In a unit cell, atoms `(A)` are present at all corner lattices, `(B)` are present at alternate faces and all edge centres. Atoms `(C)` are present at face centres left from `(B)` and one at each body diagonal at disntance of `1//4th` of body diagonal from corner.
A tetrad axis is passed from the given unit cell and all the atoms touching the axis are removed. The possible formula of the compound left is

Options

A. `AB_(3)C_(6)` and `AB_(4)C_(5)`
B. `A_(3)B_(6)C_(7)` and `A_(3)B_(6)C_(5)`
C. `A_(4)B_(5)C_(8)` and `A_(4)B_(5)C_(7)`
D. `AB_(2)C` and `ABC_(2)`

Explanation

Case I : If tetrad axis passes through the face centres were B lies then :
No. of B atoms `=4-((1)/(2)xx2)=3`
The formula of compounds left is `AB_(3)C_(6)`
Case II : If tetrad passes through the face centres where C lies, then
No. of C atoms left `= 6 -((1)/(2)xx2)=5`
The formula of compound left is `AB_(4)C_(5)`