London dispersion (LD) < dipole-dipole < H bonding < metallic bonding, covalent network, ionic.

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1 CAPTER TEN LIQUIDS AND SOLIDS For Review. Intermolecular forces are the relatively weak forces between molecules that hold the molecules together in the solid and liquid phases. Intramolecular forces are the forces within a molecule. These are the covalent bonds in a molecule. Intramolecular forces (covalent bonds) are much stronger than intermolecular forces. Dipole forces are the forces that act between polar molecules. The electrostatic attraction between the positive end of one polar molecule and the negative end of another is the dipole force. Dipole forces are generally weaker than hydrogen bonding. Both of these forces are due to dipole moments in molecules. ydrogen bonding is given a separate name from dipole forces because hydrogen bonding is a particularly strong dipole force. Any neutral molecule that has a hydrogen covalently bonded to N, O, or F exhibits the relatively strong hydrogen bonding intermolecular forces. London dispersion forces are accidental-induced dipole forces. Like dipole forces, London dispersion forces are electrostatic in nature. Dipole forces are the electrostatic forces between molecules having a permanent dipole. London dispersion forces are the electrostatic forces between molecules having an accidental or induced dipole. All covalent molecules (polar and nonpolar) have London dispersion forces, but only polar molecules (those with permanent dipoles) exhibit dipole forces. As the size of a molecule increases, the strength of the London dispersion forces increases. This is because, as the electron cloud about a molecule gets larger, it is easier for the electrons to be drawn away from the nucleus. The molecule is said to be more polarizable. London dispersion (LD) < dipole-dipole < bonding < metallic bonding, covalent network, ionic. Yes, there is considerable overlap. Consider some of the examples in Exercise 0.9. Benzene (only LD forces) has a higher boiling point than acetone (dipole-dipole forces). Also, there is even more overlap among the stronger forces (metallic, covalent, and ionic). 2. a. Surface tension: the resistance of a liquid to an increase in its surface area. b. Viscosity: the resistance of a liquid to flow. c. Melting point: the temperature (at constant pressure) where a solid converts entirely to a liquid as long as heat is applied. A more detailed definition is the temperature at which the solid and liquid states have the same vapor pressure under conditions where the total pressure is constant.

2 2 CAPTER 0 LIQUIDS AND SOLIDS d. Boiling point: the temperature (at constant pressure) where a liquid converts entirely to a gas as long as heat is applied. The detailed definition is the temperature at which the vapor pressure of the liquid is exactly equal to the external pressure. e. Vapor pressure: the pressure of the vapor over a liquid at equilibrium. As the strengths of intermolecular forces increase, surface tension, viscosity, melting point and boiling point increase, while vapor pressure decreases.. Solid: rigid; has fixed volume and shape; slightly compressible Liquid: definite volume but no specific shape; assumes shape of the container; slightly compressible Gas: no fixed volume or shape; easily compressible 4. a. Crystalline solid: Regular, repeating structure Amorphous solid: Irregular arrangement of atoms or molecules b. Ionic solid: Made up of ions held together by ionic bonding Molecular solid: Made up of discrete covalently bonded molecules held together in the solid phase by weaker forces (LD, dipole, or hydrogen bonds). c. Molecular solid: Discrete, individual molecules Network solid: No discrete molecules; A network solid is one large molecule. The forces holding the atoms together are the covalent bonds between atoms. d. Metallic solid: Completely delocalized electrons, conductor of electricity (cations in a sea of electrons) Network solid: Localized electrons; Insulator or semiconductor 5. Lattice: a three-dimensional system of points designating the positions of the centers of the components of a solid (atoms, ions, or molecules) Unit cell: the smallest repeating unit of a lattice A simple cubic unit cell has an atom, ion or molecule located at the eight corners of a cube. There is one net atom per simple cubic unit cell. Because the atoms in the cubic unit cell are assumed to touch along the cube edge, cube edge = = 2r where r = radius of the atom. A body-centered cubic unit cell has an atom, ion or molecule at the eight corners of a cube and one atom, ion, or molecule located at the center of the cube. There are two net atoms per body-centered cubic unit cell. Because the atoms in the cubic unit cell are assumed to touch along the body diagonal of the cube, body diagonal = = 4r where = cube edge and r = radius of atom. A face-centered cubic unit cell has an atom, ion, or molecule at the eight

3 CAPTER 0 LIQUIDS AND SOLIDS corners of a cube and an atom, ion, or molecule located at the six faces of the cube. There are four net atoms per face-centered unit cell. Because the atoms in the cubic unit cell are assumed to touch along the face diagonal of the cube, face diagonal = 2 = 4r. 6. Closest packing: the packing of atoms (uniform, hard spheres) in a manner that most efficiently uses the available space with the least amount of empty space. The two types of closest packing are hexagonal closest packing and cubic closest packing. In both closest packed arrangements, the atoms (spheres) are packed in layers. The difference between the two closest packed arrangements is the ordering of the layers. exagonal closest packing has the third layer directly over the first layer forming a repeating layer pattern of abab In cubic closest packing the layer pattern is abcabc The unit cell for hexagonal closest packing is a hexagonal prism. See Fig. 0.4 for an illustration of the hexagonal prism unit cell. The unit cell for cubic closest packing is the face-centered cubic unit cell. 7. Conductor: The energy difference between the filled and unfilled molecular orbitals is minimal. We call this energy difference the band gap. Because the band gap is minimal, electrons can easily move into the conduction bands (the unfilled molecular orbitals). Insulator: Large band gap; Electrons do not move from the filled molecular orbitals to the conduction bands since the energy difference is large. Semiconductor: Small band gap; The energy difference between the filled and unfilled molecular orbitals is smaller than in insulators, so some electrons can jump into the conduction bands. The band gap, however, is not as small as with conductors, so semiconductors have intermediate conductivity. a. As the temperature is increased, more electrons in the filled molecular orbitals have sufficient kinetic energy to jump into the conduction bands (the unfilled molecular orbitals). b. A photon of light is absorbed by an electron which then has sufficient energy to jump into the conduction bands. c. An impurity either adds electrons at an energy near that of the conduction bands (n-type) or creates holes (unfilled energy levels) at energies in the previously filled molecular orbitals (p-type). Both n-type and p-type semiconductors increase conductivity by creating an easier path for electrons to jump from filled to unfilled energy levels. In conductors, electrical conductivity is inversely proportional to temperature. Increases in temperature increase the motions of the atoms, which gives rise to increased resistance (decreased conductivity). In a semiconductor, electrical conductivity is directly proportional to temperature. An increase in temperature provides more electrons with enough kinetic energy to jump from the filled molecular orbitals to the conduction bands, increasing conductivity. To produce an n-type semiconductor, dope Ge with a substance that has more than 4 valence electrons, e.g., a group 5A element. Phosphorus or arsenic are two substances which will produce n-type semiconductors when they are doped into germanium. To produce a p-type semiconductor, dope Ge with a substance that has fewer than 4 valence electrons, e.g., a

