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Recently, in one of the questions in an assignment on the topic of energetics, we were asked to explain why the bond dissociation energy for the trihydrogen cation is much larger than what we would expect it to be. The value given was +849 kJ/mol. To clarify, the bond dissociation process refers to the conversion of the molecular cation into 2 individual hydrogen atoms and a proton.

My proposed explanation is that the additional interaction with the proton lowers the bonding MO of the molecular ion overall. Essentially, I viewed the interaction of the $\sigma$ bonding MO of $\ce {H2}$ with the low-energy 1s atomic orbital of the proton, giving a bonding MO which would be even lower in energy.

The answer given by the teacher is that this molecular ion is particularly stable due to the $\sigma$ aromaticity present in the ion. This results in a very large bond dissociation energy since by breaking up the molecular ion, we destroy the aromaticity of the ion.

I acknowledge that the teacher's explanation is correct but is my explanation also a valid way of looking at the problem?

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    $\begingroup$ I wouldn't bother with aromaticity at all and pointed that free proton has so hight charge density, that it attaches itself to noble gases (including helium). Also, yes, the bonding orbital of the cation is very low, but there is no need to invoke any aromaticity concept here, it is enough to say that it has no nodes and so the majority of electron density is located between the protons. $\endgroup$ – permeakra May 10 '18 at 10:28
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    $\begingroup$ The enhanced stability of the bonding orbital is aromaticity at work, no? $\endgroup$ – Oscar Lanzi May 10 '18 at 11:09
  • $\begingroup$ @OscarLanzi Yes I agree. In fact, I feel that my explanation is limited in a sense that I am looking at a localised interaction between H2 molecule and a proton, which is definitely less accurate a representation than an aromatic delocalised molecular ion. But how would MOs look like for H3+, if it were aromatic? Any ideas? $\endgroup$ – Tan Yong Boon May 10 '18 at 12:07
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    $\begingroup$ Much like the Frost circle for cyclopropenyl cation, of course. $\endgroup$ – Oscar Lanzi May 10 '18 at 12:26

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