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: 7.34d

Answer

The order in increasing size is $$Be^{2+}\lt Na^{+}\lt Ne$$

Work Step by Step

The electron configuration of $Be$ is $[He]2s^2$. To make the ion $Be^{2+}$, 2 outermost electrons would be removed. Therefore, the electron configuration of $Be^{2+}$ is $[He]$, or $1s^2$. The electron configuration of $Na$ is $[Ne]3s^1$. To make the ion $Na^{+}$, 3 outermost electrons would be removed. Therefore, the electron configuration of $Na^{+}$ is $[Ne]$, or $1s^22s^22p^6$. The electron configuration of $Ne$ is $1s^22s^22p^6$. Since the value of $n$ state of $Be^{2+}$ is lower than the other two, the distance from the nucleus of the electrons of $Be^{2+}$ is significantly smaller than that of the other two, which have already reached the higher $n$ state. In other words, the size of $Be^{2+}$ is smallest out of three. Both the remaining two have 10 electrons. But $Na^{+}$ has a higher nuclear charge than $Ne$, making the $Na^{+}$ nucleus attract the electrons more than the $Ne$ nucleus. Therefore, $Na^{+}\lt Ne$ In conclusion, the order in increasing size is $$Be^{2+}\lt Na^{+}\lt Ne$$
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