How do you write the noble-gas electron configuration for Ag?

For $\text{silver}$, $Z = 47$. It is thus 11 protons removed from the last Noble Gas, which is $\text{krypton}$, $Z = 36$.
And thus the electronic configuration of silver metal is: $\left[K r\right] 4 {d}^{10} 5 {s}^{1}$. The $A g \left(+ I\right)$ oxidation state can be rationalized on this basis.