A crystal is made up of metal iron $\text{'}{\mathrm{M}}_1\text{'}$ and $\text{'}{\mathrm{M}}_2\text{'}$ and oxide ions. Oxide ions form a $\mathrm{ccp}$ lattice structure. The cation ‘$\text{'}{\mathrm{M}}_1\text{'}$ occupies 50% of octahedral voids and the cation $\text{'}{\mathrm{M}}_2\text{'}$occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation number of $\text{'}{\mathrm{M}}_1\text{'}$ and $\text{'}{\mathrm{M}}_2\text{'}$ are, respectively:

The correct option is A

$+2,+4$

Answer:(A)

Explanation for correct option:

In $\mathrm{ccp}$ lattice oxide ion $=$corners $+$ face centered

$$\begin{gathered} =\frac{1}{8}\times 8+\frac{1}{2}\times 6 \\ =4 \end{gathered}$$

Number of ($\mathrm{O}$) $$\begin{gathered} =4 \\ \Rightarrow {\left(\mathrm{O}\right)}_4 \end{gathered}$$

In $\mathrm{ccp}$ lattice ,

Number of octahedral void$=4$

Number of tetrahedral void$=8$

$\left({\mathrm{M}}_1\right)\to 50\%$ of octahedral void

$$\begin{gathered} \mathrm{Number}\mathrm{of}\left({\mathrm{M}}_1\right)=\frac{50}{100}\times 4 \\ =2 \\ \Rightarrow {\left({\mathrm{M}}_1\right)}_2 \end{gathered}$$

$\left({\mathrm{M}}_2\right)\to 12.5\%$of tetrahedral void

$$\begin{gathered} \mathrm{Number}\mathrm{of}\left({\mathrm{M}}_2\right)=\frac{12.5}{100}\times 8 \\ =1 \\ ={\left({\mathrm{M}}_2\right)}_1 \end{gathered}$$

Hence formula must be ${\left({\mathrm{M}}_1\right)}_2\left({\mathrm{M}}_2\right){\mathrm{O}}_4$

For whole atom to be neutral(as oxidation state of $\mathrm{O}=-2$), $$\begin{gathered} \mathrm{O}.\mathrm{N}.\mathrm{of}{\mathrm{M}}_1=+2 \\ \mathrm{O}.\mathrm{N}.\mathrm{of}{\mathrm{M}}_2=+4 \end{gathered}$$

As, $$\begin{gathered} {\left({\mathrm{M}}_1\right)}_2\left({\mathrm{M}}_2\right){\mathrm{O}}_4 \\ =2\times (+2)+1\times (+4)+4\times (-2) \\ =4+4+(-8) \\ =8-8 \\ =0 \end{gathered}$$

Therefore, correct option is A