According to molecular orbital theory (MOT), $\ce{O2^2+}$ has a greater bond order than $\ce{O2}$ and two less antibonding electrons. So why is molecular oxygen $\ce{O2}$ more stable than the molecular ion $\ce{O2^2+}?$

One possible reason that comes to mind is that the antibonding (AB) orbitals, although higher in energy than the constituent atomic orbitals, are still negative in energy, so adding electrons to AB orbitals still lowers the energy. But then, by that same argument, $\ce{O2-}$ would increase stability further, which it doesn't. Am I overlooking anything?

Note: I know that from an electrostatics point of view, $\ce{O2}$ is obviously more stable as bring two negative electrons close to positive $\ce{O2^2+}$ decreases the electric potential energy of the system. I am interested in how MOT explains relative stability (or if it actually even does).

  • 2
    $\begingroup$ You have to distinguish reactivity (with commonly found substances such as water and other components in air) and bond strength. Just because something has a strong bond doesn't mean it does not react - just compare alkanes and alkenes and alkynes. $\endgroup$
    – Karsten
    Feb 18, 2021 at 22:59
  • $\begingroup$ I am strictly talking about thermodynamic stability. O2 is more thermodynamically stable than O2+2 by elementary electrostatics, but MOT predicts otherwise. I am asking for an explanation for this $\endgroup$ Feb 18, 2021 at 23:06
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    $\begingroup$ You can only compare substances when they have the same set of atoms and the same charge. How would you assign thermodynamic values to a naked electron? $\endgroup$
    – Karsten
    Feb 18, 2021 at 23:08
  • $\begingroup$ Well, I think it is not more stable. For cation enthalpy is lower, bond stronger, but... getting those electrons still gives off energy, by aforementioned electrostatics. So, why try to get answer when you already have it? $\endgroup$
    – Mithoron
    Feb 18, 2021 at 23:13
  • $\begingroup$ @Mithoron I'm afraid I don't understand. O2 is NΟΤ more (energetically) stable? $\endgroup$ Feb 18, 2021 at 23:16

1 Answer 1


You have to consider the system as a whole - you can't directly compare $\ce{O_2}$ and $\ce{O_2^2+}$ because they have different numbers of particles. To put it another way when you consider the relative stability of two interconvertible specifies you really have to write down a chemical reaction that connects them, and then consider which direction is thermodynamically favoured, and then consider whether it is kinetically feasible. Stability is a really tricky and more often than not misused word!

So here we have to look at $$ \ce{O_2 -> O_2^2+ +2e^-} $$

As you note $\ce{O_2^2+}$ has a stronger bond than $\ce{O_2}$. But that is not all that is going on. We are also binding two electrons to the Oxygens - how does that affect the energetics? Well the reaction above is simply the sum of the following processes

$$ \ce{O_2 -> O_2^+ +e^-} $$ $$ \ce{O_2^+ -> O_2^2+ +e^-} $$

In other words the energy required is the sum of the first two ionisation energies of molecular oxygen. Do we know these? Well after a good 10 seconds looking I can't find the second energy, but the first is very well known, it is after all related to the discovery of compounds of Xenon and the start of noble gas chemistry. It is 12.0697 eV (from https://webbook.nist.gov/cgi/cbook.cgi?ID=C7782447&Mask=20) which in more familiar units is 1165 kJ/mol - a huge energy, bigger than any bond energy I am aware of, and note there is at least this much to come as the second ionisation energy will be even larger, as it is pulling a negative electron away from an already positively charge centre. Thus the energy gained by attaching the two electrons will massively out weigh any increase in bonding in the molecule.

Can we relate this to MO theory? Well as Koopman's theorem tells us the ionisation energy is an approximation of the orbital energy, this is telling us that relative to having a free electron it is more energetically favourable to put the electron in the lowest energy unoccupied orbital in the molecule. That this orbital is antibonding is irrelevant. Antibonding simply means it weakens any bonding in the molecule; it doesn't say anything about the energy of that orbital relative to the free electron. And you know this must be true - many molecules have electrons in antibonding orbitals and those electrons don't just rush away to infinity, hence occupying the orbital must be a lower energy proposition for the electron compared to being unbound. Thus $\ce{O_2}$ is thermodynamically more stable than $\ce{O_2^2+}$ and two free electrons because the decrease in energy caused by the electrons occupying the highest unoccupied molecular orbitals is (much) greater than any possible decrease in the bond energy.

That's the thermodynamics. Kinetics is difficult in general, but very unlikely to be an issue here. At this point I chicken out.

  • $\begingroup$ Would it be correct to say that the bond order in MOT and the 'stability" it describes is relative to the unbonded state, NOT infinite separation of each electric particle, and O2 & O2+2 have unbonded state with different energies (infinitely separated atoms vs infinitely separated cations)? If this is correct, wouldn't that nullify the use of MOT comparing relative stability (which is pretty much all we use it for at this level) $\endgroup$ Feb 19, 2021 at 10:48
  • $\begingroup$ Secondly, what do you exactly mean by the extent of "bonding"/bond energy ? Wouldn't having more electrons in the system further lowering the electric p.e make it more difficult to pull the nuclei apart and thus increase bond energy too? $\endgroup$ Feb 19, 2021 at 10:54
  • $\begingroup$ Not totally sure I understand what you are saying, but I'm taking it as saying that you have to be careful in making sure you have a consistent zero of energy. If so yes, obviously. If you do that you can make meaningful energetic comparisons. That this is poorly taught or even omitted is a different matter - and that observation may very well be true but I have been out of the education system at least for below graduate level for so long that I don't nowadays feel qualifies to comment. $\endgroup$
    – Ian Bush
    Feb 19, 2021 at 10:57
  • $\begingroup$ To your second comment - they go into antibonding orbitals. What do antibonding orbitals do? $\endgroup$
    – Ian Bush
    Feb 19, 2021 at 10:57
  • $\begingroup$ Oh ok. I was picturing the 2 extra electrons between the nuclei for some reason. A quick refresher on antibonding electron densities explains this. So basically, while it is more difficult to pull apart each electron from the nucleus in the molecule, it is more difficult to (symmetrically?) pull apart the 2 nuclei in the dication? $\endgroup$ Feb 19, 2021 at 11:17

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