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Copyright © by Holt, Rinehart and Winston. All rights reserved. Resources Chapter menu 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 ionic-crystalline 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. Chapter 6

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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 . - PowerPoint PPT Presentation

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Page 1: Objectives

Copyright © by Holt, Rinehart and Winston. All rights reserved.

ResourcesChapter menu

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 ionic-crystalline 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.

Chapter 6

Page 2: Objectives

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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.

Chapter 6

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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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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

Section 4 Metallic BondingChapter 6

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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.

Section 4 Metallic BondingChapter 6

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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.

Section 4 Metallic BondingChapter 6

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Visual Concepts

Metallic Bonding

Chapter 6

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Visual Concepts

Properties of Metals: Surface Appearance

Chapter 6

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Visual Concepts

Properties of Metals: Malleability and Ductility

Chapter 6

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Visual Concepts

Properties of Metals: Electrical and Thermal Conductivity

Chapter 6

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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.

Section 5 Van der WaalsChapter 6

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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

Section 5 Van der WaalsChapter 6

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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).

Section 5 Van der WaalsChapter 6

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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:

H Cl

Van der WaalsChapter 6

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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.

Section 5 Van der WaalsChapter 6

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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.

Section 5 Van der WaalsChapter 6

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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.

Section 5 Van der WaalsChapter 6

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Visual Concepts

Hydrogen Bonding

Chapter 6

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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.

Section 5 Van der WaalsChapter 6

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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).

Section 5 Molecular GeometryChapter 6

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Visual Concepts

Dipole-Dipole Forces

Chapter 6

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Comparing Dipole-Dipole Forces

Section 5 Van der WaalsChapter 6

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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.

Section 5 Van der WaalsChapter 6

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Visual Concepts

Dipole-Induced Dipole Interaction

Chapter 6

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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.

Section 5 Van der WaalsChapter 6

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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.

Section 5 Van der WaalsChapter 6

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Visual Concepts

London Dispersion Force

Chapter 6

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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.

Section 5 Van der WaalsChapter 6

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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.

Section 5 Van der WaalsChapter 6

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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.

Section 5 Molecular GeometryChapter 6

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Molecular Geometry

2 methods are used to make determinations of molecular geometry

VSEPR TheoryCentral 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.

Section 5 Molecular GeometryChapter 6

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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.

Section 5 Molecular GeometryChapter 6

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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 valence-level 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: H-Be-H

Section 5 Molecular GeometryChapter 6

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VSEPR TheoryVSEPR “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” ).

Section 5 Molecular GeometryChapter 6

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Visual Concepts

Lone Pair of Electrons

Chapter 6

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Visual Concepts

VSEPR and Basic Molecular Shapes

Chapter 6

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Visual Concepts

VSEPR and Lone Electron Pairs

Chapter 6

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VSEPR Theory, continuedSample Problem E

Use VSEPR theory to predict the molecular geometry of boron trichloride, BCl3.

B

Cl

ClCl

Section 5 Molecular GeometryChapter 6

Boron trichloride has a 3/0 VSEPR combo. Its geometry therefore is trigonal planar.

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• 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:

• Ammonia’s VSEPR combo is 3/1 and therefore its geometry is trigonal pyramidal.

N HHH

Section 5 Molecular GeometryChapter 6

VSEPR Theory, continued

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• 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.

Section 5 Molecular GeometryChapter 6

VSEPR Theory, continued

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Sample Problem F

Use VSEPR theory to predict the shape of a molecule of carbon dioxide, CO2.

The VSEPR combo is 2/0 therefore the geometry is

linear.

C OO

Section 5 Molecular GeometryChapter 6

VSEPR Theory, continued

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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.

Molecular Polarity Chapter 6

Molecular Polarity

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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.)

Chapter 6

Molecular Polarity

Molecular Polarity

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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.

Chapter 6

Molecular Polarity

Molecular Polarity

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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

Chapter 6

Molecular Polarity

Molecular Polarity

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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.

Chapter 6

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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.

Chapter 6

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Central Atom “Bonding Character”

Chapter 6

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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Central Atom “Bonding Character”

Chapter 6

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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Central Atom “Bonding Character”

Chapter 6

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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Central Atom “Bonding Character”

Chapter 6

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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Geometry of Hybrid Orbitals

Section 5 Molecular GeometryChapter 6

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Visual Concepts

Hybrid Orbitals

Chapter 6

sp3d bonding, bonding 5 atoms = trigonal bipyramidal ex: select group 15 nonmetals

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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.

Chapter 6

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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.

ResonanceResonance 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.

Chapter 6

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Visual Concepts

Atomic Resonance

Chapter 6

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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.

Chapter 6

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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)

Chapter 6

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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.

Chapter 6

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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.

Chapter 6

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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.

Chapter 6

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Visual Concepts

Bond Energy

Chapter 6

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Bond Energies and Bond Lengths for Single Bonds

Properties of BondsChapter 6

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Visual Concepts

Bond Length

Chapter 6

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Bond Length and Stability

Properties of BondsChapter 6

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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.

Chapter 6

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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)

Chapter 6

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4. According to VSEPR theory, the molecular geometry for is

A. tetrahedral.

B. trigonal-pyramidal.

C. bent or angular.

D. None of the above

Standardized Test Preparation Chapter 6

Multiple Choice

+3CH

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4. According to VSEPR theory, the molecular geometry for is

A. tetrahedral.

B. trigonal-pyramidal.

C. bent or angular.

D. None of the above

Standardized Test Preparation Chapter 6

Multiple Choice

+3CH

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6. Which molecule is polar?

A. CCl4

B. CO2

C. SO3

D. none of these

Standardized Test Preparation Chapter 6

Multiple Choice

Page 71: Objectives

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6. Which molecule is polar?

A. CCl4

B. CO2

C. SO3

D. none of these

Standardized Test Preparation Chapter 6

Multiple Choice

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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.

Standardized Test Preparation Chapter 6

Multiple Choice

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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.

Standardized Test Preparation Chapter 6

Multiple Choice

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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

Standardized Test Preparation Chapter 6

Multiple Choice

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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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Multiple Choice

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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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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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Short Answer

10. What does the hybridization model help explain?

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Extended Response12. 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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Extended Response12. 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.

Standardized Test Preparation Chapter 6