I discussed in the previous post the small molecule C ₄ 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 B ₄ . 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). Since B ₄ has