Bonding, Structure, and Resonance – How To Determine Hybridization: A Shortcut

1. A Few Simple and Effective Hybridization Shortcut Examples

Sum of attached atoms plus lone pairs = 4 in sp3 hybridization.

Sum of attached atoms plus lone pairs = 3 for sp2 hybridization.

Sum of attached atoms plus lone pairs = 2 in sp hybridization.

Dealing with line diagrams containing implicit ("hidden") hydrogens and lone pairs is where things can start to get a little tricky.

The ability to quickly interpret line diagrams is as essential to understanding organic chemistry as learning the alphabet is to understanding written English. Chemists enjoy time-saving shortcuts just as much as anyone else.


Lone pairs may exist even though they aren't drawn in by oxygen, nitrogen, or fluorine.

With the following exception: a positive charge on carbon indicates that there are only six electrons surrounding it. Assume an entire octet for C, N, O, and F. Despite having a formal charge of 1, nitrogen and oxygen still have full octets.

[Advanced: Note 1 explains how to identify atom hybridization in some peculiar situations, such as free radicals, carbenes, and nitrenes

2. Two Exercises for Determining the Hybridization of an Atom

Here is a practise. Pick out which atoms in the incredibly toxic molecule created by the frog in the eccentric pyjamas below right were hybridised.

[If the molecule appears a little wacky, don't worry; just concentrate on the specific atoms that the arrows point to (A, B, C, D, and E). A and B in particular.  Maybe get some more practise with line diagrams (and "hidden" hydrogens) and come back to this later if you haven't mastered them yet.]

You can find additional hybridization practise tests here (MOC Membership is required to access them all).

3. Do Any Exceptions Apply? 

Sure.  The exceptions are explained in about 10 times as much time as the shortcut, which takes about 2 minutes to explain.

These exceptions, fortunately, can be divided into two groups. It should be noted that your course will probably be well beyond hybridization by the time it explains why these examples are exceptions.

Bottom line: You won't likely find these on your first midterm.

4. Lone Pairs Next to Pi-Bonds, Exception #1 

Atoms with lone pairs that are close to pi bonds are the main exception.

Shortcut: Instead of being in hybridised spn orbitals, lone pairs next to pi-bonds (and pi-systems) typically are in unhybridized p orbitals.

This happens most frequently with nitrogen and oxygen.

A nitrogen or oxygen that we might anticipate to be sp3 hybridised is actually sp2 hybridised (trigonal planar) in the cases below.

Why? The short answer is that the decrease in energy caused by the p-orbital's conjugation with the nearby pi-bond more than offsets the increase in energy caused by the stronger electron-pair repulsion for sp2 compared to sp3.

[See "Conjugation and Resonance" in this post]

What's the lengthy response?

5. Lone Pairs in P-Orbitals Have Better Orbital Overlap With Neighbouring Pi Systems Than Hybrid Orbitals

Let's recall the initial reasons for atom hybridization: reducing the repulsion between electron pairs.

It is worth about 5 kcal/mol [roughly 20 kJ/mol] for nitrogen to adopt a tetrahedral (sp3) geometry over a trigonal planar (sp2) geometry for a primary amine like methylamine.

Although it might not seem like much, for two species in equilibrium, a 5 kcal/mol energy difference corresponds to a ratio of approximately 4400:1. [How are we aware of this? Regarding nitrogen inversion, see this (advanced) Note 2. 

What if there was a compensating effect, making an unhybridized lone pair p-orbital more stable than one that was hybridised?

When the lone pair is close to a pi bond, this frequently turns out to be the case.  The most prevalent and significant example is amides, which are the bonds that connect amino acids. X-ray crystallography has shown that the nitrogen in amides is planar (sp2), not trigonal pyramidal (sp3).

The energy difference varies greatly, but a typical value favouring the trigonal planar geometry is around 10 kcal/mol.  [We know this because we discussed the topic of measurement-based barriers to rotation in the post on Conjugation and Resonance.]

Why is trigonal planar geometry used in this situation? Compared to the (hybridised) sp3 orbital, the p orbital has a better orbital overlap with the pi bond.

The illustration below attempts to illustrate how shifting from sp3 to sp2 hybridization improves orbital overlap by bringing the p-orbital of the pi bond closer to the adjacent p-orbitals. A stronger pi-bond between the nitrogen lone pair and the carbonyl p-orbital is made possible by improved orbital overlap, which lowers energy overall. 

You could consider that this results in a stronger "partial" C-N bond. Restricted rotation in amides and the fact that acid reacts with amides on the oxygen, not the nitrogen lone pair (!) are two significant effects of this interaction.

As with the nitrogen in enamines and countless other examples, the oxygen in esters and enols is also sp2 hybridised.

Some of the most dramatic situations are those in which the "de-hybridized" lone pair interacts with an aromatic system, as you will probably see in Org 2. Here, the energetic compensation for switching from sp3 to sp2 hybridization can be extremely high—more than 20 kcal/mol in some instances.

Because of this, the carbon (C-2) (!) atom rather than the nitrogen lone pair is the most basic site for pyrrole.

6. Geometric Restrictions, Exception #2

Geometric constraints are another situation where the actual hybridization deviates from what we might anticipate from the shortcut. For instance, the indicated carbon is attached to two atoms and zero lone pairs in the phenyl cation shown below.

What does hybridization entail?

We might anticipate the hybridization to be sp based on our shortcut.

The geometry surrounding the atom is actually much more similar to sp2. This is because adopting the linear (sp) geometry would result in far too much angle strain for the molecule to be stable.

"Geometry Determines Hybridization, Not the Other Way Around"

A quote Matt forwarded to me seems appropriate:

"Hybridization is determined by geometry, not the other way around"

That should cover everything you wanted to know about how to estimate atom hybridization.

It is sufficient to say that any post from this site with the word "shortcut" in the title will almost certainly contain more than 1000 words and >10 figures.

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