4 4 CAPTER 0 LIQUIDS AND SOLIDS group A element. Gallium or indium are two substances which will produce p-type semiconductors when they are doped into germanium.. The structures of most binary ionic solids can be explained by the closest packing of spheres. Typically, the larger ions, usually the anions, are packed in one of the closest packing arrangements, and the smaller cations fit into holes among the closest packed anions. There are different types of holes within the closest packed anions which are determined by the number of spheres that form them. Which of the three types of holes are filled usually depends on the relative size of the cation to the anion. Ionic solids will always try to maximize electrostatic attractions among oppositely charged ions and minimize the repulsions among ions with like charges. The structure of sodium chloride can be described in terms of a cubic closest packed array of Cl ions with Na + ions in all of the octahedral holes. An octahedral hole is formed between 6 Cl anions. The number of octrahedral holes is the same as the number of packed ions. So in the face-centered unit cell of sodium chloride, there are 4 net Cl ions and 4 net octahedral holes. Because the stoichiometry dictates a : ratio between the number of Cl anions and Na + cations, all of the octahedral holes must be filled with Na + ions. In zinc sulfide, the sulfide anions also occupy the lattice points of a cubic closest packing arrangement. But instead of having the cations in octahedral holes, the Zn 2+ cations occupy tetrahedral holes. A tetrahedral hole is the empty space created when four spheres are packed together. There are twice as many tetrahedral holes as packed anions in the closest packed structure. Therefore, each face-centered unit cell of sulfide anions contains 4 net S 2 ions and net tetrahedral holes. For the : stoichiometry to work out, only one-half of the tetrahedral holes are filled with Zn 2+ ions. This gives 4 S 2 ions and 4 Zn 2+ ions per unit cell for an empirical formula of ZnS. 9. a. Evaporation: process where liquid molecules escape the liquid s surface to form a gas. b. Condensation: process where gas molecules hit the surface of a liquid and convert to a liquid. c. Sublimation: process where a solid converts directly to a gas without passing through the liquid state. d. Boiling: the temperature and pressure at which a liquid completely converts to a gas as long as heat is applied. e. Melting: temperature and pressure at which a solid completely converts to a liquid as long as heat is applied. f. Enthalpy of vaporization ( vap ): the enthalpy change that occurs at the boiling point when a liquid converts into a gas. g. Enthalpy of fusion ( fus ): the enthalpy change that occurs at the melting point when a solid converts into a liquid.

5 CAPTER 0 LIQUIDS AND SOLIDS 5 h. eating curve: a plot of temperature versus time as heat is applied at a constant rate to some substance. Fusion refers to a solid converting to a liquid, and vaporization refers to a liquid converting to a gas. Only a fraction of the hydrogen bonds in ice are broken in going from the solid phase to the liquid phase. Most of the hydrogen bonds in water are still present in the liquid phase and must be broken during the liquid to gas phase transition. Thus, the enthalpy of vaporization is much larger than the enthalpy of fusion because more intermolecular forces are broken during the vaporization process. A volatile liquid is one that evaporates relatively easily. Volatile liquids have large vapor pressures because the intermolecular forces that prevent evaporation are relatively weak. 0. See Fig and 0.52 for the phase diagrams of 2 O and CO 2. Most substances exhibit only three different phases: solid, liquid, and gas. This is true for 2 O and CO 2. Also typical of phase diagrams is the positive slopes for both the liquid-gas equilibrium line and the solidgas equilibrium line. This is also true for both 2 O and CO 2. The solid-liquid equilibrium line also generally has a positive slope. This is true for CO 2, but not for 2 O. In the 2 O phase diagram, the slope of the solid-liquid line is negative. The determining factor for the slope of the solid-liquid line is the relative densities of the solid and liquid phases. The solid phase is denser than the liquid phase in most substances; for these substances, the slope of the solidliquid equilibrium line is positive. For water, the liquid phase is denser than the solid phase which corresponds to a negative sloping solid-liquid equilibrium line. Another difference between 2 O and CO 2 is the normal melting points and normal boiling points. The term normal just dictates a pressure of atm. 2 O has a normal melting point (0C) and a normal boiling point (00C), but CO 2 does not. At atm pressure, CO 2 only sublimes (goes from the solid phase directly to the gas phase). There are no temperatures at atm for CO 2 where the solid and liquid phases are in equilibrium or where the liquid and gas phases are in equilibrium. There are other differences, but those discussed above are the major ones. The relationship between melting points and pressure is determined by the slope of the solidliquid equilibrium line. For most substances (CO 2 included), the positive slope of the solidliquid line shows a direct relationship between the melting point and pressure. As pressure increases, the melting point increases. Water is just the opposite since the slope of the solidliquid line in water is negative. ere the melting point of water is inversely related to the pressure. For boiling points, the positive slope of the liquid-gas equilibrium line indicates a direct relationship between the boiling point and pressure. This direct relationship is true for all substances including 2 O and CO 2. The critical temperature for a substance is defined as the temperature above which the vapor cannot be liquefied no matter what pressure is applied. The critical temperature, like the boiling point temperature, is directly related to the strength of the intermolecular forces. Since 2 O exhibits relatively strong hydrogen bonding interactions and CO 2 only exhibits London dispersion forces, one would expect a higher critical temperature for 2 O than for CO 2.

