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 $\ce{N2^-}$.
$\begingroup$
$\endgroup$
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
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)+
-
1$\begingroup$ 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 $\endgroup$– AndrewAug 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-13Aug 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. $\endgroup$– AndrewAug 18, 2020 at 11:20