I discussed in the previous post the small molecule C_{4} and how of the sixteen valence electrons, eight were left over after forming CC σbonds 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)/Def2TZVPPD calculation, FAIR DOI: 10.14469/hpc/10157). Again both these sets fit the Hückel 4n+2 rule (n=0).
Since B_{4} has only two rather than six delocalized σorbitals, the contributions to the central BB bond are weaker and so the BB bond is much longer.
Bonding MOs for B_{4}. Click image to load 3D model 


σ1, 0.335 au  
π1, 0.372 au  
πBonding MOs for N_{4}. Click image to load 3D model 


π3  π2 
π1  
σBonding MOs for N_{4}. Click image to load 3D model 

σ3  σ2 
σ1  
The pattern for N_{4} is different in several aspects. Firstly the πsystem has six bonding electrons distributed over only four atoms. This makes the electron repulsions too high and the species is no longer stable, having one large imaginary force constant corresponding to an outofplane distorsion. Secondly the lowest energy σ orbital is highly localised onto two nitrogens rather than being delocalised around the ring periphery. So all those electrons crammed into a small space have taken their toll.
Thus far we have identified three species, B_{4}, C_{4} and N_{4} with interesting sets of respectively 4,8 and 12 electrons, all partitioned into 4n+2 collections. But what happens if one cannot do that; lets say 6 and 10 electrons? Hang around to find out!