Question

What is the critical radius ratio for the $CsCl$ structure?

Answer 1

$C{l}^{-}$ is present on each corner while $C{s}^{+}$ is present at the body center of the cube or vice versa.
$C{s}^{+}$ is present in the cubical void.
$\text{Number of}C{l}^{-}=8\times \frac{1}{8}=1$
$\text{Number of}C{s}^{+}=1\times 1=1$
$2(r^{+}+r^{-})=\sqrt{3}a\text{(Length of body diagonal)}.......\left(i\right)$
$2r^{-}=a\text{only for ideal crystal}..........\left(ii\right)$
Putiing (ii) in (i)
$\Rightarrow 2(r^{+}+r^{-})=\sqrt{3}\left(2r^{-}\right)$
$\Rightarrow (r^{+}+r^{-})=\sqrt{3}\left(r^{-}\right).......\left(iii\right)$
$\text{Divide}\left(iii\right)\text{by}{r}^{-}$
$\Rightarrow \frac{r^{+}}{r^{-}}+1=\sqrt{3}$
$\Rightarrow \frac{r^{+}}{r^{-}}=\sqrt{3}-1$
$\Rightarrow \frac{r^{+}}{r^{-}}=0.732$