I found a thread which discussed this which totally confused me. (refer to the part of the link which talks about $\ce{sp^2/sp^3}$)
The question "How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?" is helpful, but doesn't have aniline specifically. I understand that if the lone pair is in resonance, it becomes $\ce{sp^2}$.
Can someone help me out?

  • $\begingroup$ related chemistry.stackexchange.com/questions/29168/… $\endgroup$
    – Mithoron
    May 6, 2015 at 14:35
  • $\begingroup$ I marked it as a duplicate anyway, since the premise is (almost) the same, and the answer is general enough to address, that "Atoms that are sp²-hybridized and sp³-hybridized have differing geometries, which is not permitted in the resonance phenomenon." $\endgroup$ May 7, 2015 at 6:00
  • $\begingroup$ @Martin I don't understand your generalized statement. What is not permitted in resonance and how is it linked here? $\endgroup$
    – Wong
    May 9, 2015 at 12:42
  • $\begingroup$ An sp² hybridised atom is trigonal coordinated and an sp³ hybridised atom is tetrahedrally coordinated. When describing a compound with resonance structures, all of these structures have to have the same geometry, i.e. coordination is equal in all structures. Therefore an atom can never be sp² and sp³ hybridised in the same set of resonance structures. $\endgroup$ May 9, 2015 at 12:46
  • $\begingroup$ @Martin The resonance hybrid can never be both sp2 and sp3 while the indiviual structures(imaginary ones) are either. Is this true? And by "Therefore an atom can never be sp² and sp³ hybridised in the same set of resonance structures." do you mean to say the answer has to be sp2 or sp3 and not something in between them? $\endgroup$
    – Wong
    May 9, 2015 at 13:00

2 Answers 2


The nitrogen in aniline is somewhere between $\mathrm{sp^3}$ and $\mathrm{sp^2}$ hybridized, probably closer to the $\mathrm{sp^2}$ side. We are correctly taught that the nitrogen in simple aliphatic amines is pyramidal ($\mathrm{sp^3}$ hybridized). However in aniline, due to the resonance interaction between the aromatic ring and the nitrogen lone pair, considerable flattening of the nitrogen occurs (if it were completely flat it would be $\mathrm{sp^2}$ hybridized).

We can assess the nitrogen hybridization by measuring its barrier for pyramidal inversion. If a trigonal nitrogen is $\mathrm{sp^2}$ hybridized, the barrier will be zero. On the other hand, in aliphatic amines where the nitrogen is $\mathrm{sp^3}$ hybridized the inversion barrier is typically around $\pu{4-5 kcal/mol}$.

pyramidal inversion diagram

(pyramidal inversion diagram)

In aniline this barrier is very low, somewhere around $\pu{1-2 kcal/mol}$. This indicates that the nitrogen in aniline is not quite planar, but is much closer to being planar ($\mathrm{sp^2}$) than pyramidal ($\mathrm{sp^3}$).

  • $\begingroup$ What is pyramidal inversion? $\endgroup$
    – Wong
    May 6, 2015 at 15:15
  • $\begingroup$ Why can't it be called completely sp2? If it is in resonance, then it should be sp2. (that question link on my question says so) $\endgroup$
    – Wong
    May 6, 2015 at 15:18
  • 1
    $\begingroup$ See [here] (expertsmind.com/topic/properties-of-amines/…) for a nice description and picture of pyramidal inversion. $\endgroup$
    – ron
    May 6, 2015 at 15:29
  • $\begingroup$ And its contribution to the true representation is lesser as compared to the ones with the lone pair involved in resonance, so sp2 is dominant? $\endgroup$
    – Wong
    May 6, 2015 at 15:38
  • 1
    $\begingroup$ Hybridisation is a function of geometry. Since it's a requirement for resonance structures to have the same geometry, hybridisation in the different contributing configurations must not change. We know that the amine group in aniline is slightly pyramidal, hence hybridisation cannot be ideal sp². I would rather state, that due to resonance, lesser p character is mixed in the sigma system, since some of the perpendicular p orbital is necessary for the overlap of the pi system. $\endgroup$ May 7, 2015 at 5:53

As the non-bonding electron pair of nitrogen could partially contribute in the resonance with the other $\pi$ electrons in the cycle. (I will prove this point indirectly further using an experimental data), the nitrogen atom hybridization is a linear combination of sp2 and sp3.

Now if we consider the nitrogen atom in the pyridine, it is sp2 hybridized. Because the nitrogen atom has two neighboring atoms (2 carbon atoms) and one non-bonding electron pair, i.e. three neighboring electron pairs. But, the non-bonding electron pair can not contribute in the resonance with the other four $\pi$ electrons in the cycle. In fact, it is perpendicular to the aromatic $\pi$ system.

The pKa for the pyridine systems is 5.2, for aniline it is 4.6. So pyridine is a stronger base. We can link this to the availability of the lone pair in pyridine, as it is not delocalized ( perpendicular to the aromatic $\pi$ system. While in aniline the nitrogen lone pair can be delocalised using resonance with the adjacent $\pi$ system.

  • 3
    $\begingroup$ But the lone pair of nitrogen is in resonance with the ring. Since it makes it ortho para directing. $\endgroup$
    – Wong
    May 6, 2015 at 14:58
  • $\begingroup$ All fine about pyridine. But in aniline, shouldn't it be more sp2-ish?The lone pair is not localised. $\endgroup$
    – Wong
    May 9, 2015 at 12:50

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