# Why does oxygen have a lower ionization energy than oxygen?

Jul 24, 2017

Because nitrogen does not have any paired valence $2 p$ electrons that can be removed, while oxygen does.

Coulombic repulsion energy decreases the energy input required to remove a $2 p$ valence electron from $\text{O}$ atom, i.e. decreases the ionization energy of $\text{O}$ atom compared to $\text{N}$ atom.

Write out the electron configurations.

$\text{N}$: $1 {s}^{2} 2 {s}^{2} \textcolor{red}{2 {p}^{3}}$

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ "ul(uarr color(white)(darr))" } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$\underbrace{\text{ "" "" "" "" "" "" "" }}$
$\text{ "" "" } 2 p$

$\underline{\uparrow \downarrow}$
$2 s$

$\text{O}$: $1 {s}^{2} 2 {s}^{2} \textcolor{red}{2 {p}^{4}}$

$\underline{\uparrow \textcolor{red}{\downarrow}} \text{ "ul(uarr color(white)(darr))" } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$\underbrace{\text{ "" "" "" "" "" "" "" }}$
$\text{ "" "" } 2 p$

$\underline{\uparrow \downarrow}$
$2 s$

You can see that oxygen has a paired $2 p$ valence electron. That adds electron repulsion, which makes it easier for the electron to get removed.

Thus, oxygen atom has a lower (less positive) ionization energy than nitrogen atom.