In an ionic solid,anions (B) are arranged in | vedclass

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(A) Let the number of anions $(B)$ in the $hcp$ array be $n$. Since the number of octahedral voids is equal to the number of atoms in the $hcp$ arrang

Vedclass · Accepted Answer

$A_2B_3$

Vedclass · Answer

(A) Let the number of anions $(B)$ in the $hcp$ array be $n$. Since the number of octahedral voids is equal to the number of atoms in the $hcp$ arrangement,the number of octahedral voids is $n$. Given that cations $(A)$ occupy $\frac{2}{3}$ of the octahedral voids,the number of cations $(A)$ is $\frac{2}{3}n$. The ratio of cations $(A)$ to anions $(B)$ is $\frac{2}{3}n : n$,which simplifies to $2:3$. Therefore,the formula of the ionic compound is $A_2B_3$.

In an ionic solid,anions $(B)$ are arranged in an $hcp$ array and cations $(A)$ occupy $\frac{2}{3}$ of the octahedral voids. What is the formula of the ionic compound?

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