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: 7.31a

Answer

$Na^+$ ion is smaller than $F^-$ ion.

Work Step by Step

Both $F^-$ and $Na^+$ have 10 electrons, so their electron-electron repulsions are quite the same. However, the nuclear charge of $F^-$ is $9+$, while the nuclear charge of $Na^+$ is $11+$. That means the attraction of the $Na^+$ nucleus is greater than that of the $F^-$ Therefore, the electrons of $Na^+$ are more attracted and closer to the nucleus than those of $F^-$. As a result, $Na^+$ ion is smaller than $F^-$ ion.
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