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Chapter 6
Objectives
• Describe the electron-sea model of metallic bonding, and
explain how the metallic bond accounts for the characteristics of
metallic substances.
• List and describe the properties of metals.
• Explain why metals are malleable and ductile but ioniccrystalline compound are not.
• List and describe the types of Van der Waals Forces. Describe
dipole-dipole forces, hydrogen bonding, induced dipoles, and
London dispersion forces and their effects on properties such as
boiling and melting points.
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Chapter 6
Objectives
• Explain VSEPR theory.
• Predict/Explain the shapes of molecules using
VSEPR theory.
• Explain how the shapes of molecules are accounted
for by hybridization theory.
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•Explain why scientists use resonance structures to represent some
molecules.
•Explain the relationships among potential energy, distance between
approaching atoms, bond length, bond stability, and bond energy.
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Chapter 6
Section 4 Metallic Bonding
Metallic Bonding
• Chemical bonding is different in metals than it is in ionic,
molecular, or covalent-network compounds (network solids).
• The unique characteristics of metallic bonding gives metals their
characteristic properties, as such
• electrical conductivity
• thermal conductivity
• malleability
• ductility
• Luster
• sectility
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Chapter 6
Section 4 Metallic Bonding
The Metallic-Bond Model
• In a metal, the vacant orbitals in the atoms’ outer energy levels
overlap.
• This overlapping of orbitals allows the outer electrons of the
atoms to roam freely throughout the entire metal (mobile valence
electrons).
• The electrons are delocalized, which means that they do not
belong to any one atom but move freely about the metal’s network
of empty atomic orbitals.
• These mobile electrons form a sea of electrons around the metal
atoms, which are packed together in a crystal lattice.
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Chapter 6
Section 4 Metallic Bonding
The Metallic-Bond Model, continued
• The chemical bonding that results from the attraction
between metal atoms kernel (which are positively charged) and
the surrounding sea of shared valence electrons(which are
negatively charged) is called metallic bonding.
• When metal atoms bond to other metal atoms, their valence
electrons do not seem to belong to any individual atom but rather
they are “group shared” and are free to roam from atom to atom
throughout the entire metal crystal.
• The strength of the binding forces between the “sea of shared
valence” and the kernels of the metal atoms directly impacts
properties such as ductility, malleability , and sectility. The
stronger the binding action, the harder the metal sample will be.
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Chapter 6
Visual Concepts
Metallic Bonding
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Chapter 6
Visual Concepts
Properties of Metals: Surface Appearance
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Chapter 6
Visual Concepts
Properties of Metals: Malleability and Ductility
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Chapter 6
Visual Concepts
Properties of Metals: Electrical and Thermal
Conductivity
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Chapter 6
Section 5 Van der Waals
Intermolecular Forces
• The forces of attraction between molecules are known as
intermolecular forces ,aka, the Van der Waals Forces .
• The boiling point of a liquid is a good measure of the
intermolecular forces between its molecules: the higher the
boiling point, the stronger the forces between the molecules.
• Intermolecular forces vary in strength but are weaker than
bonds between atoms within molecules, ions in ionic
compounds, or metal atoms in solid metals.
• Boiling points for ionic compounds and metals tend to be much
higher than those for molecular substances: forces between
molecules are weaker than those between metal atoms or ions.
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Chapter 6
Section 5 Van der Waals
Categories of Intermolecular Forces (IMFs)
• The forces of attraction between molecules are known as
intermolecular forces ,aka, the Van der Waals Forces .
Hydrogen Bonds- formed between polar molecules that have
hydrogen bonded directly to fluorine, oxygen, or
nitrogen in the molecule.
Dipole –Dipole Interactive Forces- formed between polar
molecules that do not have
hydrogen bonding ability.
London Dispersion Forces- formed between nonpolar molecules
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Chapter 6
Section 5 Van der Waals
Intermolecular Forces, continued
• The strongest intermolecular forces exist between
polar molecules capable of hydrogen bonding.
• Because of their uneven charge distribution, polar
molecules have dipoles. A dipole is created by equal
but opposite charges that are separated by a short
distance (i.e. an uneven distribution of charge).
