chapter 11 states of matter: liquids and solids. 11 | 2 1.comparison of gases, liquids, and solids...
TRANSCRIPT
Chapter 11States of
Matter: Liquids and Solids
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1. Comparison of Gases, Liquids, and Solids
Changes of State
2. Phase Transitions
3. Phase Diagrams
Contents and Concepts
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Liquid State
1. Properties of Liquids: Surface Tension and Viscosity
2. Intermolecular Forces; Explaining Liquid Properties
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Solid State
1. Classification of Solids by Type of Attraction of Units
2. Crystalline Solids; Crystal Lattices and Unit Cells
3. Structures of Some Crystalline Solids
4. Calculations Involving Unit-Cell Dimensions
5. Determining Crystal Structure by X-Ray Diffraction
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Learning Objectives
1. Comparison of Gases, Liquids, and Solidsa. Recall the definitions of gas, liquid, and
solid given in Section 1.4.b. Compare a gas, a liquid, and a solid using a
kinetic molecular theory description.c. Recall the ideal gas law and the van der
Waals equation for gases (there are no similar simple equations for liquids and solids).
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Changes of State2. Phase Transitions
a. Define change of state (phase transition).b. Define melting, freezing, vaporization,
sublimation, and condensation.c. Define vapor pressure.d. Describe the process of reaching a dynamic
equilibrium that involves the vaporization of a liquid and condensation of its vapor.
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2. Phase Transitions (con’t)
e. Describe the process of boiling.
f. Define freezing point and melting point.
g. Define heat (enthalpy) of fusion and heat (enthalpy) of vaporization.
h. Calculate the heat required for a phase change of a given mass of substance.
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2. Phase Transitions (con’t)
i. Describe the general dependence of the vapor pressure (ln P) on the temperature (T).
j. State the Clausius–Clapeyron equation (the two-point form).
k. Calculate the vapor pressure at a given temperature.
l. Calculate the heat of vaporization from vapor pressure.
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3. Phase Diagrams
a. Define phase diagram.
b. Describe the melting-point curve and the vapor-pressure curves (for the liquid and the solid) in a phase diagram.
c. Define triple point.
d. Define critical temperature and critical pressure.
e. Relate the conditions for the liquefaction of a gas to its critical temperature.
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Liquid State
4. Properties of Liquids; Surface Tension and Viscosity
a. Define surface tension.
b. Describe the phenomenon of capillary rise.
c. Define viscosity.
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5. Intermolecular Forces; Explaining Liquid Properties
a. Define intermolecular forces.
b. Define dipole–dipole force.
c. Describe the alignment of polar molecules in a substance.
d. Define London (dispersion) forces.
e. Note that London forces tend to increase with molecular mass.
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5. Intermolecular Forces; Explaining Liquid Properties (con’t)
f. Relate the properties of liquids to the intermolecular forces involved.
g. Define hydrogen bonding.
h. Identify the intermolecular forces in a substance.
i. Determine relative vapor pressures on the basis of intermolecular attractions.
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Solid State
6. Classification of Solids by Type of Attraction of Units
a. Define molecular solid, metallic solid, ionic solid, and covalent network solid.
b. Identify types of solids.
c. Relate the melting point of a solid to its structure.
d. Determine relative melting points based on types of solids.
e. Relate the hardness and electrical conductivity of a solid to its structure.
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7. Crystalline Solids; Crystal Lattices and Unit Cells
a. Define crystalline solid and amorphous solid.
b. Define crystal lattice and unit cell of a crystal lattice.
c. Define simple cubic unit cell, body-centered cubic unit cell, and face-centered cubic unit cell.
d. Determine the number of atoms in a unit cell.
e. Describe the two kinds of crystal defects.
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8. Structures of Some Crystalline Solids
a. Define hexagonal close-packed structure and cubic close-packed structure.
b. Define coordination number.
c. Note the common structures (face-centered cubic and body-centered cubic) of metallic solids.
d. Describe the three types of cubic structures of ionic solids.
e. Describe the covalent network structure of diamond and graphite.
