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.
