# Why are double negative ions (Hydrogen) unstable?

This is not a duplicate. I have read these questions:

Can hydrogen have 3 electrons?

where Buzz says:

The double hydrogen anion H−− does not exist as a stable species. (It occurs as a resonance, with a lifetime of 23 ns though.) If you tried to add another electron to a H− ion, the repulsion from the two core 1s electrons would be greater than the attraction due to the nucleus. The result is that there is no bound 2s energy level. In fact, there is not even a bound 1s12s1 state of H−. Even with only two electrons total, there is not a bound 2s orbital.

And Jon Custer says:

I’m not sure there are any doubly negative ions that is stable.

Is there a limit of electrons a single hydrogen atom can have?

Where Emilio Pisanty says:

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.

Now when you learn about electron configuration it is not simple. Naively you think that when you combine a single proton and a single electron, they will be stable. And they are, because you think, ok, one single electron and a single proton have same but opposite EM charge, so they equal out, and you have a stable system.

Then you learn, that the Hydrogen negative ion with an extra electron is stable. Ok, so you have to accept that it is not so simple, and it is all QM.

Even two electron's double negative EM charge can equal out a single proton's EM charge and create a stable system.

Then you learn that the doubly negative Hydrogen ion is not stable. So simply the second extra electron adds to much of a negative EM charge and the proton cannot counterbalance it.

Or, three electrons cannot create such an electron configuration as per QM so that their negative EM charge would equal out the single proton's charge.

So basically the question is, where is that point, and why is it specifically between two and three electrons, that the electron configuration becomes unable to create a stable bond with a single proton?

This link I have found states they have found stable double negative Hydrogen ions (I don't have access to the full one).

This is a contradiction, which one is correct, are they stable or not?

Question:

1. What is so special about the second extra electron, and why cannot three electrons create such a configuration to counterbalance the single proton's charge and create a stable system?
• Comments are not for extended discussion; this conversation has been moved to chat. – David Z Oct 25 '19 at 10:41

I'm posting this addendum to my main answer separately, to address the last part of your question, which I find utterly unconstructive and pretty close to an abuse of this site.

This link I have found states they have found stable double negative Hydrogen ions (I don't have access to the full one).

This is a contradiction, which one is correct, are they stable or not?

No, this is not a contradiction. The only way to describe it is as an utter lack of due diligence on your part. Frankly, the fact that this has not been edited out of the question is pretty mind-boggling.

I find it extremely implausible that you do not know how to get ahold of a copy of this news item, and that you're unaware with any of the methods, say in these Academia.SE threads. Once you do get a copy, the news item is dead clear:

Also, from other data, Anbar and Schnitzer determined the half-life of the H$$^{2-}$$ ion to be 23 ± 4 nanoseconds

This is not a stable system. Moreover, even in the publicly-available abstract there is no hint or indication that the system is stable (i.e. the abstract is neutral about the system's stability), so your claim that it reports a stable H$$^{2-}$$ ion is unjustified, and dead wrong.

In any case, in the interest of assuming good faith, I'm going to assume that you didn't follow up on this not because you're lazy, but because you didn't know how to. Given a paywalled news item like this one, how does one assess what the report is about?

Well, the abstract explicitly mentions the names of the researchers involved (Dr. Michael Anbar and Dr. Rafael Schnitzer), and it's a news item, so it's extremely unlikely to be the only publication relating to the experiment. (If it were, then you can basically write it off as unreliable.) That means that there must be a paper, including both of those researchers and from around the same year, describing the experiment. How do you look for one? You go to an academic search engine and you look for papers with both of those authors, say, as in this search.

