JEE Main 2020 — Solid State Question with Solution

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

In the ccp latice,

No. of octahedral voids $=4$

No. of tetrahedrla voids $=8$

Given ${\mathrm{M}}_1$ atoms occupies $50\%$ of octahedral voids and ${\mathrm{M}}_2$ atoms occupies $12.5$ of tetrahederal voids

No. of ${\mathrm{M}}_1$ metal atoms $=4\times \frac{50}{100}=2$

No. of ${\mathrm{M}}_2$ metal atoms $=8\times \frac{12.5}{100}=1$

$\therefore$ Formula of the compound $={\left({\mathrm{M}}_1\right)}_2\left({\mathrm{M}}_2\right){\mathrm{O}}_4$

$\therefore$ Oxidation states of metals ${\mathrm{M}}_1$ & ${\mathrm{M}}_2$ respectively are $+2$ and $+4$