Tro Ch 11 Lecture PP - Highline Community College

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Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Chapter 11
Liquids,
Solids, and
Intermolecular
Forces
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
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
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Properties of the
three Phases of Matter
of r
th la
n g cu
re le
St o
rm ns
te o
In ti
ac
?
ttr
ow
A
Fl
it
e?
ill
bl
si
W
es
pr
om
C
State Shape Volume
Solid
fixed
fixed
Liquid indefinite fixed
Gas indefinite indefinite
Density
high
high
low
No No
No Yes
Yes Yes
very strong
intermediate
weak
• Fixed = keeps shape when placed in a container
• Indefinite = takes the shape of the container
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Three Phases of Water
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.
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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
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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
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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
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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
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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
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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
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Compressibility
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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
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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 don’t 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
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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
KEavg = 1.5 kT
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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
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Kinetic–Molecular 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
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Gas Structure
Gas molecules are rapidly
moving in random straight
lines, and are free from
sticking to each other.
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Kinetic–Molecular 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
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Kinetic–Molecular 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
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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
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Phase Changes
• Because the attractive forces between the molecules are
•
•
fixed, changing the material’s 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
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Phase Changes
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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
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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
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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
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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
dipole–dipole attractions
• An especially strong dipole–dipole attraction
results when H is attached to an extremely
electronegative atom. These are called
hydrogen bonds.
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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
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Dispersion Force
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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
molecules
that are
flat
molecules
have
have
more surface
larger electron cloud = increased
more
electrons,
leading
interaction
than
polarizability
polarizability = stronger attractions to increased
spherical ones
Shape of the molecule
 more surface-to-surface contact =
+
+
++
+
larger induced dipole = stronger
+
+
+ + + + + + ++ + + +
+
attraction
− −- − − −−
−
−
−- −
−−
−
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Effect of Molecular Size
on Size of Dispersion Force
As
molar
mass
Thethe
Noble
gases
increases,
the number
are all nonpolar
of
electrons
increases.
atomic
elements
Therefore the strength
of the dispersion
forces increases.
The stronger the
attractive forces
between the
molecules, the higher
the boiling point will be.
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Properties of Straight Chain Alkanes
NonPolar Molecules
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Boiling Points of n-Alkanes
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Effect of Molecular Shape
on Size of Dispersion Force
the larger surface-tosurface contact between
molecules in n-pentane
results in stronger
dispersion force
attractions
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Alkane Boiling Points
• Branched
•
chains have
lower BPs than
straight chains
The straight
chain isomers
have more
surface-tosurface contact
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Practice – Choose the Substance in Each Pair
with the Higher Boiling Point
a) CH4
C4H10
b) C6H12
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Practice – Choose the Substance in Each Pair
with the Higher Boiling Point
a) CH4
CH3CH2CH2CH3
b) CH3CH2CH=CHCH2CH3
Both molecules
are nonpolar
larger molar
mass
cyclohexane
Both molecules
are nonpolar, but
the flatter ring
molecule has
larger surface-tosurface contact
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Dipole–Dipole 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
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Effect of Dipole–Dipole Attraction on
Boiling and Melting Points
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replace with the
figure 11.8
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Example 11.1b: Determine if dipole–dipole
attractions occur between CH2Cl2 molecules
Given: CH2Cl2, EN C = 2.5, H = 2.1, Cl = 3.0
Find: If there are dipole–dipole attractions
Conceptual
Plan:
Formula
Lewis
Structure
Bond
Polarity
EN Difference
Relationships:
Solution:
Molecule
Polarity
Shape
molecules that have dipole–dipole attractions must be polar
Cl—C
3.0−2.5areas
= 0.5
4 bonding
polarpairs =
no lone
tetrahedral
C—H shape
2.5−2.1 = 0.4
nonpolar
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polar bonds and
tetrahedral shape
= polar molecule
polar molecule;
therefore dipole–
dipole attractions
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Practice – Choose the substance in each
pair with the higher boiling point
a)
CH2FCH2F
CH3CHF2
b)
or
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Practice – Choose the substance in each
pair with the higher boiling point
a)
CH2FCH2F
CH3CHF2
more polar
b)
or
polar
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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
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H-Bonding
HF
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H-Bonding in Water
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H-Bonds
• Hydrogen bonds are very strong intermolecular
attractive forces
stronger than dipole–dipole 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
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Effect of H-Bonding on Boiling Point
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For
HF,nonpolar
H2O, andmolecules,
NH3 have unusually
such as the
hydrides
strong dipole-dipole
of Group 4, attractions,
the
intermolecular
called hydrogen
attractions
bonds. Therefore
are due to
dispersion
they have forces.
