## Chemistry: The Central Science (13th Edition)

The electron configuration for $Pb^{2+}$ is $$Pb^{2+}: [Xe]4f^{14}5d^{10}6s^2$$ $Pb^{2+}$ does not have noble-gas configuration.
*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 $Pb$ atom is $$Pb: [Xe]4f^{14}5d^{10}6s^26p^2$$ 2) Here we need to find the electron configuration for $Pb^{2+}$, an ion having 2 electrons removed from a neutral $Pb$ atom. Therefore, according to the rules, when electrons are removed, we would look for the orbital with the largest n first. Here, subshells $6s$ and $6p$, with $n=6$, have the largest n. However, among these 2 subshells, $6p$ has a larger $l$ $(l=1)$, so electrons in this subshell would be removed first. 3) There are 2 electrons occupying subshell $6p$ in a neutral $Pb$ atom. Both these two electrons would be removed and then $Pb^{2+}$ can be made. That means the electron configuration for $Pb^{2+}$ is $$Pb^{2+}: [Xe]4f^{14}5d^{10}6s^2$$ The above configuration is not the configuration of any noble gases, so $Pb^{2+}$ does not have noble-gas configuration.