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I want to ask a question about the symmetry of an FO for $\ce{[CuL2]^-}$ where L is a $1e^-$ donor sigma bonding ligand (e.g. Me).

For the ligand orbitals, I produced the following orbitals shown below:

Ligand Orbitals of CuL2

and also used the character table to assist me in assigning symmetry.

Character table for Dinfh

I assigned the axes according to the principal axes, which was in the z-direction:

Axes of the ligands

This is the working I did for the bonding FO:

  • Under E, no change so give $1$
  • Under $\ce{2C_{\infty}}$ no change so give $1$
  • Under $\ce{\infty \sigma_{v}}$ no change so give $1$
  • Under $i$ no change so give $1$
  • Under $2S_{\infty}$ no change so give $1$
  • Under $\ce{\infty C_2}$ no change so give $1$

Hence the overall symmetry is $\ce{\sigma _g^+}$

However, applying the same working for the antibonding combination above:

  • Under E, no change so give $1$
  • Under $\ce{2C_{\infty}}$ no change so give $1$
  • Under $\ce{\infty \sigma_{v}}$ no change so give $1$
  • Under $i$ change in phase so give $-1$
  • Under $2S_{\infty}$ change in phase so give $-1$
  • Under $\ce{\infty C_2}$ no change so give $1$

The overall symmetry I got was $\ce{\sigma _u^-}$ but according to the diagram, it is $\ce{\sigma _u^+}$.

Where was I mistaken?

The entire MO diagram is shown below for reference:

MO diagram from Patricia Hunt, Imperial College London obtained 29/12/19

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    $\begingroup$ Under inversion 'i' the antibonding orbital is -1, so it has to be 'u' , also the $C_2$ is -1 so this points to $\sigma_u^+$ this being the only possible combination. $\endgroup$ – porphyrin Dec 30 '19 at 9:35
  • $\begingroup$ @porphyrin please explain briefly in an answer if possible. I thought that because the principal axes is the $\ce{z}$ axes going to the right, rotating in a $\ce{C2}$ fashion i.e. 180 degrees on it, there isn't a change in phase. I agree with the inversion as passing through the centre of the molecule and out the other side with the same distance, I discover a change in phase but I cannot see the $\ce{C2}$ being $-1$ Is it that the $\ce{C2}$ lies along the $x$ and $y$ axes and this is the $\ce{C2}$ you are referring to? $\endgroup$ – vik1245 Dec 30 '19 at 12:20
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    $\begingroup$ yes; the $C_2$ is in the xy plane as shown in the answer by vik1245 below. $\endgroup$ – porphyrin Dec 30 '19 at 16:30
  • $\begingroup$ @porphyrin yes yes I answered my own question! Thanks anyway for the help to make me find it! Upvote my answer if you feel it's sufficient to help other users as well. $\endgroup$ – vik1245 Dec 30 '19 at 17:56
  • $\begingroup$ What are you using FO as an abbreviation for? I've never come across that abbreviation before and I can't find it online. $\endgroup$ – atbm Dec 31 '19 at 7:42
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This diagram from Molecular Orbitals by David Willock will be sufficient to explain the matter:

enter image description here

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