# Why is there an exception in the ionization energy trend in the second-row p-block elements?

Jun 8, 2017

I think you are wondering why nitrogen has an ionization energy that locally peaks in the ionization energy chart:

(search for the atom one left of $\text{O}$, nitrogen, which is atomic number $7$.)

Nitrogen's atomic electron configuration is:

$\left[H e\right] 2 {s}^{2} 2 {p}^{3}$,

or more explicitly:

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

$\underline{\uparrow \downarrow}$

$2 s$

All the electrons are unpaired, so that's not the exception here.

The exception arises with $\text{O}$ atom, which has a lower first ionization energy even though it is more electronegative, is expected to hold onto its valence electrons more tightly, and is also smaller in radius.

This is due to the atomic electron configuration for $\text{O}$:

$\left[H e\right] 2 {s}^{2} 2 {p}^{4}$

or more explicitly:

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

$\underline{\uparrow \downarrow}$

$2 s$

One of the $2 p$ valence electrons is paired, and that one experiences repulsions with the other $2 p$ electrons, since it is in the same orbital as another electron.

That is enough to lower the first ionization energy of oxygen atom to be below that of nitrogen atom by about $\text{86 kJ/mol}$.