In a crystalline solid ,having formula AB₂O₄ , oxide ions are arranged in cubic close packed lattice while cations A are present in tetrahedral voids and cationsB are present in octahedral voids.
(a)What percentage of tetrahedral void is occupied by A ?
(b) What percentage of octahedalvoid is occupied by B ?

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Solution

In a cubic close packed lattice of oxide ions ( fcc) there would be two tetrahedral voids and one octahedral void. Since the given formula is AB₂O₄_.For four oxide ions there would be 8 tetrahedral and four octahedral voids in which two are occupied by B.

% of td voids occupied by A = $\frac{1}{8} \times 100$ = 12.5 %

% of Oh voids occupied by B = $\frac{2}{4} \times 100$ = 50 %

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