Chapter 11
Practice Exercises
11.1 (a) CH
3
CH
2
CH
2
(b) CH
3
–O–CH
3
CH
2
CH
3
< CH
3
< CH
3
CH
2
NH
2
CH
2
OH < Ca(OH)
< HOCH
2
CH
2
CH
2
2
CH
2
OH
11.2 Propylamine would have a substantially higher boiling point because of its ability to form hydrogen bonds
(there are N–H bonds in propylamine, but not in trimethylamine.)
11.3 The piston should be pushed in. This will decrease the volume and increase the pressure, and when equilibrium is re–established, there will be fewer molecules in the gas phase.
11.4 The number of molecules in the vapor will decrease, and the number of molecules in the liquid will increase, but the sum of the molecules in the vapor and the liquid remains the same.
11.5 The boiling point is most likely (a) less than 10 °C above 100 °C.
11.6 We use the curve for water, and find that at 330 torr, the boiling point is approximately 75 °C.
11.7 Adding heat will shift the equilibrium to the right, producing more vapor. This increase in the amount of vapor causes a corresponding increase in the pressure, such that the vapor pressure generally increases with increasing temperature.
11.8 Boiling Endothermic
Melting Endothermic
Condensing Exothermic
Subliming Endothermic
Freezing Exothermic
No, each physical change is always exothermic, or always endothermic as shown.
11.9 For
For
8 corners × 1/8 Ca
6 faces × 1/2 Ca fluoride:
2+
2+
8 inside the unit cell × 1 F –
Thus, the ratio is 4 Ca 2+
per corner = 1 Ca 2+
per face = 3 Ca
to 8 F – .
= 8 F –
2+
11.10 For
8 corners × 1/8 Cs + per corner = 1 Cs +
For
1 Cl − in center, Total: 1 Cl −
Thus, the ratio is 1 to 1.
11.11 The compound is an organic molecule and the solid is held together by dipole–dipole attractions and
London forces. It is also a soft solid with a low melting point, so it is a molecular crystal.
11.12 Because this is a high melting, hard material, it must be a covalent or network solid. Covalent bonds link the various atoms of the crystal.
11.13 Since the melt does not conduct electricity, it is not an ionic substance. The softness and the low melting point suggest that this is a molecular solid, and indeed the formula is most properly written S
8
.
11.14 The line from the triple point to the critical point is the vapor pressure curve, see Figure 11.21.
233

Chapter 11
11.15 Refer to the phase diagram for water, Figure 11.46. We "move" along a horizontal line marked for a pressure of 2.15 torr. At –20 °C, the sample is a solid. If we bring the temperature from –20 °C to 50 °C, keeping the pressure constant at 2.15 torr, the sample becomes a gas. The process is thus solid J gas, i.e. sublimation.
11.16 As diagramed in Figure 11.46, this falls in the liquid region.
Review Questions
11.1 The transfer of a gas from one container to another may be accompanied by either a change in shape or volume, or both. Only the shape of a liquid may change when its container is altered; the volume of a liquid does not change when the liquid is transferred to a new container. A solid changes neither its shape nor its volume when it is transferred into a new container.
11.2 Intermolecular forces in liquids and solids are more important than in gases because the molecules and atoms of liquid and solid samples are so much closer together than they are in a gas.
11.3 The intermolecular attractive forces are strongest for a solid and weakest for a gas with the liquid state in between.
11.4 Dipole–dipole interactions arise from the attraction of the permanent dipole moment of one molecule with that of an adjacent molecule, the positive end of one dipole being drawn to the negative end of the other dipole. This is diagrammed in Figure 11.3 of the text.
11.5 It is the intramolecular forces, the bonds, that are responsible for the chemical properties, not the intermolecular forces. For the physical properties, the intermolecular forces are: dipole-dipole attractions, hydrogen bonds, London forces, and ion–dipole attractions.
11.6 Polarizability is a measure of the ease with which the electron cloud is distorted. If an electron cloud is easily polarizable, that is large and easily deformed, then instantaneous dipoles and induced dipoles form without much difficulty and stronger London forces are experienced by that molecule.
11.7 Whereas the ether has no O–H linkage, ethanol does. Therefore, ethanol can have hydrogen bonding between molecules and ether cannot. Ethanol thus has stronger intermolecular forces, and its boiling point is consequently higher.
11.8 These are fluorine, oxygen, and nitrogen, which have small atomic size and high electronegativities.
11.9 (a) CaO would have stronger ion–induced dipole because the O 2– radius is smaller than S 2– , so the charge density will be higher.
