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

To solve the problem step by step, we will analyze the arrangement of atoms in the unit cell and calculate the number of each type of atom present after removing those that touch the tetrad axis.

### Step 1: Determine the number of A atoms
- Atoms A are located at all corners of the unit cell.
- There are 8 corners in a cube, and each corner atom contributes \( \frac{1}{8} \) to the unit cell.

\[ \text{Number of A atoms} = 8 \times \frac{1}{8} = 1 \]

### Step 2: Determine the number of B atoms
- Atoms B are present at alternate faces and at all edge centers.
- There are 6 faces in a cube, and since B is present at alternate faces, there are 3 face-centered B atoms contributing \( \frac{1}{2} \) each.
- Additionally, there are 12 edges in a cube, and each edge center atom contributes \( \frac{1}{4} \).

\[ \text{Number of B atoms} = 3 \times \frac{1}{2} + 12 \times \frac{1}{4} = 1.5 + 3 = 4 \]

### Step 3: Determine the number of C atoms
- Atoms C are present at face centers left from B and at each body diagonal at a distance of \( \frac{1}{4} \) of the body diagonal from the corner.
- There are 6 face centers (1 for each face) contributing \( \frac{1}{2} \) each.
- For the body diagonals, there are 4 body diagonals in a cube, and each contributes 1 atom.

\[ \text{Number of C atoms} = 6 \times \frac{1}{2} + 4 \times 1 = 3 + 4 = 7 \]

### Step 4: Adjust for the tetrad axis
- A tetrad axis is passed through the unit cell, and we need to remove the atoms that touch this axis.
- The B atoms are affected by the tetrad axis. Since there are 4 B atoms, we remove \( \frac{1}{2} \times 2 = 1 \) B atom.

\[ \text{Remaining B atoms} = 4 - 1 = 3 \]

- The C atoms are also affected. We remove \( \frac{1}{2} \times 2 = 1 \) C atom.

\[ \text{Remaining C atoms} = 7 - 1 = 6 \]

### Step 5: Write the final formula
- After removing the atoms that touch the tetrad axis, we have:
- A: 1
- B: 3
- C: 6

Thus, the formula of the compound left is:

\[ \text{Final formula} = A_1B_3C_6 \]