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I'm trying to understand how organic molecules get their colour.

  1. One major factor are conjugated double bonds which create delocalized pi-orbitals (e.g. https://en.wikipedia.org/wiki/Conjugated_system). When a photon hits an electron in the HOMO, and the photon carries the right amount of energy, the electron gets lifted to the LUMO. Other photons don't get absorbed, they are reflected/transmitted/whatever and create the colour of the pigment/object.
    Since everything down there is a wobbly probability distribution, the energy of the proton doesn't have to be at exactly a single value, but can the close to it. But, as far as I understood it, it is a very narrow energy band that gets absorbed.

  2. When light hits the cones in human eyes, we can model the activation of the different cones by scalar-multiplying the spectral power distribution of the incoming light with the sensitivity functions of the cones (e.g. https://en.wikipedia.org/wiki/LMS_color_space and https://en.wikipedia.org/wiki/CIE_1931_color_space). If we assume a uniform power distribution of the light and shine it on a colour pigment, only a small band of wavelengths should be absorbed.

Now... My problem is, that if the last sentence is true, the pigment should still appear white/grey or, at best, have a very desaturated colour. The incoming light appears white/grey because the scalar products of the incoming power spectrum with the CIE 1931 colour matching functions all produce the same value. Introducing a narrow, deep dip in the incoming power distribution via photon absorption won't change these values enough to give a strong, saturated colour. (For the actual human cone sensitivity functions, white/grey light might not be represented by three equal values, but the argument is the same.)

To get a strong colour, the pigment would need to absorb a large portion of the wavelength spectrum. And looking at the absorption spectrum of, for example, beta-carotene, this seems to be the case: https://www.photochemcad.com/compound-detail.php?name=Beta-carotene
If the $\lambda_{max} = 451\mathrm{nm}$ stated there represents the energy required to cross the HOMO-LUMO gap, why are so many other wavelengths, which are "far away" from $451\mathrm{nm}$ also absorbed?

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    $\begingroup$ Nice interactive link for beta-carotene! So it absorbs pretty much all light below about 510 nm, which means it transmits, from incident white light, red, orange, yellow and green. Then the color we perceive depends on our eyes: most of us are trichromats, some people have one of the color blindnesses, etc. $\endgroup$
    – Ed V
    Dec 27, 2021 at 15:57
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    $\begingroup$ By the way, beta-carotene has broad absorbance, so it does not really support your query about what would happen if a narrow absorbance removed just a narrow band of wavelengths from a white light source. An example of narrow band absorbances in the visible is here. And the color of the holmium chloride aqueous solution dramatically changes, depending on the source of the illuminating light: white LED lamp, white fluorescent lamp or sunlight. So, not so simple. $\endgroup$
    – Ed V
    Dec 27, 2021 at 16:07
  • $\begingroup$ Beta-carotine having broad absorbtion is exactly the point that confused me! It has to be that way to appear orange and the linked measurements confirm it. But looking at explanations how pigments such as beta-carotene get their colour, I only ever find the argument that the single HOMO-LUMO jump is responsible. $\endgroup$ Dec 27, 2021 at 18:12
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    $\begingroup$ In introductory chemistry courses, students may be introduced to a tiny quantum mechanics section, which includes the particle in a box solution of the Schrodinger equation. Then beta-carotene, with its conjugated double bonds, is modeled as a particle in a box. What could be sadly misleading later on? ;-) $\endgroup$
    – Ed V
    Dec 27, 2021 at 19:43
  • $\begingroup$ Being a pigment and absorbing just around single wavelength mutually excludes each other. You would need atomized matter in gaseous state for that. Molecules and/or condensed phases have quite broad absorption bands or even wavelength regions. $\endgroup$
    – Poutnik
    Dec 28, 2021 at 13:07

4 Answers 4

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You are right that an individual transition from one state to another by absorption of a photon is connected to a very well defined energy. However, an absorption band in UV/VIS results from transitions between various energy states.

For example, you might start from the electronic ground state before absorption. However, your molecules are in various different vibrational states. So your starting levels are different. The excited state you reach after absorption again has different vibrational levels, which have a different spacing than in the ground state. Thus, your absorption band is a superposition of many individual transitions between different states with different energy differences, and consequently is rather broad.

