Liquids and Solids: Structure, Bonding, and Properties
This chapter provides an introduction to the condensed states of matter (liquids and solids) and
the forces responsible for the physical properties of these phases. In addition an introduction to
equilibrium is provided through a study of phase changes that can occur between solids, liquids,
and gases. The following should provide some guidance to the important topics that will be
covered in this chapter.
Objectives
1. Recognize that macropscopic properties (melting and boiling points, viscosities, vapor
pressures, etc.) can provide information on molecular level interactions.
2. Understand the dependence of vapor pressure on temperature (Clausius-Clapeyron
equation). This is important as this subject provides an introduction to an important
concept (temperature dependence of equilibria) and also to an important mathematical
function.
3. Be comfortable with reading phase diagrams and understand the energetics involved as
you proceed from on point to another in these diagrams.
4. Understand the nature of the cohesive forces acting to hold the condensed phases
(liquids and solids) of matter together.
5. Be capable of classifying the different types of condensed matter (molecular, ionic,
metallic, covalent network).
6. Understand the types of forces holding together molecular, ionic, metallic, and covalent
network substances.
7. Understand the relationship between structure and intermolecular forces in molecular
substances.
8. Become acquainted with the crystalline structures for metals and ionic compounds
discussed in our text (and recognize that there are others).
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Introduction
Interactions in Condensed Phases
Earlier we learned about the behavior of gases. At that time we learned that all gases behave in
essentially the same way . . . that is, they obey the ideal gas law reasonably well under most
conditions. That behavior can be explained via the kinetic molecular theory of gases. If you do
not recall that theory, you may want to take the time to review that material. It will have
applications in the material presented in this chapter.
However, when condensed phases of matter (solids and liquids) are considered, one quickly
realizes that many different types of behavior are observed for the different kinds of substances.
The properties of each of these will be discussed in more detail later. For now it will suffice to
recognize that there are four major categories of substances:
Molecular substances: these are composed of discrete molecules (ex. hydrogen, water, sucrose,
ethanol, vegetable oils, etc.)
Covalent network substances: these are composed of atoms covalently bonded together in a
continuous network, i.e. a lattice (ex. diamond, quartz, graphite, etc.)
Ionic substances: these are composed of ions held together by electrostatic attractions in a
continuous network, i.e. a lattice (ex. NaCl, MgO, CaSO4, etc.)
Metallic substances: these are composed of positive nuclei bonded together by a sea of electrons
in a continuous network, i.e. a lattice (ex. iron, aluminum, calcium, etc.)
Pause for a moment to consider the examples of each type of substance above. These substances
have very different properties. In the next two sections we will examine some of the macroscopic
phenomena that result from forces in the condensed phases and then we will seek to obtain a
better understanding of the atomic (or molecular) level explanation for the differences in the last
three sections of this chapter.
Liquid-Vapor Equilibrium
What is this thing called Dynamic Equilibrium?
A dynamic equilibrium occurs when two processes (one the opposite of the other) occur at
exactly the same rate. When this occurs, the system will seem to be static with no change in the
properties related to the processes. Of particular importance now is the process of vaporization.
This can be understood by taking a look at the kinetic molecular theory.
Figure showing the process of dynamic equilibrium
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When a liquid is placed in a sealed container, a certain number of liquid molecules will be
converted into gas phase molecules per unit time. As the number of gas phase molecules
increases, the rate at which gas phase molecules are converted back into liquid (condensation)
increases. At some point, the number of molecules being vaporized per unit time is equal to the
number of molecules being condensed per unit time. When this occurs a state of dynamic
equilibrium exists. Molecules are still be vaporized but an equal number are being condensed.
When this occurs, the pressure exerted by the vapor over the liquid is called the vapor pressure
(recall the discussion of water vapor pressure from the ideal gas law chapter).
NOTE: the vapor pressure of a substance depends on the nature of the intermolecular forces
within the liquid and the temperature.
The temperature dependence can be understood by extending the kinetic molecular theory,
introduced earlier for gases, to the liquid state. As the temperature of a liquid increases, a greater
fraction of the liquid molecules will have kinetic energy great enough to overcome the
intermolecular forces holding the molecules together as a liquid. The amount of heat required to
vaporize one mole of a given substance is called a molar heat of vaporization (ΔHvap). Note
these are positive heats since vaporization is an endothermic process.
