# Why is the first ionization enthalpy of molecular O2 less than atomic O?

A cousin of mine asked for my help to answer one of his homework questions. It stated-

Explain why- The first ionization energy of molecular oxygen $$\ce{O2}$$ ($$\pu{1175 kJ mol-1}$$), is lesser than the first ionization energy of $$\ce{O}$$ ($$\pu{1314 kJ mol-1}$$)?

It took me some time trying to find out perfect reasons for it, because generally, bonded electrons are harder to remove than valence electrons in atoms. But indeed, not just with $$\ce{O2}$$, but $$\ce{F2}$$ ($$\pu{1515 kJ mol​-1}$$) also has a first ionization enthalpy lesser than atomic $$\ce{F}$$ which is $$\pu{1681 kJ mol​-1}$$. (I wonder why there aren't any more talked of!). So I ended up taking a "class" on the basics of MOT, trying to reason that "the energy required to remove an HOMO electron from an ABMO in $$\ce{O2}$$ is lesser than removing an electron from a bonding orbital. That could probably lead to this anomaly."

But this wasn't convincing enough for me. I could sense a lack of conviction, a whole list of factors that could go wrong, also with the possibility of a much simpler explanation. But what made me post it up here, despite its dubious look, is the lack of convincing information addressing this issue even on the internet. Hence, a proper answer would be helpful for the community at large.

• A similar explanation (kinda supporting mine), can be found here. Better answers are still awaited. Sep 17, 2020 at 14:02
• Did you check the values for $\ce{N2}$, or other such atoms? Sep 17, 2020 at 14:02
• Also, the fact that linear combination of atomic orbitals leads to a lower energy state(bonding orbitals) and a higher energy state (ABMO) seems to be pretty well defined since this would mean that you need more energy to reach zero potential for a bonding orbital w.r.t atomic orbital compared to the anti-bonding orbital where the actual energy of electron is higher than the atomic orbitals. In terms of throwaway numbers. If AO had an energy of -12 eV, then BMO would have a lower value(E$<$ 12 eV) and ABMO would be closer to the zero value than the AO.. Sep 17, 2020 at 14:09
• The highest-energy electrons in O2 are antibonding. Think about what antibonding means: it doesn't just mean higher in energy than a bonding orbital. It is antibonding with respect to the constituent AOs. Sep 17, 2020 at 14:10
• "generally, bonded electrons are harder to remove than valence electrons in atoms.". Maybe. But in O2 are the highest electrons actually bonding? Look again at the molecular orbital structure. Sep 17, 2020 at 14:12

In the molecular orbital diagrams shown in textbooks, the highest occupied molecular orbitals of $$\ce{O2}$$ and of $$\ce{F2}$$ are shown as antibonding. If the molecules and their cations had the same bond length, and the electrons in a species didn't "talk to each other", and the orbital energies of an isolated atom were not influenced by the approach of another atom, you could predict that the ionization energy of the molecular species is lower than that of the atomic species. The argument would be that these HOMOs have higher energy than the corresponding atomic orbitals (see comments to question).
According to the MO diagrams, the bond order of $$\ce{O2}$$ is 2 and that of the cation $$\ce{O2+}$$ is 2.5 (after all, we are removing an antibonding electron). Indeed, the bond length of the cation is shown as smaller in this source. I did not find a bond length for the $$\ce{F2+}$$ species. This shows that you have to at least adjust the MO diagram for the change in bond length.
To get a rationale for the ionization energies in a quantitative manner, you could calculate the energies of the four relevant species ($$\ce{O, O+, O2, O2+}$$) and look at the differences.
• So to summarize, you do concord with my views as well as fears. Well maybe then there really isn't any significant impact with those electronic forces and changes in bond length, as I thought. As these assumptions are not quite true, it is more complicated, but luckily the quick and dirty analysis yields the correct answer- that is probably the whole essence of it. I also came up with an interesting comparison, the IE1 of $\ce{Xe}$ is actually less than atomic $\ce{O}$. Never thought of that one before... Sep 18, 2020 at 2:57