Chapter 2
Molecular Representations
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 2. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• In bond-line structures, _______atoms and most ________ atoms are not drawn.
• A ________________ is a characteristic group of atoms/bonds that show a
predictable behavior.
• When a carbon atom bears either a positive charge or a negative charge, it will
have ___________, rather than four, bonds.
Review of Skills
Fill in the blanks and empty boxes below. To verify that your answers are correct, look
in your textbook at the end of Chapter 2. The answers appear in the section entitled
SkillBuilder Review.
SkillBuilder 2.1 Converting Between Different Drawing Styles
DRAW THE LEW IS ST RUCTURE OF T HE FOLLOW ING COMPOUND
CHAPTER 2
17
SkillBuilder 2.2 Reading Bond-Line Structures
N
N
O
Cl
CIRCLE ALL CA RBON ATOMS IN THE COMPOUND BELOW DRAW ALL HYDRO GEN AT OMS IN T HE COMPOUND BELOW
N
N
O
Cl
SkillBuilder 2.4 Identifying Lone Pairs on Oxygen Atoms
O
OO
AN OXYGEN ATOM
WITH A NEGATIVE
CHARGE WILL HAVE
____ LONE PAIR(S)
AN OXYGEN ATOM
WITH NO FORMAL
CHARGE WILL HAVE
____ LONE PAIR(S)
AN OXYGEN ATOM
WITH A POSITIVE
CHARGE WILL HAVE
____ LONE PAIR(S)
SkillBuilder 2.6 Identifying Valid Resonance Arrows
RULE 1: THE TAIL OF A CURVED ARROW
CANNOT BE PLACED ON A ______________
RULE 2: THE HEAD OF A CURVED ARROW
CANNOT RESULT IN _____________________
________________________________________
TAIL HEAD
SkillBuilder 2.7 Assigning Formal Charges in Resonance Structures
18
CHAPTER 2
O O H O O H O O H O O H
IDENTIFY WHICH RESONANCE STRUCTURES BELOW ARE SIGNIFICANT AND WHICH ARE INSIGNIFICANT
Solutions
2.1.
a)
C C
O
HH
H C C
C
C
H
H
H
HHH
HH
H
b)
H C C C
C
C
H
H H
H
O H
H
HC
H
HC
H
HH
H
HH
c)
OCCC
C
C
H
HH
H
H
H
H C
H
H C
H
H H
H
H
H
H
CHAPTER 2
19
2.2
(CH
3
)
3
COCH
3
and
(CH
3
)
2
CHOCH
2
CH
3
2.5.
a)
C C
C
C
C
C
C C
H
H
H
H
HH
H
H
H
H
H
H H
H
b)
CCC NCC
H
OH H
H
H HH H
H
H H
c)
C
C
C
C
C
C
H
HH
H
H
H
H
H
H
H
H
H
H
H
j)
CCC
C
C
C
H H
H
H H
H H
H
H H
k)
C C
CO
C
C
C
C
H H
H
H
H
H
HH
H
HH
H
l)
C
H
2
C
H
2
C
C
H
2
C
H
2
CH
2
C
O
2.6
a) decrease (76) b) no change (88)
c) no change (88) d) increase (57)
2.7
20
CHAPTER 2
e) f)
OH
g)
OH
h)
O
2.9.
2.10.
Br
O
O
2.11.
e
s
t
e
amide
2.12.
a)
N
b) No charge c)
N
d) No charge
CHAPTER 2
21
2.13.
a)
O
b)
O
c) No charge d)
H
O
H
H
2.14.
2.15. There are no hydrogen atoms attached to the central carbon atom. The carbon
atom has four valence electron. Two valence electrons are being used to form bonds, and
the remaining two electrons are a lone pair. This carbon atom is using the appropriate
number of valence electrons.
2.16.
2.17.
N
OH
C NO
NO O
2.18.
a) one b) zero c) one d) five
22
CHAPTER 2
2.19 Five lone pairs:
O
O
R
NH
3
2.20
2.21
a) Violates second rule by giving a fifth bond to a nitrogen atom.
b) Does not violate either rule.
c) Violates second rule by giving five bonds to a carbon atom.
2.22.
O
CHAPTER 2
23
c)
O
O
O
O
d)
NN
2.24.
O O
O O
2.25.
a)
O
O
O
O
b)
24
CHAPTER 2
e)
N
OONO
O
f)
OO
h)
2.26.
a) b)
c)
CHAPTER 2
25
2.27.
a)
N N
b)
N N
2.28.
NN
OO
2.29.
OH
O
OH
O
26
CHAPTER 2
2.31.
