13
Liquids and Solids
Chapter Goals
1. Kinetic-Molecular Description of Liquids and Solids
2. Intermolecular Attractions and Phase Changes
分子間引力及相的改變
The Liquid State
3. Viscosity 黏度
4. Surface Tension 表面張力
5. Capillary Action 毛細現象
6. Evaporation 蒸發
7. Vapor Pressure 蒸氣壓
8. Boiling Points and Distillation 沸點和蒸餾
9. Heat Transfer Involving Liquids 液體之熱傳導
2
Chapter Goals
The Solid State
10. Melting Point 熔點
11. Heat Transfer Involving Solids 固體之熱傳導
12. Sublimation and the Vapor Pressure of Solids 昇
華及固體之蒸氣壓
13. Phase Diagrams (P versus T) 相圖
14. Amorphous Solids and Crystalline Solids非晶固
態及結晶固態
15. Structures of Crystals 結晶的結構
16. Bonding in Solids 固體之鍵結
17. Band Theory of Metals金屬能帶理論
3
Kinetic-Molecular Description
of Liquids and Solids
• Solids and liquids are condensed states 壓縮狀態
– The atoms, ions, or molecules in solids and
liquids are much closer to one another than
in gases.
– Solids and liquids are highly incompressible
不能壓縮的
• Liquids and gases are fluids 流動的.
– They easily flow.
• The intermolecular attractions in liquids and
solids are strong. 液體及固體之分子間的引力很強
4
5
6
Kinetic-Molecular Description
of Liquids and Solids
• Schematic representation of the three
common states of matter.
The process in which a liquid changes to a solid
• Solidification
• Crystallization
• A more specific term
• The formation of a very ordered solid material
7
Kinetic-Molecular Description of
Liquids and Solids
• If we compare the strengths of interactions among
particles and the degree of ordering of particles, we
see that
Gases< Liquids < Solids
• Miscible可溶混的liquids are soluble in each other.
– Examples of miscible liquids:
• Water dissolves in alcohol.
• A drop of red ink in the water
Diffusion
• Gasoline dissolves in motor oil.
The miscibility of two liquids refers to their ability
to mix and produced a homogeneous solution 均
質溶液
8
Diffusion in solids: very slowly
9
Kinetic-Molecular Description of
Liquids and Solids
• Immiscible liquids are insoluble in each
other.
– Two examples of immiscible liquids:
• Water does not dissolve in oil.
• Water does not dissolve in cyclohexane 環已烷
10
Intermolecular Attractions and
Phase Changes
• Intermolecular forces 分子間作用力
– The forces between individual particles (atoms,
molecules, ions) of a substance
• Intramolecular forces 分子內作用力
– Covalent and ionic bonds within compounds
Covalent bonds
共價鍵結
hydrogen bonding
氫鍵
Intramolecular forces
Intermolecular forces
11
• 分子間吸引力是物質中個別粒子 (包括
原子、分子或離子) 間的作用力，此種
作用力遠小於分子內作用力，即化合物
內原子間所形成的共價鍵與離子鍵結。
12
Intermolecular Attractions and
Phase Changes
• There are four important intermolecular
attractions.
– This list is from strongest attraction to the weakest attraction.
1.Ion-ion interactions 離子- 離子作用力
– The force of attraction between two oppositely charged ions
is governed by Coulomb’s law 庫侖定律.
(q+)(q-)
F
d2
q+ and q- are the ion charges.
d is the distance between the e ions
F is the force of attraction
Energy has the units of forces x
distance, F x d
The energy of attraction, E
(q+)(q-)
E
d
※ 庫侖定律：
在真空中，兩個靜止點電荷之間的相互作用力的大小與兩點電荷
電量大小乘積成正比，與距離平方成反比。
13
Intermolecular Attractions and
Phase Changes
• Most ionic bonding is strong
have relatively high melting points
• Ionic substances containing multiply charged ions, such
as Al3+, Mg2+, O2-, and S2-, usually have higher melting
and boiling points than ionic compounds containing
only singly charged ions, such as Na+, K+, F-, Cl-.
• For a series of ions of similar charges, the closer
approached ions of smaller ions result in stronger
interionic attractive forces and higher melting points
(NaF, NaCl, NaBr)
14
Intermolecular Attractions
and Phase Changes
• Coulomb’s law determines:
1.The melting and boiling points of ionic compounds.
