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CHAPTER
Chemical Bonding
12
Section 12.1 The Chemical Bond Concept
2.
Atom
H
Br
Valence Electrons
1 valence e–
7 valence e–
4.
Atom
H
Br
Covalent Bond
2 valence e–
8 valence e–
6.
8.
10.
(a)
(c)
Compound
ZnBr2
IF7
Bond
ionic
covalent
Compound
(b) CO2
(d) PbSO4
Bond
covalent
ionic
(a)
(c)
Substance
CH3 OH
CH3 COCH3
Particle
molecule
molecule
Substance
(b) CoCl2
(d) SnCO3
Particle
formula unit
formula unit
(a)
(c)
Substance
Xe
XeO3
Particle
atom
molecule
Substance
(b) P4
(d) Fe3 O4
Particle
molecule
formula unit
2014 © Pearson Education, Inc.
Chemical Bonding
89
Section 12.2 Ionic Bonds
12.
14.
16.
18.
20.
22.
24.
90
(a)
(c)
Ion
Li ion
Al ion
Ionic Charge
1+ (Group 1)
3+ (Group 13)
Ion
(b) Sr ion
(d) Pb ion
Ionic Charge
2+ (Group 2)
4+ (Group 14)
(a)
(c)
Ion
Cl ion
S ion
Ionic Charge
1– (Group 17)
2– (Group 16)
Ion
(b) I ion
(d) P ion
Ionic Charge
1– (Group 17)
3– (Group 15)
(a)
(b)
(c)
(d)
Ion
Sc3+
K+
Ti4+
Ba2+
Electron Configuration
1s2 2s2 2p6 3s2 3p6
1s2 2s2 2p6 3s2 3p6
1s2 2s2 2p6 3s2 3p6
1s2 2s2 2p6 3s2 3p6 4s2 3d1 0 4p6 5s2 4d1 0 5p6
Ion
(a) Br–
(b) O2 –
(c) Se2 –
(d) N3 –
Electron Configuration
1s2 2s2 2p6 3s2 3p6 4s2 3d1 0 4p6
1s2 2s2 2p6
1s2 2s2 2p6 3s2 3p6 4s2 3d1 0 4p6
1s2 2s2 2p6
(a)
(b)
(c)
(d)
Ion
Se2 –
Br–
Rb+
Sr2+
Isoelectronic Noble Gas
Kr
Kr
Kr
Kr
(a)
(b)
(c)
(d)
Ion
Sc3+
K+
Ti4+
Ba2+
Isoelectronic Noble Gas
Ar
Ar
Ar
Xe
(a)
(b)
(c)
(d)
Ion
Br–
O2 –
Se2 –
N3 –
Isoelectronic Noble Gas
Kr
Ne
Kr
Ne
Chapter 12
2014 © Pearson Education, Inc.
26.
(a)
(b)
(c)
(d)
Al atom radius > Al ion radius
Pb atom radius > Pb ion radius
Se atom radius < Se ion radius
N atom radius < N ion radius
28.
The true statements are (b) and (c). The corrected statements are:
(a) Aluminum and nitride ions bond by electrostatic attraction.
(d) Breaking an ionic bond between aluminum and nitrogen requires energy.
Section 12.3 Covalent Bonds
30.
(a)
(b)
The bond length in H–Br is less than the sum of the atomic radii.
The bond length in S=O is less than the sum of the atomic radii.
32.
The true statements are (b) and (c). The corrected statements are:
(a) Valence electrons are shared between carbon and oxygen atoms.
(d) Energy is required to break a C—O covalent bond.
Section 12.4 Electron Dot Formulas of Molecules
34.
Molecule
Valence Electrons
(a)
Cl2
7 + 7 = 14 e–
(b)
O2
6 + 6 = 12 e–
(c)
HI
1 + 7 = 8 e–
(d) PH 3
5 + 3(1) = 8 e–
Electron Dot
Structural Formula
Cl Cl
Cl
O
O
O
H I
H
H P H
2014 © Pearson Education, Inc.
Cl
O
H
I
H
H
P
H
Chemical Bonding
91
36.
38.
