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INTERMOLECULAR FORCES
AND CONDENSED STATES
I. Intermolecular Forces A. Types of Intermolecular Forces. 1.
Van der Waals forces = attractive forces that exist between neutral
molecules. a. Are much weaker than the intramolecular force of the
covalent bond.( 150 - 1100 kJ/mol) b. Weaker than ion-ion forces
(Lattice energies, 200 - 4000 kJ/mol) 2. Dipole-dipole forces. a.
Exist between polar molecules. Due to the attraction of the
negative end of one dipole and the positive end of another. b. Vary
as 1/d2 ( d = distance ). Fairly strong ( 5-25 kJ/mol), should
depend on the dipole moments of the molecules. c. Especially
important in understanding the behavior of small polar molecules.
3. Dipole-induced dipole forces. a. The permanent dipole in one
molecule can distort, or polarize, the electron density in a
neighboring molecule creating, or inducing, a dipole. The resulting
force of attraction is called dipole-induced dipole attraction. It
can take place between a polar and a nonpolar molecule or between
two polar molecules.
polar molecule nonpolar molecule
Polarization
+-
+-
induced dipoledipole
dipole - induced dipole ! attraction
- +
b. These forces of attraction will increases as the
polarizability of the molecule increases. The polarizability is a
measure of the ease of distortion of the electron density of a
molecule. c. The polarizability increases as
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1) the number of electrons increases. 2) the size of the
molecule increases d. Forces ∝ 1/r6 and are fairly weak (2-10
kJ/mol). They are important in that they are responsible for the
solubility of nonpolar gases,
such as O2 in polar solvents, such as water. 4. Induced
dipole-induced dipole forces (London dispersion forces). a. A
dipole created by an instantaneous asymmetry in the electron
density of one molecule can
induce a dipole in a neighboring molecule, giving rise to a
force of attraction. b. Example. Consider two He atoms. Each He has
two electron in spherically symmetric 1s orbitals. However, it is
possible to have an instantaneous asymmetry in the electron density
of
one of the He atoms. This will induce an asymmetry in
neighboring He atoms. If the electrons move in a synchronous
manner, there will always be a weak attractive force between the He
atoms.
+2
+2
Instantaneous Dipole
--
-
-
-++
+
+
+
Induced dipole
+2+
+
+
+ -
-
-
-
Induced Dipole - Induced Dipole Attraction
He 1s2
Symmetric
This type of attractive forces are called London Dispersion
Forces or Induced Dipole- Induced Dipole forces.
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c. Such forces should increase as the polarizabilities of the
molecules increase. It is found in all molecules, polar and
nonpolar. It is the only type of attractive forces that exist
between nonpolar molecules. Forces ∝ 1/d6 (0.05-40 kJ/mol).
5. Hydrogen Bonding. a. Special force of attraction that exists
between unshielded nuclear charge on a H,
when it is covalently bonded to an O, N or F and the lone pair
of electrons on a neighboring O, N, or F.
Range is 10-40 kJ/mol. This is in addition to the other van der
Waals forces. b. F, O, and N are the three most electronegative
elements in the Periodic Table and are also the smallest members of
their respective groups. Therefore any lone pairs on
these atoms will be in small concentrated orbitals. 1) When
these electronegative elements form covalent bonds with hydrogen,
the
electron density in the bonds will be greatly polarized away
from the H atom. 2) Since H has only one electron (no core
electrons) the polarization of the bonds will mean that the side of
the H opposite a bond to F, O, or N will be greatly depleted of
electron density, leaving an unshielded, "bare", nucleus. 3) The
hydrogen bond arises from and attraction of a lone pair on F, O, or
N with the unshielded nuclear charge on hydrogen (due to its
covalent bonding to another F, O, or N). c. Hydrogen bonding is
unique for H bonded to F, O, or N. It is in addition to any dipole-
dipole, dipole-induced dipole or London forces. There are geometric
constraints on the atoms involved in hydrogen bonding.
