I had a test this morning which had the following question:
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?