So a charge transfer transition is unlike the charge delocalization?
In common parlance, charge transfer is a process
, whereas charge delocalization is a state
. I.e., the stability of the phenoxide
anion is often attributed to the ability of the negative charge to be delocalized over the phenyl ring. On the other hand, charge transfer can result in charge delocalization. For instance, it is well known that when bipheny
, is reduced, the two phenyl rings become coplanar, resulting in delocalization of the transfered charge over a larger area. Which is to say, the two phenyl groups in neutral biphenyl are largely independent, but in the anion (or cation) radical (or, for that matter, in the photoexcited neutral molecule), they are well-conjugated. (Ref: Almennigen, et al, J. Mol. Struct. 1985
, 59-76.) To put that in a photophysical context, we can turn again to the metal polypyridyl literature. [Ru(bpy)3
is a metal polypyridyl complex in which three bipyridine ligands are coordianted to a ruthenium dication. When photoexcited, an electron is transfered from the ruthenium metal to one of the three bipyridine ligands, whereupon the charge is delocalized over the entire ligand. This metal-to-ligand-charge-transfer (or MLCT) state gradually relaxes predominantly by luminescence back to the ground state on a microsecond timescale. Now what happens if we functionalize the bipyridine ligands with phenyl groups? In the ligand, the phenyl group is canted with respect to the bipyridine plane due to steric-hindrance of the adjacent hydrogens, much like the two phenyl groups in biphenyl. When [Ru(4,4'-diphenyl-bpy)3
is photoexcited, an electron is again transfered to the ligand. Initially this transfered charge is only delocalized over the bipyridine ring. Over a time scale of ~2 picoseconds, however, the phenyl groups rotate, allowing the charge to become further delocalized. Evidently, the energy gained by delocalizing the charge is larger than the energy cost of having those hydrogens bumping up against each other. You can read more about this at the following reference, if you wish: Damrauer and McCusker, J. Phys. Chem. A
, 103, 8440-6.
Any part of a molecular structure that facilitates charge transfer transition will negatively influence triplet yield.
With the usual acknowledgement that there are probably exceptions out there, and as long as we are understanding the triplet yield refers to a triplet pi-pi* state, yes.
In reality, particle-on-a-ring treatments of porphyrins using a 20 electron-model is incorrect, because it would imply that porphyrin is paramagnetic in its ground state. That's absolutely wrong because virtually all organic molecules are diamagnetic in their ground state.
I would soften the language. I wouldn't call it incorrect - I'd call it a poorer model than the 18 electron model. Also, I wouldn't say "virtually all", I'd say a "vast majority". In chemistry, you learn quickly to always leave room for exceptions. As soon as you say "all" or "none", someone will point out a counter example in about 5 seconds.
we need complicated mathematics to do that.
The mathematics are not so complicated to write down, but exact answers do not exist, so many approximations have to be made to find workable solutions. We can easily understand why there is a splitting, but quantitating them with any accuracy usually requires computers and sophisticated modelling programs. Semiempirical methods (parameter models based on experimental data) have can also be used with some success.
Dioxygen molecule is a triplet in its ground state. It means the molecule is paramagnetic when it is not excited. If porphyrin were also paramagnetic in its ground state, a spin-allowed reaction would take place between ground-state dioxygen and ground-state porphyrin. This doesn't happen in real-life situations.
Well, it does. The hemoglobin in your body binds oxygen, doesn't it?
I guess to be pedantic, this is oxygen reacting with the open-shell iron center, not the porphyrin itself, but the porphyrin does play a role. Beyond that, it's better to speak in terms of fast rates and slow rates (or, alternatively, reaction probabilities) rather than what does or does not happen. Singlets can and do react with triplets - the reactions are just slow. "
This point about softening your language also carries through to your other statements as well. Generally your summary is good - you just have to be careful about speaking in absolutes.