Answer
The condensed electron configuration of $Sb$ is $$Sb:[Kr]4d^{10}5s^25p^3$$
There are 3 unpaired electrons.
Work Step by Step
1) The nearest noble gas of lower atomic number of $Sb$ is $Kr$.
2) Looking the periodic table, the atomic number of $Sb$ is 51, which means it has 51 electrons. The last row ends with the atomic number 36, so $Sb$ has 36 inner-shell electrons. That leaves it with 15 outer-shell electrons.
$Sb$ belongs to the 5th row, so its outer shell is the 5th shell.
3) These 15 outer-shell electrons are distributed as follows:
- The first two go to the $5s$ subshell.
- The next 10 go to the $4d$ subshell.
- The last 3 go to the $5p$ subshell.
Therefore, the condensed electron configuration of $Sb$ is $$Sb:[Kr]4d^{10}5s^25p^3$$
4) All the inner shells, the $5s$ subshell, and the $4d$ subshell are completely filled, so all of their electrons are paired.
There are 3 electrons in subshell $5p$, and subshell $5p$ 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.
