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

Published by Prentice Hall

# Chapter 7 - Periodic Properties of the Elements - Exercises: 7.35d

#### Answer

$K^+$ is larger than $Ca^{2+}$ due to the fact that the electrons in $K^+$ experience less attraction to the nucleus than those in $Ca^{2+}$. Therefore, they have more chance to be further from the nucleus, and as a result, increases the size of $K^+$ larger than $Ca^{2+}$.

#### Work Step by Step

Both $Ca^{2+}$ 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 $Ca^{2+}$ is greater than that of $K^{+}$. That means the attraction force of the $Ca^{2+}$ nucleus to the electrons is greater than that of the $K^{+}$ nucleus. Therefore, though having the same number of electrons, the electrons in $Ca^{2+}$ are more attracted and as a result, closer to the nucleus than those in $K^{+}$. That means the ionic radium of $K^+$ is larger than the ionic radium of $S^{2-}$. In other words, $K^{+}$ is larger than $Ca^{2+}$.

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