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9. Calculations Involving Unit-Cell Dimensions
a. Calculate atomic mass from unit-cell dimension and density.
b. Calculate unit-cell dimension from unit-cell type and density.
10. Determining Crystal Structure by X-Ray Diffraction
a. Describe how constructive and destructive interference give rise to a diffraction pattern.
b. Note that diffraction of x rays from a crystal gives information about the positions of atoms in the crystal.
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Comparing Gases, Liquids, and Solids
Gases are compressible fluids.
Liquids are relatively incompressible fluids.
Solids are nearly incompressible and rigid.
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A change of state or phase transition is a change of a substance from one state (solid, liquid, gas) to another. The specific name or names for each of these transitions are given below.
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The vapor pressure of a liquid at a particular temperature is the partial pressure of the vapor over the liquid measured at equilibrium.
When a liquid is placed in a closed vessel, the partial pressure of its vapor increases over time until it reaches equilibrium. At equilibrium, evaporation and condensation continue to occur, but do so at the same rate. This situation, which is called a dynamic equilibrium, is illustrated on the next slide.
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The vapor pressure depends on the liquid and on the temperature. This relationship is illustrated on the next slide for four substances. Note that as temperature increases, vapor pressure increases.
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The boiling point is the temperature at which the vapor pressure is equal to the pressure on the liquid, usually atmospheric pressure. At this temperature, bubbles of gas form within the liquid, as illustrated in the figure. The normal boiling point is measured at 1 atmosphere pressure.
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When the pressure on the liquid increases, as is the case with a pressure cooker, the boiling point increases.
Conversely, when the pressure on the liquid decreases, as is the case at high altitude, the boiling point decreases.
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The freezing point is the temperature at which a pure liquid changes to a crystalline solid (or freezes).
The melting point is the temperature at which a crystalline solid changes to a liquid (or melts).
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Any change of state requires that energy be added to or removed from the system. As shown in the graph, for water, there are two regions that remain at the same temperature even as heat is added. Those represent regions of phase change.
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The heat of fusion, Hfus, is the heat needed to melt a solid. It is given in kJ/mol.
For water, the phase change is represented by
H2O(s) H2O(l); Hfus = 6.01 kJ/mol
The heat of vaporization, Hvap, is the heat needed to vaporize a liquid. It is given in kJ/mol.
For water, the phase change is represented by
H2O(l) H2O(g); Hvap = 40.7 kJ/mol
?
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Shown to the right is a representation of a closed container in which you have just placed 10 L of H2O. This is the starting point in time, t = 0.
Assume that all of the water is in the liquid phase at this point. The dots represent a few of the H2O molecules.
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Consider a time t = 1, when some time has passed but the system has not reached equilibrium.
1. How will the level of the liquid H2O compare to that at t =0?
2. How will the vapor pressure in the flask compare to that at t = 0?
3. How will the number of H2O molecules in the vapor state compare to that at t = 0?
4. How does the rate of evaporation in this system compare to the rate of condensation?
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Time t = 1:
1. The liquid level will be lower than at t = 0.
2. The vapor pressure will be higher than at t = 0.
3. There will be more molecules of H2O in the vapor state than at t = 0.
4. The rate of evaporation is greater than the rate of condensation.
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Consider a time t = 2, when enough time has passed for the system to reach equilibrium.
1. How will the level of the liquid H2O compare to that at t = 1?
2. How will the vapor pressure in the flask compare to that at t = 1?
3. How will the number of H2O molecules in the vapor state compare to that at t = 1?
4. How does the rate of evaporation in this system compare to the rate of condensation?
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Time t = 2 (the system is at equilibrium):
1. The liquid level will be lower than at t = 1.
2. The vapor pressure will be higher than at t = 1.
3. There will be more molecules of H2O in the vapor state than at t = 1.
4. The rate of evaporation is equal to the rate of condensation.
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When vapor pressure is plotted versus temperature, a curve is obtained.
When the natural log (ln P) of the vapor pressure is plotted against the reciprocal of temperature (1/T), a straight-line graph is obtained.
The next slide shows this graph for four liquids.