(b) Al
2
O
3
will have the stronger ion–induced dipole because Al than Mg 2+ , so it will have a higher charge density.
3+ has a higher charge and smaller size
11.10 Covalent bonds are normally about 100 times stronger than normal dipole–dipole attractions; hydrogen bonds are about 5–10 times stronger than dipole–dipole attractions.
11.11 Since these are both nonpolar molecular substances, the only type of intermolecular force that we need to consider is London forces. The larger molecule has the greater London force of attraction and hence the higher boiling point: C
8
H
18
.
11.12 London forces are diagrammed in Figure 11.6. These weak forces of attraction are caused by instantaneous dipoles that attract induced dipoles in neighboring molecules. London forces increase in strength with increasing molecular size, as illustrated in Figure 11.7 and as shown by the data of Table 11.1 and Table 11.2. London forces increase in strength as the number of atoms in a molecule increases, as
234

Chapter 11 illustrated in Figure 11.8. London forces decrease the more compact the molecule is compared to a more chainlike molecule with the same number and types of atom as illustrated in Figure 11.9.
11.13 Physical properties that depend on tightness of packing: compressibility and diffusion. Physical properties that depend on the strengths of intermolecular interactions: retention of volume and shape, surface tension, wetting of a surface by a liquid, viscosity, evaporation, and sublimation.
11.14 The rate of diffusion should increase because the molecules move faster at the higher temperature.
11.15 The particles of a gas are free to move randomly, and thus to diffuse readily. Diffusion in liquids is comparatively slower because of the more numerous collisions that a molecule in a liquid sample must undergo in traveling from place to place. The particles of a solid are not free to move from place to place in a solid sample.
11.16 Water should have the greater surface tension because it has the stronger intermolecular forces, i.e., hydrogen bonding.
11.17 There is no intermolecular force common to both polyethylene and water that can allow for wetting. The surface tension of water, which is high, is not disrupted by any effective interaction between water and polyethylene.
11.18 Surface tension is related to the energy needed to decrease the surface area of a liquid. Molecules at the surface of a liquid have no other molecules above them, and they consequently are attracted only to those molecules that are next to them – namely, those in the interior of the liquid. This is illustrated in Figure 11.13.
11.19 Wetting – spreading a liquid across a surface to form a thin film.
Surfactant – a substance that lowers surface tension in a liquid and thereby promotes wetting.
11.20 Since it is the high energy molecules in a sample that are the first to evaporate, the remaining molecules have a lower average kinetic energy. A reduction in kinetic energy corresponds to a decrease in temperature.
11.21 Glycerol ought to wet the surface of glass quite nicely, because the dipolar bonds at the surface of glass can interact strongly with the polar O–H groups of glycerol.
11.22 The snow dissipates by sublimation. Even at low temperatures (by human standard) many compounds have measurable vapor pressures.
11.23 Raising the temperature of the sample increases the fraction of molecules in the sample that have enough kinetic energy to escape by evaporation.
11.24 An increase in surface area causes an increase in the rate of evaporation because, when the surface area is increased, there are more molecules in position at the surface of the liquid sample, where they are capable of evaporation. The stronger the intermolecular forces, the less readily a substance can evaporate.
11.25 (a) sublimation
(b) vaporization
(c) condensation
(d) melting
(e) freezing
11.26 This happens because of the loss in kinetic energy that the colliding molecule experiences when it hits the surface molecules. The colliding molecule has less kinetic energy after striking the surface, and its ability to escape subsequently from the liquid is momentarily diminished by the presence of intermolecular forces of attraction generated upon mixing.
235

Chapter 11
11.27 A dynamic equilibrium is achieved when a solid is held at its melting temperature. At this point, particles are melting and freezing at an equivalent rate. This is the melting or freezing point of a substance.
11.28 A dynamic equilibrium is established if the liquid evaporates into a sealed container. It is termed a dynamic equilibrium because opposing processes (evaporation and condensation) continue to take place, once the condition of equilibrium has been achieved. At equilibrium, the rate of condensation is equal to the rate of evaporation, and there is consequently no net change in the number of molecules in the vapor or in the liquid.
11.29 Yes. This is the sublimation process.
11.30 After the molecule is in the vapor phase for a while its kinetic energy will be less because it collided with other molecules in the vapor phase and transferred some of its energy to the other molecules. This molecule is not likely to bounce out of the surface of the liquid unless it regains some kinetic energy from the liquid.
