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.
Formula of given solid is

Options

A. `A_(3)B_(8)C_(7)`
B. `AB_(4)C_(6)`
C. `A_(6)B_(4)C_(8)`
D. `A_(2)B_(9)C_(11)`

Explanation

To determine the formula of the solid based on the given arrangement of atoms in the unit cell, we will follow these steps:

### Step 1: Count the number of atoms of A
Atoms A are located at all corners of the unit cell. In a cubic unit cell, there are 8 corners. Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell.

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

### Step 2: Count the number of atoms of B
Atoms B are located at alternate face centers and at all edge centers.

- **Face Centers**: There are 6 faces in a cube, and B occupies 2 alternate faces. Each face center contributes \( \frac{1}{2} \) of an atom.

\[ \text{Contribution from face centers} = 2 \times \frac{1}{2} = 1 \]

- **Edge Centers**: There are 12 edges in a cube, and each edge center contributes \( \frac{1}{4} \) of an atom.

\[ \text{Contribution from edge centers} = 12 \times \frac{1}{4} = 3 \]

Adding these contributions together gives:

\[ \text{Total number of atoms of B} = 1 + 3 = 4 \]

### Step 3: Count the number of atoms of C
Atoms C are located at the face centers left from B and at the body diagonals.

- **Face Centers**: Since B occupies 2 face centers, there are 4 remaining faces. Each of these contributes \( \frac{1}{2} \) of an atom.

\[ \text{Contribution from remaining face centers} = 4 \times \frac{1}{2} = 2 \]

- **Body Diagonals**: There are 4 body diagonals in a cube, and each body diagonal has one atom at \( \frac{1}{4} \) of the distance from the corner.

\[ \text{Contribution from body diagonals} = 4 \times 1 = 4 \]

Adding these contributions together gives:

\[ \text{Total number of atoms of C} = 2 + 4 = 6 \]

### Step 4: Write the empirical formula
Now that we have the counts for each atom type:

- Number of A = 1
- Number of B = 4
- Number of C = 6

The empirical formula of the solid can be written as:

\[ \text{Formula} = A_1B_4C_6 \quad \text{or simply} \quad ABC_4C_6 \]

### Final Answer
The formula of the given solid is \( A_1B_4C_6 \).

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