In spinel structure, `O^(2-)` ions are cubic-close packed, where 1/8th of the tetrahedral holes are occupied by `A^(2+)` cations and 1/2 of the octahedral holes are occupied by cations `B^(3+)`. The general formula of this compound is

Options

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

Answer

B

Explanation

Let the number of ions be N.
Then number of tetrahedral voids is 2N.
The number of octahedral voids is N.
Then `A^(2+)` occupies one-eighth of tetrahedral voids which means 1/8 `xx` 2N = (1/4)N
`B^(3+)` occupies half of octahedral voids which means (1/2) `xx` N = N/2.
Therefore, the formula becomes `1//4N:1//2N:N=AB_(2)O_(4)`.
Then number of tetrahedral voids is 2N.
The number of octahedral voids is N.
Then `A^(2+)` occupies one-eighth of tetrahedral voids which means 1/8 `xx` 2N = (1/4)N
`B^(3+)` occupies half of octahedral voids which means (1/2) `xx` N = N/2.
Therefore, the formula becomes `1//4N:1//2N:N=AB_(2)O_(4)`.