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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:

Ammonia orbital digagram

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?

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    $\begingroup$ 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?!? $\endgroup$ – MaxW Feb 1 at 4:44
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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.

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  • $\begingroup$ Note that the lone pair in PH3 is not "in the s-orbital", nor is it correct that the s and p orbitals of NH3 are closer together in energy than the s and p orbitals of PH3. What is correct is that the bonding MOs of NH3 have more s character than in PH3 because there is less s/p mixing because s and p are farther apart in NH3. NH3 is thus closer to sp2 in its bonding MOs than PH3 is, and its structure is correspondingly closer to a flat plane than PH3's structure (though still quite far from flat). $\endgroup$ – Andrew Feb 1 at 14:54
  • $\begingroup$ @Andrew I guess you have a typo there, but put that aside, you say it is not correct that s and p orbitals are closer together in nitrogen than in phosphorus; then you say there is less sp mixing in NH3 (where I assume you mean PH3) because s and p are further apart. From how I understood it, that is a contradictory statement. Could you please elaborate. Also from electronic structure theory, the lone pair orbital in PH3 has about 90% s character; I'd say you could consider this as an s orbital. $\endgroup$ – Martin - マーチン Feb 8 at 9:20
  • $\begingroup$ @Martin - there's no typo - less mixing in NH3 means less s character in the lone pair, since the starting state of the lonepair orbital is pure p. It's well described here: chemistry.stackexchange.com/questions/38599/…. I don't know where you're getting your 90% number for PH3 lone pair s character. Jan says 50% in his answer, and I have an (admittedly quite old) review that has it at 14% (Dixon, Adv Inorg Chem Radiochem vol 29 p.41). 50% would be the maximum with an essentially sp type orbital. $\endgroup$ – Andrew Feb 8 at 12:36
  • $\begingroup$ @Andrew I apologise, I had a wrong number in my mind. Coulson's law would predict about sp4 orbitals for the PH bond, ergo 40% s for the lone pair. G09/NBO6 DF-BP86/def2-svp results in 57% s / 43% p character of the lone pair (or sp0.75 if you prefer); and 14% s / 85% p (or sp6) for P in the PH bonds. For comparison: NH3: LP 30% s / 70%p (sp2.4); BD 23% s / 76% p (sp3.3). Now you might call that less mixing, I wouldn't. But then again... that's all not real anyway. $\endgroup$ – Martin - マーチン Feb 8 at 13:41

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