J.D.Lee writes in his book Concise Inorganic Chemistry:

[...] An alternative method for determining the symmetry of the molecular orbital is to rotate the orbital about the line joining the two nuclei and then rotate the orbital about the line perpendicular to this. If the sign of the lobes remains the same, the orbital is gerade, and if the sign changes, the orbital is ungerade.

Now, it can be detected easily which MO, formed due to the overlapping of s-s or p-p orbitals, is gerade or ungerade as shown in these pics:


enter image description here

But what about the molecular orbitals formed due to the overlapping of s-p orbitals or p-d orbitals? mo due to to overlapping of s-p orbitals:

enter image description here

mo due to overlapping of p-d orbitals:

enter image description here

If I rotate about the first orbital about an axis perpendicular to the inter-nuclear axis, the smaller lobe goes to the right only; the sign always remains same; just the position of the lobe has been changed. But in the first picture, only the bonding mo is gerade while the anti-bonding mo is ungerade; I'm not getting how by rotating along the axis perpendicular to the inter-nuclear axis, the sign of the antibonding mo changes making it ungerade.

Can anyone please help me how to apply the procedure for the molecular orbitals formed due to the overlapping of s-p & p-d orbitals?

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    $\begingroup$ Those MOs simply cannot be labelled as g or u. MOs of non-centrosynmetric molecules (i.e. molecules that do not possess a centre of inversion) cannot be labelled g or u. Look at the MO diagram of water (C2v) - none of the orbitals have g or u labels. $\endgroup$ – orthocresol Dec 28 '15 at 17:48
  • $\begingroup$ @orthocresol: So, do you mean to say those in the pic cannot be labelled g or u? $\endgroup$ – user5764 Dec 28 '15 at 17:53
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    $\begingroup$ That's right, they cannot be labelled as such. $\endgroup$ – orthocresol Dec 28 '15 at 17:55
  • $\begingroup$ @orthocresol: But the author did label those orbitals as g & u :/ $\endgroup$ – user5764 Dec 28 '15 at 17:57
  • $\begingroup$ I think the point of all this is about MO orbital symmetry, nodes, bonding and antibonding. For these asymmetric examples, consider the orbital lobes between the nuclei if they are sigma- or pi-overlap, and if the orbital formed between them has genade or ungerade symmetry. $\endgroup$ – Spontification Dec 28 '15 at 22:49

The better way to do it is to check what happens under inversion ($i$ or $\bar 1$). If the orbital stays the same, it is g, otherwise u.

However, as Orthocresol mentioned in the comments, checking that is only possible if the entire molecule contains inversion symmetry. Not all point groups do, and the vast majority of molecules does not (partly because $C_1$ is likely the most prevalent point group out there).

For example, consider transition metals’ d orbitals. In octahedral complexes, they are labelled $\mathrm{t_{2g}}$ and $\mathrm{e_g}$. In tetrahedral complexes, which do not have inversion symmetry, they are $\mathrm{t_2}$ and $\mathrm{e}$.

  • $\begingroup$ Hmmm... could you tell what point group is? Also, if there is no inversion symmetry in those orbitals, why did Mr. Lee labelled those orbitals as $g$ and $u\;?$ $\endgroup$ – user5764 Dec 29 '15 at 12:04
  • $\begingroup$ @user36790 If you check out which symmetry features a molecule has, you realise that these happen to form a mathematical group, i.e. there’s identity, there’s an inverse element to each element and when adding two operations together, they exhibit commutativity. Only a handful of different groups actually exist, and since they help so much with symmetry considerations, they are taught in second or so year of chemistry studies. I’ll check out if I find a question here on chem.SE about them ^^ $\endgroup$ – Jan Dec 29 '15 at 12:17
  • $\begingroup$ So, should I say the author is not right in marking those orbitals as $g$ and $u$? $\endgroup$ – user5764 Dec 29 '15 at 12:27
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    $\begingroup$ @user36790 Yes, it does not make sense labelling something as g or u if there is no inversion present in the molecule. $\endgroup$ – Jan Dec 29 '15 at 12:29
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    $\begingroup$ Thanks for the confirmation. He confused me for 2 days! I'm going to read about point group in order to understand it better. Thanks, again. $\endgroup$ – user5764 Dec 29 '15 at 12:30

If opposite lobes have inversion centre i.e. the centre of symmetry with respect to phase of wave functions then the orbitals are gerade otherwise ungerade.


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