Fluorescence is the property which is exhibited when electrons emit wavelength of light lower than the one they absorbed so does that mean we can make every molecule to become fluorescent? I wanted to know how does the molecular structure contribute in giving rise to this property and how it's different from substances that are non fluorescent I searched but couldn't find anything related to molecular structure.
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2$\begingroup$ Look up the structures of phenolphthalein, rhodamine B, anthracene, rubrene, and fluorescein. The first one, which is a common pH indicator, is non-fluorescent. All the others are. See any aromatic rings in those? Any planar portions with rings? Just take a look and see if anything suggests itself! $\endgroup$– Ed VCommented Feb 4, 2022 at 13:14
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$\begingroup$ You may be interested by the book Molecular Fluorescence by Bernard Valeur and Mario Berberan-Santos (Wiley library), especially chapter 4 : "Structural Effects on Fluorescence Emission" which should answer your question perfectly. DOI : [10.1002/9783527650002.ch4](dx.doi.org/10.1002/9783527650002.ch4). $\endgroup$– mranvickCommented May 6, 2022 at 14:35
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2 Answers
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Keep in mind a few things that must happen for an absorption process to result in fluorescence:
(1) the initial transition is to an excited electronic state that observes certain rules regarding the structure of the excited state (Franck-Condon principle). That requires that the light used to excite the molecule have a sufficient frequency to result in electronic excited states.
(2) The excited electronic state must not lead to dissociation, otherwise you end up with photodissociation, not fluorescence.
(3) after excitation the molecule must relax to a state lower in energy with the same electron spin (unlike in phosphorescence, where a conversion occurs), so a transition occurs to a different vibrational excited state.
(4) the molecule must now relax from the low lying vibrational state in the excited electronic manifold to the ground state. Note that this may not happen efficiently, for reasons that have to do with transition rules. The emission process competes with non-radiative relaxation processes. The extent to which the emission represents the pathway to the ground state is the quantum yield of the fluorescence process.
So in short, provided the molecule doesn't fall apart or react when you excite it, and doesn't find other ways to dissipate energy more efficiently, then you should observe fluorescence, although this may occur outside of the visible spectrum of course.
answered Feb 4, 2022 at 7:50
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1$\begingroup$ It is not an answer to the question. The question is : why are there no non-radiative relaxation processes in fluorescent molecules ? Or : why are most molecules able to find non-radiative processes to dissipate energy, whereas fluorescent molecules do not find these relaxation modes ? Fluorescent molecules have something special. What ? $\endgroup$– MauriceCommented Feb 4, 2022 at 9:51
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2$\begingroup$ @Maurice I agree that my answer is not elaborate. I do not mention for instance that low energy $\pi\rightarrow\pi*$ transitions are important in many fluorescing molecules, as in many aromatics. Fluorescent molecules can however dissipate energy in non-radiative ways. $\endgroup$– Buck Thorn ♦Commented Feb 4, 2022 at 12:05
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$\begingroup$ Thankyou i did not know about photodissociation so i suppose this is the factor that makes other molecules non fluorescent and it arises due to their molecular arrangement, is it right? $\endgroup$ Commented Feb 6, 2022 at 9:24
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$\begingroup$ @25SimranTiwari You may want to look into the phenomenon of photobleaching, which can compete with fluorescence, and results in bond rupture or reactions. In order to make a molecule fluoresce you need to bring it into an electronic excited state. For many molecules that requires so much energy that it results in changes to structure or bonding in the molecule, rather than fluorescence. $\endgroup$– Buck Thorn ♦Commented Feb 6, 2022 at 12:10
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The fluorescence yield is $\varphi =k_k/(k_f+k_n)$ where $k_f$ is the rate constant for fluorescence and $k_n$ that for non-radiative processes. If $k_n \gg k_f$ then the fluorescence yield dis very low and we might consider the molecule to be non-fluorescent.
The question is then what makes $k_n$ large and this is answered in large part here What makes a species "fluorescence quencher"?
Edit:
In addition, symmetry of the excited state vs. ground state is important as is internal molecular motion.
Symmetry restriction is most obvious in the carotenes for when the lowest excited singlet state has the wrong symmetry for absorption/emission to the ground state no fluorescence is observed.
Instead absorption occurs to the second excited singlet state which then non-radiatively crosses to the first singlet (which is close by in energy) and as this has a forbidden transition to the ground state beta carotene, for example, if effectively non-fluorescent. The lowest singlet is then rapidly deactivated non-radiatively to the ground state by twisting of the long single /double bonded chain.
In the triphenylmethanes, such as malachite green, fluorescence is allowed by symmetry, but as the phenyl groups rotate they open up a non-radiative channel to the ground state and this quenches the fluorescence. Consequently in viscous solvents the fluorescence yield is high as molecular motion is slow, but the yield is low in mobile solvents when phenyl twisting is rapid.
