A solid has an hcp lattice. Atoms of Z (anion | vedclass

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(B) In an $hcp$ lattice,the number of atoms per unit cell is $6$. Let the number of atoms of $Z$ be $N = 6$. The number of octahedral voids is equal t

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$XYZ$

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(B) In an $hcp$ lattice,the number of atoms per unit cell is $6$. Let the number of atoms of $Z$ be $N = 6$. The number of octahedral voids is equal to the number of atoms,so the number of $X$ atoms $= N = 6$. The number of tetrahedral voids is $2N = 12$. Atoms of $Y$ occupy half of the tetrahedral voids,so the number of $Y$ atoms $= \frac{1}{2} 	imes 12 = 6$. The ratio of atoms $X : Y : Z$ is $6 : 6 : 6$,which simplifies to $1 : 1 : 1$. Therefore,the molecular formula of the solid is $XYZ$.

$A$ solid has an $hcp$ lattice. Atoms of $Z$ (anions) form the $hcp$ lattice. Atoms of $X$ (cations) occupy all the octahedral voids in the lattice. Atoms of $Y$ (cations) occupy half of the tetrahedral voids. What is the molecular formula of the solid?

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