Chemistry - Are the bonding orbitals in methane equivalent - photoelectron spectrum
Solution 1:
This question goes along the line of what does it mean when it is said that an sp3 orbital has 25% s-character. It also intrigued me so I have tried to find answer, which would not break my hybridised orbital view.
What you clearly see from the spectrum is that there are two bands, corresponding to different energy levels in the molecule and their intensity is 3:1. You can say, it is $3\times 2p : 1 \times 2s$ and you are done. Not at all. The most important thing to deal with is the $T_{d}$ symmetry of the molecule. That means, all four $\ce{C-H}$ bonds are equivalent.
But it is this symmetry which has the answer. Leaving out the $1s$ orbital of carbon, the occupied orbitals must have the following symmetries: $1\times A_1$ and $3\times T_2$. And this is what you see in the spectrum.
Now how it comes with the hybridisation? Upon simple HF/STO-3G calculation, the orbital populations are following:
- $A_1$: 0.62 C 2s + 4x 0.18 H 1s ... fully symmetric, great
- $T_2$: 0.57 C pz + 0.3 1H 1s + 0.3 2H 1s - 0.3 3H 1s - 0.3 4H 1s
- $T_2$: 0.57 C py + 0.3 1H 1s - 0.3 2H 1s - 0.3 3H 1s + 0.3 4H 1s
- $T_2$: 0.57 C px + 0.3 1H 1s - 0.3 2H 1s + 0.3 3H 1s - 0.3 4H 1s
Well, as if you would copy the symmetry table ;) The downside of this picture is that all the bonds are completely delocalised (as you would expect from canonical MOs). Although observable through PES, they do not allow you to draw line between any C and H atom.
Where do we get the hybridised orbitals? If we want to localize the orbitals, whatever scheme you choose, you just form linear combinations of the existing MOs so that you have least number of atoms involved in any bond. One of such methods is NBO and as a result you obtain:
C - H1 ( 52.55%) 0.7249* C s( 25.00%)p 3.00( 75.00%)
( 47.45%) 0.6888* H 1 s(100.00%)
C - H2 ( 52.55%) 0.7249* C s( 25.00%)p 3.00( 75.00%)
( 47.45%) 0.6888* H 2 s(100.00%)
C - H3 ( 52.55%) 0.7249* C s( 25.00%)p 3.00( 75.00%)
( 47.45%) 0.6888* H 3 s(100.00%)
C - H4 ( 52.55%) 0.7249* C s( 25.00%)p 3.00( 75.00%)
( 47.45%) 0.6888* H 4 s(100.00%)
All with 2 electron occupation and energy -0.63479 Eh.
So where is the truth? ... $\Psi$ is the answer and you can massage it to your will, depending what you want to see. You want localized bonds? You will get them. You want orbitals with the proper symmetry required by spectroscopy? As you wish.
Does the hybridisation concept have sense? In localized bond picture yes. Is it observable? No, and was never meant to be.
Solution 2:
The short answer is that's not how photoelectron spectroscopy works. It's a one-photon spectroscopy.
You have a sample of $\ce{CH4}$ in this case, and you shoot different energy photons at the sample. At certain energies, the photon is absorbed and you get an electron off. Note, that's a single electron from one photon.
For you to have any significant amount of $\ce{[CH4]+}$ around, you'd have to hit it with two photons. The first would do:
$$\ce{CH4 + ~energy -> [CH4]^{+} + ~e^-}$$
Only then would you be able to hit that one single $\ce{[CH4]+}$ ion with another photon.
That's a two-photon process. Assuming the incident light intensity is small, that's extremely unlikely. It's much more likely that that second photon hits a different $\ce{CH4}$.
In other words, normal photoelectron spectroscopy gives a set of intensities when the species absorbs a single photon with enough energy to remove a single electron (i.e., orbital energies). This process happens on the femtosecond timescale, so it's essentially a "frozen orbital" technique. There's no time to sample the electronic structure of the resulting cation.
Also, the PES of $\ce{[CH4]+}$ is completely different (seen here), because of the Jahn-Teller distortion. The geometry is $C_{2v}$ and thus there are many more electronic transitions - it actually gets quite complicated.
Solution 3:
To begin with, I am not really well-versed in this area, so would appreciate any input/corrections.
