A crystal is made up of metal ions ' M _ 1 ' and ' M _ 2 ' and...

In the ccp latice of oxide ions effective number of ${O}^{- 2}ions$ $= 8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} = 4$

In the ccp latice,

No. of octahedral voids $= 4$

No. of tetrahedrla voids $= 8$

Given ${M}_{1}$ atoms occupies $50$ of octahedral voids and ${M}_{2}$ atoms occupies $12 . 5$ of tetrahederal voids

No. of ${M}_{1}$ metal atoms $= 4 \times \dfrac{50}{100} = 2$

No. of ${M}_{2}$ metal atoms $= 8 \times \dfrac{12 .5}{100} = 1$

$∴$ Formula of the compound $= {\left({M}_{1}\right)}_{2}\left({M}_{2}\right){O}_{4}$

$∴$ Oxidation states of metals ${M}_{1}$ & ${M}_{2}$ respectively are $+ 2$ and $+ 4$