In nitrogen: $ce 2s^2 2p^3$ In oxygen: $ce 2s^2 2p^4$

This speak me that it need to be easier to remove an electron native oxygen than it is because that nitrogen as the electron in oxygen is slightly further away indigenous the nucleus an interpretation lesser nuclear charge.

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




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Oxygen has a lower an initial ionization power as the electron that is removed is coming from a paired orbital.
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Electrons within the same orbital suffer maximum repulsion together the circulation of their wavefunctions is the same, for this reason the probability thickness distribution is the same and also the electrons have the right to be thought of as occupying the very same space. This maximizes your repulsion and also increases the potential power of the electrons in that orbital, make the electrons easier to remove. This is despite the increased efficient nuclear charge experienced by the electron in oxygen and also the decreased radius the the orbital.

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


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You see from the electronic configurations:

nitrogen: $ce 2s^2 2p^3$ oxygen: $ce 2s^2 2p^4$

In reality, the first ionisation energy of nitrogen is higher than the first ionisation energy of oxygen due to the fact that nitrogen, in a stable fifty percent filled orbital state, is comparatively more stable than oxygen. Oxygen, top top the various other hand, would have tendency to shed an electron quickly to accomplish it"s more stable half filled orbital state.

Also, as a rule, half to fill and totally filled orbital claims are more stable as compared to other configurations due to the fact that they attribute to maximum exchange energies.


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