11.31 Equilibrium vapor pressure is the pressure exerted by a vapor that is in equilibrium with its liquid. It is a dynamic equilibrium because events have not ceased. Liquid continues to evaporate once the state of equilibrium has been reached, but the rate of evaporation is equal to the rate of condensation. These two opposing processes occur at equal rates, such that there is no further change in the amount of either the liquid or the gas.
11.32 The equilibrium vapor pressure is governed only by the strength of the attractive forces within the liquid and by the temperature.
11.33 Changing the volume only upsets the equilibrium for a moment, provided the volume is not increased to a point that all the liquid evaporates at which no equilibrium would exist. After sufficient time has elapsed, the rates of evaporation and condensation again become equal to one another, and the same condition of equilibrium is achieved. The vapor pressure (or the ease of evaporation) only depends on the strength of intermolecular forces in the liquid sample.
11.34 Raising the temperature increases the vapor pressure by imparting enough kinetic energy (for evaporation) to more of the liquid molecules.
11.35 In humid air, the rate of condensation on the skin is more nearly equal to the rate of evaporation from the skin, and the net rate of evaporation of perspiration from the skin is low. The cooling effect of the evaporation of perspiration is low, and our bodies are cooled only slowly under such conditions. In dry air, however, perspiration evaporates more rapidly, and the cooling effect is high.
11.36 At the temperature of the cool glass, the equilibrium vapor pressure of the water is lower than the partial pressure of water in the air. The air in contact with the cool glass is induced to relinquish some of its water, and condensation occurs.
11.37 The boiling point is the temperature at which the liquid boils and the vapor pressure of the liquid is equal to the atmospheric pressure. The normal boiling point is the boiling point of a liquid when the atmospheric pressure is 760 mm Hg.
11.38 Even at higher temperatures, the contents of the radiator do not boil, because the pressure in the system increases with temperature, since the system is closed. The boiling point of the liquid is higher because the pressure is higher.
11.39 This happens because boiling is a process that is a function of pressure. Since the vapor pressure varies with temperature, the boiling point must also change as the pressure changes.
11.40 Inside the lighter, the liquid butane is in equilibrium with its vapor, which exerts a pressure somewhat above normal atmospheric pressure. This keeps the butane as a liquid.
236

Chapter 11
11.41 The hydrogen bond network in HF is less extensive than in water, because it is a monohydride not a dihydride, that is HF can only donate one hydrogen for a hydrogen bond while water can donate two hydrogens.
11.42 At about 77 °C
11.43 Ethanol vapor is present inside the bubbles of boiling ethanol.
11.44 Since
2
Se is larger than H
2
S, its London forces are stronger than in H
2
S. Because water is capable of hydrogen bonding, whereas H
2
S is not, its boiling point is higher than that of H
2
S.
11.45 (a)
(b)
1, 3, and 5
2 and 4
2
4
(e)
(f)
(g)
(h)
The heat of vaporization is larger.
This is the temperature of line 4.
This is the temperature of line 2.
Line 3 would descend lower in temperature than line 4, before rising to the temperature of line 4.
11.46 The substance with the larger molar heat of vaporization has the stronger intermolecular forces. This is ethanol, which has hydrogen bonding, whereas ethyl acetate does not.
11.47 The heat of condensation is exothermic, and it is equal in magnitude, but opposite in sign to the heat of vaporization (which is endothermic).
11.48 The heat of vaporization of a molecular substance is generally larger than the heat of fusion, because, in vaporization, the molecules undergo much larger changes in their distance of separation (and require the disruption of much stronger intermolecular forces) than is true of melting. The heat of sublimation is typically larger than the heat of vaporization of a liquid because sublimation involves a greater change in intermolecular separation, a larger disruption of intermolecular forces of attraction, and hence a larger change in potential energy.
11.49 CH
4
< CF
4
< HCl < HF
11.50 The moisture which powers the hurricane condenses when the hurricane travels over cold water. As the amount of vapor in the storm decreases, its energy decreases.
11.51 Steam releases a considerable amount of energy in the form of condensation energy as opposed to simply cooling liquid water.
11.52 When a system at equilibrium is modified so as to upset the equilibrium, the system will respond in a manner which enables the equilibrium to be reestablished. See also Section 11.8.
11.53 By "position of equilibrium" we mean the relative amounts of the various reactants and products that exist in the equilibrium mixture.
11.54 Changing the temperature disrupts the sublimation, deposition equilibrium that was established. By lowering the temperature, the kinetic energy of the molecules will decrease and more deposition will occur, giving rise to a lower vapor pressure. This will continue until a new equilibrium is established in which the rate of sublimation and deposition are again equal.
