Animated Solution for Chemistry - States of Matter: Sodium metal crystallises in a body centred cubic lattice with a unit cell edge of $4.29\text{A˚}$. The radius of sodium atom is approximately

When we dive into the microscopic world of solid-state chemistry, visualizing the arrangement of atoms is half the battle won. In this problem, we are dealing with sodium metal, which crystallizes in a Body-Centred Cubic (BCC) lattice.

Imagine a simple cube. Now, place a sodium atom at every single one of the eight corners. But we are not done yet! To make it 'body-centred', we place one more atom right in the dead center of the cube's interior. The most critical physical constraint in a BCC lattice is that the atoms are packed tightly such that they touch each other along the body diagonal of the cube. They do not touch along the edges because the central atom pushes the corner atoms slightly apart.

To find the relationship between the radius of the atom ($r$) and the edge length of the unit cell ($a$), we need to calculate the length of this body diagonal. Let's use the Pythagorean theorem.

First, consider the bottom face of the cube. It's a square with side $a$. The diagonal across this face (the face diagonal) is given by:

Now, look at the right triangle formed by this face diagonal, a vertical edge of the cube ($a$), and the body diagonal. Applying the Pythagorean theorem again:

Now, let's look at the atoms lying on this body diagonal. We have one corner atom at the start, the central atom in the middle, and another corner atom at the end.

The length of the body diagonal is made up of: - The radius of the first corner atom ($r$) - The full diameter of the central atom ($2r$) - The radius of the opposite corner atom ($r$)

Adding these up, the total length is $r+2r+r=4r$. Since this must equal the geometric length we just calculated, we arrive at our master equation for a BCC lattice:

$$4r=\sqrt{3}a$$

The problem provides us with the edge length $a=4.29\text{A˚}$. We need to find the radius $r$. Let's rearrange our master equation:

Substituting the given value of $a$:

We know that $\sqrt{3}\approx 1.732$. Let's plug that in:

Rounding off to two decimal places, we get $1.86\text{A˚}$. This perfectly matches option (a). A beautiful and straightforward application of solid-state geometry!