Why is copper sulfate blue and zinc sulfate colorless? Why are some iron complexes attracted to a magnet and others not? Crystal field theory answers both. It builds on coordination compounds, orbital diagrams and light energy.
Key idea
In an octahedral complex, the ligands raise the energy of the two d orbitals that point at them ($e_g$) above the other three ($t_{2g}$), by the splitting energy $\Delta_o$. Strong-field ligands give a large $\Delta_o$ (electrons pair up in $t_{2g}$ first: low spin); weak-field ligands give a small one (electrons spread out first: high spin). Absorbing light of energy $\Delta_o$ lifts an electron across the gap, and the complex shows the complementary color of the light absorbed.
Why it works
The ligands' lone pairs act like negative charges. The $d_{z^2}$ and $d_{x^2-y^2}$ orbitals point straight at the six ligands on the axes, so electrons in them are repelled more; the $d_{xy}$, $d_{xz}$ and $d_{yz}$ orbitals point between the ligands. Whether a fourth electron pairs in $t_{2g}$ or climbs to $e_g$ depends on which costs less: the pairing energy or $\Delta_o$. White light minus the absorbed color is what reaches your eye.
The order of ligands by $\Delta_o$, measured from absorption spectra, is the spectrochemical series. From weakest to strongest field: $\mathrm{I^-}$, $\mathrm{Br^-}$, $\mathrm{Cl^-}$, $\mathrm{F^-}$, $\mathrm{OH^-}$, $\mathrm{H_2O}$, $\mathrm{NH_3}$, en, $\mathrm{CN^-}$, CO.
Worked examples
Worked example: Same ion, different ligands
How many unpaired electrons are in $\mathrm{[Fe(H_2O)_6]^{2+}}$ and in $\mathrm{[Fe(CN)_6]^{4-}}$?
- $\mathrm{Fe^{2+}}$: iron is in group 8 and has lost two electrons, leaving $d^6$.
- Water is weak-field: high spin, one electron in each of the five orbitals, then the sixth pairs in $t_{2g}$: $t_{2g}^4e_g^2$, 4 unpaired.
- Cyanide is strong-field: low spin, all six pair in $t_{2g}$: $t_{2g}^6$, 0 unpaired. $\mathrm{[Fe(CN)_6]^{4-}}$ is diamagnetic.
Answer: 4 and 0.
Worked example: Color and the gap
$\mathrm{[Ti(H_2O)_6]^{3+}}$ absorbs most strongly near $500\ \text{nm}$. Find $\Delta_o$ in kJ/mol and predict its color.
- $$E = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{500 \times 10^{-9}} = 3.98 \times 10^{-19}\ \text{J}$$ per ion.
- Per mole: $$(3.98 \times 10^{-19})(6.022 \times 10^{23}) = 2.39 \times 10^5\ \text{J/mol} = 239\ \text{kJ/mol}$$
- It absorbs green-yellow light, so it looks the complementary color: violet-purple.
Answer: $239\ \text{kJ/mol}$; violet.
- $d^0$ and $d^{10}$ ions ($\mathrm{Ti^{4+}}$, $\mathrm{Zn^{2+}}$) have no electron that can jump the gap, so their compounds are usually colorless.
- Tetrahedral complexes have a smaller splitting (about 4/9 of $\Delta_o$) and are nearly always high spin.
- Paramagnetic (attracted by a magnet) means at least one unpaired electron; diamagnetic means none.
Common mistake
Thinking a complex shows the color of the light it absorbs. It shows the complementary color: red–green, orange–blue and yellow–violet are the opposite pairs. Check: a solution that absorbs red light looks green, not red.