Chemistry - Solid State Question with Solution | TestHub

$\mathrm{x}+\mathrm{y}=\left[\frac{(\text{ edge length })\sqrt{3}}{4}\right]$

∵ A ${}^{+2}$ occupy corners and face centres;

${\mathrm{B}}^{-}$All tetrahedral voids

Max. distance between TVs

(two ${\mathrm{B}}^{-})=\frac{\text{ (edge length) }}{2}(\sqrt{3})$

$=\frac{4}{\sqrt{3}}(X+Y)\frac{\sqrt{3}}{2}$

$=2(\mathrm{X}+\mathrm{Y})$

Min distance (two ${\mathrm{B}}^{-}$)

$=\frac{\text{ Edge length }}{2}=\frac{2}{\sqrt{3}}(\mathrm{X}+\mathrm{Y})$