Structure of a mixed oxide is cubic close packed the cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal.A and the octahedral voids are occupied by a monvalent metal `B`. The formula of the oxide is :

Options

A. `AB_(2)O_(2)`
B. `ABO_(2)`
C. `A_(2)BO_(2)`
D. `A_(2)B_(3)O_(4)`

Answer

1

Explanation

oxide ions`(O^(2-))` are arranged in fcc manner. Therefore number of oxide ions per unit cell is
`8xx(1)/(8)+6xx(1)/(2)=4`
Therfore the number of octahedral voids is `4` and number of tetrahedral voids is `8`.
One fourth of the tetrahedral voids are occupied by divalent metal `A`. Therfore number of `A^(2+)` ions per unit cell is
`(1)/(4)xx8=2`
All the octahedral voids are occupied by a monovalent metal `B`.Therfore number of `B^(+)` ions per unit cell is `4`. Consequently the emipirical formula of mixed oxide is
`A_(2)B_(4)O_(4)` or `AB_(2)O_(2)`