How To Determine Hybridization: A Shortcut

A Shortcut For Determining The Hybridization Of An Atom In A Molecule

Here’s a shortcut for how to determine the hybridization of an atom in a molecule that will work in at least 95% of the cases you see in Org 1. 

For a given atom:

• Count the number of atoms connected to it (atoms – not bonds!)
• Count the number of lone pairs attached to it.
• Add these two numbers together.

• If it’s 4, your atom is sp³.
• If it’s 3, your atom is sp².
• If it’s 2, your atom is sp. 

(If it’s 1, it’s probably hydrogen!)

The main exception is atoms with lone pairs that are adjacent to pi bonds, which we’ll discuss in detail below.

Table of Contents

1. Some Simple Worked Examples Of The Hybridization Shortcut
2. How To Determine Hybridization Of An Atom: Two Exercises
3. Are There Any Exceptions?
4. Exception #1: Lone Pairs Adjacent To Pi-bonds
5. Lone Pairs In P-Orbitals (Versus Hybrid Orbitals) Have Better Orbital Overlap With Adjacent Pi Systems
6. Exception #2. Geometric Constraints
7. “Geometry Determines Hybridization, Not The Other Way Around”
8. Notes
9. Quiz Yourself!

1. Some Simple Worked Examples Of The Hybridization Shortcut

sp³ hybridization: sum of attached atoms + lone pairs = 4

sp² hybridization: sum of attached atoms + lone pairs = 3

sp hybridization: sum of attached atoms + lone pairs = 2

Where it can start to get slightly tricky is in dealing with line diagrams containing implicit (“hidden”) hydrogens and lone pairs.

Chemists like time-saving shortcuts just as much as anybody else, and learning to quickly interpret line diagrams is as fundamental to organic chemistry as learning the alphabet is to written English.

Remember:

• Just because lone pairs aren’t drawn in on oxygen, nitrogen, and fluorine doesn’t mean they’re not there.
• Assume a full octet for C, N, O, and F with the following one exception: a positive charge on carbon indicates that there are only six electrons around it. [Nitrogen and oxygen bearing a formal charge of +1 still have full octets].

[Advanced: Note 1 covers how to determine the hybridization of atoms in some weird cases like free radicals, carbenes and nitrenes ]

2. How To Determine Hybridization Of An Atom: Two Exercises

Here’s an exercise. Try picking out the hybridization of the atoms in this highly poisonous molecule made by the frog in funky looking pyjamas, below right.

[Don’t worry if the molecule looks a little crazy: just focus on the individual atoms that the arrows point to (A, B, C, D, E). A and B especially.  If you haven’t mastered line diagrams yet (and “hidden” hydrogens) maybe get some more practice and come back to this later.]

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Here are some more examples.

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More practice quizzes for hybridization can be found here (MOC Membership unlocks them all)

3. Are There Any Exceptions? 

Sure.  Although as with many things, explaining the shortcut takes about 2 minutes, while explaining the exceptions takes about 10 times longer.

Helpfully, these exceptions fall into two main categories. It should be noted that by the time your course explains why these examples are exceptions, it will likely have moved far beyond hybridization.

Bottom line: these probably won’t be found on your first midterm.

4. Exception #1: Lone Pairs Adjacent To Pi-bonds 

The main exception is for atoms bearing lone pairs that are adjacent to pi bonds.

Quick shortcut: Lone pairs adjacent to pi-bonds (and pi-systems) tend to be in unhybridized p orbitals, rather than in hybridized spⁿ orbitals.

This is most common for nitrogen and oxygen.

In the cases below, a nitrogen or oxygen that we might expect to be sp³ hybridized is actually sp² hybridized (trigonal planar).

Why? The quick answer is that lowering of energy from conjugation of the p-orbital with the adjacent pi-bond more than compensates for the rise in energy due to greater electron-pair repulsion for sp² versus sp³

[see this post: “Conjugation and Resonance“]

What’s the long answer?

5. Lone Pairs In P-Orbitals (Versus Hybrid Orbitals) Have Better Orbital Overlap With Adjacent Pi Systems

Let’s think back to why atoms hybridize in the first place: minimization of electron-pair repulsion.

For a primary amine like methylamine, adoption of a tetrahedral (sp³) geometry by nitrogen versus a trigonal planar (sp²) geometry is worth about 5 kcal/mol [roughly 20 kJ/mol].

That might not sound like a lot, but for two species in equilibrium, a difference of 5 kcal/mol in energy represents a ratio of about 4400:1] . [How do we know this? See this (advanced) Note 2 on nitrogen inversion] 

What if there was some compensating effect whereby a lone pair unhybridized p-orbital was actually more stable than if it was in a hybridized orbital?

This turns out to be the case in many situations where the lone pair is adjacent to a pi bond!  The most common and important example is that of amides, which constitute the linkages between amino acids. The nitrogen in amides is planar (sp²), not trigonal pyramidal (sp³), as proven by x-ray crystallography.

