I'm looking at a molecule of [Ru(bpy)3]2+ in water and the absorption and emission bands are not symmetrical. I want to know why the mirror image rule doesn't apply here. Is this because of the MLCT, is this something to do with the solvent? Are there more likely to be processes involving non-radiative decay?
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2 Answers
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Mirror image spectra are only observed in solution and then only if the ground and excited state potential energies have almost exactly the same shape. Thus in rigid molecules such as anthracene a good mirror image is observed but not in more flexible molecules such as stilbene ($\phi-C=C-\phi$). The spectra of each vibrational level is wide in solution due to solvent interactions altering energy levels v. slightly, this does not happen in the collisionless gas phase where lines are sharp. In solution the width of lines from different vibrations can overlap making a broad largely structureless spectrum.
There are different effects, a stokes shift due to solvent causing lowering of excited state energy shortly after absorption, a displacement in position of minimum energy, due to different bond lengths in each state and anharmonicity in one potential vs the other. Even slightly different Franck Condon factors between states can mean that a phosphorescence spectrum is similar, but different to its fluorescence spectrum, see figure below, when correction is made for detector sensitivity vs wavelength.
The figure below (not to scale) shows how ideal mirror image spectra are formed. Notice that displacement of potential energy is needed and that vibrational frequency in ground and excited state can be slightly different.
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$\begingroup$ So it's a combination of the structural factors (how "rigid"), solvent effects and electronic factors (Franck Codon)? Very well explained, thank you! $\endgroup$– AbuCommented Jun 2, 2020 at 18:19
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From the Jablonski diagram,
The mirror image is only true if you are talking about transitions from $S_0$ to $S_1$ (absorption) and $S_1$ to $S_0$. Also read about Kasha's rule.
Quinine is the most famous fluorescent molecule, its $complete$ absorption spectrum is not a mirror image of its emission spectrum.
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$\begingroup$ In this case, is there a reason that the metal complex is excited to a higher state and not just to the S1 level? $\endgroup$– AbuCommented Jun 1, 2020 at 19:45
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$\begingroup$ The metal complex gets excited to all possible states, because why not. $\endgroup$ Commented Jun 1, 2020 at 19:58
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$\begingroup$ @Abu, That is exactly what Kasha's rule says. Excitation can occur to various levels but emission always occurs from S1. $\endgroup$– ACRCommented Jun 2, 2020 at 0:22