Transcript INTERMOLECULAR FORCES Chap. 13

Metallic and Ionic Solids
Section 13.4
METALLIC AND IONIC SOLIDS
13.4
• Table 13.6, gives a brief over view of various
types of solids and the forces holding the
unit together.
• This table is a very good summary.
• We will now consider crystal lattice solids.
Network Solids
Diamond
Graphite
Network Solids
A comparison of diamond (pure carbon) with silicon.
Properties of Solids
1. Molecules, atoms or
ions locked into a
CRYSTAL
LATTICE
2. Particles are CLOSE
together
3. STRONG IM forces
4. Highly ordered,
rigid,
incompressible
ZnS, zinc sulfide
Figure 13.25
2-D model
Crystal Lattices and
Unit Cells of Metal Atoms
• The crystal lattice is made up of unit
cells, the smallest repeating unit of
the solid structure.
• There are many different types of
unit cells, but we will consider only
the cubic unit cells.
• Figure 13.27the seven crystal
system unit cells.
Figure 13.27
Crystal Lattices
Regular 3-D arrangements of
equivalent LATTICE POINTS in space.
The lattice points define UNIT CELLS,
the smallest repeating internal unit
that has the symmetry characteristic
of the solid.
There are 7 basic crystal systems, but
we are only concerned with CUBIC.
Cubic Unit Cells
All sides
equal length
All angles
are 90 degrees
Unit Cells of Metal Atoms
• The three important cubic unit cells are:
• simple cubic (sc)
• body-centered cubic (bcc)
• face-centered cubic (fcc).
Cubic Unit Cells
Figure 13.28
Metals have unit cells that are
• simple cubic (SC)
• body centered cubic (BCC)
• face centered cubic (FCC) or CCP
Figure 13.28
Figure 13.29
Face Atom
Corner Atom
Simple Cubic Unit Cell
Figure 13.28
• Simple cubic unit cell.
• Note that each atom is at a corner of a
unit cell and is shared among 8 unit
cells.
Body-Centered Cubic
Unit Cell
Definitions: for a Cube (x=y=z)
cell edge
cell diagonal
face diagonal
BCC cell
Body Diagonal
WHY?
Diagonal =4 atom radius = 3 edge
 cell diag 
2
  cell edge    face diag 
2
2
face diag = 2  cell edge so
 cell edge 
 cell diag 

2  cell edge

2

2
= 3  cell edge  
2
 3  cell edge  or
2
2
cell edge
cell diag = 3  cell edge
 face diag 
2
face
diag


2
=  cell edge    cell edge 
2
= 2  cell edge 
face diag = 2  cell edge
cell diagonal
2
2
face diagonal
Face Centered Cubic
Unit Cell
Atom at each cube corner plus atom
in each cube face.
NOTE:
FCC = CCP
Face diagonal
?
Crystal Lattices—Packing of
Atoms or Ions
Assume atoms
are hard spheres
and that crystals
are built by
PACKING of these
spheres as
efficiently as
possible.
FCC is more
efficient than
either BCC or SC.
Page 541
Crystal Lattices—Packing
of Atoms or Ions
Packing of C60
molecules.
They are
arranged at the
lattice points of a
FCC lattice.
Unit Cells
of Metal Atoms
• Be sure to study Examples 13.6 and
13.7, and try Exercise 13.7 to develop
skills in solving problems involving
unit cells, density, atomic mass, and
atomic radii.
• Remember that in metallic solids the
atoms touch along and edge or
diagonal depending on the unit cell.
Finding the Lattice Type
To find out if a metal is SC, BCC or
FCC, use the known radius and density
of an atom to calculate number of
atoms per unit cell.
PROBLEM Al has density = 2.699 g/cm3
and Al radius = 143 pm.
Verify that Al is FCC.
Finding the Lattice Type
SOLUTION
1. Calculate unit cell volume (FCC)
V = (cell edge)3
Edge distance comes from face diagonal.