The difference in energy varies widely, but a typical value is about 10 kcal/mol favouring the trigonal planar geometry.  [We know this because many amides have a measurable barrier to rotation a topic we also talked about in the Conjugation and Resonance post]

Why is trigonal planar geometry favoured here? Better orbital overlap of the p orbital with the pi bond vs. the (hybridized) sp³ orbital.

The drawing below tries to show how a change in hybridization from sp³ to sp² brings the p-orbital closer to the adjoining p-orbitals of the pi bond, allowing for better orbital overlap. Better orbital overlap allows for stronger pi-bonding between the nitrogen lone pair and the carbonyl p-orbital, which results in an overall lowering of energy. 

You can think of this as leading to a stronger “partial” C–N bond. Two important consequences of this interaction are restricted rotation in amides, as well as the fact that acid reacts with amides on the oxygen, not the nitrogen lone pair (!)

The oxygen in esters and enols is also also sp² hybridized, as is the nitrogen in enamines and countless other examples.

As you will likely see in Org 2, some of the most dramatic cases are those where the “de-hybridized” lone pair participates in an aromatic system. Here, the energetic compensation for a change in hybridization from sp³ to sp² can be very great indeed – more than 20 kcal/mol in some cases.

For this reason, the most basic site of pyrrole is not the nitrogen lone pair, but on the carbon (C-2) (!).

6. Exception #2. Geometric Constraints

Another example where the actual hybridization differs from what we might expect from the shortcut is in cases with geometric constraints. For instance in the phenyl cation below, the indicated carbon is attached two two atoms and zero lone pairs.

What’s the hybridization?

From our shortcut, we might expect the hybridization to be sp.

In fact, the geometry around the atom is much closer to sp². That’s because the angle strain adopting the linear (sp) geometry would lead to far too much angle strain to be a stable molecule.

7. “Geometry Determines Hybridization, Not The Other Way Around”

A quote passed on to me from Matt seems appropriate:

“Geometry determines hybridization, not the other way around”

Well, that’s probably more than you wanted to know about how to determine the hybridization of atoms.

Suffice to say, any post from this site that contains shortcut in the title is a sure fire-bet to have over 1000 words and >10 figures.

Thanks to Matt Pierce of Organic Chemistry Solutions  for important contributions to this post.  Ask Matt about scheduling an online tutoring session here.

Notes

Note 1. Some weird cases.

Sometimes you might be asked to determine the hybridization of free radicals and of carbenes (or nitrenes)

Although you’re unlikely to encounter these, let’s still have a look.

• Free radicals exist in a shallow pyramidal geometry, not purely sp² or sp³.
• However, if they are adjacent to a pi system (e.g. a C-C double or triple bond) then the shallow pyramid will re-hybridize to give it an sp² geometry, which allows for full resonance delocalization of the free radical.
• Carbenes and nitrenes would give us sp² geometry by the hybridization shortcut. However their actual structures can vary depending on whether or not the electron pair exists in a single orbital (a singlet carbene) or is divided into two singly-filled orbitals (a triplet carbene). That’s really beyond the scope of introductory organic chemistry.

What about higher block elements like sulfur and phosphorus?

Third row elements like phosphorus and sulfur can exceed an octet of electrons by incorporating d-orbitals in the hybrid.  This is more in the realm of inorganic chemistry so I don’t really want to discuss it. Here’s an example for the hybridization of SF₄ from elsewhere.  (sp³d orbitals).

Note 2: For the 5 kcal/mol figure, see here. [Tetrahedron Lett, 1971, 37, 3437]. (Kurt Mislow, RIP. )

An amine connected to three different substituents (R¹ R² and R³) should be chiral, since it has in total 4 different substituents (including the lone pair). However, all early attempts to prepare enantiomerically pure amines met with failure. It was later found that amines undergo inversion at room temperature, like an umbrella being forced inside-out by a strong wind.

In the transition state for inversion the nitrogen is trigonal planar. One can thus calculate the difference in energy between the sp³ and sp² geometries by measuring the activation barrier for this process (see ref).

Note 3:A fun counter-example might be Coelenterazine .

One would not expect both nitrogen atoms to be sp² hybridized, because that would lead to a cyclic, flat, conjugated system with 8 pi electrons : in other words, antiaromatic. I can’t find a crystal structure of the core molecule to confirm (but would welcome any additional information!)

NOTE – (added afterwards) If you draw the resonance form where the nitrogen lone pair forms a pi bond with the carbonyl carbon, then the ring system has 10 electrons and would therefore be “aromatic”.

Quiz Yourself!

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1. Barrier to pyramidal inversion of nitrogen in dibenzylmethylamine
   Michael J. S. Dewar and W. Brian Jennings
   Journal of the American Chemical Society 1971 93 (2), 401-403
   DOI: 10.1021/ja00731a016

2. Pyramidal inversion barriers: the significance of ground state geometry
   Joseph Stackhouse, Raymond D.Baechler, Kurt Mislow
   Tetrahedron Letters  Volume 12, Issue 37, 1971, Pages 3437-3440
   DOI: doi.org/10.1016/S0040-4039(01)97199-0