Copyright 2011 Pearson Education, Inc. Chapter 11 Liquids,
Solids, and Intermolecular Forces Roy Kennedy Massachusetts Bay
Community College Wellesley Hills, MA Chemistry: A Molecular
Approach, 2nd Ed. Nivaldo Tro
Slide 2
Copyright 2011 Pearson Education, Inc. Climbing Geckos Geckos
can adhere to almost any surface Recent studies indicate that this
amazing ability is related to intermolecular attractive forces
Geckos have millions of tiny hairs on their feet that branch out
and flatten out on the end setae = hairs, spatulae = flat ends This
structure allows the gecko to have unusually close contact to the
surface allowing the intermolecular attractive forces to act
strongly 2Tro: Chemistry: A Molecular Approach, 2/e
Slide 3
Copyright 2011 Pearson Education, Inc. Properties of the three
Phases of Matter Fixed = keeps shape when placed in a container
Indefinite = takes the shape of the container 3Tro: Chemistry: A
Molecular Approach, 2/e
Slide 4
Copyright 2011 Pearson Education, Inc. Notice that the
densities of ice and liquid water are much larger than the density
of steam Notice that the densities and molar volumes of ice and
liquid water are much closer to each other than to steam Notice
that the densities of ice is larger than the density of liquid
water. This is not the norm, but is vital to the development of
life as we know it. Three Phases of Water 4Tro: Chemistry: A
Molecular Approach, 2/e
Slide 5
Copyright 2011 Pearson Education, Inc. Degrees of Freedom
Particles may have one or several types of freedom of motion and
various degrees of each type Translational freedom is the ability
to move from one position in space to another Rotational freedom is
the ability to reorient the particles direction in space
Vibrational freedom is the ability to oscillate about a particular
point in space 5Tro: Chemistry: A Molecular Approach, 2/e
Slide 6
Copyright 2011 Pearson Education, Inc. Solids The particles in
a solid are packed close together and are fixed in position though
they may vibrate The close packing of the particles results in
solids being incompressible The inability of the particles to move
around results in solids retaining their shape and volume when
placed in a new container, and prevents the solid from flowing
6Tro: Chemistry: A Molecular Approach, 2/e
Slide 7
Copyright 2011 Pearson Education, Inc. Solids Some solids have
their particles arranged in an orderly geometric pattern we call
these crystalline solids salt and diamonds Other solids have
particles that do not show a regular geometric pattern over a long
range we call these amorphous solids plastic and glass 7Tro:
Chemistry: A Molecular Approach, 2/e
Slide 8
Copyright 2011 Pearson Education, Inc. 8 Liquids The particles
in a liquid are closely packed, but they have some ability to move
around The close packing results in liquids being incompressible
But the ability of the particles to move allows liquids to take the
shape of their container and to flow however, they don t have
enough freedom to escape or expand to fill the container Tro:
Chemistry: A Molecular Approach, 2/e
Slide 9
Copyright 2011 Pearson Education, Inc. 9 Gases In the gas
state, the particles have complete freedom of motion and are not
held together The particles are constantly flying around, bumping
into each other and the container There is a large amount of space
between the particles compared to the size of the particles
therefore the molar volume of the gas state of a material is much
larger than the molar volume of the solid or liquid states Tro:
Chemistry: A Molecular Approach, 2/e
Slide 10
Copyright 2011 Pearson Education, Inc. 10 Gases Because there
is a lot of empty space, the particles can be squeezed closer
together therefore gases are compressible Because the particles are
not held in close contact and are moving freely, gases expand to
fill and take the shape of their container, and will flow Tro:
Chemistry: A Molecular Approach, 2/e
Slide 11
Copyright 2011 Pearson Education, Inc. Compressibility 11Tro:
Chemistry: A Molecular Approach, 2/e
Slide 12
Copyright 2011 Pearson Education, Inc. Kinetic Molecular Theory
What state a material is in depends largely on two major factors
1.the amount of kinetic energy the particles possess 2.the strength
of attraction between the particles These two factors are in
competition with each other 12Tro: Chemistry: A Molecular Approach,
2/e
Slide 13
Copyright 2011 Pearson Education, Inc. States and Degrees of
Freedom The molecules in a gas have complete freedom of motion
their kinetic energy overcomes the attractive forces between the
molecules The molecules in a solid are locked in place, they cannot
move around though they do vibrate, they dont have enough kinetic
energy to overcome the attractive forces The molecules in a liquid
have limited freedom they can move around a little within the
structure of the liquid they have enough kinetic energy to overcome
some of the attractive forces, but not enough to escape each other
13Tro: Chemistry: A Molecular Approach, 2/e
Slide 14
Copyright 2011 Pearson Education, Inc. Kinetic Energy
Increasing kinetic energy increases the motion energy of the
particles The more motion energy the molecules have, the more
freedom they can have The average kinetic energy is directly
proportional to the temperature KE avg = 1.5 kT 14Tro: Chemistry: A
Molecular Approach, 2/e
Slide 15
Copyright 2011 Pearson Education, Inc. Attractive Forces The
particles are attracted to each other by electrostatic forces The
strength of the attractive forces varies, some are small and some
are large The strength of the attractive forces depends on the
kind(s) of particles The stronger the attractive forces between the
particles, the more they resist moving though no material
completely lacks particle motion 15Tro: Chemistry: A Molecular
Approach, 2/e
Slide 16
Copyright 2011 Pearson Education, Inc. KineticMolecular Theory
of Gases When the kinetic energy is so large it overcomes the
attractions between particles, the material will be a gas In an
ideal gas, the particles have complete freedom of motion especially
translational This allows gas particles to expand to fill their
container gases flow It also leads to there being large spaces
between the particles therefore low density and compressibility
16Tro: Chemistry: A Molecular Approach, 2/e
Slide 17
Copyright 2011 Pearson Education, Inc. Gas Structure Gas
molecules are rapidly moving in random straight lines, and are free
from sticking to each other. 17Tro: Chemistry: A Molecular
Approach, 2/e
Slide 18
Copyright 2011 Pearson Education, Inc. KineticMolecular Theory
of Solids When the attractive forces are strong enough so the
kinetic energy cannot overcome it at all, the material will be a
solid In a solid, the particles are packed together without any
translational or rotational motion the only freedom they have is
vibrational motion 18Tro: Chemistry: A Molecular Approach, 2/e
Slide 19
Copyright 2011 Pearson Education, Inc. KineticMolecular Theory
of Liquids When the attractive forces are strong enough so the
kinetic energy can only partially overcome them, the material will
be a liquid In a liquid, the particles are packed together with
only very limited translational or rotational freedom 19Tro:
Chemistry: A Molecular Approach, 2/e
Slide 20
Copyright 2011 Pearson Education, Inc. Explaining the
Properties of Liquids Liquids have higher densities than gases and
are incompressible because the particles are in contact They have
an indefinite shape because the limited translational freedom of
the particles allows them to move around enough to get to the
container walls It also allows them to flow But they have a
definite volume because the limit on their freedom keeps the
particles from escaping each other 20Tro: Chemistry: A Molecular
Approach, 2/e
Slide 21
Copyright 2011 Pearson Education, Inc. Phase Changes Because
the attractive forces between the molecules are fixed, changing the
materials state requires changing the amount of kinetic energy the
particles have, or limiting their freedom Solids melt when heated
because the particles gain enough kinetic energy to partially
overcome the attactive forces Liquids boil when heated because the
particles gain enough kinetic energy to completely overcome the
attractive forces the stronger the attractive forces, the higher
you will need to raise the temperature Gases can be condensed by
decreasing the temperature and/or increasing the pressure pressure
can be increased by decreasing the gas volume reducing the volume
reduces the amount of translational freedom the particles have
21Tro: Chemistry: A Molecular Approach, 2/e
Slide 22
Copyright 2011 Pearson Education, Inc. Phase Changes 22Tro:
Chemistry: A Molecular Approach, 2/e
Slide 23
Copyright 2011 Pearson Education, Inc. Intermolecular
Attractions The strength of the attractions between the particles
of a substance determines its state At room temperature, moderate
to strong attractive forces result in materials being solids or
liquids The stronger the attractive forces are, the higher will be
the boiling point of the liquid and the melting point of the solid
other factors also influence the melting point 23Tro: Chemistry: A
Molecular Approach, 2/e
Slide 24
Copyright 2011 Pearson Education, Inc. Why Are Molecules
Attracted to Each Other? Intermolecular attractions are due to
attractive forces between opposite charges + ion to ion + end of
polar molecule to end of polar molecule H-bonding especially strong
even nonpolar molecules will have temporary charges Larger charge =
stronger attraction Longer distance = weaker attraction However,
these attractive forces are small relative to the bonding forces
between atoms generally smaller charges generally over much larger
distances 24Tro: Chemistry: A Molecular Approach, 2/e
Slide 25
Copyright 2011 Pearson Education, Inc. Trends in the Strength
of Intermolecular Attraction The stronger the attractions between
the atoms or molecules, the more energy it will take to separate
them Boiling a liquid requires we add enough energy to overcome all
the attractions between the particles However, not breaking the
covalent bonds The higher the normal boiling point of the liquid,
the stronger the intermolecular attractive forces 25Tro: Chemistry:
A Molecular Approach, 2/e
Slide 26
Copyright 2011 Pearson Education, Inc. Kinds of Attractive
Forces Temporary polarity in the molecules due to unequal electron
distribution leads to attractions called dispersion forces
Permanent polarity in the molecules due to their structure leads to
attractive forces called dipoledipole attractions An especially
strong dipoledipole attraction results when H is attached to an
extremely electronegative atom. These are called hydrogen bonds.
