Chapter 1
Electrons, Bonds and Molecular Properties
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 1. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• _____________ isomers share the same molecular formula but have different
connectivity of atoms and different physical properties.
• Second-row elements generally obey the _______ rule, bonding to achieve noble
gas electron configuration.
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 1. The answers appear in the section entitled
SkillBuilder Review.
SkillBuilder 1.1 Determining the Constitution of Small Molecules
STEP 1 – DETERMINE THE VALENCY (NUMBER OF EXPECTED
BONDS) FOR EACH ATOM IN C
2
H
5
Cl
STEP 2 – DRAW THE STRUCTURE OF C
2
H
5
Cl BY PLACING
ATOMS WITH THE HIGHEST VALENCY AT THE CENTER,
AND PLACING MONOVALENT ATOMS AT THE PERIPHERY
Each carbon atom is expected to form ___ bonds.
2
CHAPTER 1
SkillBuilder 1.2 Drawing the Lewis Dot Structure of an Atom
STEP 1 – DETERMINE THE NUMBER
OF VALENCE ELECTRONS
STEP 2 – PLACE ONE ELECTRON
BY ITSELF ON EACH SIDE OF
THE ATOM
STEP 3 – IF THE ATOM HAS MORE THAN FOUR
VALENCE ELECTRONS, PAIR THE REMAINING
ELECTRONS WITH THE ELECTRONS ALREADY DRAWN
Nitrogen is in Group ___ of the
periodic table, and is expected
to have ___ valence electrons.
SkillBuilder 1.4 Calculating Formal Charge
H
H
N
H
H
HN
H
H
H
STEP 1 – DETERMINE THE APPROPRIATE
NUMBER OF VALENCE ELECTRONS
STEP 2 – DETERMINE THE NUMBER OF
VALENCE ELECTRONS IN THIS CASE
STEP 3 – ASSIGN A FORMAL
CHARGE TO THE NITROGEN ATOM
IN THIS CASE
Nitrogen is in
Group ___ of the
periodic table,
and is expected
to have ___
valence electrons.
In this case, the
nitrogen atom is
using only ___
valence electrons.
SkillBuilder 1.6 Identifying Electron Configurations
STEP 1 – IN THE ENERGY
DIAGRAM SHOWN HERE,
DRAW THE ELECTRON
CONFIGURATION OF
NITROGEN (USING
ARROWS TO REPRESENT
ELECTRONS).
STEP 2 – FILL IN THE BOXES BELOW WITH THE
NUMBERS THAT CORRECTLY DESCRIBE THE
ELECTRON CONFIGURATION OF NITROGEN
1s
2s
2p
1s
2s
2p
Nitrogen
CHAPTER 1
3
SkillBuilder 1.8 Predicting Geometry
H N H
H
# of single bonds =
Steric Number =
tetrahedral
arrangement of
electron pairs
ONE
LONE
PAIR
TWO
LONE
PAIRS
NO
LONE PAIRS
STEP 1 – DETERMINE THE
STERIC NUMBER OF THE
NITROGEN ATOM BELOW
BY ADDING
THE NUMBER
OF SINGLE
BONDS AND
LONE PAIRS
STEP 2 – THE STERIC NUMBER
DETERMINES THE HYBRIDIZATION STATE
AND ELECTRONIC GEOMETRY. FILL IN
THE CHART BELOW:
STEP 3 – IGNORING LONE PAIRS, IDENTIFY THE
GEOMETRY IN EACH CASE BELOW
# of lone pairs =
Steric
#
Hybridization
State
Electronic
Geometry
4
3
2
SkillBuilder 1.9 Identifying Molecular Dipole Moments
C
O
CCH
3
H
H
H
3
CH
H
STEP 1 – IDENTIFY THE GEOMETRY
OF THE OXYGEN ATOM BELOW
STEP 2 – REDRAW THE COMPOUND.
