A sample of $\ce{CaCl2}$ is placed in the flame of a Bunsen burner. The component of the substance which is responsible for the production of the flame colour is:

  1. $\ce{Ca^2+}$
  2. $\ce{Ca}$
  3. $\ce{Cl-}$
  4. $\ce{Cl}$

Why is the textbook answer $\ce{Ca}$ and not $\ce{Ca^2+}$ ?

  • 1
    $\begingroup$ What are your own thoughts on the matter? Why do you think it should be $\ce{Ca^{2+}}$ rather than $\ce{Ca}$? Do you know which process is responsible for the flame colour? Please, have a look at our homework policy. $\endgroup$
    – Philipp
    Commented Apr 10, 2014 at 2:29
  • 1
    $\begingroup$ isnt it the presence of the cation that causes the change in colour in the flame test? $\endgroup$
    – confused
    Commented Apr 10, 2014 at 2:46

2 Answers 2


The phenomenon that is witnessed during a flame test is an "atomic emission". This statement may seem inappropriate, since it is a solution of metal ions (and not atoms) that is tested. The reason for calling it atomic emission lies in the process occurring in the flame. One of the steps of the process is an atomization step. That is, the flame converts the metal ions into atoms. When a solution of sodium chloride is placed in a flame, for example, the solvent evaporates, leaving behind solid crystalline sodium chloride. This evaporation is then followed by the dissociation of the sodium chloride crystals into individual ground state atoms - a process that is termed atomization. Thus sodium atoms are actually present in the flame at this point rather than sodium ions, and the process of light emission actually involves these atoms rather than the ions.

enter image description here

  • $\begingroup$ While it is an atomic emission in the case of alkali metal salts, for Ca-Ba monochloride molecules are also of consideration, and, in fact, for blueish copper flame CuCl molecules are of essence (greenish of yellowish flame is formed otherwise) $\endgroup$
    – permeakra
    Commented Nov 7, 2015 at 12:59
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    $\begingroup$ Actually, screw it. Here is an article that discuss the question of Ca(NO3) and CaCl2 solutions in flame test. The majority of the visible spectra is derived from molecular particles www.jpyro.com/wp-content/uploads/2012/08/Kos-710-731.pdf $\endgroup$
    – permeakra
    Commented Nov 7, 2015 at 13:04
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    $\begingroup$ But alas it appears the answer appears to been plagiarised [link]faculty.sdmiramar.edu/fgarces/labmatters/instruments/aa/…. $\endgroup$ Commented Mar 12, 2021 at 7:38
  • 2
    $\begingroup$ Or it could be that the faculty copied your answer, or maybe you being part of the faculty $\endgroup$ Commented Mar 12, 2021 at 7:40
  • 5
    $\begingroup$ @Philipp It would be nice if you could address the issue with possible plagiarism. This might be a serious issue. That PDF appears to be compiled way prior to your post, in 2007, and the author's name doesn't look like your nickname. You should either properly attribute cited work, or claim the authorship in convincing way, or delete the post altogether as a plagiarism. $\endgroup$
    – andselisk
    Commented Jul 16, 2021 at 14:09

Apart from Philipp’s practical answer, we can also approach this problem theoretically. Both the $\ce{Li+}$ ion and the $\ce{Li}$ atom have electrons in lower energy levels that can be excited to higher levels given adequate heat and that can re-relaxe under photon emission to their original energy level.

In the case of a $\ce{Li}$ atom, we see an emission at $670~\mathrm{nm}$ corresponding to an energy of $1.85~\mathrm{eV}$. This relaxation corresponds to a $\ce{2s<-2p}$ transition; i.e. two quantum states with the same principal quantum number $n$. We need to add at least this transition’s energy if we want to excite an electron, since it is the lowest possible excitation energy.

In the case of a $\ce{Li+}$ ion, we can at first approximation assume the energy levels of a helium atom. Thus, the lowest excitation should be $\ce{1s->2p}$ ($\ce{1s->2s}$ is forbidden), including a change in the principal quantum number $n$. This corresponds to an energy of $\approx 20~\mathrm{eV}$. That corresponds to a wavelength of $62~\mathrm{nm}$ — well in the ultraviolet range, already approaching x-ray wavelengths. That would mean two things:

  • Undergraduate labs would likely disallow this experiment due to the very harmful wavelength (the longest-wavelength line of the Lyman series of hydrogen is $121~\mathrm{nm}$, just for comparison).

  • You would need much more energy to excite the electrons than a simple lab flame would make available. Remember that a bunsen burner flame has a temperature of $\approx 1300~\mathrm{K}$; according to this image on Wikipedia that is not really hot enough to truely emit ultraviolet photons in any reasonable concentration let alone come close to $62~\mathrm{nm}$

By comparing this to the experimental result (‘we can colour the relatively cold flame with lithium salts’), we see that the assumption ‘$\ce{Li+}$ ions are responsible for the flame colouring’ must be rejected.

Note: All calculations and predictions presented herein are based on some kind of layman’s simplification and may be physically inaccurate.


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