 Solution

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]