Solid State MCQ Questions & Answers in Physical Chemistry | Chemistry

$$C{a^{2 + }}\,ions$$   form $$ccp$$  structure while $${F^ - }\,ions$$   are in all tetrahedral holes and coordination numbers are 8 and 4.

$$\eqalign{ & \rho = \frac{{ZM}}{{{N_A}V}} \cr & Z = \frac{{\rho {N_A}V}}{M} \cr & = \frac{{8.92 \times 6.02 \times {{10}^{23}} \times {{\left( {362} \right)}^3} \times {{10}^{ - 30}}}}{{63.55}} \cr & = 4 \cr & \therefore \,\,{\text{It has }}fcc{\text{ unit cell}} \cr} $$

If in an ionic crystal of the type $${A^ + },{B^ - },$$  equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained. The defect is called Schottky defect.

$$F$$ - centres are electron trapped in anion site which are responsible for colour.

No. of atoms of $$P = 8 \times \frac{1}{8} = 1$$
No. of atoms of $$Q = 1 \times 1 = 1$$
The formula of the compound is $$PQ.$$
Coordination no. of $$P$$ and $$Q = 8 : 8$$

When equal number of cations $$\left( {N{a^ + }} \right)$$  and anions $$\left( {C{l^ - }} \right)$$  are missing from their regular lattice positions, we have schottky defect.

Cubic close packing is also called face centred cubic packing.
No. of atoms per unit cell in $$ccp$$  ( or $$fcc$$  ) = 4
∴ No. of tetrahedral voids = 2 × 4 = 8

$$\eqalign{ & {\text{Packing effieincy}} = \frac{{{\text{Area occupied by effective circles}}}}{{{\text{Area of square}}}} \cr & = \frac{{2\pi {r^2}}}{{{L^2}}} \times 100 \cr & = \frac{{2\pi {r^2}}}{{2{{\left( {\sqrt {2r} } \right)}^2}}} \times 100 \cr & = \frac{\pi }{4} \times 100 \cr & = 78.54\% \cr} $$

Octahedral voids occupy the position of edge centre and body centre.

Schottky defect arises due to missing of equal number of atoms or ions ( cation and anion ) from their normal lattice point or site.