A Hybridization Shortcut

by James

in Chemical Bonds, Organic Chemistry 1

Here’s a shortcut for figuring out the hybridization of an atom in a molecule. This will save you a lot of time.

–BEGIN SHORTCUT–

Here’s what you do:

Look at the atom.

Count the number of atoms connected to it (not bonds – atoms).

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!)

This works in at least 95% of the cases you will see in Org 1.

Here’s some simple examples.

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: a quick note on some weird cases. ]

–END SHORTCUT–

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.]

[Answers here]

Here are some more examples.

[Answers here.]

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.

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 sporbitals. So when a nitrogen that you might expect to be trigonal pyramidal sp3 is adjacent to a pi bond, its hybridization is actually sp2 (trigonal planar).

Why? The quick answer is that lowering of energy from conjugation more than makes up for any gain in energy through increased steric hindrance. [see this post: “Conjugation and Resonance“]

What’s the long answer?

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 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) (!).

[Are there exceptions to this exception? Rare ones – see below]

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.

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.


Answers to Q1: A) sp3  B) sp3  C) sp  D) sp2  E) sp3

Answers to Q2: A) sp3 B) sp2  C) sp2  D) sp2  E) sp  F) sp  G) sp  H) sp

Notes

Note 1. Some weird cases.

Unlikely to encounter these, but here goes:

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].

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 (notably by the work of Kurt Mislow (RIP)].

Note 3:A fun counter-example would 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!)

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{ 4 comments… read them below or add one }

anon

“Third row elements like phosphorus and sulfur can exceed an octet of electrons by incorporating d-orbitals in the hybrid.”

This is incorrect, and was proven wrong years ago. See http://pubs.acs.org/doi/abs/10.1021/ja00273a006 and citing references therein. It’s more accurate (and more intuitive) to continue to follow the octet rule for sulfur, phosphorus, and other heavy main group elements. SF4, for example, can be represented as four equal-weight resonance structures of the form [SF3]+[F]-, giving an overall bond order of 0.75 for each S-F bond. This way, every atom follows the octet rule in each resonance structure. Of course you could always use molecular orbital theory in conjunction with symmetry-adapted linear combination of atomic orbitals, and then you wouldn’t need to deal with “expanded octets” in hypercoordinate molecules.

Reply

Victor

Yes and no.
There’s nothing intrinsically wrong in the phrase itself as the “hypervalent” atoms DO use the higher orbitals to some extent. It is more to the point of what orbitals are involved in the overall bonding scheme. And no, nobody, who has at least some understanding of the concept of the hybridization, will insist that by saying that sulfur in SF4 has the sp3d hybridization will strictly mean that we have 100% involvement of 1 s, 3 p, and 1 d orbital in the bonding structure.
It’s the same kind of argument we can bring when discussing, say, cyclopropanone. What is the hybridization of the carbonyl carbon there? Is it sp2? Is it sp2+? Is it sp2-? Is it somewhere in between? What about the hybridization in di-central rhenium complexes with quaternary bond? Or riddle me out, for instance, the exact iodine’s hybridization in every form of periodic acid ;)
When we acknowledge the limitations of the theories we use, they are in a pretty good agreement with each other ;) And while using the MO is the best way to go, it is not what is being taught at the general chemistry or organic chemistry level, nor it is what students are facing on the test.

Reply

Victor

The 1H-NMR of coelenterazine (DOI: 10.1021/ct300356j) shows two signals at 9.13 (s, 1 H), and 6.44 (bs, 1 H), which suggests that the system is conjugated with the carbonyl making it all planar and aromatic when considering the entire bicyclic system. You can observe similar deshielding effects in, say, azulene for the protons on the 7-membered ring.

Reply

James

Now that I look at it again, you’re absolutely right. Thanks Victor.

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