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Hybridization

What is hybridization?

Organic chemists knew that methane (CH4) had a tetrahedral geometry and that all the C-H bonds were the same length and the angle between each H-C-H was 109.5o.  Since all the C-H bonds and angles are the same, the electronic configuration of the C atom is inconsistent with this geometry.

Recall from general chemistry that we can describe the ground state and valence shell electronic configurations by using Aufbau's principle in which the orbitals (i.e. 1s, 2s, 2px, 2py, 2pz, etc.) are filled from the bottom up (i.e. lowest energy first), Pauli exclusion principle (i.e. electrons are paired up in orbitals with opposite spin) and Hund's rule of maximum multiplicity (i.e. degenerate orbitals are given 1 electron until all are half filled then we pair them up).  You should have learned this in high school chemistry and then again in general chemistry as a freshman.

If you do this for C you'd get the valence shell configuration as shown below on the left.  You'll notice there are only two unpaired electrons (red) in the valence shell configuration.  THERE IS NO WAY FOR IT TO FORM FOUR BONDS!  Something is wrong!  We could add energy to the atom and bump one of the 2s electrons into the empty 2p orbital to give an atom with 4 unpaired electrons (excited state below).  The energy required would easily be compensated for by the extra bond formed.  However, in the excited state one electron is lower in energy than the other three.  If this atom were to form a methane molecule one C-H bond would be much shorter than the other three.  AGAIN THIS IS INCONSISTENT WITH THE TRUE STRUCTURE!  However, if the 2s and 2p(x,y,z) orbitals could mix (i.e. hybridize) we could potentially make four (4) sp3 hybrid orbitals all of equal energy.  This would give a methane structure that is consistent with the true structure of methane.  Hybridization occurs because the resulting atom can form 4 bonds which is much more stable than forming only two 2 bonds.  Note:  The ground state configuration as you learned in general chem only applies to bare "naked" atoms with nothing around them.  It's the presence of the H atoms that makes the orbitals hybridize.

Why bother learning hybridization?

Ask any 2nd-semester organic chemistry student and they will tell you the importance of understanding and determining hybridization.  There will be many instances when you will need to rely on your knowledge of hybridization to rationalize or determine some property of a molecule or reaction.  Some examples include the following.

  1. The strength of C-H bonds depends on the hybridization of the C. (i.e. bond strength follows sp C-H > sp2 C-H > sp3 C-H)
  2. Geometry is determined by hybridization (sp3 tetrahedral or trigonal pyramidal, sp2 trigonal planar, sp linear).
  3. The acidity of hydrogens on carbon is rationalized by considering hybridization. (Acidity follows sp C-H > sp2 C-H > sp3 C-H)
  4. The stability of substituted alkenes depends on hybridization.

How to determine hybridization?

To determine the hybridization you need to be able to count to at least 5.  You also need to know the difference between sigma (σ) and pi (π) bonds.  If there is one bond between two atoms it's a σ bond.  If there are two bonds, one is a σ and one is a π.  If there are three bonds, one is a σ and two are π bonds.  Now each second-row element can only have (1) s orbitals and (3) p orbitals (i.e. px, py, and pz).  This is four orbitals, and if each can hold two electrons then the maximum is 8 electrons total (i.e. the octet rule).  Atoms in rows beyond the second row can potentially have (1) s, (3) p's, and (5) d's, and we have what's called the 18 electron guideline.  So you need to count the total number of lone pairs and sigma bonds about the atom of interest and then simply count up through the available orbitals.  Let us try some simple examples.

Example 1

The carbon atoms in methane and other saturated hydrocarbons are sp3 hybrid.  This means that there are 4 sp3 hybrid orbitals on the C atom.  Each one is directed toward the hydrogen atoms which reside is a 1s orbital.  Using valence bond theory we would say that the σ bonds are composed of an overlapping C sp3 hybrid orbital and a H 1s orbital.  Remember sigma (σ) bonds come from s-type orbitals while pi (π) bonds come from p-type orbitals.

Example 2

The carbon atoms in ethene and other unsaturated hydrocarbons are sp2 hybrid.  This means that there are 3 sp2 hybrid orbitals on the C atom.  Each one is directed toward the hydrogen atoms and the other C.  Using valence bond theory we would say that the C-H σ bonds are composed of an overlapping C sp2 hybrid orbital and a H 1s orbital.  The remaining p orbital on each carbon forms the π bond.

More Examples