6 6 CAPTER 0 LIQUIDS AND SOLIDS Questions 2. C has the stronger intermolecular forces because it has the higher boiling point. Even though C is nonpolar, it is so large that its London dispersion forces are much stronger than the sum of the London dispersion and hydrogen bonding interactions found in 2 O.. Atoms have an approximately spherical shape (on the average). It is impossible to pack spheres together without some empty space among the spheres. 4. Critical temperature: The temperature above which a liquid cannot exist, i.e., the gas cannot be liquified by increased pressure. Critical pressure: The pressure that must be applied to a substance at its critical temperature to produce a liquid. The kinetic energy distribution changes as one raises the temperature (T 4 > T c > T > T 2 > T ). At the critical temperature, T c, all molecules have kinetic energies greater than the intermolecular forces, F, and a liquid can't form. Note: The distributions above are not to scale. 5. Evaporation takes place when some molecules at the surface of a liquid have enough energy to break the intermolecular forces holding them in the liquid phase. When a liquid evaporates, the molecules that escape have high kinetic energies. The average kinetic energy of the remaining molecules is lower, thus, the temperature of the liquid is lower. 6. A crystalline solid will have the simpler diffraction pattern because a regular, repeating arrangement is necessary to produce planes of atoms that will diffract the X-rays in regular patterns. An amorphous solid does not have a regular repeating arrangement and will produce a complicated diffraction pattern. 7. An alloy is a substance that contains a mixture of elements and has metallic properties. In a substitutional alloy, some of the host metal atoms are replaced by other metal atoms of similar size, e.g., brass, pewter, plumber s solder. An interstitial alloy is formed when some of the interstices (holes) in the closest packed metal structure are occupied by smaller atoms, e.g., carbon steels.. Equilibrium: There is no change in composition; the vapor pressure is constant. Dynamic: Two processes, vapor liquid and liquid vapor, are both occurring but with equal rates so the composition of the vapor is constant.

7 CAPTER 0 LIQUIDS AND SOLIDS 7 9. a. As the strength of the intermolecular forces increase, the rate of evaporation decreases. b. As temperature increases, the rate of evaporation increases. c. As surface area increases, the rate of evaporation increases. 20. C 2 5 O(l) C 2 5 O(g) is an endothermic process. eat is absorbed when liquid ethanol vaporizes; the internal heat from the body provides this heat which results in the cooling of the body. 2. Sublimation will occur allowing water to escape as 2 O(g). 22. The phase change, 2 O(g) 2 O(l), releases heat that can cause additional damage. Also steam can be at a temperature greater than 00 C. 2. The strength of intermolecular forces determines relative boiling points. The types of intermolecular forces for covalent compounds are London dispersion forces, dipole forces, and hydrogen bonding. Because the three compounds are assumed to have similar molar mass and shape, the strength of the London dispersion forces will be about equal between the three compounds. One of the compounds will be nonpolar so it only has London dispersion forces. The other two compounds will be polar so they have additional dipole forces and will boil at a higher temperature than the nonpolar compound. One of the polar compounds will have an covalently bonded to either N, O, or F. This gives rise to the strongest type of covalent intermolecular forces, hydrogen bonding. The compound which hydrogen bonds will have the highest boiling point while the polar compound with no hydrogen bonding will boil at a temperature in the middle of the other compounds. 24. a. Both forms of carbon are network solids. In diamond, each carbon atom is surrounded by a tetrahedral arrangement of other carbon atoms to form a huge molecule. Each carbon atom is covalently bonded to four other carbon atoms. The structure of graphite is based on layers of carbon atoms arranged in fused sixmembered rings. Each carbon atom in a particular layer of graphite is surrounded by three other carbons in a trigonal planar arrangement. This requires sp 2 hybridization. Each carbon has an unhybridized p atomic orbital; all of these p orbitals in each sixmembered ring overlap with each other to form a delocalized electron system. b. Silica is a network solid having an empirical formula of SiO 2. The silicon atoms are singly bonded to four oxygens. Each silicon atom is at the center of a tetrahedral arrangement of oxygen atoms which are shared with other silicon atoms. The structure of silica is based on a network of SiO 4 tetrahedra with shared oxygen atoms, rather than discrete SiO 2 molecules. Silicates closely resemble silica. The structure is based on interconnected SiO 4 tetrahedra. owever, in contrast to silica, where the O/Si ratio is 2:, silicates have O/Si ratios greater than 2: and contain silicon-oxygen anions. To form a neutral solid silicate, metal cations are needed to balance the charge. In other words, silicates are salts containing metal cations and polyatomic silicon-oxygen anions.

8 CAPTER 0 LIQUIDS AND SOLIDS When silica is heated above its melting point and cooled rapidly, an amorphous (disordered) solid called glass results. Glass more closely resembles a very viscous solution than it does a crystalline solid. To affect the properties of glass, several different additives are thrown into the mixture. Some of these additives are Na 2 CO, B 2 O, and K 2 O, with each compound serving a specific purpose relating to the properties of the glass. 25. a. Both CO 2 and 2 O are molecular solids. Both have an ordered array of the individual molecules, with the molecular units occupying the lattice points. A difference within each solid lattice is the strength of the intermolecular forces. CO 2 is nonpolar and only exhibits London dispersion forces. 2 O exhibits the relatively strong hydrogen bonding interactions. The differences in strength is evidenced by the solid phase changes that occur at atm. CO 2 (s) sublimes at a relatively low temperature of 7C. In sublimation, all of the intermolecular forces are broken. owever, 2 O(s) doesn t have a phase change until 0C, and in this phase change from ice to water, only a fraction of the intermolecular forces are broken. The higher temperature and the fact that only a portion of the intermolecular forces are broken are attributed to the strength of the intermolecular forces in 2 O(s) as compared to CO 2 (s). Related to the intermolecular forces are the relative densities of the solid and liquid phases for these two compounds. CO 2 (s) is denser than CO 2 (l) while 2 O(s) is less dense than 2 O(l). For CO 2 (s), the molecules pack together as close as possible, hence solids are usually more dense than the liquid phase. For 2 O, each molecule has two lone pairs and two bonded hydrogen atoms. Because of the equal number of lone pairs and O bonds, each 2 O molecule can form two hydrogen bonding interactions to other 2 O molecules. To keep this symmetric arrangement (which maximizes the hydrogen bonding interactions), the 2 O(s) molecules occupy positions that create empty space in the lattice. This translates into a smaller density for 2 O(s) as compared to 2 O(l). b. Both NaCl and CsCl are ionic compounds with the anions at the lattice points of the unit cells and the cations occupying the empty spaces created by anions (called holes). In NaCl, the Cl anions occupy the lattice points of a face-centered unit cell with the Na + cations occupying the octahedral holes. Octahedral holes are the empty spaces created by six Cl ions. CsCl has the Cl ions at the lattice points of a simple cubic unit cell with the Cs + cations occupying the middle of the cube. 26. Because silicon carbide is made from Group 4A elements and because it is extremely hard, one would expect SiC to form a covalent network structure similar to diamond. 27. The mathematical equation that relates the vapor pressure of a substance to temperature is: Δ vap ln P vap = + C R T y m x + b As shown above, this equation is in the form of the straight line equation. If one plots ln P vap vs. /T, the slope of the straight line is vap /R. Because vap is always positive, the slope of the straight line will be negative.

