Orbital Hybridization

When you look at the orbitals in a molecule, you notice that sometimes you have unpaired electrons in an orbital. It is these unpaired electrons that allow for covalent bonds to form. That is because in a covalent bond, the electrons are shared, meaning that both electrons need to occupy the same spot in the orbital. To do so, the new one joins the unpaired electrons in the host atom.

However, let’s look at Carbon’s electron diagram.

$\begin{aligned} \underline{\uparrow\downarrow} && \underline{\uparrow\downarrow}\ && \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}}\ \underline{\phantom{\uparrow\downarrow}} \\ 1s^2 \phantom{1}&& 2s^2 \phantom{1}&& 2p^2 \phantom{\uparrow\downarrow1} \end{aligned}$

Now if we try to add an electron, it won’t want to pair with any electron we have here. Instead, it would try to fit in the empty p orbital. You can’t make a bond like that!

So what happens in this case?

You may have guessed from the section title, but this is where orbital hybridization occurs. In orbital hybridization, two orbitals mix to some degree, allowing for bonds to form as expected.

Hybridization can take on many forms, and which one in particular depends on two things: the central atom, and the number of attached atoms in the molecule.

Let’s take carbon for example. Specifically let’s look at the carbon as it exists in methane (CH4).

We need to have 4 unpaired electrons, as the carbon is making 4 bonds. To do this, the s orbital will hybridize with all three p orbitals, forming:

$\begin{aligned} \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}} \\ sp^3 \phantom{2345675} \end{aligned}$

This allows it to accept the electrons from all 4 hydrogen atoms equally. This type of hybridization is called sp3, as all three p orbitals combine with the one s orbital.

The gallery below shows what these hybridized orbitals would look like compared to the initial orbital set-up.

In the introductory hybridization model, four electron domains around a central atom correspond to sp³. Domains include lone pairs as well as bonds, and each single, double or triple bond counts as one domain for geometry. CH₄ has four bonds; NH₃ has three bonds and one lone pair; H₂O has two bonds and two lone pairs. All three have four domains, but their molecular geometries differ.

In the case of CH2O, formaldehyde, the Carbon atom is attached to three other atoms, two hydrogen and an oxygen. If we want to look at this Carbon’s configuration, we’d see:

$\begin{aligned} \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}}\ &&\underline{\uparrow\phantom{\downarrow}} \\ sp^2 \phantom{23775} && p \phantom{2} \end{aligned}$

This is because each of the bonds formed initially between the Carbon and its attached atoms are on the same energy level. There is an extra bond between Carbon and Oxygen, though. That bond relies on the half-filled p orbital that remains.

The first bonds formed in the hybridized orbitals are called sigma (σ) bonds. These bonds are the first bonds formed between any two atoms. Should those atoms be doubly or triple bonded, the extra bond(s) are composed of pi (π) bonds. These bonds utilize the p orbital instead of the hybridized orbital.

Let’s go through and show the (unfilled) diagrams for each type of hybridization sp, sp2, and sp3.

sp3: $\begin{aligned} \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}} \\ sp^3 \phantom{2345675} \end{aligned}$
sp2: $\begin{aligned} \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ && \underline{\phantom{\uparrow\downarrow}} \\ sp^2 \phantom{23675} && p \phantom{1} \end{aligned}$
sp: $\begin{aligned} \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ &&\underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}} \\ sp \phantom{2354} && p \phantom{111} \end{aligned}$

So what happens when you toss in d-orbitals? Well, you get further types of hybridization, namely the sp3d and the sp3d2 hybridizations.

The electron diagrams for these look like this:

sp3d: $\begin{aligned} \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ && \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}} \\ sp^3d \phantom{25645^53675} && d \phantom{84723455^6} \end{aligned}$

sp3d2: $\begin{aligned} \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}&& \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}} \\ sp^3d^2 \phantom{2564^53675} && d \phantom{84725^6} \end{aligned}$

Example Problem 5.1.1

Identify the hybridization of Phosphorus in PCl5, and draw the electron configuration of the valence electrons.

That solution had us using a ‘magic number’, which was really just the steric number of the central atom, if you remember from section 4.4. To re-iterate, the steric number is the sum of the number of attached atoms and the number of lone pairs. This determines the hybridization, as when we hybridize the orbitals, we do so in order to fill each orbital with our bonded atoms. If we have too many electrons, we will have already paired electrons in the hybridized orbitals. This means we need to account for the lone pairs in our central atom when determining the hybridization pattern.

The π bond forms after the σ bond, and is higher in energy. That is because it forms using p or d orbitals, instead of lower energy hybridized ones. The π bond only exists between atoms that are doubly or triply bonded. A triple bond consists of two π bonds and one σ bond.

Also, another aspect of orbitals you need to know is the shielding effect. This effect explains why we only use valence electrons when we bond two atoms. You already know that valence electrons are on the outside of the atom, meaning there are electrons in lower orbitals between those and the nucleus. This fact, combined with simple like-attracts-like magnetism, explains the shielding effect.

In essence, those electrons between the valence electrons and the nucleus are exerting a repulsive force on the valence electrons, keeping them from being held to the nucleus as tightly as they should. This is why the valence electrons are most easily influenced by other atoms.