Question

A compound AB crystallises in bcc lattice with the unit cell edge length of $380$ pm . Calculate
1. the distance between oppositely charged Ions in the lattic.
2. radius of B${}^{-}$ if the radius of A${}^{+}$ is $190$ pm.

Answer 1

For bcc lattice,

$r=\frac{\sqrt{3}a}{4}$

$\Rightarrow r=\frac{\sqrt{3}\times 380}{4}=164.35pm$

The distance between oppositely charged ions =$2r=2\times 164.35=328.7pm$

$\Rightarrow A^{+}+B^{-}=328.7pm$

$\Rightarrow B^{-}=328.7-190=138.7pm$