10.3: VSPER Theory: The Effect of Lone Pairs

Learning Objectives

• To use the VSEPR model to predict molecular geometries.
• To predict whether a molecule has a dipole moment.

The Lewis electron-pair approach can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. We continue our discussion of structure and bonding by introducing the valence-shell electron-pair repulsion (VSEPR) model (pronounced “vesper”), which can be used to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds.

The VSEPR Model

The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The premise of the VSEPR theory is that electron pairs located in bonds and lone pairs repel each other and will therefore adopt the geometry that places electron pairs as far apart from each other as possible. This theory is very simplistic and does not account for the subtleties of orbital interactions that influence molecular shapes; however, the simple VSEPR counting procedure accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach.

We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing only on the number of electron pairs around the central atom, ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form groups, which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in Figures $$\PageIndex{1}$$ and $$\PageIndex{2}$$.

In the VSEPR model, the molecule or polyatomic ion is given an AXmEn designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles. Using this information, we can describe the molecular geometry, the arrangement of the bonded atoms in a molecule or polyatomic ion.

VESPR Produce to predict Molecular geometry

This VESPR procedure is summarized as follows:

1. Draw the Lewis electron structure of the molecule or polyatomic ion.
2. Determine the electron group arrangement around the central atom that minimizes repulsions.
3. Assign an AXmEn designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations from ideal bond angles.
4. Describe the molecular geometry.

We will illustrate the use of this procedure with several examples, beginning with atoms with two electron groups. In our discussion we will refer to Figure $$\PageIndex{2}$$ and Figure $$\PageIndex{3}$$, which summarize the common molecular geometries and idealized bond angles of molecules and ions with two to six electron groups.

Two Electron Groups

Our first example is a molecule with two bonded atoms and no lone pairs of electrons, $$BeH_2$$.

AX2 Molecules: BeH2

1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is

3. Both groups around the central atom are bonding pairs (BP). Thus BeH2 is designated as AX2.

4. From Figure $$\PageIndex{3}$$ we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH2 is linear.

AX2 Molecules: CO2

1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is

2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH2, the arrangement that minimizes repulsions places the groups 180° apart.

3. Once again, both groups around the central atom are bonding pairs (BP), so CO2 is designated as AX2.

4. VSEPR only recognizes groups around the central atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO2 is linear (Figure $$\PageIndex{3}$$). The structure of $$\ce{CO2}$$ is shown in Figure $$\PageIndex{1}$$.

Three Electron Groups

AX3 Molecules: BCl3

1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is

2. There are three electron groups around the central atom. To minimize repulsions, the groups are placed 120° apart (Figure $$\PageIndex{2}$$).

3. All electron groups are bonding pairs (BP), so the structure is designated as AX3.

4. From Figure $$\PageIndex{3}$$ we see that with three bonding pairs around the central atom, the molecular geometry of BCl3 is trigonal planar, as shown in Figure $$\PageIndex{2}$$.

AX3 Molecules: CO32−

1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as

2. The structure of CO32− is a resonance hybrid. It has three identical bonds, each with a bond order of $$1 \frac{1}{3}$$. We minimize repulsions by placing the three groups 120° apart (Figure $$\PageIndex{2}$$).

3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX3.

4. We see from Figure $$\PageIndex{3}$$ that the molecular geometry of CO32− is trigonal planar with bond angles of 120°.

In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time.

AX2E Molecules: SO2

1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below.

2. There are three electron groups around the central atom, two double bonds and one lone pair. We initially place the groups in a trigonal planar arrangement to minimize repulsions (Figure $$\PageIndex{2}$$).

3. There are two bonding pairs and one lone pair, so the structure is designated as AX2E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure $$\PageIndex{4}$$). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO2, we have one BP–BP interaction and two LP–BP interactions.

4. The molecular geometry is described only by the positions of the nuclei, not by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is bent, or V shaped, which can be viewed as a trigonal planar arrangement with a missing vertex (Figures $$\PageIndex{2}$$ and $$\PageIndex{3}$$). The O-S-O bond angle is expected to be less than 120° because of the extra space taken up by the lone pair.

As with SO2, this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom.

Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH2O (AX3), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°).

Four Electron Groups

One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions.

AX4 Molecules: CH4

1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is

2. There are four electron groups around the central atom. As shown in Figure $$\PageIndex{2}$$, repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°.

3. All electron groups are bonding pairs, so the structure is designated as AX4.

4. With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure $$\PageIndex{3}$$).

AX3E Molecules: NH3

1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure

2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.

3. With three bonding pairs and one lone pair, the structure is designated as AX3E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.

