What is the mechanism involved in the generation of singlet oxygen when ground state triplet oxygen is irradiated in the presence of rose bengal?

  • $\begingroup$ Dear Homework-closers: What would you have expected from OP to keep this question from being closed? Voting to reopen. $\endgroup$ – Jan Aug 28 '16 at 13:46
  • $\begingroup$ I would expect the questioner to state what they know about the mechanism so far, or at least what sources they have consulted so far. $\endgroup$ – Curt F. Sep 6 '16 at 3:09
  • $\begingroup$ Well now we've come full circle... In light of our experiment I would not exercise my close vote here. [Although in my opinion this is a borderline case. While the intention is clearly not so, it does read like a homework dump.] $\endgroup$ – orthocresol Sep 6 '16 at 4:29
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    $\begingroup$ Just because a question is not actual homework doesn't mean you can be lazy and leave out your current findings and anything else that would help anwering it, like for example showing the structure of rose bengal. It's highly impolite, you are pushing your own work off to people on the internet to do it for you. $\endgroup$ – Karl Sep 10 '16 at 9:03

Rose bengal (pictured below) is a member of the fluorescein family.

structure of rose bengal (image source: wikipedia)

There are 3 steps in the process where rose bengal sensitizes the formation of singlet oxygen..

\begin{gather} \ce{RB ->[$h\nu$]\ ^1RB}\tag1\\ \ce{^1RB ->T[ISC]\ ^3RB}\tag2\\ \ce{^3RB + ^3O2 ->T[sensitization]\ ^1RB + ^1O2}\tag3\\ \end{gather}

In the first step, ground state rose bengal (a ground state singlet) absorbs light producing an excited singlet state. Because of the heavy iodine atoms in the molecule intersystem crossing (ISC) to the excited triplet state rapidly occurs (heavy atom effect). In the presence of ground state triplet dioxygen, a process called "sensitization" occurs. In this step the excited triplet rose bengal transfers its energy to the dioxygen promoting it to the excited singlet state while the rose bengal is deactivated and returns to its single ground state. In this sensitization step, 2 triplets are converted to 2 singlets so, importantly, spin is conserved.


Although the question is about Rose Bengal the mechanism is quite general and is the basis for the first steps in photodynamic therapy as used to attempt cure certain cancers. The answer given by @ron shows the three possible steps in producing excited singlet oxygen molecules. The first condition for reaction is that the triplet of the excited dye molecule is above the energy of the oxygen $\Delta E > 0$ in figure. This is often the case, but not for molecules such as carotenes ( xanthophylls ) which have low lying excited state and so do not undergo these reactions. They are used by photosynthetic organisms to prevent reactive oxygen molecules from doing damage, but in photodynamic therapy we want reaction.

As two triplet states collide in solution and react the total angular momentum for the complexes formed must be conserved and can have quintet, triplet or singlet character but with a different probability for each type. The total spin on each molecule 1 is S$_1$ =1 and similarly for molecules 2, thus the total spin for two molecules is found from the (Clebsch-Gordon)series |S$_1$+S$_2$| ..|S$_1$-S$_2$| with each term separated by unity. Thus the terms give total spin 2, 1, 0 with multiplicities (2S+1) of 5, 3 and 1, or quintets, triplets and singlets, making 9 types of complexes all together. Thus quintets are formed with 5/9 probability, triplets with 3/9 and singlets with 1/9.

If $^3M$ is the excited triplet state of the molecule, and $^1M_0$ the ground state, the schemes for these types of quenching are

$^3M$+$^3$O$_2 \rightarrow ^1(^3M ..^3$O$_2$)$\rightarrow ^1M_0 +^1$O$_2$ (as $^1\Delta +^1\Sigma$) (Energy transfer)

$^3M$+$^3$O$_2 \rightarrow ^3(^3M ..^3$O$_2$)$\rightarrow ^1M_0 +^3$O$_2$ (as $^3\Sigma ^+_g$) (intersystem crossing)

$^3M$+$^3$O$_2 \rightarrow ^5(^3M ..^3$O$_2$)$\rightarrow $ no reaction

The last equation does not occur as it is not possible to form products from a quintet complex as each orbital in M or O$_2$ can have a maximum of only 2 electrons.

Thus both $^1\Delta$ and $^1\Sigma$ excited state can be produced, with total probability of 3/9 of collisions and the process is by energy transfer of the Dexter or exchange type rather than by the Forster of dipole-dipole type.

In solution the $^1\Sigma$ excited state has a shortish lifetime (many nanoseconds) relative to the $^1\Delta$ which is longer lived. Both lifetimes depend heavily on the solvent used. This triplet to singlet intersystem crossing (in O$_2$) occurs under the influence of solvent collisions causing spin-orbit coupling. The scheme below shows the energy levels.

singlet triplet

  • $\begingroup$ "The last equation does not occur as it is not possible to form products from a quintet complex as each orbital in M or O2 can have a maximum of only 2 electrons." Theoretically, why can't one product be in a singlet state and the other in a quintet? The total number of electrons in the system is conserved. $\endgroup$ – pentavalentcarbon Aug 27 '16 at 14:07
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    $\begingroup$ To form a quintet state then four unpaired electrons are needed, so four molecular orbitals in the same molecule would have to have unpaired electrons, (instead of two for triplet), in effect a doubly excited state, theoretical yes, in practice no, the energy required would just be too large. $\endgroup$ – porphyrin Sep 1 '16 at 21:45

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