Question

A crystal is made up of metal ions 'M $_{1}$ ' and 'M $_{2}$ ' and oxide ions. Oxide ions. form a ccp lattice structure. The cation '$M_1$ occupies $50 \%$ of octahedral voids and the cation 'M $_{2}$ ' occupies $12.5 \%$ of tetrahedral voids of oxide lattice. The oxidation numbers of ' $M _{1}$ ' and ' $M _{2}$ ' are, respectively:

• A. $+2, +4$
• B. $+3, +1$
• C. $+1, +3$
• D. $+4, +2$

Answer 1

Answer: (A)

$+2, +4$

Answer 2

$O ^{-2}$ ions form $ccp$

$M _{1}=50 \%$ of $O . V .$
$\Rightarrow \frac{50}{100} \times 4=2:\left( M _{1}\right)_{2}$
$M _{2}=12.5 \%$ of $T.V .$
$\Rightarrow \frac{12.5}{100} \times 8=1:\left( M _{2}\right)$
So formula is : $(M_{1})_{2} (M_{2})_{1} O_{4}$
So , the correct answer is (A) : +2,+4