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: 7.28c

Answer

$$Fe\gt Fe^{2+}\gt Fe^{3+}$$ The radii of $Fe$ is larger than those of $Fe^{2+}$ and $Fe^{3+}$ since cations are smaller than the corresponding neutral atoms. The radii of $Fe^{2+}$ is larger than that of $Fe^{3+}$ since it has more electrons, so there is more electron-electron repulsion to screen the attraction from the nucleus.

Work Step by Step

$$Fe\gt Fe^{2+}\gt Fe^{3+}$$ Cations have fewer electrons than their corresponding neutral atoms, but still have the same nuclear charge. That means in cations, there is less electron-electron repulsion, making the electrons more attracted to the nucleus and as a result, move closer to the nucleus. Therefore, cations are smaller than their corresponding neutral atoms. That explains why the radii of $Fe$ is larger than the radii of $Fe^{2+}$ and $Fe^{3+}$. $Fe$ has 26 electrons. In the case of $Fe^{2+}$, only 2 electrons are removed, leaving back 24 electrons. In the case of $Fe^{3+}$, 3 electrons are removed, leaving back 23 electrons. 24 electrons would have more electron-electron repulsions than 23 electrons. That means $Fe^{2+}$ would be less attracted to the nucleus than $Fe^{3+}$, due to its having more electron-electron repulsions to screen the attraction. As a result, the radii of $Fe^{2+}$ would be larger than the radii of $Fe^{3+}$.
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