Figure Showing Temperature Dependence of KE as it relates to vaporization.
A curve of vapor pressure vs. Kelvin temperature is not a straight line.
Figure of Pvap vs. Temperature.
Figure of ln Pvap vs. 1/Temp.
However, a plot of the natural logarithm of vapor pressure vs. the inverse of Kelvin temperature
does. (see above)
This yields a straight line equation y = mx + b that looks like the following at two different
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temperatures:
When the difference between these two points are taken, the result is called the ClausiusClapeyron equation. This equation can be used to determine the molar heat of vaporization for a
substance by measuring the vapor pressure of the substance at two temperatures. The
temperatures must be expressed in Kelvin. R and ΔH must be in compatible units (typically J or
kJ are used for ΔH and R = 8.314 J/Kmole or R = 8.314x10-3 kJ/Kmole). Both values of P need
only to be expressed in the same units for consistency (those units can be mm Hg, psi, atm, Pa,
bar, etc.).
Note: the equations above assume that ΔHvap is a constant. This is not strictly correct. However, over relatively
narrow temperature ranges, the assumption produces minor errors.
Quick Practice:
A certain molecular substance has a vapor pressure of 120 mm Hg at 25ºC and at 20ºC the vapor
pressure is only 100 mm Hg. What is ΔHvap for this substance in kJ per mole?
Answer: +26,470 J/mole OR about +26.5 kJ/mole since our data has only 3 sig figs.
Note: Two items are notable about this answer. First, the sign is positive indicating that
vaporization is an endothermic process (i.e. it costs energy to vaporize a substance). The
magnitude of this answer is also meaningful. Most bond energies are above 150 kJ/mole (a few
are around 1,000 kJ/mole). This is significant. If bond energies are weaker than heats of
vaporization, molecular substances do not boil . . . they decompose . . . and this actually occurs
for some substances!
A substance boils when its vapor pressure becomes equal to the external (atmospheric pressure).
The temperature required to produce a vapor pressure of 1 atm (760 mm Hg) is called the
normal boiling point of a substance (see figure above).
Substances can boil at temperatures lower or higher than this. If the pressure exerted by the
atmosphere is less than 1 atm then a substance will boil at a lower temperature (or if the
atmospheric pressure is higher then the boiling point is elevated).
Vapor pressure depends upon ΔHvap, that is, when a substance has a large heat of vaporization (in
this context we mean it will be highly positive), it will have a lower vapor pressure at any given
temperature than a substance with a smaller heat of vaporization. As a consequence of this fact,
substances with large heats of vaporization will exhibit high boiling points (i.e. more energy is
required to produce a vapor pressure of 1 atm).
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Lets examine some data that supports this conclusion:
ΔHvap (kJ/mole) at 25ºC
CS2
27.51
CH3OH
37.43
CH3CH2OH
42.32
H2O
43.98
C6H5NH2
55.83
Normal Tbp (ºC)
46.3
64.6
78.29
100.0
184.17
From this statement and our understanding of kinetic molecular theory as it applies to
vaporization, we can make the following leap . . . substances with higher boiling points
experience larger forces holding the substance together in the liquid state.
Quick Practice:
For the molecular substance above, determine the normal boiling point. That is at what
temperature will the vapor pressure be equal to 1 atm (or 760 mm Hg).
Answer: Using 26,470 J/mole and the data for 25ºC. We find that the required temperature is
360 K OR 87ºC. This is reasonable since water boils at 100ºC and has a ΔHvap = 44 kJ/mole at
298 K.
Critical Temperature and Pressure
These will be considered along with the section on "Phase Diagrams" below.
Other Properties of Condensed Phases
We learned above that heats of vaporization and boiling points can be used as indicators of the
forces holding together a substance in its liquid state. It is worth noting that many other
macroscopic properties provide insights into the molecular level world.
Surface Tension, Viscosity, and Heats of Fusion (or normal melting temperatures)
Molecules (or atoms) that reside at the surface of a liquid are surrounded by fewer neighbors
than are molecules (or atoms) that reside inside the bulk of the liquid. As a result, the molecules
at the surface experience a force that pulls them toward the interior of the liquid (where they
would experience greater attractive forces). As a result it costs energy to increase the surface area
of a liquid. This energy is referred to as the surface tension and is normally expressed in units of
J/m2.
