CsBr has bcc lattice with the edge length `9.4` Å, the shortest inter- ionic distance in between `Cs^(+)` & `Br^(-)` is:

Options

A. `8.14` Å
B. `7.14` Å
C. `7.50` Å
D. `4.07`Å

Answer

To find the shortest inter-ionic distance between Cs⁺ and Br⁻ in a bcc lattice of CsBr with an edge length of 9.4 Å, we can follow these steps:

### Step 1: Understand the BCC Structure
In a body-centered cubic (bcc) lattice, there are atoms at each corner of the cube and one atom at the center of the cube. The shortest distance between the ions will be along the body diagonal of the cube.

### Step 2: Calculate the Body Diagonal
The formula for the body diagonal (d) of a cube with edge length (a) is given by:
\[ d = \sqrt{3} \times a \]

### Step 3: Substitute the Edge Length
Given that the edge length (a) is 9.4 Å, we can substitute this value into the formula:
\[ d = \sqrt{3} \times 9.4 \]

### Step 4: Calculate the Body Diagonal
Now, calculate the body diagonal:
\[ d = \sqrt{3} \times 9.4 \approx 1.732 \times 9.4 \approx 16.26 \, \text{Å} \]

### Step 5: Find the Shortest Inter-Ionic Distance
The shortest inter-ionic distance (r) between Cs⁺ and Br⁻ is half of the body diagonal:
\[ r = \frac{d}{2} = \frac{16.26}{2} \approx 8.13 \, \text{Å} \]

### Final Answer
Thus, the shortest inter-ionic distance between Cs⁺ and Br⁻ in CsBr is approximately:
\[ \boxed{8.13 \, \text{Å}} \]

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