[a confused chiral girl]

I am confused about p-orbitals. I know that all double bonds have 1 p orbitals, but are all lone paris also in the p orbital or in s orbital?  I am reviewing my notes, and apparently both pyridine and pyrrole are aromatic compounds. Pyridine fits all: cyclic, conjugated at every atom in the ring, planar, and also Huckel's Rule. For pyrrole, isn't the N that is attached to the hydrogen sp3?? For some reason, I have made special notes saying that it is sp2 and not sp3, which doesn't make sense at all.

ps. are those lone pairs that I drew around the pyridine ring correct, or is there suppose to be Hydrogens bonded there even not it is not explicitly written in the structure?
thank you.

[Yggdrasil]

The lone pairs in pyridine are not correct, they should be hydrogens.  Note that hydrogens generally aren't explicitly written in most organic structures.

Anyway, lone pairs usually occupy a hybrid orbital.  For example, in NH₃, the N is sp³ hybridized and the lone pair occupies one of the sp³ orbitals.  So, one would expect the nitrogen in pyrrole to be sp³ hybridized.  However, in pyrrole, the nitrogen actually takes on an sp² hybridization in order to allow its lone pair to become delocalized in the aromatic system.  Since aromaticity stabilizes the molecule by so much, there is a huge driving force to put the lone pair in a p orbital instead of an sp³ orbital.

[a confused chiral girl]

  However, in pyrrole, the nitrogen actually takes on an sp² hybridization in order to allow its lone pair to become delocalized in the aromatic system.  Since aromaticity stabilizes the molecule by so much, there is a huge driving force to put the lone pair in a p orbital instead of an sp³ orbital.

thanks. but then, how come the lone pair on the N in the pyridine structure is part of the p orbital conjugation to test for aromaticity? How were you able to tell right away that the lone pair in the Pyrrole is not part of the rules for determining whether the structure is aromatic or not?

[Yggdrasil]

In pyridine, the nitrogen is sp² hybridized because it has a double bond.  The one electron in its p-orbital contributes 1 electron to the 4n+2 electrons in the aromatic ring.  In the case of pyridine, the lone pair occupies a sp₂ orbital which lies perpendicular to the conjugated pi system.  Therefore, the lone pair in the nitrogen of pyridine does not contribute to the aromatic system.

In pyrrole, one would expect the nitrogen to be sp³ hybridized because it has three single bonds.  If this were the case, the lone pair would lie in a sp³ orbital.  However, it is favorable for the nitrogen to change its hybridization to sp² and place the lone pair in a p-orbital.  In the p-orbital, the lone pairs can become delocalized and form a conjugated pi system with the other pi-bonds in pyrrole.