X−H ------- : X (X = F, O, or N) Hydrogen Bond Lone Pair d. Some
examples of molecules that undergo hydrogen bonding in their pure
states (the hydrogen atoms capable of undergoing hydrogen bonding
are in bold). O H O || | || CH3C-O-H CH3N-H H-C-O-H H2O HF NH3.
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Liquid State
I. Properties of Liquids. A. Kinetic Molecular Theory and its
Application to Liquids. 1. In liquid state molecules cannot move
independently of one another. Molecules condense together and can
move only in the body of the liquid. a. Because of the motion
collisions take place and there is a distribution of kinetic
energies much like that in gases. b. The average kinetic energy
per mole is much less than that required to
overcome attractive forces and go into the gas phase. However,
at any instant in time a certain fraction of molecules will have
sufficient kinetic energy to escape into the gaseous state If such
a high energy molecule is on the surface, it can escape from the
liquid and go into the gas phase. c. This process is called
evaporation. 1) Evaporation is a surface phenomenon. 2) Evaporation
is a cooling process. The high kinetic energy molecules leave which
lowers the average kinetic energy per mole, and hence the
temperature. 2. Evaporation in a closed container--Vapor Pressure.
a. In a closed container, as evaporation puts more molecules in the
vapor phase above the liquid, the probability that a gas molecule,
through random collision, will end up back in the liquid state
increases. b. Therefore, as evaporation continues, the rate of
condensation will increase. A point
will be reached when the rate of evaporation = rate of
condensation. At this point a dynamic equilibrium is established
and the concentration of vapor molecules above the liquid will
reach a constant value.
c. The partial pressure of vapor above the liquid will therefore
reach a constant value, called the equilibrium vapor pressure of
the liquid (or the vapor pressure).
This is the same pressure that is required to liquefy the gas at
the particular temperature.
It is independent of the size of the container but increases as
the temperature increases. d. Boiling point = the temperature at
which the vapor pressure equals atmospheric
pressure. At that temperature evaporation can take place in the
body of the liquid and the phenomenon of boiling takes place.
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Normal boiling point = temperature at which the vapor pressure
is equal to 1 atm (760 Torr) e. Substances having strong
intermolecular forces of attraction (having large values of a
in the van der Waals equation, high critical temperatures and
critical pressures) will have low vapor pressures and high normal
boiling points.
B. Variation of Vapor Pressure with Temperature -- The
Clausius-Clapeyron Equation. 1. Consider the equilibrium at
constant pressure
a. ΔH for this process = Molar heat of vaporization = ΔHvap b.
The partial pressure of the vapor in equilibrium with the liquid
is
lnP = - ΔHvap
RT + B
where B is a temperature independent constant. This equation is
called the Clausius-Clapeyron Equation.
2. ΔHvap is always positive (evaporation is endothermic),
therefore, a plot of lnP vs. 1T
will be a straight line with a slope of - ΔHvap
R , that is, a negative slope.
a. A convenient form of the equation for calculations is
ln ( P2P1 ) = -
ΔHvapR [
1T2 -
1T1 ]
b. Example. The molar heat of vaporization of H2O is 40.79
kJ/mol and its normal boiling point is 100.0 °C. Calculate the
vapor pressure of H2O at 27.0 °C. P2 = 760 Torr T2 = 100.0 + 273.1
= 373.1 K P1 = ? T1 = 27.0 + 273.1 = 300.1 K
ln(760 P1 ) = - 40.79
8.314x10-3 [ 1
373.1 - 1
300.1 ] = 3.199 760 P1 = e
3.199 = 24.5
P1 = 76024.5 = 31.0 Torr.
The actual vapor pressure of H2O at 27 °C = 27.0 Torr. The
discrepancy is due to the fact that ΔHvap is not exactly
temperature independent. C. Other Properties.