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The Clausius–Clapeyron equation describes these graphs:
R = 8.314510 J/(mol K)
Hvap must be given in J/mol
This equation can be used to find the vapor pressure, the heat of vaporization, or the temperature.
21
vap
1
2 11Δln
TTR
H
P
P
?
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The fuel requirements of some homes are supplied by propane gas, C3H8, which is contained as a liquid in steel cylinders. If a home uses 2.40 kg of propane in an average day, how much heat must be absorbed by the propane cylinder each day to evaporate the liquid propane, forming the gas that is subsequently burned? The heat of vaporization of propane is 16.9 kJ/mol.
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We want to know how much heat it takes to evaporate 2.40 kg of propane, which has a heat of vaporization of 16.9 kJ/mol.
We will convert kilograms to grams, then grams to moles, and finally moles to kilojoules.
mol
kJ16.9
g44.10
mol1
kg1
g10kg2.40
3
9.20 × 102 kJ
?
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The vapor pressure of diethyl ether (commonly known as simply ether) is 439.8 mmHg at 20.0°C. The heat of vaporization of ether is 29.2 kJ/mol. What is the vapor pressure of ether at 34.0°C?
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This problem uses the Clausius–Clapeyron equation. When setting up your variables, it simplifies your math if you let P2 be the unknown vapor pressure.
P1 = 439.8 mmHgT1 = 20.0°CP2 = ?T2 = 34°CHvap = 29.2 kJ/mol
21
vap
1
2 11Δln
TTR
H
P
P
Units are critical when working with this equation: Temperatures must be in K, and Hvap in J/mol.
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P1 = 439.8 mmHgT1 = 293°CP2 = ?T2 = 307°CHvap = 29.2 x 103 J/mol
21
vap
1
2 11Δln
TTR
H
P
P
K307
1
K293
1
KmolJ8.315
molJ1029.2
mmHg439.8ln
3
2P
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P2 =
Solve the right side first:
K307
1
K293
1
KmolJ8.315
molJ1029.2
mmHg439.8ln
3
2P
760 mmHg
0.5466mmHg439.8
ln 2
P
1.727mmHg439.8
2
P
Use ex or inverse ln
(3 significant figures)
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A phase diagram is a graphical way to summarize the conditions under which the various states of a substance are stable.
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In a phase diagram, phases are separated by lines that represent equilibrium between those phases.
The triple point is the point where all three phases are in equilibrium.
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The phase diagram for water illustrates the components described on the previous slide.
The slope of the AB line, which shows the equilibrium between solid and liquid, slants to the left. This is because the density of the liquid is greater than the density of the solid.
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The phase diagram for carbon dioxide is given at right. Note that the AB line slants to the right.
At STP, carbon dioxide is never a liquid. That is why carbon dioxide sublimes directly from solid to gas.
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The phase diagram for sulfur shows some interesting characteristics. The existence of two different solid structures for sulfur, rhombic and monoclinic, gives rise to more than one triple point.
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The critical point gives the temperature (critical temperature, TC) at which the liquid state can no longer exist and the pressure at that temperature (critical pressure). Above this temperature and pressure, there is only one state, a supercritical fluid. This is illustrated below for CO2, for which TC = 31°C.
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At 3200 m, the pressure on the liquid is less than at sea level, which means the boiling point is lower. The eggs will need to be cooked longer at the higher altitude.
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Now we will look at some properties of liquids: surface tension and viscosity.
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Surface tension is the energy required to increase the surface area of a liquid by a unit amount. The values are given in J/m2.
Surface tension arises because the molecules at the surface of a liquid experience a net force toward the center of the liquid.
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Surface tension is the property that allows insects to walk on the surface of water or a pin to float.
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Surface tension is also related to capillary rise. In this phenomenon, when a very-small-diameter glass tube (a capillary) is placed in water, the water rises.
The water is attracted to the glass, so a thin film moves up the surface of the glass. This action expands the surface area, so the water column rises to counteract it. The process then repeats itself.
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Surface tension also explains the development of a meniscus, the curvature of the surface of a liquid in a column.
