If the anions (A) form hexagonal closest pack | vedclass

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Question

(C) In a hexagonal closest packing $(HCP)$ lattice,the number of octahedral voids is equal to the number of atoms present in the lattice.Let the numbe

Vedclass · Accepted Answer

$C_2A_3$

Vedclass · Answer

(C) In a hexagonal closest packing $(HCP)$ lattice,the number of octahedral voids is equal to the number of atoms present in the lattice.Let the number of anions $(A)$ be $N$.Then,the number of octahedral voids is also $N$.The cations $(C)$ occupy $2/3$ of the octahedral voids,so the number of cations is $(2/3)N$.The ratio of $C:A$ is $(2/3)N : N$,which simplifies to $2:3$.Therefore,the formula of the compound is $C_2A_3$.

If the anions $(A)$ form hexagonal closest packing and cations $(C)$ occupy only $2/3$ of the octahedral voids in it,then the general formula of the compound would be

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