Oxygen dip in ionization energy trend

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SUMMARY

The discussion centers on the dip in ionization energy observed when moving from nitrogen to oxygen, attributed to electron-electron repulsion in doubly occupied p-orbitals. The presence of two electrons with opposing spins in the same orbital leads to increased shielding from the nucleus, making it easier to remove an electron from oxygen compared to nitrogen. This phenomenon complicates the comparison with fluorine, which has a higher ionization energy despite increased electron repulsion. The complexity of electronic configurations in multi-electron atoms is emphasized, highlighting the need for nuanced explanations beyond simple heuristics.

PREREQUISITES
  • Understanding of electron configurations in multi-electron atoms
  • Knowledge of ionization energy trends across the periodic table
  • Familiarity with concepts of electron shielding and repulsion
  • Basic grasp of spin and orbital theory
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  • Research the concept of electron shielding in multi-electron atoms
  • Study the implications of spin-orbit coupling on ionization energy
  • Explore the stability of half-filled and fully filled orbitals
  • Investigate the role of nuclear charge in ionization energy trends
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Chemistry students, educators, and researchers interested in atomic structure, ionization energy trends, and the complexities of electron interactions in multi-electron systems.

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I want a better answer than these three ones I found.
How does spin-orbit splitting also affect this or anomalous zeeman effect or decrease in spin when you move from Nitrogen to Oxygen?

Bad Teacher: The addition of the second electron into an already occupied orbital introduces repulsion between the electrons, thus it is easier to remove. that is why there is a dip in the ionization energy.

Bad Teacher: In oxygen, the last electron shares a doubly occupied p-orbital with an electron of opposing spin. The two electrons in the same orbital are closer together on average than two electrons in different orbitals, so that they shield each other from the nucleus more effectively and it is easier to remove one electron, resulting in a lower ionization energy.[2][14]

The problem is then you can't use the same explanation to explain Fluorine is higher ionization energy than Oxygen because you're also adding repulsions.

Bad Teacher:Stability of half-filled orbitals or completely filled orbitals.

This quote is using is arguing a point not building evidence for it.

I've tried studying spin and that doesn't explain it.

I tried using term symbols. Can you compare ionization energies of different atoms using that via Highest spin highest J?
 
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It appears that you are looking for a simple explanation where one doesn't exist.

Electronic configuration of multi-electron atoms is complex. While in many cases, one can come up with simple heuristics to explain what is observed, there are anomalous configurations that don't fit the pattern.

Going from N to O to F, you are not simply changing the number of electrons, but also the charge of the nucleus.
 

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