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: 7.17

Answer

An electron in $n=3$ of Kr would experience greater effective nuclear charge than that of Ar and therefore, is closer to the nucleus.

Work Step by Step

The nuclear charge of Ar is $Z(Ar)=18+$ and that of Kr is $Z(Kr)=36+$. So $Z(Kr)$ is much greater than $Z(Ar)$. An electron in a particular shell can only be screened mostly by electrons in the inner shells and a little by electrons in the same shell, and cannot be screened by electrons in the outer shell. Therefore, an electron in $n=3$ are mostly screened by electrons in $n=1$ and $n=2$ and a little screened by electrons in $n=3$. However, we see that both Kr and Ar have the same number of electrons in $n=1$ (2), $n=2$ (8) and $n=3$ (8). Therefore, $S(Kr)\approx S(Ar)$. According to the formula, $$Z_{eff}=Z-S$$ we can deduce that $Z_{eff}(Kr)$ is much greater than $Z_{eff}(Ar)$ for an electron in $n=3$. Therefore, since an electron experiencing greater effective nuclear charge would experience more attraction from the nucleus and stay closer to the nucleus, an electron in $n=3$ of Kr would be closer to the nucleus.
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