11.55 This is an endothermic system, and adding heat to the system will shift the position of the equilibrium to the right, producing a new equilibrium mixture having more liquid and less solid. Some of the solid melts when heat is added to the system.
237

Chapter 11
11.56 Crystalline solids have an ordered internal structure while amorphous solids do not have the long-range repetitive order of crystalline solids.
11.57 The entire crystal lattice of the solid can be generated by repeated use of the unit cell only.
11.58 A lattice is a set of points that have the same repeat distances and are arranged along lines oriented at the same angles. A unit cell is the smallest repeating unit of the lattice that can be used to define the lattice.
11.59 a) b)
11.60 No, it is not a unit cell of the substance. If the unit cell is move in one direction, to the right for example, then the smaller atom would be replace by a larger atom. This is not repeating unit.
11.61 These structures are both of the face–centered cubic variety. They differ only in the length of an edge for a unit cell, that is, the length of the cube edge is different in the two metals. Silver might be expected to have a face–centered unit cell, also.
11.62 Zinc sulfide has a face centered cubic lattice. Calcium fluoride also has a face centered cubic lattice.
11.63 See Figure 11.33 and 11.36.
11.64 n λ = 2d sin θ n = an integer (1, 2, 3, . . .)
λ = wavelength of the X–rays d = the interplane spacing in the crystal
θ = the angle of incidence and the angle of reflectance of X–rays to the various crystal planes.
11.65 Although there are only fourteen different kinds of lattice geometries that can fill space, there are essentially an infinite variety of cell dimensions that can be adopted by substances.
11.66 These are not 1:1 ionic substances. The cubic unit cell of NaCl contains the same number of sodium and chloride ions.
11.67 Covalent crystals are also termed network solids because they are constructed of atoms that are covalently bonded to one another, giving a giant interlocking network.
11.68 The lattice positions are occupied by metal cations, which are then surrounded by the core electrons of the metal. A sea of valence electrons encompasses the entire metallic solid.
11.69 (a) dipole–dipole, London forces, or hydrogen bonds electrostatic
(c)
11.70 Amorphous means, literally, without form. It is taken here to represent a solid that does not have the regular, repeating geometrical form normally associated with a crystal lattice.
238

Chapter 11
11.71 An amorphous solid is a noncrystalline solid. It is a solid that lacks the long–range order that characterizes a crystalline substance. When cooled, a liquid that will form an amorphous solid gradually becomes viscous and slowly hardens to give a glass, or a supercooled liquid. When crystalline solids are broken, the angles are regular and the faces are flat. When amorphous solids are broken, the faces are smooth and flat.
11.72 A supercritical fluid is a substance at a temperature above its critical temperature. Supercritical CO
2
is used to decaffeinate coffee because it replaces organic solvents such as methylene chloride and ethyl acetate which cannot be completely removed from the coffee bean. Supercritical CO
2 as 97% of the caffeine.
can remove as much
11.73 An increase in pressure should favor the system with the lower volume, i.e. the solid. Therefore, if the substance is at its melting point at a pressure of one atmosphere, and then if the pressure were to be increased, more solid would form at the expense of liquid – that is, more of the substance would freeze. If melting were to be accomplished at the higher pressure, it would require a temperature that is higher than the normal melting temperature. The phase diagram would be similar to the one for carbon dioxide, see
Fig. 11.48.
11.74 Critical temperature – the temperature above which the substance can not exist as a liquid, regardless of the applied pressure. It is, therefore, the temperature above which a gas cannot be made to liquefy, regardless of the amount of pressure that is applied.
Critical pressure – the vapor pressure of a liquid at the liquid's critical temperature.
A critical temperature and critical pressure together constitute a substance's critical point.
11.75 Carbon dioxide does not have a normal boiling point because its triple point lies above one atmosphere.
Thus, the liquid–vapor equilibrium that is taken to represent the boiling point does not exist at the pressure
(1 atm) conventionally used to designate the "normal" boiling point.
11.76 The critical temperature of hydrogen is below room temperature because, at room temperature, it cannot be liquefied by the application of pressure. The critical temperature of butane is above room temperature, because butane can be liquefied by the application of pressure.
11.77 Solid, liquid, and gas are all in equilibrium at the triple point.
Review Problems
11.78 London forces are possible in them all. Where another intermolecular force can operate, it is generally stronger than London forces, and this other type of interaction overshadows the importance of the London force. The substances in the list that can have dipole–dipole attractions are those with permanent dipole moments: (a), (b), and (d). SF
6
, (c), is a non–polar molecular substance. HF, (a), has hydrogen bonding.
