# How do you predict the hybridisation of an atom in a molecule?

## For example, the $N$ atom in Pyridine is $s {p}^{2}$ hybridised. But how do you find that?

Jun 24, 2016

You look at the number of electron groups. There's more to it than that, but at a basic level...

The number of electron groups equals the number of orbitals used to construct the hybridized orbitals.

You might find it useful to look over this answer as well.

PYRIDINE HYBRIDIZATION

In pyridine, what you have on the $\text{N}$ atom is that the major resonance structure of pyridine has $\text{N}$ with three electron groups:

• one connected to a carbon via a single bond
• one connected to a carbon via a double bond
• one that is a lone pair of electrons

With three electron groups, you have the $2 s$ and two $2 p$ orbitals hybridizing, for a total of three $s {p}^{2}$ hybridized orbitals.

ORBITAL PERSPECTIVE OF PYRIDINE

Actually, the $\text{N}$ atom has:

• two hybridized $s {p}^{2}$ atomic orbital lobes partially occupied with one electron (both used to $\sigma$ bond).
• one unhybridized $2 {p}_{z}$ atomic orbital lobe partially occupied with one electron (used to $\pi$ bond within the ring), perpendicular to the ring. This is not included in the counting of electron groups, because it isn't hybridized.
• one hybridized nonbonding $s {p}^{2}$ atomic orbital lobe (the third one of three) that holds the lone pair, parallel to the ring.

Let's fill in the missing $s {p}^{2}$ orbitals and see how they would look.

The $s {p}^{2}$ hybridization occurs because carbon needs to bond with nitrogen in a diagonal (off the Cartesian coordinate axes) manner in 2 dimensions (on the $x y$-plane).

The $2 {p}_{x}$ and $2 {p}_{y}$ orbitals of neither nitrogen nor carbon were pointed in that diagonal direction to begin with, and in a compound like pyridine, a $2 {p}_{x}$ orbital is not compatible with a $2 {p}_{y}$ orbital.

So, they, along with the $2 s$ orbital, hybridized with each other, to make three $s {p}^{2}$ hybridized orbitals.

The hybridization must occur to make diagonal bonds that are all identical in energy and look.

The $2 {p}_{z}$ orbitals of nitrogen and carbon are compatible with each other, so there was no need to hybridize with the third $2 p$ orbital (the $2 {p}_{z}$) as well. There was no need to achieve $s {p}^{3}$ hybridization with only three electron groups required.