Molecule
Valence Electrons
(a)
2(5) = 10 e–
N2
(b)
PI3
5 + 3(7) = 26 e–
(c)
CS 2
4 + 2(6) = 16 e–
(d) HOCl
1 + 6 + 7 = 14 e–
Molecule
Valence Electrons
(a)
(b)
(c)
CCl4
HONO2
CH 3 OH
(d) HOCN
92
Chapter 12
4 + 4(7) = 32 e–
1 + 3(6) + 5 = 24 e–
4 + 4(1) + 6 = 14 e–
1 + 6 + 4 + 5 = 16 e–
Electron Dot
N
Structural Formula
N
N
I
I P I
S
C
N
I
I
P
I
S
C
S
H
O
Cl
S
H O Cl
Electron Dot
Cl
Cl
Cl C Cl
Cl
H O N
O
Cl
2014 © Pearson Education, Inc.
C
Cl
Cl
O
H O N
O
O
H
H
H C O H
H
H O C
Structural Formula
H
C
O
H
C
N
H
N
H
O
Section 12.5 Electron Dot Formulas of Polyatomic Ions
40.
Polyatomic Ion Valence Electrons
(a)
IO–
7 + 6 + 1 = 14 e–
(b)
IO2 –
7 + 2(6) + 1 = 20 e–
(c)
IO3 –
7 + 3(6) + 1 = 26 e–
Electron Dot
I
Structural Formula
O –
O I O –
[O
I
O]–
O I O –
O
[O
I
O]–
O
O
O
–
O I O
(d) IO4 –
42.
7 + 4(6) + 1 = 32 e–
Polyatomic Ion Valence Electrons
[ O
O
PO4 3 –
5 + 4(6) + 3 = 32 e–
Electron Dot
HPO4 2 –
1 + 5 + 4(6) + 2 = 32 e–
O
PO3 3 –
(d) HPO3 2 –
5 + 3(6) + 3 = 26 e–
1 + 5 + 3(6) + 2 = 26 e–
[ O
H O P O
O
2–
O ] 3–
O
[H–O
3–
H O P O
O
P
O ] 2–
O
[ O
O
2014 © Pearson Education, Inc.
P
O
O P O
(c)
O ]–
Structural Formula
O
O
(b)
I
O
O
3–
O P O
(a)
O]–
[I
P
O ] 3–
O
2–
[H–O
P
O ] 2–
O
Chemical Bonding
93
44.
Polyatomic Ion Valence Electrons
Electron Dot
H
(a)
(b)
PH 4 +
SeO3 2 –
H P H
H
5 + 4(1) – 1 = 8 e–
Structural Formula
H
[ H – P – H ]+
+
H
[ O – Se – O ] 2–
O Se O 2–
O
6 + 3(6) + 2 = 26 e–
O
[ O – C – O ] 2–
O C O 2–
(c)
CO 3 2 –
(d) BO3 3 –
4 + 3(6) + 2 = 24 e–
O
O
[ O – B – O ] 3–
O B O 3–
O
5 + 3(6) + 3 = 26 e–
O
Section 12.6 Polar Covalent Bonds
46.
Within a period of elements, the electronegativity increases from left to right in
the periodic table.
48.
Nonmetals are more electronegative than semimetals.
50.
(a)
(c)
More Electronegative
Se < Br
Te < S
More Electronegative
(b) C > B
(d) Ba < Be
(Note: The more electronegative element is in bold.)
52.
54.
(a)
(c)
Bond
H—Cl
N—O
Polarity
3.0 – 2.1 = 0.9
3.5 – 3.0 = 0.5
Bond
(b) H—Br
(d) C—O
(a)
(b)
(c)
(d)
Polar Bonds Using Delta Notation
δ– C—H δ+
δ+ Se—O δ–
δ+ P—I δ–
δ+ H—Br δ–
Polarity
2.8 – 2.1 = 0.7
3.5 – 2.5 = 1.0
(Note: δ– indicates the more electronegative atom and δ+ indicates the more
electropositive atom.)
94
Chapter 12
2014 © Pearson Education, Inc.
Section 12.7 Nonpolar Covalent Bonds
56.
(a)
(b)
(c)
(d)
Bond
I—C
C—S
S—H
H—Br
Polarity
2.5 – 2.5
2.5 – 2.5
2.5 – 2.1
2.8 – 2.1
=
=
=
=
0
0
0.4
0.7
Classification
nonpolar
nonpolar
polar
polar
Thus, (a) and (b) are nonpolar.
