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Name:         Jasmine
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Date: 4/10/2003

What is orbital hybridization?

Orbital hybridization is a mathematical procedure used to describe the geometry of molecules. In many cases it is an "after the fact" explanation because often the geometry is already known. The challenge here is to "explain" hybridization without the math, but here goes:

1. The electrons in the "valence" shell of an atom are the only ones that participate in the formation of chemical bonds. They are the electrons with the largest principle quantum number. Inner electrons can "shield" the outer electrons from the positive charge of the nucleus, but they do not participate in bond formation directly.

2. Each valence electron has a "wave function" associated with it. The "wave functions" are solutions to Schroedinger's equation for that electron. In the simpler treatment these "wave functions" are taken to be hydrogen-like -- that is the solution(s) of Schroedinger's equation are taken to be that of a single electron, and the fact that this simplification is not really the case is taken into account by one or more approximations. One of the most common being that the "core" electrons just reduce the positive charge an outer valence electrons "feel". In the case of the carbon atom there are 4 valence electrons whose "wave functions" are written using a shorthand notation: 2S, 2Px, 2Py, 2Pz. The "2", the "P", and the "x,y,z" actually stand for the radial and angular dependence of the solutions. The shorthand notation is used because writing out each expression would be inconvenient, and you only need it if you are going to actually "calculate" something.

3. Naively, one would think that the four electrons of carbon would form one bond that involved the 2S electron, and three bonds that involved the 2Px,y,z electrons. These wave functions are identical except for their angular dependence which reflects their different orientation about the "x,y,z" axes, respectively. Experimentally, however we know that the four bonds are exactly equivalent. Thus CH4 forms 4 identical bonds forming a symmetric tetrahedron, not one bond that is centro-symmetric and three bonds that are 90 degrees apart for one another. OH DEAR!!! WHAT TO DO??.

4. The mathematics comes to the rescue. There is a theorem of the equations, called differential equations, of which the Schroedinger equation is one type, that: If the differential equations has several solutions (many if not most do) then any linear combination of those solutions is also a solution to the differential equation, subject to certain conditions called "boundary conditions". In the present example the "boundary condition" is that all 4 bonds have to be identical, and the molecule (in this case CH4) has no dipole moment so the 4 bonds must be symmetric with respect to rotation. Note here that the mathematical boundary conditions have been set by the experimental results.

5. ImposcuIprtdDFFPPDbTP6cM8scUsg6qking these conditions results in four solutions to the Schroedinger equation that are symmetric with respect to rotation and identical. One can write out the actual solutions, but it is customary to write the solutions in a qualitative shorthand: SP^3 (read that "s p cubed") that is the solution to Schroedinger's equation has the form : F(x,y,z) = [a(2S)+b1(2Px)+b2(2Py)+b3(2Pz)] where the constants "a, b1, b2, b3" can be computed. The allowed values of the constants are such that the four solutions F(x,y,z) are identical except that they form a tetrahedral angle (~ 109 degrees) with respect to the carbon atom which lies at the center of the tetrahedron.

6. Different hybridizations call for different values of "a, b1, b2, and b3" to match the known geometry. You might think that this is a bit self serving, since the geometry of the molecule is specified, and in some ways that is true. However, one can use these wave functions "F(x,y,z)" to predict some other properties of the molecule and the predictions do match the experimental results, so the solutions are self-consistent, even though some experimental information was used in arriving at the proper solution. It turns out that the specification of the geometry a priori is not necessary, but it makes the calculation very much easier.

This is no doubt more about "elephants" than you cared to know; however, your innocent-looking question is really very fundamental. For more detail see: Pauling & Wilson "Intro. to Quantum Mechanics" or L. Pauling "General Chemistry" for more detailed discussion.

Vince Calder

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