CHAPTER 8 | Chemical Bonding and Climate Change
8.45. Collect and Organize
Using the method in the textbook, we are to draw Lewis structures for Cl 2O and ClO3–.
Analyze
To draw the Lewis structures, we first must determine the number of valence electrons, then arrange the atoms
to show the bonding in the molecules by connecting the atoms with single covalent bonds. Finally, we
complete the octets of the atoms bonded to the central atoms, then complete the octet of the central atom.
Solve
Cl2O
(Step 1) The number of valence electrons in Cl2O is
Element
2Cl
O
Valence electrons per atom
(2  7)
+
6 = 20
(Step 2) We are given that one of the chlorine atoms is the central atom in this structure.
Cl
Cl O
(Step 3) We complete the octets on the oxygen and terminal chlorine atoms by adding three lone pairs to
each.
Cl
Cl
O
(Step 4) In this structure there are 16 electrons from six lone pairs and two bond pairs. We need 4 more
electrons (two pairs) to match the valence electrons determined in step 1. We add the lone pairs to the
central chlorine atom.
Cl
Cl
O
(Step 5) The central chlorine atom is satisfied with its octet so this Lewis structure is complete.
ClO3–
(Step 1) The number of valence electrons in ClO3 is
Element
Cl
Valence electrons per atom
7
Gain of electron due to charge
Total valence electrons
+
3O
(3  6)
= 18
+1
26
(Step 2) We are given that the chlorine atom is the central atom in this structure.
O
O Cl O
(Step 3) We complete the octets on the oxygen atoms by adding three lone pairs to each.
O
O
Cl
O
(Step 4) In this structure there are 24 electrons from nine lone pairs and three bond pairs. We need 2 more
electrons (one pair) to match the valence electrons determined in step 1. We add the lone pair to the central
chlorine atom.
O
O
Cl
O
(Step 5) The central chlorine atom is satisfied with its octet so this Lewis structure is complete. We add
brackets and the charge to indicate the ion.
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Chemical Bonding and Climate Change | 395
O
O
Cl
O
Think about It
In ClO3–, Cl is the central atom as we would guess from the formula, but in Cl 2O, oxygen could be the central
atom. As shown below, the arrangement of the molecule with Cl as the central atom gives nonzero formal
charges for the atoms (see section 8.6). In section 8.6, we will also learn that to reduce the formal charge on Cl
in ClO3–, we can form a double bond between Cl and one of the O atoms.
8.74. Collect and Organize
Pyridine has five carbons and one nitrogen in its six-membered ring, whereas pyrazine, another six-membered
ring, has four carbon atoms and two nitrogen atoms. We are to draw the resonance structures for these
compounds.
Analyze
To draw the Lewis structures we must first determine the number of covalent bonds in each structure and then
complete the octets (duets for hydrogen) as necessary and check the structure with electron bookkeeping. Once
one Lewis structure is drawn, we can then consider alternate structures in resonance with the first.
Solve
Pyridine, C5H5N
(Step 1) The number of valence electrons in C5H5N is
Element
5C
5H
N
Valence electrons per atom
5 = 30
(5  4)
+
(5  1)
+
(Step 2) We are told that the five carbon atoms and the nitrogen atom form a six-membered ring.
H
H
C
C
C
H
C
C
H
N
H
(Step 3) The duplets on the terminal H atoms are complete.
(Step 4) In this structure there are 22 electrons from 11 bond pairs. We need 8 more electrons (four pairs) to
match the valence electrons determined in step 1. The nitrogen and carbon atoms do not have octets so we add
the lone pairs to some of these atoms.
H
H
H
C
C
C
N
C
C
H
H
(Step 5) We can complete the octet for each carbon by forming double bonds between atoms in the ring.
H
H
H
C
C
C
N
H
C
C
H
H
H
H
C
C
C
N
C
C
H
H
396 | Chapter 8
The double bonds could have been drawn for the other ring atoms as well, so the two resonance forms of
C5H5N are
H
H
H
C
C
C
H
H
C
C
N
H
H
C
C
C
H
H
C
C
N
H
Pyrazine, C4H4N2
(Step 1) The number of valence electrons in C4H4N2 is
Element
4C
4H
2N
Valence electrons per atom
(4  4)
+
(4  1)
+
(2  5) = 30
(Step 2) We are told that the our carbon atoms and the nitrogen atoms form a six-membered ring.
H
N
H
C
C
C
C
H
N
H
(Step 3) The duplets on the terminal H atoms are complete.
(Step 4) In this structure there are 20 electrons from ten bond pairs. We need 10 more electrons (five pairs) to
match the valence electrons determined in step 1. The nitrogen and carbon atoms do not have octets, so we add
the lone pairs to some of these atoms.
