5 Steps to a 5 AP Chemistry 2019

Bonding ❮ 153

Molecular Geometry—VSEPR

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 VSEPR (valence-shell electron-pair repulsion) 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 geom-
   etry is determined by the remaining arrangement of atoms (as determined by the bonding
   electron groups) around the central atom.
   Figure 11.5, on the next page, 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, CO 2 , and water, H 2 O. 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

KEY IDEA

KEY IDEA

the number of electrons needed, the carbon should be the central atom. We will

work this example using both the incorrect atom arrangement and the correct atom

arrangement. Notice that in both structures all atoms have a complete octet.

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 

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O::N::C:−−O::C::N:

Number of valence electrons 6 5 4 6 4 5

• Number of nonbonding electrons –4 –0 –4 –4 –0 –4


• 1/2 Number of bonding electrons –2 –4 –2 –2 –4 –2
  Formal charges 0 + 1 –2 0 0 –1
  The formal charges make the OCN arrangement the better choice.