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Chapter 6
Van der Waals
Intermolecular Forces, continued
• A dipole is represented by an arrow with its head
pointing toward the negative pole and a crossed tail at
the positive pole. The dipole created by a hydrogen
chloride molecule is indicated as follows:
d+
d-
H Cl
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Chapter 6
Section 5 Van der Waals
Hydrogen Bonding
• Some hydrogen-containing compounds have
unusually high boiling points. This is explained by a
particularly strong type of dipole-dipole force.
• In compounds containing H–F, H–O, or H–N bonds,
the large electronegativity differences between
hydrogen atoms and the atoms they are bonded to
make their bonds highly polar.
• This gives the hydrogen atom a positive charge that is
almost half as large as that of a bare proton.
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Chapter 6
Section 5 Van der Waals
Hydrogen Bonding
• The small size of the hydrogen atom allows the atom
to come very close to an unshared pair of electrons in
an adjacent molecule.
• The intermolecular force in which a hydrogen atom
that is bonded to a highly electronegative atom is
attracted to an unshared pair of electrons of an
electronegative atom in a nearby molecule is known
as hydrogen bonding.
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Chapter 6
Section 5 Van der Waals
Hydrogen Bonding
• Hydrogen bonds are represented by dotted lines
connecting the hydrogen-bonded hydrogen to the
unshared electron pair of the electronegative atom to
which it is attracted.
• An excellent example of hydrogen bonding is that
which occurs between water molecules. The strong
hydrogen bonding between water molecules accounts
for many of water’s characteristic properties.
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Chapter 6
Visual Concepts
Hydrogen Bonding
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Chapter 6
Section 5 Van der Waals
Impacts of Hydrogen Bonding
• molecules capable of Hydrogen Bonding generally
have….
• Higher boiling points
• Higher melting points
• Decreased vapor pressures
• Higher heat of vaporization
• Lower density in solid state than in liquid
state
compared to other molecules.
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Chapter 6
Section 5 Molecular Geometry
Dipole-Dipole Interactive Forces
• The negative region in one polar molecule attracts the positive
region in adjacent molecules. So the molecules all attract each
other from opposite sides.
• Such forces of attraction between polar molecules are known as
dipole-dipole forces.
• Dipole-dipole forces , like in all IMFs, act at short range, only
between nearby molecules.
• In general, dipole-dipole forces are weaker than hydrogen bonds
but stronger than London bonds.
• Dipole-dipole forces explain, for example the difference between
the boiling points of iodine chloride, I–Cl (97°C), and bromine,
Br–Br (59°C).
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Chapter 6
Visual Concepts
Dipole-Dipole Forces
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Chapter 6
Section 5 Van der Waals
Comparing Dipole-Dipole Forces
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Chapter 6
Section 5 Van der Waals
Induced Dipoles
• A polar molecule can induce a dipole in a nonpolar
molecule by temporarily attracting its electrons.
• The result is a short-range intermolecular force that is
somewhat weaker than the dipole-dipole force.
• Induced dipoles account for the fact that a
nonpolar molecule, oxygen, O2, is able to dissolve
in water, a polar molecule.
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Chapter 6
Visual Concepts
Dipole-Induced Dipole Interaction
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Chapter 6
Section 5 Van der Waals
London Dispersion Forces
• Even noble gas atoms and nonpolar molecules can
experience weak intermolecular attraction.
• In any atom or molecule—polar or nonpolar—the
electrons are in continuous motion.
• As a result, at any instant the electron distribution may be
uneven. A momentary uneven charge can create a positive pole
at one end of an atom of molecule and a negative pole at the
other. This phenomenon is known as a momentary or temporary
dipole.
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Chapter 6
Section 5 Van der Waals
London Dispersion Forces, continued
• This temporary dipole can then induce a dipole in an adjacent
atom or molecule. The two are held together for an instant by the
weak attraction between temporary dipoles.
• The intermolecular attractions resulting from the constant motion
of electrons and the creation of momentary dipoles are called
London dispersion forces.
• London Bonds are the weakest of the IMFs and all bonds.
• Fritz London first proposed their existence in 1930.
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Chapter 6
Visual Concepts
London Dispersion Force
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Chapter 6
Section 5 Van der Waals
Comparing Strengths of IMFs
• Hydrogen bonds are the strongest of the IMFs.
• The strength of the hydrogen bond increases as the
electronegativity of the atom to which hydrogen is covalently
bonded to in a molecule increases.
