The geometry of a molecule includes a
description of the arrangements of the atoms in the molecule. At a
simple level, the molecular structure tells us which atoms are
bonded to which. At a more detailed level, the geometry includes
the lengths of all of these bonds, that is, the distances between
the atoms which are bonded together, and the angles between pairs
of bonds. For example, we find that in water,
H2OH2O,
the two hydrogens are bonded to the oxygen and each O-H bond length
is 95.72pm (where
1pm=10-12m1pm10-12m).
Furthermore,
H2OH2O
is a bent molecule, with the H-O-H angle equal to 104.5°.
(The measurement of these geometric properties is difficult,
involving the measurement of the frequencies at which the molecule
rotates in the gas phase. In molecules in crystalline form, the
geometry of the molecule is revealed by irradiating the crystal
with x-rays and analyzing the patterns formed as the x-rays
diffract off of the crystal.)

Not all triatomic molecules are bent, however.
As a common example,
CO2CO2
is a linear molecule. Larger polyatomics can have a variety of
shapes, as illustrated in Figure 1.
Ammonia,
NH3NH3,
is a pyramid-shaped molecule, with the hydrogens in an equilateral
triangle, the nitrogen above the plane of this triangle, and a
H-N-H angle equal to 107°. The geometry of
CH4CH4
is that of a tetrahedron, with all H-C-H angles equal to
109.5°. (See also Figure 2(a).)
Ethane,
C2H6C2H6,
has a geometry related to that of methane. The two carbons are
bonded together, and each is bonded to three hydrogens. Each H-C-H
angle is 109.5° and each H-C-C angle is
109.5°. By contrast, in ethene,
C2H4C2H4,
each H-C-H bond angle is 116.6° and each H-C-C bond angle is
121.7°. All six atoms of ethene lie in the same plane. Thus,
ethene and ethane have very different geometries, despite the
similarities in their molecular formulae.

We begin our analysis of these geometries by
noting that, in the molecules listed above which do
*not* contain double or triple bonds
(H2OH2O,
NH3NH3,
CH4CH4and
C2H6C2H6),
the bond angles are very similar, each equal to or very close to
the tetrahedral angle 109.5°. To account for the observed
angle, we begin with our valence shell electron pair sharing model,
and we note that, in the Lewis structures of these molecules, the
central atom in each bond angle of these molecules contains four pairs
of valence shell electrons. For methane and ethane, these four
electron pairs are all shared with adjacent bonded atoms, whereas
in

ammonia or water, one or two (respectively) of
the electron pairs are not shared with any other atom. These
unshared electron pairs are called
lone pairs . Notice that, in the two molecules with no
lone pairs, all bond angles are
*exactly* equal to the tetrahedral angle, whereas
the bond angles are only close in the molecules with lone
pairs

One way to understand this result is based on
the mutual repulsion of the negative charges on the valence shell
electrons. Although the two electrons in each bonding pair must
remain relatively close together in order to form the bond,
different pairs of electrons should arrange themselves in such a
way that the distances between the pairs are as large as possible.
Focusing for the moment on methane, the four pairs of electrons
must be equivalent to one another, since the four C-H bonds are
equivalent, so we can assume that the electron pairs are all the
same distance from the central carbon atom. How can we position
four electron pairs at a fixed distance from the central atom but
as far apart from one another as possible? A little reflection
reveals that this question is equivalent to asking how to place
four points on the surface of a sphere spread out from each other
as far apart as possible. A bit of experimentation reveals that
these four points must sit at the corners of a tetrahedron, an
equilateral triangular pyramid, as may be seen in Figure 2(b). If the carbon atom is at the
center of this tetrahedron and the four electron pairs at placed at
the corners, then the hydrogen atoms also form a tetrahedron about
the carbon. This is, as illustrated in Figure 2(a), the correct geometry of a methane
molecule. The angle formed by any two corners of a tetrahedron and
the central atom is 109.5°, exactly in agreement with the
observed angle in methane. This model also works well in predicting
the bond angles in ethane.

