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My book writes:

When white light shines through a solution of a complex ion of a transition metal, photons of a particular frequency are absorbed and their energy promotes an electron from lower energy level to upper energy level. We perceive the complementary color to the light absorbed.

However, I have a question about this. Once the electron is promoted to a higher energy level, won't it fall back down and give off a photon?

Surely the electron cannot remain at the higher energy level forever, as that would mean that no further electronic transitions could taking place, and the solution would turn colourless.

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Absorption of a photon typically results in a vibrationally excited higher electronic state of the same multiplicity.

$$\ce{S_0 ->[$h\nu_\mathrm{ex}$] S_1}$$

In most cases, the excited state deactivates through internal conversion in a radiationless process via vibrational energy exchange with solvent molecules. No light is emitted here, but the solvent is actually heated up, although it might not be measurable.

There are however cases, in which the initially formed excited singlet state species first relaxes to its lowest vibrational mode and then emits a photon to reach the ground state.

$$\ce{S_1 -> S_0 + $h\nu_\mathrm{f}$}$$

This deactivation via radiation is known as fluorescence. Fluorescence lifetimes are typically rather short (in the ns range) and the emitted photon has a lower energy than the absorbed one. This means that the emission spectrum is shifted to the red end of the uv spectrum ($\lambda_\mathrm{f} > \lambda_\mathrm{ex}$). The effect is known as Stokes shift.

A classical case for a fluorescent transition metal complex is $\ce{Ru(bpy)3^{2+}}$.

In acetonitrile, it absorbs around $\lambda_\mathrm{ex} = 460~\mathrm{nm}$. The fluorescence is observed around $\lambda_\mathrm{em} = 620~\mathrm{nm}$.

Ru(bpy)3 2+

Other bidentate ligands - more rigid than 2,2'-bipyridine (bpy) - often used are 1,10-phenanthroline (phen), dipyrido-[3,2-a:2',3'-c]phenazine (dppz), and benzo[i]dipyrido-[3,2-a:2',3'-c]phenazine (dppn).

ligands

In other cases, the excited singlet state undergoes a change in the spin multiplicity (Intersystem crossing) to a triplet state.

$$\ce{S_1 ->[ISC] T_1}$$

These excited triplet states are usually much longer-lived. When they eventually deactivate via emission of a photon, this process is known as phosphorescence.

$$\ce{T_1 -> S_0 + $h\nu_\mathrm{p}$}$$


Keeping all these processes in mind, one question arises - particular from younger students or pupils:

If our metal complex absorbs a photon and forms an excited state with such a short lifetime, why can we see a nevertheless observe permanent colour?

The answer is in the enormous numbers!

Continuous irradiation with white light means that we send a huge number of photons. Moreover, the solution of our transition metal complexes in a cuvette (or a test tube) also contains more than one molecule. In fact, much more!

As a result, we have a huge number of incidents

  • The complex absorbs a photon of a certain energy, which means that one colour is "swallowed". The complex forms a short-lived "excited state"
  • the solution reflects the remaining light
  • the excited state complex deactivates again

Since this happens all the time while we shine the light on our sample and for a lot of molecules, we observe a summary of the events: a coloured solution.

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  • $\begingroup$ I am really sorry to say this but this answer was too complicated for a high school student like me. If you do have the time, could you please add a more simple answer as to why we seen the solutions of complex ions of transition metals to be continuously colorful...I learned that color is absorbed from white light during a $d$-$d$ transition... but once this transition occurs and complementary color seen, how do we continue to see a color as the electron stays in the upper energy level... $\endgroup$ – Eliza Jan 28 '14 at 5:03
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    $\begingroup$ My bad, I didn't realize that you attend school. Keep the difficult explanations for later in your scientific career as a photochemist ;-). I added a bit to my answer to make it useful for now :) $\endgroup$ – Klaus-Dieter Warzecha Jan 28 '14 at 6:50
  • $\begingroup$ Thank you so much. This was much easier to understand. Just one thing, when the excited electron falls back to the lower energy level, is the energy of the electron released as heat?I know this is a stupid question but the thing is that I mix up this concept with the concept of flame colors of different metal ions :/ $\endgroup$ – Eliza Jan 28 '14 at 9:33
  • $\begingroup$ Yes, exactly :) $\endgroup$ – Klaus-Dieter Warzecha Jan 28 '14 at 9:36
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    $\begingroup$ Flame colours are nice! But there's no absorption of photons here. It's the other way around. With the heat of the flame, an excited state is generated! This one deactivates to the ground state by releasing the energy as light. So, flame colours result from emission of light. $\endgroup$ – Klaus-Dieter Warzecha Jan 28 '14 at 9:43
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Here's a simple explanation of why that happens:-

First the electron gets promoted to the higher level after absorbing one photon corresponding to the visible light. This makes you see the complementary color, as you already said.

BUT when the electron comes back to the unexcited state, it does not do that in a single step! The electron comes back to the unexcited state taking smaller intermediate steps. Each of these smaller transitions corresponds to the emission of a low frequency photon belonging to the Infra Red region. This is outside the human visible range and thus we don't "see" these photons. But these Infra Red photons do heat up the solution though the change may not be measurable.

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