• The strength of the hydrogen bond decreases as the size of the
atom to which hydrogen is covalently bonded increases.
• Dipole –dipole forces are, in general, weaker than hydrogen
bonds but stronger than London Bonds.
• London Bonds are the weakest of the IMFs and all bonds.
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Chapter 6
Section 5 Van der Waals
Comparing Strengths of IMFs
• In a group of related molecules ( polar or nonpolar)
the strength of the Van der Waals forces increases as
molecular mass increases.
• This is due to more electrons being available to
increase the intensity of any dipoles present in a
molecule.
• The strength of any Van der Waals Force increases
as the distance between bonding molecules
decreases.
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Chapter 6
Section 5 Molecular Geometry
Molecular Geometry
• The properties or behaviors of molecules depend not only
on the bonding of atoms but also on molecular geometry:
the three-dimensional arrangement of a molecule’s atoms.
• The polarity of each bond, along with the geometry of the
molecule, determines molecular polarity, or the uneven
distribution of charges due to molecular shape.
• Molecular polarity strongly influences the forces that act
between molecules in liquids and solids.
• A chemical formula, by itself, reveals little information about
a molecule’s geometry.
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Chapter 6
Section 5 Molecular Geometry
Molecular Geometry
2 methods are used to make determinations of
molecular geometry
VSEPR Theory
Central atom bonding character (Hybridization
Theory)
Either of these methods can be used, in conjunction
or independently, to correctly determine a
molecule’s 3-dimensional shape.
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Chapter 6
Section 5 Molecular Geometry
VSEPR Theory
• As shown at right, diatomic
molecules, like those of
(a) hydrogen, H2, and
(b) hydrogen chloride, HCl,
can only be linear because
they consist of only two
atoms.
• To predict the geometries of more-complicated
molecules, one must consider the locations of all
electron pairs surrounding the bonding atoms. This
is the basis of VSEPR theory.
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Chapter 6
Section 5 Molecular Geometry
VSEPR Theory
• The abbreviation VSEPR (say it “VES-pur”) stands for
“valence-shell electron-pair repulsion.”
• VSEPR theory states that repulsion between the sets of valencelevel electrons surrounding a “central” atom causes these sets to
be oriented as far apart as possible. (Repelled to maximum
distance)
• example: BeH2
• The central beryllium atom is surrounded by only the two
electron pairs it shares with the hydrogen atoms.
• According to VSEPR, the shared pairs will be as far away from
each other as possible, so the bonds to hydrogen will be 180°
apart from each other.
• The molecule will therefore be linear:
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Chapter 6
Section 5 Molecular Geometry
VSEPR Theory
VSEPR “combos” ( # of Atoms bonded to central atom / # of lone
pairs on central Atom)
•
•
•
•
•
•
•
•
2/0 = linear
2/1= bent or angular
2/2= bent or angular
3/0= trigonal planar (triangular)
3/1= trigonal pyramidal (pyramid)
4/0= tetrahedral
5/0= trigonal bipyramidal ******
6/0= octahedral ******
*****Note- These geometries involve the expanded valence shell
( expanded “octet” ).
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Chapter 6
Visual Concepts
Lone Pair of Electrons
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Chapter 6
Visual Concepts
VSEPR and Basic Molecular Shapes
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Chapter 6
Visual Concepts
VSEPR and Lone Electron Pairs
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Chapter 6
Section 5 Molecular Geometry
VSEPR Theory, continued
Sample Problem E
Use VSEPR theory to predict the molecular geometry of
boron trichloride, BCl3.
Cl
Cl
B
Cl
Boron trichloride has a 3/0 VSEPR combo.
Its geometry therefore is trigonal planar.
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Chapter 6
Section 5 Molecular Geometry
VSEPR Theory, continued
• VSEPR theory can also account for the geometries of
molecules with unshared electron pairs.
• examples: ammonia, NH3, and water, H2O.
• The Lewis structure of ammonia shows that the
central nitrogen atom has an unshared electron pair:
HNH
H
• Ammonia’s VSEPR combo is 3/1 and therefore its
geometry is trigonal pyramidal.
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Chapter 6
Section 5 Molecular Geometry
VSEPR Theory, continued
• The shape of a molecule refers to the positions of
atoms only (lone pairs are NOT included in the
geometry).
