Solid State MCQ Questions & Answers in Physical Chemistry | Chemistry

When insulators (non metal atoms) interact to form a solid, their atomic orbitals mix to form two bunch of orbitals, separated by a large band gap. Electrons cannot therefore be promoted to an empty level, where they could move freely.

$$\eqalign{ & 2\left( {{r_ + } + {r_ - }} \right) = a \cr & 2\left( {{r_{N{a^ + }}} + {r_{C{l^ - }}}} \right) = 564\,pm \cr & {r_{N{a^ + }}} = \frac{{564}}{2} - 181 = 101\,pm \cr} $$

$$\eqalign{ & {\text{Number of atoms}}\,\,A = \frac{1}{8} \times 8 = 1 \cr & {\text{Number of atoms}}\,B = \frac{1}{2} \times 5 = \frac{5}{2} \cr} $$
( Since atom $$B$$ is missing from one of the face centered point )
∴ Formula of ionic compound $$ = A{B_{\frac{5}{2}}} = {A_2}{B_5}$$

$$\eqalign{ & d = \frac{{Z \times M}}{{{a^3} \times {N_A}}} \cr & {\text{For }}bcc{\text{ structure,}}Z = 2 \cr & d = \frac{{2 \times 50}}{{6.023 \times {{10}^{23}} \times {{\left( {300 \times {{10}^{ - 10}}} \right)}^3}}} \cr & \,\,\,\, = 6.15\,g\,c{m^{ - 3}} \cr} $$

$$\eqalign{ & d = \frac{{Z \times M}}{{{N_A} \times {a^3}}} \cr & Z = \frac{{{a^3} \times d \times {N_A}}}{M} \cr & Z = \frac{{{{\left( {3.5 \times {{10}^{ - 8}}} \right)}^3} \times 0.53 \times 6.023 \times {{10}^{23}}}}{{6.94}} \cr & \,\,\,\,\, = 1.97 \approx 2 \cr} $$

$$d = \frac{{Z \times M}}{{{N_A} \times V}} \Rightarrow 2.70 $$     $$ = \frac{{Z \times 27}}{{6.023 \times {{10}^{23}} \times {{\left( {405 \times {{10}^{ - 10}}} \right)}^3}}}$$
$$\therefore Z = 4$$
Hence, structure of aluminium unit cell is $$fcc$$  or face-centred.

Octahedral sites are created by overlapping two equilateral triangles with apices in opposite directions.

For $$bcc$$  structure, atomic radius, $$\,r = \frac{{\sqrt 3 }}{4}a$$
$$ = \frac{{\sqrt 3 }}{4} \times 4.3 = 1.86$$
Since, $$r =$$  half the distance between two nearest neighbouring atoms.
∴ Shortest interionic distance = 2 × 1.86 = 3.72

$$\eqalign{ & {\text{For }}bcc{\text{ lattice,}} \cr & \sqrt 3 a = 4R \Rightarrow a = \frac{{4 \times 96}}{{\sqrt 3 }}pm = 221.7\,pm \cr & {\text{(where }}R{\text{ is the radius of }}Ca{\text{ atom)}} \cr & {\text{Colume of unit}} \cr & {\text{cell}} = {a^3} = {\left( {211.7 \times {{10}^{ - 12}}} \right)^3}{m^3} \cr & = 10.9 \times {10^{ - 30}}{m^3} \cr} $$

$$Ti{O_3}$$  shows metallic or insulating property depending upon temperature. As temperature varies metallic or insulting property varies. This is due to variation in energy gap between conduction band and valence band.