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(A) Let the number of atoms of $Y$ in the $ccp$ lattice be $n = 4$.Number of octahedral voids = $n = 4$.Number of tetrahedral voids = $2n = 8$.Atoms o

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$X_3 Y_2$

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(A) Let the number of atoms of $Y$ in the $ccp$ lattice be $n = 4$.Number of octahedral voids = $n = 4$.Number of tetrahedral voids = $2n = 8$.Atoms of $X$ occupy half of the octahedral voids and half of the tetrahedral voids.Number of $X$ atoms = $\frac{1}{2} 	imes 4 + \frac{1}{2} 	imes 8 = 2 + 4 = 6$.The ratio of $X : Y$ is $6 : 4$,which simplifies to $3 : 2$.Therefore,the formula of the compound is $X_3 Y_2$.

$A$ compound is formed from elements $X$ and $Y$. The atoms of $Y$ (anions) form a $ccp$ lattice. The atoms of $X$ (cations) occupy half of the octahedral voids and half of tetrahedral voids. What is the formula of the compound?

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