26Tro: Chemistry: A Molecular Approach, 2/e
Slide 27
Copyright 2011 Pearson Education, Inc. Dispersion Forces
Fluctuations in the electron distribution in atoms and molecules
result in a temporary dipole region with excess electron density
has partial () charge region with depleted electron density has
partial (+) charge The attractive forces caused by these temporary
dipoles are called dispersion forces aka London Forces All
molecules and atoms will have them As a temporary dipole is
established in one molecule, it induces a dipole in all the
surrounding molecules 27Tro: Chemistry: A Molecular Approach,
2/e
Slide 28
Copyright 2011 Pearson Education, Inc. Dispersion Force 28Tro:
Chemistry: A Molecular Approach, 2/e
Slide 29
Copyright 2011 Pearson Education, Inc. Size of the Induced
Dipole The magnitude of the induced dipole depends on several
factors Polarizability of the electrons volume of the electron
cloud larger molar mass = more electrons = larger electron cloud =
increased polarizability = stronger attractions Shape of the
molecule more surface-to-surface contact = larger induced dipole =
stronger attraction + + +++ + + - - --- - - + + + + + - - - - -
molecules that are flat have more surface interaction than
spherical ones + + + ++ + + + + + + + + + larger molecules have
more electrons, leading to increased polarizability 29Tro:
Chemistry: A Molecular Approach, 2/e
Slide 30
Copyright 2011 Pearson Education, Inc. The Noble gases are all
nonpolar atomic elements As the molar mass increases, the number of
electrons increases. Therefore the strength of the dispersion
forces increases. Effect of Molecular Size on Size of Dispersion
Force The stronger the attractive forces between the molecules, the
higher the boiling point will be. 30Tro: Chemistry: A Molecular
Approach, 2/e
Slide 31
Copyright 2011 Pearson Education, Inc. 31Tro: Chemistry: A
Molecular Approach, 2/e
Slide 32
Copyright 2011 Pearson Education, Inc. Properties of Straight
Chain Alkanes NonPolar Molecules 32Tro: Chemistry: A Molecular
Approach, 2/e
Slide 33
Copyright 2011 Pearson Education, Inc. Boiling Points of
n-Alkanes 33Tro: Chemistry: A Molecular Approach, 2/e
Slide 34
Copyright 2011 Pearson Education, Inc. 34Tro: Chemistry: A
Molecular Approach, 2/e
Slide 35
Copyright 2011 Pearson Education, Inc. Effect of Molecular
Shape on Size of Dispersion Force 35 the larger surface-to- surface
contact between molecules in n-pentane results in stronger
dispersion force attractions Tro: Chemistry: A Molecular Approach,
2/e
Slide 36
Copyright 2011 Pearson Education, Inc. Alkane Boiling Points
Branched chains have lower BPs than straight chains The straight
chain isomers have more surface-to- surface contact 36Tro:
Chemistry: A Molecular Approach, 2/e
Slide 37
Copyright 2011 Pearson Education, Inc. a)CH 4 C 4 H 10 b)C 6 H
12 C 6 H 12 Practice Choose the Substance in Each Pair with the
Higher Boiling Point 37Tro: Chemistry: A Molecular Approach,
2/e
Slide 38
Copyright 2011 Pearson Education, Inc. Practice Choose the
Substance in Each Pair with the Higher Boiling Point a)CH 4 CH 3 CH
2 CH 2 CH 3 b)CH 3 CH 2 CH=CHCH 2 CH 3 cyclohexane Both molecules
are nonpolar larger molar mass Both molecules are nonpolar, but the
flatter ring molecule has larger surface-to- surface contact 38Tro:
Chemistry: A Molecular Approach, 2/e
Slide 39
Copyright 2011 Pearson Education, Inc. DipoleDipole Attractions
Polar molecules have a permanent dipole because of bond polarity
and shape dipole moment as well as the always present induced
dipole The permanent dipole adds to the attractive forces between
the molecules raising the boiling and melting points relative to
nonpolar molecules of similar size and shape 39Tro: Chemistry: A
Molecular Approach, 2/e
Slide 40
Copyright 2011 Pearson Education, Inc. Effect of DipoleDipole
Attraction on Boiling and Melting Points 40Tro: Chemistry: A
Molecular Approach, 2/e
Slide 41
Copyright 2011 Pearson Education, Inc. 41 replace with the
figure 11.8 Tro: Chemistry: A Molecular Approach, 2/e
Slide 42
Copyright 2011 Pearson Education, Inc. Example 11.1b: Determine
if dipoledipole attractions occur between CH 2 Cl 2 molecules
molecules that have dipoledipole attractions must be polar CH 2 Cl
2, EN C = 2.5, H = 2.1, Cl = 3.0 If there are dipoledipole
attractions Solution: Conceptual Plan: Relationships: Given: Find:
EN DifferenceShape Lewis Structure Bond Polarity Molecule Polarity
Formula ClC 3.02.5 = 0.5 polar CH 2.52.1 = 0.4 nonpolar polar
molecule; therefore dipole dipole attractions 42 polar bonds and
tetrahedral shape = polar molecule 4 bonding areas no lone pairs =
tetrahedral shape Tro: Chemistry: A Molecular Approach, 2/e
Slide 43
Copyright 2011 Pearson Education, Inc. or Practice Choose the
substance in each pair with the higher boiling point a)CH 2 FCH 2
FCH 3 CHF 2 b) 43Tro: Chemistry: A Molecular Approach, 2/e
Slide 44
Copyright 2011 Pearson Education, Inc. Practice Choose the
substance in each pair with the higher boiling point more polar
a)CH 2 FCH 2 FCH 3 CHF 2 b) 44 or polarnonpolar Tro: Chemistry: A
Molecular Approach, 2/e
Slide 45
Copyright 2011 Pearson Education, Inc. Hydrogen Bonding When a
very electronegative atom is bonded to hydrogen, it strongly pulls
the bonding electrons toward it O H, N H, or F H Because hydrogen
has no other electrons, when its electron is pulled away, the
nucleus becomes deshielded exposing the H proton The exposed proton
acts as a very strong center of positive charge, attracting all the
electron clouds from neighboring molecules 45Tro: Chemistry: A
Molecular Approach, 2/e
Slide 46
Copyright 2011 Pearson Education, Inc. H-Bonding HF 46Tro:
Chemistry: A Molecular Approach, 2/e
Slide 47
Copyright 2011 Pearson Education, Inc. H-Bonding in Water
47Tro: Chemistry: A Molecular Approach, 2/e
Slide 48
Copyright 2011 Pearson Education, Inc. H-Bonds Hydrogen bonds
are very strong intermolecular attractive forces stronger than
dipoledipole or dispersion forces Substances that can hydrogen bond
will have higher boiling points and melting points than similar
substances that cannot But hydrogen bonds are not nearly as strong
as chemical bonds 2 to 5% the strength of covalent bonds 48Tro:
Chemistry: A Molecular Approach, 2/e
Slide 49
Copyright 2011 Pearson Education, Inc. Effect of H-Bonding on
Boiling Point 49Tro: Chemistry: A Molecular Approach, 2/e
Slide 50
Copyright 2011 Pearson Education, Inc. For nonpolar molecules,
such as the hydrides of Group 4, the intermolecular attractions are
due to dispersion forces. Therefore they increase down the column,
causing the boiling point to increase. HF, H 2 O, and NH 3 have
unusually strong dipole-dipole attractions, called hydrogen bonds.
