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In a crystalline solid , anion B^- are a...

In a crystalline solid , anion `B^-` are arranged in a cubic close packing. Cations `A^+` are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, what is the formula of the solids?

Text Solution

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The correct Answer is:
`A_(2)B`

Let the number of anions B = n
The number of octahedral voids = n
The number of tetrahedral voids = 2n
Since both the octahedral and tetrahedral voids are equally occupied, by cations A and also the octahedral voids are occupied (given), this means that n cations A present in octahedral voids and n cations A present in tetrahedral voids.
In other words, corresponding to n anions B, there are n + n = 2n cations.
The cations A and anions B are in the ratio = 2n : n = 2 : 1
Formula of solid = `A_(2)B`.
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