Structure of a mixed oxide is cubic closed - packed (ccp) .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 monovelent metal B .The formula of the oxide is

Options

A. `ABO_2`
B. `A_2BO_2`
C. `A_2B_3O_4`
D. `AB_2O_2`

Answer

D

Explanation

For ccp arrangement , number of `O^(2-)` ions = 4
Number of octahedral voids =4
Thus number of tetrahedral voids = 8
As `A^(2+)` (divalent matal) occupy 1/4 tetrahedral voids ,
Thus , number of `A^(2+)` ions `=1/4xx8=2` Again, `B^+` occupied all octahedral voids hence, number of B ions = 4
`therefore " Structure of oxide " = AB_2O_2`