Chapter 10: Molecular Geometry and Chemical Bonding Theory 1
chapter 10
Molecular Geometry and Chemical Bonding Theory
Chapter Terms and Definitions
Numbers in parentheses after definitions give the text sections in which the terms are explained. Starred terms are italicized in the text. Where a term does not fall directly under a text section heading, additional information is given for you to locate it.
cis* double-bonded compound with distinctive groups on the same side as the double bond (chapter introduction)
trans* double-bonded compound with distinctive groups on the opposite side as the double bond (chapter introduction)
molecular geometry general shape of a molecule, as determined by the relative positions of the atomic nuclei (10.1, introductory section)
valence-shell electron-pair repulsion (VSEPR) model predicts the shapes of molecules and ions in which valence-shell electron pairs are arranged about each atom so that electron pairs are kept as far away from one another as possible, thus minimizing electron-pair repulsions (10.1)
linear* two points at an angle of 180to one another(10.1)
trigonal planar* three points directed at 120 angles to one another (10.1)
tetrahedral* four points directed at approximately 109.5 angles to one another (10.1)
bent (angular)* two points directed to one another at an angle of less than 180 (10.1)
trigonal pyramidal* four points describing a pyramid with a triangular base (10.1)
trigonal bipyramidal arrangement* five points at the apexes of the six-sided figure formed by placing the face of one tetrahedron onto the face of another (10.1)
octahedral arrangement* six points directed to the apexes of a figure with eight triangular faces and six vertexes (10.1)
axial directions* directions 180 apart forming an axis through the central atom in a trigonal bipyramidal arrangement (10.1)
equatorial directions* directions 120 apart pointing toward the corners of the equilateral triangle lying on the plane through the central atom and perpendicular to the axial directions in a trigonal bipyramidal arrangement (10.1)
seesaw (distorted tetrahedral) geometry* four atoms arranged about a fifth atom in a tetrahedral arrangement but with three different bond angles (10.1)
T-shaped geometry* four atoms lying in one plane with one atom at the intersection of the T (10.1)
square pyramidal geometry* five atoms arranged about a sixth atom so that all bonding angles are 90 (10.1)
square planar geometry* four atoms arranged about a fifth atom so that all atoms lie in a plane and bonding angles are 90 (10.1)
dipole moment quantitative measure of the degree of charge separation in a molecule (10.2)
debyes (D)* units of dipole moment; 1 D = 3.34 10–30 C • m (10.2)
capacitance* capacity of charged plates to hold a charge (10.2)
vector* quantity having both magnitude and direction (10.2)
nonpolar* molecules in which bonds are directed symmetrically about the central atom and, therefore, have a zero dipole moment (10.2)
polar* molecules in which bonds are not symmetrical about the central atom and, therefore, have a nonzero dipole moment (10.2)
valence bond theory approximate theory to explain the electron pair or covalent bond by quantum mechanics (10.3)
overlap* occupation of the same region of space by two orbitals (10.3)
promoted* said of an electron that is raised to a higher orbital level (10.3)
hybrid orbitals orbitals used to describe the bonding in some atoms; obtained by taking combinations of atomic orbitals of the isolated atoms (10.3)
sp3* set of four hybrid orbitals constructed from one s orbital and three p orbitals (10.3)
sp2* set of three hybrid orbitals constructed from one s orbital and two p orbitals (10.3)
sp* set of two hybrid orbitals constructed from one s orbital and one p orbital (10.3)
σ (sigma) bond chemical bond with cylindrical shape about the bond axis; formed when two s orbitals overlap or by overlap along the axis of an orbital with directional character along the bond axis (10.4)
π (pi) bond chemical bond with electron distribution above and below the bond axis; formed by sideways overlap of two parallel p orbitals (10.4)
geometric (cis–trans) isomers* compounds with the same chemical formula and the same bonding but with different arrangements as about a double bond (10.4)
molecular orbital theory explanation of the electron structure of molecules in terms of molecular orbitals, which may spread over several atoms or the entire molecule (10.5, introductory section)
bonding orbitals molecular orbitals that are concentrated in regions between nuclei (10.5)
antibonding orbitals molecular orbitals that are concentrated in regions other than between nuclei (10.5)
bond order* in molecular orbital theory, for a diatomic molecule, one-half the difference between the number of electrons in bonding orbitals nb and the number in antibonding orbitals na:
(nb – na) (10.5)
homonuclear diatomic molecules molecules composed of two like nuclei (10.6)
heteronuclear diatomic molecules molecules composed of two different nuclei (10.6)
nonbonding* describes orbitals, atomic or molecular that are neither bonding nor antibonding (10.7)
photochemical smog* blend of smoke and fog produced when sunlight interacts with a mixture of nitrogen oxides and hydrocarbons in polluted air (A Chemist Looks at:Stratospheric Ozone [An Absorber of Ultraviolet Rays])
stratosphere* region of the atmosphere beginning at about 15 km (9 miles) above the surface of the earth (A Chemist Looks at:Stratospheric Ozone [An Absorber of Ultraviolet Rays])
Chapter Diagnostic Test
1.Match each of the following molecules with its corresponding geometry.
