Answer
The electron configuration for $Y^{3+}$ is $$Y^{3+}: [Kr]$$ or $$Y^{3+}:[Ar]3d^{10}4s^24p^6$$
$Y^{3+}$ has noble gas configuration of $Kr$.
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 $Y$ atom is $$Y: [Kr]4d^{1}5s^2$$
2) Here we need to find the electron configuration for $Y^{3+}$, an ion having 3 electrons removed from a neutral $Y$ 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 2 electrons occupying subshell $5s$ in a neutral $Y$ atom. These 2 electrons would be removed first. However, still 1 more is needed to remove so that $Y^{3+}$ 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 1 electron in neutral $Y$ atom. This electron would be removed so that eventually 3 electrons would be removed to create $Y^{3+}$.
That means the electron configuration for $Y^{3+}$ is $$Y^{3+}: [Kr]$$ or $$Y^{3+}:[Ar]3d^{10}4s^24p^6$$
The above configuration is the configuration of noble gas $Kr$, so $Y^{3+}$ has noble-gas configuration of $Kr$.