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 :

Solution

Chemistry ma’am · 3 steps

Count the oxide ions and both kinds of voids first, then make the charges add up to zero.

1. Voids in ccp. A ccp unit cell has 4 oxide ions. It has as many octahedral voids as atoms (4) and twice as many tetrahedral voids (8).
2. Count the cations. M₁ fills 50% of 4 octahedral voids, which is 2 ions. M₂ fills 12.5% of 8 tetrahedral voids, which is 1 ion. The formula is (M₁)₂(M₂)O₄.

   2x + y + 4(-2) = 0 ⇒ 2x + y = 8

   Option	x, y	2x + y	Neutral?
   (a)	+3, +1	7	no
   (b)	+4, +2	10	no
   (c)	+1, +3	5	no
   (d)	+2, +4	8	yes

3. So. Only M₁ = +2 and M₂ = +4 balance the -8 from four oxide ions, so the answer is (d).

The idea to keep. For N atoms in ccp or hcp there are N octahedral and 2N tetrahedral voids. Turn the percentages into ion counts, write the formula, then balance the charge.

Answer from the NTA final answer key. Solution by PrepWiser.