9 CAPTER 0 LIQUIDS AND SOLIDS 9 2. The typical phase diagram for a substance shows three phases and has a positive sloping solid-liquid equilibrium line (water is atypical). A sketch of the phase diagram for I 2 would look like this: P 90 torr s l g Exercises Statements a and e are true. For statement a, the liquid phase is always more dense than the gaseous phase (gases are mostly empty space). For statement e, because the triple point is at 90 torr, the liquid phase cannot exist at any pressure less than 90 torr, no matter what the temperature. For statements b, c, and d, examine the phase diagram to prove to yourself that they are false. Intermolecular Forces and Physical Properties 29. Ionic compounds have ionic forces. Covalent compounds all have London Dispersion (LD) forces, while polar covalent compounds have dipole forces and/or hydrogen bonding forces. For bonding forces, the covalent compound must have either a N, O or F bond in the molecule. a. LD only b. dipole, LD c. bonding, LD d. ionic e. LD only (C 4 in a nonpolar covalent compound.) f. dipole, LD g. ionic 0. a. ionic b. LD mostly; C F bonds are polar, but polymers like teflon are so large the LD forces are the predominant intermolecular forces. c. LD d. dipole, LD e. bonding, LD f. dipole, LD g. LD 5 o C T. a. OCS; OCS is polar and has dipole-dipole forces in addition to London dispersion (LD) forces. All polar molecules have dipole forces. CO 2 is nonpolar and only has LD forces. To predict polarity, draw the Lewis structure and deduce whether the individual bond dipoles cancel.

10 40 CAPTER 0 LIQUIDS AND SOLIDS b. SeO 2 ; Both SeO 2 and SO 2 are polar compounds, so they both have dipole forces as well as LD forces. owever, SeO 2 is a larger molecule, so it would have stronger LD forces. c. 2 NC 2 C 2 N 2 ; More extensive hydrogen bonding is possible. d. 2 CO; 2 CO is polar while C C is nonpolar. 2 CO has dipole forces in addition to LD forces. e. C O; C O can form relatively strong bonding interactions, unlike 2 CO. 2. Ar exists as individual atoms which are held together in the condensed phases by London dispersion forces. The molecule which will have a boiling point closest to Ar will be a nonpolar substance with about the same molar mass as Ar (9.95 g/mol); this same size nonpolar substance will have about equivalent strength of London dispersion forces. Of the choices, only Cl 2 (70.90 g/mol) and F 2 (.00 g/mol) are nonpolar. Because F 2 has a molar mass closest to that of Ar, one would expect the boiling point of F 2 to be close to that of Ar.. a. Neopentane is more compact than n-pentane. There is less surface area contact among neopentane molecules. This leads to weaker LD forces and a lower boiling point. b. F is capable of bonding; Cl is not. c. LiCl is ionic, and Cl is a molecular solid with only dipole forces and LD forces. Ionic forces are much stronger than the forces for molecular solids. d. n-exane is a larger molecule, so it has stronger LD forces. 4. Ethanol, C 2 6 O, has 2(4) + 6() + 6 = 20 valence electrons. C C O Exhibits -bonding and London dispersion forces. Dimethyl ether, C 2 6 O, also has 20 valence electrons. It has a Lewis structure of: C O C Exhibits dipole and London dispersion forces, but no hydrogen bonding since it has no covalently bonded to the O.

11 CAPTER 0 LIQUIDS AND SOLIDS 4 Propane, C 6, has (4) + 6() = valence electrons. C C C Propane only has relatively nonpolar bonds so it is nonpolar. Propane exhibits only London dispersion forces. The three compounds have similar molar mass so the strength of the London dispersion forces will be approximately equivalent. Because dimethyl ether has additional dipole forces, it will boil at a higher temperature than propane. The compound with the highest boiling point is ethanol since it exhibits relatively strong hydrogen bonding forces. The correct matching of boiling points is: ethanol, 7.5C; dimethyl ether, 2C; propane, 42.C 5. Boiling points and freezing points are assumed directly related to the strength of the intermolecular forces, while vapor pressure is inversely related to the strength of the intermolecular forces. a. Br; Br is polar, while Kr and Cl 2 are nonpolar. Br has dipole forces unlike Kr and Cl 2. b. NaCl; Ionic forces are much stronger than molecular forces. c. I 2 ; All are nonpolar, so the largest molecule (I 2 ) will have the strongest LD forces and the lowest vapor pressure. d. N 2 ; Nonpolar and smallest, so has the weakest intermolecular forces. e. C 4 ; Smallest, nonpolar molecule so has the weakest LD forces. f. F; F can form relatively strong bonding interactions unlike the others. g. C C 2 C 2 O; bonding, unlike the others, so has strongest intermolecular forces. 6. a. CBr 4 ; Largest of these nonpolar molecules so has strongest LD forces. b. F 2 ; Ionic forces in LiF are much stronger than the covalent forces in F 2 and Cl. Cl has dipole forces that the nonpolar F 2 does not exhibit; so F 2 has the weakest intermolecular forces and the lowest freezing point. c. C C 2 O; Can form bonding interactions unlike the others. d. 2 O 2 ; O-O structure produces stronger bonding interactions than F, so has greatest viscosity. e. 2 CO; 2 CO is polar so has dipole forces, unlike the other nonpolar covalent compounds.

