Question

In the literature I found compounds where gold occurs in a 1+ and 3+ oxidation state, like the perovskite CsAuCl_3, and also came across the fact that gold usually doesn't occur in a Au2+ state. However, I'm struggling to find a good explanation for this.

My current understanding is that gold ([...]6s1, 5d10) simply is a very large atom and that the 1+ state is pretty stable due to the completely empty 6s shell and the 3+ state has a d8 configuration in which all electrons are paired up with another one of the opposite spin (violating Hund's rule, but apparantly this is quite common in heavier elements). Then, since gold is so large, the outmost d orbitals are pretty delocalized and therefore make the gold2+ configuration unstable.

I've also read arguments involving relativistic effects of the s-p-shells pushing out the other orbitals, making them more delocalized, but all of this just sounds very hand-wavy to me. Maybe someone can provide me with a better answer.

Answer 1

Gold typically shows +1 and +3 oxidation states due to the stability of a completely filled d-orbital and a paired 5d8 configuration, respectively, while +2 is less stable. Increased radius and relativistic effects on 6s orbitals also contribute to the stability of these oxidation states.

Step-by-step explanation:

The explanation behind the unusual oxidation states of gold involves a combination of factors, including relativistic effects and electron orbital considerations. Gold commonly exhibits the +1 and +3 oxidation states. The +1 state arises because after losing one electron, gold has a [Xe]4f145d10 electron configuration, which is very stable due to a completely filled d-orbital.

The +3 state also reflects stability because removing all the 6s electrons and one d electron yields a 5d8 electron configuration, where all electrons are paired. The absence of the +2 oxidation state in gold is because the 5d9 configuration, which would result from the loss of two electrons, is less stable.

There is increased electron shielding and a larger radius in gold, making the removal of a second electron less favorable. Furthermore, relativistic effects significantly stabilize the 6s orbitals, making it harder to remove the second 6s electron.