# What is the shorthand electron configuration for Au?

For gold, $Z = 79$...and we gots 79 electrons to distribute.. Following the $\text{aufbau scheme}$, we build on the configuration of the last Noble gas, i.e. $X e , Z = 54$. Of course, we got a whole lanthanide series, cerium to lutetium, between xenon and goldo…
Let's try $A u : {\underbrace{\left[X e\right]}}_{\text{54 electrons}} 4 {f}^{14} 5 {d}^{10} 6 {s}^{1}$