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In a crystalline solid, anions B are arr...

In a crystalline solid, anions B are arranged in a ccp. Cations A are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, the formula of the compound is `A_2B//A_3B`.

A

`A_(2)B`

B

`AB_(2)`

C

`AB_(3)`

D

`A_(3)B`

Text Solution

Verified by Experts

The correct Answer is:
1
The unit cell of ccp structure is fcc. Therfore the number of anions `B` per unit cell is =`4`
The number of octahedral voids per unit cell =`4`
the number of tetrahedral voids per unit cell `=8`
As all the `4` octahedal voids are captured by cations `A` and cations `A` are equally distributed between octahedral and tetrahedral voids.the total number of cations `A` per unit cell is `8`.
Therfore cations `A` and cations `B` are in the ration `8:4=2:1` Consequently the empirical formula of the ionic solid will be `A_(2)B`.
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