# What is the difference between the first and second ionization energy of an element?

Jan 8, 2017

First ionization energy is for the electron removed from the neutral element, and second ionization energy is for the removal of the electron from the first ion generated due to the first ionization.

See below for clarification.

The first and second ionization energies are defined as

${\text{M" -> "M}}^{+} + {e}^{-}$, $I {E}_{1}$,
${\text{M"^(+) -> "M}}^{2 +} + {e}^{-}$, $I {E}_{2}$,

so that $I {E}_{1}$ is the energy input required to remove the first electron from $\text{M}$ and $I {E}_{2}$ is the energy input required to remove an electron from ${\text{M}}^{+}$. Thus, generally $I {E}_{i}$ is positive.

When removing an electron easily stabilizes the atom, $I {E}_{i}$ is generally small, because little energy input is needed to encourage ionization. That's why $I {E}_{1}$ for $\text{Li}$ is positive but smaller than for $\text{Be}$, as $\text{Be}$ is tougher to singly ionize.

Recall electron affinity, $E {A}_{i}$, is the opposite operation, i.e. the energy change due to adding electron $i$ (could be positive or negative). Note that because of this, for example:

• $I {E}_{1}$ for $\text{M}$ is the opposite sign to $E {A}_{1}$ for ${\text{M}}^{+}$.
• $I {E}_{2}$ for $\text{M}$ (the first ionization of ${\text{M}}^{+}$) is the opposite sign to $E {A}_{1}$ for ${\text{M}}^{2 +}$.

When adding an electron stabilizes the atom, $E {A}_{1} < 0$, generally, because energy is released upon adding the electron. That's why $E {A}_{i}$ of the noble gases is $\ge 0$.