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In a crystalline solid, anions Y^(-) are...

In a crystalline solid, anions `Y^(-)` are arranged in ccp. Cation `X^(+)` are equally distributed between octehedral and tetrahedral voids. If all the octahedral voids are occupied, what is the formula of the compound ?

Text Solution

Verified by Experts

Suppose the number of anions, `Y^(-) = N`
Then number of octahedral voids = N
Number of tetrahedral voids = 2N
Since octanhedral and tetrahedral voids are equally occupied by cations `X^(+)` and all the octahedral voids are occupied, then N cations `X^(+)` are present in octahedral voids and N cations `X^(+)` are present in tetrahedral voids. Thus,
No. of cations present = `N + N = 2N`
Ratio of cations and anions = 2: 1
`:.` Formula of the compound = `X_2Y`
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