2.The solubility of ionic compounds.
Example 13-1: Arrange the following ionic compounds in
the expected order of increasing melting and boiling
points.
NaF, CaO, CaF2
Na+F- < Ca2+F2-<Ca2+O2-
15
Intermolecular Attractions and
Phase Changes
2. Dipole-dipole interactions 偶極- 偶極作用
–在極性分子中,有正及負兩極,分子帶正電之一端吸引
另一分子帶負電的一端,即稱之
–使得極性分子間的引力大於非極性分子
–偶極─偶極吸引力一般僅是共價鍵及離子鍵強度的百
分之一，它們的強度會隨偶極之間距離的增加而變小
– BrF
16
Intermolecular Attractions and
Phase Changes
3. Hydrogen bonding 氫鍵
– Very strong dipole-dipole interaction
– Strong hydrogen bonding occurs among polar
covalent molecules containing H (hydrogen
bond donor) and one of the three small, highly
electronegative elements — F, O or N (hydrogen
bond acceptor)
– Consider H2O a very polar molecule.
17
Intermolecular Attractions
and Phase Changes
Hydrogen bond
18
Intermolecular Attractions
and Phase Changes
4. London Forces (Dispersion forces; 分散力)
– They are the weakest of the intermolecular forces.
– This is the only attractive force in nonpolar
molecules such as O2, N2 and monatomic species
such as the noble gases.
– Without dispersion forces, such substance could
not condense to form liquids or solidify to form
solids
19
Intermolecular Attractions
and Phase Changes
• In a group of Ar atoms the temporary dipole in
one atom induces other atomic dipoles.
– The positively charged nucleus
– The electron clouds of an atom in nearby
molecules
– Dispersion forces are generally stronger for
molecules that are larger
– They exist in all substances
20
倫敦分散力 (London dispersion forces)
非極性原子或分子間的作用力稱為倫敦分散力，
此種作用力是原子或分子在移動時，其電子雲會
產生瞬間偶極距 (instantaneous dipole
moments)，此時非極性原子或分子間藉由此種
瞬間偶極相互吸引，稱之為倫敦分散力，又可稱
之為瞬間偶極距-瞬間偶極距作用力。
21
分子間作用力的比較
倫敦分散力存在於所有分子中，其強度隨分子量的增加
而增加，也與分子的形狀有關。除了倫敦分散力，極性
分子間存在的另一種作用力稱為偶極-偶極作用力。
氫鍵則是鍵結在高電負度原子上之氫原子與另一高電負
度原子間之作用力，是所有分子間作用力中最強的一種
作用力。
不過上述不論那種分子間作用力，其強度皆比共價鍵或
離子鍵弱很多。
22
恆久偶極矩
23
分子間作用力
作用力的種類
作用力發生的原因
例子
偶極-偶極作用力
具有偶極矩(偶極矩大小是根
據電負度與分子結構來決定)
氫鍵
XHY
當具強極性XH鍵與帶孤對
電子的Y原子同時存在時會發
生，而這可說是一種極端的 H2O H2O
偶極-偶極作用力(當X  F、
N、O)
偶極-誘導偶極作用
力
極性分子與非極性分子同時
存在時會發生
H2OI2
倫敦分散力
發生於分子間
I2I2
H2O、HCl
分子間作用力的判斷
Intermolecular Attractions and
Phase Changes
Example 13-1: Intermolecular Forces
Identify the type of intermolecular forces that are present in a
condensed phase sample of each of the following. For each,
make a sketch, including a few molecules, that represents
the major type of force. (a) water, H2O (b) iodine, I2 (c)
nitrogen dioxide, NO2.
(a)water, H2O
•polar
•H, O hydrogen bonds
•The London forces
(b) iodine, I2
•Nonpolar the London forces
(c) nitrogen dioxide, NO2
•polar
•Dipole-dipole interactions
•The London forces
26
The Liquid State
Viscosity 黏度
• Viscosity is the resistance to flow.
– For example: molasses糖蜜, syrup糖漿or honey.
– Oil for your car is bought based on this property.
• 10W30 or 5W30 describes the viscosity of the oil at high
and low temperatures.
• Measure by Viscometer.