12 42 CAPTER 0 LIQUIDS AND SOLIDS f. I 2 ; I 2 has only LD forces while CsBr and CaO have much stronger ionic forces. I 2 has weakest intermolecular forces so has smallest Δ fusion. Properties of Liquids 7. The attraction of 2 O for glass is stronger than the 2 O 2 O attraction. The miniscus is concave to increase the area of contact between glass and 2 O. The g g attraction is greater than the g glass attraction. The miniscus is convex to minimize the g glass contact.. A molecule at the surface of a waterdrop is subject to attractions only by molecules below it and to each side. The effect of this uneven pull on the surface molecules tends to draw them into the body of the liquid and causes the droplet to assume the shape that has the minimum surface area, a sphere. 9. The structure of 2 O 2 is O O, which produces greater hydrogen bonding than water. Long chains of hydrogen bonded 2 O 2 molecules then get tangled together. 40. CO 2 is a gas at room temperature. As mp and bp increase, the strength of the intermolecular forces also increases. Therefore, the strength of forces is CO 2 < CS 2 < CSe 2. From a structural standpoint this is expected. All three are linear, nonpolar molecules. Thus, only London dispersion forces are present. Since the molecules increase in size from CO 2 < CS 2 < CSe 2, the strength of the intermolecular forces will increase in the same order. Structures and Properties of Solids 4. nλ = 2d sin θ, d = nλ 2sin θ = 54pm 0 = pm =. 0 m 2 sin d = nλ 2sin θ = 2 54pm 2 sin = 40 pm = m 4. = 2dsin θ = n 44. nλ = 2d sin θ, d = m sin 5.0 nλ = 2sin θ o = m = = 4.9 = m = 49 pm 2 sin 5.55 sin θ = n λ = 2d = 0.56, θ = A cubic closest packed structure has a face-centered cubic unit cell. In a face-centered cubic unit, there are: corners / atom / 2 atom 6 faces = 4 atoms corner face

13 CAPTER 0 LIQUIDS AND SOLIDS 4 The atoms in a face-centered cubic unit cell touch along the face diagonal of the cubic unit cell. Using the Pythagorean formula where l = length of the face diagonal and r = radius of the atom: 4r l l 2 + l 2 = (4r) 2 2 l 2 = 6 r 2 l = r l l = r = m = 5.57 Volume of a unit cell = l = ( m = cm 0 cm) 22 =.7 0 cm Mass of a unit cell = 4 Ca atoms molca atoms 40.0g Ca molca = g Ca 22 mass g density = =.54 g/cm 22 volume.70 cm 46. There are 4 Ni atoms in each unit cell: For a unit cell: mol Ni 5.69g Ni 4 Ni atoms mass density = = 6.4 g/cm atoms mol Ni = volume l Solving: l =.5 0 cm = cube edge length For a face centered cube: 4r l (4r) 2 = l 2 + l 2 = 2 l 2 r = l, r = l / r =.5 0 cm / r =.6 0 cm = 6 pm l

14 44 CAPTER 0 LIQUIDS AND SOLIDS 47. The unit cell for cubic closest packing is the face-centered unit cell. The volume of a unit cell is: V = l 0 = (492 0 cm) 22 =.9 0 cm There are 4 Pb atoms in the unit cell, as is the case for all face-centered cubic unit cells. The mass of atoms in a unit cell is: mass = 4 Pb atoms molpb atoms g Pb 2 =. 0 g molpb 2 mass.0 g density = =.6 g/cm 22 volume.90 cm From Exercise 45, the relationship between the cube edge length, l, and the radius of an atom in a face-centered unit cell is: l = r. r = l = 492pm = 74 pm = m 4. A face-centered cubic unit cell contains 4 atoms. For a unit cell: mass of X = volume density = ( cm) 0.5 g/cm 22 = 7. 0 g mol X = 4 atoms X molx 24 = mol X atoms Molar mass = g X = 0 g/mol; The metal is silver (Ag). molx / Ti 49. For a body-centered unit cell: corners + Ti at body center = 2 Ti atoms corner All body-centered unit cells have 2 atoms per unit cell. For a unit cell: molti 47.g Ti 2 atomsti density = 4.50 g/cm atoms molti = l Solving: l = edge length of unit cell =.2 0 cm = 2 pm Assume Ti atoms just touch along the body diagonal of the cube, so body diagonal = 4 radius of atoms = 4r.

15 CAPTER 0 LIQUIDS AND SOLIDS 45 The triangle we need to solve is: l 4r -.2 x 0 cm l 2 - (.2 x 0 cm) 2 (4r) 2 = (.2 0 cm) 2 + [(.2 0 cm) 2 ] 2, r =.42 0 cm = 42 pm For a body-centered unit cell (bcc), the radius of the atom is related to the cube edge length by: 4r = l or l = 4r/. 50. From Exercise 0.49: 4r l 2 l 6 r 2 = l l 2 l = 4r/ = 2.09 r l = 2.09 (222 pm) = 5 pm = 5. 0 cm In a bcc, there are 2 atoms/unit cell. For a unit cell: molba 7. g Ba 2 atomsba 2 mass atoms molba.g density = volume (5.0 cm) cm 5. If a face-centered cubic structure, then 4 atoms/unit cell and from Exercise 0.45: 2 l 2 = 6 r 2 4r l l = r = (44 pm) = 407 pm l = m = cm l molau 97.0 g Au 4 atomsau atoms molau density = (4.070 cm) = 9.4 g/cm

16 46 CAPTER 0 LIQUIDS AND SOLIDS If a body-centered cubic structure, then 2 atoms/unit cell and from Exercise 0.49: 4r l 2 l 6 r 2 = l l 2 l = 4r/ = pm =. l =. 0 cm 0 0 m molau 97.0 g Au 2 atomsau atoms molau density = (.0 cm) = 7.7 g/cm The measured density is consistent with a face-centered cubic unit cell. 52. If face-centered cubic: l = r = (7 pm) = 7 pm =.7 0 cm mol.9 g W 4 atomsw atoms mol density = (.70 cm) If body-centered cubic: = 2. g/cm l = 4r = 4 7pm = 6 pm =.6 0 cm mol.9 g W 2 atomsw atoms mol density = = 9.4 g/cm (.60 cm) The measured density is consistent with a body-centered unit cell. 5. In a face-centered unit cell (ccp structure), the atoms touch along the face diagonal: 4r l (4r) 2 = l 2 + l 2 l = r V cube = l = ( r ) = 22.6 r There are four atoms in a face-centered cubic cell (see Exercise 0.45). Each atom has a volume of 4/ πr.