H
H
C
C
N
N
C
C
H
H
(Step 5) We can complete the octet for each carbon by forming double bonds between atoms in the ring.
H
H
C
C
N
N
C
C
H
H
H
H
C
C
N
N
C
C
H
H
The double bonds could have been drawn for the other ring atoms as well, so the two resonance forms of
C4H4N2 are
H
H
C
C
N
N
C
C
H
H
H
H
C
C
N
N
C
C
H
H
Think about It
Notice that the only difference between pyrazine and pyridine is that a C—H in pyridine (5 valence e–) is
replaced by N (also 5 valence e–) in pyrazine. These compounds have the same total number of valence
electrons.
8.76. Collect and Organize
For S2O2 and S2O3 we are to draw Lewis structures and show all possible resonance forms.
Analyze
To draw the Lewis structure, we must first determine the number of covalent bonds in the structure and then
complete the octets (duets for hydrogen) as necessary and check the structure with electron bookkeeping. Once
one Lewis structure is drawn, we can then consider alternate structures in resonance with the first. Since the
Chemical Bonding and Climate Change | 397
sulfur atoms are the first elements in the molecular formulas, we place them as the central atoms in the
structures.
Solve
For S2O2
(Step 1) The number of valence electrons is
Element
2S
2O
Valence electrons per atom
(2  6)
+
(2  6) = 24
(Step 2) Sulfur is less electronegative than oxygen and so the two sulfur atoms are the central atoms in the
structure.
O
S
S
O
(Step 3) We complete the octets on the oxygen atoms by adding three lone pairs to each.
O
S
S
O
(Step 4) In this structure there are 18 electrons from six lone pairs and three bond pairs. We need 6 more
electrons (three pairs) to match the valence electrons determined in step 1. The sulfur atoms do not have octets
yet so we will add lone pairs to the S atoms.
O
S
S
O
(Step 5) We can complete the octet for the sulfur atom on the left by forming a double bond between the lefthand oxygen and that sulfur atom.
O
S
S
O
O
S
S
O
The electrons could also be distributed in two additional resonance forms that also complete the octet on all the
atoms.
For S2O3
(Step 1) The number of valence electrons is
Element
2S
3O
Valence electrons per atom
(2  6)
+
(3  6) = 30
(Step 2) Sulfur is less electronegative than oxygen and so the two sulfur atoms are the central atoms in the
structure.
O
O
S
S
O
(Step 3) We complete the octets on the oxygen atoms by adding three lone pairs to each.
O
O
S
S
O
(Step 4) In this structure there are 26 electrons from nine lone pairs and four bond pairs. We need 4 more
electrons (two pairs) to match the valence electrons determined in step 1. The sulfur atoms do not have octets
yet so we will add a lone pair to each of the S atoms.
O
O
S
S
O
(Step 5) We can complete the octet for the sulfur atom on the left by forming a double bond between the lefthand oxygen and that sulfur atom.
O
O
S
S
O
O
O
S
S
O
398 | Chapter 8
The electrons could also be distributed in three additional resonance forms that also complete the octet on all
the atoms.
Think about It
Like N2O3 in Problem 8.75, the octet rule does not allow these molecules to have a triple bond between the
sulfur atoms.
5 bonding pairs
6 bonding pairs
7 lone pairs
9 lone pairs
24 e–
30 e–
In both of the structures shown, one sulfur atom has more than an octet. We will see that sulfur may indeed
expand its octet through use of its d orbitals.
8.78. Collect and Organize
Hydroazoic acid has a linear structure with atom connectivity as described in the molecular formula
H—N—N—N
From this framework we are to draw valid resonance structures of HN 3.
Analyze
We first draw one of the valid Lewis structures by the method described in the textbook and then redistribute
the bonding pairs and lone pairs in the structure to draw resonance forms.
Solve
For HN3, hydroazoic acid
(Step 1) The number of valence electrons is
Element
H
3N
Valence electrons per atom
1
+ (3  5)
= 16
(Step 2) We are given that hydroazoic acid is a linear molecule with the connectivity of the atoms as
H N N N
(Step 3) We complete the octets on the rightmost nitrogen atom by adding three lone pairs. The duplet on the
terminal H atom is already satisfied.
H
N
N
N
(Step 4) In this structure there are 12 electrons from three lone pairs and three bond pairs. We need 4 more
electrons (two pairs) to match the valence electrons determined in step 1. The middle nitrogen atoms do not
have octets yet so we will add lone pairs.
H
N
N
N
(Step 5) We can complete the octet for the nitrogen atoms by forming a triple bond between them.
H
N
N
N
H
N
N
N
The electrons could also be distributed in two additional resonance forms that also complete the octet on all the
atoms.
