# Rank the following elements by increasing atomic radius: carbon, aluminum, oxygen, potassium.

Jan 6, 2015

The correct order with respect to increasing atomic radius is

$\text{O" < "C" < "Al" < "K}$

An important thing to notice about these four elements is the fact that only two of them, $\text{C}$ and $\text{O}$, are in the same row (the same period) of the periodic table; the other two, $\text{Al}$ and $\text{K}$, don't share a period neither with $\text{C}$ and $O$, nor with each other, as $\text{Al}$ is in the third period and $\text{K}$ is in the fourth.

This aspect will allow you to easily rank them with respect to increasing atomic radius, since both of the periodic trends in atomic size are on display here.

Let's look at $\text{C}$ and $\text{O}$ first. We know that atomic radius decreases from left to right within a period due to the increase in effective nuclear charge.

As you move to the right within a period, the number of protons elements have increases simultaneously with the number of electrons; however, electrons are being added to the same energy level across a period, which means that the increasing number of protons allows the nucleus to exert more pull on these electrons, which in turn makes the atomic radius smaller.

Since $\text{C}$ comes before $\text{O}$ in the second period, $\text{O}$ will have a smaller atomic radius than $\text{C}$.

As mentioned before, $\text{Al}$ and $\text{K}$ don't share a period. This is where the other periodic trends comes into play. As you move down a colomn (a group) of the periodic table, atomic radius increases.

This happens because electrons are now being added to higher energy levels, further away from the nucleus, which weakens the nucleus' pull. Another reason for the increase in atomic radius is the electron shielding effect, which states that the electrons in the higher energy levels are being shielded by those closer to the nucleus, further reducing the nucleus' pull on them.

This means that $\text{Al}$'s atomic radius will be smaller than that of $\text{K}$, but bigger than that of $\text{C}$ and $\text{O}$.

$\text{O" < "C" < "Al" < K}$