handout chm102 chapter 12
TRANSCRIPT
Intermolecular Forces: Liquids, Solids, and Phase Changes
Chapter 12
12-1
Objectives
Our goal in this chapter to learn about the interplay of forces that give rise to the three states of matter and their changes.
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• Seek to Answer Many Questions :
– Why does ice float on liquid water?
– Why is diamond, one form of carbon, so hard that it can scratch glass, whereas graphite, another form of carbon, is soft enough to use in pencils?
Chapter Outlines
• Physical States and Phase Changes
• Quantitative Aspects of Phase Changes
• Types of Intermolecular Forces
12-3
• Properties of the Liquid State
• The Uniqueness of Water
• The Solid State: Structure, Properties, and Bonding
Phases of Matter
• A physically distinctive form of matter is called a phase.
• The properties of each phase are determined by thebalance between the potential and kinetic energyof the particles.
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• Potential energy in the form of intermolecular forcestends to draw the molecules together. Why?
• Kinetic energy associated with the random motion ofthe molecules tends to disperse them.
Potential energy vs. Kinetic energy
Properties
Gas Potential energy (energy of attraction) is small relative to kinetic energy.
Particles are far apart. A gas has no fixed shape or volume.
Kinetic Molecular View of the Three States
12-5
Liquid Attractive forces are stronger because particles are touching.
A liquid can flow and change shape, but has a fixed volume.
Solid Attractions dominate the motion. Particles are fixed in place relative to each other.
A solid has a fixed shape and volume.
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 the charge = stronger attraction
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• larger the charge = stronger attraction• longer the distance = weaker attraction• however, these attractive forces are small relative to the
bonding forces between atoms– generally smaller charges– generally over much larger distances
6
Types of Phase Changes and their Enthalpies
melting vaporizingendothermic
sublimation
• As temperature increases, the average KE does too, so the fastermoving particles overcome attractions more easily.
• As temperature decreases, particles slow, so attractions can pull them together.
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solid liquid gas
freezing condensing
endothermic
exothermic
deposition
Two familiar Phase Changes
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Heats of Vaporization and Fusion for several common substances
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Phase Changes and their Enthalpy Changes
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Heat Involved in Phase Changes
A cooling curve for the conversion of gaseous water to ice
Consider: 2.5 mol of gaseous water in a closed container
Liquid water coolsLiquid water freezes
Gaseous water condenses
Solid water cools
Gaseous water cools
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• Within a phase, heat flow is accompanied by a change in temperature, since the average Ek of the particles changes.
Heat Involved in Phase Changes
q = (amount) x (heat capacity) x ΔT
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q = (amount)(ΔH of phase change)
• During a phase change, heat flow occurs at constant temperature, as the average distance between particles changes.
Problem 12.1 Finding the Heat of a Phase Change Depicted by Molecular Scenes
PROBLEM: The scenes below represent a phase change of water. Find the heat (in kJ) released or absorbed when 24.3 g of H2O undergoes this change.
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SOLUTION: There are 3 stages involved in this process:1) heating of the liquid to its boiling point2) the phase change from liquid to gas3) heating the gas to the final temperature
mol H2O = 24.3 g H2O x1 mol H2O
18.02 g H2O= 1.35 mol H2O
For Stage 1:q1 = n x Cwater(l) x ΔT
= (1.35 mol)(75.4 J/mol·°C)(100. - 85.0°C) = 1527 J = 1.53 kJ
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= 1527 J = 1.53 kJ
For Stage 2:q2 = n(ΔH°vap) = (1.35 mol)(40.7 kJ/mol) = 54.9 kJ
qtotal = q1 + q2 + q3= 1.53 + 54.9 + 0.760 kJ = 57.2 kJ
For Stage 3:q3 = n x Cwater(g) x ΔT
= (1.35 mol)(33.1 J/mol·°C)(117 – 100.°C) = 759.6 J = 0.760 kJ
Liquid-Gas Equilibrium
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In a closed flask, the system reaches a state of dynamic equilibrium, where molecules are leaving and entering the liquid at the same rate.
