If the anion (A) forms a hexagonal closest pa | vedclass

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Question

(C) In a hexagonal closest packing $(hcp)$,the effective number of atoms per unit cell is $6$.The number of octahedral voids is equal to the number of

Vedclass · Accepted Answer

$C_{2}A_{3}$

Vedclass · Answer

(C) In a hexagonal closest packing $(hcp)$,the effective number of atoms per unit cell is $6$.The number of octahedral voids is equal to the number of atoms in the packing,which is $6$.The cation $(C)$ occupies $2/3$ of the octahedral voids,so the number of $C$ atoms $= 6 	imes (2/3) = 4$.The number of anion $(A)$ atoms is $6$.Thus,the ratio of $C:A$ is $4:6$,which simplifies to $2:3$.Therefore,the formula of the compound is $C_{2}A_{3}$.

If the anion $(A)$ forms a hexagonal closest packing and the cation $(C)$ occupies only $2/3$ of the octahedral voids in it,then the general formula of the compound is :-

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