CHAPTER 10
Molecular Geometry and Chemical Bonding
Theory
■ SOLUTIONS TO EXERCISES
10.1. a. A Lewis structure of ClO3− is
-
O
O Cl O
There are four electron pairs in a tetrahedral arrangement about the central atom. Three
pairs are bonding, and one pair is nonbonding. The expected geometry is trigonal
pyramidal.
b. The Lewis structure of OF2 is
F O F
There are four electron pairs in a tetrahedral arrangement about the central atom. Two pairs
are bonding, and two pairs are nonbonding. The expected geometry is bent.
c. The Lewis structure of SiF4 is
F F
Si
F F
There are four bonding electron pairs in a tetrahedral arrangement around the central atom.
The expected geometry is tetrahedral.
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, Chapter 10: Molecular Geometry and Chemical Bonding Theory 339
10.2. First, distribute the valence electrons to the bonds and the chlorine atoms. Then distribute the
remaining electrons to iodine.
Cl I Cl
Cl
The five electron pairs around iodine should have a trigonal bipyramidal arrangement with two
lone pairs occupying equatorial positions. The molecule is T-shaped.
10.3. Both trigonal pyramidal (b) and T-shaped (c) geometries are consistent with a nonzero dipole
moment. In trigonal planar geometry, the Br–F contributions to the dipole moment would cancel.
10.4. On the basis of symmetry, SiF4 (b) would be expected to have a dipole moment of zero. The
bonds are all symmetrical about the central atom.
10.5. The Lewis structure for ammonia, NH3, is
H N H
H
There are four pairs of electrons around the nitrogen atom. According to the VSEPR model, these
are arranged tetrahedrally around the nitrogen atom, and you should use sp3 hybrid orbitals. Each
N–H bond is formed by the overlap of a 1s orbital of a hydrogen atom with one of the singly
occupied sp3 hybrid orbitals of the nitrogen atom. This gives the following bonding description
for NH3:
2p
3 3
sp sp
N - H bonds
2s
lone pair
Energy
1s 1s 1s
N atom N atom N atom
(ground state) (hybridized) (in NH3)
10.6. The Lewis structure of CO2 is
O C O
The double bonds are each a bond plus a bond. Hybrid orbitals are needed to describe the two
bonds. This suggests sp hybridization. Each sp hybrid orbital on the carbon atom overlaps with
a 2p orbital on one of the oxygen atoms to form a bond.
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, 340 Chapter 10: Molecular Geometry and Chemical Bonding Theory
The bonds are formed by the overlap of a 2p orbital on the carbon atom with a 2p orbital on one
of the oxygen atoms. Hybridization and bonding of the carbon atom are shown as follows:
C – O bonds
2p
2p 2p
sp sp
2s
bonds
Energy
1s 1s 1s
C atom C atom C atom
(ground state) (hybridized) (in CO2 )
10.7. The structural formulas for the isomers are as follows:
F
N N N N
F F F
cis trans
These compounds exist as separate isomers with different properties. For these to interconvert,
one end of the molecule would have to rotate with respect to the other end. This would require
breaking the bond and expending considerable energy.
10.8. There are 2 6 = 12 electrons in C2. They occupy the orbitals as shown below.
1s * 2s * 2p
1s 2s
The electron configuration is KK(2s)2(2s*)2(2p)4 . There are no unpaired electrons; therefore,
C2 is diamagnetic. There are eight bonding and four antibonding electrons. The bond order is ½(8
− 4) = 2.
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, Chapter 10: Molecular Geometry and Chemical Bonding Theory 341
10.9. There are 6 + 8 = 14 electrons in CO. The orbital diagram is
1s * 2s 2s * 2p 2p
1s
The electron configuration is KK(2s)2(2s*)2(2p)4(2p)2. There are ten bonding and four
antibonding electrons. The bond order is ½(10 − 4) = 3. There are no unpaired electrons; hence,
CO is diamagnetic.