4. There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal. In essence, this is a tetrahedron with a vertex missing (Figure $$\PageIndex{3}$$). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions (Figure $$\PageIndex{3}$$ and Figure $$\PageIndex{4}$$).

AX2E2 Molecules: H2O

1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure

2. There are four groups around the central oxygen atom, two bonding pairs and two lone pairs. Repulsions are minimized by directing the bonding pairs and the lone pairs to the corners of a tetrahedron Figure $$\PageIndex{2}$$.

3. With two bonding pairs and two lone pairs, the structure is designated as AX2E2 with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles.

4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is bent, or V shaped, with an H–O–H angle that is even less than the H–N–H angles in NH3, as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one. This molecular shape is essentially a tetrahedron with two missing vertices.

Five Electron Groups

In previous examples it did not matter where we placed the electron groups because all positions were equivalent. In some cases, however, the positions are not equivalent. We encounter this situation for the first time with five electron groups.

AX5 Molecules: PCl5

1. Phosphorus has five valence electrons and each chlorine has seven valence electrons, so the Lewis electron structure of PCl5 is

2. There are five bonding groups around phosphorus, the central atom. The structure that minimizes repulsions is a trigonal bipyramid, which consists of two trigonal pyramids that share a base (Figure $$\PageIndex{2}$$):

3. All electron groups are bonding pairs, so the structure is designated as AX5. There are no lone pair interactions.

4. The molecular geometry of PCl5 is trigonal bipyramidal, as shown in Figure $$\PageIndex{3}$$. The molecule has three atoms in a plane in equatorial positions and two atoms above and below the plane in axial positions. The three equatorial positions are separated by 120° from one another, and the two axial positions are at 90° to the equatorial plane. The axial and equatorial positions are not chemically equivalent, as we will see in our next example.

AX4E Molecules: SF4

1. The sulfur atom has six valence electrons and each fluorine has seven valence electrons, so the Lewis electron structure is

With an expanded valence, this species is an exception to the octet rule.

2. There are five groups around sulfur, four bonding pairs and one lone pair. With five electron groups, the lowest energy arrangement is a trigonal bipyramid, as shown in Figure $$\PageIndex{2}$$.

3. We designate SF4 as AX4E; it has a total of five electron pairs. However, because the axial and equatorial positions are not chemically equivalent, where do we place the lone pair? If we place the lone pair in the axial position, we have three LP–BP repulsions at 90°. If we place it in the equatorial position, we have two 90° LP–BP repulsions at 90°. With fewer 90° LP–BP repulsions, we can predict that the structure with the lone pair of electrons in the equatorial position is more stable than the one with the lone pair in the axial position. We also expect a deviation from ideal geometry because a lone pair of electrons occupies more space than a bonding pair.

At 90°, the two electron pairs share a relatively large region of space, which leads to strong repulsive electron–electron interactions.

4. With four nuclei and one lone pair of electrons, the molecular structure is based on a trigonal bipyramid with a missing equatorial vertex; it is described as a seesaw. The Faxial–S–Faxial angle is 173° rather than 180° because of the lone pair of electrons in the equatorial plane.

AX3E2 Molecules: BrF3

1. The bromine atom has seven valence electrons, and each fluorine has seven valence electrons, so the Lewis electron structure is

Once again, we have a compound that is an exception to the octet rule.

2. There are five groups around the central atom, three bonding pairs and two lone pairs. We again direct the groups toward the vertices of a trigonal bipyramid.

3. With three bonding pairs and two lone pairs, the structural designation is AX3E2 with a total of five electron pairs. Because the axial and equatorial positions are not equivalent, we must decide how to arrange the groups to minimize repulsions. If we place both lone pairs in the axial positions, we have six LP–BP repulsions at 90°. If both are in the equatorial positions, we have four LP–BP repulsions at 90°. If one lone pair is axial and the other equatorial, we have one LP–LP repulsion at 90° and three LP–BP repulsions at 90°:

Structure (c) can be eliminated because it has a LP–LP interaction at 90°. Structure (b), with fewer LP–BP repulsions at 90° than (a), is lower in energy. However, we predict a deviation in bond angles because of the presence of the two lone pairs of electrons.

4. The three nuclei in BrF3 determine its molecular structure, which is described as T shaped. This is essentially a trigonal bipyramid that is missing two equatorial vertices. The Faxial–Br–Faxial angle is 172°, less than 180° because of LP–BP repulsions (Figure $$\PageIndex{2}$$.1).

Because lone pairs occupy more space around the central atom than bonding pairs, electrostatic repulsions are more important for lone pairs than for bonding pairs.

