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Question

In a compound, oxides ions form fcc lattice, whereas 14th of tetrahedral voids are occupied by cations A2+ and 18th of octahedral voids are occupied by cations B3+ ions. The simplest formula of the compound is

A
A4BO8
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B
AB2O4
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C
AB4O8
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D
A2BO4
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Solution

The correct option is A A4BO8
In the compound , since oxide ions forms the ccp arrangements, it occupies the corners and face centres of the cubic unit cell.
No. of O2ions in unit cell =N=4

Number of tetrahedral voids =2N=8
Number of octahedral voids =N=4

A2+ ions occupies 14th of the tetrahedral voids

Number of A2+ ions in an unit cell =14×8=2

B3+ ions occupies 18th of the octahedral voids

Number of B3+ ions in an unit cell =18×4=12

The ratio of A2+:B3+:O2 is =2:12:4
Formula of compound is A2B12O4
Simplest formula :
A4BO8

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