I discussed in the previous post the small molecule C4 and how of the sixteen valence electrons, eight were left over after forming C-C σ-bonds which partitioned into six σ and two π. So now to consider B4. This has four electrons less, and now the partitioning is two σ and two π (CCSD(T)/Def2-TZVPPD calculation, FAIR DOI: 10.14469/hpc/10157). Again both these sets fit the Hückel 4n+2 rule (n=0).
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