Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 2 / b

Both dimers have two electrons in a valence sigma-bonding orbital and a formal valence bond order of one. For hydrogen the minimal-basis configuration is $(1{\sigma }_g{)}^2$, with the $1s$-derived $1{\sigma }_g$ as highest occupied molecular orbital and $1{\sigma }_u^{\ast }$ as LUMO. The bonding combination has no internuclear node; the antibonding combination does.

For lithium, include the filled core pair of levels as well as the valence levels:The first two orbitals are predominantly core $1s$ combinations whose bonding and antibonding occupations cancel in a simple bond-order count. The highest occupied molecular orbital is predominantly ${\sigma }_g(2s)$, denoted $2{\sigma }_g$; the LUMO is $2{\sigma }_u$, predominantly the antibonding $2s$ combination but modified by mixing with compatible $2p$ contributions. The nearby empty $2p$-derived sigma and pi levels must also be shown. Only orbitals of matching symmetry mix, so this does not mean that every $2s$ combination mixes with every $2p$ combination.

On a common qualitative energy reference, hydrogen $1s$ is much lower than lithium's valence $2s$, while lithium's core $1s$ levels lie much deeper. The Li valence bonding–antibonding separation is smaller, consistent with its longer and weaker bond. The figure uses a compressed core region and schematic gaps, not measured orbital energies. The role of symmetry-compatible s–p orbital mixing in the first-row diatomics is illustrated in institutional molecular-orbital notes.^[ref]

Two electrons can occupy a fully bonding orbital delocalized equally over all three hydrogen atoms. Three $1s$ atomic orbitals on an equilateral triangle form one totally symmetric $a_1^{\prime }$ combination and a degenerate pair $e^{\prime }$ of higher orbitals. In an orthonormal equal-interaction model with diagonal energy $\alpha$ and off-diagonal $\beta <0$,The lower orbital is proportional to $1s_A+1s_B+1s_C$, with constructive overlap along every edge. Its two electrons give a three-centre two-electron bond and equivalent interactions between every pair of nuclei. At the equilibrium distance, this bonding competes with nuclear Coulomb force to favour the symmetric triangular arrangement. The occupied shell is nondegenerate and closed, so there is no first-order Jahn-Teller incentive to distinguish the three edges. A straight-line or localized single-bond description would miss that equally shared bonding pattern.

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 $(a_1^{\prime }{)}^2(e^{\prime }{)}^1$. 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 ${\mathrm{H}}_2+\mathrm{H}$, 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.

Li${}_3$ can retain a bound triangular framework, but an equilateral triangle with three valence electrons is Jahn-Teller unstable. The six core electrons are effectively spectators. In the simplest triangular valence model, two electrons occupy the fully bonding level and one occupies the degenerate upper pair. Making one edge different splits that pair; the single electron occupies the lower component, giving an electronic stabilization that can outweigh the cost of a small distortion. An isosceles triangle is sufficient to lift the degeneracy, so three different edge lengths are not required.

The two 2.8-Å contacts are close to, but slightly longer than, the 2.7-Å dimer bond, while the 3.5-Å edge is much weaker. Delocalization and partial antibonding occupation explain why these cannot be treated as three independent full-strength Li–Li single bonds. lithium's nearby valence $2p$ functions provide additional polarization and mixing channels absent from a hydrogen $1s$-only description, allowing a more flexible multicentre electronic structure. The existence of a bound minimum, unlike ground-state H${}_3$, is an energetic fact rather than a consequence of three-electron counting alone. Calculated Li${}_3$ surfaces show bound states, a conical intersection and Jahn-Teller coupling.^[ref]