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: 7.28a

Answer

$$I^-\gt I\gt I^+$$ Since anions are bigger than the corresponding neutral atoms, and cations are smaller than the neutral atoms, the order of atomic and ionic radii of $I$ is as above.

Work Step by Step

*For the case of cations: 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. *For the case of anions: In constrast, anions have more electrons than their corresponding neutral atoms, but still have the same nuclear charge. That means in anions, there is more electron-electron repulsion, screening the attraction from the nucleus to the electrons. Overall, the electrons, due to less attraction, will be further from the nucleus than the corresponding neutral atoms, increasing the radii of anions. Therefore, anions are bigger than their corresponding neutral atoms. $$I^-\gt I\gt I^+$$ $I^-$ is the anion, $I$ is the neutral atom and $I^+$ is the cation. As shown above, anions are bigger than the neutral atoms, and cations are the smaller than the neutral atoms. So, obviously, the radii of $I^-$ would be largest, next is the radii of $I$ and the smallest would be the radii of $I^+$.
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