# Why is the lone pair of pyridine's nitrogen atom not part of the aromatic ring?

In pyrrole, the lone pair of electrons belonging to the nitrogen is part of the aromatic ring. However, in pyridine it is part of an sp2-hybridized orbital. Why can't it be in the p-orbital and take part in the aromatic ring?

I mean, why can't the lone pair in pyrrole be in the p-orbital, with the sp2-hybridized orbital (which is not bonded to anything) sporting a single electron?

I'm not talking about the nitrogen atom having two p-orbitals participating in the aromatic ring. That would be strange. I wonder why the existing p-orbital of N cannot carry two electrons, the way it does in pyrrole.

P.S. A related question, with a very interesting answer: sp2-hybpidization in pyridine and pyrrole

• Because there is already a $p$ orbital of nitrogen which takes part in the aromatic ring. You can't have two $p$ orbitals on one atom pointing in the same direction. And you can't have orbital take part in the ring if it is oriented differently. – Ivan Neretin Apr 27 '16 at 15:59
• @IvanNeretin You should convert that to an answer. It sums up pretty much everything there is to say here. – bon Apr 27 '16 at 16:01
• Huckel's rule demands that there be 4n+2 pi electrons in the ring. Each carbon contributes one, making for a total of 5 from the carbons. If your nitrogen lone pair is in the ring, then you will have seven pi electrons. $7 \neq 4n +2$ for $n \in \mathbb{Z}$ – orthocresol Apr 27 '16 at 16:06
• Yes, the fact that you put one electron in the p orbital and two in the sp2 orbital is solely to satisfy Huckel's rule. For pyrrole, there are only four other carbons which contribute one electron each, and the nitrogen therefore has to give two of its own in order for it to reach $4n+2$. If the lone pair sticks out, the aromatic ring would not be "disrupted". The ring would just not be aromatic. Also, the sp2 orbital in pyrrole is bonded to a hydrogen 1s. – orthocresol Apr 27 '16 at 16:13
• Well, then it's Huckel's rule alright. Also, if you'd have a pair on $p$, you'd have an unpaired electron on that $sp^2$, which is not nice. – Ivan Neretin Apr 27 '16 at 16:15

Huckel's rule demands that there be 4n+2 pi electrons in the ring. Each carbon contributes one, making for a total of 5 from the carbons. If your nitrogen lone pair is in the ring, then you will have seven pi electrons. $7≠4n+27≠4n+2$ for $n ∈Z$