Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 6 - Electronic Structure of Atoms - Additional Exercises: 6.102c

Answer

The condensed electron configuration of $Hf$ is $$Hf:[Xe]4f^{14}5d^26s^2$$ and there are 2 unpaired electrons.

Work Step by Step

1) The nearest noble gas of lower atomic number of $Hf$ is $Xe$. 2) Looking the periodic table, the atomic number of $Hf$ is 72, which means it has 72 electrons. The last row ends with the atomic number 54, so $Hf$ has 54 inner-shell electrons. That leaves it with 18 outer-shell electrons. $Hf$ belongs to the 6th row, so its outer shell is the 6th shell. 3) These 18 outer-shell electrons are distributed as follows: - The first two go to the $6s$ subshell. - The next 14 go to the $4f$ subshell. - The last 2 go to the $5d$ subshell. Therefore, the condensed electron configuration of $Hf$ is $$Hf:[Xe]4f^{14}5d^26s^2$$ 4) All the inner shells, the $4f$ subshell, and the $6s$ subshell are completely filled, so all of their electrons are paired. There are 2 electrons in subshell $5d$, and subshell $5d$ has 5 orbitals. According to Hund's rule, these 2 electrons each occupies a different orbital. In the end, there are 2 one-electron orbitals. Therefore, there are 2 unpaired electrons.
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