•H2O has a 2/2 VSEPR combo, and therefore its
molecular geometry is “bent,” or angular.
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Chapter 6
Section 5 Molecular Geometry
VSEPR Theory, continued
Sample Problem F
Use VSEPR theory to predict the shape of a molecule of
carbon dioxide, CO2.
O
C
O
The VSEPR combo is 2/0 therefore the geometry is
linear.
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Chapter 6
Molecular Polarity
Molecular Polarity
In terms of their overall electron distribution (negative charge),
molecules are considered to be either polar (uneven distribution)
or nonpolar ( even distribution).
Diatomic Molecules (molecules consisting of only two atoms)
In any diatomic molecule , the polarity of the bond between the two
atoms will determine the polarity of the molecule.
If the bond is polar, the molecule will be polar.
If the bond is nonpolar, the molecule will be nonpolar.
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Chapter 6
Molecular Polarity
Molecular Polarity
Molecules consisting of more than 2 atoms
In these molecules, the type(s) of covalent bonds within the molecule
along with the molecular geometry (in most cases) must be taken
into consideration in determining the polarity of the molecule.
If all the bonds within a molecule are nonpolar, the molecule will
be nonpolar. (Molecular shape plays no role in determining
polarity.)
If only one of the bonds in a molecule is polar, the molecule will
be polar. (Molecular shape plays no role in determining
polarity.)
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Chapter 6
Molecular Polarity
Molecular Polarity
When 2 or more of the bonds in a molecule are polar, the molecule’s
geometry or shape plays a major role in determining the molecule’s
polarity.
Symmetrical Geometries that generally lead to nonpolarity in
molecules:
Linear, trigonal planar, tetrahedral, trigonal bipyramidal
octahedral.
These shapes allow the dipoles in a molecule to cancel each other
out as long as the molecule retains its symmetry.
***Note- If all of the bonds in the molecule are not exactly the same
then the shape will become distorted and no longer allow polar
bonds to cancel resulting in the molecule being polar.
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Chapter 6
Molecular Polarity
Molecular Polarity
Asymmetrical Geometries that generally lead to polarity in molecules:
Bent/Angular, trigonal pyramidal .
These shapes will NOT allow any dipoles in a molecule to cancel
each other.
***Note- Distortion effects are inconsequential as these shapes are
already inherently asymmetrical.
REMEMBER molecular polarity must originate with polar
bonds!!!!!
In other words, NO polar bonds in a molecule = nonpolar molecule
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Chapter 6
Hybridization
• Hybridization Theory is used to explain how the
valence shell orbitals of an atom are rearranged when
some atoms form covalent bonds.
• Hybridization involves the blending of two or more
valence shell orbitals of similar energies to produce
new hybrid atomic orbitals of equal energies.
• Hybridization involves two events: promotion followed
immediately by hybridization.
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Chapter 6
Hybridization (cont.)
Promotion involves the “unpairing “ of any valence shell orbitals
that contain an orbital pairing as one of the electrons is sent to an
empty orbital in the same valence shell.
Hybridization immediately follows promotion as the valence shell
orbitals are rearranged (blended) to create new equal energy
hybrid orbitals.
The new hybrid orbitals are named after the types and numbers of
valence shell orbitals that were directly involved in the
hybridization.
ex. Carbon is sp3 hybridized in covalent bondings.
One s orbital and three p orbitals are involved
in creating the four equal energy sp3 hybrid
orbitals.
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Chapter 6
Central Atom “Bonding Character”
The bonding character of any central atom in a molecule can be
used to predict molecular geometries of molecules.
This method of predicting the shape of molecules involves
describing the kinds of orbitals directly involved in the
formation of covalent bonds. Because many of the central
atoms involve hybrid orbitals in covalent bond formation, it is
also known as hybridization theory.
s bonding= linear ex: hydrogen
p bonding= linear ex: group 17 elements
***Please note that these are not playing role of central atom.
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Chapter 6
Central Atom “Bonding Character”
p2 bonding= bent/angular
ex: group 16 nonmetals
p3 bonding, bonding 2 atoms = bent/angular ex: group 15
nonmetals
p3 bonding, bonding 3 atoms = trigonal pyramidal ex: group 15
nonmetals
***Please note that these are not involving hybridized central
atoms.