Chemical Bonding and Climate Change | 399
H
N
N
N
H
N
N
N
H
N
N
N
Think about It
The following resonance form is not valid because it has more than a duet for the H atom and has less than an
octet for the N atom.
8.85. Collect and Organize
After drawing the Lewis structures for HNC and HCN and assigning the formal charges to the atoms, we are
asked to analyze the differences in their formal charges (and choose the best, most stable, arrangement for the
atoms).
Analyze
After drawing the Lewis structures for both HNC and HCN, we assign the formal charge (FC) for each atom
based on the formula
FC = (number of valence e– for the atom) – [(number of e– in lone pairs) +
( 12  number of e– in bonding pairs)]
Solve
For both HNC and HCN there are 10 valence electrons and the Lewis structures with formal charges are
The formal charges are zero for all the atoms in HCN, whereas in HNC the carbon atom, with a lower
electronegativity than N, has a –1 formal charge.
Think about It
The HCN arrangement, being more stable, is the preferred structure.
8.87. Collect and Organize
We are to draw Lewis structures for cyanamide, H2NCN, and assign formal charges to each atom.
Analyze
Because we are asked to draw structures for the compound, we suspect that the compound may show
resonance. After drawing the possible resonance structures, we assign formal charges to all atoms in each
structure using
FC = (number of valence e– for the atom) – [(number of e– in lone pairs) +
( 12  number of e– in bonding pairs)]
Solve
For cyanamide, H2NCN there are 16 valence electrons and the possible structures with formal charges assigned
for the atoms are
The preferred structure is the one with the C triple bonded to N because all the formal charges on the structure
are zero.
Think about It
Be careful in drawing the resonance structures. The structure below is not valid because the octet rule for both
nitrogen atoms is violated.
8.89. Collect and Organize
For the arrangement of atoms in nitrous oxide (N2O) where oxygen is the central atom, we are to assign formal
charges and suggest why this structure is not stable.
400 | Chapter 8
Analyze
After drawing the possible resonance structures, we assign formal charges to all atoms in each structure using
FC = (number of valence e– for the atom) – [(number of e– in lone pairs) +
( 12  number of e– in bonding pairs)]
Solve
For nitrous oxide, N2O, there are 16 valence electrons, and the Lewis structures with formal charges assigned
to the atoms are
Because oxygen is more electronegative than nitrogen, none of these structures is likely to be stable because
the formal charge on O is positive, when it would be predicted by electronegativity to be negative.
Think about It
The Lewis structure for the arrangement N—N— O is far better by formal charge, particularly with the
resonance structure in which there is a N to N triple bond.
8.91. Collect and Organize
We are to draw the Lewis structures (with resonance forms) for nitromethane (CH 3NO2) and CNNO2 (which
has two possible skeletal structures), and we are to determine if the two molecules are resonance structures of
each other.
Analyze
For each Lewis structure, we have to be sure that in drawing the resonance structures, we only redistribute the
electrons and not move the atom positions.
Solve
(a) In CH3NO2 there are 24 valence electrons. Completing the octets for all the atoms (duplet for hydrogen),
drawing an alternate resonance structure, and assigning formal charges to the atoms gives
(b) In CNNO2 there are 26 valence electrons. Completing the octets for all the atoms, drawing the alternate
resonance structures, and assigning formal charges to the atoms gives
Formal charges are minimized in
these two structures with the negative
formal charge on the most
electronegative atom (oxygen), so
these are the preferred structures.
(c) The two structures of CNNO2 given in the text are not resonance structures because their atoms differ in
connectivity. When two molecules have the same number of atoms, but in a different arrangement, they are
called isomers.
Chemical Bonding and Climate Change | 401
Think about It
Be careful to make sure that all atoms in a resonance form have complete octets. The following Lewis structure
is not a valid Lewis structure for NCNO2 because the terminal nitrogen atom has less than 8 e–.
8.98. Collect and Organize
To determine which of the phosphorus molecules have an expanded octet, we need to consider the number of
electrons around the central atom required to form the compound.
Analyze
In each of these compounds phosphorus is the central atom as it is the least electronegative and has the highest
bonding capacity. If the number of electrons in bonding pairs and lone pairs on the phosphorus atom in the
Lewis structure of each compound is greater than eight, then phosphorus in that compound has an expanded
octet.
Solve
Molecule
Lewis structure
O
(a) POCl3
Cl
P
Cl
Number of electrons around P
10
Cl
F
F
(b) PF5
F
P
10
F
F
F
F
(c) PF3
P
8
F
(d) P2F4
F
F
F
P
P
F
8
POCl3 (a) and PF5 (b) require phosphorus to expand its octet.
Think about It
The Lewis structure of POCl3 has a P=O bond because this arrangement reduces the formal charge on all the
atoms to zero which is the preferred structure. Therefore, the phosphorus atom in this molecule expands its
octet.
