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I had a doubt while solving this problem.
At first glance of these compounds, I felt that all of them have sp3 hybridisation (for nitrogen), since all of them had 3 bonded pairs and 1 lone pair around the nitrogen atom.
But then I saw in a website that both 2 and 3 have sp2 hybridisation. Can you explain why this occurs?
The design of the question offers you four choices where choice 4 states nitrogen in both case 2 and 3 (formally) were $\text{sp}^2$ hybridized.
Case 3 however differs from case 1 and 2 by that the lone electron pair of nitrogen no longer is isolated (as in open chain/alkyl amines), but is in conjugation with both $\ce{C=C}$ double bonds. Because the skeleton is cyclic, of about planar geometry, and the total of participating $\pi$ electrons equals to 6, pyrrole is an electron rich Hückel aromatic system ($4n + 2$, here with $n = 1$). Thus, 3 it is the only example among the structures displayed where nitrogen is $\text{sp}^2$ hybridized.* Because the orientation of the lone orbital of nitrogen affects the chemistry (e.g., $\text{p}K_a$), I recommend an additional reading about e.g., imidazole:
(image credit to chem.libretexts.org)
* Actually, atoms do not hybridize. Hybridization is one of the mathematical concepts in chemistry, here: how to mix atomic orbitals to yield molecular orbitals of certain shape.
Only option three would have $\ce{sp^2}$ hybridization for nitrogen as it’s Nitrogen has π-bond in conjugation. Therefore the lone pair of nitrogen is delocalized thus in the newly formed resonating structure Nitrogen does not have 1 lone pair and 3 $\sigma$-bonds but 3 $\sigma$-bonds and one π-bond… Giving $\ce{sp^2}$ hybridization.
