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(D) $1$. Let the number of atoms of element $B$ in the $hcp$ structure be $n$.$2$. The number of octahedral voids is equal to the number of atoms form

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$AB_3$

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(D) $1$. Let the number of atoms of element $B$ in the $hcp$ structure be $n$.$2$. The number of octahedral voids is equal to the number of atoms forming the $hcp$ lattice,which is $n$.$3$. Atoms of element $A$ occupy $\frac{1}{3}$ of the octahedral voids,so the number of atoms of $A = \frac{1}{3}n$.$4$. The ratio of atoms $A : B = \frac{1}{3}n : n = 1 : 3$.$5$. Therefore,the formula of the compound is $AB_3$.

Atoms of element $A$ occupy $\frac{1}{3}$ of the octahedral voids,while atoms of element $B$ form an $hcp$ structure. Determine the formula of the compound formed by $A$ and $B$.

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