Therefore they have higher boiling points than you would expect
from the general trends. Polar molecules, such as the hydrides of
Groups 57, have both dispersion forces and dipoledipole
attractions. Therefore they have higher boiling points than the
corresponding Group 4 molecules. 50Tro: Chemistry: A Molecular
Approach, 2/e
Slide 51
Copyright 2011 Pearson Education, Inc. Step 3. Evaluate the
strengths of the total intermolecular attractive forces. The
substance with the strongest will be the liquid. Formaldehyde:
dispersion forces: MM 30.03, trigonal planar dipoledipole: very
polar C=O bond uncancelled H-bonding: no O H, N H, or FH therefore
no H - bond Fluoromethane: dispersion forces: MM 34.03, tetrahedral
dipoledipole: very polar C F bond uncancelled H-bonding: no O H, N
H, or FH therefore no H - bond Hydrogen peroxide: dispersion
forces: MM 34.02, tetrahedral bent dipoledipole: polar O H bonds
uncancelled H-bonding: O H, therefore H - bond Because the molar
masses are similar, the size of the dispersion force attractions
should be similar Because only hydrogen peroxide has the additional
very strong H - bond additional attractions, its intermolecular
attractions will be the strongest. We therefore expect hydrogen
peroxide to be the liquid. Example 11.2: One of these compounds is
a liquid at room temperature (the others are gases). Which one and
why? 51 Step 1. Determine the kinds of intermolecular attractive
forces MM = 30.03 Polar No H-Bonds MM = 34.03 Polar No H-Bonds MM =
34.02 Polar H-Bonds Step 2. Compare intermolecular attractive
forces Because all the molecules are polar, the size of the
dipoledipole attractions should be similar Only the hydrogen
peroxide also has additional hydrogen bond attractions Tro:
Chemistry: A Molecular Approach, 2/e
Slide 52
Copyright 2011 Pearson Education, Inc. a)CH 3 OHCH 3 CHF 2 b)CH
3 -O-CH 2 CH 3 CH 3 CH 2 CH 2 NH 2 Practice Choose the substance in
each pair that is a liquid at room temperature (the other is a gas)
can H-bond 52Tro: Chemistry: A Molecular Approach, 2/e
Slide 53
Copyright 2011 Pearson Education, Inc. Attractive Forces and
Solubility Solubility depends, in part, on the attractive forces of
the solute and solvent molecules like dissolves like miscible
liquids will always dissolve in each other Polar substances
dissolve in polar solvents hydrophilic groups = OH, CHO, C=O, COOH,
NH 2, Cl Nonpolar molecules dissolve in nonpolar solvents
hydrophobic groups = C-H, C-C Many molecules have both hydrophilic
and hydrophobic parts solubility in water becomes a competition
between the attraction of the polar groups for the water and the
attraction of the nonpolar groups for their own kind 53Tro:
Chemistry: A Molecular Approach, 2/e
Slide 54
Copyright 2011 Pearson Education, Inc. Immiscible Liquids 54
Pentane, C 5 H 12 is a nonpolar molecule. Water is a polar
molecule. The attractive forces between the water molecules is much
stronger than their attractions for the pentane molecules. The
result is the liquids are immiscible. Tro: Chemistry: A Molecular
Approach, 2/e
Slide 55
Copyright 2011 Pearson Education, Inc. Dichloromethane
(methylene chloride) Polar Solvents 55 Water Ethanol (ethyl
alcohol) Tro: Chemistry: A Molecular Approach, 2/e
Slide 56
Copyright 2011 Pearson Education, Inc. Nonpolar Solvents 56Tro:
Chemistry: A Molecular Approach, 2/e
Slide 57
Copyright 2011 Pearson Education, Inc. IonDipole Attraction In
a mixture, ions from an ionic compound are attracted to the dipole
of polar molecules The strength of the iondipole attraction is one
of the main factors that determines the solubility of ionic
compounds in water 57Tro: Chemistry: A Molecular Approach, 2/e
Slide 58
Copyright 2011 Pearson Education, Inc. a)CH 3 OHCH 3 CHF 2 b)CH
3 CH 2 CH 2 CH 2 CH 3 CH 3 Cl Practice Choose the substance in each
pair that is more soluble in water can H-bond with H 2 O more polar
58Tro: Chemistry: A Molecular Approach, 2/e
Slide 59
Copyright 2011 Pearson Education, Inc. Summary Dispersion
forces are the weakest of the intermolecular attractions Dispersion
forces are present in all molecules and atoms The magnitude of the
dispersion forces increases with molar mass Polar molecules also
have dipoledipole attractive forces 59Tro: Chemistry: A Molecular
Approach, 2/e
Slide 60
Copyright 2011 Pearson Education, Inc. Summary (contd) Hydrogen
bonds are the strongest of the intermolecular attractive forces a
pure substance can have Hydrogen bonds will be present when a
molecule has H directly bonded to either O, N, or F atoms only
example of H bonded to F is HF Iondipole attractions are present in
mixtures of ionic compounds with polar molecules. Iondipole
attractions are the strongest intermolecular attraction Iondipole
attractions are especially important in aqueous solutions of ionic
compounds 60Tro: Chemistry: A Molecular Approach, 2/e
Slide 61
Copyright 2011 Pearson Education, Inc. 61Tro: Chemistry: A
Molecular Approach, 2/e
Slide 62
Copyright 2011 Pearson Education, Inc. Liquids Properties &
Structure Tro: Chemistry: A Molecular Approach, 2/e
Slide 63
Copyright 2011 Pearson Education, Inc. Surface Tension Surface
tension is a property of liquids that results from the tendency of
liquids to minimize their surface area To minimize their surface
area, liquids form drops that are spherical as long as there is no
gravity 63Tro: Chemistry: A Molecular Approach, 2/e
Slide 64
Copyright 2011 Pearson Education, Inc. Surface Tension The
layer of molecules on the surface behave differently than the
interior because the cohesive forces on the surface molecules have
a net pull into the liquid interior The surface layer acts like an
elastic skin allowing you to float a paper clip even though steel
is denser than water 64Tro: Chemistry: A Molecular Approach,
2/e
Slide 65
Copyright 2011 Pearson Education, Inc. Surface Tension Because
they have fewer neighbors to attract them, the surface molecules
are less stable than those in the interior have a higher potential
energy The surface tension of a liquid is the energy required to
increase the surface area a given amount surface tension of H 2 O =
72.8 mJ/m 2 at room temperature surface tension of C 6 H 6 = 28
mJ/m 2 65Tro: Chemistry: A Molecular Approach, 2/e
Slide 66
Copyright 2011 Pearson Education, Inc. Factors Affecting
Surface Tension The stronger the intermolecular attractive forces,
the higher the surface tension will be Raising the temperature of a
liquid reduces its surface tension raising the temperature of the
liquid increases the average kinetic energy of the molecules the
increased molecular motion makes it easier to stretch the surface
66Tro: Chemistry: A Molecular Approach, 2/e
Slide 67
Copyright 2011 Pearson Education, Inc. 67Tro: Chemistry: A
Molecular Approach, 2/e
Slide 68
Copyright 2011 Pearson Education, Inc. Viscosity Viscosity is
the resistance of a liquid to flow 1 poise = 1 P = 1 g/cms often
given in centipoise, cP H 2 O = 1 cP at room temperature Larger
intermolecular attractions = larger viscosity 68Tro: Chemistry: A
Molecular Approach, 2/e
Slide 69
Copyright 2011 Pearson Education, Inc. Factors Affecting
Viscosity The stronger the intermolecular attractive forces, the
higher the liquids viscosity will be The more spherical the
molecular shape, the lower the viscosity will be molecules roll
more easily less surface-to-surface contact lowers attractions
Raising the temperature of a liquid reduces its viscosity raising
the temperature of the liquid increases the average kinetic energy
of the molecules the increased molecular motion makes it easier to
overcome the intermolecular attractions and flow 69Tro: Chemistry:
A Molecular Approach, 2/e
Slide 70
Copyright 2011 Pearson Education, Inc. 70 Insert Table 11.6
Tro: Chemistry: A Molecular Approach, 2/e
Slide 71
Copyright 2011 Pearson Education, Inc. Capillary Action
Capillary action is the ability of a liquid to flow up a thin tube
against the influence of gravity the narrower the tube, the higher
the liquid rises Capillary action is the result of two forces
working in conjunction, the cohesive and adhesive forces cohesive
forces hold the liquid molecules together adhesive forces attract
the outer liquid molecules to the tubes surface 71Tro: Chemistry: A
Molecular Approach, 2/e
Slide 72
Copyright 2011 Pearson Education, Inc. Capillary Action The
adhesive forces pull the surface liquid up the side of the tube,
and the cohesive forces pull the interior liquid with it The liquid
rises up the tube until the force of gravity counteracts the
capillary action forces The narrower the tube diameter, the higher
the liquid will rise up the tube 72Tro: Chemistry: A Molecular
Approach, 2/e
Slide 73
Copyright 2011 Pearson Education, Inc. Meniscus The curving of
the liquid surface in a thin tube is due to the competition between
adhesive and cohesive forces The meniscus of water is concave in a
glass tube because its adhesion to the glass is stronger than its
cohesion for itself The meniscus of mercury is convex in a glass
tube because its cohesion for itself is stronger than its adhesion
for the glass metallic bonds are stronger than intermolecular
attractions 73Tro: Chemistry: A Molecular Approach, 2/e
Slide 74
Copyright 2011 Pearson Education, Inc. The Molecular Dance
Molecules in the liquid are constantly in motion vibrational, and
limited rotational and translational The average kinetic energy is
proportional to the temperature However, some molecules have more
kinetic energy than the average, and others have less 74Tro:
Chemistry: A Molecular Approach, 2/e
Slide 75
Copyright 2011 Pearson Education, Inc. Vaporization If these
high energy molecules are at the surface, they may have enough
energy to overcome the attractive forces therefore the larger the
surface area, the faster the rate of evaporation This will allow
them to escape the liquid and become a vapor 75Tro: Chemistry: A
Molecular Approach, 2/e
Slide 76
Copyright 2011 Pearson Education, Inc. Distribution of Thermal
Energy Only a small fraction of the molecules in a liquid have
enough energy to escape But, as the temperature increases, the
fraction of the molecules with escape energy increases The higher
the temperature, the faster the rate of evaporation 76Tro:
Chemistry: A Molecular Approach, 2/e
Slide 77
Copyright 2011 Pearson Education, Inc. Condensation Some
molecules of the vapor will lose energy through molecular
collisions The result will be that some of the molecules will get
captured back into the liquid when they collide with it Also some
may stick and gather together to form droplets of liquid
particularly on surrounding surfaces We call this process
condensation 77Tro: Chemistry: A Molecular Approach, 2/e
Slide 78
Copyright 2011 Pearson Education, Inc. Evaporation vs.
