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In a cubic close packed structure (ccp) ...

In a cubic close packed structure (ccp) of mixed oxides, it is found that lattice has `O^(2-)` ions and one half of the octahedral voids are occupied by trivalent cations `(A^(3+))` and one-eighth of the tetrahedral voids are occupied by divalent cations `(B^(2+))`. Derive the formula of the mixed oxide.

Text Solution

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We know that there are two tetrahedral voids and one octahedral void per ion in the given arrangement.
Therefore,
The no. of octahedral voids per ion in lattice = 1
No. of trivalent cations `(A^(3+))=1xx1//2=1//2`
The no. of tetrahedral voids per ion in lattice = 2
No. of divalent cations `(B^(2+))=2xx1//8=1//4`
The formula of mixed oxide is `A_(1//2)B_(1//4)O` or `A_(2)BO_(4)`
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