A compound is formed by cation C and anion A. | vedclass

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Question

(C) In a hexagonal close packed $(hcp)$ lattice,the number of anions $(A)$ per unit cell is $6$.The number of octahedral voids in an $hcp$ lattice is

Vedclass · Accepted Answer

$C_{3}A_{4}$

Vedclass · Answer

(C) In a hexagonal close packed $(hcp)$ lattice,the number of anions $(A)$ per unit cell is $6$.The number of octahedral voids in an $hcp$ lattice is equal to the number of atoms,which is $6$.The cations $(C)$ occupy $75 \%$ of these octahedral voids.Number of cations $(C) = 6 	imes \frac{75}{100} = 6 	imes \frac{3}{4} = \frac{18}{4} = 4.5$ or $\frac{9}{2}$.The ratio of cations $(C)$ to anions $(A)$ is $\frac{9}{2} : 6$.Multiplying both sides by $2$,we get $9 : 12$.Simplifying the ratio by dividing by $3$,we get $3 : 4$.Therefore,the formula of the compound is $C_{3}A_{4}$.

$A$ compound is formed by cation $C$ and anion $A$. The anions form hexagonal close packed $(hcp)$ lattice and the cations occupy $75 \%$ of octahedral voids. The formula of the compound is

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