# Why is gold unreactive when only one electron is in the outer shell?

I've been trying to answer my (high school) daughter's questions about the periodic table, and the reactivity series, but we keep hitting gaps in my knowledge.

So I showed that the noble gases have a full outer shell, which is why they don't react with anything. And then over the other side of the periodic table we have potassium and sodium, which have only one electron in their outer shell, which is what makes them so reactive, and at the top of our reactivity list. (And the bigger they get, the more reactive, which is why we were not allowed to play with caesium in class...)

But then we looked up gold, which is at the bottom of the reactivity series, and found it also has only one electron in its outermost shell (2-8-18-32-18-1).

Is there an easy explanation for why gold doesn't fizz like potassium when you drop it in water?

(This question could be rephrased as "What properties of each element decide their ranking in the metal reactivity series?" if you prefer; that was the original question we were trying to answer.)

First off, gold does react. You can form stable gold alloys and gold compounds. It's just hard, mostly for reasons explained by the other answer

The reason bulk gold solid is largely unreactive is because the electrons in gold fall at energies which few molecules or chemicals match (i.e., due to relativistic effects).

A nice summary of some work by Jens K. Norskov can be found here: http://www.thefreelibrary.com/What+makes+gold+such+a+noble+metal%3F-a017352490

In their experiments, they distinguished between gold atoms' ability to break and form bonds and the ease with which they form new compounds, such as gold oxides. The two qualities are related: To make a compound, gold atoms must bond with other atoms, yet they cannot do so until they have sundered their bonds with neighboring gold atoms.

I think this is a nice succinct explanation. You always have this trade-off in reactions, but in gold, you don't get much energy in the new compound formation, and you're losing the gold-gold interactions.

You can, of course, react gold with aggressive reagents like aqua regia, a 3:1 mix of $\ce{HCl}$ and $\ce{HNO3}$.

If properly done, the product is $\ce{HAuCl4}$ or chloroauric acid.

Relativistic effects account for gold's lack of reactivity. Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus. This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals. These relativistic electrons gain mass and as a consequence, their orbits contract. As these s and (to some degree) p orbits are contracted, the other electrons in d and f orbitals are better screened from the nucleus and their orbitals actually expand.

Since the 6s orbital with one electron is contracted, this electron is more tightly bound to the nucleus and less available for bonding with other atoms. The 4f and 5d orbitals expand, but can't be involved in bond formation since they are completely filled. This is why gold is relatively unreactive.

If you'd like to see the formulas and math behind this (it's not all that complicated) see here. Also note that similar arguments explain mercury's anomalous properties.

• It's not good to talk about "relativistic mass" as it implies a change in the intrinsic properties that doesn't exist. – gsurfer04 Dec 18 '14 at 0:56
• Why should the 5p orbitals contract? They are empty anyway, plus they have a node at the core. – Karl Aug 28 '16 at 21:14
• @Karl The $\ce{5p}$ orbital is occupied in gold ($\ce{5p^6}$). Unlike d and f orbitals, s orbitals, and to a lesser degree p orbitals, have an appreciable electron density near the nucleus. As a consequence, in heavier elements where they are strongly attracted by the nucleus they can attain relativistic speeds and experience other relativistic effects such as orbital contraction. This earlier answer may be useful. – ron Aug 28 '16 at 22:15
• (I meant 6p, sorry.) Ah well, 5p6 belongs to the [Xe] core of the gold atom. The 6s shrinks a lot, and actually lets the 5p expand, so they become of similar energy, which means it can take part in the chemistry. Which is what you say in that earlier answer. I see no reason to say 5p shrinks, it is well below the 4f14 and 5d10 anyway. – Karl Aug 28 '16 at 22:47
• @Karl p orbitals shrink due to relativistic effects, albeit not as much as s orbitals - see here or google something like "relativistic contraction of p orbitals" and see some of the links provided – ron Aug 29 '16 at 0:27

## protected by orthocresol♦Dec 17 '18 at 23:02

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