Chapter 8
Bonding: General
Concepts
Section 8.1
Types of Chemical Bonds
A Chemical Bond
• Forces that hold groups of atoms together and
make them function as a unit.
• A bond will form if the energy of the molecule is
lower than that of the separated atoms.
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Section 8.1
Types of Chemical Bonds
The Interaction of
Two Hydrogen
Atoms
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Section 8.1
Types of Chemical Bonds
The Interaction of Two Hydrogen Atoms
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Section 8.1
Types of Chemical Bonds
Key Ideas in Bonding
• Ionic Bonding – electrons are transferred
• Covalent Bonding – electrons are shared
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Section 8.1
Types of Chemical Bonds
Polar Covalent Bond
• Unequal sharing of electrons between atoms in
a molecule.
• Results in a charge separation in the bond
(partial positive and partial negative charge).
Nonpolar Covalent Bond
• Equal sharing of electrons
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Section 8.1
Types of Chemical Bonds
The Effect of an Electric Field on Hydrogen Fluoride Molecules
 or  indicates a positive or negative fractional charge.
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Section 8.1
Types of Chemical Bonds
Polar Molecules
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Section 8.2
Electronegativity
• The ability of an atom in a molecule to attract
shared electrons to itself.
• On the periodic table, electronegativity generally
increases across a period and decreases down
a group.
• The range of electronegativity values is from 4.0
for fluorine (the most electronegative) to 0.7 for
cesium and francium (the least electronegative).
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Section 8.2
Electronegativity
The Pauling Electronegativity Values
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Section 8.2
Electronegativity
Nonpolar covalent polar covalent, and ionic
Based on electronegativity difference
Difference 0 - 0.4
Nonpolar Covalent
0.5 – 1.6 Polar Covalent
≥ 1.7
Ionic
Using electronegativity values, classify as nonpolar, polar,
or ionic.
P-Cl
Polar
Mg-Cl
Ionic
Cl-Cl nonpolar
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Section 8.2
Electronegativity
Concept Check
If lithium and fluorine react, which has more
attraction for an electron? Why?
In a bond between fluorine and iodine, which
has more attraction for an electron? Why?
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Section 8.2
Electronegativity
The Relationship Between Electronegativity and Bond Type
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Section 8.2
Electronegativity
Concept Check
What is the general trend for electronegativity
across rows and down columns on the
periodic table?
Explain the trend.
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Section 8.2
Electronegativity
Exercise
Arrange the following bonds from most to least
polar:
a) N–F
b) C–F
c) Cl–Cl
O–F
N–O
B–Cl
C–F
Si–F
S–Cl
a)
b)
c)
N–F,
C–F,
S–Cl,
O–F
N–O
Cl–Cl
C–F,
Si–F,
B–Cl,
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Section 8.2
Electronegativity
Concept Check
Which of the following bonds would be the
least polar yet still be considered polar
covalent?
Mg–O C–O O–O Si–O N–O
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Section 8.2
Electronegativity
Concept Check
Which of the following bonds would be the
most polar without being considered ionic?
Mg–O C–O O–O Si–O N–O
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Section 8.3
Bond Polarity and Dipole Moments
Dipole Moment
• Property of a molecule whose charge
distribution can be represented by a center of
positive charge and a center of negative charge.
(polar molecules)
• Use an arrow to represent a dipole moment.
 Point to the element with the higher
electronegativity
 Nonpolar molecules have zero dipole moment
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Section 8.3
Bond Polarity and Dipole Moments
Dipole Moment
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Section 8.3
Bond Polarity and Dipole Moments
No Net Dipole Moment (Dipoles Cancel)
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Section 8.5
Energy Effects in Binary Ionic Compounds
Ionic Compounds
•
•
•
•
Crystalline solids
High melting points
Brittle
Conduct electricity in molten form or when dissolved in
water
• Form a crystal lattice
• Metal + nonmetal or metal + polyatomic ion
• Examples: KBr, Na2SO4
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Section 8.5
Energy Effects in Binary Ionic Compounds
Crystal Lattice
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Section 8.5
Energy Effects in Binary Ionic Compounds
• What are the factors that influence the stability
and the structures of solid binary ionic
compounds?
