CsBr has BCC structure. The length of its side is `4.3 Å`. The minimum distance between `Cs^(+)` and `Br^(-)` ion will be

Options

A. `0.897 Å`
B. `3.72 Å`
C. `1.749 Å`
D. None of the above

Explanation

To find the minimum distance between \( Cs^+ \) and \( Br^- \) ions in a body-centered cubic (BCC) structure of cesium bromide (CsBr), we can follow these steps:

### Step 1: Understand the BCC Structure
In a BCC structure, there are atoms located at the corners of the cube and one atom at the center of the cube. For CsBr:
- The \( Cs^+ \) ion is located at the body center of the cube.
- The \( Br^- \) ions are located at the corners of the cube.

### Step 2: Identify the Unit Cell Length
The length of the side of the BCC unit cell is given as \( a = 4.3 \, \text{Å} \).

### Step 3: Calculate the Body Diagonal
In a BCC unit cell, the minimum distance between the \( Cs^+ \) ion (at the body center) and a \( Br^- \) ion (at one of the corners) can be found using the body diagonal of the cube. The body diagonal \( d \) can be calculated using the formula:
\[ d = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} \]

### Step 4: Calculate the Length of the Body Diagonal
Substituting the value of \( a \):
\[ d = 4.3 \, \text{Å} \times \sqrt{3} \approx 4.3 \, \text{Å} \times 1.732 \approx 7.45 \, \text{Å} \]

### Step 5: Find the Minimum Distance
The minimum distance between the \( Cs^+ \) and \( Br^- \) ions is half of the body diagonal because the distance from the center of the \( Cs^+ \) ion to the corner of the cube where the \( Br^- \) ion is located is half the body diagonal:
\[ \text{Minimum Distance} = \frac{d}{2} = \frac{7.45 \, \text{Å}}{2} \approx 3.725 \, \text{Å} \]

### Step 6: Conclusion
Thus, the minimum distance between \( Cs^+ \) and \( Br^- \) ions in CsBr is approximately:
\[ \text{Minimum Distance} \approx 3.72 \, \text{Å} \]