Symmetry of an FO

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: and also used the character table to assist me in assigning symmetry. I assigned the axes according to the principal axes, which was in the z-direction: 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: • 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. Dec 30 '19 at 9:35
• @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? Dec 30 '19 at 12:20
• yes; the $C_2$ is in the xy plane as shown in the answer by vik1245 below. Dec 30 '19 at 16:30
• @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. Dec 30 '19 at 17:56
• What are you using FO as an abbreviation for? I've never come across that abbreviation before and I can't find it online.
– atbm
Dec 31 '19 at 7:42 