• How strongly the ions attract each other in the
solid state is indicated by the lattice energy.
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Section 8.5
Energy Effects in Binary Ionic Compounds
Lattice Energy
• The change in energy that takes place when
separated gaseous ions are packed together to
form an ionic solid.
 Q1Q2 
Lattice energy = k 

 r 
k = proportionality constant
Q1 and Q2 = charges on the ions
r = shortest distance between the centers of the
cations and anions
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Section 8.5
Energy Effects in Binary Ionic Compounds
Lattice energy (and melting point) increases as charges
increase and r (distance) decreases
Which compound would have the higher lattice energy
(higher melting point)?
1. NaF, MgO
MgO
2. MgO, MgS
MgO
3. CaS, KCl
CaS
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Section 8.5
Energy Effects in Binary Ionic Compounds
Arrange the following ionic compounds in order of
increasing melting point.
MgO, NaBr, NaCl
NaBr, NaCl, MgO
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Section 8.5
Energy Effects in Binary Ionic Compounds
Lattice Energy
• The amount of energy needed for 1 mole of an ionic
compound to be separated into ions
NaCl (s) → Na+(g) + Cl-(g)
endothermic
• When ions combine to form the ionic compound it is
exothermic
• Explain the differences in lattice energies
Compound
Lattice Energy (kJ/mol)
NaCl
788
NaBr
736
NaI
686
MgCl2
2527
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Section 8.5
Energy Effects in Binary Ionic Compounds
Formation of an Ionic Solid
1.Sublimation of the solid metal.
• M(s)  M(g) [endothermic]
2. Ionization of the metal atoms.
• M(g)  M+(g) + e [endothermic]
3. Dissociation of the nonmetal.
• 1/2X2(g)  X(g) [endothermic]
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Section 8.5
Energy Effects in Binary Ionic Compounds
Formation of an Ionic Solid (continued)
4. Formation of X ions in the gas phase.
• X(g) + e  X(g) [exothermic]
5. Formation of the solid MX. (Lattice energy
negative value.)
• M+(g) + X(g)  MX(s)
[quite exothermic]
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Section 8.5
Energy Effects in Binary Ionic Compounds
Example
Calculate the energy change for;
Li(s) + ½ F2(g) → LiF(s)
1. Sublimation of Li
Li(s) → Li(g)
Enthalpy of sublimation = 161 kJ/mol
2. Ionization of Li(g)
Li(g) → Li+(g) + eIonization Energy = 520 kJ/mol
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Section 8.5
Energy Effects in Binary Ionic Compounds
Example
3. Dissociation of F2
F2 → 2F
½ F2 → F
Bond Energy F2 = 154 kJ/mol
½ (154) = 77 kJ
4. Forming F- ions
F + e- → FElectron Affinity = -328 kJ/mol
5. Forming solid LiF from ions
Li+ + F-→ LiF
Lattice Energy = -1047 kJ/mol
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Section 8.5
Energy Effects in Binary Ionic Compounds
Example
Hess’s Law: add all steps together
Li(s) → Li(g)
Li(g) → Li+(g) + e1/2F2(g) → F(g)
F(g) + e- → F-(g)
Li+(g) + F-(g) → LiF(s)
Li(s) + ½ F2(g) → LiF(s)
161 kJ
520 kJ
77 kJ
-328 kJ
-1047 kJ
-617 kJ
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Section 8.8
Covalent Bond Energies and Chemical Reactions
Bond Energies
• To break bonds, energy must be added to the
system (endothermic).
• To form bonds, energy is released (exothermic).