higher boiling
Therefore
points
they
than
increase
you would
down
expect
the from
column,
the causing
general
the
trends.
boiling point to increase.
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Polar molecules, such as the
hydrides of Groups 5–7, have both
dispersion forces and dipole–dipole
attractions. Therefore they have
higher boiling points than the
corresponding Group 4 molecules.
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Example 11.2: One of these compounds is a
liquid at room temperature (the others are
gases). Which one and why?
MM = 30.03
Polar
No H-Bonds
MM = 34.03
Polar
No H-Bonds
MM = 34.02
Polar
H-Bonds
Step 2.
3. Compare
1.
Evaluate the
Determine
intermolecular
thestrengths
kinds of of
intermolecular
attractive
the total intermolecular
forces
attractive forces
attractive forces. The substance with the strongest will be
Because
the
molar masses are similar, the size of
the liquid.
Formaldehyde:
Fluoromethane:
Hydrogen
peroxide:
the
dispersion
force
attractions
shouldplanar
bebent
similar
dispersion
forces:
MM
30.03, tetrahedral
34.03,
34.02,
trigonal
Because only hydrogen peroxide has the additional very
dipole–dipole:
very
polar
O–H C–F
bonds
uncancelled
uncancelled
C=O
bond
uncancelled
Because
thepolar
molecules
arebond
polar,
the size of the
strong
H-bond all
additional
attractions,
its intermolecular
H-bonding:
noattractions
O–H,
O–H,
therefore
N–H,should
orH-bond
F–Hbetherefore
no H-bond
dipole–dipole
similarexpect
attractions
will be the
strongest.
We therefore
hydrogen
to be peroxide
the liquid.also has additional
Onlyperoxide
the hydrogen
hydrogen bond attractions
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Practice – Choose the substance in each pair that is
a liquid at room temperature (the other is a gas)
a) CH3OH
CH3CHF2
can H-bond
b) CH3-O-CH2CH3
CH3CH2CH2NH2
can H-bond
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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, NH2, 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
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Immiscible Liquids
Pentane, C5H12 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.
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Polar Solvents
Dichloromethane
(methylene chloride)
Water
Ethanol
(ethyl alcohol)
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Nonpolar Solvents
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Ion–Dipole Attraction
• In a mixture, ions from an ionic compound are
•
attracted to the dipole of polar molecules
The strength of the ion–dipole attraction is one
of the main factors that determines the
solubility of ionic compounds in water
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Practice – Choose the substance in each pair
that is more soluble in water
a) CH3OH
CH3CHF2
can H-bond with H2O
b) CH3CH2CH2CH2CH3
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CH3Cl
more polar
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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 dipole–dipole
attractive forces
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Summary (cont’d)
• 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
• Ion–dipole attractions are present in mixtures of
•
•
ionic compounds with polar molecules.
