Animated Solution for Chemistry - States of Matter: An element crystallises in a face-centred cubic (fcc) unit cell with cell edge $a$. The distance between the centres of two nearest octahedral voids in the crystal lattice is

Imagine you are shrinking down to the atomic level and stepping inside a Face-Centered Cubic (FCC) unit cell. It is a beautifully symmetric structure, but the atoms don't take up all the space. There are hidden pockets of empty space called voids.

In an FCC lattice, the octahedral voids are located in two distinct types of positions. First, there is one massive void sitting right at the body center of the cube. Second, there are voids located at the exact center of each of the 12 edges of the cube.

Our mission is to find the shortest distance between any two of these octahedral voids.

The Cross-Section

To make this 3D problem easier to visualize, let's take a 2D slice. Imagine taking a sharp knife and slicing the unit cell perfectly in half, parallel to one of its faces.

This cross-section plane passes directly through the body center and cuts through four of the edge centers. What we get is a perfect square with a side length equal to the edge length of the unit cell, $a$.

In this square slice, the body center octahedral void is exactly at the center of the square. The four edge center octahedral voids are located exactly at the four corners of this square.

Now, the problem is reduced to finding the distance from the center of a square to one of its corners. Let's call this distance $x$.

We can construct a simple right-angled triangle to solve this. We draw a horizontal line from the center to the right edge, and a vertical line from that point up to the top-right corner.

Because the center is exactly halfway across the square, the base of our triangle is $\frac{a}{2}$. Similarly, the height of our triangle is also $\frac{a}{2}$. The hypotenuse of this triangle is our unknown distance, $x$.

Now, we bring in our trusty mathematical tool: the Pythagorean theorem.

The square of the hypotenuse is equal to the sum of the squares of the base and the height. We can write this as:

Let's expand those squares. Squaring both the numerator and the denominator gives us:

Adding these two identical fractions together, we get:

Which beautifully simplifies to:

Finally, to find $x$, we take the square root of both sides.

And there we have it! The minimum distance between two nearest octahedral voids in an FCC lattice is exactly $\frac{a}{\sqrt{2}}$.

I know 3D geometry can sometimes feel intimidating, but by taking a smart 2D cross-section, we turned a complex spatial problem into a simple triangle. Always look for the symmetries!