Chapter 9
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
Chapter 9
Bonding and Molecular Structure:
Orbital Hybridization and Molecular Orbitals
INSTRUCTOR’S NOTES
This chapter is a natural complement to the previous chapter’s introduction to bonding and related subjects.
The emphasis in this chapter is on valence bond theory, with only about one and a half lectures on molecular orbital
theory. Although MO theory is not always included in general chemistry, its wide application in organic chemistry
argues for at least minimally addressing the topic. The consistent answers for molecular geometry which arise from
VSEPR and valence bond theory is a good point to make.
There are many molecular modeling programs that can be used to build molecular models and hybrid and molecular
orbitals such as the CACheTM software mentioned in the previous chapter’s notes.
About three or four lectures are scheduled for this chapter.
SUGGESTED DEMONSTRATIONS
1.
Orbital Overlap

For an overhead demonstration of orbital overlap, see Rothchild, R. “Efficiency of Orbital Overlap: Visual
Demonstration,” Journal of Chemical Education 1981, 58, 757.
2.
Hybrid Orbitals

In these lectures it is most useful to show models of molecules with the hybrid orbitals in place. Suitable
models can be obtained from Aldrich Chemical Company.

Emerson, D. W. “A Colorful Demonstration to Simulate Orbital Hybridization,” Journal of Chemical
Education 1988, 65, 454.

Samoshin, V. V. “Orbital Models Made of Plastic Soda Bottles,” Journal of Chemical Education 1998, 75,
985.
3.
Molecular Orbital Theory

Shakhashiri, B. Z.; Dirreen, G. E.; Williams, L. G.; Smith, S. R. “Paramagnetism and Color of Liquid
Oxygen: A Lecture Demonstration,” Journal of Chemical Education 1980, 57, 373.

Saban, G. H.; Moran, T. F. “A Simple Demonstration of O 2 Paramagnetism. A Macroscopically Observable
Difference Between VB and MO Approaches to Bonding Theory,” Journal of Chemical Education 1973,
50, 217.

174
See the ChemistryNow website for a demonstration of the paramagnetism of liquid oxygen.
Chapter 9
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
SOLUTIONS TO STUDY QUESTIONS
Cl
9.1
Cl
C
Cl
H
The electron-pair and molecular geometries are tetrahedral. The C atom is sp3 hybridized. Three of these
hybrid orbitals each overlap with a chlorine 3p orbital to form three C—Cl sigma bonds. One hybrid
orbital overlaps with a hydrogen 1s orbital to from a C—H sigma bond.
F
9.2
F
N
F
The electron-pair geometry is tetrahedral and the molecular geometry is trigonal pyramidal. The N atom is
sp3 hybridized. Three of these hybrid orbitals each overlap a fluorine 2p orbital to form three N—F sigma
bonds.
9.3
The nitrogen atom is sp3 hybridized. The bond is formed by overlap of a nitrogen sp3 orbital and an oxygen
sp3 orbital.
9.4
The nitrogen atoms are sp3 hybridized. The bond between the nitrogens is formed by overlap of sp3 hybrid
orbitals.
9.5
The electron-pair geometry and molecular geometry are both trigonal planar. The hydridization of the
carbon atom is sp2. The two bonds are made by overlap of carbon sp2 and oxygen sp2 for the sigma bond
and carbon 2p and oxygen 2p overlap for the pi bond.
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Chapter 9
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
9.6
The electron-pair geometry and molecular geometry are trigonal planar at the central carbon; the central
carbon is sp2 hybridized. The sigma bond is formed between a carbon sp2 orbital and an oxygen sp2 orbital.
The pi bond is formed between a carbon 2p orbital and an oxygen 2p orbital.
