Why does NH3 hybridize at all?

Question

So in Chemistry class I've been taught that hybridization is a way we can explain things such as how $\ce{CH4}$, for example, forms four, equally strong bonds.

However at the same time I'm told that the Nitrogen in $\ce{NH3}$ hybridizes and forms sp³ orbitals. When I draw the orbital diagram for $\ce{NH3}$ this is what I get:

and so looking at this, I see that $\ce{NH3}$ should be able to form 3 equal bonds in the 2p subshell which matches our observations.

So my question is— why do we then explain the bonding orbitals of nitrogen in $\ce{NH3}$ using hybridization? It seems that all of our observations of $\ce{NH3}$ can be explained without the idea of hybridization?

Answer 1

Let's consider the alternatives to the pseudo-pyramidal structure we observe in reality (not sure how, probably neutron diffraction or rotational spectroscopy). First alternative: a $\text{sp}^2$-hybridization with a lone pair in a p-orbital. This already assumes hybridization, but is not good energetically because a lone pair in a p-orbital is higher in energy than one in an orbital with partial or full s-character.

Second alternative: a $\ce{PH3}$-like with H-N-H angles close to 90°, putting the lone pair in the s-orbital. This does not occur because the s and p-orbitals are close enough in energy that they will participate in the valence bonds, that is, they will hybridize. The net repulsion between the N-H bonds would outweigh the energy gained by putting the lone pair in the s-orbital.

Of course, these are post-fact explanations and they are not backed by numbers here. However, using electronic structure theory methods (computational/quantum chemistry) we can calculate the relative energies of the structures described and even carry out a search for (local) minima of energy with respect to molecular structure. They would show the same thing, albeit without explanation.