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The hint from my instructor for the answer is related to Jablonski Diagram.

The scenario: using fluorometer to measure fluorescence of a compound. How would my fluorescence results differ when dissolve analyte in water vs. water + glycerol (1:1)?

Glycerol is viscous, so the analyte will have less freedom of motion. This I got it right. But the answer is using water + glycerol would icrease intensity of my fluorescence compared to just water.

I am stuck with understanding why it increases?

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    $\begingroup$ Freedom of species to move and recombine before releasing energy as light? $\endgroup$ Commented Feb 19 at 23:50
  • $\begingroup$ What did you search? ChatGPT 4 provides a reasonable answer. I asked "Act as an expert in fluorescence spectroscopy " as a prompt, and then wrote "If we add glycerol to an aqueous solution of a fluorescent compound, should we expect higher fluorescence or lower?" See what answer and an equations you get. $\endgroup$
    – AChem
    Commented Feb 20 at 5:43

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What is being favored in water is probably the internal conversion (non-radiative decay), therefore, when going to a less polar mixture or a non-polar solvent, this pathway is less favored. Therefore, although fluorescence is not directly favored, as these two are competitive processes, the fluorescence quantum yield increases (hence its intensity as well).

Depending on what your fluorophore is, its chemical structure, polarity and the charge distribution in the excited state; different processes can occur. It is easier to rationalize it between what happens between a polar solvent and a non-polar solvent.

Polar solvents, such as water, have a high capacity to interact with polar molecules, such as fluorophores. These interactions can dissipate the energy of the excited state of the fluorophore through nonradiative mechanisms such as collisional deactivation.

It also influences what the charge distribution of your flurophore is in the excited state. In polar solvents, the solvent molecules rapidly reorient around the fluorophore in their excited state. This reorientation can generate a fluctuating electric field that perturbs the excited state of the fluorophore and facilitates internal conversion.

Additionally, if your fluorophore is very low polar, interaction with the polar solvent can increase the ground state energy of the fluorophore, reducing the energy difference between the excited state and the ground state. This lower energy difference facilitates internal deactivation by mechanisms such as internal conversion.

Since these processes are very fast, I do not think that the viscosity of the solvent plays a relevant role. Still, this parameter could be influencing the nonradiative decay constant of the solvent where the effectiveness of the collisions of the solvent and the excited state of the fluorophore are crucial.

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It really is a type of trick question because the answer depends on what the fluorescing molecule is, and whether it has high/low fluorescence yield in water/ organic solvent. Some aromatic sulphonates (e.g.ANS) have low yield in water, high in organic solvent, many other compounds hardly change yield at all with solvent. The viscosity really does not come into it unless for special cases of triphenylmethanes (malachite green for example). Glycerol it turns out is microscopically not as viscous as its bulk viscosity would suggest. A small amount of water decreases glycerol's viscosity very greatly from that in its pure state.

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