Condensation Vaporization and condensation are opposite processes
In an open container, the vapor molecules generally spread out
faster than they can condense The net result is that the rate of
vaporization is greater than the rate of condensation, and there is
a net loss of liquid However, in a closed container, the vapor is
not allowed to spread out indefinitely The net result in a closed
container is that at some time the rates of vaporization and
condensation will be equal 78Tro: Chemistry: A Molecular Approach,
2/e
Slide 79
Copyright 2011 Pearson Education, Inc. Effect of Intermolecular
Attraction on Evaporation and Condensation The weaker the
attractive forces between molecules, the less energy they will need
to vaporize Also, weaker attractive forces means that more energy
will need to be removed from the vapor molecules before they can
condense The net result will be more molecules in the vapor phase,
and a liquid that evaporates faster the weaker the attractive
forces, the faster the rate of evaporation Liquids that evaporate
easily are said to be volatile e.g., gasoline, fingernail polish
remover liquids that do not evaporate easily are called nonvolatile
e.g., motor oil 79Tro: Chemistry: A Molecular Approach, 2/e
Slide 80
Copyright 2011 Pearson Education, Inc. Energetics of
Vaporization When the high energy molecules are lost from the
liquid, it lowers the average kinetic energy If energy is not drawn
back into the liquid, its temperature will decrease therefore,
vaporization is an endothermic process and condensation is an
exothermic process Vaporization requires input of energy to
overcome the attractions between molecules 80Tro: Chemistry: A
Molecular Approach, 2/e
Slide 81
Copyright 2011 Pearson Education, Inc. Heat of Vaporization The
amount of heat energy required to vaporize one mole of the liquid
is called the heat of vaporization, H vap sometimes called the
enthalpy of vaporization Slways endothermic, therefore H vap is +
Somewhat temperature dependent H condensation = H vaporization
81Tro: Chemistry: A Molecular Approach, 2/e
Slide 82
Copyright 2011 Pearson Education, Inc. Example 11.3: Calculate
the mass of water that can be vaporized with 155 kJ of heat at 100
C 82 because the given amount of heat is almost 4x the H vap, the
amount of water makes sense 1 mol H 2 O = 40.7 kJ, 1 mol = 18.02 g
155 kJ g H 2 O Check: Solution: Conceptual Plan: Relationships:
Given: Find: kJmol H 2 Og H 2 O Tro: Chemistry: A Molecular
Approach, 2/e
Slide 83
Copyright 2011 Pearson Education, Inc. Practice Calculate the
amount of heat needed to vaporize 90.0 g of C 3 H 7 OH at its
boiling point ( H vap = 39.9 kJ/mol) 83Tro: Chemistry: A Molecular
Approach, 2/e
Slide 84
Copyright 2011 Pearson Education, Inc. Practice Calculate the
amount of heat needed to vaporize 90.0 g of C 3 H 7 OH at its
boiling point 84 because the given amount of C 3 H 7 OH is more
than 1 mole the amount of heat makes sense 1 mol C 3 H 7 OH = 39.9
kJ, 1 mol = 60.09 g 90.0 g kJ Check: Solution: Conceptual Plan:
Relationships: Given: Find: gmolkJ Tro: Chemistry: A Molecular
Approach, 2/e
Slide 85
Copyright 2011 Pearson Education, Inc. Dynamic Equilibrium In a
closed container, once the rates of vaporization and condensation
are equal, the total amount of vapor and liquid will not change
Evaporation and condensation are still occurring, but because they
are opposite processes, there is no net gain or loss of either
vapor or liquid When two opposite processes reach the same rate so
that there is no gain or loss of material, we call it a dynamic
equilibrium this does not mean there are equal amounts of vapor and
liquid it means that they are changing by equal amounts 85Tro:
Chemistry: A Molecular Approach, 2/e
Slide 86
Copyright 2011 Pearson Education, Inc. Dynamic Equilibrium
86Tro: Chemistry: A Molecular Approach, 2/e
Slide 87
Copyright 2011 Pearson Education, Inc. Vapor Pressure The
pressure exerted by the vapor when it is in dynamic equilibrium
with its liquid is called the vapor pressure remember using Daltons
Law of Partial Pressures to account for the pressure of the water
vapor when collecting gases by water displacement? The weaker the
attractive forces between the molecules, the more molecules will be
in the vapor Therefore, the weaker the attractive forces, the
higher the vapor pressure the higher the vapor pressure, the more
volatile the liquid 87Tro: Chemistry: A Molecular Approach,
2/e
Slide 88
Copyright 2011 Pearson Education, Inc. VaporLiquid Dynamic
Equilibrium If the volume of the chamber is increased, it will
decrease the pressure of the vapor inside the chamber at that
point, there are fewer vapor molecules in a given volume, causing
the rate of condensation to slow Therefore, for a period of time,
the rate of vaporization will be faster than the rate of
condensation, and the amount of vapor will increase Eventually
enough vapor accumulates so that the rate of the condensation
increases to the point where it is once again as fast as
evaporation equilibrium is reestablished At this point, the vapor
pressure will be the same as it was before 88Tro: Chemistry: A
Molecular Approach, 2/e
Slide 89
Copyright 2011 Pearson Education, Inc. Changing the Containers
Volume Disturbs the Equilibrium Initially, the rate of vaporization
and condensation are equal and the system is in dynamic equilibrium
When the volume is increased, the rate of vaporization becomes
faster than the rate of condensation When the volume is decreased,
the rate of vaporization becomes slower than the rate of
condensation 89Tro: Chemistry: A Molecular Approach, 2/e
Slide 90
Copyright 2011 Pearson Education, Inc. Dynamic Equilibrium A
system in dynamic equilibrium can respond to changes in the
conditions When conditions change, the system shifts its position
to relieve or reduce the effects of the change 90Tro: Chemistry: A
Molecular Approach, 2/e
Slide 91
Copyright 2011 Pearson Education, Inc. Vapor Pressure vs.
Temperature Increasing the temperature increases the number of
molecules able to escape the liquid The net result is that as the
temperature increases, the vapor pressure increases Small changes
in temperature can make big changes in vapor pressure the rate of
growth depends on strength of the intermolecular forces 91Tro:
Chemistry: A Molecular Approach, 2/e
Slide 92
Copyright 2011 Pearson Education, Inc. Vapor Pressure Curves
760 mmHg normal BP 100 C BP Ethanol at 500 mmHg 68.1C 92Tro:
Chemistry: A Molecular Approach, 2/e
Slide 93
Copyright 2011 Pearson Education, Inc. a)water b)TiCl 4 c)ether
d)ethanol e)acetone Practice Which of the following is the most
volatile? a)water b)TiCl 4 c)ether d)ethanol e)acetone 93Tro:
Chemistry: A Molecular Approach, 2/e
Slide 94
Copyright 2011 Pearson Education, Inc. a)water b)TiCl 4 c)ether
d)ethanol e)acetone Practice Which of the following has the
strongest Intermolecular attractions? a)water b)TiCl 4 c)ether
d)ethanol e)acetone 94Tro: Chemistry: A Molecular Approach,
2/e
Slide 95
Copyright 2011 Pearson Education, Inc. Boiling Point When the
temperature of a liquid reaches a point where its vapor pressure is
the same as the external pressure, vapor bubbles can form anywhere
in the liquid not just on the surface This phenomenon is what is
called boiling and the temperature at which the vapor pressure =
external pressure is the boiling point 95Tro: Chemistry: A
Molecular Approach, 2/e
Slide 96
Copyright 2011 Pearson Education, Inc. Boiling Point The normal
boiling point is the temperature at which the vapor pressure of the
liquid = 1 atm The lower the external pressure, the lower the
boiling point of the liquid 96Tro: Chemistry: A Molecular Approach,
2/e
Slide 97
Copyright 2011 Pearson Education, Inc. a)water b)TiCl 4 c)ether
d)ethanol e)acetone Practice Which of the following has the highest
normal boiling point? a)water b)TiCl 4 c)ether d)ethanol e)acetone
97Tro: Chemistry: A Molecular Approach, 2/e
Slide 98
Copyright 2011 Pearson Education, Inc. Heating Curve of a
Liquid As you heat a liquid, its temperature increases linearly
until it reaches the boiling point q = mass x C s x T Once the
temperature reaches the boiling point, all the added heat goes into
boiling the liquid the temperature stays constant Once all the
liquid has been turned into gas, the temperature can again start to
rise 98Tro: Chemistry: A Molecular Approach, 2/e
Slide 99
Copyright 2011 Pearson Education, Inc. The logarithm of the
vapor pressure vs. inverse absolute temperature is a linear
function A graph of ln(P vap ) vs. 1/T is a straight line The slope
of the line x 8.314 J/molK = H vap in J/mol ClausiusClapeyron
Equation The graph of vapor pressure vs. temperature is an
exponential growth curve 99Tro: Chemistry: A Molecular Approach,
2/e
Slide 100
Copyright 2011 Pearson Education, Inc. Example 11.4: Determine
the H vap of dichloromethane given the vapor pressure vs.
