Anions are in fcc positions and half of the tetrahedral holes are occupied by cations.There are H anions and 8 tetrahedral hols per unit cell

Here,face diagonal =$\sqrt 2a$

$\therefore 4r=\sqrt 2a$

$a=2\sqrt 2r$

$r_-$=radius of anion.

$r_+$=radius of cation.

$a$=edge length of the cell

Volume of unit cell =$a^3$

$\Rightarrow (2\sqrt 2 r_-)^3=16\sqrt 2r_-^3$

$\therefore$ Packing fraction =$\large\frac{4\times (\Large\frac{4}{3}\normalsize \pi r_-^3)+\Large\frac{1}{2}\normalsize \times 8\times (\Large\frac{4}{3}\normalsize \pi r_+^3)}{16\sqrt 2 r_-^3}$

$\Rightarrow \large\frac{\pi}{3\sqrt 2}$$\left\{1+(\large\frac{r_+}{r_-})^3\right\}$

Since,for tetrahedral holes

$\large\frac{r_+}{r_-}$$=0.225$

$\therefore PF=\large\frac{\pi}{3\sqrt 2}$$[1+(0.225)^3]$

$\Rightarrow 0.7485$ /unit volume of unit cell

$\therefore$ Void space =1-0.7485

$\Rightarrow 0.2515$ /unit volume of unit cell

$\therefore$ Percentage of void space =25.15% $\simeq 25\%$