… 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.
The concept of electronic orbital has become such a useful and engraved tool in understanding chemical structure and reactivity that it has almost become one of those things whose original meaning has been lost and replaced for a utilitarian concept, one which is not bad in itself but that may lead to some wrong conclusions when certain fundamental facts are overlooked.
Last week a wrote -what I thought was- a humorous post on this topic because a couple of weeks ago a viewpoint in JPC-A was published by Pham and Gordon on the possibility of observing molecular orbitals through microscopy methods, which elicited a ‘seriously? again?‘ reaction from me, since I distinctly remember the Nature article by Zuo from the year 2000 when I just had entered graduate school. The article is titled “direct observation of d-orbital holes.” We discussed this paper in class and the discussion it prompted was very interesting at various levels: for starters, the allegedly observed d-orbital was strikingly similar to a dz2, which we had learned in class (thanks, prof. Carlos Amador!) that is actually a linear combination of d(z2-x2) and d(z2-y2) orbitals, a mathematical -lets say- trick to conform to spectroscopic observations.
Pham and Gordon are pretty clear in their first paragraph: “The wave function amplitude Ψ*Ψ is interpreted as the probability density. All observable atomic or molecular properties are determined by the probability and a corresponding quantum mechanical operator, not by the wave function itself. Wave functions, even exact wave functions, are not observables.” There is even another problem, about which I wrote a post long time ago: orbitals are non-unique, this means that I could get a set of orbitals by solving the Schrödinger equation for any given molecule and then perform a unit transformation on them (such as renormalizing them, re-orthonormalizing them to get a localized version, or even hybridizing them) and the electronic density derived from them would be the same! In quantum mechanical terms this means that the probability density associated with the wave function internal product, Ψ*Ψ, is not changed upon unit transformations; why then would a specific version be “observed” under a microscope? As Pham and Gordon state more eloquently it has to do with the Density of States (DOS) rather than with the orbitals. Furthermore, an orbital, or more precisely a spinorbital, is conveniently (in math terms) separated into a radial, an angular and a spin component R(r)Ylm(θ,φ)σ(α,β) with the angular part given by the spherical harmonic functions Ylm(θ,φ), which in turn -when plotted in spherical coordinates- create the famous lobes we all chemists know and love. Zuo’s observation claim was based on the resemblance of the observed density to the angular part of an atomic orbital. Another thing, orbitals have phases, no experimental observation claims to have resolved those.
Now, I may be entering a dangerous comparison but, can you observe a 2? If you say you just did, well, that “2” is just a symbol used to represent a quantity: two, the cardinality of a set containing two elements. You might as well depict such quantity as “II” or “⋅⋅” but still cannot observe “a two”. (If any mathematician is reading this, please, be gentle.) I know a number and a function are different, sorry if I’m just rambling here and overextending a metaphor.
Pretending to having observed an orbital through direct experimental methods is to neglect the Born interpretation of the wave function, Heisenberg’s uncertainty principle and even Schrödinger’s cat! (I know, I know, Schrödinger came up with this gedankenexperiment in order to refute the Copenhagen interpretation of quantum mechanics, but it seems like after all the cat is still not out of the box!)
So, the take home message from the viewpoint in JPC is that molecular properties are defined by the expected values of a given wave function for a specific quantum mechanical operator of the property under investigation and not from the wave function itself. Wave functions are not observables and although some imaging techniques seem to accomplish a formidable task the physical impossibility hints to a misinterpretation of facts.
I think I’ll write more about this in a future post but for now, my take home message is to keep in mind that orbitals are wave functions and therefore are not more observable (as in imaging) than a partition function is in statistical mechanics.
Pauli’s Exclusion Principle is a paramount concept in Quantum Mechanics which has implications from statistical mechanics to quantum chemistry, consequently, there are many different statements to summarize it depending on the forum. I occasionally joke with my students about how we learnt it in kindergarten an how we state it now at the end of our computational chemistry course.
So, are you a toddler or high up there with W. Pauli predicting the existence of sub-atomic particles at CERN? Which statement of Pauli’s Exclusion Principle sounds more familiar to you?
LOL just feeling a little humorous this morning!
Is the C atom in methane sp3 hybridized because it’s tetrahedral or is it tetrahedral because it’s sp3 hybridized? It’s funny how many students think to this date that the correct answer is the latter; specially those working in inorganic chemistry. I ignore the reason for such trend. What is true is that most chemistry teachers seem to have lost links to certain historical facts that have shaped our scientific discipline; most of those lay in the realm of physics, maybe that’s why.
What Linus Pauling, in a very clever way, stated was that once you have a set of eigenvectors (orbitals) of the atomic Hamiltonian any combination of them will also be an eigenvector (which is normal since one of the properties of Hermitian operators is that they are linear); so why not making a symmetry adapted one? Let’s take the valence hydrogenoid orbitals (hydrogenoid being the keyword here) and construct a linear combination of them, in such a way that the new set transforms under the irreducible representations of a given point group. In the case of methane, the 2s and 2p orbitals comprise the valence set and their symmetry-adapted-linear-combination under the Td point group constitutes a set of new orbitals which now point into the vertexes of a tetrahedron. Funny things arise when we move to the next period of the table; it has been a controversy for a number of years the involvement of empty d orbitals in pentacoordinated P(V) compounds. Some claim that they lay too high in energy to be used in bond formation; while others claim that their involvement depends on the nature (electronegativity mainly) of the surrounding substituents.
In many peer reviewed papers authors are still making the mistake of actually assigning a type of hybridization to set of valence orbitals of an atom based on the bond angles around it. Furthermore, it is not uncommon to find claims of intermediate hybridizations when such angles have values in between those corresponding to the ideal polyhedron. Symmetry is real, orbitals are not; they are just a mathematical representation of the electron density distribution which allows us to construct mind images of a molecule.
Linus Pauling is one of my favourite scientific historical figures. Not only did he build a much needed at the time bridge between physics and chemistry but he also ventured into biochemistry (his model of an alpha-helix for the alanine olygopeptide became the foundation to Watson & Cricks later double helix DNA model), X-ray diffractometry, and humanities (his efforts in reducing/banning the proliferation of nuclear weapons got him the Nobel Peace Prize long after he had already received the Nobel Price in Chemistry). He was a strong believer of ortho-molecular nutrition, suggesting that most illnesses can be related to some sort of malnutrition. Linus Pauling and his book On the Chemical Bond will remain a beacon in our profession for the generations to come.
Disclaimer: The question above, with which I opened this post, was taken from an old lecture by Dr. Raymundo Cea-Olivares at UNAM back in the days when I was an undergraduate student.