• The stronger of the intermolecular forces of
attraction the more viscosity
• Increasing the size and surface area of molecule
genernally results in increased viscosity
• Temperature increase
Pentane C5H12戊烷
viscosity decrease
黏度是液體內部抵抗液體流動的阻力，當
液體的黏度愈大，液體就愈不容易流動。
液體的黏度大小與液體分子間作用大小有
關，分子間作用力愈大，黏度愈大。
Dodecane C12H26十二烷
27
The Liquid State
Surface Tension 表面張力
• Surface tension is a measure of the
unequal attractions that occur at the
surface of a liquid.
• The molecules at the surface are
attracted unevenly.
液體表面分子僅受到液面下方的分子吸引！此現象導致液體表
面分子感受到往液內的淨吸引力。因此，液面分子隨時受到一
個向下的拉力，此力即稱『表面張力』。
當分子間的作用力愈強時，表面張力愈大
28
The Liquid State
Capillary Action 毛細作用
• Capillary action is the ability of a liquid to rise (or
fall) in a glass tube or other container 當一根直徑細小
的玻璃管或毛細管放在水中時，水會在細管中往上升，此
種液態在細管中爬升的現象稱為毛細作用
• Cohesive forces內聚力 are the forces that hold
liquids together. 內聚力表示:液體分子間的作用力
• Adhesive forces 吸附力 are the forces between a
liquid and another surface. 吸附力表示:液體分子與管
壁間的附著力
– Capillary rise implies that the:
• Adhesive forces > cohesive forces
– Capillary fall implies that the:
• Cohesive forces > adhesive forces
• The smaller the bore, the higher the liquid climbs
29
The Liquid State
• Capillary action also affects the meniscus of
liquids.毛細現象會影響圓筒內液體的凹凸面
concave
convex
Adhesive forces > cohesive forces Adhesive forces < cohesive forces
水與玻璃管的吸附力大於水
分子間的內聚力，因此呈現
下凹的液面
汞分子間的內聚力大於汞 30
與玻璃管的吸附力，因此
呈現上凸的液面
The Liquid State
Evaporation 蒸發
•Evaporation is the process in which molecules
escape from the surface of a liquid and become a
gas.物質從液體表面變為氣體狀態的過程
31
The Liquid State
•Evaporation is temperature dependent.
The rate of evaporation increases as
temperature increases
Only the higher-energy molecules
can escape from the liquid phase
32
The Liquid State
•Condensation 凝結; 冷凝
•In a closed container
Dynamic equilibrium
liquid
evaporation
condensation
vapor
•LeChatelier’s Principle 勒沙特列原理
•If the vessel were left open to the air,
the equilibrium could not be reached.
•A liquid can eventually evaporated
entirely
33
The Liquid State
Vapor Pressure 蒸氣壓
• Vapor pressure is the pressure exerted by a
liquid’s vapor on its surface at equilibrium.
• Vapor Pressure (torr) and boiling point for three
liquids at different temperatures.
0oC 20oC
diethyl ether乙醚 185 442
Ethanol 酒精
12 44
water 水
5 18
30oC
647
74
32
normal boiling point
36oC
78oC
100oC
• Vapor pressures of liquids always increase as
temperature increase
液體的沸點是液體的表面蒸氣壓等於外界壓力
時的溫度
34
The Liquid State
Vapor Pressure 蒸氣壓
• Easily vaporized liquids are called volatile liquids
揮發性液體
• Stronger cohesive forces tend to hold molecules
in the liquid state
– Methanol molecules are strongly linked by
hydrogen bond lower vapor pressure
• Dispersion forces increase with increasing molecular
size larger molecules have lower vapor pressure
35
The Liquid State
Vapor Pressure 蒸氣壓
• Vapor pressure can be measured with
manometers壓力計
36
The Liquid State
Vapor Pressure as a function of
temperature
液體的沸點是液體的表面蒸氣壓等於外界壓力
時的溫度
37
The Liquid State
Boiling Points 沸點 and Distillation蒸餾
•The boiling point is the temperature at which the
liquid’s vapor pressure is equal to the applied
pressure (usually atmospheric大氣壓).
•The normal boiling point正常沸點 is the boiling point
when the pressure is exactly 1 atm (760 torr). 當外界
壓力等於一大氣壓時，此溫度稱為正常沸點
– If the applied pressure is lower than 1 atom
water boil below 100oC.