17 CAPTER 0 LIQUIDS AND SOLIDS 47 V atoms = 4 4 πr = 6.76 r Vatoms 6.76r So, = or 74.06% of the volume of each unit cell is occupied by Vcube 22.6r atoms. In a simple cubic unit cell, the atoms touch along the cube edge (l): 2(radius) = 2r = l 2 r l V cube = l = (2r) = r There is one atom per simple cubic cell ( corner atoms / atom per corner = atom/unit cell). Each atom has an assumed volume of 4/ πr = volume of a sphere. 4 V atom = πr = 4.9 r Vatom 4.9r So, = = or 52.6% of the volume of each unit cell is occupied by Vcube r atoms. A cubic closest packed structure packs the atoms much more efficiently than a simple cubic structure. 54. From Exercise 0.49, a body-centered unit cell contains 2 net atoms, and the length of a cube edge (l) is related to the radius of the atom (r) by the equation l = 4r/. Volume of unit cell = l = (4 r/ ) = 2.2 r Volume of atoms in unit cell = 2 4 π r =.7 r Vatoms.7r So: = = 6.00% occupied V 2.2r cube To determine the radius of the Fe atoms, we need to determine the cube edge length (l). molfe 55.5g Fe Volume of unit cell = 2 Fe atoms atoms molfe Volume = l = 2.6 l = 4r/, r = l / 4 = cm, l = cm 0 cm / 4 = cm cm 7.6g = cm

18 4 CAPTER 0 LIQUIDS AND SOLIDS 55. Doping silicon with phosphorus produces an n-type semiconductor. The phosphorus adds electrons at energies near the conduction band of silicon. Electrons do not need as much energy to move from filled to unfilled energy levels so conduction increases. Doping silicon with gallium produces a p-type semiconductor. Because gallium has fewer valence electrons than silicon, holes (unfilled energy levels) at energies in the previously filled molecular orbitals are created, which induces greater electron movement (greater conductivity). 56. A rectifier is a device that produces a current that flows in one direction from an alternating current which flows in both directions. In a p-n junction, a p-type and an n-type semiconductor are connected. The natural flow of electrons in a p-n junction is for the excess electrons in the n-type semiconductor to move to the empty energy levels (holes) of the p- type semiconductor. Only when an external electric potential is connected so that electrons flow in this natural direction will the current flow easily (forward bias). If the external electric potential is connected in reverse of the natural flow of electrons, no current flows through the system (reverse bias). A p-n junction only transmits a current under forward bias, thus converting the alternating current to direct current. 57. In has fewer valence electrons than Se, thus, Se doped with In would be a p-type semiconductor. 5. To make a p-type semiconductor we need to dope the material with atoms that have fewer valence electrons. The average number of valence electrons is four when mixtures of group A and group 5A elements are considered. We could dope with more of the Group A element or with atoms of Zn or Cd. Cadmium is the most common impurity used to produce p-type GaAs semiconductors. To make an n-type GaAs semiconductor, dope with an excess group 5A element or dope with a Group 6A element such as sulfur. 59. E gap = 2.5 ev J/eV = J; We want E gap = E light, so: hc λ = E ( J s)( J m/s) = m = nm 60. E = hc λ 4 (6.60 J s)(2.990 m/s) = m 9 0 J = energy of band gap / Cl 6. a. corners corner + 6 faces / 2 Cl = 4 Cl ions face 2 edges / 4 Na edge + Na at body center = 4 Na ions; NaCl is the formula. / Cl b. Cs ion at body center; corners = Cl ion; CsCl is the formula. corner c. There are 4 Zn ions inside the cube. / S corners corner / 2 S + 6 faces face = 4 S ions; ZnS is the formula.

19 CAPTER 0 LIQUIDS AND SOLIDS 49 / Ti d. corners + Ti at body center = 2 Ti ions corner 4 faces / 2 O + 2 O inside cube = 4 O ions; TiO 2 is the formula. face / Ni 62. Both As ions are inside the unit cell. corners corner + 4 edges / 4 Ni = 2 Ni ions edge The unit cell contains 2 ions of Ni and 2 ions of As which gives a formula of NiAs. 6. There is one octahedral hole per closest packed anion in a closest packed structure. If half of the octahedral holes are filled, there is a 2: ratio of fluoride ions to cobalt ions in the crystal. The formula is CoF There are 2 tetrahedral holes per closest packed anion. Let f = fraction of tetrahedral holes filled by the cations. 2 2f Na 2 O: cation to anion ratio = = 2f CdS: cation to anion ratio = = 2f ZrI 4 : cation to anion ratio = = 4, f = ; All of the tetrahedral holes are filled by Na + cations., f = ; /2 of the tetrahedral holes are filled by 2 Cd 2+ cations., f = ; / of the tetrahedral holes are filled by Zr 4+ cations. 65. In a cubic closest packed array of anions, there are twice the number of tetrahedral holes as anions present and an equal number of octahedral holes as anions present. A cubic closest packed array of sulfide ions will have 4 S 2 ions, tetrahedral holes, and 4 octahedral holes. In this structure we have /() = Zn 2+ ion and /2(4) = 2 Al + ions present along with the 4 S 2 ions. The formula is ZnAl 2 S The two-dimension unit cell is: Assuming the anions A are the larger circles, there are 4 anions completely in the unit cell. The corner cations (smaller circles) are shared by 4 different unit cells. Therefore, there is cation in the middle of the unit cell plus /4(4) = net cation from the corners. Each unit cell has 2 cations and 4 anions. The empirical formula is MA 2.

