Chapter 5 Liquids and Solids
Intermolecular Forces (Sections 5.1–5.5)
Key Concepts
intermolecular forces, phase, condensed phase, hydration, ion-dipole interaction,
hydrated salts, anhydrous salts, trends in hydration, dipole-dipole interaction,
instantaneous partial charges and dipole, induced dipole moment, London force,
polarizability, dipole–induced-dipole interaction, hydrogen bond, dimers
5.1 The Formation of Condensed Phases
Example 5.1
Account for the formation of condensed phases.
Answer
Condensed phases (solid or liquid) form when attractive forces between particles pull them
together.
5.2 Ion-Dipole Forces
Example 5.2a
Use the ion-dipole interaction energy to predict whether Li+ or Tl3+ has the stronger
attraction to water molecules. Assume that the distance r of closest approach between
the ion and water is given by the radius of the ion (see Fig. 1.43 in the text).
Solution
The relationship V ∝ −
V2 (Tl3+ )
+
V1 (Li )
=
z2 r12
z1 r2
2
=
z µ
r2
leads to the following ratio for the same polar molecule:
3 × (58 pm)2
1 × (88 pm)
2
=
10 092 pm 2
7 744 pm2
= 1.3
Thus, the energy of interaction between an H2O molecule and a Tl3+ ion is about 30% greater
than that between an H2O molecule and a Li+ ion. Because Li+ forms hydrated salts, we
predict that Tl3+ will also form hydrated salts.
Example 5.2b
Calculate the radius of a cation with a charge of +2 that has exactly the same interaction
energy with water molecules as does Na+ (r = 102 pm).
Answer
r2 = 144 pm, which is slightly larger than Ba2+ (r = 136 pm)
5.3 Dipole-Dipole Forces
Example 5.3
If cis-dibromoethene has a nonzero dipole moment (≈2.4 D) and
trans-dibromoethene has a value of 0 because its bond dipoles cancel, predict which
substance has the higher boiling temperature.
Solution
The cis substance is expected to have the higher boiling temperature because of its nonzero
dipole moment. The actual boiling temperatures, however, are not very different: cis
(112.5°C) and trans (108°C).
Note: The boiling-point rule is not always successful. The dipole moment of 1,1-dibromoethene
(≈1.4 D) is significantly smaller than the value for cis-dibromoethene, and the boiling point
1
Chapter 5
is lower [1,1 (92°C) and cis (112.5°C)]. The trans-molecule (µ = 0) is expected to have a
lower boiling temperature than the 1,1-form, but it actually has a higher one. Although there is
no dipole moment in trans-dibromoethene, localized regions of partial charge in one molecule
strongly attract regions of partial charge with the opposite sign in another molecule. The ease
of molecular packing may also play an important role as it does in solids. This exception holds
for the analogous series of dichloroethene isomers (see text examples).
5.4 London Forces
Example 5.4a
Predict the relative boiling points of the following nonpolar molecules: F2 (18 electrons),
Cl2 (34 electrons), Br2 (70 electrons), and I2 (106 electrons).
Solution
The greater the polarizability of the substance, the higher the boiling temperature. The
polarizability is proportional to the number of electrons in the molecule. The boiling points are
expected to increase in the order of F2 , Cl2 , Br2 , and I2 . The actual boiling temperatures
follow this pattern: F2 (−188°C), Cl2 (−34°C), Br2 (59°C), and I2 (184°C).
Example 5.4b
Predict the relative boiling points of the following polar molecules: H2S (18 electrons),
H2Se (36 electrons), and H2Te (54 electrons).
Solution
The greater the polarizability of the substance, the higher the boiling temperature. The
polarizability is proportional to the number of electrons in the molecule. The boiling points are
expected to increase in the order of H2S, H2Se, and H2Te. The actual boiling temperatures
follow this pattern: H2S (−60°C), H2Se (−42°C), and H2Te (−2°C).
Note: Dipole-dipole interactions predict trends opposite to those cited. Also, H2O (100°C,
10 electrons) does not conform to the general correlation, because it engages in hydrogen
bonding (Section 5.5).
Example 5.4c
Predict the relative boiling points of propane, C3H8 , and dimethyl ether, CH3OCH3 .
Solution
Both molecules have 26 electrons and are expected to have nearly the same polarizability.
Propane is essentially a nonpolar molecule (µ = 0.084 D), because of the low polarity of C−H
bonds. Dimethyl ether has a Lewis structure similar to H2O, with two methyl groups, CH3 ,
replacing two H atoms. The molecular shape is angular around the O atom, and the molecule
is polar (µ = 1.30 D). In this case, the dipole-dipole interaction is the determining factor,
because the London interactions are expected to be similar. The boiling point of dimethyl
ether is expected to be greater than that of propane. The actual boiling temperatures are −25°C
for dimethyl ether and −42°C for propane.