Figure showing surface tension forces.
Real world situations normally result in a balance of cohesive forces (forces between like
molecules) experienced inside a liquid and the adhesive forces (forces between unlike
molecules) between the liquid and the container in which it is held. Examples can be found in the
wetting of a surface, capillary action, the formation of a meniscus in a buret, etc.
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Figure showing adhesive vs. cohesive forces.
In order for liquids to flow, molecules of same type must move past one another. However, these
molecules feel attractive forces for the surrounding molecules. As such all liquids (at least the
ones that are not classified as super liquids) exhibit a resistance to flow. Viscosity is a measure of
a liquids resistance to flow and is normally expressed in Poise (P) or the SI unit of viscosity
Ns/m2 which is equal to 10 P.
Note that both surface tension and viscosity are related to forces between molecules in the liquid
state. As such we would expect the variation of viscosity and surface tension to follow the same
trend as was observed in heats of vaporization and normal boiling points.
In addition to these properties of liquids one might expect solids to exhibit properties that depend
upon the strength of forces holding that condensed phase together. Consider temperature at
which a substance melts (at 1 atm this is called the normal melting point, Tmp). The amount of
heat required to melt a given quantity of substance (a heat of fusion, ΔHfus) is related to how
much more strongly the molecules in the solid are bond together than those in the liquid (there is
also another consideration, but we will get to that in a later chapter). Thus, it might make sense
to expect some substances to require more heat than others. Not surprisingly, the amount of heat
available as energy is a function of temperature (a hot oven has more energy in it than does a
cold oven). So some substances melt at high temperatures and have high heats of fusion (ΔHfus)
while others have lower heats of fusion and melt at lower temperatures.
Now we can expand our table from above to include some more properties . . .
ΔHvap (kJ/mole) at 25ºC
CS2
27.51
CH3OH
37.43
CH3CH2OH
42.32
H2O
43.98
C6H5NH2
55.83
Normal Tbp (ºC)
46.3
64.6
78.29
100.0
184.17
Surface Tension (J/m2)
0.03158
0.02207
0.02197
0.07199
0.04212
Viscosity (cP) at 25ºC
0.352
0.544
1.074
0.890
3.847
Normal Tmp (ºC)
-110.8
-97.6
-114.1
0.0
-6.0
ΔHfus (kJ/mole) at 25ºC
4.40
3.18
5.02
6.01
10.56
at 25ºC
You will note that the trends are not perfect. However, these parameters do provide for some
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insight into the relative level of forces experienced in the condensed phases of these substances.
For an additional comparison you can add a column in the table above and include data for
mercury (a well known metal that is liquid at room temperature): ΔHvap 59.11 kJ/mole, normal
Tbp 356.73ºC, surface tension is 0.485 J/m2, viscosity is 1.526 cP, Tmp -38.83ºC, ΔHfus 2.29
kJ/mole. Note that the surface tension is very high which explains the typical behavior of
mercury rolling around in beads when spilled.
Phase Diagrams
In an earlier section we discussed the dynamic equilibrium that can exist between the liquid and
vapor phases (refer to the discussion on vaporization and vapor pressure). In the discussion we
examined a plot of vapor pressure vs. temperature that displayed the equilibrium:
Substance (l) ⇔ Substance (g)
Not too surprisingly, equilibria may also exist between: (a) solid and liquid, (b) solid and vapor,
and (c) solid, liquid, and vapor. If we generalize our plot of pressure vs. temperature to extend
over these other possible equilibrium, the resulting figure is called a phase diagram.
Key
A = solid region
B = liquid region
C = gaseous region
Tt = triple point temperature
Pt = triple point pressure
Tb.p. = normal boiling point
1.00 atm = Pb.p.
Tc = critical point temperature
Pc = critical point pressure
NOTE: the normal melting point is defined as the melting point of
the solid at 1 atm (the melting point varies much less with pressure
than does the boiling point. Why?).
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In phase diagrams the "phase boundaries", the lines that separate two phases represent the
conditions of P and T at which two phases can exist in equilibrium.