Ion–dipole attractions are the strongest
intermolecular attraction
Ion–dipole attractions are especially important in
aqueous solutions of ionic compounds
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Liquids
Properties &
Structure
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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
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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
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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 H2O = 72.8 mJ/m2
 at room temperature
 surface tension of C6H6 = 28 mJ/m2
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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
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Viscosity
• Viscosity is the resistance of a liquid to flow
 1 poise = 1 P = 1 g/cm∙s
 often given in centipoise, cP
 H2O = 1 cP at room temperature
• Larger intermolecular attractions = larger viscosity
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Factors Affecting
Viscosity
• The stronger the intermolecular attractive forces, the
•
higher the liquid’s 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
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Insert Table 11.6
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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 tube’s surface
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Heat of Vaporization
• The amount of heat energy required to vaporize
one mole of the liquid is called the heat of
vaporization, DHvap
 sometimes called the enthalpy of vaporization
• Slways endothermic, therefore DHvap is +
• Somewhat temperature dependent

DHcondensation = −DHvaporization
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Example 11.3: Calculate the mass of water that
can be vaporized with 155 kJ of heat at 100 °C
Given:
Find:
Conceptual
Plan:
155 kJ
g H2O
kJ
mol H2O
g H2O
Relationships: 1 mol H2O = 40.7 kJ, 1 mol = 18.02 g
Solution:
Check: because the given amount of heat is almost 4x
the DHvap, the amount of water makes sense
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Practice – Calculate the amount of heat needed to
vaporize 90.0 g of C3H7OH at its boiling point
(DHvap = 39.9 kJ/mol)
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Practice – Calculate the amount of heat needed to
vaporize 90.0 g of C3H7OH at its boiling point
Given:
Find:
Conceptual
Plan:
90.0 g
kJ
g
mol
kJ
Relationships: 1 mol C3H7OH = 39.9 kJ, 1 mol = 60.09 g
Solution:
Check:
because the given amount of C3H7OH is more
than 1 mole the amount of heat makes sense
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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
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Dynamic Equilibrium
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Vapor Pressure
• The pressure exerted by the vapor when it is in
dynamic equilibrium with its liquid is called the
vapor pressure
 remember using Dalton’s 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
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Vapor–Liquid 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
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Changing the Container’s Volume
Disturbs the Equilibrium
Initially, the rate of
vaporization and
condensation are equal
and the system is in
dynamic equilibrium
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When the volume is
increased, the rate of
vaporization becomes
faster than the rate of
condensation
89
When the volume is
decreased, the rate of
vaporization becomes
slower than the rate of
condensation
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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
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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
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Vapor Pressure Curves
normal BP
100 °C
760 mmHg
BP Ethanol at 500 mmHg
68.1°C
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Practice – Which of the following is the
most volatile?
a)
b)
c)
d)
e)
water
TiCl4
ether
ethanol
acetone
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Practice – Which of the following has the
strongest Intermolecular attractions?
a)
b)
c)
d)
e)
water
TiCl4
ether
ethanol
acetone
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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
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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
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Practice – Which of the following has the
highest normal boiling point?
a)
b)
c)
d)
e)
water
TiCl4
ether
ethanol
acetone
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Heating Curve of a Liquid
• As you heat a liquid, its
temperature increases
linearly until it reaches the
boiling point
 q = mass x Cs x DT
• 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
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Clausius–Clapeyron Equation
• A graph of ln(Pvap) vs. 1/T is a straight line
••• The
of
vapor
pressure
vs.
temperature
is
an
Thegraph
logarithm
of
the
vapor
pressure
vs.
inverse
slope of the line x 8.314 J/mol∙K = DHvap
exponential
growth curve
absolute
temperature
is a linear function
 in J/mol
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Example 11.4: Determine the DHvap 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
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Example 11.4: Determine the DHvap of
dichloromethane given the vapor pressure vs.
temperature data
• Graph the inverse of the absolute temperature vs.
the natural log of the vapor pressure
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Example 11.4: Determine the DHvap of
dichloromethane given the vapor pressure vs.
temperature data
• Add a trendline, making sure the display equation
on chart option is checked off
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Example 11.4: Determine the DHvap of
dichloromethane given the vapor pressure vs.
