http://www.ck12.org Chapter 9. Covalent Bonding
Now, four bonds are possible. The promotion of the electron “costs” a small amount of energy, but recall that the
process of bond formation is accompanied by a decrease in energy. The two extra bonds that can now be formed
result in a lower overall energy, and, thus, a greater stability, for the CH 4 molecule. Carbon normally forms four
bonds in most of its compounds.
The number of bonds is now correct, but the geometry is wrong. The threeporbitals (px,py,pz) are oriented at 90°
angles relative to one another. However, as we saw in our discussion of VSEPR theory, the observed H−C−H bond
angle in the tetrahedral CH 4 molecule is actually 109.5°. Therefore, the methane molecule cannot be adequately
represented by simple overlap of the 2sand 2porbitals of carbon with the 1sorbitals of each hydrogen atom.
To explain the bonding in methane, it is necessary to introduce the concept of hybridization and hybrid atomic
orbitals.Hybridizationis the mixing of the atomic orbitals in an atom to produce a set of hybrid orbitals. When
hybridization occurs, it must do so as a result of the mixing of nonequivalent orbitals. For example,sandporbitals
can hybridize, butporbitals cannot hybridize with otherporbitals.Hybrid orbitalsare the atomic orbitals obtained
when two or more nonequivalent orbitals from the same atom combine in preparation for bond formation. In the
current case of carbon, the single 2sorbital hybridizes with the three 2porbitals to form a set of four hybrid orbitals,
calledsp^3 hybrids.
Thesp^3 hybrids are all equivalent to one another. Spatially, the hybrid orbitals point toward the four corners of a
tetrahedron (Figure9.38).
The large lobe from each of thesp^3 hybrid orbitals then overlaps with normal unhybridized 1sorbitals on each
hydrogen atom to form the tetrahedral methane molecule.
Another example ofsp^3 hybridization occurs in the ammonia (NH 3 ) molecule. The electron domain geometry
of ammonia is also tetrahedral, meaning that there are four groups of electrons around the central nitrogen atom.
Unlike methane, however, one of those electron groups is a lone pair. The resulting molecular geometry is trigonal
pyramidal. Just as in the carbon atom, the hybridization process starts as a promotion of a 2selectron to a 2porbital,
followed by hybridization to form a set of foursp^3 hybrids. In this case, one of the hybrid orbitals already contains a
pair of electrons, leaving only three half-filled orbitals available to form covalent bonds with three hydrogen atoms.
The methane and ammonia examples illustrate the connection between orbital hybridization and the VSEPR model.
The electron domain geometry predicted by VSEPR leads directly to the type of hybrid orbitals that must be formed
to accommodate that geometry. Both methane and ammonia have tetrahedral electron domain geometries and, thus,
both undergosp^3 hybridization.
Likewise, the bent-shaped water molecule (H 2 O) also involves the formation ofsp^3 hybrids on the oxygen atom. In
this case, however, there are two hybrid orbitals which have lone pairs and two which bond to the hydrogen atoms.
Because the electron domain geometry for H 2 O is tetrahedral, the hybridization issp^3.