I have read in a book (NCERT India) about flame color for salts of alkali metal and color of transition metal complexes.

In the first case, it says that during excitation of electrons to higher energy it absorbs energy and then when it falls back it is emitted again giving color to the flame.

In the second case, it says that during d-d transition it absorbs a wavelength of photon and complementary color is shown. So, why can't there be emission in the second case like the first case.

  • $\begingroup$ Because the s and p orbitals always have more effective nuclear charge than d orbitald for a fixed value of principal quantum number... Moreover, the d orbitals are prone to splitting in the presence of applied field,so these d-d transitions are now across a bandgap of energy,which leads to light of certain wavelengths being emitted in the visible range.. Due to more penetration towards the nucleus, p-p transitions are merely releasing exchange energy from degenrate transitions(although p orbitals can split under some special conditions) $\endgroup$ Apr 8, 2019 at 12:48
  • 6
    $\begingroup$ There could be emission from the transition metal complexes, but usually there are non-radiative pathways that compete with emissions and remove the energy before it has much of a chance to emit. However, in some crystals, e.g. Ti sapphire(Ti/Al2O3), Ruby (with Cr ions) and Nd YAG (Yttrium Aluminium Garnet) among others, there is sufficient emission from some levels to make them into lasers. $\endgroup$
    – porphyrin
    Apr 8, 2019 at 13:35

1 Answer 1


These at two different scenarios and two different phenomena. Flame color is due to atomic emission and complexes are colored because of light absorption in the solid state and in solution.

In a high temperature flame, an atom is considered to be a free and independent entity. With heat, the electrons in the atoms are excited to higher energy levels. However, selection rules determine where an electron can land e.g. s electron goes to p, p electron can go to d, however s electron cannot go to d.

In case of transition metals, Hans Bethe (Nobel Prize) developed a theory called called crystal field splitting. In a free gaseous atom, all five d orbitals orbitals have the same energy (called degenerate orbitals), so electronic transitions do not make any sense. It is also forbidden. However in the presence of ligands, the d- orbitals are no longer of the same energy. Electronic transitions are possible within d orbitals. Search crystal field theory if you are interested...

  • $\begingroup$ I know about Crystal field splitting but I am not able to understand why can't a d-d transition be reversed and the same wavelength that was absorbed be re-emitted. $\endgroup$ Apr 9, 2019 at 2:17
  • 1
    $\begingroup$ There are two ways an excited species lose energy (i) re-emission and (ii) non-radiative loss of energy. Re-emission is somewhat common in excited gas phase atoms and even molecules, on the other hand non-radiative process are quite common in solution. Very few molecules are fluorescent. Transition metals ion also lose energy by non-radiative processes. $\endgroup$
    – AChem
    Apr 9, 2019 at 3:05

Not the answer you're looking for? Browse other questions tagged or ask your own question.