Answer
The electron configuration for $Ag^{+}$ is $$Ag^{+}: [Kr]4d^{10}$$
$Ag^+$ does not have noble-gas configuration.
Work Step by Step
*RULES TO WRITE ELECTRON CONFIGURATIONS FOR IONS:
- When electrons are removed to form cations, those of the orbitals that have the largest principal quantum number $n$ will be removed first.
- If there are more than one subshell for a given $n$, electrons of the subshell with the highest value of $l$ will be removed first.
- When electrons are added to form anions, they would be added to the empty or partially occupied orbital with the lowest value of $n$.
1) The electron configuration of a neutral $Ag$ atom is $$Ag: [Kr]4d^{10}5s^1$$
2) Here we need to find the electron configuration for $Ag^{+}$, an ion having 1 electron removed from a neutral $Ag$ atom.
Therefore, according to the rules, when electrons are removed, we would look for the orbital with the largest $n$ first.
Here, subshell $5s$, with $n=5$, have the largest $n$.
There are 1 electron occupying subshell $5s$ in a neutral $Ag$ atom. So that electron would be removed to created ion $Ag^+$.
That means the electron configuration for $Ag^{+}$ is $$Ag^{+}: [Kr]4d^{10}$$
The above configuration is not the configuration of any noble gases, so $Ag^{+}$ does not have noble-gas configuration.