# Is the hybridization only related to atomic orbitals but not to the molecular orbitals?

What is Hybridization?

(Spelling: Hybridisation (British English) / Hybridization (American English))
The hybridization is a concept that describes atomic orbitals. In other words: Hybridization is an approach to model hybrid orbitals mathematically. The geometry of a molecule determines the hybridization and not vice versa.

IUPAC Gold Book's definition of hybridization:

Linear combination of atomic orbitals on an atom. Hybrid orbitals are often used in organic chemistry to describe the bonding molecules containing tetrahedral ($sp^3$), trigonal ($sp^2$) and digonal ($sp$) atoms.

For a certain molecule structure we can allocate the atomic orbitals of one atom to hybrid orbitals which are also localized on this atom and which cannot be redistributed over the molecule to over atoms, e.g. a hydrogen can only have a s orbital but no sp hybrid orbitals because it has no p atomic orbitals.

The allocation can be done using the following relations:

• $\ce{1s + 3p ->4sp^3}$
• $\ce{1s + 2p + p ->3sp^2 + p}$
• $\ce{1s + 1p + 2p -> 2sp + 2p}$

We say: "The atom X is best described by a $\ce{s^\text{x}p^yd^z...}$ hybridization."

Combining two orbitals / two hybrid orbitals / one hybrid orbital (e.g. $\ce{sp^2}$) with a orbital (e.g. $\ce{s}$) we obtain bonds. $\sigma$-bonds with $\ce{sp}$ orbitals (ethyne) of the atomic centers have lower energy than $\sigma$-bonds with $\ce{sp^2}$ orbitals (ethylene), lower in energy means: they are better acceptors.

What is my question now?

I'm thinking if we can also describe a bond (a molecular orbital) with a hybridization, e.g. if we can combine a $\ce{sp^2}$ hybridized carbon with a $s$ atomic orbital of hydrogen to make two $\ce{sp}$ molecular orbitals (a bonding $\sigma$ orbital with $sp$ character and an antibonding $\sigma^\star$ orbital with $\ce{sp}$ character; just be saying for a $\ce{C-H}$ bond: $\ce{s} + \ce{sp^2} = 2 \ce{sp}$, what's probably wrong).

I guess this question has to be answered with No, because we cannot say a bond has hybridization.

The question might arise why I'm thinking of the hybridization of molecular orbitals.

Over several years I've heard statements like this: The $\ce{C#C}$ bond of ethyne ($\ce{HCCH}$) has higher s character (50%) than the $\ce{C=C}$ bond of ethylene ($\ce{H2CCH2}$) (33%) that's why ethyne has a lower $\mathrm{p}K_\mathrm{a}$ than ethylene (25 versus 44 at standard conditions in water as solvent) because a bond with higher s character can better stabilize a negative charge, because the s orbitals are penetrating deeper into the shell as the p orbitals do.

Again, I think hybridization is only a good model for atomic orbitals (=localized on atom) not for linear combinations of atomic orbitals (=bonds).

Can we really say the $\sigma$ bond in ethylene has $sp^2$ character?

Hybridisation is a concept devised by man to ease explanation of chemical observations. It is not a 1:1 equivalent of anything seen in molecules but rather a way of rationalising what one has seen — one that is as mathematically correct as molecular orbitals (that span the entire molecule) are.

There are two fundamentally different approaches on how to describe ‘bonds’. The first one is the molecular orbital approach. It considers electrons that can freely move around the entire molecule, in a so-called potential energy space: the potential energy derives from the (negatively charged) electron being close to (a series of positively charged) nuclei which attract each other. As you may remember from your very first chemistry classes ever, the mass of even a proton is $2000$ times that of an electron, so a carbon nucleus is some $24000$ times heavier than an electron. According to Newton, the nuclei have a lot more inertia than the electrons do, so it is reasonable to assume that whichever fixed positions the nuclei take, all electrons can then assume favourable movement patterns.

Quantum mechanics, and further quantum chemistry, give us methods to calculate what movement patterns these electrons might assume — although technically the equation is not solveable, because it also must contain electron-electron interactions, there are good approximations out there. The wave functions that one can derive from all this information correspond to molecular orbitals you might have seen before: They abide by the molecule’s symmetry (will always fit into one irreduceable representation of the molecule’s symmetry point group) but extend over the entire molecule (even the orbitals representing atomic core electrons can be thought of being delocalised somewhat).