The line between the liquid and vapor (gas) phase represents the vapor pressure curve of the
liquid (see the section on the Clausius-Clapeyron Equation above).
The line between solid and liquid represents the entire range of P and T values at which the solid
and liquid can co-exist at equilibrium.
Solid and vapor co-exist at equilibrium along the third line. Thus, for any substance there exist
many sets of T and P at which two phases can co-exist in equilibrium. However, at only one P
and T can all three phases co-exist in equilibrium, the triple point.
Also, consider the significance of the slopes of the lines. At a fixed temperature, increasing the
pressure normally will take you from a phase of lower density into a phase of higher density.
Therefore, increasing P will convert a vapor into either a solid or liquid depending on the
temperature (see note below). Or increasing the pressure on a liquid can make it solidify.
Examine the phase boundaries above and see if you can visualize these processes.
Water is unusual in this regard as it is more dense as a liquid than as a solid. Increasing the
pressure on solid water (ice) will cause it to melt. How will the slope of the solid liquid phase
boundary for water differ from the one in the diagram above?
Above the critical temperature, no amount of pressure will cause the vapor to condense as a
liquid. The phase that exists above the critical temperature is often referred to as a supercritical
fluid.
Moving Through a Phase Diagram: Heating (or Cooling) Curves and Changes in Pressure
If you draw a straight vertical line across a phase diagram, you can imagine that this corresponds
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to varying the pressure while maintaining constant temperature. OR if you draw a straight
horizontal line through the phase diagram you can imagine that this corresponds to varying
temperature while maintaining constant pressure.
On following the line from point 1 to point 2, the following will occur:
• from pt. 1 to the vapor-liquid boundary, the vapor increases in P at constant T.
• at the vapor-liquid phase boundary, the vapor condenses to form liquid and P and T are
both constant (note this would require removing heat from the system as condensation is
exothermic)
• once all the vapor condensed to liquid, increasing P would merely result in the liquid
being pressurized once P reached the level at the liquid -solid phase boundary, solid
would begin to form and P and T would be constant (again, this would require extraction
of heat energy)
• once all the liquid had been frozen, the P would build on the solid (NOTE: at some higher
P, the solid may be converted into a more dense phase of solid - another phase boundary
would be encountered between the solid phases)
Now you try figuring out what happens as temperature is increased at constant pressure from
point 3 to point 4 (don't forget the phase changes that occur at the phase boundaries). Make a plot
of T vs. heat added. It should look like this . . .
In the space below, describe the process occurring along each straight line segment of this figure:
• in the first sloped segment, the solid is being heated and therefore is getting warmer (i.e.
the temperature of a solid increases as it is heated)
• in the first horizontal segment,
•
•
•
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Quick Review: Use the phase diagram below to answer the following questions.
a. which letter in the diagram corresponds to the region of liquid
b. which letter in the diagram corresponds to the region of solid
c. the normal boiling point is (in ºC)
d. the normal melting point is (in ºC)
e. the temperature at the triple point is (in ºC)
f. the temperature at the critical point is (in ºC)
g. the pressure at the critical point is (in atm)
h. the pressure at the triple point is (in atm)
i. which has the lower density the solid or liquid phase
Draw arrows for the following processes in the phase diagram above AND write whether each
process is endo (or exo)-thermic:
melting, freezing, vaporization, condensation, sublimation, deposition
Q: If the triple point is above 1.00 atm, what is the normal melting point? Normal boiling point?
Answer: They do not exist b/c neither liquid or vapor can exist in equilibrium in this case . . .
Describe all the process that occur in following path (1→ 2→ 3→ 4) through the phase diagram.
Q: A compound is found to have a ΔHvap of 3.46 kJ/g and a ΔHsub of 4.60 kJ/g, calculate the ΔHfus
in kJ /g. Explain why you can use this data to provide an accurate result. Think about Hess’s
Law and state functions. A: 1.14 kJ/g b/c sub = fus + vap, so ΔHsub = ΔHfus + ΔHvao
Q: If you were to eat 1.5 kg of ice at 0.0ºC, your body would have to provide how much energy
to (a) melt the ice [for water ΔHfus = 333.5 J/g] AND (b) to raise the temperature of the water to
37.0ºC (body temperature) [specific heat of water = 4.184 J/gºC ]? A: 732 kJ
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Molecular Substances; Intermolecular Forces
At this time it would make some since to look at the accepted scientific explanations for the
behavior of liquids and solids by examining the forces that hold substances together in these
phases.
