A body centered cubic solid is made up of two elements A and B. Atoms of 'A' occupy two corners of the cube. Remaining positions of the unit cell are occupied by the atoms of 'B'. The formula of the compound in

Options

A. `A_(6)B_(7)`
B. `A_(7)B_(4)`
C. `AB_(7)`
D. `A_(7)B`

Answer

To find the formula of the compound formed by elements A and B in a body-centered cubic (BCC) solid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of BCC**:
In a body-centered cubic unit cell, there are 8 corners and 1 atom at the center of the cube.

2. **Determine the Contribution of Atom A**:
According to the question, atoms of element A occupy 2 corners of the cube. Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell.
- Contribution from 2 corners:
\[ \text{Contribution of A} = 2 \times \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \]

3. **Determine the Contribution of Atom B**:
The remaining positions in the unit cell are occupied by atoms of element B. In a BCC structure, there is 1 atom at the body center.
- Contribution from the body center:
\[ \text{Contribution of B} = 1 \]
- Total contribution of B in the unit cell:
Since there are 8 corners and 1 body center, the total contribution of B can be calculated as:
\[ \text{Total contribution of B} = 1 + 0 \text{ (from corners, as they are occupied by A)} = 1 \]

4. **Calculate the Total Contribution of B**:
Since the corners are occupied by A, we only consider the body center for B. Therefore, the total contribution of B remains:
\[ \text{Contribution of B} = 1 \]

5. **Express the Contributions in Simplified Form**:
Now we have:
- Contribution of A = \( \frac{1}{4} \)
- Contribution of B = \( 1 \)

To express these contributions in a simple whole number ratio, we can multiply both by 4 (the least common multiple):
\[ A: 1 \quad \text{and} \quad B: 4 \]

6. **Write the Formula**:
The formula of the compound can be expressed as:
\[ \text{Formula} = A_1B_4 \text{ or simply } AB_4 \]

### Final Answer:
The formula of the compound is \( AB_4 \).

---