Chapter 9 Intermolecular Forces, Liquids, and Solids.

Chapter 9Intermolecular Forces, Liquids, and

Solids

States of MatterThe fundamental difference between states of matter is the distance between particles.

States of MatterBecause in the solid and liquid states particles are closer together, we refer to them as condensed phases.

The States of Matter

• The state a substance is in at a particular temperature and pressure depends on two antagonistic entities:

– The kinetic energy of the particles

– The strength of the attractions between the particles

Intermolecular Forces

The attractions between molecules (intermolecular) are not nearly as strong as the intramolecular attractions that hold compounds together.


They are, however, strong enough to control physical properties such as boiling and melting points, vapor pressures, and viscosities.


These intermolecular forces as a group are referred to as van der Waals forces.

Van der Waals Forces

• London dispersion forces• Dipole-dipole interactions• Hydrogen bonding

Ion-Dipole Interactions

• A fourth type of force, ion-dipole interactions are an important force in solutions of ions.

• The strength of these forces are what make it possible for ionic substances to dissolve in polar solvents!

London Dispersion Forces

While the electrons in the 1s orbital of helium would repel each other (and, therefore, tend to stay far away from each other), it does happen that they occasionally wind up on the same side of the atom.


At that instant, then, the helium atom is polar, with an excess of electrons on the left side and a shortage on the right side.


Another helium nearby, then, would have a dipole induced in it, as the electrons on the left side of helium atom 2 repel the electrons in the cloud on helium atom 1.


London dispersion forces, or dispersion forces, are attractions between an instantaneous dipole and an induced dipole.


• These forces are present in all molecules, whether they are polar or nonpolar.

• The tendency of an electron cloud to distort in this way is called polarizability.

Factors Affecting London Forces

• The shape of the molecule affects the strength of dispersion forces: long, skinny molecules (like n-pentane tend to have stronger dispersion forces than short, fat ones (like neopentane).

• This is due to the increased surface area in n-pentane and greater polarizability.

Factors Affecting London Forces

• The strength of dispersion forces tends to increase with increased molecular weight or increasing number of electrons.

• Larger atoms have larger electron clouds, which are easier to polarize.

Dipole-Dipole Interactions

• Molecules that have permanent dipoles are attracted to each other.– The positive end of one is

attracted to the negative end of the other and vice-versa.

– These forces are only important when the molecules are close to each other.

Dipole-Dipole Interactions

The more polar the molecule, the higher is its boiling point.

Which Have a Greater Effect:Dipole-Dipole Interactions or Dispersion Forces?

• If two molecules are of comparable size and shape, dipole-dipole interactions will likely be the dominating force.

• If one molecule is much larger than another, dispersion forces will likely determine its physical properties.

How Do We Explain This?

• The nonpolar series (SnH4 to CH4) follow the expected trend.

• The polar series follows the trend from H2Te through H2S, but water is quite an anomaly.

Hydrogen Bonding

• The dipole-dipole interactions experienced when H is bonded to N, O, or F are unusually strong!

• We call these interactions hydrogen bonds.

Hydrogen Bonding

Hydrogen bonding arises in part from the high electronegativity of nitrogen, oxygen, and fluorine.

Also, when hydrogen is bonded to one of those very electronegative elements, the hydrogen nucleus is exposed.

Summarizing Intermolecular Forces

IntermolecularForces

Intermolecular Forces Affect Many Physical Properties

The strength of the attractions between particles can greatly affect the properties of a substance or solution.


The molecular basis of surface tension.

hydrogen bondingoccurs in three

dimensions

hydrogen bondingoccurs across the surface

and below the surface the net vectorfor attractive

forces is downward

Viscosity• Resistance of a liquid to

flow is called viscosity.• It is related to the ease

with which molecules can move past each other.

• Viscosity increases with stronger intermolecular forces and decreases with higher temperature.

Surface Tension

Surface tension results from the net inward force experienced by the molecules on the surface of a liquid.


