A solid `AB` has `CsCl`-type structure. The edge length of the unit cell is `404` pm. Calculate the distance of closest approach between `A^(o+)` and `B^(Θ)` ions.

Explanation

To solve the problem of finding the distance of closest approach between \( A^{+} \) and \( B^{-} \) ions in a solid \( AB \) with a CsCl-type structure, we can follow these steps:

### Step 1: Understand the Structure
In a CsCl-type structure:
- The \( B^{-} \) ions (let's assume these are \( Cl^{-} \) ions) are located at the corners of the cube.
- The \( A^{+} \) ions (let's assume these are \( Cs^{+} \) ions) are located at the body center of the cube.

### Step 2: Identify the Unit Cell Edge Length
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