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Chapter 6
Central Atom “Bonding Character”
sp bonding= linear
ex: Beryllium
sp2 bonding, bonding 2 atoms = linear ex: Boron
sp2 bonding, bonding 3 atoms = trigonal planar ex: Boron
sp3 bonding, bonding 2 atoms = linear ex: group 14 nonmetals
sp3 bonding, bonding 3 atoms = trigonal planar ex: group 14
nonmetals
sp3 bonding, bonding 4 atoms = tetrahedral ex: group 14
nonmetals
***Please note that these are involving hybridized central atoms.
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Chapter 6
Central Atom “Bonding Character”
sp3d bonding, bonding 5 atoms = trigonal bipyramidal ex:
select group 15 nonmetals
sp3d2 bonding, bonding 6 atoms = octahedral ex: select group
16 nonmetals
***Please note that these are involving hybridized central atoms
with an expanded valence shell.
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Chapter 6
Section 5 Molecular Geometry
Geometry of
Hybrid
Orbitals
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Chapter 6
Visual Concepts
Hybrid Orbitals
sp3d bonding, bonding 5 atoms =
trigonal bipyramidal ex: select
group 15 nonmetals
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Chapter 6
The Octet Rule
Exceptions to the Octet Rule
• The octet rule can not be used to explain the bonding in all
molecules. Some atoms are able to achieve stability by having
fewer than 8 valence electrons or more than 8 valence electrons .
• Hydrogen forms bonds in which it is surrounded by only
two electrons.
• Beryllium forms bonds in which it is surrounded by only
four electrons. (Normally forms ionic bonds with
nonmetals but forms covalent bonds with hydrogen.)
• Boron forms bonds in which it is surrounded by only six
electrons.
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Chapter 6
The Octet Rule
Exceptions to the Octet Rule
• Select elements in Groups 15, 16, & 17 and up can form
bonds with expanded valence, involving more than eight
electrons.
Resonance
Resonance Theory is used to explain the bonding in
molecules that exist but the bonding cant be explained
by normal means.
Resonance theory is used to explain bonding under the
following conditions:
1) Molecules with an odd number of valence electrons.
2) Molecules in which experimental evidence about the
covalent bonds making the molecule conflicts with a
reasonable Lewis diagram.
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Chapter 6
Visual Concepts
Atomic Resonance
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Chapter 6
Bond Properties
Chemical bonds, like matter, have measureable
characteristics or properties such as bond energy, bond
length (distance), bond strength, or bond stability.
Bond Energy – the amount of energy released by atoms
when they form a chemical bond with one another.
Bond energy can also be viewed as the amount of energy
required to break a chemical bond.
Bond energy can be related to other bond properties such
as bond length, bond strength, and bond stability.
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Chapter 6
Bond Properties
Bond Length (distance)- the actual distance between the
nuclei of any two bonded atoms
Bond strength- a relative measure of how a bond resists
being broken.
Bond stability- a relative measure of how a bond resists
chemical change (e.g. a measure of chemical reactivity)
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Chapter 6
Bond Properties
Bond energy vs. bond length (inversely related)
As bond energy increases, bond length decreases.
As bond energy decreases, bond length increases.
Bond energy vs. bond strength (directly related)
As bond energy increases, bond strength increases.
As bond energy decreases, bond strength decreases.
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Chapter 6
Bond Properties
Bond energy vs. bond stability (directly related)
As bond energy increases, bond stability increases.
As bond energy decreases, bond stability decreases.
****Please note that these statements only hold true for
comparisons of single bonds to one another
Multiple bonds must be considered on a case by case
scenario as each of the electron pairs form at different
bond energies.
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Chapter 6
Bond Properties
Bond energy and Chemical Change
In exergonic (exothermic ) reactions energy is released as
the reactants decrease their potential energy to form more
stable products. The products will be more stable and
have higher bond energies than the reactants from which
they were formed.
In endergonic (endothermic ) reactions energy must be
absorbed as the reactants increase their potential energy
to form less stable products. The products will be less
stable and have lower bond energies than the reactants
from which they were formed.
