An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centers of the faces of the cube .The empirical formula for this compound would be

Options

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

Explanation

To determine the empirical formula of the ionic compound with A ions at the corners of a cube and B ions at the centers of the faces of the cube, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Position of Ions in the Unit Cell**:
- A ions are located at the corners of the cube.
- B ions are located at the centers of the faces of the cube.

2. **Calculate the Contribution of A Ions**:
- Each corner of the cube contributes \( \frac{1}{8} \) of an A ion to the unit cell.
- There are 8 corners in a cube.
- Therefore, the total contribution of A ions is:
\[ \text{Total A ions} = 8 \times \frac{1}{8} = 1 \]

3. **Calculate the Contribution of B Ions**:
- Each face center contributes \( \frac{1}{2} \) of a B ion to the unit cell.
- There are 6 faces in a cube.
- Therefore, the total contribution of B ions is:
\[ \text{Total B ions} = 6 \times \frac{1}{2} = 3 \]

4. **Determine the Empirical Formula**:
- From the calculations, we have:
- A ions = 1
- B ions = 3
- Thus, the empirical formula can be written as:
\[ \text{Empirical Formula} = A_1B_3 \text{ or simply } AB_3 \]

### Final Answer:
The empirical formula for the compound is \( AB_3 \).

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