I heard that in order for two atomic orbitals to form a bonding molecular orbital, they need:

Similar energy

Similar symmetry

and Possibility of overlap with the same sign

but, for example in methane, the carbon's P orbitals, with T1u symmetry, form molecular bonds with the hydrogen's S orbital, with A1g symmetry. If the orbitals need the same symmetry to bond, how does this work? Those are different mulliken labels.

  • 4
    $\begingroup$ This statement is somewhat inaccurate. If you're looking at the tetrahedral point group, then you're not really looking at the atomic orbitals of hydrogen in isolation. You're looking for linear combinations of those 4 orbitals that map onto the irreducible representation of the group. After that, then yes, they need the same symmetry to interact. $\endgroup$
    – Zhe
    Commented Nov 8, 2023 at 18:26
  • 1
    $\begingroup$ If the H atoms are represented as 4 vectors at the corners to the centre of a tetrahedron they combined produce A1 & T2 symmetry species. The H atoms can combine (as LCAO's) with the 2s C orbital (A1) and the three 2p orbitals, which in Td belong to T2 sym. species. $\endgroup$
    – porphyrin
    Commented Nov 8, 2023 at 19:33
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    $\begingroup$ (1) There are no g and u labels in tetrahedral geometry. (2) There are four H 1s orbitals, and collectively they are A1 + T2, and only the T2 bit of it overlaps with the carbon p orbitals. $\endgroup$ Commented Nov 8, 2023 at 23:18
  • $\begingroup$ I'm counting these as answers, thanks. $\endgroup$
    – AdamT
    Commented Nov 8, 2023 at 23:38
  • $\begingroup$ This post appeared to require clarification, which it obtained in comments. However this could have received a proper answer rather than comments. $\endgroup$
    – Buck Thorn
    Commented Nov 9, 2023 at 10:04


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