The situation can be much more complex than this short description. For example, also different electronic states can participate.

You might want to search for a Jablonski diagram for a visualization, e.g. here.

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  • $\begingroup$ Thank you! That makes a lot of sense, and I suspected something similar. The only thing that bothers me is that in my mind, the different vibrational levels lead to a small number of narrow energy bands. I have a hard time imagining that they are numerous enough and close enough to each other to give a broad, "smooth" absorption spectrum overall. But I guess they do? $\endgroup$ Dec 28, 2021 at 16:34
  • $\begingroup$ Yes, in solution, there are so many molecules, with slightly different energies that there are so many transitions, that it looks like a continuous curve. $\endgroup$
    – AChem
    Dec 28, 2021 at 16:42
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    $\begingroup$ Yes, you could expect some fine structure, and look at the iodine gas phase spectrum in @M.Farooq's answer, there you have it! It is a matter of resolution of the spectrometer if you can see it. However, in condensed phase (liquid) you have additional interactions between the molecules that influence the energy states of the molecules. Thus, so many different transitions occur that no spectrometer can resolve this, and you get your typical smooth absorption bands. You can observe similar effects e.g. in infrared spectroscopy. $\endgroup$ Dec 28, 2021 at 16:43
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Color perception is a complex topic and one cannot make generalized statements.

If you are seriously investigating and want to have a deeper understanding, then get hold of a beautiful book called The Physics and Chemistry of Color: Fifteen Causes of Color by Kurt Nassau. It could be available on Internet Archive. It is a readible book without involving too much math.

The sky is blue for one reason, a Bunsen burner flame is blue for another reason, and the carrot juice is orange for another reason. The key criterion for a colored solid/ liquid or gas is to interact with visible light (~ 400 to 700 nm) in some way. There is no heavy restriction on how many wavelengths must be absorbed in for something to appear colored. There are a very few exception which something appears colored to us because of interaction with high energy particles, which are not visible us, but the interaction of those particles with matter leads to emission of light in the visible region.

When chemists talk about colors, they are usually talking about molecules in solution or in gas phase. Unlike gaseous atoms where only discrete wavelengths can be absorbed, like sodium vapor, molecules have vibrational and rotational excited energy levels besides electronic energy levels. If you excite molecules with visible light a broad range of wavelengths can be absorbed just like your beta carotene example. Consequently, this assumption

But, as far as I understood it, it is a very narrow energy band that gets absorbed.

is not right. Molecular absorption in solution or gas phase is pretty broad. I am avoiding solids. There you have to deal with their reflectance spectra. We are focussing on absorption/ transmission spectra.

One of the best examples of colored gas is that of molecular iodine vapor. It has a striking violet color. Photo credit here enter image description here.

Since the vapor is appearing colored to us, it must be interacting (absorbing in this case) with light wavelengths in the range 400-700 nm. Have a look at this absorption spectrum here. The iodine molecule in the gas phase absorbs green to orange portion of the visible spectrum. As the result, transmitted light from iodine vapor looks violet.

enter image description here

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  • $\begingroup$ Thank you for your answer! You wrote "Molecular absorption in solution or gas phase is pretty broad", but can you explain why that is? $\endgroup$ Dec 28, 2021 at 16:35
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    $\begingroup$ Because unlike atoms, molecules can also vibrate in the excited state. There are plenty of vibrational energy levels for a molecules. This is why it is not a single sharp line like atoms, but a series of lines, as you can see in the iodine spectrum. Each peak corresponds to a vibrational level. $\endgroup$
    – AChem
    Dec 28, 2021 at 16:40
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Some types of light absorbance show very narrow lines while others show broad peaks.

Atomic spectra have sharp lines when the atoms are in the gas phase. One example is the absorption of sunlight by gases on the surface of the sun. The sunlight appears white, especially when observed in outer space. However, there a various species absorbing the light at specific wavelengths, allowing spectroscopists to infer the composition of the sun's surface.

enter image description here

Source: https://calgary.rasc.ca/solarspectrum.htm

On the other hand, visible and UV absorption of samples in the liquid or solid phase are often broad. Typically, these also arise from electronic transitions, but those of valence electrons. The energy of the electronic states of valence electrons depends on bond lengths and angles (i.e. couples with vibrational states) and intermolecular interactions (i.e. will be different from molecule to molecule).

