# Non-integer hybridization

I've known that hybridization in distorted geometries is not exactly $sp^3$ or $sp^2$ or whatever. For example, $\ce{PH3}$ has nearly pure $p$ orbitals in the $\ce{P-H}$ bond, and the lone pair is in a nearly pure $s$ orbital.

Basically, since hybridisation is an addition of wavefunctions, instead of a perfect symmetrical addition of the kets, we get something else.

While answering this question, I realised that it's not that easy predicting the numbers.

Take the same image of $\ce{B2H6}$:

At first, seeing the 97°, I thought "well, the inner $\ce{B-H}$ bonds will be almost pure $p$". But, with that, I couldn't figure out where the 120° came from, because that is for perfect $sp^2$.

I then realised that I was being stupid and it wasn't that simple--just because you have a 120° doesn't mean pure $sp^2$. But, I was at a loss trying to find out approximate hybridizations for $\ce{B2H6}$.

How does one generally go about predicting such "nonuniform hybridizations" if one knows the bond angles?

Approximations are OK--I believe the exact mixture ratios will require some knowledge of the exact wavefunctions.

• I never encountered non-integer hybridization, and I don't think hybridization is so much useful as a concept…
– F'x
May 5 '12 at 13:42
• @Manishearth With the bond angle of 120, I think sp2 and then proper MO theory on the P orbitals and bridging hydrogens makes sense.
– Nick
Jun 1 '12 at 12:16
• @Nick: Yeah, maybe this system is outside of VSEPR's capabilities due to the banana bond (even I think so now). But that was just an example. Jun 1 '12 at 12:25

## 3 Answers

Classical hybridization theory does not allow for noninteger hybridizations. However, ab initio calculations can be interpreted using a bond order analysis method such as NBO, where the MO coefficients are used to provide the closest analogue possible to a classical hybridization picture.

For example, one of the pure $sp$ orbitals in Pauling's valence bond theory (where hybridization was first introduced) has wavefunction

$$\phi_{sp} = \frac{1}{\sqrt 2} \phi_{s} + \frac{1}{\sqrt{2}} \phi_{p_x}$$

Taking the square of the coefficients, this orbital is $\frac 1 2$ s character and $\frac 1 2$ p character, i.e. it is an $sp$ orbital.

The basic idea of a bond order analysis method is to reexpress a molecular orbital into a form similar to

$$\phi = c_1 \phi_{As} + c_2 \phi_{Ap_x} + c_3 \phi_{Ap_y} + c_4 \phi_{Ap_z} + \dots$$

If the other coefficients are very small, the ratio

$$n = \frac {c_2^2 + c_3^2 + c_4^2}{c_1^2}$$

would yield a number that could be used interpret $\phi$ as a $sp^n$ orbital on atom $A$.

Edit: As for inferring hybridization states from a direct inspection of nuclear geometries, there is in principle no such direct relationship in electronic structure theories that are more sophisticated than VSEPR. The former is an electronic property whereas the spatial arrangement of atomic nuclei are not, and the relationship between the two becomes much more complicated.

• "As for inferring hybridization states from a direct inspection of nuclear geometries, there is in principle no such direct relationship in electronic structure theories that are more sophisticated than VSEPR." I would have hoped you could explain that further. For example Bent's rule gives an explanation for correlation between geometry and hybridisation, and I find that by far more sophisticated than VSEPR. Feb 28 '17 at 7:25

Wave functions can be used (as described in the answer by AcidFlask) but for something that has just s and p hybridization (tetrahedral, trig planar, equatorial positions of TBP), I find it easier to use the formula $\cos x = \frac{S}{S-1} = \frac{P-1}{P}$, where $x$ is the angle between the central atom and two identical bonding atoms (in degrees), $S \times 100~\%$ is the percentage of s-character in each of the bonds, and $P \times 100~\%$ is the percentage of p-character in the bonds.

It works really well to explain things like why $\ce{PH3}$ and $\ce{NH3}$ act differently as bases, as it can be used to show that the bonds are mainly p-character (~$95~\%$), so the lone pair on $\ce{PH3}$ is mainly in an s-orbital, so is a weaker base. It could also be applied to your molecule, like the bonds in the 93 degree angle have $13~\%$ s-character, so $86~\%$ p.

For angles less than 90 degrees, the formula does not work (negative answers), but angles less than 90 have to be all p because they represent bent (or banana) bonds when they appear in molecules (like yours) with no d orbitals open.

• The formula only works for very symmetric molecules and it would assign the same hybridisation to the bond you are comparing the angle to. It completely ignores the other bonds in the molecule, see Bent's rule for a phenomenological explanation. Feb 28 '17 at 7:20

I have stumbled on this years later, and since this page comes up high on a google search for non-integer hybridization, I thought I should correct some things.

Non-integer hybridization has been around from its beginning, as its math falls out trivially from the coefficients of the orbitals used to generate the hybrid orbitals. It predates NBO by many decades.

Non-integer hybridization is a useful model for bonding because it provides simple explanations for many things. Examples include the IR stretch for cyclopropanes and cyclic ketones, the acidity of bicyclobutanes, the basicity of hindered amines, HCH bond angles in small rings. It provides a simple orbital basis for ideas from VSEPR such as the HXH angle in H3CCl vs ammonia. As with any model, it will fail outside of its purview, and your mileage may vary on whether you find it the easiest way to understand things. One does not see the math of it often in recent textbooks but its ideas are implicit in the way that many things are described.

There is a very simple mathematical relationship between the angle between two orbitals and the hybridization. If you describe the hybridization of an orbital as sp^(lambda^2), then lambda^2 = -1 / cos (angle). For example, the HCH angle in cyclopropane is 116° and you can find from this that the hybridization for the CH bonds is sp^2.28.

So for the original question, the 120° angles mean that the orbitals used for those BH bonds are essentially pure sp2. The hybridization for the orbitals for the internal BH bonds is slightly more complicated. One cannot use the angles in a ring to determine hybridization because one often has bent bonds when the atoms are constrained, but the total s character for each boron has to be 1. Since the external BH bonds use up 2/3 of the s, the remaining 1/3 is split between the two orbitals and their hybridization is about sp5 (1/6 s).

There is a tremendous amount of old literature on all of this.