58.
O2 , F2 , and I2 occur naturally as diatomic molecules.
Section 12.8 Coordinate Covalent Bonds
(Note: Coordinate covalent bonds are indicated by a dash, —.)
60.
62.
Molecule
Valence Electrons
HIO
1 + 7 + 6 = 14 e–
Molecule
Valence Electrons
Electron Dot
H
64.
1 + 7 + 2(6) = 20 e–
Polyatomic Ion Valence Electrons
O
Electron Dot
H
HIO2
I
I
O
66.
5 + 4(1) – 1 = 8 e–
Polyatomic Ion Valence Electrons
5 + 4(6) + 3 = 32 e–
I
O
Coord. Cov. Bond
H
I
O
O
Electron Dot
Coord. Cov. Bond
H P H
H
H
+
Electron Dot
O
3–
O P O
PO4 3 –
H
O
H
PH 4 +
Coord. Cov. Bond
O
2014 © Pearson Education, Inc.
H P H
+
H
Coord. Cov. Bond
O
O P O
3–
O
Chemical Bonding
95
Section 12.9 Hydrogen Bonds
68. H–F: – – – H–F:
70. A hydrogen bond is ~50% longer than a polar covalent bond.
Section 12.10 Shapes of Molecules
72.
Formula
CBr4
Electron Pair
tetrahedral
Molecular Shape
tetrahedral
Bond Angle
109.5°
74.
Formula
PH 3
Electron Pair
tetrahedral
Molecular Shape
trigonal pyramidal
Bond Angle
107°
76.
Formula
Cl2 O
Electron Pair
tetrahedral
Molecular Shape
bent
Bond Angle
104.5°
78. In CF 4 each of the four C–F bonds is polar, but the four fluorine atoms pull equally
in opposite directions to give a nonpolar molecule. The four fluorine atoms are at
the corners of a tetrahedron, and the molecular shape is described as tetrahedral.
General Exercises
80.
82.
84.
96
(a)
(b)
(c)
(d)
Substance
Pu
O2
Pu2 O3
H2 O2
Particle
atom
molecule
formula unit
molecule
(a)
(b)
(c)
(d)
Ions
1 Sc3+ and 1 N3 –
1 Ti4+ and 2 O2 –
2 NH4 + and 1 CO3 2 –
3 Hg2 2+ and 2 PO4 3 –
Chemical Formula
ScN
TiO2
(NH4 )2 CO3
(Hg2 )3 (PO4 )2
A chloride ion has one more electron than a chlorine atom. As a result, the
negative electrons in the 3p subenergy level repel each other, and the ion
becomes larger.
Chapter 12
2014 © Pearson Education, Inc.
86.
Bond
Sb—Cl
Polarity
3.0 – 1.9 = 1.1 (polar)
88.
Bond
S—I
Polarity
2.5 – 2.5 = 0 (nonpolar)
90.
Polar Bond and Delta Notation
δ+
92.
Molecule
Electron Dot
Valence Electrons
As—Cl δ–
H Sb H
SbH3
94.
5 + 3(1) = 8 e–
Polyatomic Ion Valence Electrons
4 + 3(6) + 2 = 24 e–
H
H
Electron Dot
O Si O 2–
SiO3 2 –
Structural Formula
H Sb H
Structural Formula
[ O Si
O ] 2–
O
O
96.
Formula
SiO2
Electron Pair
linear
Molecular Shape
linear
Bond Angle
180°
98.
Formula
Electron Pair
Molecular Shape
Bond Angle
CO 3 2 –
trigonal planar
trigonal planar
120°
Electron Dot
Structural Formula
O Xe O
Challenge Exercises
100.
Molecule
Valence Electrons
XeO3
8 + 3(6) = 26 e–
O Xe O
O
2014 © Pearson Education, Inc.
O
Chemical Bonding
97
Online Exercises
102.
In addition to water and ammonia, hydrogen fluoride, HF, can also form strong
hydrogen bonds between molecules. A hydrogen atom in one HF molecule can
bond to a nonbonding electron pair in another HF molecule.
104.
The two strands of a DNA molecule that form the double helix are held together
by two hydrogen bonds between the DNA bases adenine(A) and thymine(T).
98
Chapter 12
2014 © Pearson Education, Inc.
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