– 物質的沸點高低與分子間作用力有直接的關係。當分
子間作用力愈大，沸點愈高。
•Distillation 蒸餾is a method we use to separate
mixtures of liquids based on their differences in
boiling points.
38
The Liquid State
Distillation蒸餾
• Different liquids have different vapor pressure and
boil at different temperature.
• Distillation is a process in which a mixture or solution
is separated into its components on the basis of the
differences in boiling points of the components.
• Distillation is another vapor pressure phenomenon.
39
The Liquid State
Distillation蒸餾
40
The Liquid State
Heat Transfer Involving Liquids
• Heat must added to a liquid to raise its temperature
• The amount of heat that must be added to the
stated mass of liquid to raise its temperature by one
degree
Specific heat (J/goC) 比熱
• molar heat capacity (J/moloC)莫耳熱容量 (使1mol物
質上升1oC所需的熱量
• Molar heat (enthalpy焓) of vaporization莫耳汽化熱
改變1 mol 物質由液態形成氣態所需的能量
– 水在100oC 的莫耳汽化熱為40.7 kJ/mol
41
The Liquid State
40.7KJ/mol= 40.7x 1000J x mol = 2.26x103 J/g
KJ
18g
42
The Liquid State
Heat Transfer Involving Liquids
• Condensation
• Heat of condensation 凝結熱
liquid + heat
evaporation
vapor
condensation
43
The Liquid State
Example 13-2:
How much heat is released by 2.00 x 102 g of H2O as
it cools from 85.0oC to 40.0oC? The specific heat of
water is 4.184 J/goC.
q=mCT
?J = 2.00 x 102 g x (4.184J/goC) x( 85.0-40oC)
?J = 3.76x104 J = 37.6kJ
44
The Liquid State
Example 13-3:
The molar heat capacity of ethyl alcohol, C2H5OH, is
113 J/moloC. How much heat is required to raise
the T of 125 g of ethyl alcohol from 20.0oC to
30.0oC?
1 mol C2H5OH = 46.0 g
1mol
?mol C2H5OH = 125g x
46g = 2.72 mol C2H5OH
?J = 2.72mol x
= 3.07 KJ
113J x (30.0-20.0oC)
moloC
45
The Liquid State
• The calculations we have done up to now tell us the
energy changes as long as the substance remains in a
single phase.
• Next, we must address the energy associated with phase
changes.
– For example, solid to liquid or liquid to gas and the
reverse.
• Heat of Vaporization is the amount of heat required to
change 1.00 g of a liquid substance to a gas at constant
temperature.
– Heat of vaporization has units of J/g.
• Heat of Condensation is the reverse of heat of
vaporization, phase change from gas to liquid.
1.0g H2O(l) at
100oC
+2260J
-2260J
1.0g H2O(g) at 100oC
46
The Liquid State
Molar heat of vaporization or DHvap
• The DHvap is the amount of heat required to
change 1.00 mole of a liquid to a gas at
constant temperature.
DHvap has units of J/mol.
Molar heat of condensation
• The reverse of molar heat of vaporization is
the heat of condensation.
1.0mol H2O(l) at
100oC
+40.7kJ
-40.7kJ
1.0mol H2O(g) at 100oC
47
The Liquid State
加熱／冷卻曲線 (heating-cooling curve)
48
The Liquid State
Example 13-4:
How many joules of energy must be absorbed by 5.00 x 102 g of
H2O at 50.0oC to convert it to steam at 120oC? The molar heat
of vaporization of water is 40.7 kJ/mol and the molar heat
capacities of liquid water and steam are 75.3 J/mol oC and
36.4 J/mol oC, respectively.
50oC H2O(l)
100oC H2O(l) 100oC H2O(g) 120oC H2O(g)
1mol
?mol H2O = 500g x
18g = 27.8 mol H2O
?J = 27.8mol x 75.3J
x (100.0-50.0oC) = 1.05x105 J
o
mol C
3J
40.7x10
?J = 27.8mol x
= 11.31x105 J
mol
?J = 27.8mol x 36.4J
X (120.0-100.0oC) = 0.20x105 J
o
mol C
Total J = 1.05x105 + 11.31x105 + 0.2x105
49
= 12.56 x105 J or 1.26x103 kJ
The Liquid State
Example 13-5:
If 45.0 g of steam at 140oC is slowly bubbled into 450 g
of water at 50.0oC in an insulated container, can all
the steam be condensed?
