A crystal is made up of metal ions 'M1' and 'M2' and oxide ions. Oxide ions form a ccp lattice structure. The cation 'M1' occupies 50% of octahedral voids and the cation 'M2' occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation numbers of 'M1' and 'M2' are respectively:

We are told that oxide ions build a cubic close-packed (ccp, i.e. face-centred cubic) lattice. In a ccp lattice:

• The number of anions (oxide ions, here) present in one unit cell is 4.
• The number of octahedral voids present in one unit cell is equal to the number of lattice points, so it is also 4.
• The number of tetrahedral voids present in one unit cell is twice the number of lattice points, so it is 8.

Let us now count how many metal ions are inserted into these voids.

The cation $$\mathrm{M_1}$$ occupies 50 % of the octahedral voids. We have

$$\text{Number of octahedral voids}=4, \qquad 50\%=\dfrac{50}{100}=0.5.$$

So

$$\text{Number of }\mathrm{M_1}\text{ ions}=4 \times 0.5 = 2.$$

The cation $$\mathrm{M_2}$$ occupies 12.5 % of the tetrahedral voids. We have

$$\text{Number of tetrahedral voids}=8, \qquad 12.5\%=\dfrac{12.5}{100}=0.125.$$

So

$$\text{Number of }\mathrm{M_2}\text{ ions}=8 \times 0.125 = 1.$$

Now we summarise the contents of one unit cell:

$$\mathrm{O^{2-}} : 4, \qquad \mathrm{M_1} : 2, \qquad \mathrm{M_2} : 1.$$

Hence the empirical formula of the solid is

$$\mathrm{M_1}_2\mathrm{M_2}\mathrm{O}_4.$$

Next we use charge neutrality. Let the oxidation numbers of $$\mathrm{M_1}$$ and $$\mathrm{M_2}$$ be $$x$$ and $$y$$ respectively. We write the total charge balance equation:

$$\underbrace{2x}_{\text{from }\mathrm{M_1}} \;+\; \underbrace{y}_{\text{from }\mathrm{M_2}} \;+\; \underbrace{4(-2)}_{\text{from }4\mathrm{O^{2-}}}=0.$$

Simplifying, we obtain

$$2x + y - 8 = 0,$$ $$2x + y = 8.$$

We now test the given options one by one.

Option A gives $$x=+2,\;y=+4.$$ Substituting,

$$2(2) + 4 = 4 + 4 = 8,$$

which satisfies $$2x + y = 8.$$

Option B gives $$x=+1,\;y=+3$$ giving $$2(1)+3 = 5 \neq 8.$$
Option C gives $$x=+3,\;y=+1$$ giving $$2(3)+1 = 7 \neq 8.$$
Option D gives $$x=+4,\;y=+2$$ giving $$2(4)+2 = 10 \neq 8.$$

Only Option A fulfils the electro-neutrality condition.

Hence, the correct answer is Option A.