Question

In a solid, oxide ions are arranged in hcp. One third of octahedral voids are occupied by the cations A and one sixth of the tetrahedral voids are occupied by the cations B. What is the formula of the compound?

Answer 1

Given: Oxide ions (O²⁻) are arranged in a hexagonal close-packed (hcp) structure.

Cation A occupies 1/3 of the octahedral voids.

Cation B occupies 1/6 of the tetrahedral voids.

In a hexagonal close-packed (hcp) arrangement:

Number of octahedral voids = number of atoms = 1 (assuming 1 O²⁻)

Number of tetrahedral voids = 2 × number of atoms = 2 (for 1 O²⁻)

Cation A = `1/3 xx 1 = 1/3`

Cation B = `1/6 xx 2 = 1/3`

So we have,

A = `1/3`, B = `1/3`, O = 1

Multiply all values by 3, we get,

A : B : O = `1/3 xx 3 : 1/3 xx 3 : 1 xx 3`

= 1 : 1 : 3

∴ The formula of the given compound is ABO₃.