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The vapor pressure is the pressure exerted by the vapor on the liquid. The pressure increases until equilibrium is reached; at equilibrium the pressure is constant.
Factors affecting Vapor Pressure
As temperature increases, the fraction of molecules with enough energy to enter the vapor phase increases, and the vapor pressure increases.
higher T higher P
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The weaker the intermolecular forces, the more easily particles enter the vapor phase, and the higher the vapor pressure.
higher T higher P
weaker forces higher P
The effect of temperature on the distribution of molecular speeds
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Vapor Pressure as a function of Temperature and Intermolecular Forces
• Vapor pressure increases astemperature increases.
• Vapor pressure decreases as
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• Vapor pressure decreases asthe strength of the intermolecularforces increases.
The Clausius-Clapeyron Equation
CTR
HP +Δ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ 1- = ln vap
This equation relates vapor pressure to temperature.
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⎟⎟⎠
⎞⎜⎜⎝
⎛ −Δ
12
vap
1
2 11- = ln
TTR
H
P
P
The two-point form is used when the vapor pressures at two different temperatures are known.
Linear plots of the relationship between vapor pressure and temperature
CTR
HP +Δ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ 1- = ln vap
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RHvap-Δslope =
Problem 12.2 Applying the Clausius-Clapeyron Equation
SOLUTION:
PROBLEM: The vapor pressure of ethanol is 115 torr at 34.9°C. If ΔHvap of ethanol is 40.5 kJ/mol, calculate the temperature (in °C) when the vapor pressure is 760 torr.
P2P1
ln = - ΔHvap
R1T2
1T1
−
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T1 = 34.9°C + 273.15 = 308.0 K
ln760 torr115 torr
=- 40.5 x103 J/mol
8.314 J/mol·K1T2
1308.0 K
−
T2 = 350.00 K
= 350.00 K– 273.15
= 76.85°C
R T2 T1
Vapor Pressure and Boiling Point
The boiling point of a liquid is the temperature at which the vapor pressure equals the external pressure.
The normal boiling point of a substance is observed at standard atmospheric pressure or 760 torr.
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As the external pressure on a liquid increases, the boiling point increases.
standard atmospheric pressure or 760 torr.
Real world examples:
What is Phase Diagram ?
A phase diagram is a graphical representation of the temperature and pressure conditions under which a substance exists as a solid, liquid, gas, or some combination of these in equilibrium.
At the critical point, the densities of the liquid and gas phases become equal.
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A typical phase diagram. The dotted green line gives the anomalousbehavior of water. The Clausius–Clapeyron relation can be used to find therelationship between pressure and temperature along phase boundaries.
gas phases become equal.
At the triple point, all three phases are in equilibrium.
Phase Diagram for CO2
The solid-liquid line has positive slope, because the solid is denser than the liquid.
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The critical temperature, Tc, is the highest temperature at which a liquid and vapor can coexist in equilibrium as physically distinct states of matter
liquid.
Phase Diagram for H2O
The solid-liquid line has negative slope, because the solid is less dense than the liquid. Water expands on
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liquid. Water expands on freezing.
Types of Intermolecular Forces
Intermolecular forces are relatively weak compared to
Intermolecular forces arise from the attraction between molecules with partial charges, or between ions and molecules.
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Intermolecular forces are relatively weak compared to bonding forces because they involve smaller charges that are farther apart.
Comparison of Bonding (intramolecular) and Nonbonding (Intermolecular) Forces
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Comparison of Bonding (intramolecular) and Nonbonding (Intermolecular) Forces ……cont’d
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Polar molecules and dipole-dipole forces
Dipole–dipole forces arise when permanent dipoles align themselves with the positive end of one dipole directed toward the negative ends of neighboring dipoles.
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Dipole moment and Boiling point
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For compounds of similar molar mass, the greater the molecular dipole moment, the greater the dipole-dipole forces, so the more energy it takes to separate the molecules, thus the boiling point is higher.