AX2E3 Molecules: I3

1. Each iodine atom contributes seven electrons and the negative charge one, so the Lewis electron structure is

2. There are five electron groups about the central atom in I3, two bonding pairs and three lone pairs. To minimize repulsions, the groups are directed to the corners of a trigonal bipyramid.

3. With two bonding pairs and three lone pairs, I3 has a total of five electron pairs and is designated as AX2E3. We must now decide how to arrange the lone pairs of electrons in a trigonal bipyramid in a way that minimizes repulsions. Placing them in the axial positions eliminates 90° LP–LP repulsions and minimizes the number of 90° LP–BP repulsions.

The three lone pairs of electrons have equivalent interactions with the three iodine atoms, so we do not expect any deviations in bonding angles.

4. With three nuclei and three lone pairs of electrons, the molecular geometry of I3 is linear. This can be described as a trigonal bipyramid with three equatorial vertices missing. The ion has an I–I–I angle of 180°, as expected.

Six Electron Groups

Six electron groups form an octahedron, a polyhedron made of identical equilateral triangles and six identical vertices (Figure $$\PageIndex{2}$$.)

AX6 Molecules: SF6

1. The central atom, sulfur, contributes six valence electrons, and each fluorine atom has seven valence electrons, so the Lewis electron structure is

With an expanded valence, this species is an exception to the octet rule.

2. There are six electron groups around the central atom, each a bonding pair. We see from Figure $$\PageIndex{2}$$ that the geometry that minimizes repulsions is octahedral.

3. With only bonding pairs, SF6 is designated as AX6. All positions are chemically equivalent, so all electronic interactions are equivalent.

4. There are six nuclei, so the molecular geometry of SF6 is octahedral.

AX5E Molecules: BrF5

1. The central atom, bromine, has seven valence electrons, as does each fluorine, so the Lewis electron structure is

With its expanded valence, this species is an exception to the octet rule.

2. There are six electron groups around the Br, five bonding pairs and one lone pair. Placing five F atoms around Br while minimizing BP–BP and LP–BP repulsions gives the following structure:

3. With five bonding pairs and one lone pair, BrF5 is designated as AX5E; it has a total of six electron pairs. The BrF5 structure has four fluorine atoms in a plane in an equatorial position and one fluorine atom and the lone pair of electrons in the axial positions. We expect all Faxial–Br–Fequatorial angles to be less than 90° because of the lone pair of electrons, which occupies more space than the bonding electron pairs.

4. With five nuclei surrounding the central atom, the molecular structure is based on an octahedron with a vertex missing. This molecular structure is square pyramidal. The Faxial–B–Fequatorial angles are 85.1°, less than 90° because of LP–BP repulsions.

AX4E2 Molecules: ICl4

1. The central atom, iodine, contributes seven electrons. Each chlorine contributes seven, and there is a single negative charge. The Lewis electron structure is

2. There are six electron groups around the central atom, four bonding pairs and two lone pairs. The structure that minimizes LP–LP, LP–BP, and BP–BP repulsions is

3. ICl4 is designated as AX4E2 and has a total of six electron pairs. Although there are lone pairs of electrons, with four bonding electron pairs in the equatorial plane and the lone pairs of electrons in the axial positions, all LP–BP repulsions are the same. Therefore, we do not expect any deviation in the Cl–I–Cl bond angles.

4. With five nuclei, the ICl4− ion forms a molecular structure that is square planar, an octahedron with two opposite vertices missing.

The relationship between the number of electron groups around a central atom, the number of lone pairs of electrons, and the molecular geometry is summarized in Figure $$\PageIndex{6}$$.

Figure $$\PageIndex{6}$$: Overview of Molecular Geometries

Example $$\PageIndex{1}$$

Using the VSEPR model, predict the molecular geometry of each molecule or ion.

1. PF5 (phosphorus pentafluoride, a catalyst used in certain organic reactions)
2. H3O+ (hydronium ion)

Given: two chemical species

Strategy:

1. Draw the Lewis electron structure of the molecule or polyatomic ion.
2. Determine the electron group arrangement around the central atom that minimizes repulsions.
3. Assign an AXmEn designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations in bond angles.
4. Describe the molecular geometry.

Solution:

1. A The central atom, P, has five valence electrons and each fluorine has seven valence electrons, so the Lewis structure of PF5 is

B There are five bonding groups about phosphorus. The structure that minimizes repulsions is a trigonal bipyramid (Figure $$\PageIndex{6}$$).

C All electron groups are bonding pairs, so PF5 is designated as AX5. Notice that this gives a total of five electron pairs. With no lone pair repulsions, we do not expect any bond angles to deviate from the ideal.