We conclude that molecular geometry is
determined by minimizing the mutual repulsion of the valence shell
electron pairs. As such, this model of molecular geometry is often
referred to as the
valence shell electron pair repulsion (VSEPR) theory .
For reasons that will become clear, extension of this model implies
that a better name is the
Electron Domain (ED) Theory .

This model also accounts, at least
approximately, for the bond angles of
H2OH2O
and
NH3NH3.
These molecules are clearly not tetrahedral, like
CH4CH4,
since neither contains the requisite five atoms to form the
tetrahedron. However, each molecule does contain a central atom
surrounded by four pairs of valence shell electrons. We expect from
our Electron Domain model that those four pairs should be arrayed
in a tetrahedron, without regard to whether they are bonding or
lone-pair electrons. Then attaching the hydrogens (two for oxygen,
three for nitrogen) produces a prediction of bond angles of
109.5°, very close indeed to the observed angles of
104.5° in
H2OH2O
and 107° in
NH3NH3.

Note, however, that we do not describe the
geometries of
H2OH2O
and
NH3NH3
as "tetrahedral," since the
*atoms* of the molecules do not form
tetrahedrons, even if the valence shell electron pairs do. (It is
worth noting that these angles are not exactly equal to
109.5°, as in methane. These deviations will be discussed
later.)

We have developed the Electron Domain model to
this point only for geometries of molecules with four pairs of
valence shell electrons. However, there are a great variety of
molecules in which atoms from Period 3 and beyond can have more
than an octet of valence electrons. We consider two such molecules
illustrated in Figure 3.

First,
PCl5PCl5
is a stable gaseous compound in which the five chlorine atoms are
each bonded to the phosphorous atom. Experiments reveal that the
geometry of
PCl5PCl5
is that of a
trigonal bipyramid : three of the chlorine atoms form
an equilateral triangle with the P atom in the center, and the
other two chlorine atoms are on top of and below the P atom. Thus
there must be 10 valence shell electrons around the phosphorous
atom. Hence, phosphorous exhibits what is called an
expanded valence in
PCl5PCl5.
Applying our Electron Domain model, we expect the five valence
shell electron pairs to spread out optimally to minimize their
repulsions. The required geometry can again be found by trying to
place five points on the surface of a sphere with maximum distances
amongst these points. A little experimentation reveals that this
can be achieved by placing the five points to form a trigonal
bipyramid. Hence, Electron Domain theory accounts for the geometry
of
PCl5PCl5.

Second,
SF6SF6 is
a fairly unreactive gaseous compound in which all six fluorine
atoms are bonded to the central sulfur atom. Again, it is clear
that the octet rule is violated by the sulfur atom, which must
therefore have an expanded valence. The observed geometry of
SF6SF6,
as shown in Figure 3, is highly
symmetric: all bond lengths are identical and all bond angles are
90°. The F atoms form an
octahedron about the central S atom: four of the F
atoms form a square with the S atom at the center, and the other
two F atoms are above and below the S atom. To apply our Electron
Domain model to understand this geometry, we must place six points,
representing the six electron pairs about the central S atom, on
the surface of a sphere with maximum distances between the points.
The requisite geometry is found, in fact, to be that of an
octahedron, in agreement with the observed geometry.

As an example of a molecule with an atom with
less than an octet of valence shell electrons, we consider boron
trichloride,
BCl3BCl3.
The geometry of
BCl3BCl3 is
also given in Figure 3: it is
trigonal planar , with all four atoms lying in the same
plane, and all Cl-B-Cl bond angles equal to 120°. The three Cl
atoms form an equilateral triangle. The Boron atom has only three
pairs of valence shell electrons in
BCl3BCl3.
In applying Electron Domain theory to understand this geometry, we
must place three points on the surface of a sphere with maximum
distance between points. We find that the three points form an
equilateral triangle in a plane with the center of the sphere, so
Electron Domain is again in accord with the observed
geometry.

We conclude from these predictions and
observations that the Electron Domain model is a reasonably
accurate way to understand molecular geometries, even in molecules
which violate the octet rule.

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