# Why is the first ionization energy of "O" lower than for "N", but the second ionization energy for "O" higher?

Aug 12, 2017

Well, I would use NIST to check these. I looked up the first two ionization energies (by typing the atomic symbol) and got...

• $\text{IE"_1 ("N") = "14.534 eV"" "" "" ""IE"_1 ("O") = "13.618 eV}$
• $\text{IE"_2 ("N") = "29.601 eV"" "" "" ""IE"_2 ("O") = "35.121 eV}$

or...

• $\text{IE"_1 ("N") ~~ "1402 kJ/mol"" "" ""IE"_1 ("O") ~~ "1314 kJ/mol}$
• $\text{IE"_2 ("N") ~~ "2856 kJ/mol"" "" ""IE"_2 ("O") ~~ "3389 kJ/mol}$

Here, you can see that the first ionization energy of oxygen atom is lower. Consider the electron configurations.

$\text{N} : \left[H e\right] 2 {s}^{2} 2 {p}^{3}$

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

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

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

Since oxygen has a paired $2 p$ electron, that one will repel the others, making it easier to remove than any of the others.

Easier to remove $\to$ lower ionization energy.

The second ionization energy of oxygen is higher than for nitrogen. In principle, they should be the same for two seemingly identical $2 p$ electrons, BUT we would neglect an important factor...

It is because oxygen atom is smaller due to a higher effective nuclear charge ${Z}_{e f f} = Z - S$, where $S$ is approximated to be the number of core electrons and $Z$ is the atomic number.

${Z}_{e f f , N} \approx {\text{7 protons" - "2 core e}}^{-} \approx 5$
${Z}_{e f f , O} \approx {\text{8 protons" - "2 core e}}^{-} \approx 6$

So, it is more difficult to remove the electron from the smaller atom, whose electrons are more tightly bound to the nucleus.

Harder to remove $\to$ higher ionization energy.