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In a cubic packed structure of mixed oxi...

In a cubic packed structure of mixed oxides the lattice is made up of oxide ions. One fifth of tetrahedral voids are occupied by divalent `(X^(2+))` ions, while one - half of the octahedral voids are occupied by trivalent ions `(Y^(3+))` then the formula of the oxide is:

A

`XY_(2)O_(4)`

B

`X_(2)YO_(4)`

C

`X_(4)Y_(5)O_(10)`

D

`X_(3)YZ`

Text Solution

Verified by Experts

The correct Answer is:
C
In ccp anions occupy primitives of the cube while cations occupied voids in ccp there are two tetrahedral voids and one octahedral holes per anion
For one oxygen atom there are two tetrahedral holes and one octahedral hole
Since one fifth of the tetrahedral voids are occupied by divalent cations `(X^(2+))`
`:.` number of divalent cations in tetrahedral voids
`=2xx1/5`
Since half of the octahedral voids are occupied by trivalent cation `(Y^(3+))`
`:.` number of trivalent cations `=1xx1//2`
So the formula of the compounds is `(X)_(2xx1/5)(Y)_(1/2)(O)_(1)`
or `X_(2/5)Y_(1/2)O_(1) or X_(4)Y_(5)O_(10)`
Hence, (c) is correct
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