Intermolecular vs. Intramolecular
It is worth mentioning that we have learned a good deal about intramolecular forces, the
covalent bonds that hold atoms together in molecules. Of considerable importance is the fact that
covalent bonds have strengths that are on the order of 100 - 1000 kJ per mole of bonds (Recall
the section on bond energies being used to find heats of gas phase reactions and the discussion of
using covalent homonuclear bond energies to estimate the percent ionic character of
heteronuclear bonds.).
Intermolecular forces, the attractive forces between molecules that hold them together in the
condensed phases, are much weaker. They are on the order to 0.1 - 40 kJ per mole.
A natural consequence of this fact is that the condensed phases of molecular substances can be
melted and/or vaporized without decomposing the molecules into atoms (NOTE: some molecular
substances do decompose before they vaporize, however, they typically are broken down into
simpler molecular substances; not individual atoms).
It should also be recognized that the extremely weak nature of some of the intermolecular forces
allow some molecular substances to exist as gases at normal ambient conditions (about 1 atm
pressure and 25ºC). The intermediate intermolecular forces allow other substances to exist as
liquids, while the stronger intermolecular forces allow yet other substances to exist as solids at
ambient conditions.
Types of Intermolecular Forces
There are basically two types of Intermolecular forces: Dispersion Forces and Dipole Forces.
We will also mention a third type of intermolecular force, Hydrogen Bonding, but it is really
just a specific type of dipole-dipole force.
Dipole Forces
When we discussed Lewis structures and molecular geometry, we learned to distinguish between
polar and nonpolar molecules. Polar molecules have polar covalent bonds. These polar covalent
bonds produce dipole moments along the bonding axes of the molecule. When these dipole
moments cancel one another, the molecule is nonpolar. However, when the dipole moments do
not cancel, the molecule is polar.
A polar molecule has one portion that carries a partial positive charge and another portion that
carries a partial negative charge. That is the molecule has two charged poles (dipole). These
dipoles can interact with one another providing attractive forces between the polar molecules
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Refer to the figures in your text and sketch permanent dipole - dipole interactions in this space:
Figure: Permanent dipole - dipole interactions
Hydrogen Bonding
When H is bonded to a very electronegative and small atom (N, O, or F) a particularly strong
dipole develops. The interaction of these dipoles is referred to as hydrogen bonding. The
existence of hydrogen bonding typically makes the condensed phases of molecules more stable.
That is, they vaporize less easily (more will be said about this shortly).
(Note: some theoretical and experimental studies done in the last few years indicate that H-bonding may actually
contain a covalent contribution along with the electrostatic attraction . . . this actually fits in well with what we
learned about the continuum of bond types that range from pure covalent to very highly ionic bonds)
Refer to the figures in your text and sketch hydrogen boding interactions in this space:
Figure: Hydrogen bonding interactions
Dispersion Forces (non-permanent, or temporary, dipoles)
Dipole forces and hydrogen bonding occur only between polar molecular species. However,
dispersion forces are found in all molecular species: polar and nonpolar alike. This of course
means that nonpolar species have only dispersion forces to hold them together in condensed
phases. Dispersion forces are most easily described as attractive forces between weak dipoles set
up in nonpolar molecules by temporary fluctuations in electron distribution. When a nonpolar
species like H2 or He has greater than one half of its electron density shift to one side (or end) of
the molecule or atom, an instantaneous dipole forms. If no other molecules are nearby, the dipole
can disappear. However, if another molecule (or atom) is nearby, the instantaneous dipole can
affect the other molecule (or atom) by creating an induced dipole.
Refer to the figures in your text and sketch dispersion type interactions in this space:
Figure: dispersion forces or temporary dipolar interactions
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If the molecules (or atoms) are traveling too fast, their kinetic energy prevents them from
condensing. However, if the attractive force between the instantaneous and induced dipole is
great enough to overcome the kinetic energy of the species, it will condense.
