Question

Why is imidazole $ ({C_3}{H_4}{N_2}) $ aromatic?

Answer 1

Hint: Imidazole is an organic compound with the formula $ {C_3}{H_4}{N_2} $ . It is a white or colorless solid that is soluble in water, producing a mildly alkaline solution. In chemistry, it is an aromatic heterocycle, classified as a diazole, and has non-adjacent nitrogen atoms.

Complete answer:
For a compound to be considered aromatic, it must be flat, cyclic, and conjugated and it must obey Huckel's rule. Huckel's rule states that an aromatic compound must have pi electrons in the overlapping $ p - $ orbitals in order to be aromatic ( $ n $ in this formula represents any integer).
Localized vs delocalized: The upper right nitrogen has an $ s{p^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.
Huckel’s rule: 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 six electrons. Therefore, $ 4n + 2 = 6 $ and $ n = 1 $ , and it follows Huckel's Rule.
Planarity: we know this is a planar ring because we have two $ s{p^2} $ carbons adjacent to the bottom nitrogen, thus locking that nitrogen in a planar configuration, even though it is $ s{p^3} $ . And it has a conjugated $ \pi $ system.

Note:
An aromatic molecule or compound is one that has special stability and properties due to a closed loop of electrons. Not all molecules with ring structures are aromatic. Molecules that are not aromatic are termed aliphatic. If a molecule contains an aromatic sub-unit, this is often called an aryl group.