Liquid Vapor
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1. Surface Tension, γ. a. A molecule in the bulk liquid is
stabilized by the attraction of the other liquid
molecules. Therefore, a molecule on the surface will be at a
higher energy than one in the bulk of the liquid.
b. Surface tension = excess energy per unit area of the surface
= force required to stretch the surface of the liquid per unit
area. Dimensions of γ are Newton/meter. c. Surface tension is
responsible for the rise of a liquid such as H2O in a thin glass
capillary (water "wets" or adheres to glass). This also gives rise
to a meniscus when water or an aqueous solution is in a pipette or
a burette. d. Surface tension should be large for a liquid having
high attractive forces. 2. Viscosity. a. Viscosity is the
resistance to flow of a liquid. b. When a liquid flows down a tube,
the molecules slide over one another. The higher the forces of
attraction, the less of a tendency the liquid will have to flow,
and the higher will be the viscosity. c. Viscosity will also depend
on the size and shape of the molecules. 3. The following Table
lists some properties as some common liquids.
Properties of Liquids. Liquid NBP,°C ΔHvap(kJ/mol) γ (N/m)
Viscosity (centipoise) Ethyl Ether 34.5 26.0 17.0x103 0.222
(CH3OCH3) Methanol 65.0 37.4 22.6x103 0.547 (CH3OH) Ethanol 78.5
39.2 22.8x103 1.100 (CH3CH2OH) Benzene 80.1 33.9 28.9x103 0.608
(C6H6) Water 100.0 40.7 72.8x103 0.890 (HOH) II. Intermolecular
Attraction. A. General intermolecular attractive forces. 1. The
intermolecular forces that exist in the liquid state are the same
as found among gases. a. Dipole-dipole forces. b. Dipole-induced
dipole forces. c. Induced dipole-induced dipole (London dispersion)
forces 2. The first two are important in polar liquids while the
last one exists in all liquids.
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3. In some liquids, such as H2O, hydrogen bonding is also
important. B. Effects of Hydrogen Bonding on the Properties of
Water. 1. Abnormally high melting points, boiling points, and
ΔHvap. a. Compared to the other group 16 hydrides, H2O has
unusually high melting and boiling points. b. NH3 and HF also show
these abnormalities. 2. The density of H2O(s) is less than that of
H2O(l). a. In H2O(s) each O is tetrahedrally surrounded by four
H's; two are covalently bonded and two are hydrogen bonded. b.
Because of the orientation requirements imposed by hydrogen
bonding, the H2O molecules are not packed together in ice in the
spatially most efficient way. c. When ice melts, the hydrogen bonds
are broken and the ice lattice structure collapses in on itself.
Therefore, the density increases. 3. Between 0°C and 4°C, the
density of H2O(l) increases as the temperature increases. a. For
most substances density decreases as the temperature increases (
things expand on heating). b. In H2O(l) between 0 and 4°C there are
still large domains of water molecules
involved in ice-like hydrogen bonded structures. As the
temperature increases from 0 to 4°C,these domains are progressively
broken down and the liquid contracts. The maximum density occurs at
3.98°C (1.000 g/ml), as the temperature increases from this point,
the natural tendency of a substance to expand on heating dominates
and the density decreases.
4. Other Effects. a. Protein Structure. Hydrogen bonding is one
of the main forces stabilizing the α-helix structure of proteins.
b. Genetic Code. Base pair recognition in DNA and RNA is through
hydrogen bonding.
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Phase Diagrams I. Phase Diagram. A. Diagrams on a pressure vs.
temperature scale that show the conditions under which a phase will
be stable or in equilibrium with other phases.
B. Phase diagram of H2O (not to scale). Consists of three areas
that are divided by lines (A-O, O-B, and O-C) that intersect at
a
point (point O).
A
Solid
Liquid
O
B
C
Vapor
Pre
ssure
, T
orr
760
4.58
0.01 100Temperature, oC0 1. Areas. At any temperature and
pressure that falls in an area, only one phase is stable. Any other
phase will spontaneously convert to the particular stable phase. 2.
