Why does NCl3 use sp³ hybrid orbitals? I have a question regarding the hybridization of $\ce{NCl3}$.
From what I've understood, hybridization occurs when electrons belonging to different orbitals mix.
For example an electron belonging to the orbital 's' moves to one of the 'p' orbitals creating a certain number of hybrid orbitals. This occurs when more individually occupied orbitals are needed in order to complete the bonds and form the molecule.

Now, considering the fact that in $\ce{NCl3}$, nitrogen already has three individually occupied 'p' orbitals, it doesn't need to hybridise these orbitals in order to create three bonds with chlorine.
So why does $\ce{NCl3}$ utilise sp3 hybridization?
Let us suppose that your assumption is correct and nitrogen has already $\ce{2p^3}$ configuration. Let's further assume that it doesn't undergo hybridisation. Instead, the pure atomic orbitals undergo overlap with 3 $\ce{Cl}$ atoms and the filled $\ce{2s}$ shell electron pair acts as a lone pair and thus $\ce{NCl3}$ is formed. This consideration creates problems.

If pure atomic orbitals of nitrogen, i.e. $\ce{2p_x}$, $\ce{2p_y}$ and $\ce{2p_z}$ undergo &sigma;-bond formation with the chlorine atoms, then the bonds should be $90^\circ$ to each other, as $\ce{p_x}$, $\ce{p_y}$ and $\ce{p_z}$ orbitals are mutually perpendicular. But $\ce{NCl3}$ has 3 &sigma;-bonds with bond angles between $\ce{Cl-N-Cl}$ close to $107.1^\circ$, which is close to the tetrahedral bond angle of $109.5^\circ$. Therefore the pure orbital undergoing bond formation doesn't support this observation.

If we consider $\ce{sp^3}$ hybrid orbitals of $\ce{NCl3}$, this tetrahedral arrangement of three &sigma;-bonds and a lone pair (disregarding  the lone pair, it's pyramidal) is nicely explained. That's why, the hybridisation is $\ce{sp^3}$ which supports every observed chemical property of $\ce{NCl3}$.