Ask A Scientist Chemistry Archive

name         Moj S.
status       educator
age          20s

Question -   Why doesn't Mg have the flame test?
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Your inquiry is far more complicated than I suspect you anticipated. Here
goes:

1.    In order to have a positive flame test, the test element must be
present in the flame, which means that the compound containing the element
must vaporize to some extent at the flame temperature. So platinum,
tungsten, or other refractory materials do not give a positive flame test
because their volatility is so small at the temperature of ordinary gas/air
flames that there is none of the element present to give a positive test.

2.    The flame color produced in a flame test results from the emission of
VISIBLE light when an electron in an excited electronic state of the species
drops to an electronic state of lower energy emitting a photon corresponding
to the energy difference of the two electronic states. If the species has no
such pair of state with the required energy difference, it will not emit a
VISIBLE photon of light. It may, or may not, emit in the infrared and/or
ultraviolet part of the electromagnetic spectrum.

The wavelength range of visible light is about 400-800 nanometers. This
corresponds to an energy difference of about 25,000 - 12,500 cm^-1
respectively. So there must exist a species with electronic states within
this energy range. This means that the emitting species is a neutral atom or
molecule because, the ionization energy of most species is greater than this
"window" of visible light. Atomic hydrogen is an example of a species which
does not have electronic energy differences that are in the range of visible
light. The lowest pair of energy levels of hydrogen atoms (1s<-----2p ) has
an energy of abour 82,260 cm^-1 which is well into the ultraviolet range of
the electromagnetic energy spectrum. This is also the reason why H2/O2
flames are almost invisible or faintly blue (due to emissions of oxygen).

3.    In addition, not all transitions between electronic energy levels are
"allowed". The "allowed" transitions are determined by quantum mechanics,
which is beyond the scope of this forum; however, the bottom line is that
these "forbidden" transitions, even though they have an energy difference
between 25,000 and 12,500 cm^-1 are absent, or at least extremely weak. For
example, in the element K, the transition between the
states (4s<------5s), which has an energy difference in energy of 21,027
cm^-1 is not observed, even though the energy difference lies within the
visible range of the electromagnetic spectrum. It is "forbidden". In
addition, even among "allowed" transitions, the intensity of emissions
varies over many orders of magnitude -- see #4. below for example.

4.    A weaker emission may be masked by a stronger emission. This is
typically the case of substances containing Na salts as an impurity. The
intense yellow transition in Na (3s <------3p) at about 17,000 cm^-1
obscures almost all other emissions unless it is filtered out or otherwise
removed (by passing the emitted light through a prism or spectrograph for
example).

5.    It is not always self-evident WHAT species are present in a flame.
Atomic Mg has an allowed transition (3s2 <------3s3p) at about 21,900 cm^-1,
which should be visible -- and may be under proper conditions. However, if a
magnesium halide such as MgF2 or MgCl2 is heated at a fairly low
temperature, of the order of 1200-1300 C., the predominant vapor species is
the molecular form MgX2, and not the free atoms. An interesting aside,
though not relevant to your question, is that CaF2, SrCl2, all Ba halides,
and possibly even MgF2 are bent triatomic molecules, not linear, which a
simple ionic theory would suggest.

Vince Calder
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