coordination compounds - In an octahedral complex, what happens to the electrons donated by the ligand?

I am trying to understand how complexes are coloured. After some reading, I found out this was due to the d-d splitting induced by the coordinate bonds of ligands to the central metal ion. The electrons in the d-orbitals then transition between each other, and absorb light in the process.

However, I do not understand what happens to the electrons donated by the ligand. Don't they form a sigma bond with the d-orbitals of the metal ion? In which case, wouldn't they go into the d-orbitals and 'fill it up', thus disallowing transitions?

Or is it the case that the d-orbitals of the central metal ion are not involved in the bonding at all?

Answer

This is a very good question, I must say. It requires the understanding of the very fundamentals.

You're right, if the electrons from the ligand pair up with the electrons of the metal, the electrons cannot undergo $d$-$d$ transitions, and no color would be seen.

But do they pair up? Think carefully. Recall one of the most basic assumptions of the Crystal Field Theory: The attraction between the ligand and the metal is assumed to be purely electrostatic in nature. This very important assumption is the sole reason for the explanation of colors of these complexes. As the ligands approach the metal atom, they create a crystal field. These fields tend to repel the electrons of the metal and increases their energy. A perfectly symmetric spherical crystal field would increase the energy level of all the orbitals equally and to the same level. But this isn't the case in actual coordination complexes. An octahedral crystal field would approach along the axes of the $d_{x^2+y^2}$ and $d_{z^2}$ and tend to repel them to a greater extent than the other orbitals ($d_{xy}$, $d_{yz}$, $d_{zx}$). This difference in energies of the orbitals is known as crystal field splitting. We can now group these orbitals into two groups, three low energy $t_{2g}$, and two high energy $e_g$ orbitals.

Keep in mind that the ligands do not pair with the electrons of the metal. They simply repel the orbitals electrostatically and increase their potential energy in the process.

So, now how do the complexes actually get their color then? When the orbitals split, the difference in their energies is called Crystal Field Stabilization Energy (CFSE) and is denoted by $\Delta_{\text{o}}$. When photons of light are incident on the complex, it absorbs the photons which possess the energy equal to that of the value of $\Delta _{\text{o}}$. From quantum theory of electromagnetic waves, it's known that the energy of a photon is given by:

$$U = \frac{hc}{\lambda}$$

Where $h$ is known as the Planck's Constant, with a value of $\pu{6.626×10^{-34}Js}$, and $\lambda$ is the wavelength of light. If the value of the wavelength lies in the visible light spectrum, you can find the color of the light absorbed by the complex.

How do you find the color emitted out? There's a very easy and fun way to find that out. Take a look at this color wheel:

Find where the color of absorption lies. Then the color of the complex lies on the opposite side of the color of absorption.

This is a brief discussion about the colors of complexes. Let me know if you need more clarification.