# Colour intensity of transition metal complexes

Even though they are both $\mathrm{d^7}$ complexes, the colour of $\ce{[Co(H2O)6]^2+}$ is pale whereas that of $\ce{[CoCl4]^2-}$ is more intense. Does the octahedral or tetrahedral geometry matter in determining the molar absorption coefficient of the two species?

• You bet it does. Look up the crystal field theory. – Ivan Neretin Sep 13 '15 at 15:10
• @IvanNeretin unfortunately, that Wikipedia page does not explain why one colour is intense and the other is weak. – Jan Oct 3 '15 at 13:54

At first, it is necessary to ask why some transition metal complexes are colourful at all — remember that alkaline metals, alkaline earth metals and earth metals usually form colourless solutions, e.g. sodium, magnesium, aluminium. The transition metals’ colour stems from the d-electrons they have and the shape of the complexes they create. A typical orbital scheme of a 3d-metal in an octahedral complex will look something like this (image originally taken from Prof. Klüfers web scritptum to his coordination chemistry course at the LMU Munich):

On the left are the metal orbitals, from top to bottom it is 4p, 4s and 3d. On the right are the ligand group orbitals, assuming six ligands each only donating electrons in a $\unicode[Times]{x3C3}$ fashion only. Any remaining metal electrons will be filled into the $\mathrm{t_{2g}}$ orbitals which correspond to $\ce{d}_{xz}$, $\ce{d}_{xy}$ and $\ce{d}_{yz}$ and into the $\mathrm{e_g}$ orbitals corresponding to $\ce{d}_{z^2}$ and $\ce{d}_{x^2-y^2}$

The energy difference marked as $10\,\mathrm{Dq}$ in the image would then be the energy of a photon required to excite an electron from $\mathrm{t_{2g}}$ to $\mathrm{e_g}$; this photon would thereby be absorbed, which is where we get our absorption coefficient from. It is usually in the visible range of light.

There are a few quantum mechanical principles that allow or disallow excitation from one energy level to another. One of these is the spin-allowance rule: Excitations are only allowed if the spin of the entire system remains unchanged. (So excitation cannot involve a spin exchange.) This is why $\ce{Mn^{II}}$ or high-spin $\ce{Fe^{III}}$ complexes are generally colourless: all d-electrons are of identical spin so there is no freedom to be excited.

A second rule is the Laporte-rule whose formal definition I never properly understood, but which states that d to d excitation is forbidden in centrosymmetric species such as octahedra. This rule would mean that all d-element complexes should be colourless. However, complexes are not static but the ligands vibrate back and forth, often in a non-centrosymmetric way. Whenever a photon of the correct energy hits a complex that is not ideally centrosymmetric, the excitation is allowed and the photon absorbed. Because this is a rather rare phenomenon, the colour of such a complex is weaker than it could be.

The $\ce{[Co(H2O)6]^2+}$ case is more complicated again, because in a perfectly centrosymmetric world it would be degenerate: Since it is $\ce{d^7}$, two spin-down electrons share three $\mathrm{t_{2g}}$ orbitals. It would be favourable to somehow make these orbitals dissimilar. This can happen by distorting the octahedral shape in a Jahn-Teller manner: The two water ligands on the $z$ axis are moved slightly further away, creating a slight stabilisation of all orbitals which contain a $z$ component, most notably $\ce{d}_{xz}$ and $\ce{d}_{yz}$. But since the final product is still centrosymmetric, this paragraph only explains the wavelength of absorption slightly better and can be disregarded.

Next question is why the chlorido complex of cobalt(II) is not octahedric but tetrahedric. This is due to a hypothetical $\ce{[CoCl6]^4-}$ being unbalanced in terms of large chloride ions around a small cobalt ion that only has a low charge to combat the $-6$ the chlorides have to offer. The tetrahedral complex $\ce{[CoCl4]-}$ has everything that the octahedral $\ce{[Co(H2O)6]^2+}$ has except the centre of symmetry. Because centrosymmetry is missing, the Laporte rule is no longer valid and every photon of the correct energy can excite electrons. More excitation equals a stronger extinction.

Also, the colour of both complexes is notably different because the orbital scheme is, too. $\mathrm{t_2}$ and $\ce{e}$ are now much closer together and their order is flipped. Combining that with chloride ions that usually generate low values of $10\,\mathrm{Dq}$, the energy difference is even smaller therefore the excitation wavelength longer therefore the colour we see blueshifted (complementary colour).