FOR EACH POLAR COVALENT BOND,
DRAW AN ARROW THAT SHOWS THE
DIRECTION OF THE DIPOLE MOMENT
STEP 3 – REDRAW THE
COMPOUND, AND DRAW THE
NET DIPOLE MOMENT
Geometry =
SkillBuilder 1.10 Predicting Physical Properties
C O C
H
H
H
H
H
H
C C O
HH
H
H H
H
H
3
CCCH
3
CH
2
H
3
CCCH
3
O
HC
H
H
C
H
H
C C
H
H
C
H
H
H
H
H
C
H
H H
C H
H
H
H C
H
CIRCLE THE COMPOUND BELOW
THAT IS EXPECTED TO HAVE THE
HIGHER BOILING POINT
Dipole-Dipole Interactions H-Bonding Interactions Carbon Skeleton
CIRCLE THE COMPOUND BELOW
THAT IS EXPECTED TO HAVE THE
HIGHER BOILING POINT
CIRCLE THE COMPOUND BELOW
THAT IS EXPECTED TO HAVE THE
HIGHER BOILING POINT
Solutions
1.1.
a)
C O H
H
H
H
b)
C Cl
H
H
H
c)
C C H
HH
H
H
H
d)
C N
H
H
H
H
H
4
CHAPTER 1
C C O
HH
C
H
H
C
H
H
H
H
H
H
H
C
O
C C
CH
H
H
H
H
H
H
H
H
O
H
CC C
CH
H
H
H
H
H
H
H
H
H
CC C
CH
O
H
H
H
H
H
H
H
H
1.5.
a)
C
b)
O
c)
F
d)
H
1.8.
C
resembles boron because it exhibits three valence electrons.
1.9.
C
resembles nitrogen because it exhibits five valence electrons.
CHAPTER 1
5
1.12
C C
H
H H
H H
C
H
H
N
H
H
C C
H
H H
HN
C
H
H
H
H
H
CN
H
H H
H
HC
H
H
C
H
H
CN
H
HC
H
C
H
H
H
H
H
H
In all of the constitutional isomers above, the nitrogen atom has one lone pair.
1.14.
a)
BHH
H
H
Boron has a formal charge
b)
NHH
Nitrogen has a formal charge
c)
C C
H
H H
H
H
Carbon has a formal charge
1.15.
1.16.
C
H
H
C ClCC
H
H
CH
H
H
H
H
O
δ
δδ
δ+
δ
δδ
δ+
δ
δδ
δ
–
δ
δδ
δ–
1.17.
6
CHAPTER 1
1.18.
a) 1s
2
2s
2
2p
3
b) 1s
2
2s
2
2p
1
c) 1s
2
2s
2
2p
2
d) 1s
2
2s
2
2p
5
1.19. The bond angles of an equilateral triangle are 60º, but each bond angle of
cyclopropane is supposed to be 109.5º. Therefore, each bond angle is severely
1.20
a) The C=O bond of formaldehyde is comprised of one sigma bond and one pi
1.21. Rotation of a single bond does not cause a reduction in the extent of orbital
1.22.
C
CCC
C
C O CC
H
H
H
H
C
H
O
H H
O O H
sp
3
CHAPTER 1
7
1.23.
1.24.
C C
C
H C
H
C C H
HH
C
H
H
H
HHc
a
b
c < b < a
a is the longest bond
and c is the shortest bond
1.25.
a) The nitrogen atom has three bonds and one lone pair, and is therefore trigonal
pyramidal.
1.26.
(a)
C
C
N
C
C
C
O
C
H
H
H
H
H
H
H
O H
H
H
H
All carbon atoms in this molecule are tetrahedral
except for the highlighted carbon atom,
which is trigonal planar.
The oxygen atom (of the OH group)
has bent geometry,
and the nitrogen atom is trigonal pyramidal.
8
CHAPTER 1
(c)
C
C
C
C
C
C C
C
H
H
H
H
H
H
H
HAll carbon atoms are trigonal planar.
1.29.
a)
C
Cl
H
C
l
Cl
b)
H
3
C
O
CH
3
c)
N
H
H
H
d)
C
Cl
Cl
B
r
Br
1.30. CHCl
3
is expected to have a larger dipole moment than CBrCl
3
, because the
bromine atom in the latter compound serves to nearly cancel out the effects of the
other three chlorine atoms (as is the case in CCl
4
).
1.31. The carbon atom of CO
2
has a steric number of two, and therefore has linear
geometry. As a result, the individual dipole moments of each C=O bond cancel
CHAPTER 1
9
1.32.
a) The latter, because it is less branched.
1.33.