This will get you about twelve links in this case, most of them explicitly mentioning H$$^{2-}$$ in the title. These are generally paywalled, but two of the relevant ones explicitly mention the system's half-life in the (unpaywalled) abstract:

• Doubly charged negative hydrogen ions H$$^{2−}$$ and D$$^{2−}$$ have been indirectly observed in a tandem mass spectrometer using a hollow cathode duoplasmatron ion source. The half‐life of these doubly charged ions, determined in a number of drift time experiments, is 23±4 nsec. The existence of these ions is concluded from a velocity, momentum, and energy analysis of hydride ions produced in the autodetachment process: $$\mathrm X^{2-} \to \mathrm X^- + e$$ ($$\rm X=H,D$$).

• The existence of a relatively long-lived doubly charged negative atomic ion H$$^{2–}$$ (and D$$^{2–}$$), isoelectronic with the lithium atom, has been demonstrated by mass spectrometry through a combined analysis of ion energy, velocity, and momentum. This species, formed in a hydrogen plasma, has a half-life of 2.3 x 10–8 seconds before it spontaneously dissociates to produce H$$^–$$ ions.

All of this is publicly-available information that follows directly upon following the trail opened by your link, and you were pointed to this information in the comments. Why is this unconstructive claim, which detracts from the rest of your (otherwise interesting) question, still present in the post?

This is what counts as basic due diligence, at the levels of sophistication that you're asking at. Refusing to follow these basic standards, despite explicit pointers of where to follow and explicit requests that you do so, is not constructive behaviour. Please stop it.

• Hmm, although this criticism seems justified, and the answer does give relevant information, it seems this post is too negative for a StackExchange answer. – Ruslan Oct 23 '19 at 15:20
• @Ruslan The question is too unconstructive for a StackExchange question, too (and it is only one instance of an ongoing series of similarly unconstructive posts). Why isn't the community demanding more from OP? – Emilio Pisanty Oct 23 '19 at 15:21
• This is a very good answer (with the other one), specifically to what I asked. I do believe that this site had no answer for this question before, and no good description for what causes this. Now "Why is this unconstructive claim, which detracts from the rest of your (otherwise interesting) question". i will gladly delete or rephrase that part, please tell me exactly which part, or please edit it yourself if you like. I have no access to paywalled sites, otherwise I did research on this site (and somewhat outside). – Árpád Szendrei Oct 23 '19 at 15:34

Then you learn, that the Hydrogen negative ion with an extra electron is stable. Ok, so you have to accept that it is not so simple, and it is all QM.

You don't have to go to QM to have a stable system with two electrons and one proton. If you want to think within the classical electrostatic planetary model of the atom (ignoring for now that it doesn't work to describe reality), if you have two electrons on diametrically-opposite circular orbits around a proton, then the system is in equilibrium. The centripetal force on each electron is reduced, of course, but since the other electron is further away than the proton, the total force is still attractive.

That's basically what's going on in the quantum mechanical hydrogen anion (and more generally in atomic structure): shielding. To a good approximation, you can think of electrons in QM as occupying diffuse probability clouds, each of which creates an electrostatic field according to its density, which is felt by the other electrons in the system. (This self-consistent description is the essence of the Hartree-Fock approach to atomic structure. If you're not intimately familiar with it, now is the time to read up on it, in depth.)

As a standard example, for the helium atom, the two electrons occupy a $$1s$$ orbital. This means that when each electron is near the nucleus, it observes a central charge equal to the full nuclear charge, $$Z=2$$, but at the outer edges of the orbital's support region, this nuclear charge is shielded by the other electron, and the electron observes an effective central charge that's much closer to $$Z_\mathrm{eff} = 1$$.

For the hydrogen anion, the same is essentially true*, and the electronic structure is the same as in helium, so both electrons are in $$1s$$ states, and at the central part of the orbital they observe the full electric field of the central proton with charge $$Z=1$$. At the outer edges of the orbital, on the other hand, the shielding is essentially complete, so they see a vastly reduced central charge that approaches $$0$$ as you go away from the center.

This combination (near-zero effective central charge at large distances, near-$$Z=1$$ effective central charge at short distances) allows the shared $$1s$$ orbital to be bound, but it's kind of on the edge. What does "on the edge" mean, here? Basically, that there's very little room to maneuver here without breaking the system.

