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: 7.35c

Answer

$S^{2-}$ is larger than $K^+$ since the electrons in $S^{2-}$ experience less attraction to the nucleus than those in $K^+$, giving them more freedom and more chance to be further from the nucleus.

Work Step by Step

Both $S^{2-}$ and $K^+$ have 18 electrons, which means the electron-electron repulsions, or in other words, the screening constant in both ions are quite the same. However, the nuclear charge of $K^+$ is greater than that of $S^{2-}$. That means the attraction force of the $K^+$ nucleus to the electrons is greater than that of the $S^{2-}$ nucleus. Therefore, though having the same number of electrons, the electrons in $K^+$ are more attracted and as a result, closer to the nucleus than those in $S^{2-}$. That means the ionic radium of $K^+$ is smaller than the ionic radium of $S^{2-}$. In other words, $S^{2-}$ is larger than $K^+$.
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