# Question #2b845

Apr 25, 2017

Ionization energy is the energy required to remove one electron from an atom in its gaseous state.

In general, ionization energy increases across a period on the periodic table since atomic radius decreases across the period. The electrons are closer to the nucleus, and it becomes harder to remove an electron. Ionization energy decreases down a group since atomic radius increases down the group and there are additional shells that shield the attractional force of the nucleus on the valence electrons.

The trend across a period has two main exceptions, between group 2 and 3 and between group 5 and 6.

The ionization energy for group 3 is lower than the ionization energy for the element in group 2 and in the same period. This is because, for group 3, the outermost electron is in the p-orbital, while the outermost electron is group 2 is in the s-orbital.

For example, aluminum in group 3 has the electron configuration $\left[\text{Ne}\right] 3 {s}^{2} 3 {p}^{1}$ while magnesium in group 2 has the electron configuration $\left[\text{Ne}\right] 3 {s}^{2}$. The p-orbital electron is farther away from the nucleus than the s-orbital in the same shell as it has a higher energy level. Therefore, less energy is required to remove the p-orbital electron. This is why the ionization energy for group 3 is lower than the ionization energy for the element in the same period in group 2.

The ionization energy for group 6 is lower than the ionization energy for the element in the same period in group 5. This is due to the spin of electrons when they fill a shell (from the Hund's Rule and the Pauli's exclusion principle). In group 5, the spins inside the p-orbital are $| \uparrow | \uparrow | \uparrow |$. In group 6, the spins are $| \uparrow \downarrow | \uparrow | \uparrow |$. In group 6, there are two electrons that are closer together. These two electrons repel each other. It is easier to remove one of these electrons. That is why the ionization energy for group 6 is slightly lower.