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Is there a limit of electrons a single hydrogen atom can have? If so what is it? why? Is the the answer to why scalable to helium?

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  • $\begingroup$ Sounds like an exam question. $\endgroup$ – Pieter Oct 21 '19 at 21:23
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    $\begingroup$ Hahaha I’d like to see the exam.... actually its for game development!!! $\endgroup$ – hoboBob Oct 22 '19 at 12:34
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By definition, "hydrogen atom" refers to the neutral system with one proton and one electron, so it cannot hold any extra electrons.

However, protons can hold more than one electron, in which case the system is termed a hydrogen anion. This is a stable, bound system, and the reaction $$ \mathrm{H}+e^- \to \mathrm{H}^- \tag 1 $$ releases about $0.75\:\rm eV$, an energy known as the electron affinity of the hydrogen atom. (As a fun fact, the hydrogen anion is incredibly important ─ the reaction $(1)$ above is the reason why the Sun's spectrum is continuous.)

Free atoms of most elements tend to have positive electron affinities, which means that their singly-charged negative anions are stable systems, and they release energy when they capture their first extra electron. There's a few exceptions, though, starting with helium: atoms which have stable closed shells can 'reject' that extra electron, as it's forbidden from sitting in the closed valence shells and it's forced to sit at higher-energy shells that are too far uphill in energy to be stable.

If you want to up the game and go to a second extra electron, though, to get to $\rm H^{2-}$, the game runs out, and indeed it runs out for every element ─ all the second electron affinities are negative. That is, it takes work to cram a second extra electron in, and the resulting dianion will at best be in a metastable state that's ready and jumping to give that energy back out by dissociating into the single anion and a free electron. It's just too hard to try and hold two extra electrons (and their resulting mutual electrostatic repulsion) within the confines of an atomic system.

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  • $\begingroup$ Would the analogy of constantly trying to push two magnets together work for the dianion keeping its extra electron. Thank you for the answer $\endgroup$ – hoboBob Oct 21 '19 at 16:52
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    $\begingroup$ @hoboBob It's not a great analogy, but it's not too terrible. $\endgroup$ – Emilio Pisanty Oct 21 '19 at 16:55
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Your question is about the Hydrogen ion, when it gains electrons. Normally, the Hydrogen ion (we usually call the single proton without electron the Hydrogen ion) when it gains an electron, will have a negative charge.

enter image description here

Now these negative ions (with two or more extra electrons) are unstable. You are basically asking if you can bind a proton with more then two electrons.

Though, you can try to use an external magnetic field to keep it stable.

https://link.springer.com/article/10.1007/s00601-009-0018-7

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  • $\begingroup$ This answer is dead wrong ─ there is basically nothing right about it other than the platitudes in the first paragraph. Please do your due diligence and research before answering. $\endgroup$ – Emilio Pisanty Oct 21 '19 at 14:35
  • $\begingroup$ @EmilioPisanty what exactly is wrong about it? I thought he was asking whether you can bind a proton with more than one electron. $\endgroup$ – Árpád Szendrei Oct 21 '19 at 14:53
  • $\begingroup$ Precisely. You claim that it is not possible to do this in a stable manner, i.e. that the hydrogen anion (one proton, two electrons) is unstable. That's incorrect. Please do your research before continuing this conversation; I would recommend Wikipedia (together with the rest of the links in my answer) as a starting point. $\endgroup$ – Emilio Pisanty Oct 21 '19 at 15:18
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    $\begingroup$ Two extra electrons (i.e. one proton, three electrons) is indeed unstable. Your answer goes beyond that and claims that one extra electron (i.e. one proton, two electrons) is also unstable. What is it about the Wikipedia page of the hydrogen anion that gives you even the remotest impression that it's unstable? $\endgroup$ – Emilio Pisanty Oct 21 '19 at 15:46
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    $\begingroup$ If you don't understand the material, ask separately instead of posting misinformation. The comment thread of a disputed answer is nowhere close to the right place to be asking that. $\endgroup$ – Emilio Pisanty Oct 21 '19 at 21:03

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