Example 5.4d
Predict the relative boiling points of ethanol, CH3CH2OH, and dimethyl ether, CH3OCH3 .
Solution
Both molecules have identical atoms arranged differently (isomers). Ethanol has a Lewis
structure similar to H2O, with the ethyl group, C2H5 , replacing one H atom. The O−H bond is
more polar than the O−C bond, and ethanol is expected to have a greater dipole moment
(µ = 1.69 D) and boiling point (78.5°C) than dimethyl ether. The presence of hydrogen
bonding (Section 5.5), however, is actually the major reason for the higher boiling point of
ethanol.
5.5 Hydrogen Bonding
Example 5.5a
2
Use the three dot notation to show the hydrogen bonds that form in pure ammonia, NH3 ;
pure water, H2O; and pure hydrogen fluoride, HF.
Liquids and Solids
Solution
[ N–H":N ] between NH3 and NH3 molecules in pure NH3
[ O–H":O ] between H2O and H2O molecules in pure H2O
[ F–H":F ] between HF and HF molecules in pure HF
Example 5.5b
Use the three dot notation to show the additional hydrogen bonds that form in a mixture
of ammonia, water, and hydrogen fluoride.
Solution
[ N–H":O ] and [ O–H":N ]
[ N–H":F ] and [ F–H":N ]
[ O–H":F ] and [ F–H":O ]
between NH3 and H2O molecules
between NH3 and HF molecules
between H2O and HF molecules
Note: The lone pair is a requirement of hydrogen bond formation.
Liquid Structure (Sections 5.6–5.7)
Key Concepts
long-range order, short-range order, viscosity, surface tension, capillary action,
adhesion, cohesion, meniscus
Solid Structures (Sections 5.8–5.13)
Key Concepts
crystalline solid, lattice, amorphous solid, crystal faces, solids (metallic, ionic,
molecular, network), close-packed structure (hexagonal hcp and cubic ccp),
coordination number, hole (tetrahedral and octahedral), body-centered cubic structure
bcc, primitive cubic structure pc, unit cell, face-centered cubic structure fcc (ccp),
Bravais lattices, ionic structures, rock-salt (NaCl) structure, coordination number
convention, radius ratio, cesium-chloride structure, zinc-blende structure, molecular
solids, network solids, allotropes, ceramics
5.10 Unit Cells
Example 5.10a The density of lead, Pb, is 11.34 g⋅cm−3 and the atomic radius is 175 pm. Assuming the
unit cell belongs to the cubic system, determine whether it is primitive, body-centered, or
face-centered.
Solution
Calculate the density of the metal by using the relations previously developed for each of the
three cubic types. The value that best agrees with the measured density is evidence for a
particular unit cell type. Because density has units of g⋅cm–3, convert the atomic radius to units
of cm. The ratio of the molar mass to the Avogadro constant gives the mass of a single atom
in units of g.
MPb = 207.19 g⋅mol−1;
r = 175 pm × (10−10 cm⋅pm−1) = 1.75 × 10−8 cm
3
Chapter 5
(207.19 / 6.022 14 × 1023 )
Primitive:
d=
M /N A
Body-centered:
d=
2 ( M /N A)
Face-centered:
d=
8r
3
3 3/ 2
(4 /3
=
)r
4 ( M /N A )
(83/ 2 ) r 3
−8 3
8 (1.75 × 10 )
3
=
=
= 8.02 g⋅cm −3
2(207.19 / 6.022 14 × 1023 )
3 3/ 2
(4 /3
−8 3
) (1.75 × 10 )
4 (207.19 / 6.022 14 × 1023)
(83/ 2 ) (1.75 × 10−8 )3
= 10.4 g⋅cm −3
= 11.3 g⋅cm −3
The unit cell type is most likely face-centered cubic because the density for that type is
closest to the measured value of 11.34 g⋅cm−3.
Example 5.10b The density of tungsten, W (M = 183.85 g⋅mol−1), is 19.30 g⋅cm−3 and the atomic radius is
actually 137 pm. Assuming the unit cell belongs to the cubic system, determine whether it is
primitive, body-centered, or face-centered.
Answer
The unit cell type is most likely body-centered cubic because the density for that type
(19.3 g⋅cm−3) is closest to the measured value of 19.30 g⋅cm−3.
