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:
(AIPMT 2012)

A_{2} B_{3} O_{4}

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A B O_{2}

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A_{2} B O_{2}

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A B_{2} O_{2}

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Solution

The correct option is D $A B_{2} O_{2}$
$O^{2 -} makes ccp$

$So, number of O^{2 -} ions = 4$
$Tetrahedral voids = 2 \times 4 = 8$
$Octahedral voids = 4$

$\frac{1}{4} of tetrahedral void = \frac{1}{4} \times 8 = 2 occupied by A^{2 +}$

And octahedral voids = 4 occupied by $B^{+}$
$A : B : O = 2 : 4 : 4$
$So, the formula of oxide is A B_{2} O_{2}$ .

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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: (AIPMT 2012)

The correct option is D AB2O2O2− makes ccp So, number of O2− ions=4 Tetrahedral voids=2×4=8 Octahedral voids=4 14 of tetrahedral void=14×8=2 occupied by A2+ And octahedral voids = 4 occupied by B+ A:B:O=2:4:4 So, the formula of oxide isAB2O2.