# 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. – porphyrin 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? – vik1245 Dec 30 '19 at 12:20
• yes; the $C_2$ is in the xy plane as shown in the answer by vik1245 below. – porphyrin 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. – vik1245 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