5.11 Ionic Structures
Example 5.11a The ionic structure of CsF is the rock-salt structure. The ionic radii of Cs+ and F − are
170 pm and 133 pm, respectively. Show that the radius-ratio rule is not followed in this case.
Describe the properties of the unit cell. Calculate the density of CsF and compare your
answer to the experimental value of 4.12 g⋅cm−3.
Solution
Calculate the radius ratio and use the guidelines given to show that the ionic structure
predicted is not correct. Place the larger ions at lattice points within the fcc unit cell. Describe
the locations of the smaller ions within the cell. Calculate the density of the unit cell and
compare your answer to the experimental value.
Radius ratio (CsF) = r (F −)/r (Cs+) = (133 pm)/(170 pm) = 0.782 ≥ 0.70
The predicted ionic structure is the cesium-chloride structure. The rule is not followed for
the alkali fluorides in Group 1 except for LiF.
The larger ion is Cs+, and cations are located at the corners and faces of the fcc unit cell.
The F − anions are placed in the octahedral holes. The unit cell contains four cations and
four anions. The spheres touch along the edge:
a = 2 rcation + 2 ranion = 606 pm and V = a3 = 2.23 × 108 pm3 = 2.23 × 10−22 cm3.
The density of the unit cell is given by
d=
(4 CsF per cell) ( M / N A )
a
3
=
4 (151.91/ N A )
(2.23 × 10−22 )
= 4.52 g⋅cm −3 .
The calculated value is in good agreement with the experimental value.
Example 5.11b The ionic structure of NH4Cl follows the radius-ratio rule. The ionic radii of NH4+ and Cl−
are 143 pm and 181 pm, respectively. Use the radius-ratio rule to predict the ionic structure.
Describe the properties of the unit cell. Calculate the density of NH4Cl and compare your
answer to the experimental value of 1.527 g⋅cm−3.
Answer
4
Radius ratio = 0.79, conforming to the cesium-chloride structure. The larger ion is Cl−,
and anions are located at the corners of the pc unit cell. The NH4+ cation is placed in the
cubic hole. The unit cell contains one cation and one anion. The spheres touch along the
body diagonal: 3 a = 2 rcation + 2 ranion = 648 pm. a = 374 pm. The volume of the unit
cell is V = a3 = 5.23 × 107 pm3 = 5.23 × 10−23 cm3. The density of the unit cell is given by
Liquids and Solids
d=
(1 NH 4Cl per cell) ( M / NA )
a
3
=
1( 53.49/ NA )
(5.23 × 10
−23
)
= 1.70 g⋅cm −3 .
The calculated value is in good agreement with the experimental value.
Note: There are many exceptions to the radius-ratio rule. It should be used with caution.
5.12 Molecular Solids
Example 5.12
Account for the observation that ice floats on liquid water and that solid benzene sinks in
liquid benzene in terms of their solid structure.
Solution
In the structure of ice, each H2O molecule forms hydrogen bonds with four neighboring
molecules. The resulting structure is an open network of molecules that collapses when the
solid melts (Fig. 5.40 and Fig. 5.41 in the text). Other molecular solids (benzene, carbon
dioxide, etc.) are held together only by weaker and less directional London forces. The
resulting structure is a dense arrangement of molecules that expands when the solid melts
(Fig. 5.42 in the text). Solid ice floats on liquid water, whereas solid benzene sinks to the
bottom of liquid benzene.
5.13 Network Solids
Example 5.13
Describe the structure and properties of diamond and graphite, the allotropes of carbon, which
are both network solids.
Solution
Diamond is an insulator consisting of a three-dimensional network of σ-bonds formed by the
overlap of sp3 hybrid orbitals. The tetrahedral network accounts for the hardness and high
thermal conductivity of the solid (see Fig. 5.43). The unit cell is fcc with two carbon atoms
per lattice point. Carbon atoms occupy the corners, face positions, and four of the eight
tetrahedral holes in the unit cell.
Graphite consists of flat sheets of sp2 hybrid σ-bonds and delocalized π-bonds formed from
unhybridized p-orbitals overlapping side by side out of the plane of the sheets. The sheets are
held together by weak intermolecular forces, and atom impurities between the sheets allow
them to slide over one another. Graphite is a lubricant and a good conductor of electricity in
the plane of the sheets (see Fig. 5.44).
The Impact on Materials (Sections 5.14–5.16)
Key Concepts
properties in terms of electron mobility, malleability, ductility, slip planes, alloys
(homogeneous and heterogeneous), substitutional alloy, interstices, interstitial alloy,
liquid crystals, mesophase, anisotropic material, isotropic material, nematic phase,
smectic phase, cholesteric phase, thermotropic liquid crystals, lyotropic liquid crystals
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