9.7
9.8
9.9
(a) BBr3
trigonal planar
trigonal planar
sp2
(b) CO2
linear
linear
sp
(c) CH2Cl2
tetrahedral
tetrahedral
sp3
(d) CO32–
trigonal planar
trigonal planar
sp2
(a) CSe2
linear
linear
sp
(b) SO2
trigonal planar
bent
sp2
(c) CH2O
trigonal planar
trigonal planar
sp2
(d) NH4+
tetrahedral
tetrahedral
sp3
(a) C: sp3
O: sp3
(b) From left to right: sp3, sp2, sp2
(c) N: sp3
9.10
CH2: sp3
underlined atom
(a)
(b)
hybrid orbital set
both N atoms are sp3 hybridized
N
C
sp2
C of CH3
sp3
C of C=C and C=O
(c)
sp2
C of CN
sp
F
F
Si
F
electron-pair geometry
molecular geometry
hybridization
octahedral
octahedral
sp3d 2
2–
F
176
both C atoms are sp2 hybridized
C of C=C
9.11
(a)
CO2H: sp2
F
F
Chapter 9
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
F
F
(b)
Se
trigonal bipyramidal
see-saw
sp3d
trigonal bipyramidal
linear
sp3d
octahedral
square planar
sp3d 2
electron-pair geometry
molecular geometry
hybridization
octahedral
square pyramid
sp3d 2
trigonal bipyramidal
trigonal bipyramidal
sp3d
octahedral
square pyramid
sp3d 2
trigonal bipyramidal
linear
sp3d
F
F
–
(c)
Cl
I
Cl
F
F
Xe
(d)
F
F
9.12
O
F
F
Xe
(a)
F
F
F
F
O
S
(b)
F
F
F
F
F
Br
(c)
F
F
–
Br
Br
Br
(d)
F
9.13
H
–
F
P
O
P
O
O
O
F
F
tetrahedral
tetrahedral
sp3
sp3
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Chapter 9
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
F
9.14
H
S
O
S
O
O
tetrahedral
9.15
O
O
O
sp
–
F
tetrahedral
3
sp3
The C atom is sp2 hybridized. Two of the sp2 hybrid orbitals are used to form C—Cl  bonds. The third is
used to form the C—O  bond. The p orbital not used in the C atom hybrid orbitals is used to form the CO
 bond.
9.16
The C atom is sp2 hybridized. Each carbon uses two sp2 orbitals to form C-C sigma bonds with a sp2 orbital
of each adjacent carbon atom and a sp2 orbital to form a C-H sigma bond with a 1s orbital of a hydrogen
atom. Each carbon uses a 2p orbital to form a pi bond with another carbon 2p orbital.
9.17
The electron-pair geometry is tetrahedral, and molecular geometry is trigonal-pyramidal. The hybridization
of the sulfur is sp3.
9.18
The electron-pair geometry and molecular geometry are tetrahedral. The hybridization of the sulfur is sp3.
H
9.19
(a)
Cl
H
C
(b)
C
H3C
C
cis isomer
9.20
(a)
H3C
CH2CH3
cis isomer
(b) cis and trans isomers not possible
178
H
(c)
C
CH3
trans isomer
H
C
C
H
CH3
H
H
CH2OH
C
Cl
C
H
trans isomer
Chapter 9
9.21
9.22
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
H2+:
(1s)1
Bond order = 1/2(1 – 0) = 1/2
weaker H—H bond
H 2:
(1s)2
Bond order = 1/2(2 – 0) = 1
stronger H—H bond
Li2+:
(1s)2(*1s)2(2s)1(*2s)0
Bond order = 1/2(3 – 2) = 1/2
Li2–
(1s)2(*1s)2(2s)2(*2s)1
Bond order = 1/2(4 – 3) = 1/2
Li2
(1s)2(*1s)2(2s)2(*2s)0
Bond order = 1/2(4 – 2) = 1
The bond order of Li2 is greater than that of either of its ions.
9.23
2p
π*2p
2p
2p
*2s
2s

 


The C22– ion has a bond order of 1/2(8 – 2) = 3 (one  bond and two  bonds). The C2 molecule has two
fewer electrons and a bond order of 1/2(6 – 2) = 2. The C22– ion is diamagnetic.
9.24
2p
π*2p
2p
2p
*2s
2s
1and 1 ½  bonds
2 ½ bonds total
Bond order is increased ½.
Yes, it is paramagnetic.
