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

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Question

(A) Let the number of oxide ions $(O^{2-})$ in the $ccp$ lattice be $4$.Number of octahedral voids $(O.V.)$ = $4$.Number of tetrahedral voids $(T.V.)$

Vedclass · Accepted Answer

$+2, +4$

Vedclass · Answer

(A) Let the number of oxide ions $(O^{2-})$ in the $ccp$ lattice be $4$.Number of octahedral voids $(O.V.)$ = $4$.Number of tetrahedral voids $(T.V.)$ = $2 	imes 4 = 8$.$M_1$ occupies $50 \%$ of $O.V. = 0.50 	imes 4 = 2$.$M_2$ occupies $12.5 \%$ of $T.V. = 0.125 	imes 8 = 1$.The formula of the crystal is $(M_1)_2(M_2)_1O_4$.For the crystal to be electrically neutral,the sum of oxidation states must be zero:$2 	imes (	ext{O.N. of } M_1) + 1 	imes (	ext{O.N. of } M_2) + 4 	imes (-2) = 0$$2 	imes (	ext{O.N. of } M_1) + (	ext{O.N. of } M_2) = 8$.Checking option $A$: $2(+2) + (+4) = 4 + 4 = 8$. This satisfies the condition.

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