To get a strong colour, the pigment would need to absorb a large portion of the wavelength spectrum.

Yes, so atomic absorption does not change the perceived color much. On the other hand, atomic emission results in clear bright colors like in fireworks or flame tests.

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    $\begingroup$ Beautiful figure. A small note for readers. There is no instrument/spectrograph which can generate a raw solar spectrum like this. This marvellous solar spectrum is computer generated to look like a echelle spectrographs output, which also does not look like this at all. As novices, I and a good friend here were confused for a couple of years by this beautiful spectra thinking that this could be the raw image of a fancy spectrograph. $\endgroup$
    – AChem
    Dec 28, 2021 at 1:18
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    $\begingroup$ I should mention that while many of the images available for the solar spectrum might suggest that these gaps are something that need fancy instrumentation to measure indirectly and then later "reconstruct", this is not necessarily the case - the gaps are very much tangible, and can be captured even with very rudimentary tools. See for example these photos of several light spectra (including the solar spectrum) by using a CD in a cardboard box as a monochromator. $\endgroup$ Dec 28, 2021 at 1:47
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    $\begingroup$ Yes such experiments with CD spectroscopes are worth the time. $\endgroup$
    – AChem
    Dec 28, 2021 at 2:03
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As others have brought up, indeed the absorption features of most compounds with electronic transitions in the visible light range are rather broad, and this breadth is important in generating a perceived colour.

But what would happen if you somehow had a substance which acted as an "anti-laser", where somehow a very tiny sliver of a white light spectrum were absorbed? What would that look like?

As it turns out, the solar spectrum is already full of "holes", called spectral lines. These are caused by isolated atoms in the gas phase in the solar atmosphere. Because these absorption features are caused by electronic transitions in single isolated atoms, effects capable of broadening the absorption of the electronic transition lines are much more limited. For example, the solar spectrum has dark slivers at 589.158 nm and 589.756 nm, corresponding to electronic transitions of isolated sodium atoms. These "holes" in the solar spectrum are very thin, covering a wavelength region on the order of 0.001-0.01 nm wide. There are hundreds of these holes in the spectrum of sunlight, even when measured without the additional complicating effects of Earth's atmosphere. Of course, this doesn't stop sunlight from looking white, as you suspected.

To make this all seem a little less esoteric, there is a rough but rather interesting experiment which can be performed, as shown in this Youtube video. A crude gas of sodium atoms can be created simply by adding table salt to a flame. When illuminated with white light, there is nothing out of the ordinary. However, if illuminated with a sodium lamp, suddenly the fire becomes dark, almost black. What's happening is that the sodium gas in the flame is absorbing photons of exactly 589.158 nm and 589.756 nm wavelength, and if your light source contains those photons almost exclusively, then you can clearly notice their absence. But importantly, the flame was always absorbing these photons - it's just that with a white light you couldn't really see the effect, it made virtually no difference to the overall white light spectrum.

You might now be wondering whether "white" sunlight can truly be considered "white", or whether there is a "whiter" light source. In principle, you could generate an almost perfect Planck-distributed blackbody spectrum with the same colour temperature as the surface of the Sun (~5700 K) and you could imagine comparing the perceived colour of this blackbody spectrum and actual sunlight. These will almost certainly be indistinguishable to the human eye.

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    $\begingroup$ @EdV Seems like we had the same idea, I was just slightly slower at answering! But I agree, the answers are complementary. $\endgroup$ Dec 27, 2021 at 23:58
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    $\begingroup$ @NicolauSakerNeto, Beautiful experiment. $\endgroup$
    – AChem
    Dec 28, 2021 at 1:24
  • $\begingroup$ Thank you for your answer! I understand the "narrow gaps" in the solar spectrum and other spectra produced by atomic absorption. But it is still a bit unclear to me, WHY molecular absorption has a broad absorption spectrum. $\endgroup$ Dec 28, 2021 at 16:37

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