11.79 (a) London forces, dipole-diploe, H-bonding
(b)
(c)
(d)
London forces, dipole-dipole
London forces
London forces, dipole-dipole, H-bonding
11.80 Diethyl ether has the faster rate of vaporization, since it does not have hydrogen bonds, as does butanol.
11.81 Diethyl ether should have a higher vapor pressure since it has weaker intermolecular forces. Butanol has a higher boiling point since it has stronger intermolecular forces of attraction.
11.82 Ethanol, because it has H-bonding.
239

Chapter 11
11.83 The London forces are stronger in CS
2
Consequently, CS
2
because the larger S atoms are more easily polarized than O atoms.
has a higher boiling point than CO
2.
11.84 Chloroform would be expected to display larger dipole-dipole attractions because it has a larger dipole moment than bromoform. (Chlorine has a higher electronegativity which results in each C–Cl bond having a larger dipole than each C–Br bond.) On the other hand, bromoform would be expected to show stronger
London forces due to having larger electron clouds which are more polarizable than those of chlorine.
Since bromoform in fact has a higher boiling point that chloroform, we must conclude that it experiences stronger intermolecular attractions than chloroform, which can only be due to London forces. Therefore,
London forces are more important in determining the boiling points of these two compounds.
11.85 NO
2
is capable of forming a liquid at atmospheric pressure while CO
2
does not, this suggests that NO
2
has stronger intermolecular attractions than CO
2
. Since CO
2
and NO
2
cannot form hydrogen bonds, the next strongest intermolecular interactions are dipole-dipole interactions; therefore, NO interactions. This would require NO
2
to have a dipole which is only possible if it is bent, while CO
2
11.86 ether < acetone < benzene < water < acetic acid
2
probably has dipole-dipole
is linear.
11.87 diethyl ether < ethanol < water < ethylene glycol
11.88 We can approach this problem by first asking either of two equivalent questions about the system: how much heat energy (q) is needed in order to melt the entire sample of solid water (105 g), or how much energy is lost when the liquid water (45.0 g) is cooled to the freezing point? Regardless, there is only one final temperature for the combined (150.0 g) sample, and we need to know if this temperature is at the melting point (0 °C, at which temperature some solid water remains in equilibrium with a certain amount of liquid water) or above the melting point (at which temperature all of the solid water will have melted).
Heat flow supposing that all of the solid water is melted: q = 6.01 kJ/mole × 105 g × 1 mol/18.0 g = 35.1 kJ
Heat flow on cooling the liquid water to the freezing point: q = 45.0 g × 4.18 J/g °C × 85 °C = 1.60 × 10 4 J = 16.0 kJ
The lesser of these two values is the correct one, and we conclude that 16.0 kJ of heat energy will be transferred from the liquid to the solid, and that the final temperature of the mixture will be 0 °C. The system will be an equilibrium mixture weighing 150 g and having some solid and some liquid in equilibrium with one another.
The amount of solid that must melt in order to decrease the temperature of 45.0 g of water from 85 °C to 0 °C is:
16.0 kJ ÷ 6.01 kJ/mol = 2.66 mol of solid water. 2.66 mol × 18.0 g/mol = 47.9 g of water must melt.
(a)
(b)
The final temperature will be 0 °C.
47.9 g of water must melt.
11.89 The amount of heat gained by melting all the benzene is: kJ = (10.0 g benzene)
⎛
⎝
1 mol benzene ⎞⎛ 9.92 kJ
78.11 g benzene ⎠ ⎝ 1 mol
⎞
⎠
= 1.27 kJ
Assuming all of this heat is removed from the water and using:
Heat = mass × specific heat × ∆ T
∆ T = heat
=
( )
( ) = 30.4 °C
Since the final temperature cannot be lower than the initial temperature, the final temperature of the water will be 5.45 °C and there will still be some solid benzene remaining.
240

Chapter 11
11.90 kJ = (125 g H
2
O)
⎛
⎜
⎝ 18.015 g H O
⎞⎛
⎟⎜
⎠⎝
43.9 kJ
1 mol H O
11.91 kJ = (5.00 g C
3
H
6
O)
⎛
⎜
⎝
⎞
⎟
⎠
=305 kJ
58.1 g C H O
⎞⎛
⎟⎜
⎠⎝
− 30.3 kJ
1 mol C H O
⎞
⎟
⎠
= –2.61 kJ
4 surrounding center sulfide:
8 corners × 1/8 S 2 −
6 faces × 1/2 S 2 −
= 4 Zn
per face = 3 S 2–
2 +
per corner = 1 S
Total = 4 S 2–
2–
11.93 A cube has six faces and eight corners. Each of the six face atoms is shared by two adjacent unit cells:
6 × 1/2 = 3 atoms.