Lines. Delineate temperatures and pressures in which two phases are
in equilibrium. a. Line A-O. H2O(s) H2O(g) 1) A plot of the
sublimation pressure of the solid as a function of temperature. 2)
Governed by a Clausius-Clapeyron type of equation.
lnP = -!H
sub
RT+!S
sub
R
ΔHsub = ΔHvap + ΔHfus
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b. Line O-B. H2O(l) H2O(g) 1) A plot of the vapor pressure of
the liquid as a function of temperature, or a plot of the boiling
point of the liquid as a function of pressure. 2) The plot is
governed by the Clausius-Clepeyron equation. The line O-B stops at
the critical temperature and critical pressure of the substance. c.
Line O-C. H2O(s) H2O(l) 1) A plot of the freezing point of the
liquid, or melting point of the solid, as a function of pressure.
2) For H2O the freezing point decreases as the temperature
increases. This is because the density of H2O(s) is less than the
density of H2O(l). This is predicted from Le Chatelier's Principle
(when a stress is placed on a system in equilibrium the equilibrium
will shift so as to counteract the stress). The molar volume of
H2O(s) is greater than that of H2O(l). Therefore, when pressure is
applied the equilibrium H2O(s) H2O(l) will shift to the side having
the smaller molar volume, that is, to the liquid and the
temperature would have to be decreased in order to reestablish
equilibrium. 3. Point O. Triple Point. a. At this temperature and
pressure all three phases coexist in equilibrium, that is, the
melting point and the boiling point coincide. This is a unique
point for a substance. b. Note that the liquid phase can not exist
at pressures below that of the triple point. c. For H2O the
pressure of the triple point is quite low, 4.58 Torr. However, the
triple point for CO2 is at 5.2 atm and -57 °C. Therefore, while ice
melts on heating at normal atmospheric pressure, CO2(s) will
sublime, and hence the name "dry ice".
Crystalline State I. Structure of Solids. A. Crystal lattice. 1.
Crystalline solids (true solids) are characterized by both short
range and long range order. a. The particles that make up the solid
(ions, molecules, or atoms) are arranged in definite repeating
geometric patterns. b. This internal order gives rise to an
external order in that crystalline solids possess plane faces that
intersect at definite angles. 2. The internal geometric order can
be shown by the crystal lattice (or space lattice). a. The lattice
is a network of lines connecting the centers of the particles that
make up the solid. b. Shown below is a two dimensional array of
spheres forming a monolayer "solid"
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and its corresponding lattice.
Solid
Lattice
3. Unit Cell. a. It is not necessary to consider the complete
space lattice. We are only interested in the smallest part of the
space lattice which, when repeated, will generate the complete
lattice. This is called the unit cell. b. The unit cell is the
smallest part of the space lattice (that is the solid) that has all
the properties of the solid. That is, it has 1) the same geometry
or morphology as the solid. 2) the same stoichiometry as the solid.
3) the same density as the solid. c. The unit cell can be
determined experimentally from X-ray diffraction. 1) X-rays are
scattered by the electrons of the atoms in a solid. The process
acts like the X-rays were being scattered (or diffracted) by the
lines on a grating ( the "lines" are the rows of atoms in the
solid). 2) Depending on the wavelength of the X-rays, the angle of
incidence of the X-ray
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beam, and the geometry of the solid, the diffracted beam could
be in phase and are detectable, or out of phase and not detectable.
This gives rise to a diffraction pattern. 3) By analyzing the
diffraction pattern of the X-rays of a particular wavelength, the
structure of the solid, that is the unit cell, can be determined.
B. Systems of Unit Cells. 1. There are seven systems of unit cells.
These are described in the following Figure. We will be concerned
only with the Cubic System. 2. In the cubic system the basic shape
of the unit cell is that of a cube (all sides of equal length and
meeting at right angles). There are three unit cells in the cubic
system. a. Simple Cubic Unit Cell (Primitive Cubic). 1) The unit
cell consists of eight particles, one at each of the corners of the
unit cell. 2) Since each of the eight particles belong to eight
unit cells, only 1/8 of each particle is contained within a
particular unit cell. The total number of particles per simple
cubic unit cell = 1/8(8) = 1 3) Therefore, there is one unit cell
per particle. If one had a mole of a solid that crystallized in a
simple cubic unit cell, there would be 6.03x1023 unit cells in the
solid. 4) In a simple cubic unit cell the particles are in contact
along the edge of the unit cell. The radius of the particle, r, and
the length of the unit cell, a, are related by
2r = a or r = a2
5) The simple cubic unit cell is rather rare in nature. The only
metal crystallizing in this unit cell is Po.
