Animated Solution for Chemistry - States of Matter: Copper crystallises in fcc with a unit cell length of $361\text{ pm}$. What is the radius of copper atom?

Welcome to the fascinating world of Solid State Chemistry! Today, we are going to dive deep into the microscopic architecture of metals. Have you ever wondered how atoms pack themselves together to form the solid objects we interact with every day?

In this problem, we are looking at copper, which crystallizes in a Face-Centered Cubic (FCC) lattice. Our mission is to find the radius of a single copper atom, given the edge length of its unit cell. Let's break this down step by step.

Imagine you are holding a tiny, perfect cube. This is our unit cell. In a Simple Cubic lattice, atoms only sit at the eight corners of this cube. But copper is more tightly packed. In an FCC lattice, in addition to the corner atoms, there is also an atom sitting exactly at the center of each of the six faces of the cube.

Now, here is the crucial part: Where do these atoms actually touch each other?

Because of the atom sitting right in the middle of the face, the corner atoms are pushed slightly apart. They no longer touch along the edge of the cube. Instead, the atoms touch along the face diagonal.

Geometry Meets Chemistry

Let's focus on just one face of this cube. It's a square with an edge length of $a$. If we draw a diagonal across this square face, we can use the Pythagorean theorem to find its length.

The length of the face diagonal is $\sqrt{a^2+a^2}=\sqrt{2}a$.

Now, let's look at this same diagonal from the perspective of the atoms. The diagonal starts at the center of one corner atom, passes straight through the entire face-centered atom, and ends at the center of the opposite corner atom.

Mathematically, this distance covers: - The radius of the first corner atom ($r$) - The full diameter of the face-centered atom ($2r$) - The radius of the second corner atom ($r$)

Adding these up, the total length is $r+2r+r=4r$.

Since both expressions represent the exact same physical distance, we can equate them to form our master equation:

We are given that the edge length $a$ is $361\text{ pm}$. Let's substitute this value into our master equation to solve for the atomic radius $r$.

To isolate $r$, we divide both sides by 4:

We know that the value of $\sqrt{2}$ is approximately $1.414$. Let's plug that in:

Rounding to the nearest integer given in the options, we get $127\text{ pm}$.

By simply visualizing the geometry of the unit cell, we derived the formula without having to blindly memorize it.

What if the question asked about a Body-Centered Cubic (BCC) lattice? In a BCC lattice, the atoms touch along the body diagonal that cuts through the very center of the cube. The length of a body diagonal is $\sqrt{3}a$, and it also spans $4r$. So, for BCC, the master equation becomes $\sqrt{3}a=4r$.

Always rely on your spatial intuition, and solid state chemistry will become one of your strongest subjects!