9.25
(1s)2(*1s)2(2s)2(*2s)2 (2p)2(2p)4(2p)4
(a) diamagnetic vs. paramagnetic O2
(b) 1  bond vs. 1  and 1  in O2
(c) 1 vs. 2 in O2
(d) longer bond than in O2
179
Chapter 9
9.26
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
(1s)2(*1s)2(2s)2(*2s)2 (2p)2(2p)4(2p)3
(a) paramagnetic but 1 unpaired e- vs. 2 in O2
(b) 1  and ½  vs. 1  and 1  in O2
(c) 1 1/2 vs. 2 in O2
(d) longer bond than in O2
9.27
N2 (triple) < O2 (double) < C2 (double) < B2 (single) < Li2 (single)
9.28
(a) CN-, CO, NO+, C22(b) O2-, O2, NO
(c) O2-, NO
9.29
(a) ClO has 13 valence electrons. [core electrons](s)2(*s)2(p)4(p)2(p)3
(b) The HOMO is p
(c) diamagnetic
(d) Bond order = 1/2(8 – 5) = 3/2
9.30
One  bond and 0.5  bonds
(a) The NO+ ion has an even number of valence electrons (10 electrons) and so is predicted to be
diamagnetic.
(b) NO+: [core electrons](2s)2(*2s)2(2p)4(2p)2
The HOMO is 2p
(c) Bond order = 1/2(8 – 2) = 3
(d) The bond order of NO is 2.5, whereas that of NO+ is 3. Therefore, NO+ has a stronger bond than NO.
–
F
9.31
F
Al
F
F
The electron-pair and molecular geometries are tetrahedral. The Al atom is sp3 hybridized. Each of these
orbitals overlaps a fluorine 2p orbital to form four Al—F sigma bonds. Each F atom has a formal charge of
zero and the Al atom carries a –1 formal charge. This is not a reasonable charge distribution because the
less electronegative element, Al, carries a negative formal charge.
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Chapter 9
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
F
9.32
F
Cl
F
The electron-pair geometry is trigonal bipyramidal, and the molecular geometry is T-shaped. The Cl atom
is sp3d hybridized. Three of these hybrid orbitals each overlap a fluorine 2p orbital to form three Cl—F
sigma bonds.
9.33
(a) SO2
120º
sp2
(b) SO3
120º
sp2
(c) SO32–
109º
sp3
(d) SO42–
109º
sp3
SO2 and SO3 have the same bond angle and the S atom in each uses the same hybrid orbitals. SO 32– and
SO42– have the same bond angle and the S atom in each uses the same hybrid orbitals.
9.35
–
+
9.34
F
Cl
F
F
Cl
F
electron-pair geometry
tetrahedral
trigonal bipyramidal
molecular geometry
bent
linear
Cl atom hybridization
sp3
sp3d
CFS angle
~109.5o
180o
O
N
O
–
O
N
O
–
electron-pair geometry: trigonal planar
molecular geometry: bent
O—N—O bond angle = 120º
Average bond order = 3/2
N atom hybridization: sp2
–
O
9.36
O
N
O
–
O
O
N
O
–
O
O
N
O
The N atom hybridization is the same in each structure (sp2). The three sp2 hybrid orbitals are used to form
N—O  bonds. The p orbital not used in the N atom hybrid orbitals is used to form the NO  bond.
181
Chapter 9
9.37
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
N
N
O
N
N
O
N
N
O
In each structure the central N atom is sp hybridized. The other N atom hybridization changes from sp to
sp2 to sp3. The two sp hybrid orbitals on the central N atom are used to form N—N and N—O  bonds.
The two p orbitals not used in the N atom hybridization are used to form NN and NO  bonds.
9.38
9.39
O—C—O bond angle
CO bond order
C atom hybridization
CO2
180º
2
sp
CO32–
120º
4
sp2
/3
(a) All three molecules have the same molecular formula, C2H4O. They are isomers.
(b) Both carbon atoms in ethylene oxide are sp3 hybridized. The CH3 carbon in acetaldehyde is sp3
hybridized and the C=O carbon is sp2 hybridized. Both carbon atoms in vinyl alcohol are sp2
hybridized.
(c)
H—C—H angle
ethylene oxide
109º
acetaldehyde
109º
vinyl alcohol
120º
(d) All three molecules are polar.
(e) Vinyl alcohol has the strongest carbon-carbon bond, and acetaldehyde has the strongest carbon-oxygen
bond.
