An attempt to simplify (to make it a bit more comprehensible to a curious highschooler) and explain some of the terminologies used in Xahoc's answer. This is not intended to be a standalone answer and should not be considered as one; please read through the rest of the answers for a thorough explaination.
All coordination compounds have a net stabilizing energy called crystal field stabilization energy (CFSE), due to a crystal field effect with respect to a spherical symmetrical (think of ligands approaching the central metal atom (CMA) equally from all the directions, like a sphere) ligand field; CFSE always has a negative or a positive value and one of its fundamental uses is the prediction of stability of a complex (greater the magnitude of CFSE, greater is the stability). I use a simple formula to calculate CFSE:
$$ \pu{CFSE= -0.4\Delta_\circ (n_{t_2g}) + 0.6\Delta_\circ (n_{eg}) + n'P}$$
The formula doesn't hold much importance here, although the term $\Delta_\circ$ does. What's that? Glad you asked. Before we dive in, something about crystal field theory:
The CMA is assumed to be at origin and the six ligands are approaching the CMA from $+x, -x, +y, -y, +z, -z$ directions. Now the ligands are approaching along the axis, and $d_{x^2-y^2}$ and $d_{z^2}$ orbitals have their lobes along the axis, so they:
- Face more repulsion from the incoming ligands, which increases their energy content. And did I tell you that they're codenamed $e_g$?
- Form a sigma bond, which explains:
When a $\ce{Cl-}$ ligand pi-donates into the metal center, it can only interact with the $\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ $(t_\mathrm{2g})$ orbitals because the $\mathrm{d}_{x^2-y^2}$, $\mathrm{d}_{z^2}$ $(e_{g}^*)$ orbitals have only a sigma-type interaction with the ligand $\dots$
The remaining three orbitals (namely $d_{xy}, d_{yz}, d_{zx}$) are code-named $t_2g$ and are at a lower energy level.
This splitting up of the orbitals into two groups is called crystal field splitting. The energy difference between those two groups of orbitals is precisely what our $\Delta_\circ$ measures. Well, I've been talking about octahedral complexes all along, $\Delta_\circ$ has an evil brother $\Delta_t$ which goes for tetrahedral complexes. So now we know: $\Delta_\circ$ is crystal field splitting energy. Did you say energy? I think I've studied this formula somewhere:
Energy = $\displaystyle \frac{hc}{\lambda}$
Seems good, let's apply it:
$\Delta_\circ = \displaystyle \frac{hc}{\lambda_{absorbed}}$
Wait! What's up with that 'absorbed' thing? Nothing much, but it appears that complexes seem to know some unicorn magic. So what they do is: they absorb (a wavelength corresponding to) a particular color and emit a wavelength that corresponds to another color. Don't worry, some noble souls have deciphered that, they follow this:
Absorb orange, emit indigo; absorb blue, emit red, etc... Green, however is a mysterious case. Also, please keep in mind that:
If $\Delta_\circ$ is less, $\lambda_{absorbed}$ is more and $\lambda_{observed}$ and vice-versa. Good to go. When Xahoc says,
you reduce the octahedral field splitting energy $\Delta_\circ$ which redshifts the absorbance of the compound.
They mean that a higher wavelength of light (on a blue-colored tone) must've been absorbed (reddish colors have a higher wavelength).
Why does $\Delta_\circ$ decrease?
Excerpts and images that follow are directly from Miessler et al. Content is copyrighted to their respective owners.
Understanding this can be quite tricky unless you're acquainted with angular overlap method. Nevertheless, let's give it a shot.
What's angular overlap method?
This is a method of estimating the relative magnitudes of molecular orbital energies within coordination complexes. It explicitly takes into account the orbitals responsible for ligand binding as well as the relative orientation of the frontier orbitals.
Modern computational chemistry allows calculations to predict geometries, orbital shapes and energies, and other properties of coordination complexes. Molecular orbital calculations are typically based on the Born–Oppenheimer approximation, which considers nuclei to be in fixed positions in comparison with rapidly moving electrons.
I'm skipping the description of $s$, $p$ and $d$ orbitals for the sake of simplicity. Further, have a look at this orbital diagram. (click on the picture for a larger image)
The beautiful colors of many transition-metal complexes are due in part to the energy difference between the $t_2g$ and $\textit{e*g}$ orbitals in these complexes, which is often equal to the energy of photons of visible light.
As an example, if the energy levels of $\textit{e*g}$ decrease, $\Delta_\circ$ decreases. A similar thing is at play in our case, albeit this time the molecular diagrams are different and $\Delta_\circ$ corresponds to a different set of orbitals, if you're still curious I strongly recommend to read about this theory, further simplifications as such are beyond the scope of this answer.
