Why do we observe much stronger (in most cases complete) fluorescence quenching by $\ce{Pd^2+}$ than $\ce{Pt^2+}?$ I work with complexes of both metals and there is no real 100% explanation I can find why this is the case. I am looking for a full quantum mechanical explanation (if possible). If you know a piece of literature, please point me towards that.
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2$\begingroup$ Quenching of what molecules fluorescence ? Do you mean fluorescence from complexes with these metals or diffusional quenching? $\endgroup$– porphyrinCommented Mar 11, 2021 at 8:39
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$\begingroup$ @porphyrin yes I mean quenching of complexes formed with these metals $\endgroup$– TheChemistCommented Mar 11, 2021 at 9:59
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$\begingroup$ It would help if you indicated what these compounds are, but I suspect that the difference is due to relative positions of energy levels such as eg & t2g relative to those in the (presumably) aromatic ligands. $\endgroup$– porphyrinCommented Mar 11, 2021 at 14:26
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1$\begingroup$ @porphyrin yes you guessed right; if it’s due to eg and t2g positions: how exactly this plays a role would interest me. $\endgroup$– TheChemistCommented Mar 11, 2021 at 14:31
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1 Answer
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The theory of this quenching effect was developed about $70$ years ago by Förster, who showed that the quenching is not due to diffusion or collision. It is due to a resonance phenomena. The fluorescence yield is quenched by a solute whose mirror image of the adsorption spectrum overlaps the emission spectrum of fluorescence. It is is proportional to the overlap integral of these two spectra.
The quenching effect due to an agent X on the fluorescence is given by Stern-Volmer equation : ${I_0/I = 1 + \gamma [X]}$. Now T. Förster has shown that the quenching constant $\gamma$ is equal to $\gamma = \frac{2\pi ^2}{3} \frac{N}{1000} R_0^3$, where $N$ is Avogadro number, and $R_0$ is the critical Förster distance, where $R_0^6$ is proportional to the overlap integral $J(\nu )$ of the two spectra, and to the sum of all energy transfer rate constants (solvent to solutes and solute to solute).
This theory will not be developed here. But, as you are looking for a piece of literature, I recommend the following references, where this theory is explained.
See : T. Förster, Discuss. Faraday Soc. $27, 1, (1959)$ and : T. Förster, Ann. Physik $2, 55, (1947)$.
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$\begingroup$ Is this also true for the complexes formed with these metals? $\endgroup$ Commented Mar 11, 2021 at 10:00
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1$\begingroup$ @Maurice, Resonance/Forster/dipole-dipole quenching can also be diffusional controlled, and it is not the only type, there is also Dexter, electron transfer and heavy atom quenching. $\endgroup$ Commented Mar 11, 2021 at 10:11
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$\begingroup$ @porphyrin. The OP wanted references. I have given him what he wants. I should have said that Förster theory is well adapted to compounds having an absorption and/or emission spectrum in the region between$ 250$ nm to $300$ nm. Diffusion quenching also happens in the other situations, but it is weaker. Please give him references for other types of quenching. $\endgroup$– MauriceCommented Mar 11, 2021 at 10:20
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1$\begingroup$ @porphyrin yes, I so understand quenching but why the difference between Pd and Pt. There is obviously the relativistic effect which is stronger for Pt than Pd but I doubt that is the real reason. $\endgroup$ Commented Mar 11, 2021 at 10:32
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1$\begingroup$ because quenching is a before and after measurement, i.e molecule yield then molecule yield plus quencher. The OP has clarified that these are complexes so we compare different molecules not the same molecule with and without a quencher. $\endgroup$ Commented Mar 12, 2021 at 8:30