A student recently asked an interesting question What happens when KCl is introduced in the Bunsen flame?. This apparently simply question led me to an interesting debate. We all know that emission of characteristic colors of alkali metal salts especially the halides and carbonates in ordinary Bunsen flames comes from the atomic emission not their oxy compounds (like the alkaline earth salts) or their ions.

The point of debate is point number 3:

  1. KCl solution has to atomize and become KCl (solid).
  2. We would get KCl in gas phase from the heat of the flame.
  3. How does KCl (gas) convert to K atoms in a flame? What is the path? Note Thermal decomposition of salts without a flame is another story. Please do not mix, decomposition in a flame versus isolated vapor phase studies of salts.

Flame chemistry is quite complex and my understanding is that flames are electrically conducting from a very small amount of free electrons and ions. Certainly, we are not talking about cool flames like candles. Flame conductivity is a well known phenomenon and it is utilized in gas chromatography's "flame ionization detector" as well.

Herein, I am talking about spectroscopic flames used in atomic emission spectroscopy. It is known for more than 100 years that when we introduce alkali metal salts into the Bunsen flames, the conductivity of flame increases as a function of salt concentration. For alkali halides, it is a square root function of concentration. This clearly shows that there are free ions and electrons in a flame in which alkali metal salts are introduced. The amount must be really small but significant enough to alter flame properties Electrical properties of flames.

There must be certainly thermal decomposition but it is not a complete story at all. The key question is the path of KCl(s) to K atoms!

(a) KCl goes to potassium ions in the flame, by thermal decomposition, and it becomes neutral potassium atom in flame after gaining an electron (from flame components)?

(b) KCl is thermally decomposed directly to potassium atoms and there are no intermediate potassium ions?

  • 1
    $\begingroup$ The Saha ionization equation may be of interest. $\endgroup$
    – Ed V
    Commented Sep 12, 2020 at 22:52
  • $\begingroup$ Dear Ed, This is just one step before the Saha equation. How does the KCl (particle) turn into a free K atom in a flame? What is the mechanism? $\endgroup$
    – AChem
    Commented Sep 12, 2020 at 23:12
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    $\begingroup$ If you look at my KF answer here, it is energetically favorable for the ions to become neutral atoms. Same for KCl. So once a KCl particle melts and vaporizes, you get the neutral atoms, some electrons, some ions and some polyatomic species (neutral and ionic). That is my crude picture of it. ;) $\endgroup$
    – Ed V
    Commented Sep 12, 2020 at 23:27
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    $\begingroup$ It is complicated. Assuming local thermodynamic equilibrium (LTE) and reasonable flame temperature (without LTE, temperature is not even well defined), the chlorine species is predominantly in the form of neutral chlorine atoms. That sort of thing is what all those Boltzmann plots are about. My simplistic assumption is that any KCl moiety in the flame will be collisionally dissociated predominantly into K and Cl atoms, variously excited, and their ions will be present in lesser number densities. The electrons are just bit players, so to speak. $\endgroup$
    – Ed V
    Commented Sep 13, 2020 at 1:34
  • 2
    $\begingroup$ The spectroscopy of some of these halides has been studied in detail and the potential energy profiles determined for the ground state and 1 or 2 electronically excited states. In the case of NaI the ground state has a lot of ionic character but given enough energy Na +I atoms are produced and only at higher energy Na$^+$ + I$^-$ ions. Time resolved experiments show how dissociation occurs and for which A. Zewail received the Nobel Prize, see Faraday Disc. Chem. Soc. 1991, v91, p201 for a review. A few energetic electrons in the flame could easily produce atoms, in a similar way to photons. $\endgroup$
    – porphyrin
    Commented Sep 13, 2020 at 17:30


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