Answer
The electron configuration of $Zr^{4+}$ is $$[Ar]3d^{10}4s^24p^6$$ or $$[Kr]$$
$Zr^{4+}$ possesses noble-gas configuration of $Kr$.
Work Step by Step
*RULES FOR WRITING ELECTRON CONFIGURATION FOR CATIONS:
1) Write the electron configuration of the neutral atom.
2) Find out the number of electrons to be removed from the neutral atom to create the cation.
3) Removing the electrons according to the following rules:
- The electrons occupying the subshells with the largest $n$ value will be removed first.
- If there are more than one subshell having the largest $n$ value, the electrons of the subshell with the highest $l$ value will be removed first.
1) The electron configuration of a neutral $Zr$ atom is $$[Kr]4d^{2}5s^2$$
2) To make $Zr^{4+}$, 4 electrons need to be removed from a neutral $Zr$ atom.
3) Here we find that subshell $5s$ has the largest $n$ value $(n=5)$. So the electrons occupying subshell $5s$ would be removed first.
Subshell $5s$ is occupied by 2 electrons in a neutral $Zr$ atom. Both of these 2 electrons would be removed in the process to make the cation. However, still 2 more electrons need to be removed to make $Zr^{4+}$
So we move on to find the next largest $n$ value, which is $n=4$. 3 occupied subshells $4s$, $4p$ and $4d$ have $n=4$. However, subshell $4d$ has the largest $l$ value $(l=2)$. So the electrons there would be removed first.
Subshell $4d$ is occupied by 2 electrons in neutral $Zr$ atom. Still 2 more electrons need to be removed, so these 2 electrons in subshell $4d$ would be removed to make $Zr^{4+}$
So, the electron configuration of $Zr^{4+}$ is $$[Ar]3d^{10}4s^24p^6$$ or $$[Kr]$$
The above configuration is the electron configuration of noble gas $Kr$. So $Zr^{4+}$ possesses noble-gas configuration of $Kr$.
