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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?

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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 σ-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 σ-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 σ-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 $\ce{NCl3}$.

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  • $\begingroup$ Although the above-written answer is true, one can always ask Why doesn't $\ce{NCl3}$ form 3 $\sigma$ bonds out of the pure p orbitals? What is the reason that $\ce{NCl3}$ adopts such tetrahedral geometry? $\endgroup$
    – Apoorv Potnis
    Feb 7, 2018 at 12:19
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    $\begingroup$ @ApoorvPotnis I would not say that the above is correct. It assumes hybridisation prior to bond formation, which is dead wrong. Hybridisation is a mathematical concept, it only helps us understand certain aspects of the wave function. You are absolutely correct in asking for the reason of the mixing of s and p orbitals to allow more flexible bond angles, because this is essentially the reason why a description with sp³ orbitals is possible in the first place. This has partially be answered here. $\endgroup$
    – Martin - マーチン ♦
    Feb 7, 2018 at 12:49
  • $\begingroup$ @Martin-マーチン Yes, my mistake. I did not read the last paragraph properly. $\endgroup$
    – Apoorv Potnis
    Feb 7, 2018 at 13:02

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