b. Body Centered Cubic (bcc) Unit Cell. 1) The unit cell
consists of the eight particles at the corners plus another
particle
directly in the center of the unit cell. 2) There are 1/8(8) + 1
= 2 particles per body centered cubic unit cell. 3) The particles
are in contact along the diagonal of the cube. There are three
particles stacked along this diagonal. Let d = the length of the
diagonal; r = the radius of a particle and a = the length of
the edge of the unit cell. Then
d = a2 + a
2 + a
2 = 3 a
d = 4r
! 4r = 3 a or r = 4
3 a
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4) Iron, among other metals, crystallize in a bcc unit cell. c.
Face Centered Cubic (fcc) Unit Cell. 1) The unit cell consists of
the eight particles at the corners of the unit cell plus a
particle in the center of each of the six faces of the cube.
Each particle on a face is in common with two unit cells and
contributes 1/2 per unit cell.
2) The unit cell contains 1/8(8) + 1/2(6) = 4 particles. 3)
Contact between the particles is along the diagonal of the face of
the cube. Let d = the diagonal of a face; r = the radius of the
particle; and a = the length of the edge of the unit cell. Then
d = a2 + a
2 = 2 a
4r = d
! 4r = 2 a or r = 4
2 a
4) Al, Cu, Au, Ag, and many other substances, crystallize in a
fcc unit cell.
C. Calculations on Cubic unit cells. Aluminum (At. wt. = 27.0)
crystallizes in a fcc unit cell and has a density of 2.7 g/cc.
Calculate, 1. the molar volume V .
V = 27.0 g/mol
2.7 g/cc = 10.0 cm3/mol
2. the cm3 per atom of Al.
cm3 per atom = 10.0 cm3/mol
6.02x1023 atoms/mol = 1.67x10-23 cm3/atom
3. the cm3 per unit cell, that is the volume of the unit cell,
Vuc. Vuc = (1.67x10-23 cm3/atom)(4 atoms/uc) = 6.7x10-23 cm3/uc 4.
the length of the edge of the unit cell, a.
Vuc
= a3 ! a = V
uc3
a = 6.7x10-23
cm33
= 4.05x10-8
cm = 405 pm 5. the radius of an Al atom, r. In a fcc unit
cell,
r = 4
2 a =
4
2 (405 pm) = 142 pm
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II. Close Packing. Many common metals and salts crystallize in
structures that can be understood by considering how spheres of
equal size can be packed together in the spatially most efficient
way possible. This is called close packing. A. They will stack in
Layers. 1. First Layer (layer a). a. Each sphere in the layer is
touching six other spheres in that layer. b. This layer is shown in
Figure 1. 2. Second Layer (layer b). a. Each sphere in this layer
will come to rest in the crevice formed by the juncture of
three spheres in layer a (see Figure 2). b. Each sphere is then
in contact with 12 other spheres, 6 in the same layer + 3 in the
layer below + 3 in the layer above. 1) The number of nearest
neighbor spheres (the number of spheres touching a
particular sphere) is called the coordination number (CN). 2) In
close packing the coordination number = 12. c. All space is not
occupied by the spheres. There are holes, or cavities, between the
two layers. There are two types of such cavities. 1) Tetrahedral
Cavities are formed by the juncture of three spheres in one layer
and one sphere in an adjacent layer (see Figure 3). - If a particle
were in this cavity it would be tetrahedrally coordinated by the
four spheres. - They are the smaller of the two types of cavities.
