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Why does the first energy of ionization of oxygen is lesser than of nitrogen?

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Blueberries116:
The following question arises from a question I found in my book.

The problem states the following:

Experimentally it has been determined that the value of the first energy of ionization of oxygen is lesser than the first energy of ionization of nitrogen. Select the statement which explains this:

The alternatives given were:

[itex]\begin{array}{ll}
1.&\textrm{The higher electronegativity of oxygen}\\
2.&\textrm{The lesser electron affinity of nitrogen}\\
3.&\textrm{The lesser radius of oxygen}\\
4.&\textrm{The higher stability of nitrogen}\\
5.&\textrm{The higher electroaffinity of nitrogen}\\
\end{array}[/itex]

According to my book the answer is the fourth option, but I couldn't find a way to justify this. What sort of explanation can do this?.

My guess is that nitrogen forms a triple bond with itself which is higher than the oxygen and this makes the molecule more stable but I don't know if this can be translated into a higher energy of ionization nitrogen rather than the oxygen which does have a sigma and pi bonds. Am I right?. Can someone illustrate this or bring some sort of diagram with some values to justify this?. It would help a lot if an answer could help me why should I discard the second option?. Doesn't nitrogen has lesser electron affinity than the oxygen?.

Wouldn't it mean that having a lesser electron affinity cause to be easier to strip out an electron from its outher shell rather than the opposite as the question indicates?. Can someone really help me with this?.

Mr. Deeds:
Edit*

The e-config of Oxygen is paired as opposed to Nitrogen. This pairing allows for less energy required to ionize gaseous Oxygen atoms. I would have this makes Nitrogen more stable than Oxygen, but this is a special case. Look at Hund's Rule for more information about this special case.

Enthalpy:
Let's evacuate first the electron affinity. It characterises the energy that binds an additional electron. The ionization energy characterises the loss of an electron. These quantities differ, and are not clearly linked.

Do you know molecular orbitals? N2 has all valence electrons on bonding orbitals, deep, hence is very stable and non-reactive, while O2 has two more electrons that must go on anti-bonding orbitals, shallow.
https://en.wikipedia.org/wiki/Molecular_orbital_diagram#Dinitrogen

mjc123:
As the question was stated, I assume it refers to atomic O and N, not the molecules, especially as the OP is also asking other questions about atomic electron configurations.

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