# Why does NH3 hybridize at all?

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?

• Re: "It seems that all of our observations of NH3 can be explained without the idea of hybridization?" // So how do you explain the geometry?!? – MaxW Feb 1 '19 at 4:44

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.