[ostudent]

See attachment for problem.

In this question, I am confused about the effect of rigidity on effects of UV radiation.
I agree that C is the answer because it is the most highly conjugated. However, why does C being "the most rigid due to its fused aromatic ring system" make it more likely to have intense yellow fluorescence. Can this be explained?

[Irlanur]

The argument is a bit hand-waving, but one can argue than in order to be fluoresecent, the light energy that was taken up also has to be released via a radiative process. If a molecule has a lot of vibrational degrees of freedom, i.e. if it can wiggle a lot, most energy will be released as heat, not as visible light.

[rolnor]

I think A and B have to little conjugation to be fluorescent?

[zarhym]

We can change the question into this way.
Which one has the lowest excited singlet(S1) state?

[Enthalpy]

We can change the question into this way.
Which one has the lowest excited singlet(S1) state?

And how would that relate with vibrational degrees of freedom?

[zarhym]

I was thinking about the homo lumo gap. Better conjugation means lower S1/T1 state, which leads to less energy between the T1 and S0 state. Thus, the fluorescence will start to shift from UV to yellow with the growth of the conjugation.  Since C has the highest number of Pi electrons stack together, I would guess it should more likely to give yellow fluorescence among the first three choice.

I may be wrong about this.
Actually, I didn't think about vibrational degrees of freedom.
My first instinct is to use the jablonski diagram to explain this question.

[clarkstill]

The argument is a bit hand-waving, but one can argue than in order to be fluoresecent, the light energy that was taken up also has to be released via a radiative process. If a molecule has a lot of vibrational degrees of freedom, i.e. if it can wiggle a lot, most energy will be released as heat, not as visible light.

Its going way WAY back for me, but isn't it that the electronic transitions are formally spin-forbidden, but when coupled with a vibrational change the process becomes allowed (so the electronic state relaxes more quickly)?