Why is nitrogen $sp^3$ hybridized in a compound such as NH3 but $sp^2$ hybridized in something like pyrrole, which has a ring structure? In both cases there are three bonds and one lone pair.
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2$\begingroup$ There are ring structures which do have sp3 nitrogen hybridisation. Being part of a ring is not the critical thing here. $\endgroup$– matt_blackJul 2, 2022 at 11:37
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1$\begingroup$ Hybridisation is a mathematical description which usually supports our view on bonding and structure. It's is defined by structure. It's also not restricted to whole numbers. The orbitals in ammonia are close to sp³, but they are not exactly. However, in zeroth order approximation is still close enough. (In this scenario 3.14 would be a good approximation for pi.) So that arrangement of bonds is crucial for the description within the restrictions of hybridisation. $\endgroup$– Martin - マーチン ♦Jul 2, 2022 at 14:12
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1$\begingroup$ That second duplicate target has a terrible title, but anyway, the question of hybridisation in pyrrole/pyridine has been addressed before. $\endgroup$– orthocresolJul 2, 2022 at 14:13
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One might argue that the nitrogen in pyrrole does not really have a lone pair we can identify like that in ammonia or for that matter pyrrolidene ($\ce{C5H9N}$, adding enough hydrogen atoms to saturated the five-membered ring in pyrrole). In the pyrrole molecule, what "should be" the lone pair on nitrogen is instead coupled with the other pi electrons to form a continuous ring of conjugated pi bonding. Such a coupling is commonplace when it forms an aromatic ring, meaning in the most common cases $4n+2$ electrons in this ring (like the six pi electrons in pyrrole). There are also cases where what seems to be an empty orbital really isn't empty because of a similar coupling (like the cyclopropenyl cation). You should see more examples of this effect as you study aromatic compounds.
answered Jul 2, 2022 at 13:39