Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The number of $s{p}^2$-hybrid orbitals in a molecule of benzene is

Imagine you are looking at a single molecule of benzene, $C_6{H}_6$. It is one of the most iconic structures in all of chemistry. The molecule forms a perfect, flat hexagonal ring made up of six carbon atoms.

Before we can count the orbitals, we need to understand the environment of each individual carbon atom within this ring. Every carbon atom in benzene is bonded to exactly three other atoms: two adjacent carbon atoms and one hydrogen atom pointing outward.

To determine the hybridization of an atom, we look at its steric number, which is the sum of the number of atoms bonded to it and the number of lone pairs it possesses. For a carbon atom in benzene, it is bonded to $3$ atoms and has $0$ lone pairs.

A steric number of $3$ dictates that the atom must undergo $s{p}^2$ hybridization. But what does this physically mean? It means that one $s$-orbital and two $p$-orbitals from the carbon's valence shell mix together. This mixing creates exactly three equivalent $s{p}^2$ hybrid orbitals.

These three hybrid orbitals arrange themselves in a trigonal planar geometry, spreading out as far as possible to minimize repulsion, resulting in bond angles of exactly $120^{\circ }$.

Now, the math becomes beautifully simple. We have established that a single carbon atom in the benzene ring utilizes $3$ $s{p}^2$ hybrid orbitals to form its sigma ($\sigma$) bonds.

Since the benzene ring is perfectly symmetrical and consists of $6$ identical carbon atoms, we simply multiply the number of orbitals per carbon by the total number of carbons:

Thus, there are exactly $18$ $s{p}^2$ hybrid orbitals in a single molecule of benzene.

While the question only asks for the hybrid orbitals, a true chemist always visualizes the full 3D picture. If carbon uses one $s$ and two $p$ orbitals for hybridization, it has one $p$-orbital left over.

This unhybridized $p$-orbital stands perpendicular to the flat plane of the hexagonal ring. With six carbon atoms, there are six parallel $p$-orbitals. These orbitals overlap sideways, merging into a continuous ring of electron density above and below the molecule. This is the famous delocalized $\pi$-electron cloud that grants benzene its extraordinary aromatic stability!