Shape of water or mercury meniscus in glass.

adhesive forces

stronger cohesive

forces

H2O

capillarity

Hg

Phase Changes

Energy Changes Associated with Changes of State

• Heat of Fusion: Energy required to change a solid at its melting point to a liquid.


• Heat of Vaporization: Energy required to change a liquid at its boiling point to a gas.


• The heat added to the system at the melting and boiling points goes into pulling the molecules farther apart from each other.

• The temperature of the substance does not rise during the phase change.

Vapor Pressure• At any temperature, some molecules in a liquid

have enough energy to escape.• As the temperature rises, the fraction of

molecules that have enough energy to escape increases.

The effect of temperature on the distribution of molecular speed in a liquid.

Vapor Pressure

Vapor Pressure

As more molecules escape the liquid, the pressure they exert increases.

Liquid-gas equilibrium.

The liquid and vapor reach a state of dynamic equilibrium: liquid molecules evaporate and vapor molecules condense at the same rate.

Vapor Pressure

The liquid and vapor reach a state of dynamic equilibrium: liquid molecules evaporate and vapor molecules condense at the same rate.

kEtOH=K[EtOH]= Ptequil

Vapor Pressure

• The boiling point of a liquid is the temperature at which its vapor pressure equals atmospheric pressure.

• The normal boiling point is the temperature at which its vapor pressure is 760 torr.

Vapor pressure as a function of temperature and intermolecular forces.

A linear plot of vapor pressure- temperature

relationship.

The Clausius-Clapeyron Equation

ln P =

-HvapR

1T

C

ln P2P1

= -Hvap

R1T2

1T1

Phase DiagramsPhase diagrams display the state of a substance at various pressures and temperatures and the places where equilibria exist between phases.

Phase Diagrams• The AB line is the liquid-vapor interface.• It starts at the triple point (A), the point at which

all three states are in equilibrium.

Phase DiagramsIt ends at the critical point (B); above this critical temperature and critical pressure the liquid and vapor are indistinguishable from each other.

Phase DiagramsEach point along this line is the boiling point of the substance at that pressure.

Phase Diagrams• The AD line is the interface between liquid and

solid.• The melting point at each pressure can be found

along this line.

Phase Diagrams• Below A the substance cannot exist in the liquid

state.• Along the AC line the solid and gas phases are in

equilibrium; the sublimation point at each pressure is along this line.

Phase Diagram of Water

• Note the high critical temperature and critical pressure:– These are due to the strong

van der Waals forces between water molecules.

Phase Diagram of Water

• The slope of the solid–liquid line is negative.– This means that as the

pressure is increased at a temperature just below the melting point, water goes from a solid to a liquid.

Phase Diagram of Carbon Dioxide

Carbon dioxide cannot exist in the liquid state at pressures below 5.11 atm; CO2 sublimes at normal pressures.

Phase Diagram of Carbon Dioxide

The low critical temperature and critical pressure for CO2 make supercritical CO2 a good solvent for extracting nonpolar substances (such as caffeine).

CO2 as a critical fluid!

Iodine subliming.

iodine solid

iodine vapor

iodine solid

test tube with ice

Phase diagrams for CO2 and H2O.

CO2 H2O

Solids

• We can think of solids as falling into two groups:

– Crystalline—particles are in highly ordered arrangement.

Solids

– Amorphous—NO regular geometric order in the arrangement of particles at the molecular level.

Crystalline SolidsBecause of the regular order in a crystalline solid, we can focus on the three dimensional repeating pattern or arrangement called the unit cell.

Crystalline Solids

There are several types of basic arrangements in crystals. In the cubic class we have primitive, body-centered, and face-centered unit cells.

Crystalline Solids

We can determine the formula of an ionic solid by determining how many ions of each element fall within the unit cell.