■ ANSWERS TO CONCEPT CHECKS
10.1. The VSEPR model predicts that four electron pairs about any atom in a molecule will distribute
themselves to give a tetrahedral arrangement. Any three of these electron pairs would have a
trigonal pyramidal arrangement. The geometry of a molecule having a central atom with three
atoms bonded to it would be trigonal pyramidal.
10.2. A molecule, AX3, could have one of three geometries: it could be trigonal planar, trigonal
pyramidal, or T-shaped. Assuming the three groups attached to the central atom are alike, as
indicated by the formula, the planar geometry should be symmetrical, so even if the A–X bonds
are polar, their polarities would cancel to give a nonpolar molecule (dipole moment of zero). This
would not be the case in the trigonal pyramidal geometry. In that situation, the bonds all point to
one side of the molecule. It is possible for such a molecule to have a lone pair that points away
from the bonds and whose polarity might fortuitously cancel the bond polarities, but an exact
cancellation is not likely. In general, you should expect the trigonal pyramidal molecule to have a
nonzero dipole moment, but a zero dipole is possible. The argument for the T-shaped geometry is
similar to that for the trigonal pyramidal geometry. The bonds point in a plane, but toward one
side of the molecule. Unless the sum of the bond polarities was fortuitously canceled by the
polarities from the lone pairs, this geometry would have a nonzero dipole moment. This means
that molecule Y is likely to be trigonal planar, but trigonal pyramidal or T-shaped geometries are
possible. Molecule Z cannot have a trigonal planar geometry; it must be either trigonal pyramidal
or T-shaped.
10.3. Assuming there are no lone pairs, the atom has four electron pairs and, therefore, an octet of
electrons about it. The single bond and the triple bond each require a sigma bond orbital for a
total of two such orbitals. This suggests sp hybrids on the central atom.
■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
10.1. The VSEPR model is used to predict the geometry of molecules. The electron pairs around an
atom are assumed to arrange themselves to reduce electron repulsion. The molecular geometry is
determined by the positions of the bonding electron pairs.
10.2. The arrangements are linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral.
10.3. A lone pair is "larger" than a bonding pair; therefore, it will occupy an equatorial position, where
it encounters less repulsion than if it were in an axial position.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Molecular Geometry and Chemical Bonding
Theory
■ SOLUTIONS TO EXERCISES
10.1. a. A Lewis structure of ClO3− is
-
O
O Cl O
There are four electron pairs in a tetrahedral arrangement about the central atom. Three
pairs are bonding, and one pair is nonbonding. The expected geometry is trigonal
pyramidal.
b. The Lewis structure of OF2 is
F O F
There are four electron pairs in a tetrahedral arrangement about the central atom. Two pairs
are bonding, and two pairs are nonbonding. The expected geometry is bent.
c. The Lewis structure of SiF4 is
F F
Si
F F
There are four bonding electron pairs in a tetrahedral arrangement around the central atom.
The expected geometry is tetrahedral.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 10: Molecular Geometry and Chemical Bonding Theory 339
10.2. First, distribute the valence electrons to the bonds and the chlorine atoms. Then distribute the
remaining electrons to iodine.
Cl I Cl
Cl
The five electron pairs around iodine should have a trigonal bipyramidal arrangement with two
lone pairs occupying equatorial positions. The molecule is T-shaped.
10.3. Both trigonal pyramidal (b) and T-shaped (c) geometries are consistent with a nonzero dipole
moment. In trigonal planar geometry, the Br–F contributions to the dipole moment would cancel.
10.4. On the basis of symmetry, SiF4 (b) would be expected to have a dipole moment of zero. The
bonds are all symmetrical about the central atom.
10.5. The Lewis structure for ammonia, NH3, is
H N H
H
There are four pairs of electrons around the nitrogen atom. According to the VSEPR model, these
are arranged tetrahedrally around the nitrogen atom, and you should use sp3 hybrid orbitals. Each
N–H bond is formed by the overlap of a 1s orbital of a hydrogen atom with one of the singly
occupied sp3 hybrid orbitals of the nitrogen atom. This gives the following bonding description
for NH3:
2p
3 3
sp sp
N - H bonds
2s
lone pair
Energy
1s 1s 1s
N atom N atom N atom
(ground state) (hybridized) (in NH3)
10.6. The Lewis structure of CO2 is
O C O
The double bonds are each a bond plus a bond. Hybrid orbitals are needed to describe the two
bonds. This suggests sp hybridization. Each sp hybrid orbital on the carbon atom overlaps with
a 2p orbital on one of the oxygen atoms to form a bond.