• The ground state itself is stable, as you need to put in energy to dissociate it into $$\mathrm{H}^-\to \mathrm H+e^-$$.
• However, the excited states of the system are essentially gone.
• As a reminder, for neutral atomic systems (which can be interpreted as a positive-charge asymptotic Coulomb potential for each electron), there is an infinity of bound excited states (the Rydberg series).
• For negatively charged systems, however, this changes: they can only support a finite number of bound states. This means that there are at most a finite number of excited states, or even none at all.
• For H$$^-$$ in particular, the number of excited states depends on whether your definition of "bound excited states" allows for radiative transitions (see this thread for the details). The short story is that H$$^-$$ has at best only a single excited state, which has symmetry characteristics that make it basically inaccessible, so for all practical purposes it has a stable ground state and nothing else.
• That said, H$$^-$$ does have resonances, which is what excited states become when they become unstable and embedded in a continuum. These work similarly to energy eigenstates, which are specific energies at which electrons can sit in a stable state, except that now that stability is broken by the ready availability of a decay channel. Thus, while it's impossible to place electrons here permanently, these states still leave clear resonant structures in things like absorption spectra.
• On the other hand, it's not all doom and gloom, and the system is not completely fragile ─ you can actually reduce the charge of the nucleus down to about $$Z\simeq 0.911 <1$$ and it will remain bound (reference), despite the fact that at large radii each electron sees a negative effective charge and thus a repulsive potential.

So: yes, H$$^-$$ is indeed possible, because the two electrons can "squeeze in" and only partially shield the central nuclear charge, but it is, in a sense, on the edge of existence.

If you try to add a second electron, though, the game gives out: there isn't any room left in the $$1s$$ shell, so the third electron would have to sit out at the larger radius of the $$2s$$ orbital (roughly speaking), where it now sees (close to) the full effect of the inner two electrons at the $$1s$$ shell. These don't just shield the nucleus's positive charge ─ they make for an effective central charge which is negative, and thus repulsive to the third electron. This is no longer compatible with that third electron sticking around, and it's going to fly off in response.

This intuition also holds for larger systems: say, if you try to add an extra electron into a fluorine atom, which is one electron short of a full shell, then it will be very happy to accommodate that extra electron into that hole, making fluoride. But if you try to add a second extra electron, there's no longer any room for it, and it's going to be driven away.

The likeliest small-atom candidate to make a stable doubly-charged anion is oxygen, which is two electrons short of a full $$2p$$ shell, and which is more than happy to accept the first electron. However, when you work things out, it still isn't possible to cram in a second extra electron into that remaining hole in the $$2p$$ shell ─ the structure exists, but it's a resonance which flies apart into O$$^-$$ and a free electron.

The same is true for larger atoms, where it's feasible to think that if you already have 90 electrons, say, whizzing about a $$Z=90$$ nucleus, with a wide-open $$5f$$ shell to fill, then not only one but two extra electrons might be able to fit in and have enough room to get out of each other's way without their mutual repulsion (and their repulsion with all the other 90 electrons in the system). However, when people have looked, these doubly-charged systems just are not stable.

So, are atomic dianions possible? Probably not, but this is not a hard, rigorous result. When we say

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

this just comes from the experience of looking at all the atomic systems we know, trying to calculate their structures when a first electron is added, and then adding in a second electron. Thus far, all the systems we've looked at have produced unstable dianions, precisely because it is really hard to keep that much negative charge confined to such a small system. This doesn't rule out the possibility of a larger atomic system (say, something in the presumed island of stability?) having a stable dianion, it's just experience indicating that it's hard for this to happen.

So basically the question is, where is that point, and why is it specifically between two and three electrons, that the electron configuration becomes unable to create a stable bond with a single proton?

Between two and three. There are no such things as fractions of an electron, so the question as posed is pretty meaningless.

That said, what you can do is to smoothly crank up the inter-electron repulsion from zero up to its real value for a three-electron system ─ or, equivalently, tune the value of the nuclear charge and see where the system becomes unbound.

