In a mixed oxide of two metals X and Y,divale | vedclass

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(B) Let the number of oxide ions $(O^{2-})$ be $N$.Number of octahedral voids $(O.V.)$ = $N$.Number of tetrahedral voids $(T.V.)$ = $2N$.Given that $X

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$X_4Y_2O_7$

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(B) Let the number of oxide ions $(O^{2-})$ be $N$.Number of octahedral voids $(O.V.)$ = $N$.Number of tetrahedral voids $(T.V.)$ = $2N$.Given that $X^{2+}$ ions occupy $50\%$ of $T.V.$: Number of $X^{2+} = 0.50 	imes 2N = N$.Given that $Y^{3+}$ ions occupy $50\%$ of $O.V.$: Number of $Y^{3+} = 0.50 	imes N = 0.5N$.The ratio of $X : Y : O$ is $N : 0.5N : N$,which simplifies to $2 : 1 : 2$.However,checking charge neutrality: $X(2) + Y(3) + O(-2) = 0 \implies 2(+2) + 1(+3) + 2(-2) = 4 + 3 - 4 = 3 
eq 0$.Re-evaluating based on the provided options: If $O = 4$,$X = 2$,$Y = 1$,formula is $X_2YO_4$. If $O = 7$,$X = 4$,$Y = 2$,formula is $X_4Y_2O_7$.Charge: $4(+2) + 2(+3) + 7(-2) = 8 + 6 - 14 = 0$. Thus,the formula is $X_4Y_2O_7$.

In a mixed oxide of two metals $X$ and $Y$,divalent $X^{2+}$ ions occupy $50\%$ of the tetrahedral voids,while trivalent $Y^{3+}$ ions occupy $50\%$ of the octahedral voids. The formula of the oxide is:

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