Characteristics of the Major Types of Crystalline Solids

ParticlesInterparticle Forces

Physical Behavior Examples (mp,0C)

Atomic

Molecular

Ionic

Metallic

Network

Group 8A(18)[Ne-249 to Rn-71]

Molecules

Positive & negative ions

Atoms

Atoms

Soft, very low mp, poor thermal & electrical conductors

DispersionAtoms

Dispersion, dipole-dipole, H bonds

Fairly soft, low to moderate mp, poor thermal & electrical conductors

Nonpolar - O2[-219], C4H10[-138], Cl2

[-101], C6H14[-95]

Polar - SO2[-73], CHCl3[-64], HNO3[-42], H2O[0.0]

Covalent bond

Metallic bond

Ion-ion attraction

Very hard, very high mp, usually poor thermal and electrical conductors

Soft to hard, low to very high mp, excellent thermal and electrical conductors, malleable and ductile

Hard & brittle, high mp, good thermal & electrical conductors when molten

NaCl [801]CaF2 [1423]

MgO [2852]

Na [97.8]Zn [420]Fe [1535]

Type of Solids

hexagonal unit cell

closest packing of first and second layers

layer a

layer a

layer b

layer c

hexagonal closest packing

cubic closest packing

abab… (74%)abcabc… (74%)

expanded side views

face-centered unit cell

Closest Packing of Identical Non-penetrating Spheres

Ionic Solids

What are the formulas for these compounds?(a) Green: chlorine; Gray: cesium(b) Yellow: sulfur; Gray: zinc(c) Green: calcium; Gray: fluorine

CsCl ZnS CaF2

(a) (b) (c)

In ionic crystals, ions pack themselves so as to maximize the attractions and minimize repulsions between the ions.

Covalent-Network andMolecular Solids

• Diamonds are an example of a covalent-network solid in which atoms are covalently bonded to each other.– They tend to be hard and have high melting points.

Covalent-Network andMolecular Solids

• Graphite is an example of a molecular solid in which atoms are held together with van der Waals forces.– They tend to be softer and have lower melting points.

Metallic Solids• Metals are not covalently

bonded, but the attractions between atoms are too strong to be van der Waals forces.

• In metals, valence electrons are delocalized throughout the solid.

The band of molecular orbitals in lithium metal.

conductor

semiconductor

insulator

Electrical conductivity in a conductor, semiconductor, and insulator.

Crystal structures and band representations of doped

semiconductors.

rare earth magnet

superconducting ceramic disk

liquid nitrogen

The levitating power of a superconducting oxide.

Structures of two typical molecules that form liquid crystals.

Some Uses of Modern Ceramics and Ceramic Mixtures

Ceramic Applications

SiC, Si3N4, TiB2, Al2O3 Whiskers(fibers) to strength Al and other ceramics

Si3N4 Car engine parts; turbine rotors for “turbo” cars; electronic sensor units

Si3N4, BN, Al2O3 Supports or layering materials(as insulators) in electronic microchips

SiC, Si3N4, TiB2, ZrO2, Al2O3, BN

ZrO2, Al2O3

Cutting tools, edge sharpeners(as coatings and whole devices), scissors, surgical tools, industrial “diamond”

BN, SiC Armor-plating reinforcement fibers(as in Kevlar composites)

Surgical implants(hip and knee joints)

Unit cells of some modern ceramic materials.

SiC

silicon carbide

BN

cubic boron nitride (borazon)

YBa2Cu3O7

high temperature superconductor

Diffraction of x-rays by crystal planes.

Tools of the Laboratory

Formation of an x-ray diffraction pattern of the protein hemoglobin.



Scanning tunneling micrographs.

gallium arsenide semiconductor metallic gold

Problems

SAMPLE PROBLEM 1 Using the Clausius-Clapeyron Equation

SOLUTION:

PROBLEM: The vapor pressure of ethanol is 115 torr at 34.90C. If Hvap of ethanol is 40.5 kJ/mol, calculate the temperature (in 0C) when the vapor pressure is 760 torr.

PLAN: We are given 4 of the 5 variables in the Clausius-Clapeyron equation. Substitute and solve for T2.

ln

P2P1

= -Hvap

R1

T2

1T1

34.90C = 308.0K

ln760 torr115 torr

=-40.5 x103 J/mol8.314 J/mol*K

1T2

1308K

-

T2 = 350K = 770C

SAMPLE PROBLEM 2 Drawing Hydrogen Bonds Between Molecules of a Substance

SOLUTION:

PROBLEM: Which of the following substances exhibits H bonding? For those that do, draw two molecules of the substance with the H bonds between them.

