# How is Pi antibonding orbital more polar than Pi bonding orbital?

While reading about the solvent effect on electronic transition in ESR spectroscopy, I found it written somewhere that the pi anti-bonding orbitals are more polar compared to pi bonding orbitals.

Can somebody explain how and why is it so?

• I am also new to this theory, but I feel that $\displaystyle \pi ^*_{ABMO}$ has a nodal plane in between the nuclei of atoms, and also has greater energy than normal atomic orbital whereas $\displaystyle \pi _{BMO}$ is of lower energy and has maximum electron density between nuclei. Thus $\displaystyle \pi ^*_{ABMO}$ is more polar – Nikola Alfredi Aug 20 at 4:56
• We can also take into account, the high energy electronic repulsion in $\displaystyle \pi ^ *_{ABMO}$. – Nikola Alfredi Aug 20 at 4:57
• Yeah, this must be it. Thank you :) – Sanu_012 Aug 25 at 3:55