Correct Answer – D
No. of atoms `(O^(2-))` in ccp `=4`
No. of tetrahedral voids `=2xxN=8`
No. of `A^(2+)` ions `=8xx1/4=2`
No. of octahedral voids =No of `B^(+)` ions =4
Ratio , `O^(2-),A^(2+):B^(+)=4:2:4=2:1:2`
Formula of oxide `=AB_(2)O_(2)`

Correct Answer – 1
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)`