# Crystal field theory

We have been discussing metal complexes using the valence bond theory, but while it is a useful method for discussing simple bonds, it cannot account for the colours and magnetic properties of metal complexes. To understand why metal complexes are such beautiful colours, we have to use crystal field theory. Crystal field theory gives us information on the electronic structure of the metal atom, considering how the d orbitals will be affected by the ligands.

For an octahedral complex, we imagine the ligands are point charges that sit on the Cartesian (x,y,z) axes. The d orbitals are arranged around the nucleus as shown below, with two orbitals pointing directly towards the ligands, and the other three in-between. The d(x2-y2) and d(z2), which point at the ligands, experience electrostatic repulsion and are therefore at higher energy than the d(xy), d(yz) and d(xz) orbitals. This means the d orbitals are no longer degenerate, and the orbital diagram is split: The energy difference between the high and low energy orbital is ∆O (the ligand field splitting parameter, if you’re feeling wordy), and it is the size of ∆O that dictates the spin, colour and magnetic properties. The size of the splitting parameter is determined field strength of the ligand. A strong field ligand, such as CN, will give a large splitting parameter and a weak field ligand, I for example, will have a small value for ∆. The ligands are arranged into the spectrochemical series, indicating the strength of the ligand field.

Now that we know about splitting, the next question is how do the d electrons distribute themselves between the non-degenerate orbitals? For example, we have two Fe2+ complexes, [Fe(CN)­6]4- and [Fe(H2O)6]2+. We get two different splitting diagrams: When we go to fill the orbitals according to Hund’s rule, we should half fill each orbital before going back and pairing the electrons, as it requires energy to put two electrons in an orbital. This is fine, if ∆ is low, as with [Fe(H2O)6]2+, and we get a high spin complex, with a higher number of unpaired electrons. However, if ∆ is large, as with [Fe(CN)­6]4-­, the pairing energy is lower than the energy required to get an electron into the high energy orbitals, so the complex is low spin. High spin complexes are paramagnetic, as the unpaired electrons give the complex an inherent magnetism. The magnetic moment can be calculated according to the following equation: μs.o. = √{4S(S+1)}. This is the spin-only formula, where S is the number of unpaired electrons.

The colour of transition metal complexes also depends on the magnitude of ∆, as the energy of absorbed light will correspond to the energy difference between the low and high energy orbitals. If ∆ is small, the complex will absorb lower energy light, (red) and appear blue,  the opposite is true for complexes with high ∆.

http://www.chemguide.co.uk/inorganic/complexions/colour.html

http://mooni.fccj.org/~ethall/cobalt/cobalt.htm (warning – animated gifs)

http://chem1180.blogspot.com/2010/12/227-crystal-field-theory.html