C2H6(a) CH3OH(b) CH3C NH2

O

(c)

PLAN: Find molecules in which H is bonded to N, O or F. Draw H bonds in the format -B: H-A-.

(a) C2H6 has no H bonding sites.

(c)(b)C O H

H

H

H

COH

H

H

H

CH3C N

O

H

H

CH3CN

O

H

H

CH3CN

O

H

H

CH3CN

O

H

H

SAMPLE PROBLEM 3 Predicting the Type and Relative Strength of Intermolecular Forces

PROBLEM: For each pair of substances, identify the dominant intermolecular forces in each substance, and select the substance with the higher boiling point.

(a) MgCl2 or PCl3

(b) CH3NH2 or CH3F

(c) CH3OH or CH3CH2OH

(d) Hexane (CH3CH2CH2CH2CH2CH3)

or 2,2-dimethylbutaneCH3CCH2CH3

CH3

CH3PLAN: Use the formula, structure and Table 2.2 (button).

•Bonding forces are stronger than nonbonding(intermolecular) forces.

•Hydrogen bonding is a strong type of dipole-dipole force.

•Dispersion forces are decisive when the difference is molar mass or molecular shape.

SOLUTION:

SAMPLE PROBLEM 3 Predicting the Type and Relative Strength of Intermolecular Forces

continued

(a) Mg2+ and Cl- are held together by ionic bonds while PCl3 is covalently bonded and the molecules are held together by dipole-dipole interactions. Ionic bonds are stronger than dipole interactions and so MgCl2 has the higher boiling point.

(b) CH3NH2 and CH3F are both covalent compounds and have bonds which are polar. The dipole in CH3NH2 can H bond while that in CH3F cannot. Therefore CH3NH2 has the stronger interactions and the higher boiling point.

(c) Both CH3OH and CH3CH2OH can H bond but CH3CH2OH has more CH for more dispersion force interaction. Therefore CH3CH2OH has the higher boiling point.(d) Hexane and 2,2-dimethylbutane are both nonpolar with only dispersion forces to hold the molecules together. Hexane has the larger surface area, thereby the greater dispersion forces and the higher boiling point.

SAMPLE PROBLEM 4 Determining Atomic Radius from Crystal Structure

PROBLEM: Barium is the largest nonradioactive alkaline earth metal. It has a body-centered cubic unit cell and a density of 3.62 g/cm3. What is the atomic radius of barium?

(Volume of a sphere: V = 4/3r3)

PLAN: We can use the density and molar mass to find the volume of 1 mol of Ba. Since 68%(for a body-centered cubic) of the unit cell contains atomic material, dividing by Avogadro’s number will give us the volume of one atom of Ba. Using the volume of a sphere, the radius can be calculated.

density of Ba (g/cm3)

volume of 1 mol Ba metal volume of 1 Ba atom

radius of a Ba atom

multiply by packing efficiency

reciprocal divided by M V = 4/3r3

volume of 1 mol Ba atoms

divide by Avogadro’s number

SAMPLE PROBLEM 4 Determining Atomic Radius from Crystal Structure

SOLUTION:

continued

Volume of Ba metal =137.3 g Ba

mol Ba= 37.9 cm3/mol Ba

37.9 cm3/mol Ba x 0.68 = 26 cm3/mol Ba atoms

mol Ba atoms

6.022x1023 atoms= 4.3x10-23 cm3/atom

r3 = 3V/4 = 2.2 x 10-8cm

1 cm3

3.62 gx

26 cm3

mol Ba atomsx

r =3V

43

3(4.3x10-23cm3 )

4 x 3.143

Chapter 9 Intermolecular Forces, Liquids, and Solids.

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Transcript of Chapter 9 Intermolecular Forces, Liquids, and Solids.