Say I promote a molecule to a triplet state via repeated electronic excitation events. Here, this is going to mean that an electron from the HOMO level, originally with a spin-UP and spin-DOWN electron, will be promoted to the LUMO level such that its spin is the same as the electron left-over in the HOMO level (we can then use the Pauli exclusion principle to partially understand that the triplet state will generally have a longer lifetime than the typically higher energy singlet state). The classic example of a molecule in a triplet state is oxygen (though here, this is also the ground state, which is somewhat unusual):


Prior to returning to the ground state via phosphorescence or a reverse intersystem crossing event, let's say that the triplet $T_1$ state molecule reacts with a reducing agent via an exothermic bond formation process, and that the reaction is reversible. Consider that there are numerous instances where triplet state reactivity is higher than ground state reactivity.

The following two things are unclear to me:

(1) When the reducing agent dissociates, is the molecule in the $T_1$ triplet state or the $G_0$ ground state? If the molecule is still in the triplet state, can it enter the ground state while remaining bound to the reducing agent?

(2) Call the energy gap between the relevant $T_1$ triplet state and the $G_0$ ground state $E_1$. Call the energy released into a surrounding thermal bath by the reducing agent reacting with the molecule in the $T_1$ triplet state $E_2$. Must $||E_2|| > ||E_1||$?

My guess is that the pair of electrons donated by the reducing agent will necessarily set $||E_2|| > ||E_1||$ and that the molecule will be left in the ground state upon dissociation of the reducing agent since the electrons will be reallocated to properly satisfy spin-pairing requirements.

  • $\begingroup$ Since you stated it is an exothermic reaction, then ||E2||>||E1||. $\endgroup$
    – LDC3
    Commented Mar 29, 2014 at 5:04
  • $\begingroup$ @LDC3 Sorry, I should clarify that I meant "exothermic" with respect only to the reaction between the molecule in the $T_1$ state and the reducing agent. I.e. if we had a population of molecules in the triplet state and introduced the reducing agent, a thermal bath containing these molecules would heat up. We're not accounting here for the energy required to originally "push up" the molecule to the triplet state (i.e. $E_1$). $\endgroup$
    – 1Q86
    Commented Mar 29, 2014 at 5:06
  • $\begingroup$ As to your terminology and notation: What do you mean with "fluorescent excitation"? Emission from $S_1$ or higher does not play a role here: The excitation is mostly just $S_0 \rightarrow S_1$, followed by ISC to a triplet state with lower energy. And what do you mean with "repeated events". I've never seen $G_0$ to denote a ground state. $\endgroup$ Commented Mar 29, 2014 at 6:51
  • $\begingroup$ @KlausWarzecha I should have just said "optical excitation", I didn't mean to imply that there was a relevant emission event. We're just pushing something like a dye from the ground state to the excited state repeatedly until we have an intersystem crossing event to the triplet state. $\endgroup$
    – 1Q86
    Commented Mar 29, 2014 at 8:14
  • 1
    $\begingroup$ @KlausWarzecha Done and done. :) $\endgroup$
    – 1Q86
    Commented Mar 29, 2014 at 8:24

1 Answer 1


We're looking at a case of PET (photoinduced electron transfer) with electron back transfer between an (electron) acceptor A and a donor D here.

For organic dyes, the ground state typically is a singlet state ($S_0)$, excitation of A yields an excited singlet state (typically $S_1$).

$\ce{A ->[h\nu]\ ^1A^{\ast}}$

Quinone based dyes (or aza-BODIPY derivatives) often have a high triplet quantum yield, the following ISC (intersystem crossing) to a triplet state is efficient then.

$\ce{^1A^{\ast} ->[ISC]\ ^3A^{\ast}}$

In the excited (triplet) state, electron transfer from a (ground state) donor may occur by filling the lower SOMO of the excited species:

$\ce{^3A^{\ast} + D -> A^{\bullet-} + D^{\bullet+}}$

Is the acceptor radical anion still in a triplet state? No, since there no longer an electron pair with opposite spins.

Q: So, what is this good for?

A: By electronic excitation, we managed to facilitate a redox reaction that would not have been possible in the ground state and generated a radical ion pair.

Radical ions are reactive intermediates.

For a spectroscopist, it can be favourable if the only reaction of the radical ion pair is electron back transfer:

$\ce{A^{\bullet-} + D^{\bullet+} -> A + D}$

This regenerates acceptor and donor in their former spin states, which - as mentioned before - are typically singlet states in the case of organic molecules.

Practically, this means that no permanent bleaching occurs. You can excite the sample with laser pulses again and again to measure the transients in time-resolved uv-vis spectrosopy. + From a preparative perspective, this is the worst case: there is no product being formed.

We have examined more productive cases of PET in the context of PET-initiated radical cyclizations of isoprenoid polyalkenes a while ago and published about these, e.g. in J. Phys.Chem A, in J.Chem. Soc., Faraday Trans. and in Pure Appl. Chem.. (The last two should be freely accessible).

PET cyclization of 1,1-dicarbonitrile

In the course of these reactions, an alkene is oxidized to the corresponding radical cation by the excited state of an electron acceptor. The radical cation traps a nucleophile (water, alkanol), immediately deprotonates and thus form a radical, which undergoes 5-exo-trig cyclization to a stabilized dicyanomethyl radical. The latter is reduced to its corresponding anion by $A^{\bullet-}$ and eventually protonated to yield the product (as a pair of separatable diastereomers).

  • $\begingroup$ So in any case, we lose the triplet state, meaning that $||E_2|| > ||E_1||$ right? Could you explain a little bit more about what you mean by: "From a preparative perspective"? As you might guess, my background is not in chemistry. $\endgroup$
    – 1Q86
    Commented Mar 29, 2014 at 9:00
  • $\begingroup$ @1Q86 I will explain the spectroscopic and preparative aspects a bit later - at the moment I'm busy enjoying the wonderful weekend weather :) $\endgroup$ Commented Mar 29, 2014 at 16:55
  • $\begingroup$ Please enjoy the nice weather! It is raining here unfortunately. $\endgroup$
    – 1Q86
    Commented Mar 29, 2014 at 22:02
  • $\begingroup$ Also, I think I know what you mean by preperative in this case, you mean the situation where we're aiming to use PET to generate a population of radical dye/etc. anions by irradiating this dye groups in the presence of a reducing agent. As such, electron back transfer will reduce the yield. Is this right? Just to say, my interest is mostly to understand what's going on, I don't have a particular reaction in mind that I'd like to optimize. As such, given what you've said, a confirmation for the answer $G_0$ / "no" for my 1st question, and "Yes" for my 2nd question would be very helpful. $\endgroup$
    – 1Q86
    Commented Mar 29, 2014 at 22:10
  • $\begingroup$ Can we assume that the radical anion acceptor has lower free energy (i.e. is more thermodynamically stable) than the original singlet ground state of the acceptor prior to interacting with the donor? $\endgroup$
    – 1Q86
    Commented Mar 30, 2014 at 21:22

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