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Q. Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is

Solution:

According to $ccp$, structure,
Number of $O^{2-}$ ions = 4
$\therefore $ tetrahedral void = 8
and octahedral void = 4
$\because $ A ions occupy l/4th of tetrahedral voids
$\therefore $ Number of A ions = $ \frac{1}{4} \times 8$ = 2
Again B ions occupy all the tetrahedral voids
$\therefore $ Number of B ion = 4
A : B : O = 2 : 4 : 4 = 1 : 2 : 2
$\therefore $ the formula of oxide = $AB_2O_2$