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 292: 7.33c


The size of $K$ atom is largest in the atomic order due to the higher state $n=3$ that it possesses. The size of $K^+$ ion is smallest in the ionic order due to the higher nuclear charge of $K^+$ compared to the other two.

Work Step by Step

In the atomic radii order, the size of $K$ is largest since $K$ is in the higher $n$ state than $Cl$ and $S$, which enables the opening of the chance for the electrons to be further from the nucleus, making its radium larger than the other two. In the ionic radii order, the size of $K^+$ is smallest since $K^+$ has a higher nuclear charge than the other two ions, while its number of electrons is the same. That means the nucleus attracts the electrons better, pulling the electrons towards the nucleus, therefore decreasing the radium of $K^+$. Also, the same number of electrons means $K^+$ is in the same state as the other two ions. The reverse position of $S$ and $S^{2-}$ can be explained similarly.
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