In a compound AB_2O_4,the oxide ions form a c | vedclass

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(C) In a $CCP$ lattice,the number of oxide ions $(O^{2-})$ per unit cell is $4$.The number of octahedral voids $(O.V.)$ is $4$ and the number of tetra

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$1/2$

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(C) In a $CCP$ lattice,the number of oxide ions $(O^{2-})$ per unit cell is $4$.The number of octahedral voids $(O.V.)$ is $4$ and the number of tetrahedral voids $(T.V.)$ is $8$.Given the formula $AB_2O_4$,there are $1$ bivalent ion $(A^{2+})$ and $2$ trivalent ions $(B^{3+})$ per $4$ oxide ions.$A^{2+}$ ions occupy octahedral voids: $1$ $O.V.$ is occupied by $A^{2+}$.$B^{3+}$ ions occupy both $T.V.$ and $O.V.$ in equal numbers. Let $x$ be the number of $B^{3+}$ ions in $T.V.$ and $x$ be the number of $B^{3+}$ ions in $O.V.$.Total $B^{3+} = x + x = 2x = 2$,so $x = 1$.Thus,$B^{3+}$ ions occupy $1$ $T.V.$ and $1$ $O.V$.Total $O.V.$ occupied = (occupied by $A^{2+}$) + (occupied by $B^{3+}$) = $1 + 1 = 2$.Total $O.V.$ available = $4$.Unoccupied $O.V.$ = $4 - 2 = 2$.Fraction of unoccupied $O.V.$ = $2/4 = 1/2$.

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