# Why do elements in columns 6 and 11 assume 'abnormal' electron configurations?

When I look around for why copper and chromium only have one electron in their outermost s orbital and 5/10 in their outermost d orbital, I'm bombarded with the fact that they are more stable with a half or completely filled d orbital, so the final electron enters that orbital instead of the 'usual' s orbital.

What I'm really looking for is why the d orbital is more stable this way. I assume it has to do with distributing the negative charge of the electrons as evenly as possible around the nucleus since each orbital of the d subshell is in a slightly different location, leading to a more positive charge in the last empty or half-filled d orbital. Putting the final electron in the s orbital would create a more negative charge around the atom as a whole, but still leave that positive spot empty.

Why does this not happen with the other columns as well? Does this extra stability work with all half or completely filled orbitals, except columns 6 and 11 are the only cases where the difference is strong enough to 'pull' an electron from the s orbital? It seems like fluorine would have a tendency to do do this as well, so I suppose the positive gap left in the unfilled p orbital isn't strong enough to remove an electron from the lower 2s orbital.

• In my paper academia.edu/18391675/… I have had some doubts about the role of the electrons dipole moment distribution in chromium atom which is quite symmetrical if I use my octahedral model instead of s- and p-orbitals. ... – HolgerFiedler Nov 16 '15 at 11:33
• ...Now a have read on Wikipedia about the magnetic characteristics of pure chromium: "Chromium is remarkable for its magnetic properties: it is the only elemental solid which shows antiferromagnetic ordering at room temperature (and below). Above 38 °C, it transforms into a paramagnetic state." This matches perfect into my model. – HolgerFiedler Nov 16 '15 at 11:33
• Sometimes the "why" questions are really hard to answer. The universe was built in a specific way and we may have to accept it as it is. It's really annoying that humanity cannot explain everything about the universe. – Pritt says Reinstate Monica Apr 20 '17 at 3:15

As I understand this, there are basically two effects at work here.

When you populate an $$\mathrm{s}$$-orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the $$\mathrm{d}$$-orbitals, making them higher in energy (and more radially diffuse). The difference in energy between putting all the electrons in $$\mathrm{d}$$-orbitals and putting one in an $$\mathrm{s}$$-orbital increases as you fill the $$\mathrm{d}$$-orbitals.

Additionally, pairing electrons in one orbital (so, adding the second $$\mathrm{s}$$ electron) carries a significant energy cost in terms of Coulombic repulsion because you're adding an electron essentially in exactly the same space as there's already an electron.

I'm assuming that the effect isn't strong enough to avert fluorine having a $$\mathrm{2s^2}$$ occupation, and if you look at gadolinium, the effect there isn't strong enough to stop the $$\mathrm{s}$$ from filling (large nuclear charge and orbital extent at the nucleus is a good combination energy-wise), it does manage to make it more favourable to add the electron into the $$\mathrm{5d}$$ instead of the $$\mathrm{4f}$$ orbitals.

Also, if you take a look at tungsten vs gold, there the effect isn't strong enough for tungsten to avoid a $$\mathrm{6s^2}$$ occupation, but is for gold - more $$\mathrm{d}$$ electrons making the screening effect overcome the strong nuclear charge and enhanced nuclear penetration of an $$\mathrm{s}$$-orbital.

• Isn't there also the fact that for large atoms, relativistic effects pull the outer shells closer in which could be a factor for more shielding? – Ali Caglayan May 28 '15 at 17:46
• @Alizter: Yes, relativistic effects especially shrink the s orbitals, leading to more shielding for the higher angular momentum orbitals which thus become more diffuse and higher in energy. I can't remember exactly how the magnitudes for the different contributions shake out. – Aesin May 30 '15 at 0:39

This is just a confirmation to Aesin's answer...

Say, we take copper. The expected electronic configuration (as we blindly fill the $$\mathrm{d}$$-orbitals along the period) is $$\ce{[Ar]}\mathrm{3d^9 4s^2}$$, whereas the real configuration is $$\ce{[Ar]}\mathrm{3d^{10} 4s^1}$$. There is a famous interpretation for this, that $$\mathrm{d}$$-orbitals are more stable when half-filled and completely-filled. That's a complete myth. There are very few pages explaining this myth, like chemguide.co.uk.

