Is a molecule more stable with less bonding electrons or more anti-bonding electrons? This question arose when I was asked the stability comparison between $\ce{N2^+}$ and $\ce{N2^-}$.


1 Answer 1


Both (N2)+ and (N2)- have same bond order of 2.5 but (N2)+ would be more stable as it has lesser number of electrons leading to lower inter-electronic repulsions.It can also be said that (N2)- has more antibonding electrons leading to a lower stability than (N2)+

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    $\begingroup$ Elaborating a bit- intro texts often show bonding and anti bonding orbitals equidistant from the starting atomic orbitals, but when you do the math, bonding orbitals are stabilized by a smaller amount than antibonding orbitals are destabilized $\endgroup$
    – Andrew
    Aug 14, 2020 at 0:42
  • $\begingroup$ @Andrew, Is that really true? I don't know much about this topic but I always thought that energy is partitioned equally such that the potential lost by BMO is same as that gained by ABMO. $\endgroup$
    – ba-13
    Aug 18, 2020 at 6:44
  • $\begingroup$ Yes. The equal separation is a very useful approximation, but not quite quantititatively correct. The extra destabilization of the antibonding orbitals is why, for example, He2 is not as stable as He. $\endgroup$
    – Andrew
    Aug 18, 2020 at 11:20

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