- They are more numerous than the other type of cavity. The number
of tetrahedral
holes twice is equal to the number of spheres in the close
packed array (one cavity above and below each sphere)
2) Octahedral Cavities are formed by the juncture of three
spheres in one layer and three spheres in an adjacent layer (see
Figure 4). - If a particle were in this cavity it would be
octahedrally surrounded by the six spheres. - They are larger but
less numerous than tetrahedral cavities. - The number of octahedral
cavities is equal to the number of spheres in the close packed
array.
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Figeue 2. Layers a and b in a close packed array.
Figure 3. Tetrahedral cav ities.
Figure 4. An octahedral cavity.
First Layer Sphere
Second Layer Sphere
Figure 1. Layer a in a close packed array.
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3. Third Layer. There are two possibilities. a. Spheres in the
third layer cover tetrahedral holes. 1) When one third layer comes
to rest over a tetrahedral hole, all other spheres in that layer
must cover tetrahedral holes. Every sphere in the third layer will
be directly over a sphere in the first layer. Therefore, the third
layer is just a repeat of the first layer, layer a. 2) When each
succeeding layer covers tetrahedral cavities, the layer sequence is
ababab … 3) This type of close packing is called Hexagonal Close
Packing (hcp). The unit cell of the resulting solid is in the
hexagonal system. 4) Examples of hcp solids are Zn, Cd, and H2(s).
b. Spheres in the third layer cover octahedral holes. 1) When one
sphere comes to rest over an octahedral cavity, all other spheres
in that layer must cover octahedral cavities. The spheres in the
third layer are not over spheres in either the first or the second
layer. This in a unique layer, called layer c. 2) When octahedral
holes are covered by each succeeding layers, the layer sequence is
abcabcabc… 3) This type of close packing is called Cubic Close
Packing (ccp). The unit cell is face centered cubic. 4) Examples:
Al, Cu, Ag, Au, and many others. B. Structures of Ionic Compounds.
1. The structures of many ionic compounds can be understood in
terms of one ion forming a close packed array and the other ion
occupying the one of the cavities in that array. a. Since anions
are larger than cations, most of the time, but not all the time,
the anions form the close packed array and the smaller cations are
in the cavities. b. In ionic compounds, The coordination number of
an ion is defined as the number of nearest neighbor counter-ions.
1) The coordination number of an ion in a tetrahedral cavity = 4.
2) The coordination number of an ion in a octahedral cavity = 6. 2.
Examples. a. NaCl (rock salt) 1) The Cl- ions form a ccp array and
the Na+ ions occupy all the octahedral cavities in the Cl- close
packed array. 2) The coordination number of Na+ = 6 Since there are
just as many Na+ ions as Cl- ions, there also must six Na+ ions
around each Cl-. Therefore, the coordination number of Cl- = 6.
(NaCl could
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also be described as an array of close packed Na+ ions with the
Cl- ions occupying octahedral holes.) 3) Per unit cell there are 4
Cl- ions (the Cl- ions crystallize in a fcc unit cell) and 4
Na+ ions. 4) Many MX type salts crystallize in the NaCl
structure. This is called the rock salt structure or a 6:6
structure. The 6:6 refers to the coordination numbers of the cation
and anion, respectively. b. Li2O 1) The O2- ions form a ccp array
with the Li+ ions occupying all the tetrahedral
cavities in the O2- array. 2) The coordination number of Li+ =
4. Since there are twice as many Li+ ions as there are O2- ions,
there must be twice as many Li+'s around an O2- as there are O2-'s
around a Li+. Therefore the coordination number of O2- = 8. 3) Per
unit cell there are 4 O2- ions (O2- forms a fcc unit cell) and 8
Li+ ions. 4) This structure is called an antifluorite structure.
The mineral fluorite, CaF2, has a structure in which the Ca2+ ions
are ccp and the F- ions occupy the tetrahedral cavities. c. MgAl2O4
(Spinel). 1) The O2- ions form a ccp array with the Mg2+ ions in
the tetrahedral holes and the Al3+ ions in octahedral holes. 2) Not
all the cavities are occupied. Per formula unit there are four O2-
ions which
gives rise to eight tetrahedral and four octahedral holes.
