`A_(2)B` has antifluorite structure (B forms FCC lattice and A occupies tetrahedral voids).If all ions along any one body diagonal are removed, then new formula of compound will be :

Options

A. `A_(4)B_(5)`
B. `A_(8)B_(5)`
C. `A_(7)B_(6)`
D. `A_(8)B_(4)`

Answer

To solve the problem, we need to analyze the antifluorite structure of the compound \( A_2B \) and determine the new formula after removing ions along one body diagonal.

### Step-by-Step Solution:

1. **Understanding the Antifluorite Structure**:
- In the antifluorite structure, \( B \) ions form a face-centered cubic (FCC) lattice, and \( A \) ions occupy the tetrahedral voids.
- In an FCC lattice, there are 8 atoms at the corners and 6 atoms at the face centers.

2. **Calculating the Number of \( B \) Ions**:
- The total number of \( B \) ions in the unit cell can be calculated as:
\[ \text{Number of } B = \left(8 \times \frac{1}{8}\right) + \left(6 \times \frac{1}{2}\right) = 1 + 3 = 4 \]
- Therefore, there are 4 \( B \) ions in the unit cell.

3. **Calculating the Number of Tetrahedral Voids**:
- The number of tetrahedral voids in a structure is twice the number of atoms in the FCC lattice:
\[ \text{Number of tetrahedral voids} = 2 \times \text{Number of } B = 2 \times 4 = 8 \]
- Thus, there are 8 \( A \) ions in the unit cell.

4. **Removing Ions Along One Body Diagonal**:
- When we remove all ions along one body diagonal, we remove 2 \( B \) ions (located at the corners) and 2 \( A \) ions (located in the tetrahedral voids).
- After removal, the remaining \( B \) ions:
\[ \text{Remaining } B = 4 - 2 = 2 \]
- The remaining \( A \) ions:
\[ \text{Remaining } A = 8 - 2 = 6 \]

5. **Finding the New Ratio of \( A \) to \( B \)**:
- The new ratio of \( A \) to \( B \) is:
\[ \frac{A}{B} = \frac{6}{2} = 3 \]
- Thus, the new formula can be expressed as \( A_3B_1 \) or simply \( A_3B \).

6. **Final Formula**:
- Therefore, the new formula of the compound after removing the ions along one body diagonal is \( A_3B \).

### Conclusion:
The new formula of the compound after removing ions along one body diagonal is \( A_3B \).