_____ / (1)SeCl2 / (a)square planar_____ / (2)CCl4 / (b)angular
_____ / (3)(CH3) 3 As / (c)linear (possesses zero dipole moment)
_____ / (4)XeF4 / (d)tetrahedral (possesses zero dipole moment)
_____ / (5)CO2 / (e)pyramidal
2.Determine whether each of the following statements is true or false. If a statement is false, change it so that it is true.
a.In valence bond theory, chemical bonding is the result of the overlap of outer-shell atomic orbitals. True/False: ______
______
______
b.In sp2 hybridization, the third p orbital disappears. True/False: ______
______
______
c.When the twisting motion of atoms about a bond is restricted, geometric (cis–trans) isomers become possible for a molecule. True/False: ______
______
______
d.The C2 molecule should be stable and diamagnetic, whereas the Mg2 molecule should be unstable. True/False: ______
______
______
3.In molecular orbital theory, each molecular orbital
a.can accommodate a maximum of two electrons.
b.can accommodate the number of electrons in the valence shell of the atoms.
c.is formed from the atomic orbitals on the less electronegative atom.
d.is concentrated in the region of a specific chemical bond.
e.defines the spectroscopic characteristics of the atoms forming a chemical bond.
4.In the compound C2H2, the expected value for the HCC bond angle is
a.90°.
b.120°.
c.109.5°.
d.180°.
e.none of the above because the molecule should not exist.
5.Draw and name the molecular geometry for each of the following species.
a.ICl2–
b.GeF4
c.SbF52–
d.SF6
e.NH2–
f.BCl3
6.Write the structural formulas showing the geometric isomers possible for the compound with the condensed formula CH3CHCHCH2CH3. (The carbons are joined to each other, and there is a double bond between the second and third carbons.) Also, name the prefix for each isomer.
7.Answer the three questions below given the following information: The electron configuration of sulfur is [Ne] 3s23p4.
a.What type of hybridization or bonding exists in the sulfur hexafluoride molecule?
b.What type of hybridization or bonding exists in the sulfur tetrafluoride molecule?
c.What type of hybridization or bonding would be expected in a SF2 molecule?
8.Describe in your own words just what orbital hybridization is.
Answers to Chapter Diagnostic Test
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer.
1.
(1)b
(2)d
(3)e
(4)a
(5)c (10.1, 10.2; PS Sk. 1, 2)
2.
a.True. (10.3)
b.False. The third p orbital remains in position and can contain electrons. It is also available for bonding with other nonhybridized p orbitals. (10.4)
c.True. (10.4)
d.True. (10.5, 10.6, PS Sk. 4)
3.a (10.5)
4.d (10.3, 10.4, PS Sk. 3)
5.
a. / / or /(linear ion)
b. / / or /
(tetrahedral molecule)
c. / / or /
(square pyramidal ion)
d. / / or /
(octahedral molecule)
e.
(bent ion)
f.
(trigonal planar molecule)
6.
/ and /(cis) / (trans) / (10.4)
7.
a.sp3d2would create six orbitals for bonding.
b.sp3d would be expected, with one lone pair.
c.sp3 might occur, with two lone pairs. (10.3, PS Sk. 3)
8.You are on your own. (10.3)
Summary of Chapter Topics
10.1 The Valence-Shell Electron-Pair Repulsion (VSEPR) Model
Learning Objectives
- Define molecular geometry.
- Define valence-shell electron-pair repulsion model.
- Note the difference between the arrangement of electron pairs about a central atom and molecular geometry.
- Note the four steps in the prediction of geometry by the VSEPR model.
