A heteroaromatic molecule with six -electrons in S0 (three of each spin) will in its n,* state, where n is an in-plane orbital, have four -electrons of one spin ( spin) and three of the other ( spin), and this applies to both the singlet and triplet n,* state (Figure ). To understand their aromatic, nonaromatic or antiaromatic characters, we utilize Mandado's 2n+1 rule for aromaticity of separate spins. With this rule, Hückel's 4n+2 rule for closed-shell singlet state aromaticity is fractioned into a 2n+1 -electron part and a 2n+1 -electron part (Figure ), while Baird's 4n rule for the lowest * triplet state of [4n]annulenes is fractioned into a 2n+1 -electron and a 2n-1 -electron part, all four numbers corresponding to aromaticity for separate spins. Conversely, the lowest * triplet state of a species with 4n+2 -electrons is antiaromatic, having 2n+2 and 2n -electrons. A similar rule, derived by Valiev et al., tells that molecules are aromatic (antiaromatic) when having an odd (even) number of doubly and singly occupied -orbitals, suggesting that Mandado's rule should apply also to singlet excited states with the same electron configuration as the * triplet state. We have recently revealed the difficulty in proper assessment of excited state aromaticity through molecules which instead of Baird-aromatic are Hückel-aromatic or Hückel-Baird hybrid-aromatic in their lowest excited states. The situation also becomes more complex in heteroaromatics with in-plane lone-pairs (n, herein labelled n). The n,* state of such a (4n+2)-electron molecule has 2n+2 and 2n+1 -electrons (or vice versa), and would at first glance be nonaromatic if the aromatic -component exactly cancels the antiaromatic -component. Yet, can the two parts combined instead lead to aromatic or antiaromatic residuals?