In a ccp structure of a mixed oxide,1/8 of th | vedclass

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

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$X{Y_2}{O_4}$

Vedclass · Answer

(A) Let the number of oxide ions $(O^{2-})$ in the $ccp$ lattice be $N = 4$.Number of octahedral voids $(OV)$ = $N = 4$.Number of tetrahedral voids $(TV)$ = $2N = 8$.Given that $1/8$ of the $TV$ are occupied by $X^{+2}$ ions:Number of $X^{+2}$ ions = $\frac{1}{8} 	imes 8 = 1$.Given that $50\%$ of the $OV$ are occupied by $Y^{+3}$ ions:Number of $Y^{+3}$ ions = $\frac{50}{100} 	imes 4 = 2$.Thus,the ratio of $X : Y : O$ is $1 : 2 : 4$.The formula of the oxide is $X{Y_2}{O_4}$.

In a $ccp$ structure of a mixed oxide,$1/8$ of the tetrahedral voids are occupied by divalent $X^{+2}$ ions and $50\%$ of the octahedral voids are occupied by trivalent $Y^{+3}$ ions. The formula of the oxide is:

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