temperature data Enter the data into a spreadsheet and calculate
the inverse of the absolute temperature and natural log of the
vapor pressure 100Tro: Chemistry: A Molecular Approach, 2/e
Slide 101
Copyright 2011 Pearson Education, Inc. Example 11.4: Determine
the H vap of dichloromethane given the vapor pressure vs.
temperature data Graph the inverse of the absolute temperature vs.
the natural log of the vapor pressure 101Tro: Chemistry: A
Molecular Approach, 2/e
Slide 102
Copyright 2011 Pearson Education, Inc. Example 11.4: Determine
the H vap of dichloromethane given the vapor pressure vs.
temperature data Add a trendline, making sure the display equation
on chart option is checked off 102Tro: Chemistry: A Molecular
Approach, 2/e
Slide 103
Copyright 2011 Pearson Education, Inc. Example 11.4: Determine
the H vap of dichloromethane given the vapor pressure vs.
temperature data Determine the slope of the line 3776.7 3800 K
103Tro: Chemistry: A Molecular Approach, 2/e
Slide 104
Copyright 2011 Pearson Education, Inc. Example 11.4: Determine
the H vap of dichloromethane given the vapor pressure vs.
temperature data Use the slope of the line to determine the heat of
vaporization slope 3800 K, R = 8.314 J/molK 104Tro: Chemistry: A
Molecular Approach, 2/e
Slide 105
Copyright 2011 Pearson Education, Inc. ClausiusClapeyron
Equation 2-Point Form The equation below can be used with just two
measurements of vapor pressure and temperature however, it
generally gives less precise results fewer data points will not
give as precise an average because there is less averaging out of
the errors o as with any other sets of measurements It can also be
used to predict the vapor pressure if you know the heat of
vaporization and the normal boiling point remember: the vapor
pressure at the normal boiling point is 760 torr 105Tro: Chemistry:
A Molecular Approach, 2/e
Slide 106
Copyright 2011 Pearson Education, Inc. T 1 = BP = 64.6 C, P 1 =
760 torr, H vap = 35.2 kJ/mol, T 2 = 12.0 C P 2, torr T 1 = BP =
337.8 K, P 1 = 760 torr, H vap = 35.2 kJ/mol, T 2 = 285.2 K P 2,
torr Solution: Example 11.5: Calculate the vapor pressure of
methanol at 12.0 C 106 the units are correct, the size makes sense
because the vapor pressure is lower at lower temperatures Check:
Conceptual Plan: Relationships: Given: Find: P 1, T 1, H vap P2P2
T(K) = T(C) + 273.15 Tro: Chemistry: A Molecular Approach, 2/e
Slide 107
Copyright 2011 Pearson Education, Inc. Practice Determine the
vapor pressure of water at 25 C (normal BP = 100.0 C, H vap = 40.7
kJ/mol) 107Tro: Chemistry: A Molecular Approach, 2/e
Slide 108
Copyright 2011 Pearson Education, Inc. T 1 = BP = 100.0 C, P 1
= 760 torr, H vap = 40.7 kJ/mol, T 2 = 25.0 C P 2, torr T 1 = BP =
373.2 K, P 1 = 760 torr, H vap = 40.7 kJ/mol, T 2 = 298.2 K P 2,
torr Practice Determine the vapor pressure of water at 25 C 108 the
units are correct, the size makes sense because the vapor pressure
is lower at lower temperatures Check: Solution: Conceptual Plan:
Relationships: Given: Find: P 1, T 1, H vap P2P2 T(K) = T(C) +
273.15 Tro: Chemistry: A Molecular Approach, 2/e
Slide 109
Copyright 2011 Pearson Education, Inc. Supercritical Fluid As a
liquid is heated in a sealed container, more vapor collects,
causing the pressure inside the container to rise and the density
of the vapor to increase and the density of the liquid to decrease
At some temperature, the meniscus between the liquid and vapor
disappears and the states commingle to form a supercritical fluid
Supercritical fluids have properties of both gas and liquid states
109Tro: Chemistry: A Molecular Approach, 2/e
Slide 110
Copyright 2011 Pearson Education, Inc. The Critical Point The
temperature required to produce a supercritical fluid is called the
critical temperature The pressure at the critical temperature is
called the critical pressure At the critical temperature or higher
temperatures, the gas cannot be condensed to a liquid, no matter
how high the pressure gets 110Tro: Chemistry: A Molecular Approach,
2/e
Slide 111
Copyright 2011 Pearson Education, Inc. Sublimation and
Deposition Molecules in the solid have thermal energy that allows
them to vibrate Surface molecules with sufficient energy may break
free from the surface and become a gas this process is called
sublimation The capturing of vapor molecules into a solid is called
deposition The solid and vapor phases exist in dynamic equilibrium
in a closed container at temperatures below the melting point
therefore, molecular solids have a vapor pressure solid gas
sublimation deposition 111Tro: Chemistry: A Molecular Approach,
2/e
Slide 112
Copyright 2011 Pearson Education, Inc. Sublimation 112Tro:
Chemistry: A Molecular Approach, 2/e
Slide 113
Copyright 2011 Pearson Education, Inc. Melting = Fusion As a
solid is heated, its temperature rises and the molecules vibrate
more vigorously Once the temperature reaches the melting point, the
molecules have sufficient energy to overcome some of the
attractions that hold them in position and the solid melts (or
fuses) The opposite of melting is freezing 113Tro: Chemistry: A
Molecular Approach, 2/e
Slide 114
Copyright 2011 Pearson Education, Inc. Heating Curve of a Solid
As you heat a solid, its temperature increases linearly until it
reaches the melting point q = mass x C s x T Once the temperature
reaches the melting point, all the added heat goes into melting the
solid the temperature stays constant Once all the solid has been
turned into liquid, the temperature can again start to rise
ice/water will always have a temperature of 0 C at 1 atm 114Tro:
Chemistry: A Molecular Approach, 2/e
Slide 115
Copyright 2011 Pearson Education, Inc. Energetics of Melting
When the high energy molecules are lost from the solid, it lowers
the average kinetic energy If energy is not drawn back into the
solid its temperature will decrease therefore, melting is an
endothermic process and freezing is an exothermic process Melting
requires input of energy to overcome the attractions between
molecules 115Tro: Chemistry: A Molecular Approach, 2/e
Slide 116
Copyright 2011 Pearson Education, Inc. Heat of Fusion The
amount of heat energy required to melt one mole of the solid is
called the Heat of Fusion, H fus sometimes called the enthalpy of
fusion Always endothermic, therefore H fus is + Somewhat
temperature dependent H crystallization = H fusion Generally much
less than H vap H sublimation = H fusion + H vaporization 116Tro:
Chemistry: A Molecular Approach, 2/e
Slide 117
Copyright 2011 Pearson Education, Inc. Heats of Fusion and
Vaporization 117Tro: Chemistry: A Molecular Approach, 2/e
Slide 118
Copyright 2011 Pearson Education, Inc. Heating Curve of Water
118Tro: Chemistry: A Molecular Approach, 2/e
Slide 119
Copyright 2011 Pearson Education, Inc. Segment 1 Heating 1.00
mole of ice at 25.0 C up to the melting point, 0.0 C q = mass x C s
x T mass of 1.00 mole of ice = 18.0 g C s = 2.09 J/molC 119Tro:
Chemistry: A Molecular Approach, 2/e
Slide 120
Copyright 2011 Pearson Education, Inc. Segment 2 Melting 1.00
mole of ice at the melting point, 0.0 C q = n H fus n = 1.00 mole
of ice H fus = 6.02 kJ/mol 120Tro: Chemistry: A Molecular Approach,
2/e
Slide 121
Copyright 2011 Pearson Education, Inc. Segment 3 Heating 1.00
mole of water at 0.0 C up to the boiling point, 100.0 C q = mass x
C s x T mass of 1.00 mole of water = 18.0 g C s = 2.09 J/molC
121Tro: Chemistry: A Molecular Approach, 2/e
Slide 122
Copyright 2011 Pearson Education, Inc. Segment 4 Boiling 1.00
mole of water at the boiling point, 100.0 C q = n H vap n = 1.00
mole of ice H fus = 40.7 kJ/mol 122Tro: Chemistry: A Molecular
Approach, 2/e
Slide 123
Copyright 2011 Pearson Education, Inc. Segment 5 Heating 1.00
mole of steam at 100.0 C up to 125.0 C q = mass x C s x T mass of
1.00 mole of water = 18.0 g C s = 2.01 J/molC 123Tro: Chemistry: A
Molecular Approach, 2/e
Slide 124
Copyright 2011 Pearson Education, Inc. Practice How much heat,
in kJ, is needed to raise the temperature of a 12.0 g benzene
