sp² hybridized orbital

I do not understand why and how an atom with $\ce{sp^2}$ hybridization has only one p orbital?? 


For example in pyridine why is the lone pair of nitrogen not counted as a pi electron. 




The picture below from the [UC Davis ChemWiki](http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_07%3A_Organic_compounds_as_acids_and_bases/Section_7.4%3A_More_on_resonance_effects_on_acidity_and_basicity#7.4B:_The_basicity_of_nitrogen-containing_groups:_aniline.2C_imines.2C_pyridine.2C_and_pyrrole) summarizes why that lone pair is not available. It also explains how pyridine can function as a base.
![pyridine orbitals](https://i.stack.imgur.com/dxpOH.png)


The lone pair in the $sp^2$ orbital is perpendicular to the $p$ orbitals and $\pi$ bonds in the molecule. the electron pair from the nitrogen that participates in aromaticity is **the one in the $\ce{C=N}$ bond**.


To answer your first question about hybridization, we have a conservation law that guarantees that when orbitals mix (for any reason), we need to get the same number of orbitals out as we put in. The law is the [Conservation of Angular Momentum](http://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum), which does apply in quantum systems.


Nitrogen, in its valence shell, has four orbitals: $2s,2p\_x,2p\_y,2p\_z$. The $sp^2$ hybridization is achieved by mixing two $p$ orbitals and one $s$ orbital. Because of the conversation of angular momentum, four orbitals in means four orbitals out. Thus, the $p$ orbital that was not mixed is still there (and is the only one still there).


$$[s,p,p],p\implies[sp^2,sp^,sp^2],p$$


We cannot have two $p$ orbitals at an $sp^2$ hybridized atom because then we would have 5 orbitals (or we would not have $sp^2$).



> 
> Now, let's create a general rule for aromatic heterocycles.
> 
> 
> 


Look at pyrrole (also from ChemWiki);


![pyrrole](https://i.stack.imgur.com/o1GRx.png)


In the structure of pyrrole, the $\ce{N-H}$ bond uses the $sp^2$ orbital and the lone pair is in the $p$ orbital. In general, if there is a double bond drawn to the nitrogen atom (like in pyridine), then the $\pi$ bond uses the $p$ orbital and the lone pair is in the $sp^2$ orbital and does not count toward aromaticity. If there is no double bond drawn to the nitrogen atom, usually there is a $\ce{N-H}$ bond (or some other bond to the nitrogen). This $\ce{N-H}$ bond uses the $sp^2$ orbital and the lone pair uses the $p$ orbital and counts toward aromaticity.


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