In a face centered cubic lattice atoms A are at the corner points and atoms B at the face centered points. If atoms B is missing from one of the face centered points, the formula of the ionic compound is

Options

A. `AB_(2)`
B. `A_(5)B_(2)`
C. `A_(2)B_(3)`
D. `A_(2)B_(5)`

Answer

To determine the formula of the ionic compound in a face-centered cubic (FCC) lattice where atoms A are at the corners and atoms B are at the face-centered points, and one atom B is missing, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Contribution of Corner Atoms (A):**
- In a face-centered cubic lattice, there are 8 corner atoms.
- Each corner atom contributes \( \frac{1}{8} \) to the unit cell.
- Therefore, the total contribution from the corner atoms (A) is:
\[ \text{Total A} = 8 \times \frac{1}{8} = 1 \]

2. **Identify the Contribution of Face-Centered Atoms (B):**
- In a face-centered cubic lattice, there are 6 face-centered atoms.
- Each face-centered atom contributes \( \frac{1}{2} \) to the unit cell because it is shared between two unit cells.
- Since one of the face-centered atoms is missing, we have 5 face-centered atoms contributing:
\[ \text{Total B} = 5 \times \frac{1}{2} = \frac{5}{2} \]

3. **Combine the Contributions:**
- Now we have:
- Total A = 1
- Total B = \( \frac{5}{2} \)

4. **Determine the Empirical Formula:**
- To express the formula in the simplest whole number ratio, we can multiply both the number of A and B by 2:
\[ \text{A} = 1 \times 2 = 2 \]
\[ \text{B} = \frac{5}{2} \times 2 = 5 \]
- Thus, the empirical formula of the ionic compound is:
\[ \text{A}_2\text{B}_5 \]

### Final Answer:
The formula of the ionic compound is \( \text{A}_2\text{B}_5 \).