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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?

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Wong
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  • related http://chemistry.stackexchange.com/questions/29168/what-is-the-hybridization-of-the-carbonyl-oxygen-in-a-carboxylic-acid – Mithoron May 06 '15 at 14:35
  • 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." – Martin - マーチン May 07 '15 at 06:00
  • @Martin I don't understand your generalized statement. What is not permitted in resonance and how is it linked here? – Wong May 09 '15 at 12:42
  • 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. – Martin - マーチン May 09 '15 at 12:46
  • @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? – Wong May 09 '15 at 13:00
  • It can be something in between, but it has to be the same hybridisation in all resonance structures/ configurations. – Martin - マーチン May 09 '15 at 13:01
  • @Martin Why so? When we donate lone pair to the ring, the N has +ve charge and is planar. So all those resonance structures contribute sp2 while the initial localized lone pair structure is tetrahedral and will contribute sp3. The combined effect of this will have something in between of sp2/sp3. – Wong May 09 '15 at 13:05
  • *NO*. That is not resonance. You can go ahead and ask this as a new question, I do not have the time to discuss this in the comments. Hybridisation is a helpful concept (not necessarily required for chemistry), resonance is a necessary concept to extend the scope of Lewis structures. Both are just simplifications of the true bonding situation. – Martin - マーチン May 09 '15 at 13:12
  • @Martin I was talking about these structures. And considering contributions of different structures on hybridization. Is this wrong? If yes, I will ask another question. – Wong May 09 '15 at 13:22
  • @Wong Did you ask a new question? – ado sar Nov 16 '20 at 00:31

2 Answers2

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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 ($\mathrm{sp^3}$).

Mathew Mahindaratne
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ron
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  • What is pyramidal inversion? – Wong May 06 '15 at 15:15
  • 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) – Wong May 06 '15 at 15:18
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    See [here] (http://www.expertsmind.com/topic/properties-of-amines/pyramidal-inversion-913906.aspx) for a nice description and picture of pyramidal inversion. – ron May 06 '15 at 15:29
  • 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? – Wong May 06 '15 at 15:38
  • Yes, that's exactly correct. – ron May 06 '15 at 15:44
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    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. – Martin - マーチン May 07 '15 at 05:53
  • @Martin So in the end it is more sp2 since it is just slightly pyramidal(sp3)? – Wong May 09 '15 at 12:45
  • @Wong Yes, I assume, that is the case. And this is probably also the reason, why inversion is fast, has a low barrier, respectively. – Martin - マーチン May 09 '15 at 12:50
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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.

Yomen Atassi
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    But the lone pair of nitrogen is in resonance with the ring. Since it makes it ortho para directing. – Wong May 06 '15 at 14:58
  • All fine about pyridine. But in aniline, shouldn't it be more sp2-ish?The lone pair is not localised. – Wong May 09 '15 at 12:50