Answer
the electron configuration for $Zr^{4+}$ is $$Zr^{4+}: [Kr]$$ or $$Zr^{4+}: [Ar]3d^{10}4s^24p^6$$
$Zr^{4+}$ 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 $Zr$ atom is $$Zr: [Kr]4d^{2}5s^2$$
2) Here we need to find the electron configuration for $Zr^{4+}$, an ion having 4 electrons removed from a neutral $Zr$ 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 $Zr$ atom. So these 2 electrons would be removed first. However, to make $Zr^{4+}$, still 2 more must be removed.
3) We move on to find the subshell with the next largest $n$, that is, $n=4$.
Here, we find subshell $4s$, $4p$ and $4d$ have $n=4$. But subshell $4d$ has the highest value of $l$ $(l=2)$. So $4d$ would be considered first.
Subshell $4d$ are occupied by 2 electrons. So these 2 electrons would be removed to have eventually 4 electrons removed to make $Zr^{4+}$.
That means the electron configuration for $Zr^{4+}$ is $$Zr^{4+}: [Kr]$$ or $$Zr^{4+}: [Ar]3d^{10}4s^24p^6$$
The above configuration is the configuration of noble gas $Kr$, so $Zr^{4+}$ has noble-gas configuration of $Kr$.