Answer
The condensed electron configuration of $Br$ is $$Br: [Ar]4s^23d^{10}4p^5$$
There is 1 unpaired electron.
Work Step by Step
1) The nearest noble gas of lower atomic number of $Br$ is $Ar$.
2) Looking the periodic table, the atomic number of $Br$ is 35, which means it has 35 electrons. The last rows ends with the atomic number 18, so $Br$ has 18 inner-shell electrons. That leaves it with 17 outer-shell electrons.
$Br$ belongs to the 4th row, so its outer shell is the 4th shell.
3) These 17 outer-shell electrons are distributed as follows:
- The first two go to the $4s$ subshell.
- The next 10 go to the $3d$ subshell.
- The last 5 go to the $4p$ subshell.
Therefore, the condensed electron configuration of $Br$ is $$Br: [Ar]4s^23d^{10}4p^5$$
4) All the inner shells, the $4s$ subshell, and the $3d$ subshell are completely filled, so all of their electrons are paired.
There are 5 electrons in subshell $4p$, and subshell $4p$ has 3 orbitals. According to Hund's rule, the first 3 electrons each occupy a different orbital, then the last 2 each occupy a one-electron orbital. So, that leaves 1 orbital with just 1 electron.
Therefore, there is 1 unpaired electron.