20 50 CAPTER 0 LIQUIDS AND SOLIDS 67. F ions at corners / F /corner = F ion per unit cell; Because there is one cubic hole per cubic unit cell, there is a 2: ratio of F ions to metal ions in the crystal if only ½ of the body centers are filled with the metal ions. The formula is MF 2 where M 2+ is the metal ion. 6. Mn ions at corners: (/) = Mn ion; F ions at 2 edges: 2(/4) = F ions Formula is MnF. Assuming fluoride is - charged, the charge on Mn is From Fig. 0.7, MgO has the NaCl structure containing 4 Mg 2+ ions and 4 O 2 ions per facecentered unit cell. molmgo 4 MgO formula units atoms Volume of unit cell = g MgO 40.g MgO molmgo cm.5g Volume of unit cell = l, l = cube edge length; l = (7.4 = 7.4 = 2.67 For a face-centered unit cell, the O 2 ions touch along the face diagonal: cm 2 l 4r 2, r 2 =.49 O O cm 22 0 g MgO 2 0 cm ) / = cm 0 cm The cube edge length goes through two radii of the O 2 anions and the diameter of the Mg 2+ cation. So: l = 2 r 2, r 2 O Mg 0 cm = 2(.49 0 cm) + 2, 2 r 2 = 6.5 r Mg Mg 9 0 cm 70. CsCl is a simple cubic array of Cl ions with Cs + in the middle of each unit cell. There is one Cs + and one Cl ion in each unit cell. Cs + and Cl ions touch along the body diagonal. body diagonal = r 2r l, l = length of cube edge 2 Cs Cl In each unit cell: molcscl 6.4 g CsCl mass = CsCl formula unit formula units molcscl volume = l = l = g CsCl cm.97g CsCl = cm 2 0 cm, l = 4. 0 cm = 4 pm = length of cube edge r 2r l (4pm) = 75 pm 2 Cs Cl = g

21 CAPTER 0 LIQUIDS AND SOLIDS 5 The distance between ion centers = r = 75 pm/2 = 5 pm Cs r Cl From ionic radii: r = 69 pm and r = pm; r r = 69 + = 50. pm Cs Cl Cs Cl The actual distance is pm (2.%) greater than that calculated from values of ionic radii. 7. a. CO 2 : molecular b. SiO 2 : network c. Si: atomic, network d. C 4 : molecular e. Ru: atomic, metallic f. I 2 : molecular g. KBr: ionic h. 2 O: molecular i. NaO: ionic j. U: atomic, metallic k. CaCO : ionic l. P : molecular 72. a. diamond: atomic, network b. P : molecular c. 2 : molecular d. Mg: atomic, metallic e. KCl: ionic f. quartz: network g. N 4 NO : ionic h. SF 2 : molecular i. Ar: atomic, group A j. Cu: atomic, metallic k. C 6 2 O 6 : molecular 7. a. The unit cell consists of Ni at the cube corners and Ti at the body center, or Ti at the cube corners and Ni at the body center. b. / = atom from corners + atom at body center; Empirical formula = NiTi c. Both have a coordination number of (both are surrounded by atoms). / Xe / 4 F 74. corners + Xe inside cell = 2 Xe; edges + 2 F inside cell = 4 F corner edge Empirical formula is XeF 2. This is also the molecular formula. 75. Structure Structure 2 / Ca corners corner / Ti = Ca atom corners = Ti atom corner 6 faces / 2 O face / 4 O = O atoms 2 edges corner = O atoms Ti at body center. Formula = CaTiO Ca at body center. Formula = CaTiO In the extended lattice of both structures, each Ti atom is surrounded by six O atoms. 76. With a cubic closest packed array of oxygen ions, we have 4 O 2 ions per unit cell. We need to balance the total charge of the anions with a + charge from the Al + and Mg 2+ cations.

22 52 CAPTER 0 LIQUIDS AND SOLIDS The only combination of ions that gives a + charge is 2 Al + ions and Mg 2+ ion. The formula is Al 2 MgO 4. There are an equal number of octahedral holes as anions (4) in a cubic closest packed array, and twice the number of tetrahedral holes as anions in a cubic closest packed array. For the stoichiometry to work out, we need 2 Al + and Mg 2+ per unit cell. ence, one-half of the octahedral holes are filled with Al + ions and one-eighth of the tetrahedral holes are filled with Mg 2+ ions. 77. a. Y: Y in center; Ba: 2 Ba in center / Cu Cu: corners = Cu, edges corner / 4 O O: 20 edges edge Formula: YBa 2 Cu O 9 = 5 oxygen, faces / 4 Cu = 2 Cu, total = Cu atoms edge / 2 O = 4 oxygen, total = 9 O atoms face b. The structure of this superconductor material follows the second perovskite structure described in Exercise The YBa 2 Cu O 9 structure is three of these cubic perovskite unit cells stacked on top of each other. The oxygen atoms are in the same places, Cu takes the place of Ti, two of the calcium atoms are replaced by two barium atoms, and one Ca is replaced by Y. c. Y, Ba, and Cu are the same. Some oxygen atoms are missing. / 4 O 2 edges edge = O, faces Superconductor formula is YBa 2 Cu O 7. / 2 O = 4 O, total = 7 O atoms face 7. a. Structure (a): / Tl Ba: 2 Ba inside unit cell; Tl: corners = Tl corner Cu: 4 edges O: 6 faces Structure (b): / 4 Cu edge / 2 O face = Cu / 4 O + edges edge Tl and Ba are the same as in structure (a). Ca: Ca inside unit cell; Cu: edges O: 0 faces / 2 O face / 4 O + edges edge = 5 O; Formula = TlBa 2 CuO 5 / 4 Cu = 2 Cu edge = 7 O; Formula = TlBa 2 CaCu 2 O 7

