Answer
The condensed electron configuration of $Ga$ is $$Ga: [Ar]4s^23d^{10}4p^1$$
There is 1 unpaired electron.
Work Step by Step
1) The nearest noble gas of lower atomic number of $Ga$ is $Ar$.
2) Looking the periodic table, the atomic number of $Ga$ is 31, which means it has 31 electrons. The last rows ends with the atomic number 18, so $Ga$ has 18 inner-shell electrons. That leaves it with 13 outer-shell electrons.
$Ga$ belongs to the 4th row, so its outer shell is the 4th shell.
3) These 13 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 one go to the $4p$ subshell.
Therefore, the condensed electron configuration of $Ga$ is $$Ga: [Ar]4s^23d^{10}4p^1$$
4) All the inner shells, the $4s$ subshell, and the $3d$ subshell are completely filled, so all of their electrons are paired.
There are 1 electron in subshell $4p$ and 1 electron is not enough to make a pair in any case.
Therefore, there is 1 unpaired electron.
