# 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+ and B– ions?

**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 ^{+} and B^{–} ions? **

**Answer:**

**Given Values**

Edge length (a) = 404 pm

Type of unit cell = CsCl type = BCC

Note: In CsCl structure, chloride ions are present at corner of the cube while Cesium ion present at body center of the cube.

As we know the relation between edge length (a) and distance (d) between two nearest ions for BCC is :

\( d=\frac{\sqrt{3}.a}{2} \)

**Calculations**

**Calculate the distance (d) of closest approach:**

\( d=\frac{\sqrt{3}.a}{2} = \frac{1.732 \times 404~pm}{2} \) = 349.86 pm

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