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 $N = 4$ (for $ccp$ lattice).Number of octahedral voids $(O.V.)$ = $N = 4$.Number of tetrahedral voids $

Vedclass · Accepted Answer

$+2, +4$

Vedclass · Answer

(A) Let the number of oxide ions $(O^{2-})$ be $N = 4$ (for $ccp$ lattice).Number of octahedral voids $(O.V.)$ = $N = 4$.Number of tetrahedral voids $(T.V.)$ = $2N = 8$.Number of $M_1$ ions = $50 \% 	ext{ of } 4 = 0.5 	imes 4 = 2$.Number of $M_2$ ions = $12.5 \% 	ext{ of } 8 = 0.125 	imes 8 = 1$.The formula of the crystal is $(M_1)_2(M_2)_1O_4$.Since the crystal is electrically neutral,the total positive charge must balance the total negative charge.Total negative charge = $4 	imes (-2) = -8$.Total positive charge = $2 	imes (	ext{O.N. of } M_1) + 1 	imes (	ext{O.N. of } M_2) = +8$.Checking option $A$: $2(+2) + 1(+4) = 4 + 4 = +8$. This satisfies the condition.

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