# Why does Ruthenium only have one electron in its 5s orbital in the neutral oxidation state?

So recently I have undertaken some research with some organometallic chemistry involved, specifically regarding that of ruthenium-aryl compounds. Trying to understand the chemistry of the central metal ion, I noticed something weird a couple of weeks ago, from Wikipedia: Ruthenium only has one electron in that orbital ($4d^75s^1$), rather than the two one would expect a transition metal to have, going with a naive Aufbau across the periodic table.

I am aware, I think, in the 4th period period transition metals, of a tendency to promote their 4s electrons to the 3d orbitals, when possible, so that they have either "half-filled" (5 electrons) or completely filled (10 electron) shells. These are "magic numbers" for transition metals, apparently; just another one of those heuristic things they teach you in inorganic chemistry that I wish I understood more about. But alas, ruthenium has an asymmetric 7 electrons in its 4d orbital, blasted thing.

Does anyone know anything? I tried using Google Scholar and a few databases, but when I type in "orbital" pretty much all I see is molecular orbital calculations specific to particular molecules. I'd really love to see papers pointing at some faint ideas I may be dimly aware about...

• Look at this explanation: madsci.org/posts/archives/2005-07/1120542929.Ch.r.html – RFG Nov 13 '13 at 20:33
• The above link is about how to obtain electronic configuration from spectroscopy experiment, but not why the electronic configuration should be. One may wonder, if the ground state of Ru is $^5$D$_4$, then the configuration would be 4d$^6$5s$^2$, like Fe/Os. Then the question is why the ground state of Ru is $^5$F$_5$? I have a pure guess about the reason... see next comment (the link is too long) – user26143 Nov 13 '13 at 21:01

From Oxtoby et al, Principles of Modern Chemistry, 7th edi p219 and Levine, Quantum Chemistry, 5th edi, p314, the orbital energies of Sc and the elements beyond, 3d<4s. The electrostatic repulsion makes energy of 3d$^3$ higher than 3d$^1$4s$^2$.
Just my pure guess, Ru is bigger than Fe, electrostatic repulsion is smaller, so 5s$^1$. However, the 6s$^2$ electrons in Os are extra stable due to relativistic effect, so again 6s$^2$.
• Just to add slightly to this answer, it isn't so much that the d orbitals in $\ce{Ru}$ are large, but rather that the d orbitals in $\ce{Fe}$ are anomalously small. This happens because the 3d subshell is the first of its kind, and as such its orbital wavefunctions have no radial node (the same happens with 1s, 2p, 4f, 5g, etc). This creates unusually high interorbital and intraorbital electron repulsion. Also, in $\ce{Os}$, not only is the 6s subshell relativistically stabilized, but the 5d subshell is slightly relativistically destabilized, favouring the 6s² ground state even more. – Nicolau Saker Neto Nov 14 '13 at 10:11
• @Billare Aha, here is one of the main sources I came across regarding compact orbitals. It really is surprising how the lack of radial node for some wavefunctions explains all sorts of interesting behaviour, such as the high stability of $\pi$ bonds in the second period nonmetals, the reluctance for fourth period transition metals to display high oxidation numbers and their relatively weak bonding, and the striking chemical similarity among the lanthanides. – Nicolau Saker Neto Nov 15 '13 at 0:56