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Question

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:

A
A2BO2
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B
A2B3O4
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C
AB2O2
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D
ABO2
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Solution

The correct option is B AB2O2
It is given that the structure of mixed oxide is cubic closed packed.
Therefore, number of O2 ions according to ccp =4
So, number of tetrahedral voids =8 and octahedral voids =4
A ions occupy 14th of tetrahedral voids =2
B ions occupy all octahedral voids =4
A:B:O =2:4:4 =1:2:2
Hence, the formula is AB2O2.

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