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I looked at the flames of copper (I) chloride and of copper (II) chloride through a spectroscope and they looked the same. The flame colour is the same too. But since they have different oxidation states, and therefore different electron configurations, should their emission spectra be slightly different?

Would this logic also apply, for example, to elemental sodium in a discharge tube, vs sodium chloride burning in a flame? Would their emission spectra differ?

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  • $\begingroup$ I just read that samples of copper (I) chloride usually are contaminated with copper (II) chloride, which explains seeing the same colour flame and spectrum. However, I still do not know the answer to my question and would very much like to. $\endgroup$ – Rafael Aug 28 at 1:08
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    $\begingroup$ Once in the flame, you have atoms and ions in excited states. The oxidation states of species before they were introduced into the flame are history. $\endgroup$ – Ed V Aug 28 at 1:08
  • $\begingroup$ So the lines seen are a combination of the photons emitted by unionized atoms and whatever mixture of ions is present? Does this apply to the discharge tube case too? $\endgroup$ – Rafael Aug 28 at 1:11
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    $\begingroup$ Yes. Sample introduction involves several steps, but what you get in the flame or plasma or discharge tube is excited atoms, excited ions, perhaps some excited high temperature species, and unexcited versions of these. It is generally very complicated and lots of factors are in play. $\endgroup$ – Ed V Aug 28 at 1:15
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    $\begingroup$ A sodium atom and a sodium ion have different spectra. The sodium atom, unexcited, cannot emit light, but it can absorb light, as in atomic absorption spectroscopy. An electronically excited sodium atom can emit light. A sodium ion is simply another species that can be excited, etc. $\endgroup$ – Ed V Aug 28 at 1:42
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Interesting question. Keep in mind that the elemental emission spectrum in a flame or plasma and even a discharge does not remember its history in solution or a solid phase.

The punchline is that the emission spectrum is dependent on the elements gas phase chemistry in the flame/plasma/discharge. Prof. Ed has explained you the example of sodium. Let us say we have the following

(a) a block of element sodium (b) a block of sodium chloride crystal (c) solution of sodium chloride in water

If you introduce (a), (b) and (c) in the flame, the flame will be colored yellow in each case, which means that the emission is coming from a common emitter. That emitter is a elemental sodium atom excited by high temperature in the gas phase. Thus atomic emission spectrum is a fingerprint of the element.

You may ask that you introduced Na(+) in the flame in the case of b and c. Flames can easily reduce an ion to the elemental state.

How to see the sodium ion spectrum: As we just said, the emission spectrum is independent of this original state. You can only cause ionization by increasing the temperature. This is the way to see the spectrum of an ion. If we were using a high temperature flame/electrical discharge, we would start seeing Na(+) spectrum along with elemental sodium spectrum.

Coming to your particular example: You introduced copper (I) and copper (II) into a flame and they all colored it beautiful blue-green. The reason is that if the flame temperature is low, compounds cannot fully dissociate into atoms (not enough energy to break the bonds). In such cases, very simple diatomic or triatomic molecules are formed in the flame which emit their characteristic colors. In the case of copper, CuCl is formed in the flame. CuOH may be formed as well. Whether you introduced Cu(I) or Cu(II), as a chloride, it does not remember its solution phase or solid phase history.

If you were indeed using a high temperature flame, you will never ever such a blue green coloration, because this time, the emission is from Cu atoms (in UV). Hope that clarifies your confusion.

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  • $\begingroup$ Thank you, that is quite clarifying. I am only confused by one thing: Flames can easily reduce an ion to the elemental state. Na(+) is far more stable than Na... why would it reduce itself to the unfavourable unionized form? You and Ed said the lines in the visible emission spectrum of sodium are from excited atoms... I just don't get why not also ions if the substance being introduced to the flame is actually entirely Na(+). $\endgroup$ – Rafael Aug 28 at 12:39
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    $\begingroup$ Like @M. Farooq said, the chemistry in the flame is not the chemistry of what was introduced. If an aerosol of saltwater is nebulized into the flame, the droplets lose water and become tiny salt particles. Those melt and vaporize. Then you have a complicated mixture, highly temperature dependent, of atoms, ion and electrons: a plasma. The Saha equation deals with the relative abundances of the various species. So sodium atoms are the prevalent species and atomic emission from it is what you see. It might help to think of a related system: magnesium acetate. If this is introduced to the flame $\endgroup$ – Ed V Aug 28 at 12:46
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    $\begingroup$ then you will have magnesium atoms and singly ionized magnesium atoms, even though magnesium has 2+ oxidation number in the original magnesium acetate solution. You can easily search for the energy level diagrams for magnesium (“Mg I”) and singly ionized magnesium (“Mg II”) and see that they differ a lot. But there is very little doubly ionized magnesium ion in the flame. And the acetate ions are busted up, as expected. $\endgroup$ – Ed V Aug 28 at 12:51
  • $\begingroup$ Thank you Ed, I understand. I think I am just surprised to find out that in the very hot environment of the flame, "stable" ions would spontaneously become "unstable" atoms, due to my more pedestrian understanding of where and how atoms and ions exist (i.e. Mg II is more favourable than Mg, at all times). $\endgroup$ – Rafael Aug 28 at 13:05
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    $\begingroup$ @Rafael, ED has already explained. You have to think of the chemistry of what happens after the solvent dries up in the flame. Yes, Na+ is more thermodynamically stable at room temperature in an isolated surrounding, but flame chemistry (= hot gas phase reaction) generates neutral Na atoms, and under those reaction conditions, Na+ is not preferred. The spectra of ionized atoms (by using very temperatures) were known in the 1920s. Amazing, isn't it? $\endgroup$ – M. Farooq Aug 28 at 13:08

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