In $CsCl , Cl ^{-}$lies at corners of simple cube and $Cs ^{+}$at the body centre. Hence, along the body diagonal, $Cs ^{+}$and $Cl ^{-}$touch each other so
$r_{ Cs ^{+}}+r_{ Cl ^{-}}=2 r$
Calculation of $r$ In $\Delta E D F$,
Body centred cubic unit cell
$F D=b=\sqrt{a^{2}+a^{2}}=\sqrt{2} a$
In $\triangle A F D$,
$c^{2}=a^{2}+b^{2}=a^{2}+(\sqrt{2} a)^{2}=a^{2}+2 a^{2}$
$ c^{2} =3 a^{2} $
$ c =\sqrt{3} a $
As $\triangle A F D$ is an equilateral triangle.
$\therefore \sqrt{3} a=4 r$
$[\because C=3 r+r+r]$
$\Rightarrow r=\frac{\sqrt{3} a}{4}$
Hence, $r_{ Cs }+r_{ Cl }{ }^{-}=2 r=2 \times \frac{\sqrt{3}}{4} a=\frac{\sqrt{3}}{2} a$
