# What is happening when KCl burns in a bunsen flame? [closed]

I did the experiment to qualitatively observe the different colours given off by different salts when placed in a flame.

But what is actually happening to the salt ?

Take $$\ce {KCl}$$ for example.

Does the $$\ce{ KCl}$$ react with $$\ce {O2}$$ ?

What does the reaction equation look like ?

Other salts include $$\ce {LiCl}, \ce{BaCl2}$$, $$\ce{SrCl2}$$, $$\ce{CuCl2}$$ and $$\ce {NaCl}$$.

I wanted to elaborate a little bit more so that the OP does not develop misconceptions. It is the potassium atoms not potassium ions that color the flame purple. While I was writing the answer, another poster emphasized the same point.

Caveat: In ordinary Bunsen burner type flames, emission from calcium, Sr, and Ba is from molecular compounds not atoms. The story of atomic emission is only true for alkali metals. The flame temperature is quite low in Bunsen burners. You need special flames to "see" the atomic emission of Ca, Ba, Sr.

Assume the processes when you introduce KCl in the flame:

a) The salt has to melt KCl (l) b) It has to vaporize KCl (gas) c) It has to decompose into constituents. Flames have free electrons, so potassium ions are converted to potassium atoms. d) From the thermal energy of the flame, potassium atoms are excited. Once they de-excite, you see the violet/lilac color. It is an ultrafast process! e) The flame temperature is so "low" on the scale of universal temperatures (stars, Sun) is that only 1-2% atoms get excited.

Now you may ask, what about the chlorine atoms, well, they emit in the deep ultraviolet and we cannot see them. The flame temperature is so low that it cannot excite chlorine atoms.

Keep in mind it is not a single wavelength. If you were lucky enough, you could use a pocket spectroscope and see red and violet lines (plus a persistent yellow line of sodium contamination).

Actually the strongest lines for potassium are 764 and 769 nm which are deep red. Many humans cannot see this deep red. So your flame would rather be deep red. Potassium atoms also emit less strongly at the other extreme end of the spectrum, corresponding to deep purple lines around 404 nm. The resulting "mix" of wavelengths appears to our eyes as lilac color!

• "Flames have free electrons, so potassium ions are converted to potassium atoms." - Who has misconceptions here? There's no K cations in vapor in the first place, just molecular KCl chemistry.stackexchange.com/questions/14174/… Sep 12 '20 at 17:53
• Mr. Mithoron, Who said K cations, anywhere? Ever heard of flame ionization detector in gas chromatography? Flame is a conductor of electricity, it has a small amount of positive ions as well as free electrons. May be read this physics answer here to learn more about flames: physics.stackexchange.com/questions/45105/… Sep 12 '20 at 18:33
• @Mithoron, Have you heard of reducing or oxidizing flames in emission spectroscopy? Sep 12 '20 at 18:36
• I never realized that for some materials it's a molecular spectrum and for others its atomic, but then again I don't know anything about Chemistry :-) What about copper? In the recent Con-Ed transformer “fire”, what exactly produced the color of the huge blue glow over New York City? I have a "colorful" but somewhat speculative answer there; without a spectrum of the green glow it's hard to say for sure.
– uhoh
Sep 13 '20 at 6:24
• @uhoh, yes those are molecular copper compounds in the flame. Copper atoms resonance lines are in the ultraviolet. Sep 13 '20 at 13:45

Upon the heat of the flame, the crystals shatter and ...

The correct term is "the crystals vaporize"

is an ensemble of K+ and Cl− in regular proximity to each other. The heat of the flame separates them, and excites them, independent of each other, e.g., an outer electron of K+ is then lifted to a higher level.

When ionic crystals vaporize, there are no longer ions! The emission of spectral lines comes from potassium or sodium etc atoms. The temperature of Bunsen flames is not high enough to make a plasma of ions.

• Because solid KCl already consists of ions, I did not recognize that the flame of the Bunsen burner had to ionize the material. Nor was I aware a monoatomic ion like $\ce{K+}$, thus in nobel gas configuration, would this easily be reduced back to K, which -- in condensed state -- tends to oxidize rapidly. Even if there are the terms of «oxidizing zone» and «zone of reduction» while preparing the bead test or using the blow pipe. Sep 14 '20 at 17:44