Therefore, the Mg2+ ion
occupy only 18 of the tetrahedral and the Al3+ ions occupy
12 of the octahedral
cavities. 3) Per unit cell there are 4 O2- ions, 1 Mg2+ ion and
1 Al3+ ion. 4) This structure is called the spinel structure. There
are a number of minerals of the formula AB2O4 , where A is a
divalent ion and B is a trivalent ion, that have the spinel
structure. IV. Solid State Defects. A. Defects in chemically pure
substances. 1. Vacancies. Some of the lattice sites are not
occupied. 2. Interstitials. Due to thermal motion, a particle can
move into a type of cavity that is not ordinarily occupied, an
interstitial position. The introduction of these types of defects
on heating of a solid is important in determining the melting point
of the solid. 3. Dislocations. The planes of atoms are not
perfectly aligned.
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a. Edge dislocations. A plane of particles will not go all the
way through the crystal. b. Screw dislocation. The planes of
particles are warped about an axis such that they form a spiral
like the pitch of a screw. B. Chemical Impurities. 1.
Nonstoichiometric solids. a. These defects can occur in salts where
the metal cation can exist in several oxidation states. Some
examples are TiO1.7-1.8 and Fe0.95O. Small amounts of the higher
oxidation state metal could replace the lower oxidation state metal
in the lattice. In order to maintain electrical neutrality, some
metal sites must be vacant. For example, Fe0.95O is due to the
presence of Fe3+ being in some of the Fe2+ sites. There are also
some metal rich nonstiochiometric compounds, such as, Cu1.7S and
Cu1.65Te. b. Many of the heavier metal hydrides can have varying
amounts of hydrogen. An
example is VH0.56 Other metals, such as, Fe, Pd, and Pt are
permeable to hydrogen. It is thought that hydrogen atoms diffuse
through these metals by occupying interstitial sites in the metal
lattice.
2. Chemical doping - semiconductors. a. n-type semiconductor. 1)
Formed when small amounts of a group 15 element are introduced into
a group 14 element's lattice. The group 14 solid is said to be
doped with the group 15 element. An example is Si doped with P 2)
The Si is sp3 hybridized and forms covalent bonds with four other
silicons. When a P replaces a Si in this lattice, four of its
valence electrons are involved in covalent bonding while the extra
P electron is free to move in the crystal. The doped Si then
becomes a good conductor of electricity. b. p-type
semiconductor.
1) Formed when a group 14 crystal is doped with a group 13
element. An example is Si doped with B.
2) The B has only three valence electrons and when it replaces a
Si atom in the Si crystal there will be a vacancy of one electron
or a "hole" into which a electron
from a neighboring bond could migrate. In this way the vacancy,
or hole, could move through the crystal in a direction opposite to
that of electron motion; this will have the effect of a "positive
hole" moving through the crystal giving rise to p-type
semiconduction.
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V. Types of Solids. It is possible to classify solids not
according to their structures, but according to the particles
occupying the lattice sites. Molecular _Ionic Covalent __Metallic
Units that occupy Molecules Cations Atoms Metal cations Anions in
electron sea Binding Forces Van der Waals Electrostatic Covalent
Electrical (dipole-dipole, attraction attraction between London) +
ions and - electrons Properties Very soft Hard and Very Hard Hard
to soft brittle Low melting High melting Very High Very variable
points points points melting points Volatile Good Insula- Good
insula- Nonconduc- Good tors tors tors conductors Examples H2, H2O,
NaCl, KBr, C(diamond), Na, K, Al, CO2 Fe(NO3)2 Carborundum Fe
(SiC), Quartz (SiO2)
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Condensed Phases
Problem Set 1. Consider the following diagram of the logarithm
of the vapor pressure vs. the reciprocal
of the temperature for several substances. ( Note (l) = liquid,
(s) = solid )
ln
P(P
in t
orr
)
4
6
8
10
12
A(l)
B(l)
C(l)
C(s)
.0020 .0025 .0030 .0035 .0040 .0045
1 / T , K - 1 a. Which liquid is the most volatile at 127 °C ?