The Hydrogen Bond
Hydrogen bonding is possible for molecules that have a hydrogen atom covalently bonded to a small, highly electronegative atom with lone electron pairs, specifically N, O, or F.
An intermolecular hydrogen bond is the attraction between the H atom of one molecule and a lone pair of the N, O, or F atom of another molecule.
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the N, O, or F atom of another molecule.
Hydrogen bonding and Boiling point
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Problem 12.3
Drawing Hydrogen Bonds Between Molecules of a Substance
PROBLEM: Which of the following substances exhibits H bonding? For any that do, draw the H bonds between two of its molecules.
(a) C2H6 (b) CH3OH (c)
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SOLUTION:
(a) C2H6 has no N, O, or F, so no H-bonds can form.
(b) CH3OH contains a covalent bond between O and H. It can form H bonds between its molecules:
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(c) can form H bonds at two sites:
Polarizability and Induced Dipoles
A nearby electric field can induce a distortion in the electron cloud of an atom, ion, or molecule.
- For a nonpolar molecule, this induces a temporary dipole moment.
- For a polar molecule, the field enhances the existing
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- For a polar molecule, the field enhances the existing dipole moment.
Polarizability: the ease with which a particle’s electron cloud can be distorted.
Trends in Polarizability
Smaller particles are less polarizable than larger ones because their electrons are held more tightly.
Polarizability increases down a group because atomic size increases and larger electron clouds distort more easily.
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Polarizability decreases across a period because of increasing Zeff.
Cations are smaller than their parent atoms and less polarizable; anions show the opposite trend.
Dispersion (London) Forces
Dispersion (or London) forces arises when an instantaneous dipole in one particle induces a dipole in another, resulting in an attraction between them.
• Fluctuations in the electron distribution in atoms and molecules result in a temporary dipole.
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• Dispersion forces exist between all particles (atoms, ions, molecules), increasing the energy of attraction in all matter.
• Dispersion forces are stronger for more polarizable particles. Larger particles experience stronger dispersion forces than smaller ones.
molecules result in a temporary dipole.
Dispersion Forces among Nonpolar Particles
A. When atoms are far apart they do not influence one other.
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B. When atoms are close together, the instantaneous dipole in one atom induces a dipole in the other.
C. The process occurs throughout the sample.
Molar mass and trends in boiling point
Dispersion forces are stronger for larger, more polarizable particles.
Polarizability depends on the
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Polarizability depends on the number of electrons, which correlates closely with molar mass.
Molecular shape, intermolecular contact, and boiling point
There are more points at which dispersion forces act.
There are fewer points
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There are fewer points at which dispersion forces act.
Problem 12.4 Predicting the Types of Intermolecular Forces
PROBLEM: For each pair of substances, identify the key bonding and/or intermolecular force(s), and predict which one of the pair has the higher boiling point:
(a) MgCl2 or PCl3 (b) CH3NH2 or CH3F (c) CH3OH or CH3CH2OH
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(d) Hexane (CH3CH2CH2CH2CH2CH3) or 2,2-dimethylbutane
SOLUTION:
(a) MgCl2 consists of Mg2+ and Cl- ions held together by ionic bonding forces; PCl3 consists of polar molecules, so intermolecular dipole-dipole forces are present. The ionic bonding forces in MgCl2 are stronger than the dipole-dipole forces in PCl3..
MgCl2 has a higher boiling point than PCl3.
(b) CH3NH2 and CH3F both consist of polar molecules of about the
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(b) CH3NH2 and CH3F both consist of polar molecules of about the same molar mass. CH3NH2 has covalent N-H bonds, so it can form H bonds between its molecules. CH3F contains a C-F bond but no H-F bond, so dipole-dipole forces occur but not H bonds.
CH3NH2 has a higher boiling point than CH3F.
(c) CH3OH and CH3CH2OH are both polar molecules and both contain a covalent O-H bond. Both can therefore form H bonds.
