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. If the length of the edge of the unit cell in lead sulphide is $5.94\times {10}^{-x}$ cm, then value of $x$ is

Answer 1

The unit cell of PbS has NaCl type structure with fcc arrangement of sulphide ions (present at the corners and on the faces of the cube) and lead ions will be present in the void on edge.
Let r and R be the radii of lead ions and sulphide ions respectively.

$\text{Edge length}=R+2r+R=2(r+R)=2\times 297pm=594pm=5.94\times {10}^{-8}cm=5.94\times {10}^{-x}cm$

Hence, $x=8$