Consider the molecular orbital diagram of methane, for example found here:enter image description here

I would like to know what happens with the two 1s orbital electrons of carbon in the molecular orbitals it is forming. Is it that the sigma 1s* anti molecular orbital is having very high energy so it's not filled and hence the full figure actually would be like I have depicted below?

attempt of MO diagram for methane including 2s electrons

This (my figure) also tells intuitively why in the above shown figure on starting electrons from the valence shells are getting filled in the sigma 2s bonding orbital first?

(Note : I don't want a deep explanation related to formation of possible combinations of bonding, antibonding orbitals on this case as I just know the knowledge sufficient for molecules of the form like $\ce{N2}$, $\ce{C2}$, $\ce{O2}$, etc.. I do not know about this polyatomic type, I was just intuitively and a bit logical trying to explanain it with respect to the MO diagrams of molecules which I do know. From that I am considering to explain this filling order.)

  • $\begingroup$ I have removed the embedding of an external image and tried to fix as much of the language issues I could comprehend. I could not find any copyright information of the image, so I did not include it here. If you want to include it, you should convert it to a format that is supported by the Stack Exchange Image hoster, upload it and cite it accordingly. I have left the image for those who are willing to click links. $\endgroup$ Mar 30 at 20:39
  • $\begingroup$ I have also down-voted this question. Either you are interested in what is a complete and possible explanation, or you want to reaffirm your own believes. You have explicitly excluded the first, which makes this question utterly useless for most of our audience. Unfortunately what you think is incorrect on all levels of theory, which makes it also useless for most of our audience. $\endgroup$ Mar 30 at 20:43
  • $\begingroup$ I understood i will fix the image and @Martin-マーチン what was incorrect on all levels of theory ? May you please tell ? $\endgroup$
    – WizardMath
    Mar 31 at 0:35
  • $\begingroup$ Please cite the original source and make sure you're company with their copyright. With "all levels of theory" I mean from all angles. Your theory is unfortunately incorrect and you need a deep explanation to understand why. $\endgroup$ Apr 1 at 20:28

1 Answer 1


In order for atomic orbitals coming from two different atoms to produce a reasonable molecular orbital, they should have about the same energy. The energy of the 1s electrons of the carbon atom are rather low ($\pu{−37.8 MJ/mol}$), and much lower than the 1s electrons of the $\ce{H}$ atoms ($\pu{-1.31 MJ/mol}$). So their contributions to any molecular orbital are negligible. It looks as if they do not make any molecular orbital.

  • $\begingroup$ May you answer this too regarding this "one thing i was thinking about was even though its very negligible but with which orbitals of hydrogen will the 1s of carbon combine to product molecular orbitals as we can observe that by itself all the four h 1s are foming bonding/anti with the upper orbitals (2s...) of carbons in the MOT diagram"? $\endgroup$
    – WizardMath
    Mar 30 at 14:55
  • 1
    $\begingroup$ $99.9999...99$ % of the four atomic orbitals $1s$ of $\ce{H}$ are forming molecular orbitals with the $2s$ and $2p$ of the carbon atom. The rest, $0.000..01$ % is forming a dim molecular orbital with the $1s$ of the carbon. As a result, this molecular orbital is not different from the $1s$ of the carbon atom. $\endgroup$
    – Maurice
    Mar 30 at 18:45
  • $\begingroup$ You might not like the way the browser is presenting the post, but at the very least it is technically correct markup. That is in the sense that it will not break down with one of the next updates if ever they come. I know that this is unsatisfying, but that's the best on offer right now. In any case, MathJax would not have been necessary at all for this post. $\endgroup$ Mar 30 at 20:46
  • $\begingroup$ Okay thanks @Maurice $\endgroup$
    – WizardMath
    Mar 31 at 0:34

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