= Solution
\b[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.<A diabatic representation of the two lowest electronic states of Li3 by Elham Nour Ghassemi and others>[Abstract and potential-energy-surface analysis]{ref}
Back to article page
