If atoms are removed from half of the edge-centred `OV_(s)` in `RbBr`, then the molecular formula of unit cell is

Options

A. `Rb_(2)Br_(2)`
B. `Rb_(1.5)Br_(3)`
C. `Rb_(2.5)Br_(4)`
D. `Rb_(4)Br_(2.5)`

Explanation

To solve the problem, we need to analyze the structure of RbBr and determine how the removal of atoms from the edge-centered octahedral voids affects the molecular formula of the unit cell.

### Step-by-Step Solution:

1. **Understanding the Structure of RbBr**:
- RbBr consists of Rb⁺ and Br⁻ ions. In the crystal lattice, Br⁻ ions form a face-centered cubic (FCC) lattice.
- In an FCC lattice, the number of Br⁻ ions can be calculated as follows:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) (total contribution from corners = \( 8 \times \frac{1}{8} = 1 \)).
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) (total contribution from faces = \( 6 \times \frac{1}{2} = 3 \)).
- Therefore, total Br⁻ ions = \( 1 + 3 = 4 \).

2. **Calculating Rb⁺ Ions**:
- Rb⁺ ions occupy the octahedral voids in the structure.
- There are two types of octahedral voids: edge-centered and body-centered.
- In the unit cell, there are 12 edge-centered octahedral voids, each contributing \( \frac{1}{4} \) (total contribution from edge centers = \( 12 \times \frac{1}{4} = 3 \)).
- There is also 1 body-centered octahedral void contributing 1.
- Therefore, total Rb⁺ ions = \( 3 + 1 = 4 \).

3. **Initial Molecular Formula**:
- Initially, the molecular formula of the unit cell is \( \text{Rb}_4\text{Br}_4 \).
- This can be simplified to \( \text{RbBr} \).

4. **Effect of Removing Atoms**:
- The problem states that atoms are removed from half of the edge-centered octahedral voids. This means we will remove Rb⁺ ions from half of the 12 edge-centered voids.
- Half of 12 is 6, so we remove Rb⁺ from 6 edge-centered voids.
- After removal, the remaining edge-centered voids contribute \( 6 \times \frac{1}{4} = 1.5 \) Rb⁺ ions.

5. **Final Count of Ions**:
- The number of Br⁻ ions remains unchanged at 4.
- The number of Rb⁺ ions after removal is \( 1.5 \) from the edge-centered voids plus \( 1 \) from the body-centered void, giving a total of \( 1.5 + 1 = 2.5 \).

6. **Final Molecular Formula**:
- The final molecular formula of the unit cell after the removal of Rb⁺ ions is \( \text{Rb}_{2.5}\text{Br}_4 \).

### Conclusion:
The molecular formula of the unit cell after the removal of atoms from half of the edge-centered octahedral voids in RbBr is \( \text{Rb}_{2.5}\text{Br}_4 \).