Question

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

For example, the `N` atom in Pyridine is `sp^2` hybridised. But how do you find that?

Answer 1

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.

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DISCLAIMER: Long answer!

PYRIDINE HYBRIDIZATION

In pyridine, what you have on the `"N"` atom is that the major resonance structure of pyridine has `"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 `2s` and two `2p` orbitals hybridizing, for a total of three `sp^2` hybridized orbitals.

ORBITAL PERSPECTIVE OF PYRIDINE

Actually, the `"N"` atom has:

• two hybridized `sp^2` atomic orbital lobes partially occupied with one electron (both used to `sigma` bond).
• one unhybridized `2p_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 `sp^2` atomic orbital lobe (the third one of three) that holds the lone pair, parallel to the ring.

Let's fill in the missing `sp^2` orbitals and see how they would look.

The `sp^2` hybridization occurs because carbon needs to bond with nitrogen in a diagonal (off the Cartesian coordinate axes) manner in 2 dimensions (on the `xy`-plane).

The `2p_x` and `2p_y` orbitals of neither nitrogen nor carbon were pointed in that diagonal direction to begin with, and in a compound like pyridine, a `2p_x` orbital is not compatible with a `2p_y` orbital.

So, they, along with the `2s` orbital, hybridized with each other, to make three `sp^2` hybridized orbitals.

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

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