The ease with which a species forms these temporary dipoles is determined by the polarizability
of the species. In general, the following can be said:
(a) species with more electrons are more easily polarized, form temporary dipoles, AND
(b) long molecules are more easily polarized than compact spherical molecules because the
electrons are more easily dispersed.
The first factor above leads some authors to state that a good rule of thumb is larger molar mass
= larger intermolecular forces = higher boiling points. This rule should really only be applied
within a homologous series of substances (like He, Ne, Ar, Kr, Xe OR F2, Cl2, Br2, I2)
Carefully examine the information given in your text and the examples reviewed in class.
Consider the following regarding these data tables.
Nonpolar species (dispersion forces only) - compare the b.p of Ar, F2, and C2H6
even though C2H6 has the lowest molar mass it as the highest b.p. (stronger intermolecular
forces) because it is a longer molecule
Polar vs. Nonpolar Species - take special note of the comparison of polar species to nonpolar
species of comparable molar mass (and size / shape)
the polar species has the higher b.p. (stronger intermolecular forces) because it has both
dispersion and dipole forces working to hold the condensed phase together
Hydrogen Bonding - ex. compare the m.p. or the b.p. of HI, HBr, HCl, and HF
even though HCl is more polar than HBr which is more polar than HI the boiling points decrease
as one goes from the largest to smallest of these species, why? The increased dispersion forces in
HI more than make up for the decrease in the dipole forces. However, HF, the smallest of all the
species, has the highest b.p. (strongest intermolecular forces) because it has Hydrogen bonding.
Therefore, its weaker disperion forces are more than offset by the very strong H-Bonding dipole
force.
Final Thought
Take a look at the b.p. listed in the tables discussed above. You will note that molecular
substances can exist as gases (b.p. below 20 - 25ºC), liquids (m.p. below but b.p. above 20 25ºC), or solids (m.p. above 20 - 25ºC). Also, think about substances or mixtures that you are
very familiar with: air, gasoline, table sugar, wood, plastics, and more. Based on their properties
what can you say about the intermolecular forces present in them?
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Also consider the solubility of molecular substances in one another. In general, likes dissolve
likes. So water will dissolve polar species and will not dissolve nonpolar species well. On the
other hand nonpolar species dissolve well in one another. We will spend much more time
discussing these details in the next chapter.
Network Covalent, Ionic, and Metallic Solids
Refer to the Figure in your text for a molecular level model of the four types of substances.
•
•
•
network covalent solids, atoms are joined in continuous network with one another by
covalent bonds
ionic solids*, ions are attracted to one another by electrostatic attraction between anions
and cations (some ionic compounds are liquids at ambient conditions)
metallic solids*, metallic cations surrounded by a sea of electrons (Hg is a liquid at
normal ambient conditions)
Keep these models in mind as the following discussion progresses.
Properties of Network Covalent Solids
1. High melting points (usually 1000ºC and up). Remember covalent bonds are much
stronger than intermolecular forces. This also leads to these materials being rather hard
and exhibiting very low vapor pressures.
2. Insoluble in common solvents (unless a reaction occurs). Think about it . . . does glass or
diamond dissolve in water or gasoline?
3. Generally good electrical insulators (graphite is an exception to this rule). In most
covalent network structures, the valence electrons are all tied up in covalent bonds.
Therefore, they are not free to roam around to carry electrical current.
Consider carefully the text's description of graphite vs. diamond . . . these will be discussed in
class. Also, find some information on quartz vs. soft glass vs. hard glass. These are all important
and widely used substances.
Properties of Ionic Solids*
1. High melting due to large ionic (electrostatic) attractions . . . more on this below. Also
hard and not easily vaporized.
2. Ionic solids do not conduct electricity well because the electrons are all tightly bound to
ions and the ions are locked into fixed positions in a lattice. However, when an ionic
compound melts OR when it is dissolved in water, the ions are free to move around and
carry charge.
3. Many ionic solids are water soluble (some are not, refer to the solubility rules from
Chapter 4). Water is a very polar species and is capable of interacting with both cations
and anions (because water has a large dipole).
