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In solid oxide are arranged in ccp .One ...

In solid oxide are arranged in ccp .One - sixth of tetrabedral voids are occupied by cation A which one third of octahedral voids are occupied by cation `B` .What is the formula of compound ?

Text Solution

Verified by Experts

Number of oxide ions forming ccp = 4
(Occupy 8 comers and 6 face centres)
Number of Tetrahedral voids = 8
Number of Octahedral voids = 4
Number of cations A occupying Tetrahedral voids `= (8)/(6)`
Number of cations B occupying Octahedral voids `= (4)/(3)`
`A : B : O^(-2)`
`(8)/(6):(4)/(3):4`
`8:8:24`
`therefore 1:1:3`
And formula of compound is `ABO_(3)`.
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