H
2
N
HO OH
H
2
N
HO OH
2.32.
a)
OO O
e)
O O
f)
OHOHOHOHOH
CHAPTER 2
27
g)
N N
h)
NN N
2.33.
a)
NHNHNH
NN N
28
CHAPTER 2
e)
S
O
S
O
h)
j)
k)
C
N
C
N
C
N
l)
OH OH
CHAPTER 2
29
2.34.
2.35.
OH
δ
δδ
δ
–
δ
δδ
δ–δ
δδ
δ–
δ
δδ
δ–δ
δδ
δ–
2.36.
a)
NNH
H
H
H
delocalized
sp
2
hybridized
trigonal planar
localized
sp
3
hybridized
trigonal pyramidal
b)
NO
O
One of these lone pairs is
delocalized. The oxygen
atom is therefore sp
2
hybridized and has bent
geometry.
localized
sp
2
hybridized
geometry not relevant
(connected to only one atom)
delocalized
sp
2
hybridized
trigonal planar
30
CHAPTER 2
f)
localized
sp
2
hybridized
geometry not relevant
(connected to only one atom)
OO
O
2.37. Both lone pairs are localized and, therefore, both are expected to be reactive.
2.38.
localized
(not participating in resonance)
N
N
O
NH
2
H
localized
(not participating in resonance)
delocalized
(participating in resonance)
localized
(not participating in resonance)
2.39.
2.40.
2.41.
CHAPTER 2
31
2.43. Twelve (each oxygen atom has two lone pairs)
2.44.
O
N
N O
ON
2.46.
a)
C
4
H
10
C
6
H
14
C
8
H
18
C
12
H
26
In each of the compounds above, the number of hydrogen atoms is equal to two
times the number of carbon atoms, plus two.
32
CHAPTER 2
c)
C
6
H
10
C
9
H
16
C
9
H
16
C
7
H
12
2.47.
a) an sp
2
hybridized atomic orbital
b) a p orbital
c) a p orbital
2.48.
a)
N
OH
HN
OH
HN
OH
H
2.49.
a) (CH
3
)
3
CCH
2
CH
2
CH(CH
3
)
2
b) (CH
3
)
2
CHCH
2
CH
2
CH
2
OH
c) CH
3
CH
2
CH=C(CH
2
CH
3
)
2
CHAPTER 2
33
2.50.
2.51.
(d) is not a valid resonance structure, because it violates the octet rule. The nitrogen atom
has five bonds in this drawing, which is not possible, because the nitrogen atom only has
four orbitals with which it can form bonds.
2.52. 15 carbon atoms and 18 hydrogen atoms:
2.55.
a)
O O
b)
OO
34
CHAPTER 2
e)
NHNHNHNHNH
HHH H H
g)
O O O O
O
OO
CHAPTER 2
35
i)
OH OH OH
OH
2.56. These structures do not differ in their connectivity of atoms. They differ only in
the placement of electrons, and are therefore resonance structures.
2.57.
a) constitutional isomers
2.58.
a) b)
OH
c)
O
36
CHAPTER 2
2.59. The nitronium ion does not have any significant resonance structures because any
attempts to draw a resonance structure will either 1) exceed an octet for the nitrogen atom
or 2) generate a nitrogen atom with less than an octet of electrons, or 3) generate a
2.61. Both nitrogen atoms are sp
2
hybridized and trigonal planar, because in each case,
the lone pair participates in resonance.
2.62.
HO
OH
H
HH
HO
OH
H
HH
HO
OH
H
HH
CHAPTER 2
37
2.63.
a) The molecular formula is C
3
H
6
N
2
O
2
b) There are two sp
3
hybridized carbon atoms
c) There is one sp
2
hybridized carbon atom
g)
O
N
O
H NH
2
not relevant
(only connected to one other atom)
trigonal planar
trigonal pyramidal
tetrahedral tetrahedral
bent
trigonal planar
2.64.
a) The molecular formula is C
16
H
21
NO
2
b) There are nine sp
3
hybridized carbon atoms
38
CHAPTER 2
2.65.
H
OH
H
O
H
OH
H
O
H
OH
H
O
H
OH
H
O
H
OH
H
O
δ
δδ
δ+
δ
δδ
δ+
δ
δδ
δ+
δ
δδ
δ–δ
δδ
δ–
2.66.
a) Compound B has one additional resonance structure that Compound A lacks,
because of the relative positions of the two groups on the aromatic ring.
Specifically, Compound B has a resonance structure in which one oxygen atom has
a negative charge and the other oxygen atom has a positive charge:
CHAPTER 2
39
2.67.
The single bond mentioned in this problem has some double bond character, as a result of
resonance: