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?

  • 4
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.