Animated Solution for Chemistry - States of Matter: In a face-centred cubic lattice, atom $A$ occupies the corner positions and atom $B$ occupies the face centred positions. If one atom of $B$ is missing from one of the face centred points, the formula of the compound is

Imagine you are a microscopic detective, and your crime scene is a perfect Face-Centered Cubic (FCC) crystal lattice. Everything is usually perfectly ordered, but today, there is a twist—an atom has gone missing! Our job is to figure out the new identity, or the empirical formula, of this altered crystal.

In a standard FCC lattice, we have two prime locations for atoms to reside: the corners of the cube and the centers of each face. The problem tells us that Atom $A$ occupies the corner positions, while Atom $B$ occupies the face-centered positions.

If this were a perfect, undisturbed crystal, we would have $8$ Atom $A$'s at the corners and $6$ Atom $B$'s on the faces. But the universe is rarely perfect. We are told that exactly one Atom $B$ is missing from its face-centered spot. Let's break down how this changes the math.

Atom A

Let's start with Atom $A$. There are $8$ corners in a cube, and an Atom $A$ sits at every single one of them. However, in the world of crystal lattices, atoms are shared. A corner atom doesn't belong exclusively to one unit cell; it acts as a junction point for $8$ different unit cells meeting at that corner.

Because it is shared equally among $8$ cells, its contribution to our specific unit cell is only $\frac{1}{8}$.

Therefore, the effective number of $A$ atoms in our unit cell is:

So, we have exactly $1$ effective Atom $A$.

Atom B

Now, let's look at Atom $B$. Normally, a cube has $6$ faces, meaning we should have $6$ Atom $B$'s. But remember our missing atom? Because one is gone, we only have $5$ Atom $B$'s left on the faces.

Just like corner atoms, face-centered atoms are also shared. An atom sitting flat on the face of a cube is shared exactly in half by the unit cell right next to it. Thus, each face-centered atom contributes $\frac{1}{2}$ to our unit cell.

With $5$ atoms remaining, the effective number of $B$ atoms is:

The Empirical Formula

We now have the effective number of both atoms in our unit cell. Atom $A$ contributes $1$, and Atom $B$ contributes $\frac{5}{2}$.

This gives us a raw ratio of $A_1{B}_{5/2}$.

However, chemical formulas must be expressed in the simplest whole-number ratios. You can't have half an atom in an empirical formula! To fix this, we simply multiply the entire ratio by $2$ to clear the denominator.

And there we have it! By carefully tracking the contributions of each lattice point and accounting for the missing atom, we've successfully deduced that the formula of the compound is $A_2{B}_5$.