1. C ₂ A ₃ 2. C ₃ A ₂ 3. C ₃ A ₄ 4. C ₄ A ₃

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 :

1. C\(_2\)A\(_3\)

2. C\(_3\)A\(_2\)

3. C\(_3\)A\(_4\)

4. C\(_4\)A\(_3\)

Doubtly Asked 30 Oct 2026 NEET UG 2019 neet Save WhatsApp Report

Best answer

Answer: 3. C\(_3\)A\(_4\)

In hcp, if there are \(N\) anions then there are \(N\) octahedral voids.

Cations occupy 75% of octahedral voids, so cations = \(0.75N = \tfrac{3}{4}N\).

Ratio C : A = 3/4 : 1 = 3 : 4, so the formula is \(\mathrm{C_3A_4}\), option (3).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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