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"Electrons are negatively charged, therefore, two electrons repel each other. When electrons are paired in a 3p orbital there is electron pair repulsion which causes the energy of the electrons to increase. Increased energy makes the electron easier to remove and therefore the ionisation energy decreases."

I found this statement when studying ionisation energy. Is it correct to say that as electron energy increases, ionisation energy (EI) decreases? I understand the factors affecting EI (atomic radius, nuclear charge, shielding and electron pair repulsion) but have not seen this directly compared to electron energy before. Do each of the factors increase/decrease electron energy or is it irrelevant to think about it in that way?

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    $\begingroup$ It is very analogous to gravity. If the satellite mechanical energy ( PE + KE ) increases, is the energy needed to leave Earth gravity smaller ? Note that energy of bound electrons is referred to a free electron in rest as zero reference value, so electron energy + electron ionization energy = 0. $\endgroup$
    – Poutnik
    Dec 22, 2020 at 9:09
  • $\begingroup$ The ionisation energy is fixed, so increasing the energy of an orbital means that less energy is now needed to remove an electron from that orbital. $\endgroup$
    – porphyrin
    Mar 19 at 9:13

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First of all, the energy of any electron is decided by the energy level in which it stays. So if electron is very close to nucleus that means it is in high energy level which then corresponds to it's high energy and if it's quite away or far from nucleus then it's energy is low. So question says if energy of electron is high , then it clearly understood its close to nucleus in high energy level hence energy required to remove it, that is Ionisation energy is also .

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  • $\begingroup$ So you say electrons in the 6s orbital have lower energy than those in the 1s orbital, and the orbitals are filled from those with the highest energy. Does not it sound suspicious to you ? $\endgroup$
    – Poutnik
    Dec 22, 2020 at 22:02
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Kinda.

An atom or ion has multiple configurations that can be (very roughly) described as electrons (arrows directed up or down) put into boxes * (representing individual orbitals each capable to accomodate one up-electron and one down-electron). This description ignores many subtleties and configurations that can be tranformed into each other by repositioning 'boxes' belonging to one shell should be considered one state, but as a very rough description it goes.

Assuming this description, you can attribute several ionization energies to particular state, depending on which one electron you removes. Those different energies may be sometimes directly observed in proper experiments.

In this case, yes, removing an electron from doubly occupied cell takes less energy than from single-occupied cell of the same subshell, assuming all other things equal. The last part is important, because there are other effects affecting relative energy of states.

This, however, doesn't mean it is easier in common sense. In quantum physics processes are characterized by their probability and event of particular type, say a particular channel of ionization, may be be more or less probably than event of another type, say a different channel of ionization.

For this reason using word 'easier' is a bad idea. Use more precise words whenever possible.

* Like in this question Electron orbital diagram of vanadium

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