[CaptainQuandry]

Why on earth does Gold have an oxidation number of 3? People have told me to look at the electron configuration, but the answer is not obvious. None of the general rules seem to apply. The electrons don't seem to move to other energy levels. No energy levels are filled. Some elements like to be half- filled but that still wouldn't work. PLz enlighten me as soon as possible.

[Demotivator]

....5s2  5p6  5d10    
6s1

So it must be losing one 6s and two 5s electrons.

[jdurg]

Hmmmm....  I just realized something.  Both Gold and Cesium have 6s1 as their outermost electron shell.  Both metals are yellow in color.  I wonder if the 6s1 electron has something to do with the yellow color to these two metals?  It's also pretty interesting how cesium is such an incredibly reactive metal while gold is incredibly not reactive.

(And yes, you are correct Demotivator.  Gold's common oxidation states are +1 and +3, so the +1 is indicitive of the 6s1 electron being lost, and the +3 is because of the 5s and 6s electrons being lost.)

[shelanachium]

Gold is a highly electronegative element and its compounds are covalent, so talking about electrons being lost to make ions doesn't really fit. Gold's electronegativity is connected to its outer s electron being tightly bound, rather than loosely as in Cs, partly because it is underlain by d orbitals which are poor shielders of nuclear charge. In fact one of gold's few ionic compounds is CsAu which contains Au-! Its electron affinity is close to that of iodine, and it has much in common with that element.

Mercury which follows gold also holds its outer s electrons tightly and has some noble-gas-like properties such as low atomisation energy and low reactivity. Mercury can be called a 'pseudo-noble gas' and as with the true noble gases ions with the same configuration are stable e.g. Au-, Tl+, Pb2+. It has even been predicted that mercury should have a volatile square-planar tetrafluoride fluoride HgF4 - like XeF4.

Gold resembles iodine in easily forming linear [AuCl2]-, square planar [AuCl4]- and also [AuF6]- ions, all of which are covalently bound. Chemically it is hardly metallic at all. I suppose AuF might be ionic but it decomposes at once to free Au and AuF3, a covalent chain polymer in which Au is surrounded by a square of F atoms.

So the oxidation states (not ionic charges except in Au-) of Au are -1 (CsAu etc), +1, linear covalent; +3 (square-planar covalent) and +5 (octahedral covalent). You may note that Au (III) is isoelectronic with Pt(II) which also forms square-planar complexes, and Au(V) is isoelectronic with Pt(IV) which forms countless octahedral complexes.

[CaptainQuandry]

Thank y'all for responding to my message so quickly. However i still have some questions. Gold's MAIN oxidation number is 3 (I am positive about this). Why wouldn't it just lose one electron in the s orbital instead of losing 3? In other words.. why is gold's main oxidation 3 not 1?

[Demotivator]

True, the main O.N. is +3.
Maybe it's an interplay between the thermodynamics of bond making vs ionization (I say ionization loosely as oxidation numbers don't reflect full ionization but are related to it.). With trivalent gold, there are more bonds which is more favorable than monovalent. Silver is only monovalent, however, because ionization energies are unfavorable due to it's smaller size.