Question

$CsBr$ has $bcc$ structure with edge length $4.3\mathring{A}$ The shortest inter ionic distance in between $C{s}^{+}$ and $B{r}^{-}$ is

• A. 3.72
• B. 1.86
• C. 7.44
• D. 4.3

Answer 1

Answer: (A)

For bcc structure, atomic radius, $r=\frac{\sqrt{3}}{4}a$
$=\frac{\sqrt{3}}{4}\times 4.3=1.86$
Since, r = half the distance between two nearest neighbouring atoms.
$\therefore$ Shortest inter ionic distance = 2 $\times$ 1.86
= 3.72