• The clear starting point for this is $$Z=3$$, the neutral lithium atom, which is known to be stable.
• When you get down to $$Z=2$$, you'll be describing the helium anion, He$$^-$$, which is known to be unstable (it has a negative electron affinity). This means that the critical nuclear charge $$Z_c$$ is somewhere between $$2$$ and $$3$$.
• In other words: unlike hydrogen, helium cannot accept even a single extra electron. Shielding just doesn't work in this system because, as discussed above, the $$1s$$ shell is already full and the extra electron needs to sit out on a $$2s$$ orbital, which is doesn't have enough access to the inner regions of the system and just sees a negative effective central charge.
• Obviously, this means that getting down to $$Z=1$$, which is what you'd need to get stable H$$^{2-}$$ dianions, is just not going to happen.

So: we can rephrase your question as

what's the lowest nuclear charge $$Z$$ that's compatible with a stably bound three-electron system,

and we have a first answer in that $$Z$$ must be between $$2$$ and $$3$$. The real answer, however, is much more interesting than that, and it's a matter of ongoing research, as exemplified in particular by

which shows that the critical charge $$Z_c$$ is bounded above by $$Z_c\leq 2.000\,001.$$ In other words, there are stably bound three-electron systems all the way down to $$Z=2+10^{-6}$$, but what happens between that and $$Z=2$$ is currently an open question.

• It's entirely possible that every $$Z>2$$ admits stable three-electron systems, and that the boundary is at $$Z=2$$ (which is itself unstable).
• It's also entirely possible that there exists a $$Z_c$$ strictly between $$2$$ and $$2.000\,001$$, such that all $$Z>Z_c$$ give stable systems, but $$2 will give an unstable system.

That's a pretty tantalizing state of affairs, but it seems to be where the literature sits as of now.

* I should note that these heuristics aren't really true at full rigour. For H$$^-$$, electron correlation effects are much more important than in helium, and the Hartree-Fock method doesn't actually work; in fact, it's entirely possible that no HF ground state actually exists. To describe this system rigorously, you need to use full-dimensional methods, in which the wavefunction is a function over the six-dimensional configuration space, instead of a Slater determinant of individual three-dimensional waves.

You can get some intuition for this by using a rough approximation: treat an ion as a single point charge. In that case, it is obvious that a positive ion and an electron have a bound state, while a negative ion and an electron would not. A neutral atom and an electron would, in this approximation, have zero binding energy, and so you have to do a more precise calculation to determine the true binding energy- depending on the sign of this correction, some neutral atoms can accept a second electron and some cannot.

In other words, a positive ion attracts an electron and so will bind with it, while a negative ion repels an electron and so will not. A neutral atom and an electron do not attract or repel, so you have to actually do the quantum mechanical calculations precisely to determine if there is a bound state.

• "A neutral atom and an electron do not attract or repel, so you have to actually do the quantum mechanical calculations precisely to determine if there is a bound state." Can you please elaborate on that? – Árpád Szendrei Oct 22 '19 at 20:48
• @ÁrpádSzendrei The rough approximation tells you that the electron would have a binding energy of zero in that case. It is certainly not exactly zero in reality, so essentially the approximation doesn't tell us if the state is bound or unbound in that case. – Chris Oct 22 '19 at 20:55
• Thank you. So a neutral atom (Hydrogen) and one extra electron (together basically a negative Hydrogen ion) would not for sure be bound? – Árpád Szendrei Oct 22 '19 at 21:00
• @ÁrpádSzendrei For hydrogen specifically, it is bound. For some other atoms (all the noble gases, for instance), it is not. – Chris Oct 22 '19 at 21:03
• @ÁrpádSzendrei Every different number of electrons and protons is its own, extremely complicated system. The single electron+single proton is an easy system with a closed form solution, but systems with a higher number of particles are not. – Chris Oct 22 '19 at 21:18