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

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 $Kr$.
*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 $Kr$.