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Structure of a mixed oxide is cubic close-packed (c.c.p). 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:
A
$AB_2O_2$
B
$ABO_2$
C
$A_2BO_2$
D
$A_2B_3O_4$
Detailed Solution
$O^{2-}$ ions form the ccp lattice, so there are 4 oxide ions per unit cell.
Tetrahedral voids = 8; one fourth are occupied by $A^{2+}$: $\frac{1}{4}\times8 = 2$
Octahedral voids = 4; all are occupied by $B^+$: 4
Formula: $A_2B_4O_4$, i.e. $AB_2O_2$ (charge check: +2 + 2 − 4 = 0)
Tetrahedral voids = 8; one fourth are occupied by $A^{2+}$: $\frac{1}{4}\times8 = 2$
Octahedral voids = 4; all are occupied by $B^+$: 4
Formula: $A_2B_4O_4$, i.e. $AB_2O_2$ (charge check: +2 + 2 − 4 = 0)