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The energy of the electrostatic interaction between ions obeys Coulomb's Law.
where Q1 and Q2 are the charges on the ions and d is the distance between the centers of the ions
d = rcation + ranion
This obviously leads us to the following generalizations:
(a) the larger the charges on the ions, the stronger the forces holding the ions together,
AND,
(b) the smaller the ions, the stronger the forces holding the ions together.
Both of these concepts are illustrated by the following list.
Melting Points of a Few Ionic Compounds
NaCl
801ºC
KBr
734ºC
CaSO4
1400ºC
CaO
2927ºC
Try to explain the above differences in melting points by considering charges and sizes of ions:
NaCl vs. KBr (both have +1 cations and -1 anions . . . therefore consider size explanation)
CaSO4 vs. CaO (consider size of sulfate vs. oxide ions)
NaCl or KBr vs. CaO (consider charge as the primary factor)
Properties of Metals
1. High electrical and thermal conductivity as a result of freely mobile electrons.
2. Ductility (ability to be drawn into wire) and malleability (ability to be hammered or
rolled into sheets) since the electrons bonding the metal cations together can freely move
when the metal lattice is shifted.
3. Metallic luster (until oxidized). Most metals have a wide energy distribution for their
electrons and thus interact by absorbing and rr-emiting light across the entire visible
range (metals can also absorb and re-emit other wavelengths that our eyes do not detect).
4. Insoluble in common solvents unless a reaction occurs. However, in the spirit of the
general rule of likes dissolve likes, most metals readily are dissolved by metallic mercury
(Hg is a liquid at ambient conditions) forming solutions called amalgams.
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Metals are bonded together by their valence electrons. In general, the metals with fewer valence
electrons are softer and lower melting than the metals with more valence electrons. This trend
can be observed in the following comparison:
Melting Points of a Few Metallic Solids
Na
97.8ºC
Mg
648.8ºC
Al
660.4ºC
Fe
1535ºC
Cu
1083ºC
However, one must be careful not to over-generalize. Proceeding across the transition metals,
melting points do indeed increase. However, after reaching the middle of the d block melting
points begin to drop again. This is because the pairs of electrons in the d orbitals are less
effective at metallic bonding than unpaired electrons in the d orbitals.
Crystal Structures
Review the text sections dealing with the crystal structures of metals and ionic compounds.
What is a unit cell?
Explain the differences between simple cubic, body centered cubic, and face centered cubic.
List the common packing arrangements for metal atoms and know which leave the smallest
amount of space vacant. What is the difference between hexagonal closest packing and
cubic closest packing? Which is the same as fcc?
What is the reason that ionic compounds of the same type (ex. +1, -1 compounds like NaCl and
CsCl) have the ions arranged in different packing configurations (i.e. why do they exhibit
different crystal structures)? Hint: what is the radius ratio rule??
Do not worry about memorizing formula for doing calculations based on packing arrangements
like those shown in the text (examples 11.9 and 11.10).
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Keywords
(you should add any missing key words or phrases from the chapter)
adhesive
Deposition
melting
boiling point
diploe force
melting point
critical temperature
dispersion (London) forces
network covalent
cohesive
electron-sea model
phase diagram
condensation
heats of
sublimation
cubic cell
- fusion
triple point
- body centered cubic (BCC)
- sublimation
unit cell
- face centered cubic (FCC)
- vaporization
vapor pressure
- simple
hydrogen bonding
vaporization
Review Questions:
1.
Identify each of the following statements as true or false (T or F). If the statements are
false change them to make them true!
a. As molecules get larger, dipole-dipole interactions increase.
b. All polar molecules have higher boiling points than all nonpolar molecules.
c. Ionic solids have low melting points due to their weak electrostatic forces.
d. London forces are the only intermolecular forces in covalent networks.
e. Covalent network substances are hard, brittle, & have high melting point.
f. The reason CH3OH has a higher normal boiling point than CH3CH2CH3 is
that the CH3OH has only dipole forces acting between its molecules in the
liquid phase whereas CH3CH2CH3 has only London forces acting between its
molecules in the liquid phase.
2.
In each of the following pairs, which should have the higher melting point? Explain the
reasons for your choice in terms of the types of forces involved and the magnitude of the
forces.
a. C
vs.
CH4
b. MgO
vs.
KCl
c. H2O
vs.
H2S
d. C3H8
vs.
