# Why do ionisations from non-bonding orbitals result in little/no vibrational structure in a photoelectron spectrum?

I understand that ionisations from bonding and antibonding orbitals create vibrational structure from the Franck-Condon principle, resulting in a greater overlap integral between different vibrational levels, but surely if a non bonding orbital is ionised the equilibrium bond length will not change (much), resulting in a good overlap integral between the vibrational levels?

If someone could explain the principle to me in this context I would be very grateful.

• The idea is that your molecule, prior to ionisation, is nearly always in a vibrational ground state. The intensity of the line is therefore governed by the overlap of the vibrational level of the ionised molecule with the ground-state vibrational level of the unionised molecule. For ionisation from a nonbonding orbital, this is largest for $v' = 0$ i.e. transition to vibrational ground state of ionised molecule; from then on the overlap (and intensity) just decreases – orthocresol May 13 '16 at 15:55
• whereas for ionisation from bonding/antibonding the Franck-Condon factor is largest for $v' \neq 0$ - this means that your intensities go up to a certain point, and then only decrease – orthocresol May 13 '16 at 15:56