Question

$CsBr$ has bcc structure with edge length $4.3A^o$. The shortest interionic distance ( in $A^o$) in between $C{s}^{+}$ and $B{r}^{-}$ is:(Give your answer upto two decimal only)

Answer 1

For a bcc structure, we have
$r=\frac{\sqrt{3}}{4}a$
Where, (r=radius and a = edge length)
$r=\frac{\sqrt{3}}{4}\times 4.3$
$r=1.86$
We know, r = hlf the distance between two nearest neighbouring atoms,
$\therefore \text{Shortest interionic distance}=2\times 1.86=3.72$