9.40
(a) carbon 1: sp2
carbon 2: sp2
(b) angle A = 120º; angle B = 120º; angle C = 120º
(c) No, cis-trans isomerism is not possible
9.41
(a) CH3 carbon atom: sp3
C=N carbon atom: sp2
N atom: sp2
(b) C—N—O angle = 120º
9.42
(a) angle A = 120º; angle B = 109º; angle C = 109º; angle D = 120º
(b) carbon 1: sp2; carbon 2: sp2; carbon 3: sp3
9.43
(a) C(1) = sp2; O(2) = sp3; N(3) = sp3; C(4) = sp3; P(5) = sp3
(b) angle A = 120º; angle B = 109º; angle C = 109º; angle D = 109º
(c) The P—O and O—H bonds are most polar ( = 1.3)
9.44
(a) 1  bond and 11  bonds.
(b) C(1) = sp3, C(2) = sp2, O(3) = sp3
(c) The C=O bond is the shortest and strongest CO bond.
(d) angle A = 109º, angle B = 109º, angle C = 120º
182
Chapter 9
9.45
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
(a) The C=O bond is the most polar bond in the molecule.
(b) There are 18  bonds and 5  bonds in the molecule.
(c) The trans isomer is shown. The cis isomer is
H
H
H
C
H
C
C
C
H
C
C
C
C
H
C
O
H
H
(d) All carbon atoms in the molecule are sp2 hybridized.
(e) All three angles are 120º.
–
O
3–
O
O
I
9.46
O
O
O
O
O
O
electron-pair geometry
tetrahedral
trigonal bipyramidal
molecular geometry
tetrahedral
trigonal bipyramidal
hybridization of I atom
9.47
I
sp
3
sp3d
(a) sp3d in SbF5, sp3d 2 in SbF6–
(b)
H
F
H
+
The geometry of H2F+ is bent, and the F atom is sp3 hybridized.
O
Xe
9.48
O
O
O
Xe
O
O
O
Electron-pair geometry
tetrahedral
tetrahedral
Molecular geometry
trigonal pyramidal
tetrahedral
3
Xe
sp hybridized
sp3 hybridized
2-–
9.49
(a)
O
O
bond order = 1
(b) [core electrons](2s)2(*2s)2(2p)4(2p)2(π*2p)4
bond order = 1/2(8 – 6) = 1
(c) Yes, the two bonding theories lead to the same magnetic character (diamagnetic) and bond order.
183
Chapter 9
9.50
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
N2 [core electrons](2s)2(*2s)2(2p)4(2p)2
N2+ [core electrons](2s)2(*2s)2(2p)4(2p)1
N2– [core electrons](2s)2(*2s)2(2p)4(2p)2(π*2p)1
N2
N2+
N 2–
(a)
diamagnetic
paramagnetic
paramagnetic
(b)
2  bonds
2  bonds
1 1⁄2  bonds
3
2 1/2
2 1/2
(c) bond order
N2 < N2+ ≈ N2–
(d)
—increasing bond length
N2+ ≈ N2– < N2
(e)
—increasing bond strength
9.51
B2 and O2 are paramagnetic, Li2, B2, and F2 have a bond order of 1, C2 and O2 have a bond order of 2, and
N2 has the highest bond order, 3.
9.52
Molecule or ion
HOMO
paramagnetic
π*
–
(b) OF
diamagnetic
π*
(c) O22–
diamagnetic
π*
(d) Ne2+
paramagnetic
*2p
(e) CN
paramagnetic
(a) NO
9.53
Magnetic behavior
CN
2
2
2p
4
[core electrons](2s) (*2s) (2p) (2p)
1
(a) The HOMO is 2p
(b) Bond order = 1/2(7 – 2) = 2 1/2
(c) One-half net  bond and two net  bonds
(d) paramagnetic
9.54
(a) C6 ring carbon atoms: sp2; side chain carbon atoms: sp3; N atom: sp3
(b) angle A = 120º; angle B = 109º; angle C = 109º
(c) 23  bonds and 3  bonds
(d) The molecule is polar
(e) The H+ ion attaches to the most electronegative atom in the molecule, N, and this is confirmed by the
electrostatic potential map (most negative region near N).
9.55
(a) All of the C atoms are sp3 hybridized
(b) C—O—H angle = 109º
(c) The molecule is polar
(d) The six-member ring is non-planar. The ring could only be planar if the carbon atoms were sp2
hybridized as in benzene.