1mol
?mol steam = 45.0g x
18g = 2.5 mol steam
?mol H2O = 450g x 1mol = 25.0 mol H2O
18g
Calculate the amount of heat required to condense the steam
2.5mol x(36.4J/moloC)x(140-100oC)+2.5 mol x (40.7kJ/mol)
=105.4kJ
Calculate the amount of heat available in the liquid water
25.0mol x (75.3J/moloC)x(100.0-50oC) = 94.1kJ
Amount of heat to condense all of the steam is 105kJ
Amount of heat that the liquid water can absorb is 94.1kJ
Thus all of the steam cannot be condensed
50
The Liquid State
Example 13-6:
Arrange the following substances in order of increasing
boiling points.
C2H6, NH3, Ar, NaCl, AsH3
Ar < C2H6 < AsH3 < NH3 < NaCl
nonpolar nonpolar polar very polar ionic
London London dipole-dipole H-bonding ion-ion
52
Example 13-2: Heat of Vaporization
Calculate the amount of heat, in joules, required to
convert 180.0grams of water at 10.0oC to steam at
105.0oC.
10oC H2O(l) 100oC H2O(l) 100oC H2O(g) 105oC H2O(g)
?J = 180.0g x 4.18J
x (100.0-10.0oC) = 0.67x105 J
o
g C
3J
2.26x10
?J = 180.0g x
= 4.07x105 J
g
oC) = 0.01820x105 J
?J = 180.0g x 2.03J
x
(105.0-100.0
goC
Total J = 0.67x105 + 4.07x105 + 0.0182x105
= 4.76 x105 J
Example 13-3: Heat of Vaporization
Compare the amount of “cooling” experienced by an individual who
drinks 400ml of ice water (0.0oC) with the amount of “cooling”
experienced by an individual who “sweats out” 400ml of water.
Assume that the sweat is essentially pure water and that all of it
evaporates. The density of water is very nearly 1.00g/ml at both
0.0oC and 37oC, average body temperature. The heat of
vaporization of water is 2.41kJ/g at 37oC.
The amount of heat lost by perspiration =
the amount of heat required to vaporize 400g of water at 37oC
Raising the temperature of 400.0g of water from 0oC to 37oC
?J = 400.0g x 4.18J
x (37.0-0.0oC) = 6.19x104 J
o
g C
or 61.9kJ
o
Evaporating 400.o ml of water at 37 C requires
?J = 400.0ml x 1.0g x (2.41x103 J/g) = 9.64x105 J
ml
or 964kJ
Sweating removes more heat than drinking ice water
Example 13-5: Boiling Points Versus Intermolecular
Forces
Predict the order of increasing boiling points for the
following: H2S; H2O; CH4; H2; KBr.
• KBr: is ionic, it boils at the highest temperature
• Hydrogen bond: H2O the next highest temperature
• polar covalent substance: H2S
• CH4; H2 are nonpolar .
CH4 is larger than H2;
so the dispersion forces are stronger in CH4
 CH4 boils at a higher temperature than H2
H2 < CH4 < H2S < H2O < KBr
The Solid State
Normal Melting Point
• The normal melting point is the temperature at
which the solid melts (liquid and solid in equilibrium)
at exactly 1.00 atm of pressure.
• Freezing point  melting point
• The melting point increases as the strength of the
intermolecular attractions increase.
solid
melting
freezing
liquid
56
The Solid State
• Which requires more energy?
NaCl(s)
H2O(s)
or
NaCl(l) Ion-ion interaction
H2O(l)
Hydrogen bond
57
Heat Transfer Involving Solids
Heat of Fusion
• Heat of fusion 熔化熱is the amount of heat required
to melt one gram of a solid at its melting point at
constant temperature.
Heat of fusion
+334J
o
1.00g H2O(l) at 0oC
1.00g H2O(s) at 0 C
-334J
Heat of solidification
• Heat of solidification固化熱 is the reverse of
the heat of fusion.
58
Specific heat
Specific heat
比熱
Heat of vaporization
Heat of fusion
溶化熱
Specific heat
比熱
59
Heat Transfer Involving Solids
Molar heat of fusion 莫耳熔化熱 or Hfusion
• The molar heat of fusion is the amount of heat
required to melt a mole of a substance at its melting
point.
