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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 subshell of the d orbital 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 flourine 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.

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5 Answers 5

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

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

Additionally, pairing electrons in one orbital (so, adding the second 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 $2s^2$ occupation, and if you look at gadolinium, the effect there isn't strong enough to stop the 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 5d instead of the 4f orbitals.

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

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

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

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

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


Here's a paper which supports that this is quite untrue...

In the case of chromium, this means that $4s^1\ 3d^5$ will be lower in energy than $4s^2\ 3d^4$, 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 4s and 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 3d orbital has dropped in energy below that of the 4s, so that it is better to have the paired electrons in the d and the unpaired one in the s. The reason why the 3d is lower than 4s is tied to the high effective nuclear charge. The high effective nuclear charge gives rise to the small size of 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.

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Perhaps this shouldn't be counted as an answer, but since this topic has been resurrected, I'd like to point to Cann (2000). 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 $p^{3}/p^{4}$, $f^{9}/f^{10}$, etc) has local maxima at half-filled and filled subshell configurations. This can be used to defend the favourability of $s^{1}d^{5}$ and $s^{1}d^{10}$ configurations relative to $s^{2}d^{4}$ and $s^{2}d^{9}$, even though there would be a slight increase of electron density in the more compact $d$ subshells.

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

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We know that the electronic configuration of chromium is $\ce{[Ar] 3d^5 4s^1}$. It is because promotion occurs in case of $3d$ and $4s$ 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.

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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

The experimental data has 6-unpaired e's for Cr and 1 for Cu, therefore, the spdf model must "adapt". It is easy to see Cu (s1d10), but less for Cr (s1d5).

The MCAS orbital model presents a different explanation as to why these are different. The MCAS atomic orbital model provides a more logical sstem for the periodic table, too. See the Cu, Cr issue and the periodic table according to this orbital model at this site You can see more about Modeling the MCAS Way at this other site which is the sharper version of the original that is in the arxiv.org database as html/physics/9902046v2.

Of course, the physics and chemistry communities stand entrenched behind the spdf model that is heavily steeped in math.

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Welcome to Chemistry.SE. Your answer would be improved if you could summarize the content of the links you present rather than just putting the url into the text. –  Ben Norris Apr 17 '13 at 1:32
    
Thanks for the advise. I wrote much as it was. The two links have abstracts. It would be difficult to "summarize" something when a reader would undoubtedly have little knowledge of the model that is different that the spdf one. The URL links let the reader know where to look for what the model is and, as I pointed out, addresses the Cu, Cr electron issue explicitedly, which is why I chose to answer the question. –  Joel M Williams Apr 17 '13 at 2:21

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