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Some compounds are colored due to HOMO-LUMO transition. We see the complimentary color to the frequency which the electron absorbs to get the high molecular orbital. But after some the electron should return to the ground state emitting the same frequency again, and, hence, after some time the compound should start appearing white since all frequencies will be present in the light reflected by it.

Or, if the electron does not return to the ground state, there must be a point where saturation occurs, that is, all the possible electrons have gone to the excited state, in which case the compound should again appear white. So why are compounds colored?

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    $\begingroup$ Most absorbing substances do not re-emit, but convert the electron energies to other forms. Those that do usually emit longer wavelengths as fluorescence. $\endgroup$
    – Poutnik
    Jun 19, 2023 at 5:39
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    $\begingroup$ Most of the electronic transitions happening after absorption of a photon happen very quickly. So there is little scope in most molecules for a "buildup" in the excited state. And there are multiple pathways for the electrons to get back where they started many of which don't involve visible light. $\endgroup$
    – matt_black
    Jun 19, 2023 at 9:56
  • $\begingroup$ Looks like poorly worded duplicate of chemistry.stackexchange.com/questions/88767/… $\endgroup$
    – Mithoron
    Jun 19, 2023 at 16:19

2 Answers 2

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If you are seriously interested in understanding why things are colored, there is a book called The Physics and Chemistry of Color: The Fifteen Causes of Color, by Kurt Nassau. It should be available in any good library, and Internet Archive has it as well. The main theme is that there are 15 separate color causes. In your case, there are a couple of very common misconceptions due to web searches.

  1. The complementary color is a very crude estimate of the reason behind the apparent color of solutions and even gases. In my opinion, this should never be applied to proper chemistry.

  2. Once a molecular species absorbs light of a certain frequency, the electrons relax to the ground state but without emitting light. Perhaps a physicist can explain the exact mechanism of those radiationless transitions. Most of the absorbed energy of light is released as heat. Certain molecules do not do that, instead they emit light of a longer wavelength. This is called fluorescence (like your highlighter marker ink).

  3. What you are describing happens only in extremely rare cases for example in hot gas phase atoms. For example, you have vapor of sodium atoms in a flame, and you shine 589 nm at the sodium vapor, the atoms absorb 589 nm and re-emit 589 nm. The is called resonance fluorescence.

Finally,

where saturation occurs, that is, all the possible electrons have gone to the excited state,

Now you are talking about lasing phenomenon (of lasers), which is another world to explore.

One can now see how complex the scenario is. This is why the suggested book is > 500 pages.

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    $\begingroup$ Nice answer I already upvoted! I never got around to getting the book by Kurt Nassau, so I should fix that. There seems to be an uptick in causes of color questions: I just answered one here. $\endgroup$
    – Ed V
    Jun 19, 2023 at 11:32
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Absorption and transmission do play a role in the color of the compound and absorption of a certain region of the visible light results in transmission of the remainder of the region, often resulting in complementary colors.

Excitation and relaxation very often occur via different mechanism due to selection rules. Thus, the same frequency of photon that was absorbed is not emitted, which would have resulted in a white colored compound.

Example of Ruby

energy diagram and illustration for color of ruby

Ruby has a brilliant red color. It occurs because of several factors:

  1. Excitation from the $^4A_2$ ground state of chromium ion (the impurity responsible for color in ruby) to the $^4T_2$ or $^4T_1$ states (broadened by vibrational and rotational energy levels) corresponds to violet and green-yellow parts of white light passing through ruby resulting in transmission of strong red which is partially responsible for the red color of ruby.

  2. The relaxation, however, does not occur directly from $^4T_2$ or $^4T_1$ to $^4A_2$ due to selection rules and instead passes through an intermediate $^2E$ state. Part of this causes slight warming, and the other part result in bright red fluorescence.

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  • $\begingroup$ Could you please add a reference for the source of the image? Thanks! $\endgroup$
    – Ed V
    Jun 19, 2023 at 20:03
  • $\begingroup$ @EdV I reconstructed the image from the source hyperlinked in the heading Example of Ruby (Britannica). $\endgroup$
    – ananta
    Jun 19, 2023 at 20:07
  • $\begingroup$ Thanks! Generally we like to have sources listed for all images, to avoid copyright issues, unless the images are made by the poster. Links, as you know, sometimes go bad. $\endgroup$
    – Ed V
    Jun 19, 2023 at 20:12

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