2
$\begingroup$

It is known that the interaction between filled orbitals from the perspective of molecular orbital theory is overall destabilising because antibonding MOs are more antibonding than bonding MOs are bonding. Consider a large molecule (e.g. $\ce {I2}$). Due to the presence of multiple net destabilising interactions between the filled inner shell atomic orbitals (e.g. $\ce {1s, 2s, 2p}$ ...), wouldn't its formation be almost impossible since the single stabilising interaction due to the valence atomic orbitals would be small, compared with the total destabilisation from all the interactions between the inner shell orbitals, which are in fact, very numerous?

$\endgroup$
  • 1
    $\begingroup$ Indeed, the single stabilising interaction of the valence orbitals is small, but the total destabilization is smaller yet. See, all those 1s, 2s... are awfully small, so from their own POW they are awfully far from each other. $\endgroup$ – Ivan Neretin Jan 10 at 13:26
  • $\begingroup$ When overlap is zero, the MOs don't have bonding or antibonding character. The overlap is not zero, but is tiny enough that it might as well be zero. $\endgroup$ – orthocresol Jan 10 at 14:08
  • $\begingroup$ @orthocresol Are you trying to say that for the iodine molecule, the overlap integral S for 1s-1s interaction is almost negligible, hence even though there is an interaction, the bonding and antibonding interactions are equally bonding and antibonding respectively? $\endgroup$ – Tan Yong Boon Jan 10 at 14:13
  • $\begingroup$ Yes, and that's what Ivan was trying to say too, I believe. $\endgroup$ – orthocresol Jan 10 at 14:14
  • $\begingroup$ @orthocresol Actually, I just read Molecular QM by Atkins & Friedman (4th ed.) concerning this topic and they mentioned in the text that interactions between compact AOs and diffuse AOs can be neglected. So I suppose that the case of compact AOs would be 1s orbitals interacting in iodine molecule. And this can be explained by the overlap integral S being small. What about "diffuse AOs", is the idea also S being small? $\endgroup$ – Tan Yong Boon Jan 10 at 14:36
0
$\begingroup$

On p. 261, Atkins & Friedman (2005) mention two criteria when considering the significance of orbital interactions. On the first criterion, they write:

First, to participate significantly in bond formation, atomic orbitals must be neither too diffuse nor too compact. In either case, there would be only weak constructive or destructive overlap between neighbouring atoms, and only feeble bonds would result. It follows that in Period 2, ($\ce {1s,1s}$)-overlap can be largely neglected in comparison with ($\ce {2s,2s}$)-overlap, for $\ce {1s}$ orbitals are too compact to have significant overlap with each other. Indeed, it is generally safe, for qualitative discussions at least, to consider only overlap between orbitals of the valence shell, for only these orbitals are neither too compact nor too diffuse to have significant overlap.

This means that it is likely that for the $\ce {I2}$ molecule I mentioned in the question, the destabilisation from the interaction of filled inner shell orbitals (e.g. $\ce {1s, 2s, 2p}$ etc.) is probably very insignificant. As orthocresol writes in the comments, this can also be related to the overlap integral $\ce {S}$.

In orthocresol's answer here, we note that the reason behind antibonding MOs being more antibonding than bonding MOs are bonding is that the value of the overlap integral is not negligible (i.e. $\ce {S}$ is not much less than $\ce {1}$ such that $\ce {S}$ $\approx$ $\ce {0}$). However, in the case of the $\ce {1s-1s}$ overlap in the $\ce {I2}$ molecule, the orbitals may be too compact to interact such that $\ce {S}$ may actually become negligible. Thus, the antibonding and bonding MOS are destabilised and stabilised to the same extent respectively. Thus, these filled orbital interactions of the inner shell orbitals do not constitute much destabilisation.

Reference

Atkins, P.; Friedman, R. Molecular Quantum Mechanics (4th ed.). Oxford University Press, 2005.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.