Animated Solution for Chemistry - States of Matter: The edge length of a face centered cubic cell of an ionic substance is $508\text{ pm}$. If the radius of the cation is $110\text{ pm}$, the radius of the anion is

In the fascinating world of solid-state chemistry, visualizing the invisible is half the battle won. When we talk about an ionic substance crystallizing in a Face-Centered Cubic (FCC) lattice, we are usually referring to the classic rock salt (NaCl) structure.

Let's embark on a journey to decode the geometry of this microscopic world and solve our problem step-by-step.

Imagine you are shrinking down to the size of an atom and standing right on the edge of an FCC unit cell. What do you see? At the corners of this cubic room, you will find the bulky anions. But the edge isn't empty! Right in the middle of the edge, nestled comfortably in what we call an octahedral void, sits a smaller cation.

Because these ions are packed as closely as possible, they physically touch each other along this very edge. This physical contact is the key to unlocking the mathematical relationship between their sizes and the size of the unit cell.

Let's translate our visual observation into a pristine mathematical equation. The total length of the edge, which we call $a$, is simply the sum of the segments that make it up.

Starting from one corner, we traverse the radius of the first anion ($r^{-}$), then we cross the entire diameter of the cation in the middle ($2r^{+}$), and finally, we traverse the radius of the second anion at the other corner ($r^{-}$).

Mathematically, this gives us:

Which elegantly simplifies to:

This is our master equation for an FCC ionic lattice of the NaCl type.

Now, let's bring in the numbers provided in the problem. We are given the edge length $a=508\text{ pm}$ and the cation radius $r^{+}=110\text{ pm}$.

Substituting these into our master equation, we get:

I know algebra can sometimes be tedious, but let's take a breath and solve this systematically. First, we divide both sides by 2 to isolate the terms inside the parenthesis:

Finally, to find the radius of the anion, we subtract the cation's radius from 254:

And there we have it! The radius of the anion is exactly $144\text{ pm}$. By simply visualizing the edge of the unit cell, a seemingly complex solid-state problem reduces to basic arithmetic. Always remember to draw or imagine the lattice first!