Question

Solid $AB$ has $CsCl$ type structure. If the radius of $A^{+}$ ion is $73.2\text{pm}$ then the minimum radius of $B^{-}$ $\left(\text{in pm}\right)$ ion will be:

Answer 1

In $CsCl$, the coordination number of both $C{s}^{+}$ and $C{l}^{-}$ is $8$.
$AB$ has similar type of structure with $CsCl$.
If $r_{A^{+}}$ is the radius of cation $A^{+}$ and $r_{B^{-}}$ is the radius of anion $B^{-}$, then
$\frac{r_{A^{+}}}{r_{B^{-}}}=0.732\Rightarrow \frac{73.2}{r_{B^{-}}}=0.732$
$\Rightarrow r_{B^{-}}=100\text{pm}$