I have an assignment question for second year inorganic which asks to rank the intensity of d-d transitions for a number of complexes. Two are $\ce{[Fe(OH2)6]^3+}$ and $\ce{[Fe(CN)6]^4-}.$ Both will have Laporte-forbidden transitions, but which will be 'more forbidden', so to speak?


3 Answers 3


You see that both cases are Laporte-forbidden, now how about spin-forbidden?

Assuming the complex vibrates so it's no longer exactly octahedral, can you make a spin-allowed transition for $\ce{[Fe(CN)_6]^{4-}}$? How about $\ce{[Fe(H_2O)_6]^{3+}}$? When you answer this you will answer your question.

You should find that $\ce{[Fe(H_2O)_6]^{3+}}$, with all orbitals occupied by electrons having the same spin, has only spin-forbidden d-d transitions. Therefore it will undergo weaker transitions than $\ce{[Fe(CN)_6]^{4-}}$ which has some spin-allowed transitions.


The chance of a transition being allowed or 'forbidden', and hence its intensity, depends on both the symmetry of the states involved and the spin change that occurs, if any.

The first step, however, is to find out how the orbitals $e_g,\; t_{2g}$ are occupied with the electrons. The Fe(II) is $d^6$ and has a 'strong' ligand in CN$^-$ compared to H$_2$O (see spectrochemical series) which means that the CN splits the orbital energies by more than the spin pairing energy. This makes the Fe(II)(CN)$_6^{4-}$ low spin so that the $t_{2g}$ are filled and the $e_g$ empty. Thus a transition on spin terms is allowed, i.e. promoting one electron into the $e_g$ orbital can be done without an overall spin change. The ground state has no unpaired electrons so the total spin $S=0$ so the multiplicity is $2S+1=1$, a singlet. The excited state has 5 electrons in $t_{2g}$ and 1 in $e_g$ this can also be spin zero. If there were a spin change to $S= 1$, thus making a triplet, then this would be spin forbidden.

In the Fe(III) complex there are 5 electrons and a 'weak' ligand so the $t_{2g}$ to $e_g$ energy gap is smaller than the spin pairing energy and this complex is high spin $S=5/2$ with multiplicity $2S+1=6$, i.e. each orbital has a single electron with the same spin. Exciting an electron can only mean that spin pairing must occur and this will change the overall spin multiplicity to $S=3/2, \;2S+1=4$ which corresponds to a change in (spin) angular momentum, and as as this must be conserved the transition is 'forbidden'.

Because the photon does not interact with the electron spin any spin changing transitions are very weak indeed. They do become partly allowed by spin-orbit coupling which is the interaction of the electron spin with the angular momentum of the electron in its orbital.

The second part of the allowed/forbidden transition involves the symmetry of the states. This occurs because the chance of a transition depends on the transition moment integral. This has a spatial and spin part, we never need to work out its value but use symmetry and spin arguments, as above, to determine which, is any, part is exactly zero, i.e. not small but exactly zero. The integral is $\int \psi_i \mu \psi_g d\tau \int \alpha_i\alpha_f ds$. The terms are $i,\;f$ for initial and final states, $\mu$ is the transition dipole operator (see later) and $\psi$ are the wavefunctions for the two states involved and $\alpha$ the electron spin wavefunction. As these latter are orthogonal if they are the same $\alpha_i=\alpha_f$ then the integral is finite and if different it is zero. This is what is explained in words above in the spin case.

The transition dipole operator is a measure of the interaction of the photon with the molecule and it is an 'odd' function. Suppose $f(x)$ is the function, if it is odd, such as $f(x)=x$, then $f(-x)=-f(x) $ and if even such as $f(x)=x^2$ then $f(-x)=f(x) $. To work out the $\int_{-\infty}^\infty \psi_i \mu \psi_g d\tau$ integral we can use the odd-even property of the wavefunctions. This is usually done by reference to their symmetry species as defined in a point group table. However, from the properties of odd even functions we can determine a lot. An odd function integrated from $-\infty \to \infty$ is always zero. The negative part always exactly cancels the positive one. If the two wavefunctions $\psi_i,\;\psi_f$ are both even, i.e. have even symmetry species, and as $\mu$ is odd then the product $\psi_i\mu\psi_f$ is odd and the integral is exactly zero and a transition is forbidden, and its intensity zero.

However, because a molecule can vibrate (or be distorted by Jahn-Teller interactions) it is quite easy to change the molecules symmetry and so make the transition a little 'allowed'. The strength of this symmetry 'forbidden' transition (The Laporte Rule) is about 100 to 1000 times stronger than any spin-forbidden transition.


(I wish I could find pictures of the UV/Vis spectra to support this answer but alas …)

In the very narrow context of your question that only considers $\mathrm{d{\rightarrow}d}$ transitions, I would like to refer entirely to Oscar Lanzi’s answer. However, there is an important aspect to the colours of both ions that $\mathrm{d{\rightarrow}d}$ transitions cannot explain.

As you correctly note, both complexes should have Laporte-forbidden transitions. This, and the expected answer Oscar Lanzi gives, can be demonstrated most beautifully by adding fluoride to a solution of $\ce{[Fe(H2O)6]^3+}$ to give the $\ce{[FeF6]^3-}$ complex—the solution will become entirely colourless as is expected based on $\mathrm{d{\rightarrow}d}$ transitions.

However, iron(III) solutions tend to be distinctly yellow as is $\ce{[Fe(CN)6]^4-}$; the latter even carries the German name gelbes Blutlaugensalz (rough translation: yellow alkaline blood salt) highlighting its yellow colour. Neither can be explained by $\mathrm{d{\rightarrow}d}$ transitions which, as they are Laporte-forbidden, should result in weak colours at best. Instead, the colours we observe originate from metal-to-ligand charge transfer transitions. For reference, please see figure 1 below which contains the full molecular orbital scheme of an octahedral $\ce{[ML6]}$ 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.

The transitions which result in the strong yellow colour of both complexes originate not from transitions between the $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ complexes (the transition labelled $\pu{10Dq}$ in figure 1) but from those labelled $\mathrm{t_{1u}}$. These correspond to orbitals mostly centred on the ligands, roughly describing a subset of how electron density is donated to the metal in dative bonds, and thus the transition could be described as oxidising the ligand and reducing the metal briefly; compare equation $(1)$.

$$\ce{[\overset{+III}{Fe}(H2\overset{-II}{O})6]^3+ -> [\overset{+II}{Fe}(H2\overset{-II}{O})5(H2\overset{-I}{O})1]^3+}\tag{1}$$

It is important to note that this should be considered a mere formality or thought experiment; it is an intramolecular transition and the species overall does not change. But this depiction may be helpful to imagine what is going on.

In any case, these transitions are from an ungerade orbital to a gerade orbital—the original $\mathrm{t_{1u}}$ orbital is antisymmetric with respect to inversion while the target $\mathrm{t_{2g}}$ or $\mathrm{e_g^*}$ orbitals are both symmetric with respect to inversion. Thus, this transition is not covered by Laporte’s rule (which only applies to gerade to gerade or ungerade to ungerade); the resulting absorption is strong.

These charge transfer transitions are orders of magnitude more intense than any Laporte-forbidden $\mathrm{d{\rightarrow}d}$ transitions and thus they dominate both the UV/Vis spectra and the resulting colour we perceive.


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