Question

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

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

Answer 1

Answer: (A)

$209.6\times {10}^{-24}$

According to NaCl structure, shortest distance between a $P{b}^{2+}$ ion and $S^{2-}$ ion is $\frac{a}{2}=297$ pm $=297\times {10}^{-10}$
So, the volume of unit cell is $a^3=209.58\ast {10}^{-24}c{m}^3$

Hence,option A is the correct answer.