CsCl has bec arrangement and its unit cell edge length is 400 pm. Calculate the inter-ionic distance in CsCl.

Explanation

To calculate the inter-ionic distance in CsCl, we need to follow these steps:

### Step 1: Understand the Structure
CsCl crystallizes in a body-centered cubic (BCC) arrangement. In this structure, Cl⁻ ions are located at the corners of the cube, while the Cs⁺ ion is located at the body center.

### Step 2: Identify the Body Diagonal
In a cubic unit cell, the body diagonal can be calculated using the formula:
\[ \text{Body Diagonal} = \sqrt{3} \times a \]
where \(a\) is the edge length of the cube.

### Step 3: Calculate the Body Diagonal
Given that the edge length \(a\) of CsCl is 400 pm, we can calculate the body diagonal:
\[ \text{Body Diagonal} = \sqrt{3} \times 400 \, \text{pm} \]
Calculating this gives:
\[ \text{Body Diagonal} = \sqrt{3} \times 400 \approx 692.8 \, \text{pm} \]

### Step 4: Relate Body Diagonal to Ionic Radii
The body diagonal in the BCC structure consists of one Cs⁺ ion and two Cl⁻ ions. Therefore, we can express the body diagonal in terms of the ionic radii:
\[ \text{Body Diagonal} = R_{\text{Cs}^+} + 2R_{\text{Cl}^-} \]
where \(R_{\text{Cs}^+}\) is the radius of the Cs⁺ ion and \(R_{\text{Cl}^-}\) is the radius of the Cl⁻ ion.

### Step 5: Solve for the Inter-Ionic Distance
Since we want to find the inter-ionic distance (which is the distance between Cs⁺ and Cl⁻ ions), we can rearrange the equation:
\[ R_{\text{Cs}^+} + R_{\text{Cl}^-} = \frac{\text{Body Diagonal}}{2} \]
Substituting the value of the body diagonal:
\[ R_{\text{Cs}^+} + R_{\text{Cl}^-} = \frac{692.8 \, \text{pm}}{2} = 346.4 \, \text{pm} \]

### Conclusion
The inter-ionic distance in CsCl is:
\[ \text{Inter-Ionic Distance} = R_{\text{Cs}^+} + R_{\text{Cl}^-} = 346.4 \, \text{pm} \]