# Why is ruthenium(III) intensely colored in water

I was preparing a solution of ruthenium(III) chloride (aqueous) when I added some (nearly black) $\ce{RuCl3 . H2O}$ to the water. I was surprised that a tiny amount could dye the water so well, I'd say you can even compare this to $\ce{KMnO4}$.

The question is why is the color so intense?

Ruthenium(III) is a $\mathrm{d^5}$-system so the transitions should be spin-forbidden if we are talking about $\ce{d\bond{->}d}$ transitions. If this is the case we can consider a CT-transition. But if I look at the chemistry of manganese(II) it is slightly pink and doesn't dye water at all although it has the same ligands (chlorido and aqua). And if we compare the colors of manganese(VII), chromium(VI) and vanadium(V) as well as manganese(VII) to renium(VII) with the same ligands, in this case oxido-ligands we see that the more stable states (vanadium(V) and renium(VII)) do not pull electrons that strong anymore and the color shifts further into something invisible to the human eye.

So ruthenium(III) should be even less colored than manganese(II) if we ignore the fact that it has 3 chlorido ligands for a second. So the only option I have left would be the π-donor ability of the chlorido-ligands.

I have read some time ago, that manganese is one of the only elements that form in as their aqueous $\ce{MCl2}$ a cis-dichlorido-coordinated species while all the other (3)d-metals form trans-coordinated ones. And I have already asked about the cis to trans difference in another post so I know now that it does indeed make a difference.

I'd like to compare this to the ruthenium(III) now but ruthenium(III) in water is somewhat complicated as a lot of possible compounds can form especially as the trihydrate will start to exchange chlorido-ligands for aqua ligands at a certain period of time and the trihydrate salt is said to be yellow-orange like iron(III) while I start with a monohydrate which is black.

I'm not sure whether this effect is now related to the rutheium only or if this is just the case because of some symmetry effect but I can't find a good answer on this anywhere.

• Ru(III) is d5, but is it high- or low-spin? – orthocresol Mar 4 '17 at 14:23
• Might be. You mean like with Co(III) where nearly every complex is low spin? Might be. They even have a diagonal relationship... – Justanotherchemist Mar 4 '17 at 21:37
• It's not about a diagonal relationship - which barely exist in the d-block - more like 4d/5d metals are nearly always low spin because of their large ligand-field splittings $\Delta$. – orthocresol Mar 4 '17 at 22:00
• Ah good to know. I always thougt this was only the case for the 5d metals. Ok question solved then, thank you. – Justanotherchemist Mar 4 '17 at 22:26
• Perhaps the other relevant thing is that there's also greater vibronic coupling in 4d/5d metals, which helps to relax the Laporte selection rule. So, 4d/5d typically have more intense absorption than 3d. – orthocresol Mar 4 '17 at 22:27

As you said, in $\ce{RuCl3}$, the charge of the Ruthenium ion is +3, meaning d5. In water, the RuCl3 becomes $\ce{[Ru(H2O)6]^3+}$, an octahedral complex. The electron configuration of the Ruthenium ion can either be $\mathrm{t_{2g}{}^5\,e_g{}^0}$ (low spin), or $\mathrm{t_{2g}{}^3\,e_g{}^2}$ (high spin), as shown in the pictures below:

You also need to know about CFSE (crystal field stabilization energy), which is the difference in the energy of the free metal ion, and the metal ion in the complex.

According to the electron configuration diagrams above, the CFSE for each configuration is:

$$\text{Octahedral }\mathrm{d^5}\text{ — low spin}\\ \text{CFSE} = -0.4\Delta_\mathrm{O} \times 5 + 2P = 2P - 2 \Delta_\mathrm{O}\\[1.5em] \text{Octahedral }\mathrm{d^5}\text{ — high spin}\\ \text{CFSE} = -0.4\Delta_\mathrm{O} \times 3 + 0.6\Delta_\mathrm{O} \times 3 = 1.2\Delta_\mathrm{O} - 1.2\Delta_\mathrm{O} = 0$$

Where $P$ is the mean pairing energy, the energy required to overcome the repulsion between the electrons and pair them together. $\Delta_\mathrm{O}$ is the octahedral splitting parameter, which I assume you already know.

As you can see, a high spin $\mathrm{d^5}$ octahedral complex will not lead to a change in energy compared to a free ion. Therefore, the spin is determined by the low spin configuration. If the pairing energy is bigger than the octahedral splitting parameter, the metal ion will gain energy, and will rather be high spin. If the pairing energy is smaller than the octahedral splitting parameter, the metal ion will lose energy and will rather be low spin.

The octahedral splitting parameter is mostly affected by the ligand type (sigma, pi acceptor, pi donor). The pairing energy is mostly affected by the size of the d orbitals — the bigger they are, the smaller it is (because the electrons have more room, the repulsion between them is smaller).

4d and 5d transition metals have large d orbitals, so that the mean pairing energy is usually smaller than the splitting parameter, which results in low spin complexes regardless to the ligand type.

Therefore, a $\ce{Ru^3+}$ ion, which is $\mathrm{4d^5}$, will form low spin octahedral complex, that has spin allowed transitions with intense colors.

About charge transfer. It's possible, but I personally don't know.

• Good first answer and welcome to chem.SE. – Nilay Ghosh Sep 1 '17 at 14:29