The previous examples of four atom systems displaying two layers of aromaticity illustrated how 4 (B4), 8 (C4) and 12 (N4) valence electrons were partitioned into 4n+2 manifolds (respectively 2+2, 6+2 and 6+6). The triplet state molecule B2C2 with 6 electrons partitioned into 2π and 4σ electrons, with the latter following Baird’s aromaticity rule.[cite]10.1021/ja00769a025[/cite],[cite]10.1021/cr300471v[/cite]. Now for the final missing entry; as a triplet C2N2 has 10 electrons, which now partition into 4 + 6. But would that be 4π + 6σ or 4σ + 6π? Well, in a way neither! Read on.