Question

Why is imidazole (C3H4N2) aromatic?

Answer 1

Well, the requirements for aromaticity are:

1. Planar ring
2. `4n + 2` `pi` electrons
3. Delocalization of electrons throughout the ring (i.e. conjugated `pi` system)

Drawing out imidazole, we get:

LOCALIZED VS. DELOCALIZED

I would examine the electron geometries to check whether or not all the lone pairs of electrons supposedly conjugated throughout the system are in an orbital that is perpendicular to the ring, NOT parallel (if electrons are in a parallel orbital, they are localized outside of the otherwise aromatic ring).

We see two lone pairs. The upper right nitrogen has an `sp^2` hybridization, since it has three electron groups, so that lone pair is localized outside the ring. That means it won't participate in the delocalization presented in imidazole's resonance structures, and hence, does not count towards Hückel's Rule.

HÜCKEL'S RULE

So, imidazole has two `pi` electrons from the left and right double bonds each. Also, as it turns out, the lone pair on the bottom nitrogen IS within the ring, making it `6` electrons. Therefore, `4n + 2 = 6` and `n = 1`, and it follows Hückel's Rule .

PLANARITY

Next, we know this is a planar ring because we have two `sp^2` carbons adjacent to the bottom nitrogen, thus locking that nitrogen in a planar configuration, even though it is `sp^3`.

CONJUGATED `\mathbf(pi)` SYSTEM?

Finally, drawing the resonance structures, we get:

thus showing full conjugation throughout the `\mathbf(pi)` system.