Wikipedia is correct on the matter; the photoelectron spectrum of methane is not a proof that hybrid orbitals are incorrect. The same can be said of water, ammonia, and so on. When one says that two peaks in the PES are observed and that it must correspond to two different levels, Koopmans' theorem is being implicitly invoked:
$$\mathrm{IP} = -\varepsilon_i$$
Here, $\mathrm{IP}$ refers to the experimental ionisation potential, and $\varepsilon_i$ refers to the energy of a canonical orbital $|\chi_i\rangle$, which are obtained by solving the Hartree–Fock equations:
$$\hat{f}|\chi_i\rangle = \varepsilon_i|\chi_i\rangle$$
The problem with applying this theorem to hybrid orbitals is that hybrid orbitals are not eigenfunctions of the Fock operator; the hybrid orbitals $\phi_i$ are obtained by some unitary transformation of the canonical orbitals $\chi_i$.
$$\begin{pmatrix} \phi_1 \\ \phi_2 \\ \vdots \\ \phi_i \end{pmatrix} = \mathbf{U}\begin{pmatrix} \chi_1 \\ \chi_2 \\ \vdots \\ \chi_i \end{pmatrix}; \qquad \mathbf{U}^{-1} = \mathbf{U}^\mathrm{T}$$
and Koopmans' theorem does not apply after a unitary transformation, as has been explained in this question. So, the PES cannot be used as proof for the non-existence of hybrid orbitals, since the peaks in the PES cannot be linked to energies of hybrid orbitals.
How can hybrid orbitals explain the PES? Can they even do so? It is instructive to consider the "incorrect" way of doing it, first. This is adapted from Section 2.9 of The Chemical Bond – Fundamental Aspects of Chemical Bonding, edited by Gernot Frenking and Sason Shaik. Let's say we have four equivalent hybrid orbitals in methane, $\{\phi_i\}$ $(i = 1,2,3,4)$. Then the wavefunction of methane, ignoring core orbitals, can be written as a Slater determinant
$$\Psi(\ce{CH4}) = |\phi_1 \phi_2 \phi_3 \phi_4 \bar{\phi}_1 \bar{\phi}_2 \bar{\phi}_3 \bar{\phi}_4|$$
(no bar indicates spin up, a bar indicates spin down). Now, according to the wrong interpretation, one can remove an electron from any of these four orbitals. Without loss of generality we will remove a spin down electron. This supposedly would lead to four possible new wavefunctions:
$$\begin{align} \Psi_1(\ce{CH4+}) &= |\phi_1 \phi_2 \phi_3 \phi_4 \bar{\phi}_2 \bar{\phi}_3 \bar{\phi}_4| \\ \Psi_2(\ce{CH4+}) &= |\phi_1 \phi_2 \phi_3 \phi_4 \bar{\phi}_1 \bar{\phi}_3 \bar{\phi}_4| \\ \Psi_3(\ce{CH4+}) &= |\phi_1 \phi_2 \phi_3 \phi_4 \bar{\phi}_1 \bar{\phi}_2 \bar{\phi}_4| \\ \Psi_4(\ce{CH4+}) &= |\phi_1 \phi_2 \phi_3 \phi_4 \bar{\phi}_1 \bar{\phi}_2 \bar{\phi}_3| \\ \end{align}$$
In Koopmans' theorem, the ionisation potential is identified as the difference in energy between the ionised wavefunction and the unionised wavefunction. The theorem states that this is simply the negative of the energy of the orbital from which the electron was removed.
The problem is that these ionised wavefunctions, $\Psi_i(\ce{CH4+})$ $(i = 1,2,3,4)$, are not acceptable wavefunctions for the molecular ion, as they do not respect the $T_\mathrm{d}$ symmetry of the molecular ion (the equilibrium geometry of the molecular ion is not $T_\mathrm{d}$, but PES is a "fast" technique and the geometry must remain the same during ionisation - in other words, Franck–Condon principle).
I can make an educated guess about what exactly "respecting the symmetry" entails, but I'm not 100% sure, so it will be the subject of a separate question, and I will edit it in here if/when it gets answered. Anyway, according to Hiberty and Shaik (who wrote the chapter), a group theoretical analysis will show that the acceptable linear combinations transform as $\mathrm{A_1} + \mathrm{T_2}$, which have different energies. Hence, the PES can still be rationalised using a hybrid orbital approach.
A refutation of the "failures" of VB theory is given in much more detail by Hiberty and Shaik, in Chapter 5 of their other book A Chemist's Guide to Valence Bond Theory. But I'll have to leave that to somebody else to describe, as I don't fully understand it myself.