[lbjohnson]

What is the best way of determining whether a compound will increase luminescence via the antenna effect on a lanthanide complex?

For example, I know organic units increase luminescence such as 2,2-bipyridine.
However if there was a cobalt ion coordinated to this 2,2-bipy and also coordinated to two bromide ions, how would this affect the luminescence? Would I have to consider crystal field splitting or ligand to metal charge transfer?

How would the luminescence change with different metal ions, for example magnesium, zinc or copper? Is it as a result of the differing d-electrons?

Picture attached.

Any help or guidance would be much appreciated! Thank you very much 

[Corribus]

If you have an open shell d-block metal coordinated to your absorber, I think there will be a good chance of quenching. Either you will get charge transfer from the metal to the ligand (or vice-versa), which can be a dark state, or simple nonradiative deactivation via empty d-orbitals.

[lbjohnson]

If you have an open shell d-block metal coordinated to your absorber, I think there will be a good chance of quenching. Either you will get charge transfer from the metal to the ligand (or vice-versa), which can be a dark state, or simple nonradiative deactivation via empty d-orbitals.

Hi thank you very much.
So a d-block metal that had a full shell of d-electrons (i.e. zinc²⁺) wouldn't cause quenching?
And when a d-block metal such as zinc²⁺ is coordinated, would it cause an increase in luminescence as the ligand is more rigid? And therefore has a greater quantum efficiency?
Can't really find much about this online unfortunately.

[Corribus]

Zinc usually won't completely quench the luminescence of a bound ligand. E.g., compare the fluorescence yield of a porphyrin bound to zinc (3.3%) vs that of a porphyrin bound to iron (<<0.1%).  Do note that the zinc does lead to some quenching. The fluorescence yield of free base (not bound to anything) porphyrin is about 13%. The quenching in the case of zinc is predominantly due to spin-orbit coupling (heavy atom effect), which increases efficiency of excited triplet state generation.

I would say that in most cases a luminescent ligand is going to be fairly rigid already (it has to be, to have a decent fluorescene yield), so the effect of increasing rigidly when binding to a metal on the fluorescence yield will be small, but I could imagine scenarios where this occurs. I could also imagine scenarios where the opposite would occur (binding of metal puts ligand in unfavorable conformation for fluorescence). These effects would have to be taken on a case by case basis - I think it's hard to speak generally. Keep in mind also in your case of interest, the luminescence comes from the bound lanthanide ion, not the ligand itself, so rigidity of the ligand wouldn't matter so much except for influencing the excited state lifetime of the ligand, which would increase the likelihood of energy transfer to the lanthanide.

Ultimately, it all comes down to relative rates. If your rate of energy transfer to the lanthanide is extremely fast, then it will remain competitive even with a nearby metal ion that can facilitate other types of excited-state relaxation. If your rate of energy transfer is slow, then the effect of a nearby metal ion would be comparatively large. It's been quite a while since I read any literature on lanthanide fluorescence sensitization from bound antennae, so I'm not sure off the top of my head what kind of rates we're dealing with, and what kinds of factors are necessarily important to increase energy transfer efficiency. But to answer your question: yes, in general a closed shell metal ion like zinc will quench the luminescence far less than an open shell metal ion.