The eight corner atoms are each shared by eight unit cells:
8 × 1/8 = 1 atom.
The total number atoms to be assigned to any one cell is thus 3 + 1 = 4.
(6 × 1/2 copper atoms) + (8 × 1/8 copper atoms) = 4 copper atoms
11.94 Using the Bragg equation (eqn. 11.2), n λ = 2d sin θ
(a)
(b) n(229pm) = 2(1,000)sin θ
0.1145n = sin θ
θ = 6.57° n(229pm) = 2(100)sin θ
0.458n = sin θ
θ = 27.3°
11.95 n λ = 2d sin θ
λ
2sin θ
At 20.0°
2 sin 20.0
°
At 27.4° d = 206 pm
×
×
2 sin 27.4
°
At 35.8° d = 153 pm
×
× ° d = 121 pm
11.96 From Figure 11.31, we can see that the length of the diagonal of the cell = 4r, where r = radius of the atom. According to the Pythagorean theorem, a 2 + b 2 = c 2
241

Chapter 11 for a right triangle. Since a = b here, we may re–write this as
2l 2 = c 2 , where l = length of the edge of the unit cell. As mentioned above, the diagonal of the unit cell = 4r, so we may say that
2l l 2
2 = (4r) 2
= (4r) 2 l 2 = 8r 2
/2 l 2 = 16r 2 /2 l = 8r
2
Finally, substituting the value provided for r in the problem, l =
( ) 2
= 3.51 Å. Using the conversion factor 1pm = 100 Å, this is 351 pm.
11.97 The following diagram is appropriate:
The face diagonal is 4 times the radius of the atom. The Pythagorean theorem is:
diagonal 2 = edge 2 + edge 2 .
Hence we have: [4(144 pm)] 2 = 2 × edge 2 . Solving for the edge length we get 407 pm.
11.98 According to the Pythagorean theorem, a 2 + b 2 = c 2 for a right triangle. First, we need to find the length of a diagonal on a face of the unit cell. Since a = b here, we may re-write this as
2l 2 = c 2 , where l = length of the edge of the unit cell and c = the diagonal length. Using the given 412.3 pm as the length of the edge, c = 583.1 pm. The diagonal length inside the cell from corner to opposite corner may now be found by the same theorem: a 2 + b 2 = c 2
(412.3) 2 + (583.1) 2 = c 2 c = 714.1 pm
This diagonal length inside the cell from corner to opposite corner is due to 1 Cs + ion and 1 Cl −
2r
Cs+
ion (see Figure 11.34). Therefore:
+ 2r = 714.1pm r
2r
Cs+
2r
Cs+
Cs+
Cl −
+ 2(181pm) = 714.1pm
= 352 pm
= 176 pm
11.99 (2 × r
Rb
) + (2 × r
Cl
) = 658 pm
(2 × r
Rb r
Rb
) + (2 × 181 pm) = 658 pm
= 148.0 pm
11.100 Each edge is composed of 2 × radius of the cation plus 2 × radius of the anion. The edge is therefore
2 × 133 + 2 × 195 = 656 pm.
11.101 2 × r
Na
+ d
Cl
= 564.0 pm
2 × 95 pm + d d
Cl
Cl
= 564.0 pm
= 374.0 pm
242

Chapter 11
11.102 This is a metallic solid.
11.103 This is a molecular solid.
(a)
(b)
11.105
(b) molecular (e) covalent
(c)
11.106 This must be a molecular solid, because if it were ionic it would be high–melting, and the melt would conduct.
11.107 This is a covalent network solid.
11.108 (a) solid (b) gas (c) liquid
11.109 The solid–liquid line slants toward the right.
11.110
(d) solid, liquid, and gas
1.0
0.30
-15.0 -10.0
90
T (
C)
11.111 Sublimation is possible only below a pressure of 0.30 atm, as marked on the phase diagram. The density of the solid is higher than that of the liquid. Notice that the line separating the solid from the liquid slopes to the right, in contrast to the diagram for water, Figure 11.46 of the text.
