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Why is the first ionization energy in oxygen slightly more than nitrogen?

  • In nitrogen: $\ce{[He] 2s^2 2p^3}$
  • In oxygen: $\ce{[He] 2s^2 2p^4}$

This tells me that it should be easier to remove an electron from oxygen than it is for nitrogen as the electron in oxygen is slightly further away from the nucleus meaning lesser nuclear charge.

But why is it harder to remove an electron from oxygen, i.e. why is the first ionization energy of oxygen higher?

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You see from the electronic configurations:

  • nitrogen: $\ce{[He] 2s^2 2p^3}$
  • oxygen: $\ce{[He] 2s^2 2p^4}$

In reality, the first ionisation energy of nitrogen is greater than the first ionisation energy of oxygen because nitrogen, in a stable half filled orbital state, is comparatively more stable than oxygen. Oxygen, on the other hand, would tend to lose an electron easily to achieve it's more stable half filled orbital state.

Also, as a rule, half filled and fully filled orbital states are more stable as compared to other configurations because they attribute to maximum exchange energies.

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    $\begingroup$ Are you sure "Oxygen, ..., would tend to lose an electron easily"? I am asking because $\ce{O+}$ is really uncommon... $\endgroup$ – Gaurang Tandon Feb 23 '18 at 12:17
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    $\begingroup$ p4 has the same exchange energy as p3. It is due to repulsion. $\endgroup$ – Michael F. Mar 29 '19 at 16:15
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    $\begingroup$ (-1) Despite being accepted, this answer is wrong. Exchange energy is only a small consideration (if it is one at all), the correct reason is repulsion between paired electrons. $\endgroup$ – orthocresol Jan 17 at 2:02
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Oxygen has a lower first ionization energy as the electron that is removed is coming from a paired orbital. enter image description here

Electrons within the same orbital experience maximum repulsion as the distribution of their wavefunctions is the same, so the probability density distribution is the same and the electrons can be thought of as occupying the same space. This maximizes their repulsion and increases the potential energy of the electrons in that orbital, making the electrons easier to remove. This is despite the increased effective nuclear charge experienced by the electron in oxygen and the decreased radius of the orbital.

See: "Physical Chemistry", Atkins, P.W. Section 13.4, p.p.370 (4th edition) - sorry, I have an old one!

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    $\begingroup$ This is the correct answer! $\endgroup$ – orthocresol Jan 17 at 2:07

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