Answer
The condensed electron configuration of $Bi$ is $$Bi:[Xe]4f^{14}5d^{10}6s^26p^3$$
There are 3 unpaired electrons.
Work Step by Step
1) The nearest noble gas of lower atomic number of $Bi$ is $Xe$.
2) Looking the periodic table, the atomic number of $Bi$ is 83, which means it has 83 electrons. The last row ends with the atomic number 54, so $Bi$ has 54 inner-shell electrons. That leaves it with 29 outer-shell electrons.
$Bi$ belongs to the 6th row, so its outer shell is the 6th shell.
3) These 29 outer-shell electrons are distributed as follows:
- The first two go to the $6s$ subshell.
- The next 14 go to the $4f$ subshell.
- The next 10 go to the $5d$ subshell.
- The last 3 go to the $6p$ subshell.
(Both the $4f$ and $5d$ are occupied with some minor variations. But in the end, 14 go to $4f$ and 10 go to $5d$)
Therefore, the condensed electron configuration of $Bi$ is $$Bi:[Xe]4f^{14}5d^{10}6s^26p^3$$
4) All the inner shells, the $6s$ subshell, the $5d$ subshell, and the $4f$ subshell are completely filled, so all of their electrons are paired.
There are 3 electrons in subshell $6p$, and subshell $6p$ has 3 orbitals. According to Hund's rule, these 3 electrons each occupies a different orbital. In the end, there are 3 one-electron orbitals.
Therefore, there are 3 unpaired electrons.