Diagonal distance = √2 • edge
(Diagonal) 2 = 2 (edge) 2
Therefore,
Diag = 2 • (edge)
 face diag 
2
 face diag 
2
=  cell edge    cell edge 
2
= 2  cell edge 
face diag = 2  cell edge
2
2
Finding the Lattice Type
PROBLEM Al has density = 2.699 g/cm3 and
Al radius = 143 pm. What is Al’s lattice type?
SOLUTION
Here diagonal = 4 • radius of Al = 572 pm
Therefore, edge = 572 pm / 2 = 404 pm
In centimeters, edge = 4.04 x 10-8 cm
So, V of unit cell = (4.04 x 10-8 cm)3
V = 6.62 x 10-23 cm3
Finding the Lattice Type
PROBLEM Al has density = 2.699 g/cm3 and
Al radius = 143 pm. Verify that Al is FCC.
SOLUTION
2. Use V and density to calculate
mass of unit cell from
DENSITY = MASS / VOL
Mass = density • volume
= (6.62 x 10-23 cm3)(2.699 g/cm3)
= 1.79 x 10-22 g/unit cell
Finding the Lattice Type
PROBLEM Al has density = 2.699 g/cm3 and
Al radius = 143 pm. Verify that Al is FCC.
SOLUTION
3. Calculate number of Al per unit cell from
mass of unit cell.
Mass 1 Al atom =
26.98 g
1 mol
•
mol
6.022 x 1023 atoms
1 atom = 4.480 x 10-23 g, so
1.79 x 10 -22 g
1 atom
•
= 3.99 Al atoms/unit cell
-23
unit cell
4.480 x 10
g
32
Number of Atoms per Unit Cell
In the previous problem we calculated 4 Al atoms
per unit cell. What does the cell look like?
1. Each corner Al is 1/8 inside the unit cell.
8 corners (1/8 Al per corner) = 1 net Al
2. Each face Al is 1/2 inside the cell
6 faces (1/2 per face) =
3 net Al’s
Number of Atoms per Unit Cell
In the previous problem
we calculated 4 Al
atoms per unit cell. What
does the cell look like?
FCC
Repeating
unit
Number of Atoms per Unit Cell
How many atoms in a SC, BCC or FCC?
SC
Repeating
unit
BCC
Repeating
unit
Number of Atoms per Unit Cell
Unit Cell Type
SC
BCC
FCC
Net Number Atoms
1
2
4
Structures and Formulas of Ionic Solids
• Sodium chloride and cesium chloride
are two common type ionic solids.
• The sodium chloride structure is FCC in
Cl- with the sodium ions in the
octahedral holes.
• The cesium chloride structure is simple
cubic (SC) in Cl- with the cesium ion in
the center hole.
• Remember, the anion/cation ratio is
determined by the chemical formula.
Simple Ionic Compounds
Lattices of many simple ionic solids are
built by taking a SC or FCC lattice of
ions of one type and placing ions of
opposite charge in the holes in the
lattice.
EXAMPLE: CsCl has a SC lattice
of Cs+ ions with Cl- in the center.
Simple Ionic Compounds
CsCl has a SC lattice
of Cs+ ions with Clin the center.
1 unit cell has 1 Cl- ion
plus
(8 corners)(1/8 Cs+ per
corner)
= 1 net Cs+ ion.
Figure 13.30
Simple Ionic Compounds
Salts with formula
MX can have SC
structure — but
not salts with
formula MX2 or
M2X.
Simple Ionic Compounds
Many common salts have FCC
arrangements of anions with cations in
OCTAHEDRAL HOLES — e.g., salts
such as CA = NaCl
• FCC lattice of anions ----> 4 A-/unit cell
• C+ in octahedral holes ---> 1 C+ at center
+ [12 edges • 1/4 C+ per edge]
= 4 C+ per unit cell
44
Figure 13.31
46
Ionic Lattice
Page 620
47
The Sodium Chloride Lattice
Na+ ions are in
OCTAHEDRAL
holes in a facecentered cubic
lattice of Cl- ions.
Comparing NaCl and CsCl
• Even though their formulas have one
cation and one anion, the lattices of CsCl
and NaCl are different.
• The different lattices arise from the fact
that a Cs+ ion is much larger than a Na+
ion.