• The molar heat of solidification is the reverse of
molar heat of fusion
Molar heat of fusion
+6012J
o
1.00mole H2O(l) at 0oC
1.00mole H2O(s) at 0 C
-6012J
Molar heat of solidification
60
Heat Transfer Involving Solids
• The heat of fusion depends on the intermlacular
forces of attraction in the solid state
– Heats of fusion are usually higher for substances
with higher melting points.
61
Heat Transfer Involving Solids
Example 13-7:
Calculate the amount of heat required to convert
150.0 g of ice at -10.0oC to water at 40.0oC.
specific heat of ice is 2.09 J/goC
-10oC H2O(s)
0oC H2O(s)
0oC H2O(l)
40oC H2O(l)
oC) = 3.14x103 J
?J = 150.0g x 2.09J
x
(10.0
goC
4
?J = 150.0g x 334J
g = 5.01x10 J
?J = 150.0g x 4.18J
x (40.0-0.0oC) = 2.51x104 J
o
g C
Total J = 3.14x103 + 5.01x104 + 7.83x104
= 7.83x104J
62
Example 13-6: Heat of Fusion
The molar heat of fusion, Hfus, of Na is 2.6kJ/mol at its
melting point, 97.5oC, How much neat must be
absorbed by 5.0g of sodium Na at 97.5oC to melt it?
?J = 5.0g x 1mol x 2.6kJ = 0.57 kJ
23g
mol
63
Example 13-2: Heat of Fusion
Calculate the amount of heat that must be absorbed by
50.0g of ice at -12.0oC to convert to water at 20.0oC.
-10oC H2O(s)
0oC H2O(s)
0oC H2O(l)
20oC H2O(l)
oC) = 1.25x103 J
?J = 50.0g x 2.09J
x
(0.0-(-12).0
goC
3J
?J = 50.0g x 334J
=
16.7x10
g
?J = 50.0g x 4.18J
x (20.0-0.0oC) = 4.18x103 J
o
g C
Total J = 1.25x103 + 16.7x103 + 4.18x103
= 22.1 kJ
64
Sublimation and the Vapor
Pressure of Solids
Sublimation昇華
• In the sublimation process the solid
transforms directly to the vapor phase
without passing through the liquid
phase.
• Solid CO2 or “dry” ice does this well.
• iodine
sublimation
gas
solid
deposition
Sublimation can be used
to purify volatile solids
65
Exothermic 放熱
昇華
蒸發
凝結
Endothermic 吸熱
Transitions among the three states of matter
66
Phase Diagrams (P versus T)相圖
• Phase diagrams are a convenient way to
display all of the different phase transitions of a
substance.
• The equilibrium pressure-temperature (壓力-溫
度) relationships among the different phases of
a given pure substance in a closed system (密
閉系統).
67
Phase Diagrams (P versus T)相圖
• This is the phase diagram for water.
Negative slope of line AB
AB
Melting curve
AD
Sublimation
curve
•Ice is less dense than liquid
• The network of hydrogen
bonding in ice is more
extensive than that in a
liquid water
三相點
68
Phase Diagrams (P versus T)
Compare water’s phase
diagram to carbon dioxide’s
phase diagram.
Critical point
三相點
For CO2: critical point is at 31oC and 73 atm
For H2O: critical point is at 374oC and 218 atm
Liquid CO2 cannot exist at
atmospheric pressure
69
To illustrate the use of a phase diagram in
determining the physical state or states of a system
under different sets of pressures and temperatures
70
Amorphous無定形的Solids and
Crystalline結晶狀的Solids
• Amorphous solids do not have a well
ordered molecular structure.
– Examples of amorphous solids include waxes,
glasses, asphalt柏油.
• Crystalline solids have well defined
structures that consist of extended array of
repeating units called unit cells.
– Crystalline solids display X-ray diffraction
patterns which reflect the molecular structure.
– The Bragg equation, detailed in the textbook,
describes how an X-ray diffraction pattern can
be used to determine the interatomic distances
in crystals.