- Predict the molecular geometry (two, three, or four electron pairs). (Example 10.1)
- Note that a lone pair tends to require more space than a corresponding bonding pair and that a multiple bond requires more space than a single bond.
- Predict the molecular geometry (five or six electron pairs). (Example 10.2)
Problem-Solving Skill
1.Predicting molecular geometries. Given the formula of a simple molecule, predict its geometry using the VSEPR model (Examples 10.1 and 10.2).
Molecular geometry and directional bonding are crucial topics because these phenomena are major factors in why molecules react as they do. The VSEPR model gives us a very good way to approach these topics. You must remember, however, that it is just a model. Whether a predicted geometry agrees with experiment depends on many factors. The excellent discussion of molecular-geometry models by R. S. Drago [“A Criticism of the Valence-Shell Electron-Pair Repulsion Model as a Teaching Device,” J. Chem. Educ., 50 (1973), p. 244] points out the utility and limitations of models. A review of this article is very appropriate for Chapter 10.
To be able to predict molecular geometry, you must first write the Lewis structure to determine where the valence electrons are. You then must determine the electron arrangement (geometry). To do this, you must memorize the correlations between the number of electron pairs (or groups) around the central atom and the geometry (2 pairs indicate linear; 3 pairs, trigonal planar; etc.). You will find this information in text Figure 10.3. Now you are ready to place atoms with the electron pairs (or groups) and to determine the atomic arrangement (molecular geometry), as shown in text Figure 10.4. Note that a tetrahedral electron arrangement can lead to three different molecular geometries depending on the number of lone pairs.
You should learn to draw the molecular geometries. There are two basic ways. One is to use different types of lines (e.g., simple, wedged, and dotted) for bonds to indicate three-dimensional structure. In this method, the wedged line, , indicates that the atom is coming out of the plane of the paper. A dotted line means the atom is going behind the plane of the paper. A simple straight line indicates the atom is in the plane of the paper.
The other method is to draw an outline of the geometric figure representing the electron geometry, put the central atom in the center of the figure, and place the outer atoms or electron pairs at the vertexes of the figure. Common figures are
Both methods of drawing molecular geometries are shown in the solutions to Chapter 10 exercises. Always place the lone pairs on the central atom in the drawing.
You also should memorize the angles (in degrees) that go with the particular electron geometries. Linear is 180; trigonal planar, 120; tetrahedral, 109.5; trigonal bipyramidal, 120 and 90; and octahedral, 90. This, then, gives you the expected bond angles.
If you have difficulty visualizing molecular geometry in three dimensions, the stereoscopic plots of molecules found in L. H. Hall’s book should be very helpful (Group Theory and Symmetry in Chemistry [New York: McGraw-Hill Book Co., 1969], pp. 15–31). The textual material in this book is advanced, so you may ignore it. The stereographic drawings, however, are well worth your effort in trying to locate a copy of this text in your library.
Exercise 10.1
Use the VSEPR method to predict the geometry of the following ion and molecules.
a.ClO3–
b.OF2
c.SiF4
Solution:
a.ClO3– electron-dot structure:
Electron arrangement is tetrahedral (text Figure 10.3). Molecular geometry is trigonal pyramidal:
(See text Figure 10.4 under 4 electron pairs, 1 lone pair.)
b.OF2 electron-dot structure:
Electron arrangement is tetrahedral. Molecular geometry is bent:
(See text Figure 10.4 under 4 electron pairs, 2 lone pairs.)
c.SiF4 electron-dot structure:
Electron arrangement is tetrahedral. Molecular geometry is tetrahedral:
(See text Figure 10.4 under 4 electron pairs, 0 lone pairs.)
Exercise 10.2
According to the VSEPR model, what molecular geometry would you predict for iodine trichloride, ICl3?
Solution: Electron-dot structure is as follows:
Electron arrangement is trigonal bipyramidal. Molecular geometry is T-shaped:
/ or /(See text Figure 10.8 under 5 electron pairs, 2 lone pairs.)
10.2 Dipole Moment and Molecular Geometry
Learning Objectives
- Define dipole moment.
- Explain the relationship between dipole moment and molecular geometry. (Example 10.3)
- Note that the polarity of a molecule can affect certain properties, such as boiling point.
Problem-Solving Skill
2.Relating dipole moment and molecular geometry. State what geometries of a molecule AXn are consistent with the information that the molecule has a nonzero dipole moment (Example 10.3).
