A crystal is made up of metal ions M_1 and M_ | vedclass

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(A) Let the number of oxide ions $(O^{2-})$ be $N = 4$ (since they form a $ccp$ lattice).Number of octahedral voids $(O.V.)$ = $N = 4$.Number of tetra

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$+2, +4$

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(A) Let the number of oxide ions $(O^{2-})$ be $N = 4$ (since they form a $ccp$ lattice).Number of octahedral voids $(O.V.)$ = $N = 4$.Number of tetrahedral voids $(T.V.)$ = $2N = 8$.Number of $M_1$ ions = $50 \% 	ext{ of } O.V. = 0.50 	imes 4 = 2$.Number of $M_2$ ions = $12.5 \% 	ext{ of } T.V. = 0.125 	imes 8 = 1$.The formula of the crystal is $M_1{}_2 M_2{}_1 O_4$.For the crystal to be electrically neutral,the total positive charge must equal the total negative charge.Total negative charge = $4 	imes (-2) = -8$.Therefore,the total positive charge must be $+8$.Let $x$ be the oxidation number of $M_1$ and $y$ be the oxidation number of $M_2$.$2x + 1y = 8$.Checking the options:If $x = +2$ and $y = +4$,then $2(2) + 1(4) = 4 + 4 = 8$. This satisfies the condition.Thus,the oxidation numbers are $+2$ and $+4$.

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