Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 6 - Ionic Compounds: Periodic Trends and Bonding Theory - Section Problems - Page 219: 47


$a) Rb$ $b) N^{3-}$ $c) Cr^{3+}$

Work Step by Step

$a) Rb^{+} $is smaller because since it has lost an electron, there are more protons then electrons and thus the presence of more protons will pull the electrons closer, decreasing the atomic radius $b) N^{3-}$ is larger because it has more electrons than protons. This results in more shielding and more repulsion which increases the atomic radius. $c) Cr^{3+}$ is larger because it has lost fewer electrons than $ Cr^{6+}$ Thus the shielding in $Cr^{3+}$ is larger than $ Cr^{6+}$ and thus it has more repulsion and it is bigger
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.