# What is the most common oxidation state of gold?

What is the most common oxidation state for gold?
(a) +1 (b) -1 (c) +2 (d) +3

Since the electronic configuration of $\ce{Au}$ is $\ce{[Xe] 4f^{14} 5d^10 6s^1}$. After removing the $\ce{6s^1}$ electron and gaining $+1$ charge, $\ce{Au}$ should have noble gas configuration and hence this should be the most stable configuration. So the answer has to be +1. Could someone verify my reasoning?

• The most important (natural as well as anthropogenic) oxidation state of gold is actually zero. Apparently, the question refers to the oxidation state of gold in its compounds. – user7951 May 8 '16 at 9:10
• Gold is present in $+2$ oxidation state in this complex: $\ce{[Au2Cl2\{μ-(CH2)2PPh2\}2]}$. The +2 oxidation state will stable in dimeric form. – Dinesh Kumar Dec 13 '17 at 6:58

It's not obvious, but common oxidation state for gold is +3. It caused by destabilization of the $5d^{10}$ orbital. Detailed explanation you can find in The Chemistry of Gold, in Chapter 1.1.3.
• @Abhishek Mhatre Well, it says in Wiki "Au(I), [...] is the most common oxidation state with soft ligands such as thioethers, thiolates, and tertiary phosphines." And 2 sentences after: "Au(III) (auric) is a common oxidation state, [...]". For example, salts of Au(I) (except insoluble (AuI) or extremle stable $[Au(CN)_{2}]^{-}$) disproportionate in water: $3Au^{+} = 2Au + Au^{3+}$. – Vadim Shkaberda May 8 '16 at 14:32