Question

A crystal of lead $\left(II\right)$ sulphide has a $NaCl$ structure. In this crystal, the shortest distance between a $P{b}^{2+}$ ion and $S^{2-}$ ion is $297\text{pm}$. What is the volume of unit cell in lead sulphide?
(Given, $(594{)}^3=209584584$)

• A. $209.6\times {10}^{-24}{\text{cm}}^3$
• B. $207.8\times {10}^{-23}{\text{cm}}^3$
• C. $22.3\times {10}^{-23}{\text{cm}}^3$
• D. $209.8\times {10}^{-23}{\text{cm}}^3$

Answer 1

The correct option is A $209.6\times {10}^{-24}{\text{cm}}^3$

The shortest distance between $P{b}^{2+}$ ion and $S^{2-}$ ions is $297\text{pm}$. This corresponds to one half the length.

This is because, along the edge, the spheres touch each other. If a is the edge length and r and r' are the radii of cation and anion, respectively.
$a=2(r+r^{\prime })$
The shortest distance between the cation and anion is $r+r^{\prime }$.
Hence,
$a=2(r+r^{\prime })=2\times 297\text{pm}=594\text{pm}$
Now,
$Volume\left(V\right)=a^3=(594\text{pm}{)}^3=209.6\times {10}^{-24}{\text{cm}}^3$