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I am having trouble wrapping my head around electron affinities. And the textbook explanations aren't very helpful.

So, the textbook says that the 1st electron affinity is generally exothermic. The reason given is that the electron added is strongly attracted by the effective nuclear charge. In contrast, the successive electron affinities are always endothermic. This is due to the additional energy needed to overcome electrostatic repulsion.

This leaves me wondering:

  1. Why is so much energy released when an electron is added to a neutral atom? I understand that the more energy released, the more stable the product is, but why does it become more stable? And why is this seemingly independent of electronic configuration?

  2. Why isn't the attraction between the electron and the nucleus a factor in the subsequent electron affinities? In the 1st electron affinity, the atom is neutral, and yet there is still attraction between the electron and the nucleus. In successive electron affinities, the ion is negatively charged, but won't there still be attraction?

  3. Isn't there also repulsion to overcome in the 1st electron affinity? Sure, the atom is neutral, but won't the added electron be repelled by the valence shell electrons or something?

Thanks.

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    $\begingroup$ 1. Forget atoms, they are complicated. Any neutral thing is attracted to any charge. $\endgroup$ Jan 30, 2023 at 15:51
  • $\begingroup$ Why is so much energy released? It is not so much. It is less than atom ionisation energies. $\endgroup$
    – Poutnik
    Jan 31, 2023 at 13:23
  • $\begingroup$ @Ivan Neretin Really? What about a beam of neutrons in an electric field? It doesn't deflect because its charge is neutral. $\endgroup$
    – John Smith
    Feb 1, 2023 at 5:18
  • $\begingroup$ I said "attracted to charge", not "affected by field". The latter is not true; if the field is uniform, then of course a neutral particle would not be affected by it. On the other hand, the former is true, because the field of a point charge is non-uniform, and the polarizability creates some net force. Sure enough, it is much smaller than the attraction between two opposite charges. $\endgroup$ Feb 1, 2023 at 7:40
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    $\begingroup$ It makes no sense to say that ions are more stable (or less stable) than neutral atoms. See, atoms can't turn into ions anyway. It is true, though, that an anion is more stable than a system of a neutral atom and a free electron far away. $\endgroup$ Feb 1, 2023 at 14:49

1 Answer 1

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Electron affinities are generally smaller than ionization energies.
$\pu{IE_{\ce{Na}} = 5.14 eV}$ but $\pu{EA_{\ce{Cl}} = 3.61 eV}$

If the electron clouds had been fully spherical symmetric and fully between the external electron and the nucleus, the effective nuclear charge would be for that electron zero.

But as individual orbitals, except $\mathrm{s}$ ones, are not spherically symmetric and as part of atomic electron probability density is behind the captured electron, the effective atom charge is slightly positive, so the electron can be bound. That is true for most of elements, except noble gases and some elements with full or half full orbital groups.


When an atom becomes an anion, it becomes a non-point-like negative charge. The same sign charges mutually repulse each other sooner than partial attraction due asymmetry or penetration of electron density decreases little the repulsion.


Yes, there is repulsion, but weaker than not fully screened off nucleus attraction, so there is still net attraction. Additionally, electrons partially repulse each other toward the nucleus.

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  • $\begingroup$ Interesting, from the table, that the inert gases and Be, N, Mg and Mn, Zn, Cd, Yb, Hg, Pu and perhaps a few more, have negative heats of electron affinity. Those atoms won't hold onto a free electron! They would rather be neutral than negative. $\endgroup$ May 17, 2023 at 14:06
  • $\begingroup$ @JamesGaidis If your review their positions in the periodic table, their electronic configuration involve full or half occupation of orbital groups, making from them kind of semi-noble gases in the electron affinity aspect. $\endgroup$
    – Poutnik
    May 17, 2023 at 14:11

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