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Chapter 9
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
F
F
B
9.56
F
F
H
H
C
C is sp2 hybridized, H—C—C = 120º
C
H
H
B is sp2 hybridized, F—B—B = 120º
B
H
H
N
N is sp3 hybridized, H—N—N = 109º
N
H
H
H
O
O is sp3 hybridized, H—O—O = 109º
O
H
9.57
(a) The geometry about the boron atom is trigonal planar in BF3, tetrahedral in H3N—BF3.
(b) Boron is sp2 hybridized in BF3, sp3 hybridized in H3N—BF3.
(c) Yes
(d) The ammonia molecule is polar with the N atom partially negative. While the BF 3 molecule is
nonpolar overall, each of the B–F bonds is polarized such that the B has a partial positive charge. The
partially negative N in NH3 is attracted to the partially positive B in BF3.
(e) One of the lone pairs on the oxygen in H2O can form a coordinate covalent bond with the B in BF3.
9.58
(a) Even though the atoms are sp3 hybridized, the bond angles in the three-member ring must be 60º.
(b) sp3
(c) Because the angles are significantly less than the expected value of 109º, the ring structure is strained
and relatively easy to break. (The molecule is reactive.)
(d) Yes, the molecule is polar. The negative partial charge is near oxygen and the hydrogen atoms carry a
partial positive charge.
9.59
(a) NH2–: electron pair geometry,
SO3: electron pair geometry, trigonal planar;
tetrahedral; molecular geometry, bent; sp3
molecular geometry trigonal planar; sp2
O
N
H
S
H
O
O
185
Chapter 9
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
–
O
H
(b)
H
S
O
The angles around N and S are approximately 109º.
N
O
(c) The hybridization of the N atom does not change (sp3 in NH2– and H2N—SO3–). The S atom
hybridization changes from sp2 in SO3 to sp3 in H2N—SO3–.
(d) SO3 is an electron pair acceptor. The S atom carries a positive partial charge so it is predicted to act as
an electron-acceptor.
9.60
(a) The keto and enol forms are not resonance structures because both electron pairs and atoms have been
rearranged.
(b) The terminal —CH3 carbon atoms are sp3 hybridized and the three central carbon atoms are sp2
hybridized in the enol form. In the keto form, the terminal —CH3 carbon atoms and the central C
atom are sp3 hybridized and the two C=O carbon atoms are sp2 hybridized.
(c) Enol form:
Keto form:
—CH3 carbon atoms
tetrahedral
central three carbon atoms
trigonal planar
—CH3 carbon atoms
tetrahedral
central —CH2— carbon atom
tetrahedral
C=O carbon atoms
trigonal planar
Only the center carbon atom changes geometry, from trigonal planar to tetrahedral
–
H
(d)
CH3
C
C
O
C
CH3
O
–
H
CH3
C
C
C
O
CH3
O
–
H
CH3
C
O
C
C
CH3
O
(e) Possible for the enol form
(f) Yes. The negative partial charges are on the oxygen atoms. The positive partial charges are on the
hydrogen atoms, especially the O—H hydrogen.
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Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
9.61
Hybridization of the carboxylate C is sp2.
C-C bond: overlap of sp2 hybrid orbital on the carbon of the carboxylate with sp3 hybrid orbital on the
methyl carbon.
C-O single bond: The Lewis structure by itself would make it appear that this bond is formed by the
overlap of an sp2 hybrid orbital on carbon with an sp3 hybrid orbital on the oxygen. In the resonance
hybrid, however, the oxygen would also have sp2 hybridization.
9.62
(a) The central atoms are sp hybridized in all four molecules/ions. The most stable resonance structures are
shown below (i.e, those that minimize formal charges and place negative charges on the most
electronegative atom).
For the C of CO2 and the central nitrogen of N3-, both  bonds are made from overlap of an sp hybrid
orbital on the central atom with an sp2 hybrid orbital from an outer atom. Both  bonds are made from
overlap of a 2p orbital on the central atom with a 2p orbital on an outer atom. For the C of OCN- and the
central nitrogen of N2O, the  bond comprising the single bond is made from an sp hybrid orbital on the
central atom and an sp3 hybrid orbital on the outer atom. The  bond of the triple bond is made from an sp
hybrid orbital on the central atom and a sp hybrid orbital on the outer atom. The  bonds of the triple
bonds are made from overlap of 2p orbitals on the central atom and 2p orbitals of the outer atom.
