In a compound, oxide ions are arranged in ccp. One-sixth of the tetrahedral voids are occupied by cations. `(A)` and one-third of `OV_(s)` are occupieed by cations `(B)`. (a) What is the formula of the compound of the compound? (b) What are the charges on `A` and `B` ?

Explanation

a. First method
Number of `O^(2-)` ions in `c cp = 4`
Number of `TV_(s) = 8`
Number of `OV_(s) = 4`
Number of cations `A = (1)/(6) xx TV_(s)`
`= (1)/(6) xx 8 = (4)/(3)`
Number of cations `B = (1)/(3) xx OV_(s)`
`= (1)/(3) xx 4 = (4)/(3)`
Formula `A_((4)/(3)) B_((4)/(3)) O_((4)^(2-)) = ABO_(3)`
Second method
Let number of `OV^(2-)` in `c cp = 1//"atom"`
Number of `TV_(s) = 2//"atom"`.
Number of `OV_(s) = 1//"atom"`.
Number of cations `A = (1)/(6) xx TV_(s) = (1)/(6) xx 2 = (1)/(3)`
Number of cations `B = (1)/(3) xx OV_(s) = (1)/(3) xx 2 = (1)/(3) xx 1 = (1)/(3)`
Thus, formula is: `A_(1//3) B_(1//3) O_(1)^(2-) implies ABO_(3)`
b. Various possibilities of charges on `A` and `B` since charge on oxide ion is `(-2)`, lt brgt `ABO_(3) implies A^(3+) B^(3+) O_((1)^(2-))`
`A^(2+) B^(4+) O_(3)^(2-)`
`A^(4+) B^(2+) O_(3)^(2-)`
`A^(5+) B^(1+) O_(3)^(2-)`
`A^(5+) B^(1+) O_(3)^(2-)`
`A^(1+) B^(5+) O_(3)^(2-)`