[Winga]

For a compound having n, pi & pi* orbitals (e.g. carbonyl group), it can undergo n-->pi* and pi-->pi* tansitions.
The absroption wavelength (lambda maximum) of n-->pi* is longer than that of pi--pi* as the energy gap between n & pi* is smaller than pi & pi*.

Under polar solvent, the absorption of n-->pi* will shift to shorter wavelength while for pi-->pi* will shift to longer wavelength because the polar solvent stabilizes these three orbitals in different extent, n > pi* > pi.

The reason for this stabilizing effect from my lecture note is that the pi* orbital is more molar than pi orbital (just concern about pi & pi* at this moment) as polar solvent stabilizes polar substances more.
However, I don't know why pi* orbital is more polar.

In my opinion, there is interaction between pi/pi* orbital and orbitals from polar solvent molecules which generates new MOs which are more stable. Moreover, the interaction is much better for pi* orbitals than pi orbital.

[Mitch]

Draw a pi* orbital you'll see its much more polar than a pi orbital.

[Winga]

I know their shapes, but I don't know what governs the dipole moment of the orbitals.
(It seems that there is no net dipole in pi* orbital)

[Mitch]

If you know the shape of pi* then you would know that carbon has a large amount of electron density and oxygen much less. So there would be a dipole moment.

[Winga]

Mmm...I know that too, but I think I am asking alkenes not C=O.
So, for C=O, how can I determine that pi* is more polar than pi as pi-orbital is also polar?

By the way, for alkene pi-->pi* transition, there should be no solvent effect, right?

[Mitch]

I don't think there would be a solvent effect for alkenes. A bonding pi orbital can't be that polarized since their shaing electron density.

[Winga]

If you know the shape of pi* then you would know that carbon has a large amount of electron density and oxygen much less. So there would be a dipole moment.

However, larger lobes of pi* orbital on carbon side doesn't mean that the electron density is higher at carbon atom, right?

[Mitch]

I thought that was what it meant.

[Winga]

For a carbonyl ligand, the reason for which atom attaching to a metal centre is that the shape of the HOMO having a larger lobe on carbon side, that mean more e- density on carbon side?

[Mitch]

Yes, thats what I thought.