C12H26
e. CH3OCH3
vs.
CH3CH2CH3
f. SiO2
vs.
NH4Cl
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3.
Of the following possible structural arrangements, identify which best matches the
statements below.
simple cubic cell
face-centered cubic cell
body centered cubic cell
a. the least stable structure (structure has too much empty space)
b. has highest packing efficiency (only 26% empty space)
It is worth mentioning that the reasons why a given metal crystallizes in a particular
structure is not well understood. However, the type of crystalline structure does impact
many properties of metals: for example metals which pack in the simple cubic structure
(like alkali metals) have low melting pts., hardness, and densities. Metals with the fcc
structure (or a closely related structure called hcp that has the same packing efficiency)
tend to have high melting pts., hardness, and densities. Explain why this is sensible.
4.
What is the primary reason for the difference in structure observed in the compounds
LiCl, NaCl, and CsCl?
5.
Which one of the following DECREASES as the strength of the attractive intermolecular
forces INCREASES?
a.
b.
c.
d.
e.
6.
Under which of the following conditions will vaporization best occur?
a.
b.
c.
d.
e.
7.
heat of vaporization
normal boiling point
the sublimation temperature of a solid
the vapor pressure of a liquid
none of the above
high mass, large surface area, and high kinetic energy
weak intermolecular forces, high kinetic energy, and large surface area
high molecular energy and small surface area
low kinetic energy, strong intermolecular forces, and large surface area
small surface area, low kinetic energy, and low molecular mass
The boiling points of the halogens increase going from F2 to I2. What intermolecular
forces are responsible for this trend?
a.
b.
c.
d.
e.
permanent dipole
hydrogen bonding
ion-ion attraction
London dispersive forces
ion-dipole attraction
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8.
In which of the following processes will energy be evolved as heat?
a. sublimation
d. melting
9.
Which of the statements a to d is INCORRECT?
a.
b.
c.
d.
e.
10.
an ionic crystal
a network covalent solid
a metallic crystal
a molecular crystal
not enough data given
A crystal does not conduct electricity, yet its melt and aqueous solutions do. It is hard
and brittle and melts at a very high temperature. What type of crystal is it?
a.
b.
c.
d.
e.
13.
an ionic crystal
a network covalent solid
a metallic crystal
a molecular crystal
not enough data given
A crystal does not conduct electricity, even after melting. It is hard and brittle and melts
at a very high temperature. What type of crystal is it?
a.
b.
c.
d.
e.
12.
molecular solids have high melting points
the binding forces in molecular solids include London forces
ionic solids have high melting points
ionic solids are insulators
all of the statements a to d are correct
A solid and its melt readily conduct electricity. The crystal also has a luster and is easily
deformed. Thus, it is (also explain why it cannot be the others):
a.
b.
c.
d.
e.
11.
b. vaporization c. crystallization
e. none of these
an ionic crystal
a network covalent solid
a metallic crystal
a molecular crystal
not enough data given
In which of the following substances would dispersion forces be the only significant
factor in determining the boiling point?
1. Cl2
2. HF
3. Ne
4. KNO2
5. CCl4
a. 1, 3, 5
b. 1, 2, 3
c. 2, 4
d. 2, 5
e. 3, 4, 5
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14.
Which of the substances below is an example of a network solid?
a.
b.
c.
d.
e.
15.
S8 (s)
SiO2 (s)
MgO (s)
NaCl (s)
C25H52 (s)
Answer each of the following as true or false and give an explanation of your answer.
(a) The role of intermolecular forces is more important in a gas than in a liquid .
(b) The process of melting is always exothermic.
(c) Metals are characterized by their high melting points, brittle nature, and good
electrical conductivity.
(d) During a phase change from solid to liquid, a substance’s temperature
remains the same as more liquid is formed and rises only after all the solid has melted.
(e) A liquid with a low vapor pressure at 25ºC has a high molar heat of
vaporization.
16.
Arrange the following substances in the order in which you would expect their boiling
points to increase. Provide a description of the types of intermolecular forces which are
important for each of the substances.
FNO
(MM = 49.00)
Cl2
(MM = 70.91)
O2
(MM = 32.00)
CH3CH2OH (MM = 46.08)
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