Exercise 10.3
Bromine trifluoride, BrF3, has a nonzero dipole moment. Indicate which of the following geometries are consistent with this information.
a.Trigonal planar
b.Trigonal pyramidal
c.T-shaped
Known:Dipole moment is nonzero in molecules where there is separation of the centers of + and – charge. This happens when bond polarities do not cancel or where lone electron pairs exist and their effect is not canceled by bond polarity in the opposite direction.
Solution: In a trigonal planar arrangement (a), bond polarities cancel and the dipole moment would be zero. In (b) and (c), bond polarities would not cancel, and depending on the bond polarities in relation to the lone pair or pairs, the molecule could have a dipole moment. Therefore, both (b) and (c) are consistent molecular geometries to give a dipole moment. [Note that the electron geometry of BrF3 is trigonal bipyramidal; therefore, only (c) is a possibility for the molecular geometry of BrF3.]
Exercise 10.4
Which of the following would be expected to have a dipole moment of zero on the basis of symmetry? Explain.
a.SOCl2
b.SiF4
c.OF2
Known: The relationship between molecular geometries in an AXn species and dipole moment (from the text); bond polarities (Section 9.5)
Solution:
a.The geometry of SOCl2 is pyramidal:
On the basis of symmetry, this molecule would be expected to have a nonzero dipole moment.
b.The geometry of SiF4 is tetrahedral:
This geometry gives a dipole moment of zero as long as bonds are of equal polarity, as they are. Therefore, this molecule is expected to have a dipole moment of zero.
c.The geometry of OF2 is bent:
This geometry indicates a nonzero dipole moment because bond polarities do not cancel. However, one must consider bond polarities versus the lone pairs on oxygen and thus must look to laboratory measurements. One would predict a nonzero dipole moment for this molecule, believing the effect of the lone pairs to be greater than the opposing electronegativity of the fluorine atoms.
10.3 Valence Bond Theory
Learning Objectives
- Define valence bond theory.
- State the two conditions needed for bond formation, according to valence bond theory.
- Define hybrid orbitals.
- State the five steps in describing bonding, following the valence bond theory.
- Apply valence bond theory (two, three, or four electron pairs). (Example 10.4)
- Apply valence bond theory (five or six electron pairs). (Example 10.5)
Problem-Solving Skill
3.Applying valence bond theory. Given the formula of a simple molecule, describe its bonding, using valence bond theory (Examples 10.4, 10.5, and 10.6).
The essence of valence bond theory is that electrons in atoms making up molecules remain in orbitals on the individual atoms, that only the valence electrons in a particular atomic orbital are involved in bonding, and that the bonding electrons spend most of their time between the two nuclei, in the space of the orbital overlap.
Many students have difficulty comprehending orbital hybridization. It will be helpful to remember that an orbital represents the volume of space in which there is a high probability of finding an electron. In some molecules, however, spectroscopic studies show that the electrons are not where we would expect them. In short, they are in different volumes of space. We call these new regions hybrid orbitals because we describe them by mathematically combining the valence orbitals of an atom to give new orbitals. (Recall hybrid plants from your biology class.)
Note that according to theory, all valence orbitals do not have to be involved in the hybridization process. When one s and two p orbitals are hybridized to form three sp2 hybrid orbitals, one valence p orbital remains unhybridized. This volume of space will remain perpendicular to the plane of the three sp2 orbitals in their trigonal planar arrangement. (See Figure A.) When one s and one p orbital combine to form two sp hybrid orbitals, two valence p orbitals are left unhybridized. They are at right angles to each other and to the plane of the linear sp hybrid orbitals. (See Figure B.)
Figure A / Figure BA few words need to be said about electron promotion as part of hybridization. As you will note from the text discussion of the bonding in carbon, boron, oxygen, and xenon, promotion is used in the case of carbon, boron, and xenon but not in oxygen. The concept of promotion is used when both filled and unoccupied orbitals are mathematically combined. In the case of oxygen, all orbitals involved in the hybridization have electrons in them, and promotion is not possible.
The following steps summarize how to describe bonding in a given molecule according to the valence bond theory. Do enough exercises and problems so that you can perform the steps in order without looking at the list.
1.Write the Lewis formula for the molecule.
2.Determine the electron arrangement about the central atom (text Figure 10.3).