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, 340 Chapter 10: Molecular Geometry and Chemical Bonding Theory
The bonds are formed by the overlap of a 2p orbital on the carbon atom with a 2p orbital on one
of the oxygen atoms. Hybridization and bonding of the carbon atom are shown as follows:
C – O bonds
2p
2p 2p
sp sp
2s
bonds
Energy
1s 1s 1s
C atom C atom C atom
(ground state) (hybridized) (in CO2 )
10.7. The structural formulas for the isomers are as follows:
F
N N N N
F F F
cis trans
These compounds exist as separate isomers with different properties. For these to interconvert,
one end of the molecule would have to rotate with respect to the other end. This would require
breaking the bond and expending considerable energy.
10.8. There are 2 6 = 12 electrons in C2. They occupy the orbitals as shown below.
1s * 2s * 2p
1s 2s
The electron configuration is KK(2s)2(2s*)2(2p)4 . There are no unpaired electrons; therefore,
C2 is diamagnetic. There are eight bonding and four antibonding electrons. The bond order is ½(8
− 4) = 2.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 10: Molecular Geometry and Chemical Bonding Theory 341
10.9. There are 6 + 8 = 14 electrons in CO. The orbital diagram is
1s * 2s 2s * 2p 2p
1s
The electron configuration is KK(2s)2(2s*)2(2p)4(2p)2. There are ten bonding and four
antibonding electrons. The bond order is ½(10 − 4) = 3. There are no unpaired electrons; hence,
CO is diamagnetic.
■ ANSWERS TO CONCEPT CHECKS
10.1. The VSEPR model predicts that four electron pairs about any atom in a molecule will distribute
themselves to give a tetrahedral arrangement. Any three of these electron pairs would have a
trigonal pyramidal arrangement. The geometry of a molecule having a central atom with three
atoms bonded to it would be trigonal pyramidal.
10.2. A molecule, AX3, could have one of three geometries: it could be trigonal planar, trigonal
pyramidal, or T-shaped. Assuming the three groups attached to the central atom are alike, as
indicated by the formula, the planar geometry should be symmetrical, so even if the A–X bonds
are polar, their polarities would cancel to give a nonpolar molecule (dipole moment of zero). This
would not be the case in the trigonal pyramidal geometry. In that situation, the bonds all point to
one side of the molecule. It is possible for such a molecule to have a lone pair that points away
from the bonds and whose polarity might fortuitously cancel the bond polarities, but an exact
cancellation is not likely. In general, you should expect the trigonal pyramidal molecule to have a
nonzero dipole moment, but a zero dipole is possible. The argument for the T-shaped geometry is
similar to that for the trigonal pyramidal geometry. The bonds point in a plane, but toward one
side of the molecule. Unless the sum of the bond polarities was fortuitously canceled by the
polarities from the lone pairs, this geometry would have a nonzero dipole moment. This means
that molecule Y is likely to be trigonal planar, but trigonal pyramidal or T-shaped geometries are
possible. Molecule Z cannot have a trigonal planar geometry; it must be either trigonal pyramidal
or T-shaped.
10.3. Assuming there are no lone pairs, the atom has four electron pairs and, therefore, an octet of
electrons about it. The single bond and the triple bond each require a sigma bond orbital for a
total of two such orbitals. This suggests sp hybrids on the central atom.
■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
10.1. The VSEPR model is used to predict the geometry of molecules. The electron pairs around an
atom are assumed to arrange themselves to reduce electron repulsion. The molecular geometry is
determined by the positions of the bonding electron pairs.
10.2. The arrangements are linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral.
10.3. A lone pair is "larger" than a bonding pair; therefore, it will occupy an equatorial position, where
it encounters less repulsion than if it were in an axial position.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
