Chemistry: The Central Science (13th Edition)

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

Chapter 8 - Basic Concepts of Chemical Bonding - Exercises - Page 334: 8.12b


The electron configuration of hafnium is $$[Xe]4f^{14}5d^26s^2$$

Work Step by Step

1) Find out the number of electrons in a hafnium atom The atomic number of hafnium is 72. That means there are 72 electrons in a hafnium atom. 2) Find the noble gas of the largest lower period $Hf$ is in period 6. The noble gas of period 5 is $Xe$. So $Xe$ would be used in the electron configuration of $Hf$. The atomic number of $Xe$ is 54, or $Xe$ has 54 electrons in its atom. Therefore, $Hf$ has 54 inner-shell electrons and $72-54=18$ outer-shell electrons. 3) Arrange the outer-shell electrons in shells and subshells of increasing levels of energy: - 14 electrons would first occupy subshell $4f$. - The next 2 electrons would then occupy subshell $6s$. - The last 2 electrons occupy subshell $5d$. Therefore, the electron configuration of hafnium is $$[Xe]4f^{14}5d^26s^2$$
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