Answer
The electron configuration for $Ru^{3+}$ is $$Ru^{3+}: [Kr]4d^5$$
$Ru^{3+}$ 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 $Ru$ atom is $$Ru: [Kr]4d^{7}5s^1$$
2) Here we need to find the electron configuration for $Ru^{3+}$, an ion having 3 electrons removed from a neutral $Ru$ 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 $Ru$ atom. That electron would be removed first. However, still 2 more are needed to remove so that $Ru^{3+}$ can be made.
3) The subshells with the next largest value of $n$ is $4s$, $4p$ and $4d$. Among those, subshell $4d$ has the largest $l$ $(l=2)$. So electrons in subshell $4d$ would be considered first.
Subshell $4d$ is occupied by 7 electrons in neutral $Ru$ atom. 2 electrons among these would be removed so that eventually 3 electrons would be removed to create $Ru^{3+}$.
That means the electron configuration for $Ru^{3+}$ is $$Ru^{3+}: [Kr]4d^5$$
The above configuration is not the configuration of any noble gases, so $Ru^{3+}$ does not have noble-gas configuration.