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Chapter 6
Visual Concepts
Bond Energy
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Chapter 6
Properties of Bonds
Bond Energies and Bond Lengths for Single
Bonds
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Chapter 6
Visual Concepts
Bond Length
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Chapter 6
Properties of Bonds
Bond Length and Stability
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Chapter 6
Network Solids
Network solids are also known as covalent solids or
covalent crystals.
These substances are not composed of separate, distinct
molecules. Instead ,they appear to be a single, giant
molecule in which covalent bonds extend from one atom
to another in a continuous network pattern throughout the
entire substance.
There is no involvement of Van der Waals Forces for
these molecules.
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Chapter 6
Network Solids
Properties of Network Solids
Extremely high melting and boiling points.
Poor heat and electrical conductors.
Extreme hardness.
Examples: Diamond and graphite (pure carbon)
Silicon Carbide
Silicon Dioxide (aka quartz)
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Chapter 6
Standardized Test Preparation
Multiple Choice
4. According to VSEPR theory, the molecular geometry
for CH+3 is
A. tetrahedral.
B. trigonal-pyramidal.
C. bent or angular.
D. None of the above
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Chapter 6
Standardized Test Preparation
Multiple Choice
4. According to VSEPR theory, the molecular geometry
for CH+3 is
A. tetrahedral.
B. trigonal-pyramidal.
C. bent or angular.
D. None of the above
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Chapter 6
Standardized Test Preparation
Multiple Choice
6. Which molecule is polar?
A. CCl4
B. CO2
C. SO3
D. none of these
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Chapter 6
Standardized Test Preparation
Multiple Choice
6. Which molecule is polar?
A. CCl4
B. CO2
C. SO3
D. none of these
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Chapter 6
Standardized Test Preparation
Multiple Choice
7. What is the hybridization of the carbon atoms in C2H2?
A. sp
B. sp2
C. sp3
D. The carbon atoms do not hybridize in C2H2.
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Chapter 6
Standardized Test Preparation
Multiple Choice
7. What is the hybridization of the carbon atoms in C2H2?
A. sp
B. sp2
C. sp3
D. The carbon atoms do not hybridize in C2H2.
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Chapter 6
Standardized Test Preparation
Multiple Choice
8. Which of the following compounds is predicted to have
the highest boiling point?
A. HCl
B. CH3COOH (Note: the two oxygen atoms bond to
the carbon)
C. Cl2
D. SO2
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Chapter 6
Standardized Test Preparation
Multiple Choice
8. Which of the following compounds is predicted to have
the highest boiling point?
A. HCl
B. CH3COOH (Note: the two oxygen atoms bond to
the carbon)
C. Cl2
D. SO2
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Chapter 6
Standardized Test Preparation
Multiple Choice
9. An unknown substance is an excellent electrical
conductor in the solid state and is malleable. What type
of chemical bonding does this substance exhibit?
A. ionic bonding
B. molecular bonding
C. metallic bonding
D. cannot determine from the information given
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 6
Standardized Test Preparation
Multiple Choice
9. An unknown substance is an excellent electrical
conductor in the solid state and is malleable. What type
of chemical bonding does this substance exhibit?
A. ionic bonding
B. molecular bonding
C. metallic bonding
D. cannot determine from the information given
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Chapter 6
Standardized Test Preparation
Short Answer
10. What does the hybridization model help explain?
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Chapter 6
Standardized Test Preparation
Extended Response
12. Naphthalene, C10H8, is a nonpolar molecule and has a boiling
point of 218°C. Acetic acid, CH3CO2H, is a polar molecule and
has a boiling point of 118°C. Which substance has the stronger
intermolecular forces? Briefly explain your answer.
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 6
Standardized Test Preparation
Extended Response
12. Naphthalene, C10H8, is a nonpolar molecule and has a boiling
point of 218°C. Acetic acid, CH3CO2H, is a polar molecule and
has a boiling point of 118°C. Which substance has the stronger
intermolecular forces? Briefly explain your answer.
Answer: Naphthalene has the stronger intermolecular forces even
though it is nonpolar, because its boiling point is higher than that
of acetic acid. Boiling point is directly correlated to strength of
intermolecular forces; the stronger the intermolecular forces, the
more energy needed to break all the intermolecular forces, and
therefore the higher the boiling point. Naphthalene is so large
that its dispersion forces are greater than the sum of the
dispersion forces and hydrogen bonding in acetic acid.
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