How does d-orbital splitting form colors for transition metals?
Why can't the electrons in the d-orbital split for the element Zinc?

  • 2
    $\begingroup$ Electrons don't split. Orbital energies do. They do it in zinc all the same, but it has completely filled d-orbital, hence no electronic transitions, hence no color. $\endgroup$ Commented Nov 15, 2016 at 15:16
  • 1
    $\begingroup$ In any form of spectroscopy (nmr, IR, microwave, uv visible etc) a photon is absorbed and the molecule/atom goes to a new state. In your case an electron is promoted from one level into an empty or partly empty level. The colour appears as the photons absorbed can no longer be observed in the light source. If the levels are maximally filled then no transitions are possible $\endgroup$
    – porphyrin
    Commented Nov 15, 2016 at 15:26

1 Answer 1


Let’s first address what leads to the colour of a given species. Colour is the absorption of one or more wavelenghts of visible light. Considering that visible light is rather energetic (but not as energetic as UV light and way less than X raays), this energy difference typically corresponds to the difference between occupied and unoccupied (spin-) orbitals in compounds. Thus, if a photon of the correct wavelenght interacts with an electron present in orbital $\Psi_1$, that electron can be excited to orbital $\Psi_2$ with the photon’s energy. Later relaxation processes are often complex, but sometimes visible light can again be liberated upon relaxation; typically as phosphorenscence or fluorescence. I have been glossing over a lot of important issues here, but realise that ‘what is colour’ is too broad a question for this margin.

d-orbital splitting does not form colour in itself nor does it happen in itself. In an uncoordinated metal ion (or atom) in vacuum, the five d-orbitals are degenerate and thus have the same energy level. It is only upon bonds being formed to other species that any type of splitting can occur. This is typically discussed in the context of coordination complexes.

The simplest model (but also the weakest) to explain d-orbital splitting on coordination complexes is the crystal field model. It basically assumes negative point charges approaching the central metal from certain axes which leads to destabilisation (increase in energy) of any orbitals that point towards that axis. In an octahedral complex, the axes are $x,y$ and $z$ so the orbitals whose energies are increased are $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$ — the ones with lobes pointing in the direction of the $x,y$ and $z$ axes.

A more sophisticated model to describe complexes and d-orbital splitting is the use of molecular orbital theory which comes in multiple levels (inclusion or ignorance of π-symmetric ligand orbitals). Scheme 1 shows the typical molecular orbital scheme of an octahedral complex.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

On the left-hand side of scheme 1, you can see the metal’s 3d, 4s and 4p orbitals, on the right-hand side six σ-symmetric ligand orbitals; in the middle their interactions are given. As we can see, a principal stabilisation derives from the $\mathrm{e_g}$ orbitals (corresponding to $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$) forming bonding-antibonding interactions with the corresponding ligand group orbitals. This leads to the metal-centred orbitals becoming antibonding ($\mathrm{e_g^*}$) and thus rising in energy. The $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}, \mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$) are nonbonding with respect to σ interactions, only.

The energy difference, marked in the scheme as $10~\mathrm{Dq}$, typically corresponds to the visible wavelength a complex will absorb and thus to the complementary colour of the complex. Chromium(III) complexes (three d-electrons; all populating $\mathrm{t_{2g}}$ orbitals) are typically green, the complementary colour of which being red, meaning they absorb red light. The wavelength can be calculated.

And finally, scheme 1 above also tells you what happens if zinc forms complexes. For a start, there is d-orbital splitting to be observed in zinc complexes. However, the overall stabilisation is very low since zinc is a $\mathrm{d^{10}}$ ion meaning that all five d-orbitals ($\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$) are fully populated — since two of those are antibonding, all bonding interactions derive solely from zinc’s 4s and 4p orbitals. Finally, since $\mathrm{e_g^*}$ is fully populated, the lowest energy difference with which a photon can be absorbed is no longer that marked $\mathrm{10~Dq}$ but that which corresponds to the $\mathrm{e_g^* \rightarrow a_{1g}^*}$ transition. This very high energy difference is firmly within the ultraviolet range and invisible to the human eye.


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