I just came across an article where they said that sodium fluoride is not quencher (in the context of fluorescence). Why is NaF not a quencher but NaBr is one?

  • $\begingroup$ quencher is any species that decreases the intensity of fluorescence when photochemically atoms collide $\endgroup$
    – kibet
    Nov 21, 2023 at 11:57

2 Answers 2


A fluorescence quencher is any species that causes your fluorophore to stop fluorescing. There are several ways this can happen, but there are two broad categories: chemical reactions and energy transfer. I will give simple examples of each category below. However, which method is operating is going to depend on the fluorophore. Bromide and fluoide will not behave the same way with every fluorophore. Some might be quenched by fluoride and not bromide. Some might be quenched by both. Some might be quenched by neither.

However, in the case of the specific fluorophore in the article, it is quenched by bromide and not be fluoride. The article likely provides an explanation or hypothesis about this behavior, or it cites something that does. Likely this article then is about a selective sensing application. Since bromide turns the sensor off (quenches fluorescence) and fluoride does not, the sensor is selective for bromide over fluoride.

Quenching by chemical reaction

Any chemical reaction will change the relative energies and energy spacing of the frontier molecular orbitals, which could lead to quenching. This chemical change could be something irreversible like oxidizing or reducing the fluorophore (probably not what is happening here). Other chemical reactions are possible, and it could be that bromide reacts and fluoride does not.

This change could also be a reversible one, like protonation/deprotonation, ligand binding, etc. If the fluorophore in your article is a transition metal complex, than it could be quenched by complexation with bromie anion (which displaces some ligand that has charge-transfer absorption/fluorescence with the metal). Fluoride might not do this.

This behavior could be more complex. Perhaps the fluorophore only fluorescences in the presence of some analyte. Maybe bromide reacts with the analyte (and fluoride does not) or bromide competes with the analyte but does not promote fluorescence.

Quenching by energy transfer

Fluorescence works as follows. The fluorophore absorbs a photon and is excited from its ground state to its excited state. When the excited state relaxes back to the ground state it emits another photon.

Fluorescence can be quenched if the excited state can be intercepted. An example is as follows. You have a system with your fluorophore and another molecule. Your fluorophore has a smaller HOMO-LUMO gap than this other molecule, so the fluorophore selectively absorbs photons of a certain frequency range and is excited. However, if the other molecule has a lower energy excited state than your fluorophore, then energy transfer can occur between them if they are close. Now your fluorophore is back in its ground state and the other molecule is in its excited state. If this molecule does not fluoresce, then fluorescence is quenched. This is probably not happening in your example. Note this phenomenon can be given various names like "resonance energy transfer" or "Forster energy transfer."

This type of quenching can get more complex as well. The energy acceptor could go and do chemistry or it could be so eager to be reduced that it not only steals the energy from the fluorophore, but it grabs an electron as well. This is not likely what is going on with bromide and fluoride.

The other type of behavior that can go on here is analyte-induced aggregation. Aggregation of the fluorophore causes the electronic properties to smear out over the aggregate, which likely means no more fluorescence (and perhaps no more absorbance) at the frequencies you care about. This phenomena could be happening in your example. Bromide could induce aggregation while fluoride does not.


Fluorescence (or phosphorescence) quenching is the process by which an excited state looses its energy to another species during a collision (in solution or vapour phase) and it always occurs in competition with fluorescence (or phosphorescence) and non-radiative transitions of internal conversion and intersystem crossing. Thus quenching always results in reduced fluorescence yield and lifetime.

Let S$_1$ be the excited singlet S$_0$ the ground state, T the triplet and Q the quencher. The mechanisms are (a) heavy atom quenching (b) energy transfer (c) electron transfer and (d) chemical reaction.

In (a) The heavy atom, bromide ions, iodide ions, krypton, xenon etc. causes increased non-radiative transitions so S$_1$ crosses to T with a greater rate constant and so the fluorescence yield falls as does the fluorescence lifetime.The quenching is caused by increased spin-orbit coupling from the quencher.
The quencher does not need to remove any energy from the fluorophore since this is left as vibrational and rotational energy in the triplet, which then transfers it to solvent. The triplet can then suffer quenching from another quencher molecules to form S$0$ the ground state. This also reduces any triplet lifetime and phosphorescence yield.

The extent of spin-orbit coupling is proportional to the quenchers mass so is far greater for heavy species hence the name. This explains the effect of Br vs F in your question.
Paramagnetic species such as O$_2$ also quench by causing extra spin-orbit coupling but O$_2$ can also quench by energy transfer as it has two low lying excited states.

(b) Energy transfer. This has two forms (i) Dexter transfer and (ii) Forster or dipole dipole transfer. Dexter is generally short range but Forster can be long range transfer in which S$_1$ can be a few nanometers away from Q for quenching to occur. The condition for transfer is one of energy conservation, the emission spectrum of the donor (S$_1$) must overlap in energy with the absorption of the acceptor. In dipole-dipole transfer the acceptor is electronically excited. This is usually written as D$^*$ + A -> D + A$^*$

(c) In electron transfer the donor and acceptor have to be stable in their ground states. Either the donor or acceptor has to be electronically excited and thus its redox potential is changed to be favourable ($\Delta G$ negative) and reaction occurs. D$^*$ + A -> D$^+$ + A$^-$ or A$^*$ + D -> A$^-$ + D$^+$. After this the D$^+$ and A$^-$ can recombine to D and A in their ground state. The rate constant of energy depends exponentially on their separation and only occurs at close to contact.

(d) In a reaction (other than electron transfer) the excited state is deactivated as bonds are broken or made. The subsequent mechanism depends on the exact chemical nature of the excited molecule and quencher.


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