Using the quantum mechanical method of calculation (operators) you get an energy for each molecular orbital. Thereby you have ordered your molecular orbitals by energy (approximately equalling the number of nodal planes) — this gives you HOMO and LUMO that you might have heard of before.

To reduce this back into ‘bonds’, we need to use mathematical tricks. Simply spoken: All wave functions can be treated as vectors, and vectors can be linear combinated to give other vectors — meaning we take $0.5$ times orbital $\Psi_1$, $0.01$ times orbital $\Psi_2$ and so on for each molecular orbital, add them up, divide them by a normalisation factor, and we get a new orbital that has a different shape (and a new energy). If we do this in a sensible way, we can manage to reduce molecular orbitals to those that only ‘span’ across one bond — because we have cancelled out all wave functional contributions from other parts of the molecule. This would correspond to your typical $\sigma$ and $\pi$ type bonding, nonbonding and anti-bonding orbitals.

The second method is the hybridisation approach. This time, you take the general shape of the molecule and consider each atom by itself. For each atom, you can assume a hybridisation based on its geometry: If there are three other atoms in a triangular shape around it, all sharing the same plane with the atom in question, we can assume it to be sp² hybridised.

Once we have done this, we need to take a look at neighbouring atoms to create ‘bonds’ again. Usually, there will be two hybrid orbitals ‘pointing’ at each other, ready to create a bond. Using the same technique we have to create the hybrid orbitals, we can mix these two hybrid orbitals to create another orbital. This orbital is not considered to be of a new sp-hybrid type, but rather gets a new type of name: It can be $\sigma$, $\pi$ or $\delta$ (to name the most common ones). These Greek letters closely resemble s, p and d, and the bond orbitals in question share a common symmetric trait with the corresponding atomic orbitals: $\sigma$ orbitals can be rotated freely around the bond axis, $\pi$ orbitals have one and $\delta$ orbitals two nodal planes along the bond axis (the bond axis is completely part of the nodal plane).

If we did this correctly, we will arrive at orbitals of similar shape and energy as with the above method, only we are much, much faster, which is why this method is often taught in schools.

Onto the question of how do the $\ce{C-C}$ (and $\ce{C-H}$) bonds differ in $\ce{C2H4}$ and $\ce{C2H2}$:
You just saw how we decided what orbitals we want to combine to form a bond. In $\ce{C2H4}$, we are combining a carbon’s sp² orbital with a hydrogen’s s-orbital to create a $\sigma$ bond. For simplicity, we can label this bond $\sigma_{\ce{sp^2-s}}$. Similarly for $\ce{C2H2}$, only that this one has to be labelled $\sigma_{\ce{sp-s}}$. You see that you created bonds with different hybrid orbitals and therefore the bonds will have different ‘shapes’ and energies. This is what is meant with

The $\ce{C#C}$ bond of ethyne ($\ce{HCCH}$) has higher s character ($50\,\%$) than the $\ce{C=C}$ bond of ethylene ($\ce{H2CCH2}$) ($33\,\%$)

Technically, this is short for:

The hybrid atomic orbitals of the carbons of ethyne have a higher s character ($50\,\%$) than those of ethene ($33\,\%$), so the $\sigma_{\ce{C-H}}$ bonds will retain that property somewhat.

Long story short: No, we cannot say that a $\sigma$ bond has sp² or sp³ character; we can (and should) only say that the atomic orbital of one side of that bond has this character.

And: It is wrong to assume $\ce{sp^2 + s} = \ce{2 sp}$, because that would recentre a p-orbital from the carbon to the hydrogen.

• Let me be the nitpicker this time: The two approaches you describe are not exclusive and lead to an identical bonding picture, if treated with the proper quantum chemical approach, where the hybridisation approach is covered by valence bond theory. In short: You can extract the VB picture from the MO ansatz and vice versa. Jun 26, 2015 at 6:23
• @Martin-マーチン would it be correct to say that MOT is a more fundamental approach and can make predictions about bond strength, geometry etc.... while hybridization of VBT is a reverse-engineering approach and can only rationalize them? Feb 11, 2021 at 0:32
• @OVERWOOTCH No. Fundamentally, at their respective limits: MOT = VBT. That is also why you can approximate one with the other. What is in the between is too complex to deal with in a comment or maybe even an answer. Feb 12, 2021 at 22:54