As we fill the electrons starting from $$\mathrm{3d^1}$$, we'd be stuck at chromium and also at copper. While filling chromium and copper, it has been observed that the energies of $$\mathrm{4s}$$ and $$\mathrm{3d}$$ orbitals are fairly close to each other. The increasing nuclear charge (as we go along the period) and the size and shape of $$\mathrm{d}$$-orbital should be a reason. This similarity makes the energy for pairing up electrons in d-orbital very less than that of pairing up in $$\mathrm{s}$$-orbital (i.e.) the energy difference between these orbitals is much less than the pairing energy required to fill the electrons in $$\mathrm{4s}$$ orbital. Moreover, the energy for the configuration $$\mathrm{3d^5 4s^1}$$ is much less than that of $$\mathrm{3d^4 4s^2}$$. Since we usually fill electrons in the order of increasing energy, the next electron (in case of manganes) goes into the $$\mathrm{4s}$$-orbital.

The same reason for effective nuclear charge makes the $$\mathrm{3d}$$-orbitals somewhat lower in energy than $$\mathrm{4s}$$-orbitals and hence, the unusual configuration of $$\ce{Cr}$$ and $$\ce{Cu}$$.

From a paper of Richard Hartshorn and Richard Rendle of the University of Canterbury (*.doc file), which supports that this is quite untrue:

In the case of chromium, this means that $$\mathrm{4s^1 3d^5}$$ will be lower in energy than $$\mathrm{4s^2 3d^4}$$, because in the second case you have to "pay" the electron pairing energy. Since this pairing energy is larger than any difference in the energies of the $$\mathrm{4s}$$ and $$\mathrm{3d}$$ orbitals, the lowest energy electron configuration will be the one which has one electron in each of the six orbitals that are available. Effectively this is Hund's rule applying not just to strictly degenerate orbitals (orbitals with the same energy), but to all orbitals that are (significantly) closer in energy than the electron pairing energy.

In the case of copper, the $$\mathrm{3d}$$-orbital has dropped in energy below that of the $$\mathrm{4s}$$, so that it is better to have the paired electrons in the $$\mathrm{d}$$ and the unpaired one in the $$\mathrm{s}$$. The reason why the $$\mathrm{3d}$$ is lower than $$\mathrm{4s}$$ is tied to the high effective nuclear charge. The high effective nuclear charge gives rise to the small size of $$\ce{Cu}$$ compared with the earlier transition metals, and also means that orbitals in inner shells are more stabilised with respect to those further out for copper than for earlier elements.

• Does that paper mean that $4s$ orbitals have less energy than $3d$ orbitals? That would be in conflict with your chemguide link, which i had incidentally read before and i am convinced of its correctness... – FreezingFire May 18 '16 at 5:04

Perhaps this shouldn't be counted as an answer, but since this topic has been resurrected, I'd like to point to Cann.[1] He explains the apparent stability of half-filled and filled subshells by invoking exchange energy (actually more of a decrease in destabilization due to smaller-than-expected electron-electron repulsions).

According to him, there is a purely quantum mechanical energy term proportional to $$\frac{n_{↑}(n_{↑}-1)}{2} + \frac{n_{↓}(n_{↓}-1)}{2},$$ where $$n_{↑}$$ and $$n_{↓}$$ represent the number of spin-up and spin-down electrons in a subshell. This term decreases the potential energy of the atom, and it can be shown that the difference in exchange energy between two consecutive subshell populations (such as $$\mathrm{p}^{3}/\mathrm{p}^{4}$$, $$\mathrm{f}^{9}/\mathrm{f}^{10}$$, etc) has local maxima at half-filled and filled subshell configurations. This can be used to defend the favourability of $$\mathrm{s}^{1}\mathrm{d}^{5}$$ and $$\mathrm{s}^{1}\mathrm{d}^{10}$$ configurations relative to $$\mathrm{s}^{2}\mathrm{d}^{4}$$ and $$\mathrm{s}^{2}\mathrm{d}^{9}$$, even though there would be a slight increase of electron density in the more compact $$\mathrm{d}$$ subshells.

However, this clashes with Crazy Buddy's reference, which seems to deny any stabilization effect. So, which is (more) true? Or is neither?

1. Cann, P. Ionization Energies, Parallel Spins, and the Stability of Half-Filled Shells. J. Chem. Educ. 2000, 77 (8), 1056 DOI: 10.1021/ed077p1056.

We know that the electronic configuration of chromium is $$\ce{[Ar] 3d^5 4s^1}$$. It is because promotion occurs in case of $$\ce{3d}$$ and $$\ce{4s}$$ orbitals — in other words, the electron is shifted from a lower energy level to a higher one (also known as excitation). Promotional energy and pairing energy both are endergonic so the process which requires less energy would preferably take place.

• Could you explain a bit more? For example, why does that process have less energy? Elaborate a bit and your post will be improved greatly :) – ManishEarth Nov 19 '12 at 18:51