Common Ionic Solids
Titanium dioxide,
TiO2
There are 2 net
Ti4+ ions and 4 net
O2- ions per unit
cell.
Find the density of MgO
Given:
• The structure is FCC in O2- with Mg2+
in the octahedral holes
• Mg2+ radius = 86 pm
• O2- radius 126 pm
O's define the lattice
1 edge =
2-
2+
2-
1 O radius + 2 Mg radii + 1 O radius
= 126 + 2(86) + 126 = 424 pm
since CD
FD= 3edge then
424pm
=edge= 245pm
3
BCC cell
Since FCC
vol = (edge)
3
3
7
=245 =1.47×10 pm
10
3
since 1 cm = 10 pm
 1.47×10 cm
-23
3
mass of MgO = 40.3g/mol
40.3g MgO
1mol MgO
-23
×
=
6.69×10
g MgO
23
1mol MgO 6.02×10 MgO molecules
4 MgO
-23
1 unit cell = 4 MgO  6.69×10 g MgO ×
1unit cell
= 2.67 ×10-22 g MgO/unit cell
3.51g

MgO
3
cm
Since FCC
BCC cell
vol = (edge)3
=2453 =1.47×107 pm 3
since 1 cm = 1010pm
 1.47×10-23cm 3
Structures and Formulas of
Ionic Solids
• See Examp. 13.8 &13.9, Exer. 13.8. & 13.9
• O.H. # 97 to det. the formula of perovskite.
• Additional packing structures are given on
page 623. The structure labeled CCP is just
another name for FCC.
• O.H. B , #70p, and # 71p.
• For ionic unit cells, the ions of opposite charge
(usually) touch along an edge or diagonal.
• See pages 621 for problem-solving tips.
Page 619
Page 623
Common Ionic Solids
• Zinc sulfide, ZnS
• The S2- ions are in
TETRAHEDRAL
holes in the Zn2+
FCC lattice.
• This gives 4 net Zn2+
ions and 4 net S2ions.
Common Ionic Solids
• Fluorite or CaF2
• FCC lattice of Ca2+
ions
• This gives 4 net Ca2+
ions.
• F- ions in all 8
tetrahedral holes.
• This gives 8 net Fions.
13.5 MOLECULAR AND
NETWORK SOLIDS
• Molecular solids are held together by
relatively weak intermolecular forces.
–The molecules themselves may be made
up of only a few atoms or up to 50 or 60
atoms.
–The bonds between the atoms within the
molecules are always covalent.
MOLECULAR AND
NETWORK SOLIDS
• Network solids (Macromolecular) are made
up of huge molecules having 105or more
atoms.
– Often the chunk we are looking at is one giant
molecule!!
• Diamond and graphite are examples of this
type of solid.
– O.H. # 56 & # 75 and 3D models.
13.6 THE PHYSICAL PROPERTIES
OF SOLIDS
• Some common properties of solids are:
»melting point
»heat of fusion
»sublimation point
• The melting point is the temperature
when the solid is in equilibrium with the
liquid, normally at a pressure of 1 atm.
PHYSICAL PROPERTIES OF SOLIDS
• Heat of fusion or enthalpy of fusion,
is the energy required to change one
mole ( molar enthalpy of fusion) of the
solid into the liquid.
»The magnitudes of these properties
can be linked to the intermolecular
forces involved.
PHYSICAL PROPERTIES OF SOLIDS
• Sublimation is the direct change from
solid to vapor without passing through
the liquid state. For certain solids at
room temperature and pressure, this is
the normal transition.
• The enthalpy of sublimation
corresponds to this process.
PHYSICAL PROPERTIES OF SOLIDS
• The lattice energy is the energy required to
convert the ionic solid to gaseous ions.
• Trends in lattice energies can be predicted
using intermolecular forces.
• O.H. Figure “13.39” and Table 13.7.
Figure 13.35
Enthalpies of Fusion
13.7 CHANGES IN STRUCTURE AND
PHASE
• Phase diagrams can be used to illustrate the
changes in phase that result from changes in
temperature and pressure.
