Animated Solution for Chemistry - States of Matter: CsCl crystallises in body centered cubic lattice. If '$a$' its edge length, then which of the following expressions is correct?

Imagine you are shrinking down to the atomic level and stepping inside a crystal of Cesium Chloride (CsCl). What do you see? You would find yourself standing inside a perfect cube. At every single corner of this cube, there is a Chloride ion ($C{l}^{-}$).

But the real star of the show is right in the middle. Sitting perfectly at the body center of this cubic room is a large Cesium ion ($C{s}^{+}$). This specific arrangement, where one type of ion is at the corners and another is at the body center, is often referred to as a Body-Centered Cubic (BCC) type lattice.

To understand the math behind this structure, we need to look at the geometry of the cube itself. Let's say the length of one edge of this cubic unit cell is $a$.

In a cube, the longest straight line you can draw inside it is the body diagonal. This line goes from one bottom corner, straight through the very center of the cube, all the way to the opposite top corner. Using the Pythagorean theorem in 3D space, the length of this body diagonal is mathematically proven to be $\sqrt{3}a$.

Where Math Meets Physics

Here is the crucial physical constraint: in this crystal lattice, the ions are packed as tightly as possible. Because the central $C{s}^{+}$ ion is quite large, it pushes the corner $C{l}^{-}$ ions slightly apart. Therefore, the ions do not touch along the edges of the cube.

Instead, they touch along that long body diagonal we just talked about. If you trace the path along the body diagonal, you start at the surface of a corner $C{l}^{-}$ ion, travel through its radius ($r_{C{l}^{-}}$), then you enter the central $C{s}^{+}$ ion, traveling through its entire diameter ($2r_{C{s}^{+}}$), and finally, you travel through the radius of the opposite corner $C{l}^{-}$ ion ($r_{C{l}^{-}}$).

Now, we just equate our geometric knowledge with our physical observation.

The total length of the body diagonal based on the touching ions is:

We also know the geometric length of the body diagonal is $\sqrt{3}a$. Setting these two equal gives us our master equation:

To find the sum of the radii, we simply divide both sides by 2:

$$r_{C{s}^{+}}+r_{C{l}^{-}}=\frac{\sqrt{3}}{2}a$$

This elegant equation perfectly bridges the gap between the macroscopic edge length of the crystal and the microscopic radii of its constituent ions.