[Lynda92]

the molecular orbital digram of methane shows that there are 4 molecular bonding orbitals.but one out the three bonding orbitals is of a lower energy and is explained to have a nodal plane.
I don't understand why only one of the molecular bonding orbitals has a node, when carbon's valence atomic orbitals has two p orbitals each with a nodal plane.

MO diagram of methane is on 2nd page:

http://www.users.csbsju.edu/~frioux/h2bond/MethaneMOBonding.pdf   

[orgopete]

I'm not sure if I understand your question. As I understand the methane MO diagrams in the pdf, Professor Rioux is arguing that if methane were sp3, then all orbitals of methane should be identical. The electron ionization spectrum of methane has two peaks. By implication, methane must have promoted one electron from a 2s² orbital to a 2p_z orbital. Therefore it is showing methane with hydrogen has s-s, s-p, s-p, and s-p bonds to explain the electron ionization spectrum.

As a parenthetical comment, I have been thinking about this type of problem for some time. A factor that I remain skeptical of is whether we really know what we think we know. Since s and p refer to sharp and principle for the emission lines, obviously hybridization should not have those lines. A problem with the supposed shapes of s and p orbitals is that they should give a different structure than exists for methane. One can take and determine the hydrogens of methane are identical (I think) and one can measure the ionization spectrum (it is data, it must be true). The question one can think about is how do you explain it. Rioux is arguing hybridization fails and I remain skeptical that methane is 2s¹ and 2p³. How do we know the ionization spectrum only shows the valence electrons? Are there any other explanations?

[Enthalpy]

I would have expected claims different from that.

I expect the lowest valence orbital to be the one without a nodal plane - if I understand the terms properly. That is, this lowest is the one that has a constant sign over the whole molecule. The three others have both signs, hence pass by zero through some curved surface, which is a plane for an isolated atom, and approximately one in methane.

An isolated carbon has three 2p orbitals, of which two are occupied, in addition to two electrons filling the 2s. Well, few people have already seen an isolated carbon atom in its ground state, as this needs difficult conditions.

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I wondered as well if the spectrum peak of higher energy were from 1s, but the clear answer is NO. Because 1 proton puts the 1s energy at -13.6eV, so 6 protons would put it (for a single electron, sure) around -500eV, not near -20eV as the spectrum shows. Clearly from valence.

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The sp3 hybridization has a serious drawback, in addition to this spectrum: the weighted sum of the 2s and 2p orbitals called sp3 is not a stationary solution. It oscillates over time and must radiate light - it does, in fact: lithium has a 2p to 2s line. Sp3 is not an eigenfunction of the energy operator. It is not a stable orbital. It won't appear on a spectrum.

In contrast, the four molecular orbital sketched in the Pdf are independent from time.

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It is important to understand that these four molecular orbitals do not correspond to one C-H bond each. It is most obvious at the lowest orbital (left drawing) which is identical in all 4 directions. The remaining three are identical in all 4 directions as well: they only change their sign at one of three "planes" of symmetry.

Did you notice? The axis of each carbon's 2p differs from the directions of the hydrogens. If you like crystals, the 2p are in 100 directions and the hydrogens in 111.

Welcome to the world of quantum mechanics... No, I'm not at home there neither. And it's disturbing for everyone.

Also useful to understand: you can't distinguish between electrons. So you don't have to (and can't) tell to which C-H bond one electron pertains. You can only tell "this orbital is occupied", which is the assertion that has experimental consequences.