# why do we see characteristic colors of alkali metals in oxidizing flame?

I am a high school student and I am a little confused in a topic, My confusion is that:

We know alkali metals and their salts imparts characteristic color in oxidizing flame, for example: Lithium gives crimson red color, sodium gives yellow , potassium gives violet and Caesium gives blue, My school textbook says that this is because the heat from the flame excites the outermost electron to a higher energy level and when it returns back it will radiate in visible range as I mentioned above, but if this is the correct reason then we know the outermost electron of lithium is the most attracted among all alkali metals .So, it means that it would take high energy to excite its outer electron so a short wavelength of light is needed i.e probably in violet region, so when it returns back it should emit the same wavelength. Then why do we see these weird pattern of colors? I think this is not the complete reason for why do we see those colors? please explain this in brief only so that I can easily understand.

I don't think it is a good idea to connect ionization energy with the color of flame emission, especially in a Bunsen burner.

Ionization energy means that you have separated the electron out of the nucleus attractive field. However, electronic transitions, corresponding to visible wavelengths, take place while the electron is still bound to the nucleus. Thus the flame color of alkali metals might correlated with the IP but not the causation.

Let us look at the resonance wavelengths of emission of Li, Na, K, Rb, and Cs. Resonance wavelengths means the transition from the $$\mathrm{1s}$$ to $$\mathrm{2p}$$ orbital or $$n\mathrm s$$ to $$(n+1)\mathrm p$$ orbital in the case of alkali metals.

$$\begin{array}{r|ccccc} \hline \text{Element} & \ce{Li} & \ce{Na} & \ce{K} & \ce{Rb} & \ce{Cs} \\ \hline \lambda/\pu{nm} & 670 & 589 & 769 & 794 & 894 \\ \hline \end{array}$$

You cannot even see these wavelengths with your eyes after 700 nm which are in the deep deep red. Some people can see the red lines of K 769 nm. However, you can see the energy gap is indeed decreasing as we go down the group. The ionization energy also goes down the group. However, you can see, the colors which you see in the Bunsen burner have nothing to do with ionization energy.

The wavelengths have been extracted from a old book titled Graphische Darstellung der Spektren von Atomen und Ionen mit ein, zwei und drei Valenzelektronen by Dr. W. Grotrian, 1928. I think NIST atomic spectra database will be better today but I am not sure if they list resonance lines clearly. This book has graphically labelled transitions for almost all elements known in 1928.

• Yes, this is the answer. First ionization means total expulsion of the outermost electron. Meanwhile, electronic transitions occur when the outermost electron is energetically excited and gets promoted to a superior level of energy and back, emitting the remaining energy under the form of photons. – TheRelentlessNucleophile Jan 6 at 21:32
• @M.Farooq It would be nice to see the source for the wavelength values. Your another answer which I suppose you've copy-pasted a good chunk of data from states "very old book by Grotrian". That's kinda cryptic. Could you please specify the title, year, edition, page number and IDs, if any? – andselisk Jan 6 at 22:40
• I extracted data from the Grotrian diagrams, it is not listed in tables as one would expect. It is indeed from 1928, really old book by Dr. Grotrian. Will add the reference shortly. Most likely readers will benefit from the NIST database rather than this ancient inaccessible book. I am not sure if they have Grotrian diagrams. – M. Farooq Jan 6 at 22:51

The relationship between ionisation energy and emission spectra is complex

You make the assumption (implicitly) that the colour you see in emission spectra is from the single complete ionisation of an electron from the highest energy electron orbital in lithium. But the colour you see is not that simple for two reasons.

One is that you only see visible light (roughly the octave from about 400 nm to about 750 nm of wavelength). Many emission lines occur outside that range in the infra red or ultra violet, so the colour you see will rarely be a product of the complete range of possible emission lines.

Another is that there are far more transitions that the one that is related to complete ionisation of the outermost electron. All atoms have many possible electron orbitals higher than the lowest occupied one. Transitions between most pairs of energy levels are possible (there are complicated but irrelevant reasons why some can't happen related to quantum mechanics). And flames often have enough energy to put some electrons in many of those orbitals leading to a large number of possible emission lines when electrons can fall from any energy level to any other energy level. (see this introduction to the concept that explains the complexity for the simplest atom, hydrogen, which shows the calculated orbital energies and the possible transitions that result).

So colours result from filtering of the possible emissions to visible light only and the complexity of the transitions that are possible. So there is no meaningful relationship between colour and the ionisation energy.

• "Transitions between every pair of energy levels are possible." This ignores the atomic spectroscopy selection rules. You are somehow implying that there is a relationship between observed flame color and ionization energy but it is complex. The point is, does the relationship exist? You might wish to reconsider these points. – M. Farooq Jan 6 at 22:53
• @M.Farooq I'm answering at a basic level appropriate to the question. Transition rules are one level of complexity too far (and they explain brightness not existence of transitions). And I imply nothing at all about the relationship with ionisation energy: I explain why you should not even expect one when trying to explain colour. – matt_black Jan 6 at 23:25
• We should not teach something to students which has to be unlearnt later. One can say that certain transitions are allowed rather than absolutes "Transitions between every pair of energy levels are possible." Selection rules in atomic spectroscopy have nothing to with the brightness of the line, it is either yes or no (see for example Laporte selection rule). Brightness of a line depends on the population of the atoms in particular excited state, and this is governed by Boltzmann's distribution and temperature only (as long as it is allowed). – M. Farooq Jan 7 at 0:37
• At this level of explanation this is not a problem. Besides selection rules are not that absolute in many systems otherwise we wouldn't have phenomena like phosphorescence. My point was not not to state an absolute rule but merely a simple explanation avoiding concepts unnecessary to the core explanation–like triplet and singlet states. Explaining those would add nothing of value to the idea that there are many energy levels and many possible transitions and the ones you that contribute to colour are unlikely to be all of the possible ones. – matt_black Jan 7 at 0:52
• There is no atomic phosphorescence phenomenon in the flames or otherwise in gas phase. Some ions may show it in a solid state. – M. Farooq Jan 7 at 1:04