \frac{{\sqrt 3 }}{2} \times 390 \cr & r_{NH_4^ + }^ + = 338pm - 180pm = 158\,pm \cr} $$

$${\text{Density of the unit cell,}}\,\rho = \frac{{Z \times M}}{{{a^3} \times {N_A}}}$$
$${\text{Given :}}\frac{{{\rho _A}}}{{{\rho _B}}} = 1.6,\frac{{{a_B}}}{{{a_A}}} = 2,$$       $${M_B} = 200\,g/mol$$
$$\eqalign{ & {\rho _A} = \frac{{Z \times {M_A}}}{{{{\left( {{a_A}} \right)}^3} \times {N_A}}};{\rho _B} = \frac{{Z \times {M_B}}}{{{{\left( {{a_B}} \right)}^3} \times {N_A}}} \cr & {\text{(}}Z{\text{ will be same for isomorphous crystals}}{\text{.)}} \cr & \frac{{{\rho _A}}}{{{\rho _B}}} = \frac{{{M_A}}}{{{M_B}}} \times {\left( {\frac{{{a_B}}}{{{a_A}}}} \right)^3} \Rightarrow 1.6 = \frac{{{M_A}}}{{200}} \times {\left( 2 \right)^3} \cr & {M_A} = \frac{{1.6 \times 200}}{8} \cr & \,\,\,\,\,\,\,\,\,\, = 40\,g/mol \cr} $$

Quartz exhibits piezoelectricity and thus can be used as a piezoelectric material .

In zinc blende structure, $${S^{2 - }}\,ions$$   are arranged in $$ccp$$  while $$Z{n^{2 + }}\,ions$$   occupy half of the tetrahedral sites.

In a unit cell, $$W$$ atoms at the corner $$ = \frac{1}{8} \times 8 = 1$$
$$O$$ - atoms at the centre of edges $$ = \frac{1}{4} \times 12 = 3$$
$$Na$$ - atoms at the centre of the cube $$ = 1$$
$$W:O:Na = 1:3:1$$
Hence, formula $$ = NaW{O_3}$$

In a hep structure atoms are located at the corners and at centres of two hexagons placed parallel to each other; three more atoms are placed in a parallel plane midway between these two planes. Here each atom is surrounded by 12 others and is said to have co-ordination number of 12 .

$$\eqalign{ & {\text{Limiting ratio}}, \cr & \frac{{{r_{{A^ + }}}}}{{{r_{{B^ - }}}}} = 0.414 \cr & {r_{{B^ - }}} = \frac{{{r_{{A^ + }}}}}{{0.414}} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{120}}{{0.414}} \cr & \,\,\,\,\,\,\,\,\,\, = 290\,pm \cr} $$

$$\eqalign{ & {\text{For }}fcc, \cr & r = \frac{{\sqrt 2 }}{4}a \cr & \,\,\,\, = \frac{{\sqrt 2 }}{4} \times 361 \cr & \,\,\,\, = \frac{{1.414 \times 361}}{4} \cr & \,\,\,\, = 127\,pm \cr} $$