(b) For each of the four molecules are ions, three resonance structures can be drawn – one with two double
bonds, one with a triple bond on the left and a single bond on the right, and one with a triple bond on the
right and a single bond on the left. The structure that minimizes formal charges, places opposite formal
charges on adjacent atoms, places like formal charges as far apart as possible, and places a negative formal
charge on the most electronegative atom or a positive formal charge on the least electronegative atom is the
resonance structure that should contribute the most to the actual structure. For CO 2, the formal charge on
each atom in the structure in part (a) is zero; the bond lengths are expected to be equivalent and to be
typical of a carbon-oxygen double bond. For N3-, the formal charge on each outer atom is -1 and on the
central atom is +1in the structure in part (a); the bond lengths are expected to be equivalent and to be
typical of a nitrogen-nitrogen double bond. For OCN-, the formal charges are oxygen, -1; carbon, 0; and
nitrogen, 0. The carbon-oxygen bond should be close to single bond length, and the carbon-nitrogen bond
should be close to triple bond length. For N2O, the central nitrogen has a formal charge of +1, the outer
187
Chapter 9
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
nitrogen has a formal charge of 0, and the oxygen has a formal charge of -1. The nitrogen-nitrogen bond
should approach triple bond length, while the N-O bond should approach a single bond length.
9.63
MO theory pictures one net  bond for each S-O linkage plus a contribution from  bonding. The 
bonding in this molecule will be similar to that in O3 discussed in the text. There will be two electrons in a
 bonding MO and two electrons in a  non-bonding MO. This gives an overall bond order of 1 for the 
bonding in the entire molecule and there a net  bond order of 0.5 for each S-O linkage. The total bond
order for each S-O linkage is therefore 1.5 (1 from  bonding and 0.5 from  bonding).
9.64
The electronic geometry at the nitrogen atoms is trigonal planar, and the molecular geometry is bent. The
nitrogen atoms are sp2 hybridized. The N=N double bond is comprised of a  bond (made from the overlap
of sp2 hybrid orbits on the nitrogen atoms) and a p bond (made from overlap of 2p orbitals on the nitrogen
atoms).
9.65
The maximum number of hybrid orbitals that a carbon atom can form is four and the minimum number that
can be formed is two. Carbon has only four valence orbitals, one s and three p, so it cannot form more than
four hybrid orbitals.
9.66
(a) CF4 is isoelectronic with BF4– (32 valence electrons)
(b) SiF4 (32 valence electrons) and SF4 (34 valence electrons) are not isoelectronic
(c) BF4–: sp3
9.67
SiF4: sp3
SF4: sp3d
(a) C atom: sp2; N atom: sp3
(b) Another resonance structure (showing only the peptide linkage and the formal charges on O and N) is
This resonance structure is less important owing to the separation of charges.
(c)
O
C
–1
+1
N
H
The fact that the amide link is planar indicates that the resonance structure shown above has some
importance. The highly electronegative oxygen (C=O) and nitrogen (–NH2 group) atoms carry partial
negative charges and the –O–H and amide N–H hydrogen atoms carry partial positive charges
188
Chapter 9
9.68
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
The higher the bond order, the shorter the bond length and the larger the bond energy. Acetylene has a
carbon-carbon triple bond (bond order = 3), so it has the shortest and strongest carbon-carbon bond. Ethane
has a carbon-carbon single bond (bond order = 1), so it has the longest and weakest carbon-carbon bond.
9.69
Molecular orbital theory correctly predicts the electronic structures for odd-electron molecules and other
molecules such as O2 that do not follow the electron-pairing assumptions of the Lewis dot structure
approach.
9.70
Valence bond theory uses resonance to explain the bond order of 1.5 in O 3. Molecular orbital theory uses
three  molecular orbitals (bonding, nonbonding, and antibonding) combined with sigma bonding and
antibonding molecular orbitals to explain the bond order.
9.71
Orbital C < Orbital B < Orbital A
9.72
(a) The number of hybrid orbitals is always equal to the number of atomic orbitals used.
(b) No. All hybrid orbital sets involve an s orbital.
(c) The energy of the hybrid orbital set is the weighted average of the energy of the combining atomic
orbitals.
