Molecular Geometry—VSEPR for AP Chemistry

By — McGraw-Hill Professional
Updated on Feb 9, 2011

Practice problems for these concepts can be found at:

The shape of a molecule has quite a bit to do with its reactivity. This is especially true in biochemical processes, where slight changes in shape in three-dimensional space might make a certain molecule inactive or cause an adverse side effect. One way to predict the shape of molecules is the valence-shell electron-pair repulsion (VSEPR) theory. The basic idea behind this theory is that the valence electron pairs surrounding a central atom, whether involved in bonding or not, will try to move as far away from each other as possible to minimize the repulsion between the like charges. Two geometries can be determined; the electron-group geometry, in which all electron pairs surrounding a nucleus are considered, and molecular geometry, in which the nonbonding electrons become "invisible" and only the geometry of the atomic nuclei are considered. For the purposes of geometry, double and triple bonds count the same as single bonds. To determine the geometry:

  1. Write the Lewis electron-dot formula of the compound.
  2. Determine the number of electron-pair groups surrounding the central atom(s). Remember that double and triple bonds are treated as a single group.
  3. Determine the geometric shape that maximizes the distance between the electron groups. This is the geometry of the electron groups.
  4. Mentally allow the nonbonding electrons to become invisible. They are still there and are still repelling the other electron pairs, but we don't "see" them. The molecular geometry is determined by the remaining arrangement of atoms (as determined by the bonding electron groups) around the central atom.

Figure 11.5 shows the electron-group and molecular geometry for two to six electron pairs.

For example, let's determine the electron-group and molecular geometry of carbon dioxide, CO2, and water, H2O. At first glance, one might imagine that the geometry of these two compounds would be similar, since both have a central atom with two groups attached. Let's see if that is true.

First, write the Lewis structure of each. Figure 11.6 shows the Lewis structures of these compounds.

Next, determine the electron-group geometry of each. For carbon dioxide, there are two electron groups around the carbon, so it would be linear. For water, there are four electron pairs around the oxygen—two bonding and two nonbonding electron pairs—so the electron-group geometry would be tetrahedral.

Finally, mentally allow the nonbonding electron pairs to become invisible and describe what is left in terms of the molecular geometry. For carbon dioxide, all groups are involved in bonding so the molecular geometry is also linear. However, water has two nonbonding pairs of electrons so the remaining bonding electron pairs (and hydrogen nuclei) are in a bent arrangement.

Molecular Geometry—VSEPR

Molecular Geometry—VSEPR

This determination of the molecular geometry of carbon dioxide and water also accounts for the fact that carbon dioxide does not possess a dipole and water has one, even though both are composed of polar covalent bonds. Carbon dioxide, because of its linear shape, has partial negative charges at both ends and a partial charge in the middle. To possess a dipole, one end of the molecule must have a positive charge and the other a negative end. Water, because of its bent shape, satisfies this requirement. Carbon dioxide does not.

Practice problems for these concepts can be found at:

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