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Posted

This anomaly is a result of the spin exchange energy for nitrogen being twice that of oxygen.
This extra exchange energy stabilises the nitrogen more than the oxygen is stabilised so it takes additional energy to strip away the first electron (ionise) the nitrogen as compared to the trend line.
The exchange energy arises due to alternative ways of populating the outer p orbitals and the resulting ways the spins pair up.

 

Posted
5 hours ago, studiot said:

Are you still interested in this?

It was late last night and my explanation was rather short.

Yes. Please explain it more.

Posted
5 minutes ago, King E said:

Yes. Please explain it more.

OK, take a walk round the block whilst it is coming up - about an hour.

:)

Posted

First let us take a look at your question in a more general context. Looking at the first ionisation energies down and across the periodic table we can see that although the periodic repetition is evident the variation variation with a period is not smoothly progressive but is broken by some reversals such as with nitrogen/oxygen and also with phosphorus/sulphur and arsenic/selenium which are in similar positions.

Exenerg1.jpg.ba572b0bfe1ddcdfc4fdfb4433c1490f.jpg

So it is worth looking for some general principles to explain this.

Now nitrogen has 3 outer p electrons and oxygen has 4. The standing rules for placing electrons in orbitals are due to Hund and are basically those of maximum separation.
They follow the idea that the placing of two electrons in one and the same orbital has higher energy than placing them in separate ones due to electrostatic repulsion.

This is true but not the whole story. Electrons have another repository for energy - quantum spin or spin/spin interaction.
The absolute value of this energy is much smaller than the orbital energy, however its presence can modify the orbital energy by a small amount.
This amount is called the exchange energy and is a function of the number of pairs of electrons in different orbitals with the same (or parallel) spin.
So if P is the number of pairs with a given spin (up or down) and K is an energy constant we have

exchange energy  = P x K

Note you cannot have parallel spins in the same orbital but spin up and spin down count separately so in a full orbital the up and down spins can each be parallel to an electron in a different orbital.

I have worked out in detail the exchange energies for nitrogen and oxygen in the next attachment.

Exererg2.thumb.jpg.2fb62ccdeed39179831a59086fbcda09.jpg

Finally it is worth taking a look at the whole period and noting that exchange energy has zero value for some atoms. (Note The Beryllium/Boron reversal is due to the change from s to p orbitals)

Here is a neater summary table for all 6 possible p electrons.

Exenerg3.jpg.8e1887e0f57f6490c29f61cff2e04602.jpg

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