Crystal Field Explanation of Colors of Coordination Compounds

Question

Crystal Field Theory explains colors of Coordination compounds as follows : A d-orbital splits into multiple orbitals, the process being called crystal field splitting. When white light falls on the compound, an electron makes a transition into a higher state thus absorbing a particular wavelength of light. The rest of the light is reflected. This reflected light is responsible for the color of the compound.

Problem: Suppose, all electrons have now transit into a higher state. If more light falls on the compound, there should be no more transitions as all the electrons are already in the highest state. This means the compounds should lose color after some time in the light.

Question: How is the light absorbed again and again? If we suppose that the electron falls back to lower state, that would emit the same wavelength it absorbed. Even in this case, how do we explain color?

P.S. Also, if they can absorb only a particular wavelength(given quantisation), that should not bring about a color change. There must be a range of wavelengths(however narrow) being absorbed. What is the reason for that?

Answer 1

Your assumption that the electron emits a photon when transitioning back to the ground state is not entirely correct.

The excited electron can also transition back by transferring the excitation energy to heat energy. The compound then warms up ever so slightly and the electron is buzzing away happily in the lower $d$ orbitals.

As to your P.S.-ed question, I'll quote Wikipedia:

The main factors that cause broadening of the spectral line into an absorption band of a molecular solid are the distributions of vibrational and rotational energies of the molecules in the sample (and also those of their excited states). [...] In gas phase spectroscopy, the fine structure afforded by these factors can be discerned, but in solution-state spectroscopy, the differences in molecular micro environments further broaden the structure to give smooth bands.