[SrV]

Pyridine exhibits a UV transition of the type n  Π* at 270 nm. In this transition, one of the unshared electrons on nitrogen is promoted from a nonbonding MO to a Π*-antibonding MO. What is the effect on this UV peak if pyridine is protonated?





The answer states that:
When the pyridinium ion is protonated, the lone pairs of electrons on nitrogen is tied up in a bond to hydrogen.  As a result, it is lower in energy compared with the unprotonated form, so it takes more energy to promote one of these electrons into Π* orbital.  Therefore, protonation shifts the absorbance peak to a lower wavelength (higher energy).

From this what I visualized in my mind was unprotonated electron was elevated to the LUMO.

Thank you

[Corribus]

When a nonbonding orbital becomes bonding, it is stabilized, because bond formation releases energy.  The antibonding pi orbital energy is not affected by this process.

The absorption wavelength of the unprotonated form is equal to the difference in energy between the π^* orbital and the nonbonding lower energy orbital.  Call the energy of the nonbonding lower energy orbital x and the energy of the π^* y.  We'll call the energy of the new bonding orbital after protonation x'.

ΔE for the unprotonated form is therefore y - x.  ΔE for the protonated form is y - x'.  y is the same in both cases, but x' < x because we've said the HOMO becomes stabilized.  Therefore ΔE (unprotonated) > ΔE (protonated), which is a blueshift during protonation.