(liquid C) b. At what temperature will liquid A abd B have the same
vapor pressure ? (≈526 K) c. Estimate the normal boiling point of
A. (350 K) d. Which liquid has the highest value of ΔHvap ? (liquid
A) e. Estimate the temperature and pressure of the triple point of
C.(303 K, 3.63x104 Torr) 2. The vapor pressure of a liquid obeys
the relation
ln P = - T
4000 + 7.881
where P is the vapor pressure in torr. a. What is the value of
ΔHvap for this liquid ? (3.33x104 J/mol) b. What is the normal
boiling point of the liquid ? (3206 K)
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3. The vapor pressure of a liquid is 5.2 torr at 25.0 °C and
35.6 torr at 40.0 °C. Calculate a. the value of ΔHvap of the
liquid.(9.95x104 J/mol) b. The vapor pressure of the liquid at 60.0
°C. (353 Torr) 4. Consider the following liquids: C6H6, CH4,
C6H5Cl, CH3Cl. Arrange the liquids in order of a. increasing normal
boiling points. (CH4
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is the more numerous one. g.Per unit cell, there are ________ I-
ions and _________ Cu+ ions. 11. Shown below is a representation of
the first two layers in a close packed array. (The solid circles
are in the first layer.)
a. Two types of cavities exist in this lattice. Identify several
of these using "O' for octahedral
and "T" for tetrahedral. b. Over what type of cavity would the
third layer be located if the lattice were cubic ? If the
lattice were hexagonal close packed? 12. The usual geometric
picture drawn of the octahedral arrangement of six groups about a
central point is shown below. For one of the octahedral holes in
the above diagram, label the surrounding spheres a, b, c, d, e,and
f to indicate how they fit the drawing below.
a
c
db
e
f
hole
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13. Consider the following substances in their solid state. NO2
N2 CaO Cu SiO2 Listed below are a number of statements. By each
statement write the formula of the above substance that is best
described by the statement. a. Is a good conductor in the liquid
state but not in the solid state. b. Has the lowest melting point.
c. Has the highest melting point. d. At any particular temperature
will have the highest vapor pressure. e. Crystal lattice is
stabilized by dipole and London forces only. f. Lattice points
occupied by ions. g. Crystal lattice stabilized by covalent bonds.
14. The mineral perovskite has a structure in which Ca2+ ions
occupy the corners of a cubic unit
cell, O2- ions are in the centers of each of the faces, and a
Ti4+ is at the center of the cell. What is the formula of
perovskite.
15. On the folowing page is a phase diagram for a hypothetical
substance. a. On the diagram label the regions in which the solid,
liquid, and vapor phases would be
stable. b. Give the temperature and pressure of the triple
point. c. Which line represents a plot of the sublimation pressure
of the solid as a function of temperature ? d. Which line
represents a plot of the vapor pressure of the liquid as a function
of temperature ? e. What is the normal boiling point of the
substance ? f. Suppose you have a sample of this substance at 0.8
atm and 20 °C. Describe what would happen if you heated the
substance to 120 °C at a constant pressure of 0.8 atm. Give the
temperatures of any phase changes that may occur during the heating
process. g. Do the same as in (f) when the pressure is 0.4 atm. h.
Which would be greater, the density of the solid or the density of
the liquid? Justify your answer.
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1.0
0.5
20 40 60 80 100
A
O
B
C
120
Temperature, oC
Pre
ssu
re,
atm
Answers 10-16 10. a. Tetrahedral b. abcabcabc... c. Octahedral
d. 4 e. Octahedral, 6 f. Octahedral, Tetrahedra; g. 4,4 11. a. see
notes. b. Octahedral. Tetrahedral. 13. a. CaO b. N2 c. SiO2 d. N2
e. NO2 f. CaO & Cu g. SiO2 14. CaTiO3 15. b. triple point: 32
°C & 0.5 atm c. AO d. OB e. 120 °C f. melts at 32 °C, boils at
96 °C g. sublimes at 23 °C h. dsolid = dliquid since OC is
essentially vertical.