When the liquid is attracted to the glass, the curvature is concave, as is the case with water.
When the liquid is not attracted to the glass, the curvature is convex, as is the case with mercury.
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Viscosity is the resistance to flow exhibited by liquids and gases.
This property differentiates the flow of syrup from the flow of water: Syrup is more viscous than water.
In the figure, steel balls were dropped into the glycerol and the water at the same time.
glycerol water
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To explain the properties of liquids that we’ve just studied, we need to learn about intermolecular forces—that is, the forces of interaction between molecules.
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We can learn about intermolecular forces by studying heats of vaporization.
Neon is normally a gas, but it liquefies at a temperature of –246°C at 1 atm of pressure. The heat of vaporization for liquid neon is 1.77 kJ/mol.
The Hvap for neon is in part the energy needed to push back the atmosphere when the vapor forms (0.23 kJ/mol) and in part the energy needed to overcome intermolecular attractions (1.54 kJ/mol).
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Three types of intermolecular forces exist between neutral molecules or atoms: dipole-dipole forces, London forces, and hydrogen bonding forces.
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The term van der Waals forces is a general term that includes both dipole-dipole forces and London forces.
Polar molecules exhibit dipole-dipole forces that result in alignment of the molecules. This phenomenon is illustrated below for solid and liquid HCl.
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In nonpolar molecules, there is no dipole-dipole force, yet there is still a force of attraction. In 1930, Fritz London explained this relationship with what we now call the London forces.
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In a nonpolar molecule, the charge is uniformly distributed over time. But in any one instant, the charge is not uniformly distributed. In that instant, there is an instantaneous dipole. To see how this works, let’s look at a neon atom.
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The instantaneous dipole of the neon atom induces an instantaneous dipole in adjacent atoms, resulting in an attractive force between them.
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In the next moments, the electrons move, and the instantaneous dipoles change. However, the instantaneous dipoles tend to change together, maintaining the attractive force between them.
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London forces are the weak attractive forces between molecules resulting from the small, instantaneous dipoles that occur because of the varying positions of the electrons during their motion about the nuclei.
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All molecules exhibit London forces.
London forces increase with increasing mass (atomic number) because the presence of more electrons causes a stronger instantaneous dipole.
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London forces are also larger for more polarizable (more easily distorted) molecules.
On the next slide, each compound has the same chemical formula, C5H12. However, in each compound, the carbon atoms are bonded in a different structure. Note how the heats of vaporization differ for the molecules.
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The straight-chain compound has the strongest intermolecular forces and the highest heat of vaporization because it is most flexible and, therefore, most polarizable.
The most compact compound has the weakest intermolecular forces and the lowest heat of vaporization.
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We have just related heat of vaporization to the strength of intermolecular forces: The stronger the intermolecular forces, the higher the heat of vaporization.
We can do the same for vapor pressure, boiling point, surface tension, and viscosity.
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The stronger the intermolecular forces, the lower the vapor pressure at any given temperature.
The stronger the intermolecular forces, the higher the boiling point.
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The stronger the intermolecular forces, the higher the surface tension.
Viscosity depends in part on intermolecular forces: The stronger the intermolecular forces, the more viscous the liquid.
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Let’s look at Table 11.2 again.
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The compounds are listed in order of increasing molar mass. We would expect that the vapor pressures would decrease from top to bottom, but this is not the case: The vapor pressures of water and glycerol are much lower than anticipated.
How can this be explained?
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Let’s look at the structures of water, glycerol, and methanol. These compounds have unusually low vapor pressures and unusually high boiling points. What do these molecules have in common?
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Each molecule has one or more –OH group.
Molecules with this group have an additional force called the hydrogen bonding force.
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Hydrogen bonding is a weak attractive force that exists between hydrogen atoms bonded to a very electronegative atom, X, and a lone pair of electrons on another small, electronegative atom, Y.
—X—H . . . Y—
Most often, X is F, O, or N, the smallest, most electronegative elements.
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Hydrogen bonding in water is illustrated here in three different formats: ball-and-stick model, electrostatic map showing electron density distributions, and Lewis formula.
