Animated Solution for Chemistry - Solid State: Element 'B' forms ccp structure and 'A' occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is

Deriving the Formula of a Bimetallic Oxide

Imagine you are an architect, but instead of bricks and mortar, you are building with atoms. Solid state chemistry is exactly that—a beautiful, highly ordered game of 3D Tetris. In this problem, we are tasked with finding the empirical formula of a bimetallic oxide by analyzing where each type of atom sits within the crystal lattice.

Let's break down the architecture step by step.

The Foundation

The foundation of our crystal is element B, which forms a cubic close-packed (ccp) structure. A ccp lattice is geometrically identical to a face-centered cubic (fcc) lattice.

To find the effective number of B atoms in one unit cell, we must count their contributions. The atoms at the 8 corners of the cube are shared by 8 adjacent unit cells, contributing $8\times \frac{1}{8}=1$ atom. The atoms at the 6 face centers are shared by 2 unit cells, contributing $6\times \frac{1}{2}=3$ atoms.

Adding these together, the effective number of B atoms per unit cell is:

Let's call this foundational number $N$. So, $N=4$.

In any close-packed structure, the empty spaces between the atoms are called voids. There is a golden rule in solid state chemistry that connects the number of lattice atoms ($N$) to the number of voids:

1. The number of Octahedral Voids is exactly equal to $N$. 2. The number of Tetrahedral Voids is exactly equal to $2N$.

Since our lattice has $N=4$ effective atoms, it must contain exactly 4 octahedral voids and 8 tetrahedral voids.

Filling the Voids

Now, let's look at element A. The problem states that A occupies exactly half of the octahedral voids.

Since there are 4 octahedral voids in total, the number of A atoms is:

Next, we place the Oxygen (O) atoms. The problem tells us that oxygen atoms occupy all the tetrahedral voids.

Since there are 8 tetrahedral voids in total, the number of oxygen atoms is:

The Empirical Formula

We now have the exact count of each atom in a single unit cell: - Element A: 2 atoms - Element B: 4 atoms - Oxygen O: 8 atoms

The ratio of atoms is $A:B:O=2:4:8$.

However, a chemical formula must always be expressed in the simplest whole-number ratio. By dividing the entire ratio by their greatest common divisor (which is 2), we get:

Therefore, the simplest empirical formula for this bimetallic oxide is $A{B}_2{O}_4$.

This elegant, symmetrical result perfectly matches option (b). By mastering the relationship between lattice atoms and voids, you can decode the structure of any complex crystal!