D The PF5 molecule has five nuclei and no lone pairs of electrons, so its molecular geometry is trigonal bipyramidal.

2. A The central atom, O, has six valence electrons, and each H atom contributes one valence electron. Subtracting one electron for the positive charge gives a total of eight valence electrons, so the Lewis electron structure is

B There are four electron groups around oxygen, three bonding pairs and one lone pair. Like NH3, repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.

C With three bonding pairs and one lone pair, the structure is designated as AX3E and has a total of four electron pairs (three X and one E). We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.

D There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal, in essence a tetrahedron missing a vertex. However, the H–O–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions:

Exercise $$\PageIndex{1}$$

Using the VSEPR model, predict the molecular geometry of each molecule or ion.

1. XeO3
2. PF6
3. NO2+

trigonal pyramidal

octahedral

linear

Example $$\PageIndex{2}$$

Predict the molecular geometry of each molecule.

1. XeF2
2. SnCl2

Given: two chemical compounds

Strategy:

Use the strategy given in Example$$\PageIndex{1}$$.

Solution:

1. A Xenon contributes eight electrons and each fluorine seven valence electrons, so the Lewis electron structure is

B There are five electron groups around the central atom, two bonding pairs and three lone pairs. Repulsions are minimized by placing the groups in the corners of a trigonal bipyramid.

C From B, XeF2 is designated as AX2E3 and has a total of five electron pairs (two X and three E). With three lone pairs about the central atom, we can arrange the two F atoms in three possible ways: both F atoms can be axial, one can be axial and one equatorial, or both can be equatorial:

The structure with the lowest energy is the one that minimizes LP–LP repulsions. Both (b) and (c) have two 90° LP–LP interactions, whereas structure (a) has none. Thus both F atoms are in the axial positions, like the two iodine atoms around the central iodine in I3. All LP–BP interactions are equivalent, so we do not expect a deviation from an ideal 180° in the F–Xe–F bond angle.

D With two nuclei about the central atom, the molecular geometry of XeF2 is linear. It is a trigonal bipyramid with three missing equatorial vertices.

2. A The tin atom donates 4 valence electrons and each chlorine atom donates 7 valence electrons. With 18 valence electrons, the Lewis electron structure is

B There are three electron groups around the central atom, two bonding groups and one lone pair of electrons. To minimize repulsions the three groups are initially placed at 120° angles from each other.

C From B we designate SnCl2 as AX2E. It has a total of three electron pairs, two X and one E. Because the lone pair of electrons occupies more space than the bonding pairs, we expect a decrease in the Cl–Sn–Cl bond angle due to increased LP–BP repulsions.

D With two nuclei around the central atom and one lone pair of electrons, the molecular geometry of SnCl2 is bent, like SO2, but with a Cl–Sn–Cl bond angle of 95°. The molecular geometry can be described as a trigonal planar arrangement with one vertex missing.

Exercise $$\PageIndex{2}$$

Predict the molecular geometry of each molecule.

1. SO3
2. XeF4

trigonal planar

square planar

Molecules with No Single Central Atom

The VSEPR model can be used to predict the structure of somewhat more complex molecules with no single central atom by treating them as linked AXmEn fragments. We will demonstrate with methyl isocyanate (CH3–N=C=O), a volatile and highly toxic molecule that is used to produce the pesticide Sevin. In 1984, large quantities of Sevin were accidentally released in Bhopal, India, when water leaked into storage tanks. The resulting highly exothermic reaction caused a rapid increase in pressure that ruptured the tanks, releasing large amounts of methyl isocyanate that killed approximately 3800 people and wholly or partially disabled about 50,000 others. In addition, there was significant damage to livestock and crops.

We can treat methyl isocyanate as linked AXmEn fragments beginning with the carbon atom at the left, which is connected to three H atoms and one N atom by single bonds. The four bonds around carbon mean that it must be surrounded by four bonding electron pairs in a configuration similar to AX4. We can therefore predict the CH3–N portion of the molecule to be roughly tetrahedral, similar to methane:

The nitrogen atom is connected to one carbon by a single bond and to the other carbon by a double bond, producing a total of three bonds, C–N=C. For nitrogen to have an octet of electrons, it must also have a lone pair:

Because multiple bonds are not shown in the VSEPR model, the nitrogen is effectively surrounded by three electron pairs. Thus according to the VSEPR model, the C–N=C fragment should be bent with an angle less than 120°.

The carbon in the –N=C=O fragment is doubly bonded to both nitrogen and oxygen, which in the VSEPR model gives carbon a total of two electron pairs. The N=C=O angle should therefore be 180°, or linear. The three fragments combine to give the following structure:

We predict that all four nonhydrogen atoms lie in a single plane, with a C–N–C angle of approximately 120°. The experimentally determined structure of methyl isocyanate confirms our prediction (Figure $$\PageIndex{7}$$).

