Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 7 - Periodic Properties of the Elements - Exercises - Page 293: 7.45c


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 $Kr$.
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