In solid oxide are arranged in ccp .One - sixth of tetrabedral voids are occupied by cation A which one third of octahedral voids are occupied by cation `B` .What is the formula of compound ?

Options

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

Answer

B

Explanation

Suppose `O^(2-)` ions = n . Then octahedral voids = n,
tetrahedral voids = 2 n
Cations A present `=(1)/(6)xx2n=(n)/(3)` .
Cations B present `=(1)/(3)xxn=(n)/(3)`.
`:.` Ratio `A:B:O^(2-)=(n)/(3):(n)/(3):n=1:1:3` .
Hence, , formula is `ABC_(3)`