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?

1 Answer
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.


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.

![http://wps.prenhall.com/](useruploads.socratic.org)

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.