sample from 10.0 C to 25.0 C? 124Tro: Chemistry: A Molecular
Approach, 2/e
Slide 125
Copyright 2011 Pearson Education, Inc. 12.0 g benzene, seg 1 (T
1 = 10.0 C, T 2 = 5.5 C), seg 2 = melting, seg 3 (T 1 = 5.5 C, T 2
= 25.0 C) kJ 12.0 g benzene, seg 1 = 0.2325 kJ, seg 2 = 1.51 kJ,
seg 3 (T 1 = 5.5 C, T 2 = 25.0 C) kJ Practice How much heat is
needed to raise the temperature of a 12.0 g benzene sample from
10.0 C to 25.0 C? 125 H fus 9.8 kJ/mol, 1 mol = 78.11 g, 1 kJ =
1000 J, q = mC s T C s,sol = 1.25 J/gC, C s,liq = 1.70 J/gC
Solution: Conceptual Plan: Relationships: Given: Find: gJkJ Seg 1
12.0 g benzene, seg 1 = 0.2325 kJ, seg 2 = melting, seg 3 (T 1 =
5.5 C, T 2 = 25.0 C) kJ gJ Seg 3 gmolkJ Seg 2 12.0 g benzene, seg 1
= 0.2325 kJ, seg 2 = 1.51 kJ, seg 3 = 0.3978 kJ kJ Tro: Chemistry:
A Molecular Approach, 2/e
Slide 126
Copyright 2011 Pearson Education, Inc. Phase Diagrams Phase
diagrams describe the different states and state changes that occur
at various temperature/pressure conditions Regions represent states
Lines represent state changes liquid/gas line is vapor pressure
curve both states exist simultaneously critical point is the
furthest point on the vapor pressure curve Triple point is the
temperature/pressure condition where all three states exist
simultaneously For most substances, freezing point increases as
pressure increases 126Tro: Chemistry: A Molecular Approach,
2/e
Slide 127
Copyright 2011 Pearson Education, Inc. Phase Diagrams Pressure
Temperature vaporization condensation critical point triple point
Solid Liquid Gas 1 atm normal melting pt. normal boiling pt. Fusion
Curve Vapor Pressure Curve Sublimation Curve melting freezing
sublimation deposition 127Tro: Chemistry: A Molecular Approach,
2/e
Slide 128
Copyright 2011 Pearson Education, Inc. 128Tro: Chemistry: A
Molecular Approach, 2/e
Slide 129
Copyright 2011 Pearson Education, Inc. Phase Diagram of Water
Temperature Pressure critical point 374.1 C 217.7 atm triple point
Ice Water Steam 1 atm normal boiling pt. 100 C normal melting pt. 0
C 0.01 C 0.006 atm 129Tro: Chemistry: A Molecular Approach,
2/e
Slide 130
Copyright 2011 Pearson Education, Inc. 130Tro: Chemistry: A
Molecular Approach, 2/e
Slide 131
Copyright 2011 Pearson Education, Inc. Morphic Forms of Ice
131Tro: Chemistry: A Molecular Approach, 2/e
Slide 132
Copyright 2011 Pearson Education, Inc. Phase Diagram of CO 2
Pressure Temperature critical point 31.0 C 72.9 atm triple point
Solid Liquid Gas 1 atm -56.7 C 5.1 atm normal sublimation pt. -78.5
C 132Tro: Chemistry: A Molecular Approach, 2/e
Slide 133
Copyright 2011 Pearson Education, Inc. 133Tro: Chemistry: A
Molecular Approach, 2/e
Slide 134
Copyright 2011 Pearson Education, Inc. 134Tro: Chemistry: A
Molecular Approach, 2/e
Slide 135
Copyright 2011 Pearson Education, Inc. 20.0 C, 72.9 atm 56.7 C,
5.1 atm 10.0 C, 1.0 atm 78.5 C, 1.0 atm 50.0 C, 80.0 atm 20.0 C,
72.9 atm liquid 56.7 C, 5.1 atm solid, liquid, gas 10.0 C, 1.0 atm
gas 78.5 C, 1.0 atm solid, gas 50.0 C, 80.0 atm scf Practice
Consider the phase diagram of CO 2 shown. What phase(s) is/are
present at each of the following conditions? 135Tro: Chemistry: A
Molecular Approach, 2/e
Slide 136
Copyright 2011 Pearson Education, Inc. Water An Extraordinary
Substance Water is a liquid at room temperature most molecular
substances with similar molar masses are gases at room temperature
e.g. NH 3, CH 4 due to H-bonding between molecules Water is an
excellent solvent dissolving many ionic and polar molecular
substances because of its large dipole moment even many small
nonpolar molecules have some solubility in water e.g. O 2, CO 2
Water has a very high specific heat for a molecular substance
moderating effect on coastal climates Water expands when it freezes
at a pressure of 1 atm about 9% making ice less dense than liquid
water 136Tro: Chemistry: A Molecular Approach, 2/e
Slide 137
Copyright 2011 Pearson Education, Inc. Solids Properties &
Structure Tro: Chemistry: A Molecular Approach, 2/e
Slide 138
Copyright 2011 Pearson Education, Inc. Crystal Lattice When
allowed to cool slowly, the particles in a liquid will arrange
themselves to give the maximum attractive forces therefore minimize
the energy The result will generally be a crystalline solid The
arrangement of the particles in a crystalline solid is called the
crystal lattice The smallest unit that shows the pattern of
arrangement for all the particles is called the unit cell 138Tro:
Chemistry: A Molecular Approach, 2/e
Slide 139
Copyright 2011 Pearson Education, Inc. Unit Cells Unit cells
are 3-dimensional usually containing 2 or 3 layers of particles
Unit cells are repeated over and over to give the macroscopic
crystal structure of the solid Starting anywhere within the crystal
results in the same unit cell Each particle in the unit cell is
called a lattice point Lattice planes are planes connecting
equivalent points in unit cells throughout the lattice 139Tro:
Chemistry: A Molecular Approach, 2/e
Slide 140
Copyright 2011 Pearson Education, Inc. 7 Unit Cells Cubic a = b
= c all 90 a b c Tetragonal a = c < b all 90 a b c Orthorhombic
a b c all 90 a b c Monoclinic a b c 2 faces 90 a b c Hexagonal a =
c < b 2 faces 90 1 face 120 c a b Rhombohedral a = b = c no 90 a
b c Triclinic a b c no 90 a b c 140
Slide 141
Copyright 2011 Pearson Education, Inc. Unit Cells The number of
other particles each particle is in contact with is called its
coordination number for ions, it is the number of oppositely
charged ions an ion is in contact with Higher coordination number
means more interaction, therefore stronger attractive forces
holding the crystal together The packing efficiency is the
percentage of volume in the unit cell occupied by particles the
higher the coordination number, the more efficiently the particles
are packing together 141Tro: Chemistry: A Molecular Approach,
2/e
Slide 142
Copyright 2011 Pearson Education, Inc. Cubic Unit Cells All 90
angles between corners of the unit cell The length of all the edges
are equal If the unit cell is made of spherical particles of each
corner particle is within the cube of each particle on a face is
within the cube of each particle on an edge is within the cube
142Tro: Chemistry: A Molecular Approach, 2/e
Slide 143
Copyright 2011 Pearson Education, Inc. 143Tro: Chemistry: A
Molecular Approach, 2/e
Slide 144
Copyright 2011 Pearson Education, Inc. Cubic Unit Cells -
Simple Cubic Eight particles, one at each corner of a cube 1/8 th
of each particle lies in the unit cell each particle part of eight
cells total = one particle in each unit cell 8 corners x 1/8 Edge
of unit cell = twice the radius Coordination number of 6 2r 144Tro:
Chemistry: A Molecular Approach, 2/e
Slide 145
Copyright 2011 Pearson Education, Inc. Simple Cubic 145Tro:
Chemistry: A Molecular Approach, 2/e
Slide 146
Copyright 2011 Pearson Education, Inc. Cubic Unit Cells -
Body-Centered Cubic Nine particles, one at each corner of a cube +
one in center 1/8 th of each corner particle lies in the unit cell
two particles in each unit cell 8 corners x 1/8 + 1 center Edge of
unit cell = (4/ 3) times the radius of the particle Coordination
number of 8 146Tro: Chemistry: A Molecular Approach, 2/e
Slide 147
Copyright 2011 Pearson Education, Inc. Body-Centered Cubic
147Tro: Chemistry: A Molecular Approach, 2/e
Slide 148
Copyright 2011 Pearson Education, Inc. Cubic Unit Cells -
Face-Centered Cubic 14 particles, one at each corner of a cube +
one in center of each face 1/8 th of each corner particle + 1/2 of
face particle lies in the unit cell 4 particles in each unit cell 8
corners x 1/8 + 6 faces x 1/2 Edge of unit cell = 2 2 times the
radius of the particle Coordination number of 12 148Tro: Chemistry:
A Molecular Approach, 2/e
Slide 149
Copyright 2011 Pearson Education, Inc. Face-Centered Cubic
149Tro: Chemistry: A Molecular Approach, 2/e
Slide 150
Copyright 2011 Pearson Education, Inc. Solution: fcc = 4
atoms/uc, Al = 26.982 g/mol, 1 mol = 6.022 x 10 23 atoms Example
11.6: Calculate the density of Al if it crystallizes in a fcc and
