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Could someone please enlighten me on what would be the theoretical effect of a counter ion on the colour of the flame emitted by a metallic salt in a flame test? Eg. NaCl - what would be the effect of the non-metal on the general colour of the flame of the compound?

I am looking into specific groups of metallic salts based on 3 metals: Sodium, Potassium and Copper II. I have searched throughout the literature but have not found an answer yet.

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Recall the flame emission experiment is basically an atomic emission experiment. We don't say it is a compound emission experiment. The job of the flame/plsama or any high temperature medium is to decompose the compound into the constituent atoms or even ions for an experiment. Once a dilute metal salt solution is aspirated into the flame, the compound decomposes into constituents and then to the atoms. For example, you start with NaBr, NaF or NaI or NaNO3, in the flame you will get one and only emission at 589 nm from the sodium atoms. All of these compounds will make the flame bright yellow. So what happened to the non-metallic anion? Provided the flame temperature is high, one should "see" emission from non-metallic atoms as well. Our eyes cannot see the emission bc it in the deep UV region.

Sometimes the anion interferes with the atomization process such as phosphate for calcium. It forms very thermally stable compounds in the flame. You will see the signal of calcium go down quickly if phosphate ion is present.

Forget about copper as atomic emission, I guess you are thinking about the green flame of copper compounds. This only happens in low temperature flames. That color originates from molecular compounds of copper in the flame, and the anion may play a role. You need a very high temperature flame and copper emits in the UV.

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You can put in the flame any compound of Sodium, Potassium or Copper. The flame will be colored. But the emitted light will probably be highest with chlorides, because chlorides are more volatile than sulfates, carbonates or any other compounds. In order for the flame to be colored, the substance must first be volatilized, then decomposed into its atoms, then excited by the flame up to an electronic level. And then, on the way back, light is emitted. The first step, the volatilization, is easy with chlorides (maybe also with other halogenides - I don't know). With copper, it seems to be more complicated, because the light can be emitted from copper metal and also from volatile chlorides like $CuCl$, before their decomposition. It depends on the temperature.

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