Answer
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 $$[Xe]4f^{14}5d^26s^2$$