temperature data
• Determine the slope of the line
 −3776.7 ≈ −3800 K
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Example 11.4: Determine the DHvap 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/mol∙K
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Clausius–Clapeyron 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
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Example 11.5: Calculate the vapor pressure of
methanol at 12.0 °C
Given:
T1 = BP = 337.8
64.6 °C,
K, P11 == 760
760 torr,
torr, DH
DHvap
35.2 kJ/mol,
kJ/mol,
vap = 35.2
T2 = 285.2
12.0 °C
K
Find: P2, torr
Conceptual
P1, T1, DHvap
Plan:
Relationships:
P2
T(K) = T(°C) + 273.15
Solution:
Check: the units are correct, the size makes sense because the
vapor pressure is lower at lower temperatures
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Practice – Determine the vapor pressure of water at
25 C (normal BP = 100.0 C, DHvap= 40.7 kJ/mol)
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Practice – Determine the vapor pressure of
water at 25 C
Given:
T1 = BP = 373.2
100.0 K,
°C,PP1 1==760
760torr,
torr,DH
DHvap
40.7kJ/mol,
kJ/mol,
vap==40.7
T2 = 298.2
25.0 °C
K
Find: P2, torr
Conceptual P , T , DH
1
1
vap
Plan:
P2
Relationships: T(K) = T(°C) + 273.15
Solution:
Check: the units are correct, the size makes sense because the
vapor pressure is lower at lower temperatures
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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
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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
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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
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deposition
111
gas
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Sublimation
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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
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Heating Curve of a Solid
• As you heat a solid, its
temperature increases linearly
until it reaches the melting point
 q = mass x Cs x DT
• 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
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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
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Heat of Fusion
• The amount of heat energy required to melt one mole of the
solid is called the Heat of Fusion, DHfus
 sometimes called the enthalpy of fusion
• Always endothermic, therefore DHfus is +
• Somewhat temperature dependent
DHcrystallization = −DHfusion
 Generally much less than DHvap
 DHsublimation = DHfusion + DHvaporization

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Heats of Fusion and Vaporization
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Heating Curve of Water
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Segment 1
• Heating 1.00 mole of ice at −25.0 °C up to the
•
melting point, 0.0 °C
q = mass x Cs x DT
 mass of 1.00 mole of ice = 18.0 g
 Cs = 2.09 J/mol∙°C
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Segment 2
• Melting 1.00 mole of ice at the melting point,
•
0.0 °C
q = n∙DHfus
 n = 1.00 mole of ice
 DHfus = 6.02 kJ/mol
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Segment 3
• Heating 1.00 mole of water at 0.0 °C up to the
•
boiling point, 100.0 °C
q = mass x Cs x DT
 mass of 1.00 mole of water = 18.0 g
 Cs = 2.09 J/mol∙°C
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Segment 4
• Boiling 1.00 mole of water at the boiling point,
•
100.0 °C
q = n∙DHvap
 n = 1.00 mole of ice
 DHfus = 40.7 kJ/mol
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Segment 5
• Heating 1.00 mole of steam at 100.0 °C up to
•
125.0 °C
q = mass x Cs x DT
 mass of 1.00 mole of water = 18.0 g
 Cs = 2.01 J/mol∙°C
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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?
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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?
kJ,°C, T2 = 5.5 °C),
Given: 12.0 g benzene, seg 1 =(T0.2325
1 = −10.0
seg 2 = melting,
1.51 kJ, seg 3 (T
= 0.3978
kJ T2 = 25.0 °C)
1 = 5.5 °C,
kJ
Find:
Conceptual Seg 231
Plan:
Relationships:
g
mol
J
kJ
DHfus 9.8 kJ/mol, 1 mol = 78.11 g, 1 kJ = 1000 J, q = m∙Cs∙DT
Cs,sol = 1.25 J/g°C, Cs,liq = 1.70 J/g°C
Solution:
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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
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Phase Diagrams
Fusion Curve
critical
point
Pressure
melting
1 atm
freezing
Liquid
Solid
normal
boiling pt.
normal
melting pt.
Sublimation
Curve
triple
point
vaporization
condensation
sublimation
Gas
deposition
Temperature
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Vapor Pressure
Curve
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Pressure
Phase Diagram of Water
1 atm
critical
point
374.1 °C
217.7 atm
Ice
Water
normal
melting pt.
0 °C
normal
boiling pt.
100 °C
triple
point
0.01 °C
0.006 atm
Steam
Temperature
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Morphic Forms of Ice
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Pressure
Phase Diagram of CO2
Liquid
Solid
-56.7 °C
5.1 atm
critical
point
31.0 °C
72.9 atm
triple
point
Gas
normal
1 atm sublimation pt.
