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I was reading about Jablonski diagrams and how they are structured. Bold lines would indicate ground electronic levels, and thin lines indicate the associated vibrational levels. Absorption only happens between the different electron energy (orbital) levels, however any additional associated vibrational level can be occupied (for example, transition from S(0) ground state to the third vibrational state of S(1)). enter image description here

Why is this the case though? Couldn't a transition happen between the vibrational levels of an energy level (for example, from ground state S(0) to third vibrational state of S(0)) or does an absoprtion always change the electronic level?

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An electronic transition e.g. from ground level $S_0$ to $S_1$ requires much more energy as between ground level $S_0$ to energetically closer vibrational levels (without changes the electronic state).

Let's assume only one photon is "in charge" to provide the relevant excitation.* The energy light provides is not a continuous, but a discrete quantity; literally, the photons of electromagnetic radiation carry energy in quanta. Because the energetic difference between the electronic states $S_0$ and $S_1$ are farther away, than the about the second (only vibrational transition), you typically observe electronic excitations of molecules with (visible) light (e.g., perception of colors, UV-Vis spectroscopy). To induce changes in vibrational states, you however typically use infrared radiation (as in IR spectroscopy); in comparison to visible light, this is a radiation of longer wavelength.

addition to answer a comment:

  • Spectra can be either absorption spectra, because you observe how much incident radiation intensity (of varying wavelength, or energy) is attenuated by the sample. This uptake of energy (excitation) can yield the release of of energy (relaxation) for example by radiation of similar, or different wavelength; in some instances launch chemical reactions (e.g., radical photochlorination and photobromination).

    Or, the spectrum is an emission spectrum. An example is the emission map in fluorescence spectroscopy; a sample is excited by radiation of a fixed wavelength, you monitor the wavelength dependent intensity of light "broadcast" by the sample.

  • A molecule typically can undergo electronic, vibrational, and rotational transitions. Most often, the gap between two electronic states (e.g. $S_0 \rightarrow S_1$, thick black lines in the figure below) is much larger than the one between two vibrational ($\nu_0 \rightarrow \nu_1$, thin black lines below) belonging to the same electronic state. Transitions between adjacent rotational states (not shown below) are even closer, energetically speaking. As already symbolized in the illustration used in your question, the states are all «there» (though not always equally well accessible, as transitions follow selection rules). Simultaneously, there is partially overlap; see e.g. some vibrational states of $S_0$ could be energetically equivalent to vibrational states of $S_1$, or $S_2$.

  • For simplicity of the representation, the straight blue arrows depart from $(S_0, \nu = 0)$ to higher $S_1$ and some $\nu$ (i.e., not only to $\nu = 0$). It equally is possible for an electronic transition to depart from an other vibrational state than $\nu = 0$:

    enter image description here

    (image credit to LibreTexts Chemistry)

    This contributes to the observation of UV-Vis absorption bands extending multiple nm width (contrasting to the line spectra e.g. in atomic absorption spectroscopy), especially with growing complexity of the investigated molecule (many energetically similar vibrations) at non-zero temperature, surrounded by neighboring molecules of solvent, or solid matrix.

* Second-harmonic generation is an example where two-photon absorptions may occur.

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  • $\begingroup$ Thank you that makes sense, as IR photons have low energy. One question though. So basically suppose if I would want to excite from S(0) to S(1), which is an electronic transition which needs a photon of 495 nm. But in theory, I could use 490 nm so I get the excitation from ground state S(0) to S(1), but in the third vibrational mode? Is that why it's called an absorption spectrum, because I can also use photons which are slightly higher in energy than for the simple electronic transition? $\endgroup$
    – Mäßige
    Jun 4, 2023 at 18:31
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If you stay in the ground state then an infra-red or vibrational transition could populate, say, $v=3$ from $v=0$ always depending on selection rules. When the transition happens to an excited state then this is an electronic transition and typically needs more energy i.e. visible/uv light, rather than infra-red as in the ground state. The electronic transition also has selection rules but now the excited state is different to the ground state.

You can also have transition in an excited state, at $S_1, (v=0 \to v=3)$ but only using time resolved infra-red methods as excited states always decay back to the ground state.

(btw. The diagram shows states, not orbitals, each state is the total energy of the electrons in each orbital, with the ground state set to zero.)

Edit: I would also add that your diagram shows the situation in solution, in the vapour/gas phase vibrational relaxation can be completely absent then fluorescence can occur from the initially excited level to the ground state vibrational levels.

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  • $\begingroup$ Because in solution, fluorescence is rare, rather de-excitation because of collision with solvent molecules? $\endgroup$
    – Mäßige
    Jun 4, 2023 at 18:32
  • $\begingroup$ Fluorescence is not rare in solution at all. $\endgroup$
    – AChem
    Jun 4, 2023 at 18:42
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    $\begingroup$ In solution fluorescence comes primarily from the lowest vibrational level, as energy is lost via collisions usually far faster than the fluorescence decay rate constant. This need not be the case in the gas phase as vibrational relaxation can be far slower or effectively absent. Fluorescence yield can be v. low in some types of molecules due to intersystem crossing forming triplets but is often large, particularly in aromatics and dyes. $\endgroup$
    – porphyrin
    Jun 4, 2023 at 20:00

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