# Why do complex ions not emit light from de-excitations of electrons?

I understand that complex ions are coloured due to d-orbital splitting which results in electrons being able to absorb wavelengths of visible light and become excited to the higher energy state meaning the transmitted light is coloured. However I don't see why the electron would not simply de-excite and emit the same wavelength of light that it absorbed which would result in no visible difference. Is there some reason why the same wavelength is not emitted or perhaps the electron does not de-excite at all?

The fate of electronically excited states is not immediately obvious.

First of all the rate at which the electron emits a photon and falls down a level is dependent upon the energy difference between the two levels, it is only an expectation value of a macroscopic observable, on an individual atomic scale it is impossible to predict when the electron in the upper level will emit a photon, however multiple instances of the same measurement will tend towards the expectation value. The direct result of this is that there is a path difference between the emitted photon and the 'rest' of the photons that were not absorbed by the complex and the result is a different interference between photons in the transmitted light compared to the incident light and hence a new colour is observed.

There is another fate of excited electrons, the Franck-Condon principle invokes the very plausible assumption that the nuclei do not respond immediately (at least on the time scale of electronic transitions) to the new equilibrium bond position that is determined by the new electronic state of the complex. This means the excited electronic state will most likely not be in the lowest vibrational state of the new equilibrium bond configuration, the splitting of vibrational energy levels is orders of magnitude lower than those of electronic transitions and hence the same law that governs electronic emission applies, the electron is much more likely to fall down these vibrational energy levels first, without emitting photons of visible light.

The electron will then be expected to, by the time it is occupying the lowest vibrational state of the excited electronic state, invoke a transition down to the ground state energy level. The energy of the excited state is described in a basic form by $$H=E^*+v$$

where $H$ is the energy of the absorbed photon, $E^*$ is the energy of the excited state in the lowest vibrational mode and $v$ is the contribution from the initial vibrational state of the excited complex.