## Chemistry: The Central Science (13th Edition)

Published by Prentice Hall

# Chapter 7 - Periodic Properties of the Elements - Exercises: 7.34c

#### Answer

The order in increasing size is $$Ti^{4+}\lt Sc^{3+}\lt Ca$$

#### Work Step by Step

The electron configuration of $Ti$ is $[Ar]3d^24s^2$. To make the ion $Ti^{4+}$, 4 outermost electrons would be removed. Therefore, the electron configuration of $Ti^{4+}$ is $[Ar]$, or $1s^22s^22p^63s^23p^6$. The electron configuration of $Sc$ is $[Ar]3d^14s^2$. To make the ion $Sc^{3+}$, 3 outermost electrons would be removed. Therefore, the electron configuration of $Sc^{3+}$ is also $[Ar]$, or $1s^22s^22p^63s^23p^6$. The electron configuration of $Ca$ is $[Ar]4s^2$. Since $Ca$ reaches a higher value of $n$ state $(n=4)$ than the other two, there is more chance for electrons in $Ca$ atom to be further from the nucleus since there are new orbitals further from the nucleus than those in $n=3$. In other words, the size of $Ca$ is largest out of three. Both the remaining two ions have 18 electrons. But $Ti^{4+}$ has a higher nuclear charge than $Sc^{3+}$, making the $Ti^{4+}$ nucleus attract the electrons more than the $Sc^{3+}$ nucleus. Therefore, $Sc^{3+}\gt Ti^{4+}$ In conclusion, the order in increasing size is $$Ti^{4+}\lt Sc^{3+}\lt Ca$$

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