• These diagrams contain the following 6
points of interest:
1. triple point
2. normal boiling point
3. melting point
4. normal sublimation point
5. critical point
6. critical temperature
TRANSITIONS
BETWEEN
PHASES
See the phase diagram for water, Figure 13.37.
Lines connect all conditions of T and P where
EQUILIBRIUM exists between the phases on
either side of the line.
(At equilibrium particles move from liquid to gas
as fast as they move from gas to liquid, for
example.)
Phase Diagrams
• There are three phases, and
equilibrium exists between any two
phases at the phase boundary.
• There are two major classes of
diagrams (1) Water and (2) carbon
dioxide are most commonly used.
• In water the normal boiling point is
present, the normal sublimation point
is because the triple point is in front
of the normal sublimation point
H2O
TRANSITIONS
BETWEEN PHASES
71
As P and T increase, you finally reach
the CRITICAL T and P
• • P critical
.
•
LIQUID
•
High Pressure
•
• Note that line
• goes straight up
•
•
GAS
High Temperature
(TC,PC) Above critical T
no liquid exists
no matter how
• • T critical
high the
pressure.
Tc, Pc
CO2
T3
Tos
73
Water
Carbon dioxide
Phase Diagram for Water
Animation of solid
phase.
Phase Diagram for Water
Animation of equilibrium between
solid and liquid phases.
Phase Diagram for Water
Animation of liquid phase.
Phase Diagram for Water
Animation of equilibrium between
liquid and gas phases.
Phase Diagram for Water
Animation of gas phase.
Phase Diagram for Water
Animation of equilibrium between
solid and gas phases.
Phase Diagram for Water
Animation of triple point.
At the TRIPLE POINT all three
phases are in equilibrium.
Phases
Diagrams—
Important Points for
Water
T(C)
P(mmHg)
Normal boil point
100
760
Normal freeze point
0
760
Triple point
0.0098
4.58
TRANSITIONS
BETWEEN PHASES
As P and T increase, you finally reach
the CRITICAL T and P
• • P critical
.
•
LIQUID
•
High Pressure
•
• Note that line
• goes straight up
•
•
GAS
High Temperature
(TC,PC) Above critical T
no liquid exists
no matter how
• • T critical
high the
pressure.
Critical T and P
COMPD
H2O
CO2
CH4
Freon-12
Tc(oC)
374
31
-82
112
Pc(atm)
218
73
46
41 (CCl2F2)
Notice that Tc and Pc depend on
intermolecular forces.
Solid-Liquid Equilibria
In any system, if you increase P the DENSITY
will go up.
Therefore — as P goes up, equilibrium favors
phase with the larger density (or
SMALLER volume/gram).
Density
cm3/gram
Liquid H2O
1.00 g/cm3
1.00
Solid H2O
0.917 g/cm3
1.09
Solid-Liquid Equilibria
ICE
favored at
low P
P
Solid
H2O
LIQUID H 2O
favored at
high P
Liquid
H2O
Normal
freezing
point
760
mmHg
0 C
T
Solid-Liquid Equilbria
Raising the pressure at
constant T causes
water to melt.
P
The NEGATIVE SLOPE
of the S/L line is
unique to H2O.
Almost everything
else has positive
slope.
Solid
H 2O
Liquid
H 2O
Normal
freezing
point
760
mmHg
0C
T
The behavior of water
under pressure is an
example of
LE CHATELIER’S
PRINCIPLE
At Solid/Liquid
equilibrium, raising P
squeezes the solid.
It responds by going to
phase with greater
density, i.e., the liquid
phase.
Solid-Liquid
Equilbria
P
Solid
H2O
Liquid
H2O
Normal
freezing
point
760
mmHg
0C
T
Solid-Vapor Equilibrium
At P < 4.58 mmHg and T < 0.0098  C
solid H2O can go directly to vapor.
This process is called SUBLIMATION
This is how a frost-free refrigerator
works.
In a closed system…Questions
1.Can we change the equilibrium in this
system?
2.Is there any reason for wanting to change
the equilibrium in this system?
Frost-free freezers
Air in the
freezer is
warmed then
dried. The
vapor pressure
of ice is 4.579
torr. Warm,
desiccated air
can remove
water vapor.
WATER
P
R
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S
S
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