Question

$PbS$ has $NaCl$ type structure. The shortest distance between $P{b}^{2+}$ and $S^{2-}$ ions is $297pm$. What is the volume of unit cell of $PbS$?

• A. $210\times {10}^{-24}c{m}^3$
• B. $150\times {10}^{-24}c{m}^3$
• C. $268\times {10}^{-24}c{m}^3$
• D. $310\times {10}^{-23}c{m}^3$

Answer 1

The correct option is A $210\times {10}^{-24}c{m}^3$

In $NaCl,N{a}^{+}$ occupies all the octahedral voids and the 8 tetrahedral voids are left vacant.
$C{l}^{-}$ occupies the corners and face centres of the cubic unit cells.

For an anion $C{l}^{-}$,
Distance of the nearest cation $N{a}^{+}$ will be $\frac{a}{2}$ (a is edge length).

This is the distance from the corner atom $C{l}^{-}$ to the edge atom $N{a}^{+}$.

Since, $PbS$ has $NaCl$ type structure. The shortest distance between $P{b}^{2+}$ and $S^{2-}$ ions is also $\frac{a}{2}$.

$\text{Edge length of unit cell}=2\times 297pm$
$=594\times {10}^{-10}cm$
$\text{Volume of unit cell}=(594\times {10}^{-10}{)}^{3}c{m}^3\approx 210\times {10}^{-24}c{m}^3$