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We can see the impact of adding hydrogen bonding forces on the boiling points of Group VI hydrides.
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The boiling points for Group IVA, VA, and VIIA hydrides are graphed to the right.
Group IVA hydrides follow the expected pattern. In Group VA, NH3 does not. Likewise, in Group VIIA, HF does not.
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a. There are several ways to determine this.
First, the products in the first reaction form an explosive mix.
Second, the volume of gas from the first reaction is three times that of the second reaction.
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b. The first reaction requires breaking bonds. The second reaction is a phase change that involves disrupting intermolecular forces. Because chemical bonds are much stronger than intermolecular forces, the enthalpy of the first reaction would be much higher than that of the second reaction.
c. The enthalpy change for the first reaction could be calculated using standard enthalpy of formation values (Chapter 6) or bond energies (Chapter 9).
?
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Identify the intermolecular forces that you would expect for each of the following substances:
a. O2
b. H2O2
c. CHBr3
a.Nonpolar molecule: London forcesb.Polar molecule with O—H bond: London forces,
dipole-dipole forces, hydrogen bonding c. Polar molecule: London forces, dipole-dipole
forces
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A solid is a nearly incompressible state of matter with a well-defined shape, because the units (atoms, molecules, ions) making up the solid are in close contact and reside in fixed positions or sites.
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We will now look at solids in more detail. We will first identify the basic unit that makes up the solid and the forces between those units. We will then use this information to classify solids into one of four types: molecular solid, ionic solid, metallic solid, or covalent network solid.
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A molecular solid consists of atoms or molecules. It is held together by intermolecular forces.
A metallic solid consists of the positive cores of metal atoms. It is held together by metallic bonding, a “sea” of delocalized electrons.
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An ionic solid is composed of cations and anions. It is held together by ionic bonds, the electrical attractions between oppositely charged particles.
A covalent network solid consists of atoms. These atoms are held together in large chains or networks by covalent bonds.
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Carbon exists as a covalent network solid in both the graphite and diamond forms.
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Physical properties such as melting point, hardness, and electrical conductivity are related to solid structure.
Table 11.1 gives melting points and boiling points for various solids, indicating the solid structure of each.
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In general, molecular solids have lower melting points, and covalent network solids and ionic solids have high melting points.
The melting points of ionic solids vary owing to differences in lattice energy that reflect the charge and sizes of the ions involved.
The melting points of metals vary widely.
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Hardness depends on how easily the structural units can be moved relative to one another.
Molecular solids tend to be soft. By contrast, three-dimensional covalent network solids, such as diamond, are very hard.
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Ionic compounds are brittle because they tend to fracture easily along crystal planes.
Metals are malleable, so that they can be easily shaped by hammering.
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Metals are good electrical conductors because of their delocalized valence electrons.
Ionic solids do not conduct electricity. When melted, however, they do conduct electricity. They also conduct electricity when they are dissolved in water.
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Solids can also be described as either crystalline or amorphous.
A crystalline solid is composed of one or more crystals with a well-defined ordered structure in three dimensions.
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An amorphous solid has a disordered structure and lacks a well-defined arrangement of basic units.
A glass is an amorphous solid obtained by cooling a liquid rapidly enough that its basic units are “frozen” in random positions before they can assume a crystalline arrangement.
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The ordered structure of a crystal is described in terms of its crystal lattice. A crystal lattice is the geometric arrangement of lattice points of a crystal in which we choose one lattice point at the same location within each of the basic units of the crystal.
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The crystal structure and lattice for copper is shown here. The copper atoms have been shrunk to make the structure more visible.
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A unit cell is the smallest boxlike unit from which you can imagine constructing a crystal by stacking the units in three dimensions.
The shapes of the unit cells for the seven different crystal systems are shown on the next slide. Each unit cell features a different angle between the faces and relationship between the dimensions of the faces.
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We will now look in more detail at the cubic crystal system.
There are three possible unit cells: simple cubic, body-centered cubic, and face-centered cubic.
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We have assumed that crystals have perfect order. In fact, real crystals have defects or imperfections.
