Question

In a closed packed structure of mixed oxides, the oxide ions are arranged in hcp array. One eighth of tetrahedral voids are occupied by divalent cations (A) while one half of octahedral voids are occupied by trivalent cations (B). What is the formula of the compound?

Answer 1

Given: Oxide ions (O²⁻) are in an hcp arrangement.

Divalent cation A occupies 1/8 of tetrahedral voids.

Trivalent cation B occupies 1/2 of octahedral voids.

In hcp arrangement:

Tetrahedral voids = 2 per O²⁻ ion

Octahedral voids = 1 per O²⁻ ion

Number of cations:

Cation A occupies `1/8`​ of tetrahedral voids = `1/8 xx 2 = 1/4` A ions

Cation B occupies `1/2`​ of octahedral voids = `1/2 xx 1 = 1/2` B ions

O²⁻ ions = 1 (assumed)

So, A : B : O = `1/4 : 1/2 : 1`

Multiply all by 4 to get whole numbers,

A : B : O = 1 : 2 : 4

∴ The formula of the compound is AB₂O₄.