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This is a nice question when you find it out, and I am really looking for a proper answer.

Take quicksilver (Hg) in the periodic table. It has one proton more than Gold (melting point 1337.33 K), and one less than Thallium (melting point 577 K). It belongs to the same group as Zinc (692.68 K) and Cadmium (594.22 K). All not very high melting points, but still dramatically higher than quicksilver (234.32 K). When his neighbors melt, quicksilver vaporizes (at 629.88 K).

What is the reason for this exceptional behavior of quicksilver ?

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I made a couple of edits for people who may not be familiar with what quicksilver is. –  David Z Feb 3 '11 at 1:23
    
I wonder, in what dictionary Stefano found that old fashioned "Quicksilver". Maybe he "extrapolated" from dutch? –  Georg Feb 3 '11 at 13:43
    
@Georg: I thought it was the layperson name. I always call it mercury, but I assumed non-scientific people may search for quicksilver on google. –  Stefano Borini Feb 3 '11 at 15:20
    
AFAIK quicksilver is oldfashioned, maybe brought to England by German miners. Being a "long word" alone is rather unenglish. –  Georg Feb 3 '11 at 16:17
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2 Answers 2

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Although the question has been partially answered, there is a superb reference on this topic which will certainly give you some of the deep, and not so deep insights needed to understand the answer to this question.

http://pubs.acs.org/doi/abs/10.1021/ed068p110

Nevertheless, both the contraction of the s(1/2) orbitals predicted by the Dirac equation, and the filled valence shell of Hg are the major causes for the odd physical properties of Hg. However, there are other effects that should be considered and the paper above is pretty clear on those.

Let me reformulate in order to make things clearer and more specific:

The reason for liquid Hg can be stated simply: the outer electrons of Hg (6s2) that participate in metallic bonds are "less available" to bonding (which might be observed for example by looking at binding energies of clusters of Hg, dimers, etc) then other common metals, and hence the interaction between Hg atoms is much less strong compared to other metal-metal bonds. The explanation for the "less availability" of the 6s electrons is the contraction of the 6s orbitals, caused by the high speeds achieved by those electrons. This effect is promptly predicted by the Dirac equation. Now, one would ask: what about Au? It has a 6s electron and it is a very stable solid metal, which means it forms strong metallic bonds with other gold atoms. The question then turns to be, what makes metallic bonds strong in gold and weak in Hg? And, here comes the indispensable valence shell argument: gold has only one electron in its 6s orbital, while Hg has two electrons in the 6s orbital. It turns out that this creates a large difference in bonding: we can use a simple qualitative molecular orbital argument to understand why. Let's imagine a simple picture of the bonding between two Au atoms, such that only the 6s electron of each atom contributes to the bonding. That being true, 2 MOs are formed in the process of bonding, and the 6s electron from each atom occupies the bonding orbital, while the antibonding MO is unoccupied. The bonding MO has a lower energy than the 6s orbitals, which stabilizes the dimer, while the antibonding has a higher energy than the 6s orbitals, making it unfavorable for a bond to form. For Au, as we just saw the 6s electrons would be in the bonding orbital and the metallic bond would be favorable. Now, this scheme is easily extendable for Hg, and the surprise here is that since the 6s of Hg has 2 electrons, in order for Hg to dimerize, 2 electrons from our simplified model would go to the bonding molecular orbital, while the other two would be in the antibonding molecular orbital. Therefore, the stabilizing effect generated by electrons in the bonding orbitals is now circumvented by the destabilization provoked by the two electrons in antibonding molecular orbitals. In these circumstances, the metallic bond would be unfavorable and the interaction between the Hg atoms will be weak, i.e., dispersion will probably dominate. By scaling these arguments to a much larger number of Hg atoms we can see that at room temperature solid Hg would not be stable, just like a bunch of other systems in which the constituents interact mainly via van der Waals forces.

A question that might arise from the analysis above is why cadmium or zinc which have 5s and 4s orbitals filled are solids at room temperature, and the reason is that in this case the s orbitals do not suffer appreciable relativistic contraction, and therefore it is not unfavorable for these atoms to generate solids in which those electrons participate in metallic bonds. We can note that Hg(2+)-Hg(2+) is a quite stable ion in which the Hg(2+) also have filled shells (though the 6s orbitals are unoccupied now). The reason for this last case is that the most external valence subshell is the 5d, with 10 electrons. This subshell is much more diffuse than the 6s shell and the relativistic effects are opposite, i.e., the d orbitals are extended compare to the non-relativistic case, and hence the electrons would be more easily engaged in interionic bonding, even if they are held by dispersion forces only, since the polarizability of Hg2+ is much larger compared to Hg.

The arguments above were derived from a simplified model, but others more complex based on those might be easily derived. I know that there are weak spots, but as noted in my comments, experimental facts and calculations converge with those explanations, and the paper referenced above shows part of this in more depth, and even explains other interesting relativistic effects such as the "inert pair effect" that general chemistry professors love to talk about, but in most of the times, do not know what it really is.

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I cannot access this link, but I can imagine what the topics are there. I am really tired of this "valence shell" chat of laymen. –  Georg Feb 3 '11 at 13:51
    
@Georg The major explanations in the link are based on calculations, experimental facts and models such as the ones given by quantum mechanical molecular orbital theory and the Dirac equation. Quantum Mechanical MO theory, or even old and modern valence bond theory, though inaccurate in not capturing full electronic correlation, are powerful ways to interpret chemistry, specially when the predictions of those and experiment are alike. The concept of filled valence shell for the properties of Hg is important in all of those (Dirac eqn, MO, and VB theory). I wonder what else would you want. –  Raphael R. Feb 3 '11 at 15:34
    
One more thing: even after disconsidering VB or MO theories, the concept of valence shell remains being important. Wavefunction methods will either be perturbation, or expansion of possible electronic configurations, in which the characteristics of the atoms (such as their valence shell) will determine the probability for distinct electron configurations in the expansion and also the most important corrections in perturbation theory. Of course if you want, you can only do the math or computation and get your results without worrying about any of this, but I believe the point here is clear now. –  Raphael R. Feb 3 '11 at 15:45
    
Of course "valence shell" is important, the problem is that it is too "coarse" to explain the fine tuning which is the true answer to the question. Here, we are dealing about some figure of, may be some dozen kcal/mole of more or less localized binding energy in the crystal lattice of mercury! I do not know the true "reason", but I know it is within that portion of deviation of (most) metals from ideal metallic bond. (the latter would imply closest packing lattices, and low melting points, and softness!) –  Georg Feb 3 '11 at 16:06
    
@Georg Any thoughts on the more specific and detailed argument I have in my answer now? –  Raphael R. Feb 4 '11 at 1:27
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Gold has its 6s valence shell unfilled, so is more reactive than Hg. thallium has a valence electron in the 6p shell which again is unfilled. The valence shell of mercury, 6s2 is filled, and is drawn closer to the nucleus because the proximity causes the electron to move at relativistic speeds thereby causing its rest mass to increase, which draws it closer yet to the nucleus. Thus it is even less reactive with other mercury atoms and thus has a lower melting point. See Mercury

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Good question--did a little digging-- –  Gordon Feb 3 '11 at 0:55
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protected by Qmechanic Mar 25 '13 at 9:19

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