-78.5 °C
Temperature
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Practice – Consider the phase diagram of CO2
shown. What phase(s) is/are present at each of the
following conditions?
•
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
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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. NH3, CH4
 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. O2, CO2
• 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
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Solids
Properties &
Structure
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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
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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
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7 Unit Cells
c
c
c
b
b
b
a
Cubic
a=b=c
all 90°
a
Tetragonal
a=c<b
all 90°
a
Orthorhombic
abc
all 90°
c
c
b
a
Hexagonal
a=c<b
2 faces 90°
1 face 120°
c
a b
Monoclinic
abc
2 faces 90°
c
b
b
a
Rhombohedral
a=b=c
no 90°
140
a
Triclinic
abc
no 90°
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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
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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
⅛ of each corner particle is within the cube
½ of each particle on a face is within the cube
¼ of each particle on an edge is within the cube
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Cubic Unit Cells Simple Cubic
• Eight particles, one at each
•
corner of a cube
1/8th of each particle lies in the
unit cell
each particle part of eight cells
total = one particle in each unit
cell
•
•
2r
8 corners x 1/8
Edge of unit cell = twice the
radius
Coordination number of 6
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Simple Cubic
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Cubic Unit Cells Body-Centered Cubic
• Nine particles, one at each
•
corner of a cube + one in center
1/8th of each corner particle lies
in the unit cell
two particles in each unit cell
•
•
4r
3
8 corners x 1/8
+ 1 center
Edge of unit cell = (4/ 3) times
the radius of the particle
Coordination number of 8
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Body-Centered Cubic
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Cubic Unit Cells Face-Centered Cubic
• 14 particles, one at each corner
•
of a cube + one in center of
each face
1/8th of each corner particle +
1/2 of face particle lies in the
unit cell
2r 2
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
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Face-Centered Cubic
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Example 11.6: Calculate the density of Al if it
crystallizes in a fcc and has a radius of 143 pm
3, m
−22
−22
face-centeredcubic,
cubic,Vr==6.618
143 pm
1.43
xx10
10−8−23cm,
cmm
= 1.792
=1.792x x1010
gg
Given: face-centered
33
Find: density,
density,g/cm
g/cm
Conceptual
Plan:
fcc
mass
r
l = 2r√2
# atoms x mass of 1 atom
m, V
Relationships:
l
V
V=l3
d
d = m/V
1 cm =
1 pm = 10−12 m V = l 3, l = 2r√2, d = m/V
fcc = 4 atoms/uc, Al = 26.982 g/mol, 1 mol = 6.022 x 1023 atoms
102 m,
Solution:
Check:
the accepted density of Al at 20°C is 2.71 g/cm3, so the
answer makes sense
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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
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Practice – Estimate the density of Rb if it
crystallizes in a bcc and has a radius of 247.5 pm
−22−8
3, m= 2.839 x 10−22−22
cubic,
V=rr=
1.868
x 10
cm
body-centered
body-centered
cubic,
cubic,
=2.475
247.5
xpm
10
cm,
m = 2.839 x 10 g g
Given: body-centered
333
Find: density,
density,
density,g/cm
g/cm
g/cm
Conceptual
Plan:
bcc
mass
r
l = 4r/√3
# atoms x mass of 1 atom
m, V
Relationships:
l
V
V=l3
d
d = m/V
1 cm =
1 pm = 10−12 m V = l 3, l = 4r/√3, d = m/V
bcc = 2 atoms/uc, Rb = 85.47 g/mol, 1 mol = 6.022 x 1023 atoms
102 m,
Solution:
Check:
the accepted density of Rb at 20°C is 1.53 g/cm3, so the answer
makes sense
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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
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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
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Closest-Packed Structures
Third Layer – with Offset 2nd 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
Cubic Closest-Packed
Hexagonal
Closest-Packed
Face-Centered Cubic
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Hexagonal Closest-Packed
Structures
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Cubic Closest-Packed Structures
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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
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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
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Molecular Solids
• The lattice sites are occupied by molecules
CO2, H2O, C12H22O11
• The molecules are held together by
intermolecular attractive forces
dispersion forces, dipole–dipole attractions, and
H-bonds
• Because the attractive forces are weak, they
tend to have low melting points
generally < 300 °C
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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
•
cation–anion 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
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Ionic Crystals
CsCl
coordination number = 8
Cs+ = 167 pm
