Why are the noble metals inert? I am wondering: The noble 'gases' are inert because they have closed shells and don't want to give that up. But the noble metals, such as Copper, Silver, Rhodium, Gold, don't seem to have this.
For example, Rhodium has electron configuration $4d^{8}\, 5s^1$. To me, this looks very similar to the alkali metals, which we all know are 'very' reactive. So why doesn't Rhodium happily give up its s-electron? Same with Silver, $4d^{10}\, 5s^1$. Why not give up the s electron?
 A: Gold, as a metal, is, like an noble gas, not very reactive. But if you take a single gold atom it is very reactive. It combines with other gold atoms to form gold metal, and it reacts chemically with other atoms. For example, the Colorado, US town of Telluride is named after a gold-tellurium mineral found there, see Calaverite:
http://www.galleries.com/minerals/sulfides/calaveri/calaveri.htm
So these metals are not noble as compared to noble gasses. They are reactive. But I think you're asking the question "why are these metals noble while the other ones aren't?"
The noble metals are the least "reactive". See
http://en.wikipedia.org/wiki/Reactivity_series 
As shown in the table in the above wikipedia article, the noble metals are those where it is very difficult to remove electrons.
Now take a look at silver, $4d^{10}5s^1$. If we are to remove one electron it's going to be the one which has the highest energy. Now normally you'd think that would be the $5s^1$ electron. But instead, on all of the heavier metals, the S levels have LOWER energies than you'd expect. And so when you singly ionize silver the result would be the loss of one of the 4d electrons. See
http://www.chemguide.co.uk/atoms/properties/atomorbs.html
for a chart showing the energies of the first few electrons.
The spreading out of the orbitals like this is due to the shielding of the nucleus by the lower energy electrons. That is, the S orbitals penetrate the closest to the nucleus and so they see a nucleus which, in effect, has a higher charge than the P, D, and F orbitals.
The overall effect is that the chemistry of heavier metals is more a function of the outermost (or most energetic) F orbital, if present. If it's not present, then the most energetic D orbital, then most energetic P orbital. The S orbital is never the most energetic.
A: Some more misconceptions:
The ChemGuide website quoted above might be a useful reference for "UK-based exam purposes" as stated there, but it certainly does not help in solving the question. The arguments given above that followed the comments on ChemGuide are inaccurate. A simple quantum chemistry calculation of gold in its ground state will give you that the electron in the s orbital (A1G) is the most energetic in this atom. Hence, ionization will most easily be accomplished by removal of this electron, and not of d electrons, and this is easily proved by another computation for ionized gold, which will show you that the 5d orbitals will remain filled while the 6s orbital is no longer occupied. Actually, it is known that if the most external d shell is filled, the energies of these orbitals will be effectively lowered, and there is a very high probability that the ionized electron will not come from it, but from more energetic s or p orbitals. (I have just done a few of these calculations in order to make sure this point is right)
In order to analyze why some metals are more inert than others, various effects come into play. Relativistic effects, such as the contraction of s orbitals, for example, are a major factor in making gold less reactive than silver, and in lowering the oxidation potential of gold. So, in addition to looking at chemical potentials when discussing the inertness of metals in different environments, it is better not to reduce the arguments to simple electron configuration trends which usually work quite well for main group elements, since, though they might generate insights for the understanding of the behavior of metals, these insights might be either right or wrong.