• Mostly correct, but there is a problem; the Laporte rule still forbids d-d transitions, even in tetrahedral complexes. The g-g or u/u requirement doesn't apply for tetrahedral complexes, but $\Delta L = \pm 1$ still does. If the transition were fully allowed then you would expect a colour as intense as that of KMnO4 (LMCT is fully allowed), but the extinction coefficient of a typical tetrahedral complex is ~2 orders of magnitude smaller than that of a complex where LMCT occurs. – orthocresol Feb 13 '18 at 21:05

If I remember this correctly (and this has been some time) in order to excite one electron from one energy level to another using a photon the transition dipole moment has to change. You can show this for the vibrational and also for the rotational excitation if you want to look this up. I think for scattering processes like Raman the polarizability has to change and for absorption and emission the transition dipole moment has to change. So if you want to describe this using quantum mechanics you will require an operator, and your interaction operator is in this case one for the transition dipole moment, which is an odd function (x). Now in quantum mechanics, when you want to describe the probability of something happening, you need the two energy levels, the operator and they are all multiplied and then integrated over the total space. So imagine now you need to integrate something that consists of three entries a * b * c, one is already an odd function because of the transition dipole moment operator being odd (x) then the transition has to be from a symmetrically odd orbital into an even one or vice versa. Because otherwise the product would be an odd function. Imagine a simple 2-dimensional function like x³, it has the same area in the negative region as in the positive one therefore the negative contributions and the positive one cancel each other out and the probability of the transition happening is zero.

The octahedron is such a case, because the lower $t_2g$ and the higher $e_g$ both have this 'g' which stands for even. In a tetrahedron this is not the case. The only reason why the octahedron is colored at all is because the ligands are still able to move or vibrate at bit and distort the octahedron taking some symmetry.

That was now a very short summary but it should contain most things.

• Yes, you remember correctly: $x,y,z$ transforms as $\mathrm t_1$ and $\mathrm e \times \mathrm t_1 \times \mathrm t_2$ contains the totally symmetric irrep, so the transition dipole moment is nonzero. There is probably a second factor, related to mixing between $n\mathrm d$ and $(n+1)\mathrm p$ orbitals, which helps to relax the $\Delta L = \pm 1$ requirement as well. In octahedral symmetry the d and p orbitals transform as different irreps and cannot mix. The other (downvoted) answer is trying to get at this, but doesn't seem to explain it very well. – orthocresol Feb 13 '18 at 21:01
• Really? I always thought there is at least a slight mixing of d and p in order to overcome that forbidden d-d-transition? – Justanotherchemist Feb 13 '18 at 21:13
• The way I understand it, at least, is that the forbidden aspect of the d-d transition only strictly applies to isolated atoms. In a metal complex you have MOs which can be considered "mostly d", but they aren't d orbitals because there is contribution from ligand orbitals as well as other AOs on the metal. How much the d-d rule applies is therefore a function of how much d character the MOs have; for Oh where d/p cannot mix this is more, but for Td where d/p can mix this is less. – orthocresol Feb 13 '18 at 21:17
• I checked some of the newer books by Mingos and you are right, they do not mix because of symmetry issues. I didn't find an explanation anywhere but it seems like this only applies to σ-ligands, π-ligands seem to have the $t_2g$-symmetry as well and then I'd expect some mixing again, therefore, allowing the d-d-transition. – Justanotherchemist Feb 15 '18 at 11:22
• On the other hand, how many pure σ-complexes exist in an octahedral geometry? For alkyl with such high oxidation states, you'd expect hydride-elimination, if you protect it by using norbornene derivates you run out of space and for pure hydrides I guess you'd favor the redox-chemistry. So perhaps these very rare complexes just don't have a d-d-transition? – Justanotherchemist Feb 15 '18 at 11:24

In octahedral there is inversion centre so centre of symmetry is there so there is no chance of p-d mixing.. So they are laporte forbidden..... But in tetrahedral complexes there is no centre of symmetry.. So there is a chance of p-d orbital mixing.. Therefore these are laporte partly allowed.. So intensity of color is more. In case of CoCl4 2- than ur octahedral complex

• Could you neaten this up a bit? The extra periods, especially, are not necessary, unless you are breaking up a direct quote into smaller pieces. – jonsca Sep 2 '17 at 2:49