in a crystalline solid, anions B are arranged in a cubic close packing. cations are equally distributed between octahedral and tetrahedral voids. if all the octahedral voids are occupied, what is the formula of the solid

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Solution

Answer : A2B

The number of atoms per unit cell in ccp = 4
Since anion B are arranged in ccp therefore 4 atom of B per uit cell.

In ccp total number of octahedral voids are 4. All the octahedral voids are occupied therefore number of cation in octahedral voids will be 4.
Given that cations A are equally distributed between octahedral and tetrahedral voids so number of cations in octahedral and tetrahedral voids are 4 + 4 = 8
So the formula of compound will be A8B4 or A2B

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