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?

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    $\begingroup$ 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 $\endgroup$ Aug 20 '20 at 4:56
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    $\begingroup$ We can also take into account, the high energy electronic repulsion in $\displaystyle \pi ^ *_{ABMO} $. $\endgroup$ Aug 20 '20 at 4:57
  • $\begingroup$ Yeah, this must be it. Thank you :) $\endgroup$
    – Sanu_012
    Aug 25 '20 at 3:55

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