A crystal is made up of metal ‘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.

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+2, +4

+4, +2

+1, +3

+3, +1

answer is A.

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Detailed Solution

If $O^{2 -}$ ions form ccp that gives 4.0v’s and 8 T.v’s

50% of O.V i.e, 2.0 Vs are occupiiied by $M_{1}$

12.5% of T.V i.e 7 T.V is occupied by $M_{2}$

So in one formule unit there is $2 M_{1} 1 M_{2} and 4 O^{- 2}$

So $M_{1} must be M_{1}^{+ 3}$ and $M_{2} must be M_{2}^{2 +}$

$\Rightarrow M_{2} {(M_{1})}_{2} O_{4}$

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