Q.

In a cubic close packed structure of mixed oxide the lattice is made up of oxide ions, one fifth of tetrahedral voids occupied by divalent (X+2)ions while one half of the octahedral voids are occupied by trivalent ion (Y+3). The simple formula of the oxide is XmYnOp. The value of m+n+p is

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answer is 19.

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Detailed Solution

Total number of oxide ions=Zeff for FCC=4

There are 8 tetrahedral voids and 4 octahedral voids in a FCC lattice .

For X2+:
Number of ions=(1/5)×8=8/5

For Y3+:
Number of ions=(1/2)×4=2

Molecular formula is : X8/5Y2O4

Simplest formula is X4Y5O10
 

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In a cubic close packed structure of mixed oxide the lattice is made up of oxide ions, one fifth of tetrahedral voids occupied by divalent (X+2)ions while one half of the octahedral voids are occupied by trivalent ion (Y+3). The simple formula of the oxide is XmYnOp. The value of m+n+p is