Certain patterns are seen in the structures of moderately complex molecules. For example, carbon atoms with four bonds (such as the carbon on the left in methyl isocyanate) are generally tetrahedral. Similarly, the carbon atom on the right has two double bonds that are similar to those in CO2, so its geometry, like that of CO2, is linear. Recognizing similarities to simpler molecules will help you predict the molecular geometries of more complex molecules.

Example $$\PageIndex{3}$$

Use the VSEPR model to predict the molecular geometry of propyne (H3C–C≡CH), a gas with some anesthetic properties.

Given: chemical compound

Strategy:

Count the number of electron groups around each carbon, recognizing that in the VSEPR model, a multiple bond counts as a single group. Use Figure $$\PageIndex{3}$$ to determine the molecular geometry around each carbon atom and then deduce the structure of the molecule as a whole.

Solution:

Because the carbon atom on the left is bonded to four other atoms, we know that it is approximately tetrahedral. The next two carbon atoms share a triple bond, and each has an additional single bond. Because a multiple bond is counted as a single bond in the VSEPR model, each carbon atom behaves as if it had two electron groups. This means that both of these carbons are linear, with C–C≡C and C≡C–H angles of 180°.

Exercise $$\PageIndex{3}$$

Predict the geometry of allene (H2C=C=CH2), a compound with narcotic properties that is used to make more complex organic molecules.

The terminal carbon atoms are trigonal planar, the central carbon is linear, and the C–C–C angle is 180°.

Molecular Dipole Moments

You previously learned how to calculate the dipole moments of simple diatomic molecules. In more complex molecules with polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine whether there is a net dipole moment. Mathematically, dipole moments are vectors; they possess both a magnitude and a direction. The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO2, a linear molecule (Figure $$\PageIndex{8a}$$). Each C–O bond in CO2 is polar, yet experiments show that the CO2 molecule has no dipole moment. Because the two C–O bond dipoles in CO2 are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO2 molecule has no net dipole moment even though it has a substantial separation of charge. In contrast, the H2O molecule is not linear (Figure $$\PageIndex{8b}$$); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H2O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H2O to hydrogen-bond to other polarized or charged species, including other water molecules.

Other examples of molecules with polar bonds are shown in Figure $$\PageIndex{9}$$. In molecular geometries that are highly symmetrical (most notably tetrahedral and square planar, trigonal bipyramidal, and octahedral), individual bond dipole moments completely cancel, and there is no net dipole moment. Although a molecule like CHCl3 is best described as tetrahedral, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments cannot cancel one another, and the molecule has a dipole moment. Due to the arrangement of the bonds in molecules that have V-shaped, trigonal pyramidal, seesaw, T-shaped, and square pyramidal geometries, the bond dipole moments cannot cancel one another. Consequently, molecules with these geometries always have a nonzero dipole moment.

Molecules with asymmetrical charge distributions have a net dipole moment.

Example $$\PageIndex{4}$$

Which molecule(s) has a net dipole moment?

1. $$\ce{H2S}$$
2. $$\ce{NHF2}$$
3. $$\ce{BF3}$$

Given: three chemical compounds

Strategy:

For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment.

Solution:

1. The total number of electrons around the central atom, S, is eight, which gives four electron pairs. Two of these electron pairs are bonding pairs and two are lone pairs, so the molecular geometry of $$\ce{H2S}$$ is bent (Figure $$\PageIndex{6}$$). The bond dipoles cannot cancel one another, so the molecule has a net dipole moment.

2. Difluoroamine has a trigonal pyramidal molecular geometry. Because there is one hydrogen and two fluorines, and because of the lone pair of electrons on nitrogen, the molecule is not symmetrical, and the bond dipoles of NHF2 cannot cancel one another. This means that NHF2 has a net dipole moment. We expect polarization from the two fluorine atoms, the most electronegative atoms in the periodic table, to have a greater affect on the net dipole moment than polarization from the lone pair of electrons on nitrogen.

3. The molecular geometry of BF3 is trigonal planar. Because all the B–F bonds are equal and the molecule is highly symmetrical, the dipoles cancel one another in three-dimensional space. Thus BF3 has a net dipole moment of zero:

Exercise $$\PageIndex{4}$$

Which molecule(s) has a net dipole moment?

• $$\ce{CH3Cl}$$
• $$\ce{SO3}$$
• $$\ce{XeO3}$$
$$\ce{CH3Cl}$$ and $$\ce{XeO3}$$