I would just like to ask if the pre-requisite for using Bent's Rule is that the bonding in the molecule involves hybridised orbitals.

I was thinking of why the bond angle in hydrogen sulfide (~ 92 degrees) is larger than the expected 90 degrees. I thought that Bent's Rule would be able to offer a good explanation for this as the s orbital character concentrated towards the electropositive hydrogen atoms results in stronger bond pair-bond pair repulsion, increasing the bond angle. However, I realised that my entire argument may fall apart if Bent's Rule does not work for bonding involving unhybridised orbitals.


1 Answer 1


Bent's rule is not as much a rule as it is an observation. It was first applied in the context of valence bond theory and therefore used for hybrid orbitals only. At the initial stages, VBT only used s, sp, sp2, sp3, and p type orbitals. Bent proposed to allow also anything in between aiming at more flexibility for the wave function and therefore better accuracy of the description. As such it is only a tool for an already approximate view on bonding.

Most importantly, and in any and all general cases, the hybridisation state of the orbitals is determined by the molecular structure. For any case in which hybrid orbitals are used, there is also a true solution which does not use them.

Though Bent's rule has some predictive character for anything that is close to hybrid orbitals, it is basically just a shorthand for a much deeper rooted phenomenon.

However, Bent's rule is not useful in any case where there is a low s-p mixing, i.e. anything below period 2 is a stretch.

  • $\begingroup$ When u say "s-p mixing", do u mean s-p hybridisation? I know s-p mixing as an MO thing, but it does mean s-p hybridisation in the VB context right? $\endgroup$ Nov 16, 2017 at 15:15
  • $\begingroup$ Could u also elaborate on your last statement? From what I read, u don't seem to have explained it $\endgroup$ Nov 17, 2017 at 0:22
  • $\begingroup$ @TanYongBoon The s-p gap in atoms of the third period and below is in most cases too large for s and p orbitals of the same atom to mix in molecular orbitals. The hybridisation approximation breaks down at that point. That is one of the reasons for the different angles in $\ce{NH3}$ versus $\ce{PH3}$, see here. $\endgroup$ Nov 17, 2017 at 2:59
  • $\begingroup$ Yes, s-p mixing is a generic term in MO theory. It refers to s and p orbitals of the same atom being a contributor to the same molecular orbital. In VB theory it used to be achieved with hybrid orbitals. $\endgroup$ Nov 17, 2017 at 3:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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