Cl─ = 181 pm
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NaCl
coordination number = 6
Na+ = 97 pm
Cl─ = 181 pm
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Lattice Holes
Tetrahedral
Hole
Octahedral
Hole
Simple Cubic
Hole
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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
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Cesium Chloride Structures
• Coordination number = 8
• ⅛ 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 ⅛), therefore
the formula is CsCl
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Rock Salt Structures
• Coordination number = 6
• Cl─ ions (181 pm) in a face-centered
cubic arrangement
 ⅛ of each corner Cl─ inside the unit cell
 ½ 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
¼ of each edge Na+ inside the unit cell
• Na:Cl = (¼ x 12) + 1: (⅛ x 8) + (½ x 6)
•
= 4:4 = 1:1,
Therefore the formula is NaCl
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Zinc Blende Structures
• Coordination number = 4
• S2─ ions (184 pm) in a face-centered
cubic arrangement
 ⅛ of each corner S2─ inside the unit cell
 ½ of each face S2─ inside the unit cell
• Each Zn2+ (74 pm) in holes between
S2─
 tetrahedral holes
 1 whole particle in ½ the holes
• Zn:S = (4 x 1) : (⅛ x 8) + (½ x 6) =
•
4:4 = 1:1,
Therefore the formula is ZnS
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Fluorite Structures
• Coordination number = 4
• Ca2+ ions (99 pm) in a face-centered
cubic arrangement
 ⅛ of each corner Ca2+ inside the unit cell
 ½ of each face Ca2+ inside the unit cell
• Each F─ (133 pm) in holes between
Ca2+
 tetrahedral holes
 1 whole particle in all the holes
• Ca:F = (⅛ x 8) + (½ x 6): (8 x 1) =
•
4:8 = 1:2,
Therefore the formula is CaF2
 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
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Practice – Gallium arsenide crystallizes in a cubic
closest-packed array of arsenide ions with gallium ions in
½ 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 = ½ (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
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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
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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
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Metallic Structure
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Metallic Bonding
• Metal atoms release their valence
•
+
electrons
Metal cation “islands” fixed in a “sea” of
mobile electrons
+
e+
+
e-
+
+
+
ee-
+
+
+
ee-
+
+
e-
+
e+
+
+
ee-
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+
+
ee-
+
+
+
ee-
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+
+
+
ee-
+
+
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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
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The Diamond Structure:
a 3-Dimensional Network
• The carbon atoms in a diamond each have four
covalent bonds to surrounding atoms
sp3
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
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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
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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
 sp2
 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
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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
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Silicates
• ~90% of Earth’s crust
• Extended arrays of SiO
sometimes with Al substituted for Si –
aluminosilicates
• Glass is the amorphous form
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Quartz
• SiO2 in pure form
impurities add color
• 3-dimensional array of Si covalently bonded
to 4 O
 tetrahedral
• Melts at ~1600 °C
• Very hard
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Micas
• There are various kinds of mica that have slightly
different compositions – but are all of the general
form X2Y4-6Z8O20(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 Fe3+ 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
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Practice – Pick the solid in each pair with the
highest melting point
a)
KCl ionic
b)
C(s, graphite) cov. network S8 molecular
c)
Kr atomic
K
d)
SrCl2 ionic
SiO2 (s, quartz) cov. network
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SCl2 molecular
183
metallic
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Band Theory
• The structures of metals and covalent network
•
solids result in every atom’s 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
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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
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Molecular Orbitals of Polylithium
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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
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Types of Band Gaps and
Conductivity
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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
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Doping Semiconductors
• Doping is adding impurities to the semiconductor’s
•
•
•
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.
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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
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