Chemistry - Non-integer hybridization

Solution 1:

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.

Solution 2:

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.