Additional Exercises
11.112 As the cooling takes place, the average kinetic energy of the gas molecules decreases, and the attractive forces that can operate among the gas molecules become able to bind the various molecules together in a process that leads to condensation. The air thus loses much of its moisture on the ascending side of the mountain. On descending the other side of the mountain, the air is compressed, and the temperature rises according to
Charles' Law. The relative humidity of the air is now very low for two reasons: much of the moisture was released on the other side of the mountain range, and now that the temperature is higher, there is far less than the maximum allowable water content. The coast of California receives the rain since the air releases its moisture to the west of the mountains and the valleys east of the mountains do not receive any rain.
11.113 The oxidation state of the chromium in CrO
3
is Cr 6+ charge, it pulls electron density from the anion, O 2–
and for Cr
2
, towards itself to a greater extent than the Cr 3+
Therefore, there will be more covalent character between the Cr
O
6+
3
, it is Cr
and O 2–
3+ . Since Cr 6+ has a higher
as compared to Cr 3+
.
and O 2– .
This will cause the melting point of CrO
3
to be lower than Cr
2
O
3
.
243

Chapter 11
11.114 Intermolecular (between different particles): The principal attractive forces are ion–ion forces and ion– dipole forces. These are of overwhelming strength compared to London forces, which do technically exist.
Intramolecular (within certain particles): The sulfite ion (SO
3
2 − ) is a polyatomic ion whose atoms are held together by covalent bonds . Although a covalent bond is not an “attraction” in itself, the attractive forces which make up these bonds are valence electron’s attractions to the nuclei of neighboring atoms in the polyatomic ion.
11.115 Yes, because of the possibility of a weak hydrogen bond between the carbonyl oxygen of acetone and an
OH group of water.
11.116 The pressure of the water is:
P
( )( )
= 13.16 torr = 13.16 mm Hg
The number of moles of water is calculated from the ideal gas equation:
PV = nRT
=
RT n =
(
13.16 torr
)
(
1 atm (
10.0 L
)
760 torr mol K
) ( )
= 7.20 × 10 –3 mol H
2
O
The number of grams of water is calculated from the molecular mass of water: g H
2
O = 7.20 × 10 –3 mol H
2
O
⎛
⎜
⎝
⎞
⎟
⎠
= 0.130 g H
2
O
11.117 a = edge of a unit cell
First determine the mass of 1 atom of silver in grams: mass of 1 atom of silver =
⎛
⎝
107.87 g Ag
1 mol Ag
⎞
⎟⎜
⎛
⎜
×
1 mol Ag
⎞
⎟ = 1.791 × 10 –22 g /atom Ag
Then determine the volume of each unit cell and the number of atoms in each unit cell.
(a) Simple Cubic Lattice:
2r = a volume of the unit cell = a
2.39 × 10 7 pm 3
⎛
⎜
3 = (2r)
1 cm
3 = (2 × 144 pm)
⎟
⎠
⎞
3
= 2.39 × 10
3
–23
= 2.39 × 10
cm 3
7 pm
/unit cell
3 number of atoms in the unit cell = 1
1 atom/unit cell
Density = mass/volume =
⎛
⎜
×
− 22
1 atom Ag
g
⎞
⎟
⎠
⎛
⎜
1 atom Ag
1 unit cell
⎞
⎟
⎛
⎝
1 unit cell
×
− 23
cm
3
(b) Body-Centered Cubic Lattice:
(4r) 2 = 3a 2
(4 × 144 pm) 2 = 3a 2 a = 333 pm
Volume of the unit cell = a
3.69 × 10 7 pm 3
⎛
⎜
3 = (333 pm)
1 cm
⎟
⎠
⎞
3
3 = (333 pm)
= 3.69 × 10 –23
3 = 3.69 × 10
cm 3 /unit cell
7 pm 3
⎞
⎠
= 7.49 g cm –3 number of atoms in the unit cell = 2
2 atom/unit cell
Density = mass/volume =
⎛
⎜
× − 22
1 atom Ag
g
⎞
⎠
⎛
⎜
2 atom Ag
1 unit cell
⎞
⎠
⎛
⎝
1 unit cell
3.69 10
− 23
cm
3
⎞
⎠
= 9.70 g cm –3
244

Chapter 11
(c) Face-Centered Cubic Lattice:
(4r) 2 = 2a 2
(4 × 144 pm) 2 a = 407 pm
= 2a 2
Volume of the unit cell = a
6.76 × 10 7 pm 3
⎛
⎜
3 = (407 pm)
1 cm
⎟
⎠
⎞
3
3 = (407 pm)
= 6.76× 10 –23
3 = 6.76 × 10
cm 3 /unit cell
7 pm 3 number of atoms in the unit cell = 4
4 atom/unit cell
Density = mass/volume =
⎛
⎜
×
− 22
1 atom Ag
g
⎞
⎟
⎠
⎛
⎜
4 atom Ag
1 unit cell
⎞
⎟
⎛
⎝
1 unit cell
×
− 23
cm
3 atom cm
3
⎞
⎠
= 3.275 10
− 22
g/atom
⎞
⎠
= 10.6 g cm –3
Silver has a face-centered cubic lattic.
