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When we heat a ionic substance/compound, the ionic bond gets weakened and the metal is ionized. Is the color imparted to the flame due to absorption of light? i.e. Do we see the remaining wavelengths? Or is it due to the falling back of electrons to their ground state?

Shouldn't we be seeing two colours one after other? Since the electrons take up energy of particular wavelengths (so we see the remaining colours) and then the wavelength which got used up earlier returning back when electrons return back to the ground state?

The coordination compounds should also show same dual-color property. Since when electrons absorb energy they go from t2g to eg and when they return, they should give back the wavelength.

This could also be said for halogens.

Why do we see only one colour? Which colour is it in which case? The absorption one or the emission one?

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You seem to be mixing up a term. Let me get that out of the way first.

Ionisation is the name for a process creating an ion from a neutral species. However, if you heat an ionic substance and break the ionic bond, you are not ionising but deionising (although that term is usually used for a different process). The reaction equation for sodium chloride would be the following:

$$\ce{NaCl ->[\Delta] Na + Cl}$$


Now to the actual question. The perception of colour is based on one of two principles (as you noted):

  1. out of white light, certain wavelengths are absorbed leaving behind a remaining spectrum and causing the perceived colour to be complementary to that of the absorbed light

  2. light of a certain wavelength is emitted, directly creating the impression of said colour.

As you also noted, this could mean that we see up to two colours in the same electronic system: One due to the absorption and the second, complementary one during emission. So what is it that we see?

For coloured compounds at room temperature — that includes, but is not limited to transition metal salts and coordination complexes — we typically see absorption colour. Take for example the hexachloridoferrate(III) anion $\ce{[FeCl6]^3-}$: It has an intense yellow colour which can be explained by the absorption of very short visible wavelengths. The shortest visible wavelength is a blue one, so the perceived colour is yellow. The colouring is due to ligand-to-metal charge transfer transitions. In the corresponding fluoridocomplex $\ce{[FeF6]^3-}$, the orbital energies of the fluoride ions is predicted to be lower, giving a higher transition energy and shifting that transition just into UV wavelengths — the hexafluorideferrate(III) complex is colourless.

Also in organic chemistry, there is a number of yellow-ish compounds which occur because they just absorb at high enough wavelengths to shift out of the UV range. (UV absorption is very common for organic compounds containing a phenyl ring.)

We only see one colour, however. This is because the absorbed energy — which immediately excites an electron into a higher orbital — is liberated into the surroundings as vibrations and ultimately heat. The electron does not fall back into the ground state in a single transition but via a number of intermediates loosing energy all the way; hence we do not see a corresponding emission colour.


In flame colouring, one typically takes white (or colourless) compounds and heats them in a flame (the flame ideally being colourless, too). Only in the heat of the flame do they show their colour — and in a typical ion-lottery (qualitative inorganic analysis) lab, this is done in a dark corner of the lab, because there is actual emitted light. What is happening here?

Through the heat of the flame we are transferring energy into the compound, first causing the bonds to break (liberating atoms) and then also causing these atoms to vapourise so that we end up with gaseous metal atoms and gaseous non-metal atoms. Further heat is introduced which — if it is enough — will finally cause the transition of an electron into a higher orbital. This is, of course, unstable, but here the relaxation is immediately back down into the ground state for a significant percentage of these atoms; each of them releasing a photon with exactly the wavelength of the corresponding transition. And what we see are those photons, corresponding exactly to the wavelength of the energy difference between two orbitals.


And finally, just to confuse you, I am going to quote an experiment I saw in my school’s biology course. We had isolated chlorophyll from leaves of a plant and had it in a solution in an Erlenmeyer flask. It appeared green. Our teacher held it into the strong light beam of an overhead projector and it appeared as red. The experiment was explained as follows:

  • Red light is absorbed by chlorophyll, causing a green impression under normal conditions.

  • in the strong light beam, so much red light was absorbed and so many chlorophyll molecules were excited that the energy had ‘nowhere to go’; thus, the same red light was emitted to all directions and we were able to see that red colour.


So to sum it up:

  • in coloured salts at room temperature, we see the complementary colour of an absorbed photon. We do not see the emission (and thus the actual colour) because it is too unfrequent; most energy absorbed is released as vibrations and heat.

  • in flame colouring, we have highly excited atoms and we do not actually see the absorption since the energy necessary is transferred by heat. However, we do see emitted photons of exactly the energy difference wavelength from a significant number of electrons falling back into a lower energy state. We see the actual colour but not the complementary one.

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  • $\begingroup$ you should add 'black body radiation color' and color temperature part to this post, I believe. $\endgroup$ – permeakra Jan 30 '16 at 18:14
  • $\begingroup$ @permeakra Unfortunately, I don’t know enough about it =C $\endgroup$ – Jan Jan 30 '16 at 18:30
  • $\begingroup$ Just to check if I got it right - when we provide too much energy, we see emission color but when the compound is at normal conditions, we see absorption color? Right? $\endgroup$ – Quark Jan 30 '16 at 18:37
  • $\begingroup$ @Quark Simply put: Yes. $\endgroup$ – Jan Jan 30 '16 at 18:56

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