H C
H
H
C
H
H
C C
H
H
C
H
H
H
H
H
H C
H
H
C
H
H
C C
H
H
O H
H
H
CCC
C CH
H
HH
H
H
H
H
H
H
H
H
HC
H
H
C
H
H
CH
C
C
H
H
H
H
H
H
H C
H
H
C
H
H
C C
H
H
C
H
H
C
H
H
H
H
O H
Increasing boiling point
1.34.
a)
CCC
HH
C
H
H
H
H
H
H
H
H
H
CC C
CH
H
H
H
H
H
H
H
H
H
C
CC C
CH
H
H
H
H
H
H
H
C
H
H
H
H
H
H
H
CC C
CH
H
H
H
H
H
C
H
C H
H
H
H H
H
d)
C C Cl
HH
H
H
H
10
CHAPTER 1
1.35.
1.36.
a)
H
B
r
δ
δδ
δ
+
δ
δδ
δ
–
b)
H
C
l
δ
δδ
δ
+
δ
δδ
δ
–
c)
δ
δδ
δ+
δ
δδ
δ
–
H
O
H
δ
δδ
δ+
d)
HOCH
H
H
δ
δδ
δ+δ
δδ
δ–δ
δδ
δ+
1.37.
a) NaBr, because the difference in electronegativity between Na and Br is greater
1.38.
a)
C C
H
H H
H
H
OH
b)
C C N
H
H
H
1.39.
H C
C
H
C C
H
H
H
All carbon atoms in this molecule are tetrahedral
except for the carbon atom bearing the negative
CHAPTER 1
11
1.40.
CN
H
HC
H
C
H
H
H
H
H
C
H
H
H
1.41.
Br
1.42.
C C
HH
H C H
H
H
1.44.
a) Oxygen b) Fluorine c) Carbon d) Nitrogen e) Chlorine
1.45.
a) ionic
1.46.
a)
H C C
H
H
OH
H
H
H C O
H
H
C
H
H
H
12
CHAPTER 1
1.47.
H C C
H
H
OH
O
H
O
H
H C C
O
H
H
OH
O
H
H
H C O
O
H
H
C
O
H
H
H
H C O
H
H
OC H
O
H
H
H C O
H
H
C
O
H
H
O
H
1.48.
1.49.
a) All bond angles are approximately 109.5º, except for the C-O-H bond angle
which is expected to be less than 109.5º as a result of the repulsion of the lone
pairs on the oxygen.
1.50.
a) sp
3
, trigonal pyramidal
1.51. Sixteen sigma bonds and three pi bonds.
1.52.
1.53.
a) yes
b) no (this compound can serve as a hydrogen bond acceptor, but not a hydrogen
bond donor)
c) no
CHAPTER 1
13
d) no
e) no (this compound can serve as a hydrogen bond acceptor, but not a hydrogen
bond donor)
f) yes
g) no
h) yes
1.55.
a)
C
C
C C
H
H
C
H
C
HThe highlighted carbon atoms are sp
2
hybridized and
trigonal planar. The remaining four carbon atoms are
sp hybridized and linear.
1.56.
C
C
C
C
N
C
C
C
N
C
N
O
C
C
C
CC C
C
CC
C
H
H
H
H
H
H
H
H
H
HH
HH
H
H
H
H
H
H
H
H
The highlighted carbon atoms are
sp
3
hybridized and
tetrahedral. The remaining carbon atoms are
sp
2
hybridized and trigonal planar.
1.57.
a) oxygen b) fluorine c) carbon
1.58.
14
CHAPTER 1
1.60. The two isomers are:
H C C
H
H
OH
H
H
H C O
H
H
C
H
H
H
The first will have a higher boiling point because it possesses an OH group which
can form hydrogen bonds.
1.61.
a)
C
C
C
C
C
C ClH
H
Cl
H
H
b)
C
C
C
C
C
C HH
H
Cl
H
Cl
Cl
Cl
1.62. The third chlorine atom in chloroform partially cancels the effects of the other two
chlorine atoms, thereby reducing the molecular dipole moment relative to
methylene chloride.
1.63.
a) Compound A and Compound B
b) Compound B
CHAPTER 1
15
1.64.
a)
C
C
C
C
C
C HH
H
H
H
H
C
C C
C
C
HH
H
CH
HH
H
H
1.65.
CCCC
H
H H
H
H
H
C
H
H
H H
CCCC
H
H
H
H
H
C
H
H
H HH
CCCC
C
H H
H
H
HHHH
H
H
CCCC
C
H
H
H
HHHH
H
H
H
CCCC
C
H
H
H
HHHH
H
H H