71
All crystal contain regularly repeating
arrangements of atoms, molecules or ions
•Unit cell 單位晶胞
– The Lengths of its edges (a,b,c)
– The angles between the edges
(a,b,g)
– Three-dimensional arrangement
– Lattice point 晶格
立方體
四角體
斜方晶
單斜體
三斜晶系
六角體
菱形六面體
岩鹽
金紅石
瀉利鹽
石膏
矽土重鉻酸鉀
方解石
Structure of Crystals
• Unit cells are the smallest repeating unit of a
crystal.
– As an analogy, bricks are repeating units for buildings.
• There are seven basic crystal systems.
螢石
立方體 黃銅礦
綠寶石
六角體
四角體
藍銅礦
霰石斜方晶
單斜體
薔薇輝石
方解石菱形六面體
三斜晶系
73
圖 14.14
三種結晶固體的範例。每一例子只顯示結構的一部分，各結構是在三次
元以相同的圖形做連續延伸。(a)原子固體。在鑽石中每個圓球代表一
個碳原子。(b)離子固體。在氯化納固體中，每個圓球分別代表Na＋ 和
Cl－ 離子。(c)分子固體。在冰中，每三個圓球單元代表一個H2O 分子。
點線表示極性水分子之間氫鍵。
P.425
不同種類固體的範例
表 14.3
P.426
Structure of Crystals
• We shall look at the three
variations of the cubic crystal
system.
• Simple cubic unit cells單立方
– The balls represent the positions of
atoms, ions, or molecules in a
simple cubic unit cell.
– In a simple cubic unit cell each
atom, ion, or molecule at a corner
is shared by 8 unit cells
– Thus 1 unit cell contains 8(1/8) = 1
atom, ion, or molecule.
76
Structure of Crystals
• Body centered cubic
(bcc)體心立方has an
additional atom, ion, or
molecule in the center of
the unit cell.
• On a body centered
cubic unit cell there are 8
corners + 1 particle in
center of cell.
– 1 bcc unit cell
• contains 8(1/8) + 1 = 2
particles.
77
Structure of Crystals
• A face centered cubic (fcc)
面心立方 unit cell has a cubic
unit cell structure with an
extra atom, ion, or molecule
in each face.
• A face centered cubic unit
cell has 8 corners and 6
faces.
– 1 fcc unit cell contains
• 8(1/8) + 6(1/2) = 4
particles.
78
• Isomorphous
– refers to crystals having the same
atomic arrangement
• Polymorphous
– Refers to the substances that crystallize
in more than one crystalline
arrangement
Bonding in Solids
• Four categories
– Metallic solids金屬固體 − Ionic solids離子固體
– Molecular solids分子固體 − Covalent solids
80
Bonding in Solids
• Molecular Solids have molecules in each of the
positions of the unit cell.
– Molecular solids have low melting points, are
volatile, and are electrical insulators.
• Examples of molecular solids include:
– water, sugar, carbon dioxide, benzene
81
Bonding in Solids
• Covalent Solids have atoms that are covalently
bonded to one another
• Some examples of covalent solids are:
• Diamond, graphite, SiO2 (sand), SiC碳化矽
石墨
82
Bonding in Solids
• Ionic Solids have ions that occupy the positions in
the unit cell.
• Examples of ionic solids include:
– CsCl, NaCl, ZnS
83
Bonding in Solids
• Metallic Solids may be thought of as positively
charged nuclei surrounded by a sea of electrons.
• The positive ions occupy the crystal lattice positions.
• Examples of metallic solids include:
– Na, Li, Au, Ag, ……..
84
• 金屬呈現另一種形式的原子固體。
• 大部分金屬的鍵結是強但無方向性。
• 電子海模型 (electron sea model)：金屬原子規則排列
於價電子海中，此電子海為所有原子所共用並且在金屬結
晶中極易運動。
• 因為金屬結晶的本性，其他元素相對容易被加入，這些含
有其他元素的金屬物質稱為合金。
85
• 合金 (alloy)的定義為一物質含有多種元
素且具有金屬性質。
• 合金一般有兩種型態：
– 取代型合金 (substitutional alloy)：
是某些金屬原子被其他金屬原子所取
代，例如黃銅，在銅內約有1/3 的銅
原子被鋅原子所取代，
– 間隙型合金 (interstitial alloy)：金屬
原子緊密堆積出結構且其內的空隙 (孔
洞) 被其他較小的原子所佔據。鋼鐵是
一種在孔隙內包含了碳原子的鐵結晶。
P. 428