Daily Archives: August 10, 2017
… but were afraid to ask
How I learned to stop worrying and not caring that much about hybridization.
The math behind orbital hybridization is fairly simple as I’ll try to show below, but first let me give my praise once again to the formidable Linus Pauling, whose creation of this model built a bridge between quantum mechanics and chemistry; I often say Pauling was the first Quantum Chemist (Gilbert N. Lewis’ fans, please settle down). Hybrid orbitals are therefore a way to create a basis that better suits the geometry formed by the bonds around a given atom and not the result of a process in which atomic orbitals transform themselves for better sterical fitting, or like I’ve said before, the C atom in CH4 is sp3 hybridized because CH4 is tetrahedral and not the other way around. Jack Simmons put it better in his book:
The atomic orbitals we all know and love are the set of solutions to the Schrödinger equation for the Hydrogen atom and more generally they are solutions to the hydrogen-like atoms for which the value of Z in the potential term of the Hamiltonian changes according to each element’s atomic number.
Since the Hamiltonian, and any other quantum mechanical operator for that matter, is a Hermitian operator, any given linear combination of wave functions that are solutions to it, will also be an acceptable solution. Therefore, since the 2s and 2p valence orbitals of Carbon do not point towards the edges of a tetrahedron they don’t offer a suitable basis for explaining the geometry of methane; even more so these atomic orbitals are not degenerate and there is no reason to assume all C-H bonds in methane aren’t equal. However we can come up with a linear combination of them that might and at the same time will be a solution to the Schrödinger equation of the hydrogen-like atom.
Ok, so we need four degenerate orbitals which we’ll name ζi and formulate them as linear combinations of the C atom valence orbitals:
ζ1= a12s + b12px + c12py + d12pz
ζ2= a22s + b22px + c22py + d22pz
ζ3= a32s + b32px + c32py + d32pz
ζ4= a42s + b42px + c42py + d42pz
to comply with equivalency lets set a1 = a2 = a3 = a4 and normalize them:
a12 + a22 + a32 + a42 = 1 ∴ ai = 1/√4
Lets take ζ1 to be directed along the z axis so b1 = c1 = 0
ζ1 = 1/√4(2s) + d12pz
since ζ1 must be normalized the sum of the squares of the coefficients is equal to 1:
1/4 + d12 = 1;
d1 = √3/2
Therefore the first hybrid orbital looks like:
ζ1 = 1/√4(2s) +√3/2(2pz)
We now set the second hybrid orbital on the xz plane, therefore c2 = 0
ζ2 = 1/√4(2s) + b22px + d22pz
since these hybrid orbitals must comply with all the conditions of atomic orbitals they should also be orthonormal:
〈ζ1|ζ2〉 = δ1,2 = 0
1/4 + d2√3/2 = 0
d2 = –1/2√3
our second hybrid orbital is almost complete, we are only missing the value of b2:
ζ2 = 1/√4(2s) +b22px +-1/2√3(2pz)
again we make use of the normalization condition:
1/4 + b22 + 1/12 = 1; b2 = √2/√3
Finally, our second hybrid orbital takes the following form:
ζ2 = 1/√4(2s) +√2/√3(2px) –1/√12(2pz)
The procedure to obtain the remaining two hybrid orbitals is the same but I’d like to stop here and analyze the relative direction ζ1 and ζ2 take from each other. To that end, we take the angular part of the hydrogen-like atomic orbitals involved in the linear combinations we just found. Let us remember the canonical form of atomic orbitals and explicitly show the spherical harmonic functions to which the 2s, 2px, and 2pz atomic orbitals correspond:
ψ2s = (1/4π)½R(r)
ψ2px = (3/4π)½sinθcosφR(r)
ψ2pz = (3/4π)½cosθR(r)
we substitute these in ζ2 and factorize R(r) and 1/√(4π)
ζ2 = (R(r)/√(4π))[1/√4 + √2 sinθcosφ –√3/√12cosθ]
We differentiate ζ2 respect to θ, and set it to zero to find the maximum value of θ respect to the z axis we get the angle between the first to hybrid orbitals ζ1 and ζ2 (remember that ζ1 is projected entirely over the z axis)
dζ2/dθ = (R(r)/√(4π))[√2 cosθ –√3/√12sinθ] = 0
sinθ/cosθ = tanθ = -√8
θ = -70.53°,
but since θ is measured from the z axis towards the xy plane this result is equivalent to the complementary angle 180.0° – 70.53° = 109.47° which is exactly the angle between the C-H bonds in methane we all know! and we didn’t need to invoke the unpairing of electrons in full orbitals, their promotion of any electron into empty orbitals nor the ‘reorganization‘ of said orbitals into new ones. Orbital hybridization is nothing but a mathematical tool to find a set of orbitals which comply with the experimental observation and that is the important thing here!
To summarize, you can take any number of orbitals and build any linear combination you want, in order to comply with the observed geometry. Furthermore, no matter what hybridization scheme you follow, you still take the entire orbital, you cannot take half of it because they are basis functions. That is why you should never believe that any atom exhibits something like an sp2.5 hybridization just because their bond angles lie between 109 and 120°. Take a vector v = xi+yj+zk, even if you specify it to be v = 1/2i that means x = 1/2, not that you took half of the unit vector i, and it doesn’t mean you took nothing of j and k but rather than y = z = 0.
This was a very lengthy post so please let me know if you read it all the way through by commenting, liking, or sharing. Thanks for reading.