H = (bonds broken) – (bonds formed)
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Section 8.8
Covalent Bond Energies and Chemical Reactions
Exercise
Predict H for the following reaction:
CH3N  C(g )  CH3C  N(g )
Given the following information:
Bond Energy (kJ/mol)
C–H
413
C–N
305
C–C
347
891
CN
H = –42 kJ
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Section 8.8
Covalent Bond Energies and Chemical Reactions
Examples
Using the table of bond energies on pg 372, calculate ∆H
for:
. N2 + 2H2 → N2H4
81 kJ
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Section 8.10
Lewis Structures
Lewis Structure
• Shows how valence electrons are arranged
among atoms in a molecule.
• Reflects central idea that stability of a compound
relates to noble gas electron configuration.
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Section 8.10
Lewis Structures
Duet Rule
• Hydrogen forms stable molecules where it shares two
electrons.
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Section 8.10
Lewis Structures
Octet Rule
• Elements form stable molecules when
surrounded by eight electrons.
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Section 8.10
Lewis Structures
Concept Check
Draw a Lewis structure for each of the
following molecules:
H2
F2
O2
N2
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Section 8.10
Lewis Structures
Draw the Lewis Structures for:
H2O, PBr3, and HCN
Br
H O H
Br
P Br
H C N
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Section 8.10
Lewis Structures
Steps for Writing Lewis Structures
1. Count the # of electrons needed for all atoms to
obey octet rule. (Need)
2. Count the total # of valence electrons. (Have)
3. Subtract and divide by 2. This equals the # of
bonds in the molecule.
4. Make sure every atom has 8 electrons. (Except
H)
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Section 8.10
Lewis Structures
Concept Check
Draw a Lewis structure for each of the
following molecules:
NH3
CO2
CO32N2O4
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Section 8.11
Exceptions to the Octet Rule
• H: 2 e• Be: 4 e• B: 6 e• EX: BH3
H
H B H
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Section 8.11
Exceptions to the Octet Rule
Expanded Octet
More than 8e- around the central atom
Examples: PCl5
SF6
XeF4
Have and Need math formula will not work for the
expanded octet!
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Section 8.11
Exceptions to the Octet Rule
• When it is necessary to exceed the octet rule for
one of several third-row (or higher) elements,
place the extra electrons on the central atom.
SF4 = 34e–
AsBr5 = 40e–
F
F S
F
F
Br Br
Br As Br
Br
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Section 8.11
Exceptions to the Octet Rule
Concept Check
Draw a Lewis structure for each of the
following molecules:
BF3
PCl5
XeF4
BrF5
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Section 8.12
Resonance
• More than one valid Lewis structure can be
written for a particular molecule.
NO3– = 24e–
O
O
N
O

O
O
O
N
O

O
N
O
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Section 8.12
Resonance
• Actual structure is an average of the resonance
structures.
• Electrons are really delocalized – they can move
around the entire molecule.
O
O
N
O

O
O
O
N
O

O
N
O
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Section 8.12
Resonance
Concept Check
Draw a Lewis structure for each of the
following molecules:
CO
CH3OH
CO2
OCN–
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Section 8.12
Resonance
Formal Charge
• Used to evaluate nonequivalent Lewis
structures.
• Atoms in molecules try to achieve formal
charges as close to zero as possible.
• Any negative formal charges are expected to
reside on the most electronegative atoms.
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Section 8.12
Resonance
Formal Charge
Formal charge (FC)
FC = (# valence e–) – (# bonds + # unshared e-).
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Section 8.12
Resonance
Concept Check
Consider the Lewis structure for POCl3.
Assign the formal charge for each atom in the
molecule.
Cl
P: 5 – (4+0) = +1
O: 6 – (1+6) = –1
Cl: 7 – (1+6) = 0
Cl P O
Cl
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Section 8.12
Resonance
Rules Governing Formal Charge
• The sum of the formal charges of all atoms in a
given molecule or ion must equal the overall
charge on that species.