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CH3CH2OH has a higher boiling point than CH3OH.
CH3CH2OH has a larger molar mass than CH3OH and its dispersion forces are therefore stronger.
(d) Hexane and 2,2-dimethylbutane are both nonpolar molecules and therefore experience dispersion forces as their only intermolecular force. They have equal molar masses but different molecular shapes.
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Cylindrical hexane molecules make more intermolecular contact than the more compact 2,2-dimethylbutane molecules.
Hexane has a higher boiling point than 2,2-dimethylbutane.
Surface Tension:
A surface molecule experiences a net attraction downward. This causes a liquid surface to have the smallest area possible.
Properties of the Liquid State
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An interior molecule is attracted by others on all sides.
Surface tension is the energy required to increase the surface area of a liquid. The stronger the forces between the particles the higher the surface tension.
Substance FormulaSurface Tension
(J/m2) at 200C Major Force(s)
Diethyl ether
Ethanol
Dipole-dipole; dispersion
H bonding
1.7x10-2
2.3x10-2
CH3CH2OCH2CH3
CH3CH2OH
Surface Tension and Forces Between Particles
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Ethanol
Butanol
Water
Mercury
H bonding
H bonding; dispersion
H bonding
Metallic bonding
2.3x10
2.5x10-2
7.3x10-2
48x10-2
CH3CH2OH
CH3CH2CH2CH2OH
H2O
Hg
Capillary Action:
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A. Water displays a concave meniscus.
B. Mercury displays a convex meniscus.
H-bonding ability of Water:
hydrogen bond donor
hydrogen bond acceptor
Properties of Water
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Each H2O molecule can form four H bonds to other molecules, resulting in a tetrahedral arrangement.
1. Solvent Properties of Water
The hexagonal structure of ice
2. Unusual density of solid water
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Ice has an open structure due to H bonding. Ice is therefore less dense than liquid water.
3. Thermal Properties of water
4. Surface Properties
Solids: Properties and structure
Solids are divided into two categories:
Crystalline solids have well defined shapes due to the orderly arrangement of their particles.
Amorphous solids lack orderly arrangement and have
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Amorphous solids lack orderly arrangement and have poorly defined shapes.
12-52Some crystalline solids
The Crystal Lattice and the Unit Cell
• 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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7 Unit Cells
Cubica = b = call 90°
a
b
c
Tetragonala = c < ball 90°
a
b
c
Orthorhombica ≠ b ≠ call 90°
a
b
c
Monoclinica ≠ b ≠ c
2 faces 90°
a b
c
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all 90° all 90° all 90°
Hexagonala = c < b
2 faces 90°1 face 120°
c
a
b
Rhombohedrala = b = cno 90°
ab
c
Triclinica ≠ b ≠ cno 90°
a
b
c
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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,
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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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Tro: Chemistry: A Molecular Approach, 2/e
Simple Cubic Unit Cell
Expanded view.
Space-filling view.
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Atoms/unit cell = (⅛ x 8) = 1
⅛ atom at 8 corners
Coordination number = 6
Space-filling view.
Body-centered cubic unit cell
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⅛ atom at 8 corners
1 atom at center
Atoms/unit cell = (⅛ x 8) + 1 = 2
Coordination number = 8
The face-centered cubic cell.
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Atoms/unit cell = (⅛ x 8) + (½ x 6) = 4
⅛ atom at 8 corners
½ atom at 6 faces
Coordination number = 12
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Packing spheres to obtain three cubic and hexagonal cells
large diamond space
2nd layer directly over 1st
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simple cubic(52% packing efficiency)
body-centered cubic(68% packing efficiency)
2nd layer over diamond spaces in 1st
3rd layer over diamond spaces in 2nd
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Problem 12.5 Determining Atomic Radius
PROBLEM: Barium is the largest nonradioactive alkaline earth metal. It has a body-centered cubic unit cell and a density of 3.62 g/cm3. What is the atomic radius of barium?
