In an ionic solid anions are arranged in hcp array and cations occupy $\frac{2}{3}$ of octahedral voids.…

In an ionic solid anions are arranged in hcp array and cations occupy $\frac{2}{3}$ of octahedral voids. What is the formula of ionic compound? [ consider $\mathrm{A}=$ cation ; $\mathrm{B}=$ anion]
  1. AB
  2. $\mathrm{A}_2 \mathrm{~B}_3$
  3. $\mathrm{A}_3 \mathrm{~B}_2$
  4. $\mathrm{AB}_3$

Solution

The hexagonal close-packed (hcp) array of anions B contains 6 atoms per unit cell.

In close-packed structures, the number of octahedral voids equals the number of atoms, giving 6 voids.

Since cations A occupy $\frac{2}{3}$ of these voids, the number of A cations is $\frac{2}{3} \times 6 = 4$.

The A:B ratio becomes 4:6, which simplifies to 2:3.

The formula is therefore $\mathrm{A}_2\mathrm{B}_3$.

Final answer: $\boxed{\text{B}}$

Asked in: MHT CET 2025 (05 May Shift 2)

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