From diagrams, it's rather obvious how $sp$ orbitals are hybridized - the hybrids are just a composite of the $s$ and the $\pm p_{(x)}$ orbitals. However, $sp^2$ orbitals are not just composites of $s, \pm p_x, \pm p_y$, and the same goes for $sp^3$.

This incongruity leads to my question.

I am aware that the general shapes are trigonal planar and pyramidal, respectively. However, I recall learning of a more mathematical approach that used the wave function (I think?)

In my notes, I found this formula. $$\Phi_i=\sum_{j=1}^nc_{ij}\phi_j, i=1,2,3...n$$ My understanding is that $n$ is the number of atomic orbitals (and thus also the number of orbitals). I believe $c$ is a constant, and $\phi$ is the orbital.

Later, for $sp^2,$ I have the following formulas:

\begin{align} \require{cancel} \Phi_1 &=c_{1,1}\cdot s+ \cancelto{0}{c_{1,2}\cdot p_x} + c_{1,3} \cdot p_y &&= c_{1,1}\cdot s+c_{1,3} \cdot p_y\\ \Phi_2 &=c_{2,1}\cdot s+ c_{2,2}\cdot p_x + c_{2,3} \cdot p_y &&= c_{2,1}\cdot s+ c_{2,2}\cdot p_x + c_{2,3} \cdot -p_y\\ \Phi_2 &=c_{3,1}\cdot s+ c_{3,2}\cdot p_x + c_{3,3} \cdot p_y &&= c_{3,1}\cdot s+ c_{3,2}\cdot -p_x + c_{3,3} \cdot -p_y \\ \end{align} I have some recollection that the wave functions zeroes out for particular quantum number values, but I can't remember how it all fits together with this equation.

I couldn't find this formula online, so I'm hoping that it might make sense to someone here.

Thanks for the help!

  • 1
    $\begingroup$ Referring to MathJax: the align environment will flush to the right at odd ampersands and to the left at even, ergo r&l&r&l&r. $\endgroup$ – Martin - マーチン Sep 4 '20 at 6:25

What we call "Hybridization" is really just a mathematical transformation of an approximated wavefunction. Your basic summation formula describes how to build some sort of customized wavefunction by forming a linear combination of some set of "basis" orbitals; you will find this equation over and over if you look up "basis set expansion". Now, of course you can't just use any random coefficients $c_{ij}$ of your liking for this custom wavefunction; you have to adhere to some physical rules. Most importantly, the overall energy that the wavefunction represents must not change. I won't go into detail about the constraints, but if you are interested, google "unitary transformation".

For orbital hybridization, you first have to find a set of basis orbitals from a physically sound starting wavefunction. If you only look at an isolated atom, then atomic orbitals are obviously a useful choice. From there, you can form combinations of those orbitals (by some criteria that really aren't uniquely defined) while adhering to the rules of the unitary transformation. In the end, you will arrive at your preferred hybrid orbitals that fulfill your chosen criterion as best as they can while keeping the wavefunction energetically equivalent to the original one. Some of the coefficients $c_{ij}$ in your equations may come out to be 0, typically for reasons of symmetry. For example, an $sp^2$ hybrid that is oriented along the $y$ axis of your coordinate system would obviously only have contributions from the $p_y$ orbital, not from $p_x$ and $p_z$. But the exact values of the cofficients will generally depend on the specifics of your problem, most importantly the definition of your coordinate system. So you really can't say per se which coefficients will drop out to be 0 and which won't -- it's one of those "it depends" situations.

General Thoughts:

Unfortunately, hybridization is a concept that is so misleading that I would consider its general use in chemistry to be outright physically incorrect. People invoke it to suggest and visualize bonding situations in molecules and to explain the resulting molecular shapes, but unfortunately always seem to forget some very basic undergraduate quantum chemistry in this process.

Hybrid orbitals are linear combinations of eigenfunctions to different energy eigenvalues of the underlying one-electron Hamiltonian; think $s$ and $p$ orbitals. As such, they are not solutions to the time-independent Schrödinger Equation anymore, which in turn means that they are time-dependent functions. This implies two things:

1.) Hybrid orbitals do not have a fixed shape; what you see visualized is just a single snapshot in time. The orbital shapes periodically "morph" back and forth and can at other points in time look nothing like those suggestive localized "bonding" orbitals anymore. In fact, the hybrid orbitals can even change places among each other over time if the symmetry of the system permits.

To better drive this point home, here are some (semi-quantitative) visualizations of the time development of the bonding and non-bonding orbitals in water.

2.) Hybrid orbitals do not have defined energies, since they are not solutions to the time-independent Schrödinger Equation. If you were to do any sort of "energy measurement" on a hybrid orbital, you would end up with a Schrödinger's Cat situation where you observe the energies of the constituent atomic orbitals with certain probabilities that are connected to the expansion coefficients $c_{ij}$.

In addition, there is a third problem. When the pure integer mixtures of s and p orbitals (e.g. $\mathrm{sp}^2$, $\mathrm{sp}^3$) were not sufficient to explain molecular shapes, people started to come up with weird fractional mixtures like $\mathrm{sp}^{1.5}$ or some such. How are those numbers obtained? - As stated above, from the wavefunction side this comes down to a "localization" procedure of the molecular orbitals; however, the criteria by which to do this are not strictly defined, and there exist multiple different schemes for it. So this raises the third problem:

3.) Hybridization is not stringently mathematically defined if one regards it as an orbital localization procedure.

So, the next time someone presents you with hybrid orbitals to explain e.g. why methane is tetrahedral, ask them about the energy and the electronic density distributions of those orbitals, and how they would determine the hybrid orbitals for e.g. chloromethane where the bond angles and energies are clearly different than in methane.

  • $\begingroup$ "an sp2 hybrid that is oriented along the y axis of your coordinate system would obviously only have contributions from the p_y orbital, not from p_x and p_z." Did you mean to say "sp", or am I misunderstanding? $\endgroup$ – electronpusher Sep 4 '20 at 6:41
  • $\begingroup$ Great points in your General Thoughts section! $\endgroup$ – electronpusher Sep 4 '20 at 6:43
  • $\begingroup$ Well, both an sp or sp2 hybrid along the y axis wouldn't have x or z components. sp3, too as a matter of fact. Simply because if they had px or pz orbitals mixed into them, they wouldn't be oriented purely along the y axis. But again, that also depends on the coordinate system, and if you change that coordinate system, the designations of your orbitals may also change. $\endgroup$ – Antimon Sep 4 '20 at 16:33
  • $\begingroup$ Short question: can I use the concept of hybridisation to rationalise the fact that methane (or chloromethane) have a tetrahedral like shape? In spite of your big warning, I would say yes, I can. $\endgroup$ – Alchimista Nov 9 '20 at 8:33
  • $\begingroup$ Well then I'd prompt you to demonstrate how. Neither do the orbitals that you propose have a fixed shape, so you can't point to them and say "See, these bonding electrons like to be between the C atom and this H atom", nor do they give you useful energy values to characterize the energetics of your bonds. The only way hybrids work to rationalize the "static" geometry of a molecule is if you use the superposition of all hybrids so that they become time-independent functions, in which case by definition you're back at MO theory. $\endgroup$ – Antimon Nov 9 '20 at 19:44

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