9.73
B surrounded by three electron pairs: electron-pair geometry and molecular geometry are both trigonal
planar; the hybridization is sp2; formal charge is zero.
B surrounded by four electron pairs: electron-pair geometry and molecular geometry are both tetrahedral;
the hybridization is sp3; the formal charge is -1.
9.74
(a) two parallel p orbitals on the first C and the central carbon overlap to form a pi bond. The last C must
be rotated so that the p orbitals for it and the central C align to form another pi bond.
(b) C(1) sp2; C(2) sp; C(3) sp2
(c) The overlaps of C1 and C3 atoms’ sp2 orbitals with the C2 atom’s sp orbitals form the sigma bonds.
The overlap of the carbon’s 2p orbitals forms the pi bonds.
9.75
HF has a bond order of 1. HF2- has a bond order of 1 for the entire three-center-four-electron bond and
thus a 0.5 bond order per H-F linkage. It should therefore be easier to break the bond in HF 2-, leading to a
smaller bond enthalpy for HF2- than for HF.
9.76
(a) Hybridization is sp3d. The Xe-F bonds are formed by overlap of an sp3d hybrid orbit on Xe with an sp3
hybrid orbital on F. The lone pairs on Xe are in sp3d hybrid orbitals.
(b) Along the z-axis, bonding, non-bonding, and antibonding orbitals can be constructed from combinations
of the 5pz orbital on the Xe and the two 2pz orbitals on the fluorines. Four electrons are placed in the
bonding and nonbonding orbitals.
189
Chapter 9
9.77
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
(a) Br 0.4569 g/(79.904 g/mol) = 0.005718 mol Br
P 0.5431 g F/(18.9984 g/mol) = 0.02859 mol F
0.02859 mol F/0.005718 mol Br = 5.000 mol F: 1 mol Br
Empirical formula is BrF 5
(b) Molecular geometry is square pyramidal; hybridization is sp3d2.
(c) Yes, the square pyramidal structure is expected to be polar.
9.78
(a) A resonance structure can be drawn with double bonds between the alternate carbons and nitrogens of
the ring; the average of the two resonance structures would result in identical bonds with a bond order of
1.5 between each C and N of the ring.
(b) H = [(1 mol)(fHo(melamine(s)) + (6 mol)(fHo(NH3(g))) + (3mol)(fHo(CO2(g)))] – (6
mol)(fHo(urea(s))) =
[(1 mol)(-66.1 kJ/mol) + (6 mol)(-45.90 kJ/mol) + (3 mol)(-393.509 kJ/mol)] – (6 mol)(-333.1 kJ/mol) =
+477 kJ
9.79
(a) Br 0.9090 g/(79.904 g/mol) = 0.01138 mol Br
0.0910 g/(15.9994 g/mol) = 0.00569 mol O
0.01138 mol Br/0.00569 mol O = 2.00 mol Br: 1 mol O
Empirical formula OBr 2
The oxygen should be sp3 hybridized.
(b) 13 valence electrons; (s)2(*s)2(p)4(p)2(*p)3; HOMO is *p
SOLUTIONS TO APPLYING CHEMICAL PRINCIPLES: PROBING MOLECULES WITH
PHOTOELECTRON SPECTROSCOPY
1.
metal
2.
E = h = hc/ = (6.626 x 10-34 J·s)(2.998 x 108 m/s)/(58.4 x 10-9 m) = 3.40 x 10-18 J/photon
(3.40 x 10-18 J/photon)(6.022 x 1023 photons/mol) = 2.05 x 106 J/mol = 2.05 x 103 kJ/mol
3.
2p
4.
h3.40 x 10-18 J from problem 2 above
IE = h3.40 x 10-18 J – 4.23 x 10-19J = 2.98 x 10-18 J/electron
(2.98 x 10-18 J/electron)(6.022 x 1023 electron/mol)(1 kJ/1000 J) = 1.79 x 10 3 kJ/mol
(2.98 x 10-18 J)(1 eV/1.60218 x 10-19 J) = 18.6 eV
190
Chapter 9
5.
Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
The 15.6 eV and 16.7 eV ejected electrons came from bonding orbitals. Removing an electron from a
bonding orbital weakens the bonds and thus results in a longer bond length. The 18.6 eV ejected electron
comes from an antibonding orbital.
191