CsCl crystallises in body centred cubic lattice. If 'a' is its edge length then which of the following expressions is correct?

We are told that cesium chloride, $$CsCl$$, crystallises in a body-centred cubic lattice. In such a lattice the chloride ions $$Cl^-$$ occupy the eight corners of the cube, while the cesium ion $$Cs^+$$ sits exactly at the body centre. Because of this arrangement the cation and anions are in contact along the body diagonal of the cube.

Let the edge length of the cube be denoted by $$a$$. For any cube, a basic result from geometry states that the length of its body diagonal equals the square root of three times the edge length. Mathematically, the formula is

$$\text{body diagonal} = \sqrt{3}\,a.$$

Now we examine how many ionic radii are strung along that same body diagonal. Starting from one corner $$Cl^-$$ ion, we move to the central $$Cs^+$$ ion and then continue on to the opposite corner $$Cl^-$$ ion. The chloride at a corner touches the cesium at the body centre, so the distance between their centres is simply the sum of their radii:

$$\text{distance (corner to centre)} = r_{Cl^-} + r_{Cs^+}.$$

In exactly the same way, the cesium ion at the centre touches the chloride ion at the opposite corner, giving an identical distance:

$$\text{distance (centre to opposite corner)} = r_{Cs^+} + r_{Cl^-}.$$

Hence the full body diagonal is obtained by adding those two identical segments:

$$\text{body diagonal} = (r_{Cl^-} + r_{Cs^+}) + (r_{Cs^+} + r_{Cl^-}).$$

Simplifying the right-hand side we get

$$\text{body diagonal} = 2\,(r_{Cs^+} + r_{Cl^-}).$$

But from the geometric formula quoted earlier, the same body diagonal is also $$\sqrt{3}\,a$$. Therefore we can equate the two expressions:

$$\sqrt{3}\,a = 2\,(r_{Cs^+} + r_{Cl^-}).$$

Now we solve for the required sum of the ionic radii. Dividing both sides by $$2$$ gives

$$r_{Cs^+} + r_{Cl^-} = \frac{\sqrt{3}}{2}\,a.$$

This matches exactly the expression presented in Option C.

Hence, the correct answer is Option C.