# Blog Archives

## All you wanted to know about Hybrid Orbitals…

#### … but were afraid to ask

#### or

#### 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 CH_{4} is sp^{3} hybridized because CH_{4} 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 *ζ _{i}* and formulate them as linear combinations of the C atom valence orbitals:

*ζ _{1}*=

*a*+

_{1}2s*b*+

_{1}2p_{x}*c*+

_{1}2p_{y}*d*

_{1}2p_{z}*ζ _{2}*=

*a*+

_{2}2s*b*+

_{2}2p_{x}*c*+

_{2}2p_{y}*d*

_{2}2p_{z}*ζ _{3}*=

*a*+

_{3}2s*b*+

_{3}2p_{x}*c*+

_{3}2p_{y}*d*

_{3}2p_{z}*ζ _{4}*=

*a*+

_{4}2s*b*+

_{4}2p_{x}*c*+

_{4}2p_{y}*d*

_{4}2p_{z}to comply with equivalency lets set *a _{1}* =

*a*=

_{2}*a*=

_{3}*a*and normalize them:

_{4}*a _{1}*

*+*

^{2}*a*

_{2}*+*

^{2}*a*

_{3}*+*

^{2}*a*

_{4}*= 1 ∴*

^{2}*a*= 1/√4

_{i}Lets take *ζ _{1}* to be directed along the

*z*axis so

*b*=

_{1}*c*= 0

_{1}*ζ _{1 }*= 1/√4(

*2s*) +

*d*

_{1}2p_{z}since *ζ _{1}* must be normalized the sum of the squares of the coefficients is equal to 1:

^{1}/_{4} + *d _{1}^{2}* = 1;

*d _{1}* =

^{√3}/

_{2}

Therefore the first hybrid orbital looks like:

*ζ _{1}* =

^{1}/

_{√4}(

*2s*) +

^{√3}/

_{2}(

*2p*)

_{z}We now set the second hybrid orbital on the xz plane, therefore *c _{2}* = 0

*ζ _{2}* =

^{1}/

_{√4}(

*2s*) +

*b*+

_{2}2p_{x}*d*

_{2}2p_{z}since these hybrid orbitals must comply with all the conditions of atomic orbitals they should also be orthonormal:

〈*ζ _{1}*|

*ζ*〉 = δ

_{2}_{1,2}= 0

^{1}/_{4} + *d _{2}*

^{√3}/

_{2}= 0

*d _{2}* = –

^{1}/

_{2√3}

our second hybrid orbital is almost complete, we are only missing the value of *b _{2}*:

*ζ _{2}* =

^{1}/

_{√4}(

*2s*) +

*b*+-

_{2}2p_{x}^{1}/

_{2√3}(

*2p*)

_{z}again we make use of the normalization condition:

^{1}/_{4} + *b _{2}^{2}* +

^{1}/

_{12}= 1;

*b*=

_{2}^{√2}/

_{√3}

Finally, our second hybrid orbital takes the following form:

*ζ _{2}* =

^{1}/

_{√4}(

*2s*) +

^{√2}/

_{√3}(

*2p*) –

_{x}^{1}/

_{√12}(

*2p*)

_{z}The procedure to obtain the remaining two hybrid orbitals is the same but I’d like to stop here and analyze the relative direction *ζ _{1}* and

*ζ*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:

_{2}ψ* _{2s}* = (1/4π)

^{½}

*R*(

*r*)

ψ* _{2px}* = (3/4π)

^{½}sinθcosφ

*R*(

*r*)

ψ* _{2pz}* = (3/4π)

^{½}cosθ

*R*(

*r*)

we substitute these in *ζ _{2}* and factorize R(r) and

^{1}/

_{√(4π)}

*ζ _{2}* = (

^{R(r)}/

_{√(4π)})[

^{1}/

_{√4}+ √2 sinθcosφ –

^{√3}/

_{√12}cosθ]

We differentiate *ζ _{2}* 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

*ζ*and

_{1}*ζ*(remember that

_{2}*ζ*is projected entirely over the

_{1}*z*axis)

d*ζ _{2}*/dθ = (

^{R(r)}/

_{√(4π)})[√2 cosθ –

^{√3}/

_{√12}sinθ] = 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 *sp ^{2.5}* hybridization just because their bond angles lie between 109 and 120°. Take a vector

*v*= x

*i*+y

*j*+z

*k*, even if you specify it to be

*v*= 1/2

*i*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.

## Pauling hybridization model

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.