# What makes tetrahedral and octahedral transition metal complexes coloured?

May 10, 2016

In solutions (i.e. as ions), they form transition metal complexes that absorb visible light.

CRYSTAL FIELD SPLITTING DIAGRAMS

It is easiest to show why the ideal octahedral and tetrahedral complexes are sometimes colored. Recall their splitting diagrams from crystal field theory?

Well, the correct version (which accounts for stabilization of the ${d}_{x y}$, ${d}_{x z}$, and ${d}_{y z}$ orbitals for an octahedral field, or for the ${d}_{{z}^{2}}$ and ${d}_{{x}^{2} - {y}^{2}}$ for the tetrahedral field), is: These splitting energies are called $\setminus m a t h b f \left({\Delta}_{o}\right)$ for the octahedral complex and $\setminus m a t h b f \left({\Delta}_{t} = \frac{4}{9} {\Delta}_{o}\right)$ for the tetrahedral complex.

HOW DO THESE SPLITTING ENERGIES COME IN?

Whenever ${\Delta}_{o}$ or ${\Delta}_{t}$ are in the range of 400~"700 nm" (or 25000 ~ "14285 cm"^(-1)), the compound can absorb visible light to excite an electron from a lower-energy $d$ orbital to the ones right above them.

Then, the relaxation of these electrons would emit light back to your eyes and allow you see a specific color.

You will see the color complementary to the one corresponding to the wavelength absorbed (e.g. absorb blue light, see orange light).

So, you would see colors resulting from ${e}_{g} \to {t}_{2 g}$ relaxations, or ${t}_{2} \to e$ relaxations. These can be fairly weak, however. These have molar absorptivities at around $\epsilon \approx \text{50 L/mol"cdot"cm}$.

Note that those complexes that are lightly-colored have electronic transitions that are forbidden, whether by having a center of symmetry (Laporte-forbidden) or by having the requirement of flipping an electron spin (spin-forbidden)

CHARGE-TRANSFER TRANSITIONS

You can also have charge-transfer transitions, if your transition metal complex also has a halogen in it (such as ${\left[{\text{IrBr}}_{6}\right]}^{3 -}$, a ${d}^{6}$ complex). These are generally high-spin complexes due to the weak-field nature of these ligands.

These halogens can interact with the solvent (such as water) and the ligands on the complex can donate electrons into the highest-energy $d$ orbitals. This is known as LMCT, or ligand-to-metal charge transfer.

You can also have something similar in a complex with many $\pi$ acceptors, such as $\text{CO}$, ${\text{CN}}^{-}$, ${\text{PR}}_{3}$, and ${\text{NO}}^{+}$. This is known as MLCT, or metal-to-ligand charge transfer.

Both of these (LMCT and MLCT) are going to give strong charge-transfer absorption bands, and since they are so strong (often the molar absorptivity can be around $\epsilon \approx \text{50000 L/mol"cdot"cm}$), they can extend into the visible region and still give vividly-colored complexes.