8.102. Collect and Organize
By drawing the Lewis structures of PO43– and NO43– we are to describe the differences in bonding between
these molecules.
Analyze
Nitrogen is a second period element that is not able to expand its octet, but phosphorus, as a third period
element, can expand its octet.
Solve
Both molecules have 32 valence electrons, and the Lewis structures with formal charges are as follows:
402 | Chapter 8
Because oxygen is more electronegative than N, the structures where the O atom has a –1 formal charge seem
reasonable. However, the formal charges on P and O in PO 43– can be reduced because P can expand its octet
to form a double bond with O.
In PO43– there is a double bond in a single resonance form, whereas in NO43– all bonds are single bonds.
Think about It
In Lewis structures, always try to minimize formal charge. For elements in the third or higher periods, formal
charge reduction can be accomplished by expanding the octets to form double bonds.
8.104. Collect and Organize
We are to draw the Lewis structures for the ions in NF 4SbF6.
Analyze
In this compound, NF4+ is the cation (analogous to the ammonium cation), and SbF6– is the anion.
Solve
NF4+ has 32 valence electrons, and its Lewis structure is
SbF6– has 48 valence electrons, and its Lewis structure is
Think about It
In this example, we see plainly that nitrogen did not need to expand its octet to bond with 4 F atoms, but Sb
did need to expand the octet to bond with 6 F atoms.
8.108. Collect and Organize
For each molecule combining N with O we are to determine which are odd-electron molecules.
Analyze
To answer this we need only add up the valence electrons for each molecule.
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Solve
(a) NO has (1N  5 e–) + (1O  6 e–) = 11 e–
(b) NO2 has (1N  5 e–) + (2O  6 e–) = 17 e–
(c) NO3 has (1N  5 e–) + (3O  6 e–) = 23 e–
(d) N2O4 has (2N  5 e–) + (4O  6 e–) = 34 e–
(e) N2O5 has (2N  5 e–) + (5O  6 e–) = 40 e–
The odd-electron molecules are NO, NO2, and NO3 (a–c).
Think about It
For these nitrogen–oxygen molecules, notice that the odd-electron species have an odd number of N atoms in
their formulas.
8.118. Collect and Organize
In order to rank the bond lengths in CO, CO2, and CO32– we need to draw the Lewis structures with resonance
forms if necessary.
Analyze
In Lewis structures single bonds are longer than double bonds, which are longer than triple bonds. We must
also consider any resonance forms that these molecules might have.
Solve
For CO, the bond order for the C — O bond is 3.0.
For CO2, the bond order for the C — O bond is 2.0.
For CO32–, the bond order for the C — O bond is 1.33 due to resonance.
In order of increasing bond length: CO < CO2 < CO32–.
Think about It
Resonance has quite an effect on the bond order and length.
8.120. Collect and Organize
In order to rank the bond energies for CO, CO2, and CO32–, we need to draw the Lewis structures, with
resonance if necessary, for these molecules.
Analyze
In Lewis structures, single bonds have the lowest bond energy. Double bonds are stronger (have higher bond
energies) than single bonds and triple bonds are stronger than double bonds. The higher the bond order, the
higher the bond energy.
Solve
For CO, the bond order for the C — O bond is 3.0.
For CO2, the bond order for the C — O bond is 2.0.
For CO32–, the bond order for the C — O bond is 1.33 due to resonance.
404 | Chapter 8
In order of increasing bond energy: CO32– < CO2 < CO.
Think about It
Notice that this ranking is the reverse of the ranking for increasing bond length in Problem 8.118.
8.131. Collect and Organize
For the reaction of ammonia with oxygen to give water and nitrogen dioxide, we can use the bond energies of
the N—H (388 kJ/mol), O O (495 kJ/mol), N O (607 kJ/mol), N— O (201 kJ/mol), and O —H
(463 kJ/mol) bonds to estimate the enthalpy of the reaction.
Analyze
To be sure which bonds are breaking and which bonds are being formed, it is helpful to draw the Lewis
structures of each of the products and reactants. The enthalpy of a reaction as estimated by bond energies is
given as
H rxn   H bond breaking +  H bond forming
where ∆Hbond breaking and ∆Hbond forming are average bond energies for the bonds in the reactants and products,
respectively.
Solve
∆Hrxn = [(12  388 kJ/mol) + (7  495 kJ/mol)] + [– (4  201 kJ/mol) – (4  607 kJ/mol) – (12  463 kJ/mol)]
∆Hrxn = –667 kJ
Think about It
The bonding in NO2 is not strictly one single bond and one double bond as shown by the resonance structures
For calculations using bond energies, however, we can use one of the “frozen” resonance structures to assign
bond energy values. s with zero formal charges on all of the atoms.