Question

A crystal of lead(II) sulphide has $NaCl$ structure. In this crystal the shortest distance between $P{b}^{+2}$ ion and $S^{2-}$ ion is 297 pm. What is the length of the edge of the unit cell in lead sulphide? Also calculate the unit cell volume.

Answer 1

In $NaCl$ type structure shortest distance of the anion and cation=$\frac{a}{2}$ (where $a$ is edge length)

$\Rightarrow \frac{a}{2}=297\Rightarrow a=594$ pm

Volume of unit cell=$a^3=209.58\times {10}^{-30}{m}^3$