11.118 Clouds form when the humid air of a warm front encounters the cool, relatively dry air of a cold front because the moisture in the air condenses.
11.119 We start by determining the volume of one unit cell:
407.86 pm = 407.86 × 10 –12
Vol = (407.86 × 10 –10 cm) 3
m = 407.86 × 10
= 6.7847 × 10 –23
–10
cm
cm
3
Next, we calculate the volume per atom, remembering that each unit cell contains a total of 4 gold atoms:
6.7847 × 10 –23 cm 3 / 4 atoms = 1.6962 × 10
This value is multiplied by the density:
–23 cm 3
⎛
⎜
× − 23
cm
3
19.31 g
Finally, it is necessary to divide this value into the atomic mass of gold:
196.97 g/mol
×
×
− 22 g/atom
11.120 Using the Bragg equation (eqn. 11.2), n λ = 2d sin θ
154= 2dsin(12.8°)
77 = d(0.232) d = 347 pm
The mass of the unit cell would be:
(4 potassium ions × 39.10) + (4 chloride ions × 35.45) = 298.20 amu and its volume would be:
= 4.18 × 10 7 pm 3
Therefore, its density would be: d = m/V = 7.14 × 10 –6 amu/pm 3
11.121 Using Hess’s Law, sublimation may be considered equivalent to melting followed by vaporization.
∆ H sublimation
= ∆ H fusion
+ ∆ H vaporization
= 10.8 kJ/mol + 24.3 kJ/mol = 35.1 kJ
245

Chapter 11
11.122 Radius = 0.5diameter = 0.50 pm = r
Simple Cubic:
Volume of cube: (2r) 3 = (2 × 0.50 pm) 3 = 1 pm 3
Volume of atoms: 1 atom in unit cell
Volume of atoms =
4
π r 3 =
4
π (0.5 pm) 3 = 0.524 pm 3
3 3
Volume of empty space = Volume of cube – Volume of atoms = 1 pm 3 – 0.524 pm 3 = 0.476 pm 3
Body-Centered Cubic:
Volume of cube: (2r)
(4r) 2
3
= 3(edge)
(4 × 0.50 pm) 2 a = 1.15 pm
2
= 3(a) 2 edge = a
Volume of cube = a 3 = (1.15 pm)
Volume of atoms: 2 atoms in unit cell
3 = 1.52 pm 3
Volume of atoms = 2(
4
π r 3 ) = 2(
4
π (0.5 pm) 3 ) = 1.05 pm 3
3 3
Volume of empty space = Volume of cube – Volume of atoms
= 1.52 pm 3 – 1.05 pm 3 = 0.47 pm 3
Face-Centered Cubic:
Volume of cube: (2r) 3
(4r) 2 = 2(edge)
(4 × 0.50 pm) 2
2
= 2(a) 2 edge = a a = 1.41 pm
Volume of cube = a 3 = (1.41 pm)
Volume of atoms: 4 atoms in unit cell
3 = 2.83 pm 3
Volume of atoms = 4(
4
π r 3 ) = 4(
4
π (0.5 pm) 3 ) = 2.09 pm 3
3 3
Volume of empty space = Volume of cube – Volume of atoms
= 2.83 pm 3 – 2.09 pm 3 = 0.74 pm 3
The efficiency of packing is determined by dividing the volume occupied by the atoms divided by the volume occupied by the cube:
Body-Centered Cubic 1.05 pm 3
Face-Centered Cubic 2.09 pm 3
÷
÷ 1 pm 3 = 0.524
1.52 pm
÷ 2.83 pm 3
3 = 0.691
= 0.739
11.123 From Figure 11.31, we can see that the length of the diagonal of the cell = 4r, where r = radius of the atom. According to the Pythagorean theorem, a 2 + b 2 = c 2 for a right triangle. Since a = b here, we may re–write this as
2l 2 = c 2 , where l = length of the edge of the unit cell. As mentioned above, the diagonal of the unit cell = 4r, so we may say that
2l 2 = (4r) 2
2l 2 = 16r
(0.125)l 2
2
= r 2
( ) 2
= r
Finally, substituting the value provided for l in the problem, r =
( )( ) 2
= 144.20 pm.
246