Answer
The electron configuration for $Sn^{2+}$ is $$Sn^{2+}: [Kr]4d^{10}5s^2$$
$Sn^{2+}$ 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 $Sn$ atom is $$Sn: [Kr]4d^{10}5s^25p^2$$
2) Here we need to find the electron configuration for $Sn^{2+}$, an ion having 2 electrons removed from a neutral $Sn$ atom.
Therefore, according to the rules, when electrons are removed, we would look for the orbital with the largest $n$.
Here, both subshell $5s$ and subshell $5p$, with $n=5$, have the largest $n$.
3) So we move to the next rule: look for the subshell with the highest value of $l$ first.
Subshell $5p$ $(l=1)$ has a higher value of $l$ than subshell $5s$ $(l=0)$. So we would consider subshell $5p$ first.
There are 2 electrons occupying subshell $5p$ in a neutral $Sn$ atom. Therefore, to make $Sn^{2+}$, these 2 electrons would be removed.
That means the electron configuration for $Sn^{2+}$ is $$Sn^{2+}: [Kr]4d^{10}5s^2$$
The above configuration is not a noble-gas configuration, so $Sn^{2+}$ does not have noble-gas configuration.
