CsBr has bcc like structures with edge length `4.3Å`. The shortest inter ionic distance in between `Cs^(+)` and `Br^(-)` is:

Options

A. (a) 3.72
B. (b) 1.86
C. (c) 7.44
D. (d) 4.3

Answer

To find the shortest interionic distance between Cs⁺ and Br⁻ in cesium bromide (CsBr) with a body-centered cubic (BCC) structure and an edge length of 4.3 Å, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**:
- CsBr has a BCC structure. In this structure, Cs⁺ ions are located at the body center of the cube, while Br⁻ ions are located at the corners of the cube.

2. **Identify the Edge Length**:
- The edge length (a) of the BCC unit cell is given as 4.3 Å.

3. **Calculate the Body Diagonal**:
- The body diagonal (d) of a cube can be calculated using the formula:
\[ d = \sqrt{3} \times a \]
- Substituting the value of a:
\[ d = \sqrt{3} \times 4.3 \, \text{Å} \approx 7.44 \, \text{Å} \]

4. **Relate Body Diagonal to Ionic Radii**:
- In a BCC structure, the body diagonal is equal to the sum of the radii of the ions involved. For Cs⁺ and Br⁻, the relationship can be expressed as:
\[ d = R_{Cs^+} + 2R_{Br^-} \]
- Here, \(R_{Cs^+}\) is the radius of the Cs⁺ ion and \(R_{Br^-}\) is the radius of the Br⁻ ion.

5. **Solve for the Ionic Radii**:
- Rearranging the equation gives:
\[ R_{Cs^+} + 2R_{Br^-} = 7.44 \, \text{Å} \]
- We need to find the shortest interionic distance between Cs⁺ and Br⁻, which is simply \(R_{Cs^+} + R_{Br^-}\).

6. **Express the Interionic Distance**:
- The shortest interionic distance (d_interionic) can be expressed as:
\[ d_{interionic} = R_{Cs^+} + R_{Br^-} \]
- From the earlier equation, we can express \(R_{Br^-}\) in terms of \(R_{Cs^+}\):
\[ R_{Br^-} = \frac{7.44 - R_{Cs^+}}{2} \]
- Substituting this back into the equation for \(d_{interionic}\):
\[ d_{interionic} = R_{Cs^+} + \frac{7.44 - R_{Cs^+}}{2} \]
- Simplifying this gives:
\[ d_{interionic} = \frac{2R_{Cs^+} + 7.44 - R_{Cs^+}}{2} = \frac{R_{Cs^+} + 7.44}{2} \]

7. **Final Calculation**:
- Assuming typical ionic radii (for example, \(R_{Cs^+} \approx 1.69 \, \text{Å}\) and \(R_{Br^-} \approx 1.96 \, \text{Å}\)), we can calculate:
\[ d_{interionic} = 1.69 + 1.96 = 3.65 \, \text{Å} \]
- However, since we are looking for the shortest interionic distance, we can directly use the body diagonal calculation to find:
\[ d_{Cs^+ \text{ to } Br^-} = 3.72 \, \text{Å} \]

### Final Answer:
The shortest interionic distance between Cs⁺ and Br⁻ in CsBr is approximately **3.72 Å**.