Question

$CsBr$ has a bcc type structure with edge length $4.3\text{pm}$. The shortest inter ionic distance between $C{s}^{+}$ and $B{r}^{-}$ is:

• A. $3.72\text{pm}$
• B. $1.86\text{pm}$
• C. $7.44\text{pm}$
• D. $4.3\text{pm}$

Answer 1

The correct option is A $3.72\text{pm}$

$CsBr$ has a bcc type structure. Here $C{s}^{+}$ ion are present at the body centre of the cube and $B{r}^{-}$ ions are present at the corners of the cube. So the relation between inter ionic distance $(r^{+}+r^{-})$ and edge length will be
$\sqrt{3}a=2(r^{+}+r^{-})$
Shortest distance would be $r^{+}+r^{-}$
Here, $r^{+}+r^{-}=\frac{\sqrt{3}a}{2}=\frac{1.73\times 4.3}{2}=3.7195\text{pm}\approx 3.72\text{pm}$