`CsBr` has `bcc` structure with edge length of `43 pm`. The shortest interionic distance between cation and anion is

Options

A. `37.2`
B. `18.6`
C. `74.4`
D. `43`

Explanation

To find the shortest interionic distance between the cation (Cs⁺) and the anion (Br⁻) in cesium bromide (CsBr) which has a body-centered cubic (BCC) structure, we can follow these steps:

### Step 1: Understand the structure
In a BCC structure:
- The anions (Br⁻) are located at the corners of the cube.
- The cation (Cs⁺) is located at the body center of the cube.

### Step 2: Identify the edge length
The edge length (a) of the cubic unit cell is given as:
- \( a = 43 \, \text{pm} \)

### Step 3: Determine the formula for interionic distance
In a BCC structure, the shortest distance between the cation and anion can be calculated using the formula:
- \( \text{Shortest interionic distance} = \frac{\sqrt{3}}{2} a \)

### Step 4: Substitute the edge length into the formula
Now, we substitute the value of \( a \):
- \( \text{Shortest interionic distance} = \frac{\sqrt{3}}{2} \times 43 \, \text{pm} \)

### Step 5: Calculate the value
Calculating this gives:
- \( \text{Shortest interionic distance} = \frac{1.732}{2} \times 43 \)
- \( = 0.866 \times 43 \)
- \( \approx 37.2 \, \text{pm} \)

### Conclusion
Thus, the shortest interionic distance between the cation and anion in CsBr is approximately:
- **37.2 pm**

### Final Answer
The answer is option A: **37.2 pm**.

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