How do I know whether s-p mixing would occur in some diatomic heteronuclear molecules like CN, CO, NO etc?


2 Answers 2


I don't think you can know, a priori, whether or not s-p mixing occurs without first having some experimental data. I'll use water, a triatomic molecule, as my case in point.

For over 50 years students have been told that water is roughly $\ce{sp^3}$ hybridized. The general description is that there are two equivalent O-H $\ce{sp^3}$ sigma bonds and two equivalent lone pairs, also in $\ce{sp^3}$ orbitals. The lone pair - lone pair repulsion is greater than the sigma bond - sigma bond repulsion, so the lone pair-O-lone pair angle opens up slightly and the $\ce{H-O-H}$ angle closes down to the observed 104.5 degrees.

With the advent of photoelectron spectroscopy it was found that the two lone pairs in water were not equivalent (2 signals were observed for the lone pairs). Now, the hybridization of water is described as follows:

  • 2 $\ce{sp^3}$ O-H sigma bonds
  • one lone pair in a p orbital
  • and the second lone pair in an $\ce{sp}$ orbital

If chemists knew how to predict s-p mixing, it wouldn't have taken them over 50 years to understand how they mix in water.

  • 1
    $\begingroup$ what are the implications of water's hybridization? Is one lone pair more reactive than the other? $\endgroup$
    – Dissenter
    Commented Dec 29, 2014 at 17:55
  • $\begingroup$ Is there any handy rule for some simpler cases, like the molecules I mentioned? $\endgroup$
    – EJC
    Commented Dec 29, 2014 at 17:59
  • $\begingroup$ @Marko Not that I'm aware of. $\endgroup$
    – ron
    Commented Dec 29, 2014 at 18:00
  • 2
    $\begingroup$ @Dissenter The two lone pairs are not interconvertible by a symmetry operation. Therefore they are different, this includes reactivity. $\endgroup$
    – ron
    Commented Dec 29, 2014 at 18:03
  • $\begingroup$ Muliken's 1935 "Electronic Structures of Polyatomic Molecules. VII. Ammonia and Water Type Molecules and Their Derivatives" givens energy values for water MOs, with the two non-bonding (lone pair) orbitals being 32 eV and 12.7 eV. UV and "electron impact" data are cited. scitation.aip.org/content/aip/journal/jcp/3/8/10.1063/1.1749715 The degree of s-p mixing for each orbital is calculated in his 1955 Electronic Population Analysis on LCAO–MO Molecular Wave Functions. I scitation.aip.org/content/aip/journal/jcp/23/10/10.1063/… Non-equivalent lone pairs are long known. $\endgroup$
    – DavePhD
    Commented Mar 22, 2015 at 15:24


Hybridization is a questionable model. It was originally introduced to consider organic molecules, and there it works brilliantly. However, it is hard to use outside organic chemistry with molecules with simple bounding. For example, adequate description of cyclopropane requires fractional indices in orbitals.

Mathematically, hybridization process is transformation of several orbitals using linear transform. Since molecular orbitals are described as linear combinations of atomic orbitals, there is no problem to use any imaginable set of atomic orbitals as long as their linear combinations can produce all 'original' atomic orbitals. Basically, hybridization is similar to change of basis vectors in geometry: it is OK as long as the transformation follows simple rules, and it may be built to preserve values of some scalar vector expressions.

Since from mathematical viewpoint hybridization is nothing more than a special transformation of basis set, it does no change 'actual' molecular orbitals, meaning that calculations using hybridized and unhybridized orbitals must provide same result (actually, the very transformation in question is purposefully built this way). So, your question does not have a proper answer as the question itself makes wrong assumption. Hybridization does not occur, it is a model people use, it does not correspond to any real-world phenomenon

Now, the proper question would be: when hybridization model is applicable and provides valuable insight into electronic structure of considered molecule?

When you are considering electronic structure of isolated 2-atom molecules it is usually better to not use hybridization. It does not provide any simple conclusions for inferring preferred geometry, but leads away from more 'accurate' MO LKAO model. Considering interaction of such molecule as a ligand with metal center, however, returns some use to hybridization for simple molecules, because it simplifies geometry consideration. For example, using electron count rules it is possible to some extent to predict geometry of $\ce{NO}$ fragment, inferring preferable hybridization state of the nitrogen atom. It is also much easier to consider interactions with central atom for $\sigma sp$ lone pair than several dinuclear molecular orbitals.

When considering geometry of organic and some inorganic molecules ($\ce{BCHNOF}$) with little angular stress and no hypercoordinated atoms, hybridization works well enough as long as you do not try to use it to predict one-electron properties, like ionisation potentials. However, stressed molecules like [1,1,1]-propellane or dehydroadamantane and hypercoordinated atoms, like methanium cation, require separate, case-by-case analysis to be integrated into framework of hybridization model. For most other atoms hybridization model cannot provide valuable insights.

Again, hybridization does not work for 1-electron properties. For example, methane, a perfect example of ideal $sp^3$ hybridization has TWO peaks in photoionization spectrum, implying it has 2 unequal sets of valence orbitals (roughly -2.25 and -1.5 eV energy levels) despite hybridization model predicting it to have 4 perfectly equal bonds. MO LKAO model, on the other hand, predicts several bonding orbitals, one significantly lower in energy than three others.

  • $\begingroup$ I didn't think about hybridisation, but s-p mixing that causes the pi MO to be above sigma MO in some molecules (like nitrogen) when I asked the question. $\endgroup$
    – EJC
    Commented Dec 29, 2014 at 19:34
  • 3
    $\begingroup$ @Marko It occurs in ALL molecules, the question is to what degree. Unfortunately, there is no way to predict if it will change relative position of orbitals in any particular case without some complex calculation. If relative position of orbitals is important, use some free package (say, us-gamess, free for personal use) to calculate orbitals and view them in some free viewer, say, avogadro. Fortunately, it is quite easy for 2-atom molecules these days. $\endgroup$
    – permeakra
    Commented Dec 29, 2014 at 19:38

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