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We all know from general chemistry that the s-orbital is spherical, and p-orbitals are dumbbell-looking orbitals oriented along the x, y, and z axes of the Cartesian system. We also know that VSEPR describes the 3D shape of the second period elements reasonably well.

And here we have a problem: the atomic orbitals are at 90° to each other, while the VSEPR theory predicts the 3D structure of, say, methane (CH_{4}) to be tetrahedral with bond angles around 109.5°. So, how can we have 109.5° bond angles made by the orbitals which are at 90° to each other? 🤯🤷♂️🤔

In 1931 the twice Nobel Laureate Linus Pauling proposed the model of “mixing” the orbitals or “hybridizing” them to account for the observed bonding pattern. Pauling proposed sort of a combination of the orbitals giving you an orbital that has partial characters. Not a complete s- or a p-orbital, but rather something with a partial s- and partial p-character.

Let me emphasize one more time that hybridization is a ** mathematical model**. There’s no actual “process” that happens to orbitals that causes the hybridization. The atomic orbitals don’t actually change before going into the bonding with other atoms. Hybridization is a mathematical model that describes how the atomic orbitals would’ve looked like based on the observable molecular orbitals.

Ok, now when we know that hybridization is a ** model** and not an actual process, let’s look at how this “process” happens. 🤣 Each bond takes 2 electrons to complete. If we look at the carbon atom atomic orbitals, we’ll see the 2 electrons on the 2s and 2 electrons on the 2p shells. This would only allow carbon to make 2 bonds since it only has 2 unpaired electrons. To make four bonds, carbon would have to “decouple” its s-electrons onto the p shell.

Rearranging the electrons in an atom in this way also makes the orbitals closer in energy making them virtually degenerate. This allows for easier “mixing” or hybridization as we know it. Thus, when we mix those orbitals together we end up with a set of “hybrids” and any leftovers that were not hybridized.

An important thing to remember: # of AO’s = # of MO. So, when we mix the atomic orbitals to make the hybrids, we will end up with the exactly the same number of the the orbitals when we’re done. Thus, by mixing 4 orbitals (one s and three p), we’ll always get 4 molecular orbitals (hybrids or not).

Mixing an s-orbital and three p-orbitals gives four sp^{3} orbitals.

The resulting four hybrid sp^{3} orbitals are all degenerate in energy, meaning they are all the same. Also, according to VSEPR theory, those orbitals need to be as symmetric around each other as possible. This gives a tetrahedral structure with bond angles around 109.5°.

When those hybrid orbitals make bonds, we get molecular orbitals oriented in the same direction. So, as I’ve mentioned earlier, while the hybridization and the hybrid orbitals might be the mathematical model, it does help us predict and illustrate the actual molecular orbitals in the molecule. BTW, the molecular orbital theory (MOT) is a mathematical model as well. 🤣 However, we can perform calculations using the MOT to predict the electron densities around the molecule congruent with the real physical observations.

When we mix one s-orbital and two p-orbitals, we get three sp^{2} hybridized orbitals. You’ll also have one leftover p-orbital that didn’t participate in the hybridization.

This hybridization gives you the trigonal planar geometry around the central atom with the p-orbital sticking in the up and down vertical direction.

The “unused” p-orbital can make a π-bond or to participate in a complex resonance conjugation.

In the case of the sp hybridization, only one s- and one p-orbital are mixed together to make hybrids. This leaves two unused p-orbitals.

The unused sp orbitals force the structure to have a linear 3D geometry.

The unused p-orbitals can make two double bonds, a triple bond, or potentially participate in resonance with other orbitals.

The easiest way to determine hybridization is to with the VSEPR theory and determine the number of electron groups around your central atom. To put it plain, I can summarize the hybridizations in the following picture:

So, the 3 groups around the central atom gives you the sp^{3} hybridization, the three groups gives you sp^{2} hybridization, and the two groups yield the sp-hybridized species. Also remember, we do count the the spare electron pairs as the electron groups too!

Unless the electron pair is next to a double or a triple bond (or an empty p-orbital), the electron pair will be on the hybrid orbital and not the p-orbital. While that is not 100% true in reality, that’s the way we treat it within the scope of a typical organic chemistry class, so we’ll stick with it too.

I’ve mentioned above, that a double or a triple bond next to an electron pair matters. When you have an electron pair next to a p-orbital or a π-bond, there’s a resonance between those.

This is a very typical “trick question” on the exam, so you wanna keep this in mind. It’s also important to remember that the electron pair has to be physically able to align with the p-orbital or a π-bond for this to happen. So, the isolated electron pairs will still be sitting on the hybrid orbitals even when they are next to double bonds.

Spotting the isolated electron pairs can be a little tricky, so you may wanna do some practice to master this skill. There is a quick rule of thumb you can use. If you have an electron pair on the atom that already has a double bond, chances are, it’s going to be isolated. It’s not a 100% foolproof trick, but it works for cyclic structures.

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