has a radius of 143 pm 150 the accepted density of Al at 20C is
2.71 g/cm 3, so the answer makes sense face-centered cubic, r = 143
pm density, g/cm 3 Check: Conceptual Plan: Relationships: Given:
Find: 1 cm = 10 2 m, 1 pm = 10 12 m lrVmassfccdm, V # atoms x mass
of 1 atom V = l 3, l = 2r2, d = m/V d = m/V l = 2r2V = l 3
face-centered cubic, r = 1.43 x 10 8 cm, m = 1.792 x 10 22 g
density, g/cm 3 face-centered cubic, V = 6.618 x 10 23 cm 3, m
=1.792 x 10 22 g density, g/cm 3 Tro: Chemistry: A Molecular
Approach, 2/e
Slide 151
Copyright 2011 Pearson Education, Inc. Practice Estimate the
density of Rb if it crystallizes in a body-centered cubic unit cell
and has an atomic radius of 247.5 pm 151Tro: Chemistry: A Molecular
Approach, 2/e
Slide 152
Copyright 2011 Pearson Education, Inc. Solution: bcc = 2
atoms/uc, Rb = 85.47 g/mol, 1 mol = 6.022 x 10 23 atoms Practice
Estimate the density of Rb if it crystallizes in a bcc and has a
radius of 247.5 pm the accepted density of Rb at 20C is 1.53 g/cm
3, so the answer makes sense body-centered cubic, r = 247.5 pm
density, g/cm 3 Check: Conceptual Plan: Relationships: Given: Find:
1 cm = 10 2 m, 1 pm = 10 12 m lrVmassbccdm, V # atoms x mass of 1
atom V = l 3, l = 4r/3, d = m/V d = m/V l = 4r/3V = l 3
body-centered cubic, r = 2.475 x 10 8 cm, m = 2.839 x 10 22 g
density, g/cm 3 body-centered cubic, V= 1.868 x 10 22 cm 3, m=
2.839 x 10 22 g density, g/cm 3 152Tro: Chemistry: A Molecular
Approach, 2/e
Slide 153
Copyright 2011 Pearson Education, Inc. Closest-Packed
Structures First Layer With spheres, it is more efficient to offset
each row in the gaps of the previous row than to line up rows and
columns 153Tro: Chemistry: A Molecular Approach, 2/e
Slide 154
Copyright 2011 Pearson Education, Inc. Closest-Packed
Structures Second Layer The second layer atoms can sit directly
over the atoms in the first layer called an AA pattern Or the
second layer can sit over the holes in the first layer called an AB
pattern 154Tro: Chemistry: A Molecular Approach, 2/e
Slide 155
Copyright 2011 Pearson Education, Inc. Closest-Packed
Structures Third Layer with Offset 2 nd Layer The third layer atoms
can align directly over the atoms in the first layer called an ABA
pattern Or the third layer can sit over the uncovered holes in the
first layer called an ABC pattern Hexagonal Closest-Packed Cubic
Closest-Packed Face-Centered Cubic 155Tro: Chemistry: A Molecular
Approach, 2/e
Slide 156
Copyright 2011 Pearson Education, Inc. Hexagonal Closest-Packed
Structures 156Tro: Chemistry: A Molecular Approach, 2/e
Slide 157
Copyright 2011 Pearson Education, Inc. Cubic Closest-Packed
Structures 157Tro: Chemistry: A Molecular Approach, 2/e
Slide 158
Copyright 2011 Pearson Education, Inc. Classifying Crystalline
Solids Crystalline solids are classified by the kinds of particles
found Some of the categories are sub-classified by the kinds of
attractive forces holding the particles together 158Tro: Chemistry:
A Molecular Approach, 2/e
Slide 159
Copyright 2011 Pearson Education, Inc. Classifying Crystalline
Solids Molecular solids are solids whose composite particles are
molecules Ionic solids are solids whose composite particles are
ions Atomic solids are solids whose composite particles are atoms
nonbonding atomic solids are held together by dispersion forces
metallic atomic solids are held together by metallic bonds network
covalent atomic solids are held together by covalent bonds 159Tro:
Chemistry: A Molecular Approach, 2/e
Slide 160
Copyright 2011 Pearson Education, Inc. 160Tro: Chemistry: A
Molecular Approach, 2/e
Slide 161
Copyright 2011 Pearson Education, Inc. Molecular Solids The
lattice sites are occupied by molecules CO 2, H 2 O, C 12 H 22 O 11
The molecules are held together by intermolecular attractive forces
dispersion forces, dipoledipole attractions, and H-bonds Because
the attractive forces are weak, they tend to have low melting
points generally < 300 C 161Tro: Chemistry: A Molecular
Approach, 2/e
Slide 162
Copyright 2011 Pearson Education, Inc. Ionic Solids Lattice
sites occupied by ions Held together by attractions between
oppositely charged ions nondirectional therefore every cation
attracts all anions around it, and vice-versa The coordination
number represents the number of close cationanion interactions in
the crystal The higher the coordination number, the more stable the
solid lowers the potential energy of the solid The coordination
number depends on the relative sizes of the cations and anions that
maintains charge balance generally, anions are larger than cations
the number of anions that can surround the cation is limited by the
size of the cation the closer in size the ions are, the higher the
coordination number is 162Tro: Chemistry: A Molecular Approach,
2/e
Slide 163
Copyright 2011 Pearson Education, Inc. Ionic Crystals CsCl
coordination number = 8 Cs + = 167 pm Cl = 181 pm NaCl coordination
number = 6 Na + = 97 pm Cl = 181 pm 163Tro: Chemistry: A Molecular
Approach, 2/e
Slide 164
Copyright 2011 Pearson Education, Inc. Lattice Holes Simple
Cubic Hole Octahedral Hole Tetrahedral Hole 164Tro: Chemistry: A
Molecular Approach, 2/e
Slide 165
Copyright 2011 Pearson Education, Inc. Lattice Holes In
hexagonal closest-packed or cubic closest- packed lattices there
are eight tetrahedral holes and four octahedral holes per unit cell
In a simple cubic lattice there is one cubic hole per unit cell
Number and type of holes occupied determines formula (empirical) of
the salt = Octahedral = Tetrahedral 165Tro: Chemistry: A Molecular
Approach, 2/e
Slide 166
Copyright 2011 Pearson Education, Inc. Cesium Chloride
Structures Coordination number = 8 of each Cl (184 pm) inside the
unit cell Whole Cs + (167 pm) inside the unit cell cubic hole =
hole in simple cubic arrangement of Cl ions Cs:Cl = 1: (8 x ),
therefore the formula is CsCl 166Tro: Chemistry: A Molecular
Approach, 2/e
Slide 167
Copyright 2011 Pearson Education, Inc. Rock Salt Structures
Coordination number = 6 Cl ions (181 pm) in a face-centered cubic
arrangement of each corner Cl inside the unit cell of each face Cl
inside the unit cell Na + (97 pm) in holes between Cl octahedral
holes 1 in center of unit cell 1 whole particle in every octahedral
hole of each edge Na + inside the unit cell Na:Cl = ( x 12) + 1: (
x 8) + ( x 6) = 4:4 = 1:1, Therefore the formula is NaCl 167Tro:
Chemistry: A Molecular Approach, 2/e
Slide 168
Copyright 2011 Pearson Education, Inc. Zinc Blende Structures
Coordination number = 4 S 2 ions (184 pm) in a face-centered cubic
arrangement of each corner S 2 inside the unit cell of each face S
2 inside the unit cell Each Zn 2+ (74 pm) in holes between S 2
tetrahedral holes 1 whole particle in the holes Zn:S = (4 x 1) : (
x 8) + ( x 6) = 4:4 = 1:1, Therefore the formula is ZnS 168Tro:
Chemistry: A Molecular Approach, 2/e
Slide 169
Copyright 2011 Pearson Education, Inc. Fluorite Structures
Coordination number = 4 Ca 2+ ions (99 pm) in a face-centered cubic
arrangement of each corner Ca 2+ inside the unit cell of each face
Ca 2+ inside the unit cell Each F (133 pm) in holes between Ca 2+
tetrahedral holes 1 whole particle in all the holes Ca:F = ( x 8) +
( x 6): (8 x 1) = 4:8 = 1:2, Therefore the formula is CaF 2
fluorite structure common for 1:2 ratio Usually get the
antifluorite structure when the cation:anion ratio is 2:1 the
anions occupy the lattice sites and the cations occupy the
tetrahedral holes 169Tro: Chemistry: A Molecular Approach, 2/e
Slide 170
Copyright 2011 Pearson Education, Inc. Practice Gallium
arsenide crystallizes in a cubic closest-packed array of arsenide
ions with gallium ions in the tetrahedral holes. What is the ratio
of gallium ions to arsenide ions in the structure and the empirical
formula of the compound? As = cpp = 4 atoms per unit cell Ga = (8
tetrahedral holes per unit cell) Ga = 4 atoms per unit cell Ga:As =
4 atoms :4 atoms per unit cell = 1:1 The formula is GaAs 170Tro:
Chemistry: A Molecular Approach, 2/e
Slide 171
Copyright 2011 Pearson Education, Inc. Nonbonding Atomic Solids