(Volume of a sphere = πr3.)43
12-62
SOLUTION:
Volume/mole of Ba metal = 1density
x M 1 cm3
3.62 g Ba= x
137.3 g Ba1 mol Ba
= 37.9 cm3/mol Ba
Volume/mole of Ba atoms = cm3/mol Ba x packing efficiency= 37.9 cm3/mol Ba x 0.68
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= 37.9 cm3/mol Ba x 0.68 = 26 cm3/mol Ba atoms
= 4.3 x 10-23 cm3/Ba atom
Volume of Ba atom = 26 cm3
1 mol Ba atomsx
1 mol Ba atoms6.022x1023 Ba atoms
Volume of Ba atom = πr343
r =4π√ 3V3
√3 3(4.3x10-23cm3)
4 x 3.14=
= 2.2 x 10-8 cm
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Edge length and atomic (ionic) radius in the three cubic unit cells
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Diffraction of x-rays by crystal planes
Tools of the Laboratory
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Formation of an x-ray diffraction pattern of the protein hemoglobin
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Problem 12.6 Determining Atomic Radius from the Unit Cell
PROBLEM: Copper adopts cubic closest packing, and the edge length of the unit cell is 361.5 pm. What is the atomic radius of copper?
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SOLUTION:
Using the Pythagorean theorem to find C, the diagonal of the cell’s face:
C = A2 + B2√ The unit cell is a cube, so A = B, Therefore
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C = 2A2√ = 2(361.5 pm)2√ = 511.2 pm
C = 4r, so r =511.2 pm
4= 127.8 pm
Types of Crystalline Solids
Atomic solids consist of individual atoms held together only by dispersion forces.
Molecular solids consist of individual molecules held together by various combinations of intermolecular forces.
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Ionic solids consist of a regular array of cations and anions.
Metallic solids have exhibit an organized crystal structure.
Network Covalent solids consist of atoms covalently bonded together in a three-dimensional network.
Characteristics of the Major Types of Crystalline Solids
Type Particle(s) Interparticle Forces
Physical Properties Examples [mp, °C]
Atomic Atoms Dispersion Soft, very low mp, poor thermal and electrical conductors
Group 8A(18)(Ne [-249) to Rn [-71])
Molecular Molecules Dispersion, Fairly soft, low to Nonpolar*
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Molecular Molecules Dispersion, dipole-dipole, H bonds
Fairly soft, low to moderate mp, poor thermal and electrical conductors
NonpolarO2 [-219], C4H10 [-138]Cl2 [-101], C6H14 [-95]P4 [44.1]PolarSO2 [-73], CHCl3 [-64]HNO3 [-42], H2O [0.0]CH3COOH [17]
*Nonpolar molecular solids are arranged in order of increasing molar mass. Note the correlation with increasing melting point (mp).
Characteristics of the Major Types of Crystalline Solids
Type Particle(s) Interparticle Forces
Physical Properties Examples [mp, °C]
Ionic Positive and negative ions
Ion-ion attraction Hard and brittle, high mp, good thermal and electrical conductors when molten
NaCl [801]CaF2 [1423]MgO [2852]
Metallic Atoms Metallic bond Soft to hard, low to very Na [97.8]
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Metallic Atoms Metallic bond Soft to hard, low to very high mp, excellent thermal and electrical conductors, malleable and ductile
Na [97.8]Zn [420]Fe [1535]
Network covalent
Atoms Covalent bond Very hard, very high mp, usually poor thermal and electrical conductors
SiO2 (quartz) [1610]C (diamond) [~3550]
Rock Salt Structures:
• Coordination number = 6
• Cl─ ions 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+ in holes between Cl─
– octahedral holes
Structures of Some Ionic Solids
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– 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
• Therefore, the formula is NaCl
73
Zinc Blende Structures
• Coordination number = 4
• S2─ ions 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+ in holes between S2
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• Each Zn2+ 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
74
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
Structures of Some Network Covalent Solids
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– due to the strong covalent bonds holding the atoms in position
– used as abrasives
• Electrical insulator
• Thermal conductor– best known
• Chemically very nonreactive
75
sp3 hybridzed C atoms, each of which is bonded with tetraheral geometry to four other carbons
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
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• bond length = 142 pm
– sp2
• The sheets are then stacked and held together by dispersion forces– sheets are 341 pm apart
76
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
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the sheets together, they can slide past each other• glide planes
– lubricants
• Electrical conductor– parallel to sheets
• Thermal insulator
• Chemically very nonreactive
77
sp2 hybridzed C atoms, each of which is bonded with trogonal planar geometry to three other carbons
Structure of Fullerene
• The third crystalline allotrope of carbon is called fullerene. It consists of spherical C60
molecules with extraordinary shape of a soccer ball. The C60 molecule has 12 pentagonal and 20 hexagonal faces, with each atom sp2-hybridzed and bonded to three
12-7878
sp2 hybridzed C atoms, each of which is bonded to three other carbons
each atom sp2-hybridzed and bonded to three other atoms.
