Animated Solution for Chemistry - States of Matter: The number of octahedral voids per lattice site in a lattice is ......... . (Rounded off to the nearest integer)

The beauty of solid-state chemistry lies in our ability to visualize the invisible. When we talk about crystal lattices, we are essentially discussing the architectural blueprints of matter. In this problem, we are tasked with finding the ratio of octahedral voids to lattice sites. To do this, we don't just count the literal atoms we see in a single box; we must think about how these atoms and spaces are shared across an infinite, repeating 3D grid.

The FCC Lattice

To make our visualization concrete, let's assume our crystal forms a Face-Centered Cubic (FCC) lattice. This is one of the most common and efficient ways nature packs spherical atoms together.

In an FCC unit cell, the lattice sites—the actual positions occupied by atoms—are located at the $8$ corners of the cube and at the centers of its $6$ faces. But here is the catch: a unit cell is not an isolated island. It is tightly packed with neighbors on all sides.

The Concept of Effective Lattice Sites

Because the unit cell shares its boundaries, the atoms sitting on those boundaries are also shared.

An atom at a corner is the meeting point of $8$ different cubes. Therefore, only $\frac{1}{8}$ of that atom actually belongs to our specific unit cell. Since there are $8$ corners, their total contribution is $8\times \frac{1}{8}=1$ atom.

Similarly, an atom sitting on a face is shared equally between $2$ adjacent cubes. Its contribution is $\frac{1}{2}$. With $6$ faces, the total contribution is $6\times \frac{1}{2}=3$ atoms.

Adding these together, we find the total number of effective lattice sites (often denoted as $Z$) in an FCC unit cell:

Locating Octahedral Voids

Now, let's hunt for the empty spaces. An octahedral void is a specific type of empty space surrounded by exactly six atoms, forming the shape of an octahedron.

In an FCC unit cell, these voids are located in two distinct types of positions: 1. The Body Center: There is exactly one void perfectly hidden in the dead center of the cube. 2. The Edge Centers: There is a void located at the exact midpoint of every single edge of the cube. Since a cube has $12$ edges, there are $12$ such voids.

The Effective Number

Just like the atoms, we must calculate the effective number of these voids.

The void at the body center is completely enclosed within our unit cell. It is not shared with anyone. So, its contribution is a full $1$.

However, the voids on the edges are a different story. Any edge of a cube is shared by $4$ adjacent cubes in a 3D lattice. Therefore, a void sitting on an edge is split four ways, contributing only $\frac{1}{4}$ to our unit cell. With $12$ edges, their total contribution is $12\times \frac{1}{4}=3$ voids.

Adding these together, we find the total number of effective octahedral voids:

The Ratio

The question asks for the number of octahedral voids per lattice site. This is simply the ratio of the effective number of voids to the effective number of lattice sites.

For every single lattice site in the crystal, there is exactly one octahedral void.

A Powerful Secret: This elegant $1:1$ ratio is not just a quirk of the FCC lattice. It is a universal truth for any close-packed structure, including the Hexagonal Close-Packed (HCP) lattice. If a lattice has $Z$ effective atoms, it will always have exactly $Z$ effective octahedral voids.

Tetrahedral Voids

While we are exploring the architecture of crystals, it is worth mentioning the other major type of empty space: the tetrahedral void. These are smaller spaces surrounded by four atoms.

In an FCC lattice, there are $8$ tetrahedral voids, all located entirely within the body of the unit cell (two on each body diagonal). Since they are not shared, the effective number of tetrahedral voids is $8$.

Notice the pattern? The number of tetrahedral voids is exactly double the number of effective lattice sites ($2Z$). So, if the question had asked for tetrahedral voids per lattice site, the answer would be $2$.

By mastering the visualization of the unit cell and understanding the concept of "effective" counting, these seemingly complex solid-state problems become beautifully simple.