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Section 8.12
Resonance
Rules Governing Formal Charge
• If more than 1 Lewis structures exist for a
species, those with formal charges closest to
zero and with any negative formal charges on
the most electronegative atoms are considered
the best structure.
O C O
O C O
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Section 8.13
Molecular Structure: The VSEPR Model
Sigma and Pi Bonds
• Single bonds are composed of 1 sigma bond.
• Double bonds are composed of 1 sigma and 1 pi bond.
• Triple bonds are composed of 1 sigma and 2 pi bonds.
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Section 8.13
Molecular Structure: The VSEPR Model
Examples
Give the number of sigma and pi bonds in the following:
1. HCN
2 sigma, 2 pi
2. C2H4
5 sigma, 1 pi
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Section 8.13
Molecular Structure: The VSEPR Model
VSEPR Model
• VSEPR: Valence Shell Electron-Pair Repulsion.
• Bonds repel each other
• Lone pairs repel more than bonds
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Section 8.13
Molecular Structure: The VSEPR Model
VSEPR
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Section 8.13
Molecular Structure: The VSEPR Model
VSEPR: Two Electron Pairs
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Section 8.13
Molecular Structure: The VSEPR Model
VSEPR: Three Electron Pairs
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Section 8.13
Molecular Structure: The VSEPR Model
VSEPR: Four Electron Pairs
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Section 8.13
Molecular Structure: The VSEPR Model
VSEPR: Iodine Pentafluoride
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Section 8.13
Molecular Structure: The VSEPR Model
Arrangements of Electron Pairs Around an Atom Yielding
Minimum Repulsion
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Section 8.13
Molecular Structure: The VSEPR Model
Arrangements of Electron Pairs Around an Atom Yielding
Minimum Repulsion
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Section 8.13
Molecular Structure: The VSEPR Model
Structures of Molecules That
Have Four Electron Pairs
Around the Central Atom
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Section 8.13
Molecular Structure: The VSEPR Model
Structures of Molecules
with Five Electron Pairs
Around the Central Atom
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Section 8.13
Molecular Structure: The VSEPR Model
Concept Check
Determine the shape for each of the following
molecules, and include bond angles:
HCN
PH3
SF4
HCN – linear, 180o
PH3 – trigonal pyramid, 109.5o
SF4 – see saw, 90o, 120o
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Section 8.13
Molecular Structure: The VSEPR Model
Concept Check
Determine the shape for each of the following
molecules, and include bond angles:
O3
XeF4
O3 – bent, 120o
XeF4 – square planar, 90o, 180o
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Section 8.13
Molecular Structure: The VSEPR Model
Nonpolar Molecules
Equal electron distribution, symmetrical around central
atom
Examples: CCl4, CO2
Must have the same atom around the central atom to be
nonpolar.
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Section 8.13
Molecular Structure: The VSEPR Model
Polar Molecules
Unequal electron distribution, not symmetrical around the
central atom
Examples: H2O, NH3
Polar molecules have a dipole moment, nonpolar: zero
dipole moment.
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Section 8.13
Molecular Structure: The VSEPR Model
Concept Check
True or false:
A molecule that has polar bonds will always
be polar.
-If true, explain why.
-If false, provide a counter-example.
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Section 8.13
Molecular Structure: The VSEPR Model
Let’s Think About It
• Draw the Lewis structure for CO2.
• Does CO2 contain polar bonds?
• Is the molecule polar or nonpolar overall? Why?
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Section 8.13
Molecular Structure: The VSEPR Model
Concept Check
True or false:
Lone pairs make a molecule polar.
-If true, explain why.
-If false, provide a counter-example.
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Section 8.13
Molecular Structure: The VSEPR Model
Let’s Think About It
• Draw the Lewis structure for XeF4.
• Does XeF4 contain lone pairs?
• Is the molecule polar or nonpolar overall? Why?
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