• Graphene is a one-atom-thick planar sheet of sp2-bondec carbon atoms that are densely packed in a honeycomb crystal lattice
• Graphite itself consists of many graphene sheets stacked together.
Structure of Graphene
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• The carbon-carbon bond length in graphene is approximately 0.142 nm.
• Properties:Graphene is as the strongest material ever measured, some 200 times stronger than structural steel.
Quartz• SiO2 in pure form
– impurities add color
• 3-dimensional array of Si covalently bonded to 4 O– tetrahedral
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– tetrahedral
• Melts at ~1600 °C• Very hard
80
Tro: Chemistry: A Molecular Approach, 2/e
Metallic Bonding
• Metal atoms release their valence electrons
• Metal cation “islands” fixed in a “sea” of mobile electrons
+ + + + + + + + +
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e-
e-e-
e-
e-
e-
e-
e-
e-
e-
e-
e-
e-
e-
e-
e-
+ + + + + + + + +
+ + + + + + + + +
+ + + + + + + + +
81
Crystal structures of metals
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A. Copper adopts cubic closest packing.
B. Magnesium adopts hexagonal closest packing.
The band of molecular orbitals in lithium metal
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Electrical conductivity in a conductor, semiconductor, and insulator
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conductor semiconductor insulator
The conducting properties of a substance are determined by the energy gap between the valence and conduction bands of MOs.
n- and p-Type SemiconductorsAn n-type semiconductor is produced when a crystal is “doped” with another substance with more valence electrons
The energy levels of
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The energy levels of these donor atoms lie quite close to the conduction band and the “extra” electron(s) are lost to the conduction band
Crystal structures and band representations of doped semiconductors
12-86
Pure silicon has an energy gap between its valence and conduction bands. Its conductivity is low at room temperature.
Doping silicon with phosphorus adds additional valence e-. These enter the conductance band, bridging the energy gap and increasing conductivity.
n- and p-Type Semiconductors
The energy levels of
A p-type semiconductor is produced when a crystal is “doped” with another substance with less valence electrons
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The energy levels of these acceptor atoms lie quite close to the valence band and electrons are easily promoted from the valence band into the acceptor level EOS
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Doping silicon with gallium removes electrons from the valence band and introduces positive ions. Si electrons can migrate to the empty orbitals, increasing conductivity.
The Photovoltaic Cell
A photovoltaic cell, or solar cell, is a semiconductor device that converts light to electricity
The cell consists of a thin layer of p-type semi-conductor, such as Si doped with Al, in contact with an n-type semiconductor, such as Si doped with P
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The p-type semiconductor in the solar cell must be very thin—about 1 × 10–4 cm to reduce the tendency for conduction electrons produced by sunlight to be captured by positive holes and immobilized in covalent bonds.
An Example Cell
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EOS
p-n junction
The p-n junction
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Placing a p-type semiconductor adjacent to an n-type creates a p-n junction. Electrons flow freely in the n-to-p direction.