Animated Solution for Chemistry - States of Matter: An ionic compound has a unit cell consisting of $A$ ions at the corners of a cube and $B$ ions on the centres of the faces of the cube. The empirical formula for this compound would be

Imagine you are standing inside a perfectly cubic room. This room represents our unit cell, the fundamental building block of a crystal lattice. The problem introduces two types of ions: $A$ and $B$.

To understand the structure, we must place these ions exactly where the problem dictates. The $A$ ions are positioned at the eight corners of our cubic room. Meanwhile, the $B$ ions are situated right at the center of each of the six walls (faces) of the room. Visualizing this arrangement is the crucial first step before we dive into the mathematics of atomic contributions.

Sharing Atoms

Here is where many students make a mistake: an ion sitting on the boundary of a unit cell does not belong entirely to that single cell. It is shared with its neighbors!

Let's look at the $A$ ions at the corners. Think of a corner as a plot of land where eight different neighborhoods intersect. Because exactly eight cubes meet at any single corner in a 3D lattice, the atom sitting there is divided equally among them. Therefore, the contribution of a single corner atom to our specific unit cell is exactly $\frac{1}{8}$.

Now, let's analyze the $B$ ions on the faces. A face is like a common wall shared between two adjacent rooms. Any atom sitting on that face is split right down the middle. Thus, the contribution of a face-centered atom to our unit cell is exactly $\frac{1}{2}$.

Now that we have our contribution rules, the math becomes incredibly straightforward.

For the $A$ ions, a cube has $8$ corners. Since each corner contributes $\frac{1}{8}$ of an ion, we calculate the effective number of $A$ ions as: So, effectively, there is only one complete $A$ ion in the entire unit cell.

For the $B$ ions, a standard cube has $6$ faces. Since each face contributes $\frac{1}{2}$ of an ion, we calculate the effective number of $B$ ions as: This means there are effectively three complete $B$ ions in our unit cell.

We have all the pieces of the puzzle! The effective number of $A$ ions is $1$, and the effective number of $B$ ions is $3$.

The empirical formula represents the simplest whole-number ratio of the atoms in the compound. Since our ratio of $A$ to $B$ is simply $1:3$, we can write the final empirical formula as:

$$A{B}_3$$

This perfectly matches option (b). Always remember to carefully read the problem to check if any atoms are missing from their standard positions, as that is a favorite twist examiners use to test your conceptual clarity!