Noble gases in solid form Solid held together by weak dispersion
forces very low melting Tend to arrange atoms in closest-packed
structure either hexagonal cp or cubic cp maximizes attractive
forces and minimizes energy 171Tro: Chemistry: A Molecular
Approach, 2/e
Slide 172
Copyright 2011 Pearson Education, Inc. Metallic Atomic Solids
Solid held together by metallic bonds strength varies with sizes
and charges of cations coulombic attractions Melting point varies
Mostly closest-packed arrangements of the lattice points cations
172Tro: Chemistry: A Molecular Approach, 2/e
Slide 173
Copyright 2011 Pearson Education, Inc. Metallic Structure
173Tro: Chemistry: A Molecular Approach, 2/e
Slide 174
Copyright 2011 Pearson Education, Inc. Metallic Bonding Metal
atoms release their valence electrons Metal cation islands fixed in
a sea of mobile electrons e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e-
e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- +++++++++ +++++++++
+++++++++ 174Tro: Chemistry: A Molecular Approach, 2/e
Slide 175
Copyright 2011 Pearson Education, Inc. Network Covalent Solids
Atoms attached to their nearest neighbors by covalent bonds Because
of the directionality of the covalent bonds, these do not tend to
form closest-packed arrangements in the crystal Because of the
strength of the covalent bonds, these have very high melting points
generally > 1000 C Dimensionality of the network affects other
physical properties 175Tro: Chemistry: A Molecular Approach,
2/e
Slide 176
Copyright 2011 Pearson Education, Inc. The Diamond Structure: a
3-Dimensional Network The carbon atoms in a diamond each have four
covalent bonds to surrounding atoms sp 3 tetrahedral geometry This
effectively makes each crystal one giant molecule held together by
covalent bonds you can follow a path of covalent bonds from any
atom to every other atom 176Tro: Chemistry: A Molecular Approach,
2/e
Slide 177
Copyright 2011 Pearson Education, Inc. Properties of Diamond
Very high melting point, ~3800 C need to overcome some covalent
bonds Very rigid due to the directionality of the covalent bonds
Very hard due to the strong covalent bonds holding the atoms in
position used as abrasives Electrical insulator Thermal conductor
best known Chemically very nonreactive 177Tro: Chemistry: A
Molecular Approach, 2/e
Slide 178
Copyright 2011 Pearson Education, Inc. The Graphite Structure:
a 2-Dimensional Network In graphite, the carbon atoms in a sheet
are covalently bonded together forming six-member flat rings fused
together similar to benzene bond length = 142 pm sp 2 each C has
three sigma and one pi bond trigonal-planar geometry each sheet a
giant molecule The sheets are then stacked and held together by
dispersion forces sheets are 341 pm apart 178Tro: Chemistry: A
Molecular Approach, 2/e
Slide 179
Copyright 2011 Pearson Education, Inc. Properties of Graphite
Hexagonal crystals High melting point, ~3800 C need to overcome
some covalent bonding Slippery feel because there are only
dispersion forces holding the sheets together, they can slide past
each other glide planes lubricants Electrical conductor parallel to
sheets Thermal insulator Chemically very nonreactive 179Tro:
Chemistry: A Molecular Approach, 2/e
Slide 180
Copyright 2011 Pearson Education, Inc. Silicates ~90% of Earths
crust Extended arrays of Si O sometimes with Al substituted for Si
aluminosilicates Glass is the amorphous form 180Tro: Chemistry: A
Molecular Approach, 2/e
Slide 181
Copyright 2011 Pearson Education, Inc. Quartz SiO 2 in pure
form impurities add color 3-dimensional array of Si covalently
bonded to 4 O tetrahedral Melts at ~1600 C Very hard 181Tro:
Chemistry: A Molecular Approach, 2/e
Slide 182
Copyright 2011 Pearson Education, Inc. Micas There are various
kinds of mica that have slightly different compositions but are all
of the general form X 2 Y 4-6 Z 8 O 20 (OH,F) 4 X is K, Na, or Ca
or less commonly Ba, Rb, or Cs Y is Al, Mg, or Fe or less commonly
Mn, Cr, Ti, Li, etc. Z is chiefly Si or Al but also may include Fe
3+ or Ti Minerals that are mainly 2-dimensional arrays of Si bonded
to O hexagonal arrangement of atoms Sheets Chemically stable
Thermal and electrical insulator 182Tro: Chemistry: A Molecular
Approach, 2/e
Slide 183
Copyright 2011 Pearson Education, Inc. a)KCl ionicSCl 2
molecular b)C(s, graphite) cov. network S 8 molecular c)Kr atomicK
metallic d)SrCl 2 ionicSiO 2 (s, quartz) cov. network Practice Pick
the solid in each pair with the highest melting point a)KClSCl 2
b)C(s, graphite) c)KrK d)SrCl 2 SiO 2 (s, quartz) 183Tro:
Chemistry: A Molecular Approach, 2/e
Slide 184
Copyright 2011 Pearson Education, Inc. Band Theory The
structures of metals and covalent network solids result in every
atoms orbitals being shared by the entire structure For large
numbers of atoms, this results in a large number of molecular
orbitals that have approximately the same energy; we call this an
energy band 184Tro: Chemistry: A Molecular Approach, 2/e
Slide 185
Copyright 2011 Pearson Education, Inc. Band Theory When two
atomic orbitals combine they produce both a bonding and an
antibonding molecular orbital When many atomic orbitals combine
they produce a band of bonding molecular orbitals and a band of
antibonding molecular orbitals The band of bonding molecular
orbitals is called the valence band The band of antibonding
molecular orbitals is called the conduction band 185Tro: Chemistry:
A Molecular Approach, 2/e
Slide 186
Copyright 2011 Pearson Education, Inc. Molecular Orbitals of
Polylithium 186Tro: Chemistry: A Molecular Approach, 2/e
Slide 187
Copyright 2011 Pearson Education, Inc. Band Gap At absolute
zero, all the electrons will occupy the valence band As the
temperature rises, some of the electrons may acquire enough energy
to jump to the conduction band The difference in energy between the
valence band and conduction band is called the band gap the larger
the band gap, the fewer electrons there are with enough energy to
make the jump 187Tro: Chemistry: A Molecular Approach, 2/e
Slide 188
Copyright 2011 Pearson Education, Inc. Types of Band Gaps and
Conductivity 188Tro: Chemistry: A Molecular Approach, 2/e
Slide 189
Copyright 2011 Pearson Education, Inc. Band Gap and
Conductivity The more electrons at any one time that a substance
has in the conduction band, the better conductor of electricity it
is If the band gap is ~0, then the electrons will be almost as
likely to be in the conduction band as the valence band and the
material will be a conductor metals the conductivity of a metal
decreases with temperature If the band gap is small, then a
significant number of the electrons will be in the conduction band
at normal temperatures and the material will be a semiconductor
graphite the conductivity of a semiconductor increases with
temperature If the band gap is large, then effectively no electrons
will be in the conduction band at normal temperatures and the
material will be an insulator 189Tro: Chemistry: A Molecular
Approach, 2/e
Slide 190
Copyright 2011 Pearson Education, Inc. Doping Semiconductors
Doping is adding impurities to the semiconductors crystal to
increase its conductivity Goal is to increase the number of
electrons in the conduction band n-type semiconductors do not have
enough electrons themselves to add to the conduction band, so they
are doped by adding electron-rich impurities p-type semiconductors
are doped with an electron-deficient impurity, resulting in
electron holes in the valence band. Electrons can jump between
these holes in the valence band, allowing conduction of
electricity. 190Tro: Chemistry: A Molecular Approach, 2/e
Slide 191
Copyright 2011 Pearson Education, Inc. Diodes When a p-type
semiconductor adjoins an n-type semiconductor, the result is an p-n
junction Electricity can flow across the p-n junction in only one
direction this is called a diode This also allows the accumulation
of electrical energy called an amplifier 191Tro: Chemistry: A
Molecular Approach, 2/e