= Solution
\b[The third electron must enter the higher, degenerate and partly antibonding orbital pair rather than the already filled fully bonding orbital.] At an equilateral geometry the configuration becomes $(a_1')^2(e')^1$. This weakens the collective bonding and creates <orbital degeneracy>, so a distortion can lower the singly occupied component and localize bonding. Along the ground-state <reaction coordinate>, formation of a normal two-electron H–H bond with a separated <hydrogen> atom becomes favourable.
The result is ground-state dissociation to $\mathrm{H_2+H}$, not loss of a <proton> from the cation or formation of three isolated atoms. One should not claim that adding one electron exactly cancels two bonding <electrons>: a simple three-orbital count alone does not predict a zero <bond order> or prove dissociation. The important qualitative changes are occupancy of an antibonding/degenerate level and access to a lower-energy dimer-plus-atom arrangement. A complete dissociation statement requires the potential-energy surface, including nuclear repulsion; electronically excited neutral H$_3$ states are a separate issue.
Back to article page
