In a cubic,`A` atoms are present on alternative corners, `B` atoms are present on alternate faces, and `C` atoms are present on alternalte edges and body centred of the cube. The simplest formula of the compound is

Options

A. `A_(2)BC_(4)`
B. `AB_(2)C_(4)`
C. `ABC_(4)`
D. `ABC_(2)`

Answer

B

Explanation

Number of `A` atoms at alternate corners
`= 4` corner `xx (1)/(8)` per corner share `= (1)/(2)`
Number of `B` atoms at alternate face
`= 2` faces `xx (1)/(2)` per face center share `= 1`
Number of `C` atoms at alternate edges and at body centre
`=` (`4` edges `xx (1)/(4)` per edge centre share)
`+ 1` atom at body centre
`= 1 + 1 = 2`
Thus, formula `A_((1)/(2))B_(1)C_(2)`
Simplifying, `AB_(2)C_(4)`.