Ruby is one example. It is primarily Al2O3, but some Al3+ ions are replaced with Cr3+ ions, giving the red color.
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We now look in more detail at the structures of several solids of different types: molecular, metallic, ionic, and covalent network.
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Molecular solids exhibit cubic and hexagonal close-packed structures.
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Closest packing of spheres begins with one layer, A, followed by layer B over spaces in layer A. The next layer can be either layer C or a repeat of layer A.
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When the pattern is ABABAB, a hexagonal close-packed (hcp) structure results.
When the pattern is ABCABC, a cubic close-packed structure results. This is identical to a structure with a face-centered cubic (fcc) unit cell.
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In any close-packed structure, each atom is surrounded by other atoms. The number of nearest neighbors is called the coordination number.
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Many metallic solids are close-packed crystals. In a body-centered cubic (bcc) lattice, 68% of the space is occupied by atoms. In a close-packed (cp) structure, 74% of the space is occupied by atoms.
The table on the next slide identifies the type of solid each metal forms.
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The crystal structure of ionic solids is complicated by the fact that we need to give both cations and anions positions within the crystal structure.
We will examine the crystal structure of three compounds with the general formula MX: CsCl, NaCl, and ZnS.
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In CsCl, the cation and the anion have approximately the same size.
Two models of the unit cell are shown on the next slide: a space-filling model (left) and a model with the ions shrunk in size to emphasize the structure (right).
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The crystal structure has a Cs+ ion in the center with a Cl- ion at each corner. We could also take the unit cell to have a Cl- ion at the center and Cs+ ions at each corner. This gives a total of one Cs+ ion and one Cl- ion (8 corners times 1/8 each) per unit cell.
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In NaCl, the Cl- ions are much larger than the Na+
ions. As a result, the Cl- ions almost touch one another.
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The crystal structure involves two overlapping lattices. Each has one ion at each corner and on each face. This gives a total of four of each ion per unit cell: 8 corners times 1/8 ion each + 6 faces times ½ ion each per unit cell.
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Zinc sulfide is polymorphic—that is, it crystallizes into two different structures. The hexagonal structure is called wurtzite; the cubic structure is called zinc blende or sphalerite. We will discuss the cubic structure.
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Each corner, each face, and the center has a S2- ion. There are also four Zn2+ ions completely within the unit cell. This gives a total of four Zn2+ ions and four S2- ions (8 corners times 1/8 Cl- each + 6 faces times ½ Cl- each) per unit cell.
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The structure of a covalent network solid is determined primarily by the direction of the covalent bonds.
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The structure of diamond is similar to that of ZnS, with a carbon atom replacing each Zn2+ position (dark spheres) and each S2- position (light spheres). Each carbon is bonded to four other carbons and has a tetrahedral geometry.
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The structure of graphite consists of large, flat sheets of carbon atoms covalently bonded to form hexagonal arrays. These sheets are stacked on top of each other. They slide over one another easily, making graphite a good lubricant.
The graphite structure has several possible resonance formulas and, therefore, delocalized electrons. That property makes graphite a good conductor.
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The figure on the left shows one resonance structure for graphite. The figure on the right shows an electron micrograph of graphite in which the layer structure is visible.
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The large atoms, A, occupy each corner and each face. Eight corners times ⅛ atom each plus six faces × ½ atom each gives a total of four A atoms. The four small atoms, B, are completely contained within the unit cell. With four A and four B, the formula is AB.
Because the four B atoms are completely within the unit cell, the unit cell is determined by the arrangement of A atoms, which is face-centered cubic (fcc).
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Knowing the unit cell type and its dimensions allows several calculations:
• Using the cell’s dimensions, you can calculate its volume.
• From the density and volume of the crystal, you can calculate its mass.
• Using the mass of the crystal and the type of unit cell, you can calculate the mass of an atom.
Alternately, from the density and the unit cell structure, you can calculate the length of the unit cell edges.
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Determining Crystal Structure by X-Ray Diffraction
The ordered structure of crystal planes can reflect x rays in a diffraction pattern. This pattern can be analyzed to determine the positions of the atoms in the unit cell.