The first chapter of Fleming's "Molecular Orbitals and Organic Interactions," describes the formation of triatomic hydrogen. In summary, the sigma MO of diatomic hydrogen interacts in phase with the third hydrogen's AO. In consequence, a new bonding and antibonding MO are formed. The antibonding MO of diatomic hydrogen does not interact due to antisymmetry.

Since two electrons will fill the bonding MO and one will fill the antibonding MO, there is a overall bonding interaction. This raises my question as to why triatomic hydrogen is relatively short-lived, a duration of one millionth of a second prior to decomposition.

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    $\begingroup$ According to Wikipedia, the molecule exists only in the excited state. If I get it right, it means that, as shown here: upload.wikimedia.org/wikipedia/commons/thumb/0/07/… The energy required for the molecule to exist (SOMO energy level) is higher than the energy of the "generating reactants": H2 and H (H3+ is shown in the image, but the energy levels involved are the same). By releasing an electron, the molecule can become the more stable H3+ cation, with a low energy HOMO. $\endgroup$ – The_Vinz Jul 15 '19 at 5:55
  • $\begingroup$ @The_Vinz Wouldn't combining the sigma MO with the third hydrogen's AO lead to an overall reduction in energy? The antibonding MO should be as "antibonding" as diatomic hydrogen's antibonding MO. $\endgroup$ – MacroGuy Jul 15 '19 at 5:58
  • $\begingroup$ Possibly, but the antibonding MO of H2 is LUMO, unoccupied, right? That means that it doesn't represent the ground state energy of H2. After combination with another H, that LUMO or the new one (which I don't know if they are isoenergetic, actually) is occupied, and the molecule is destabilized $\endgroup$ – The_Vinz Jul 15 '19 at 6:05
  • $\begingroup$ Is the short life due to high reactivity (2 H3 --> 3 H2), which should happen at EVERY collision, or to a unimolecular decomposition? $\endgroup$ – James Gaidis Jul 15 '19 at 14:06

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