A compound formed by Mg, Al and  O, is found to have cubic close array of oxide ions, in which ${\mathrm{Mg}}^{2+}$ occupying ${\frac{1}{8}}^{\mathrm{th}}$ of tetrahedral voids and ${\mathrm{Al}}^{3+}$ ions  occupying $\frac{1}{2}$ the octahedral voids. The formula for the compound is :

Oxide ions have ccp array
$\therefore$    Number of Oxide ions/unit cell = 4
Thus number of OV / unit cell = 4
        number of TV / unit cell = 8

${\mathrm{Mg}}^{+2}$

Occupied ${\frac{1}{8}}^{\mathrm{th}}$  of TV

 $\therefore$   Number of  ${\mathrm{Mg}}^{+2}$ ions/ unit cell  =  $\frac{1}{8}\left(8\right)=1$

${\mathrm{Al}}^{+3}\mathrm{\hspace{0.17em}}$

Occupied $\frac{1}{2}$  of OV

$\therefore$   Number of  ${\mathrm{Al}}^{+3}\mathrm{\hspace{0.17em}}$ ions/ unit cell  =  $\frac{1}{2}\left(4\right)=2$

Formula :

$\begin{array}{l}\frac{{\mathrm{Mg}}^{+2}}{1}\frac{{\mathrm{Al}}^{+3}}{2}\frac{{\mathrm{O}}^{-2}}{4}\\ \boxed{{\mathrm{MgAl}}_2{\mathrm{O}}_4}\end{array}$