23 CAPTER 0 LIQUIDS AND SOLIDS 5 Structure (c): Tl and Ba are the same, and two Ca are located inside the unit cell. Cu: 2 edges Formula: TlBa 2 Ca 2 Cu O 9 / 4 Cu = Cu; O: 4 faces edge / 2 O face / 4 O + edges edge = 9 O Structure (d): Following similar calculations, formula = TlBa 2 Ca Cu 4 O b. Structure (a) has one planar sheet of Cu and O atoms, and the number increases by one for each of the remaining structures. The order of superconductivity temperature from lowest to highest temperature is: (a) < (b) < (c) < (d). c. TlBa 2 CuO 5 : + 2(2) + x + 5(2) = 0, x = + Only Cu + is present in each formula unit. TlBa 2 CaCu 2 O 7 : + 2(2) (x) + 7(2) = 0, x = +5/2 Each formula unit contains Cu 2+ and Cu +. TlBa 2 Ca 2 Cu O 9 : + 2(2) + 2(2) + (x) + 9(2) = 0, x = +7/ Each formula unit contains 2 Cu 2+ and Cu +. TlBa 2 Ca Cu 4 O : + 2(2) + (2) + 4(x) + (2) = 0, x = +9/4 Each formula unit contains Cu 2+ and Cu +. d. This superconductor material achieves variable copper oxidation states by varying the numbers of Ca, Cu and O in each unit cell. The mixtures of copper oxidation states are discussed above. The superconductor material in Exercise 0.77 achieves variable copper oxidation states by omitting oxygen at various sites in the lattice. Phase Changes and Phase Diagrams 79. If we graph ln P vap vs /T, the slope of the resulting straight line will be Δ vap /R. P vap ln P vap T (Li) /T T (Mg) /T torr 0 02 K K 9 K K

24 54 CAPTER 0 LIQUIDS AND SOLIDS For Li: We get the slope by taking two points (x, y) that are on the line we draw. For a line: slope = Δy Δx y x 2 2 y x or we can determine the straight line equation using a calculator. The general straight line equation is y = mx + b where m = slope and b = y-intercept. The equation of the Li line is: ln P vap = (/T) +.6, slope = K Slope = Δ vap /R, Δ vap = slope R = K.45 J/Kmol Δ vap = J/mol = 5 kj/mol For Mg: The equation of the line is: ln P vap = (/T) +.7, slope = K Δ vap = -slope R = K.45 J/Kmol, Δ vap = J/mol = 9 kj/mol The bonding is stronger in Li since Δ vap is larger for Li. 0. We graph ln P vap vs /T. The slope of the line equals Δ vap /R.

25 CAPTER 0 LIQUIDS AND SOLIDS 55 T(K) 0 /T (K ) P vap (torr) ln P vap slope = ( ) K = 4600 K K = Δ R vap Δ vap, Δ vap =,000 J/mol = kj/mol.45j / K mol To determine the normal boiling point, we can use the following formula: P ln = P2 Δ vap = R T 2 T At the normal boiling point, the vapor pressure equals.00 atm or 760. torr. At 27 K, the vapor pressure is 4.4. torr (from data in the problem). 4.4,000J / mol ln =, J / K mol T2 27K.97 = (/T ) = /T 2 = 2.0 0, T 2 = 57 K = normal boiling point. At 00. C (7 K), the vapor pressure of 2 O is.00 atm = 760. torr. For water, Δ vap = 40.7 kj/mol. P ln P2 Δ R vap T2 T P 2 or ln P Δ R vap T T torr J / mol ln =, 760. torr.45j / K mol 7K T = 2 7K = T, 2 T 2 = , T 2 = T 2 = 62 K or 9 C P J / mol 2. ln =,.00.45J / K mol 7K 62K ln P 2 = 5.27, P 2 = e 5.27 = 94 atm

26 56 CAPTER 0 LIQUIDS AND SOLIDS P Δ Δ. vap 6torr vap ln, ln P2 R T2 T 2torr.45J / K mol K 5K Solving: vap =. 0 4 J/mol; For the normal boiling point, P =.00 atm = 760. torr torr ln = 2torr J / K mol J / mol K, T T = T = 50. K = 77C; The normal boiling point of CCl 4 is 77C. 4. P ln P2 Δ R vap T2 T P = 760. torr, T = 56.5 C = 29.7 K; P 2 = 60. torr, T 2 =? J / mol ln =, J / K mol T = T = 4. 0, T 2 =.02 0, T 2 = 24.5 K = 5. C T torr J / mol ln =, P 2.45J / K mol ln 60. ln P 2 =.05 ln P 2 = 5.40, P 2 = e 5.40 = 22 torr 5.

27 CAPTER 0 LIQUIDS AND SOLIDS X(g, 00. C) X(g, 75 C), ΔT = -25 C q = s gas m ΔT =.0 J 250. g (25 C) = 600 J = 6. kj g C X(g, 75 C) X(l, 75 C), q 2 = 250. g X(l, 75 C) X(l, -5 C), q = X(l, 5 C) X(s, 5 C), q 4 = 250. g X(s, 5 C) X(s, 50. C), q 5 = mol 20. kj = 67 kj 75.0 g mol 2.5 J 250. g (-90. C) = -56,000 J = 56 kj g C mol 5.0 kj = 7 kj 75.0 g mol.0 J 250. g (5 C) = 26,000 J = 26 kj g C q total = q + q 2 + q + q 4 + q 5 = = 72 kj 7. 2 O(s, 20. C) 2 O(s, 0 C), ΔT = 20. C 2.0J q = s ice m ΔT = g 20. C = J = 20. kj g C 2 O(s, 0 C) 2 O(l, 0 C), q 2 = g 2 O mol.02g 6.02kJ = 67 kj mol 4.2 J 2 O(l, 0 C) 2 O(l, 00. C), q = g 00. C = J = 20 kj g C 2 O(l, 00. C) 2 O(g, 00. C), q 4 = g 2 O(g, 00. C) 2 O(g, 250. C), q 5 = mol.02g 40.7 kj mol = 0 kj 2.0 J g 50. C = J g C q total = q + q 2 + q + q 4 + q 5 = = 60 kj = 50 kj. 2 O(g, 25 C) 2 O(g, 00. C), q = 2.0 J/g C 75.0 g (25 C) = -00 J =. kj 2 O(g, 00. C) 2 O(l, 00. C), q 2 = 75.0 g mol 40.7 kj = 69 kj.02g mol 2 O(l, 00. C) 2 O(l, 0 C), q = 4.2 J/g C 75.0 g (-00. C) = 2,000 J = 2 kj To convert 2 O(g) at 25 C to 2 O(l) at 0 C requires (. kj 69 kj 2 kj =) 205 kj of heat removed. To convert from 2 O(l) at 0 C to 2 O(s) at 0 C requires:

28 5 CAPTER 0 LIQUIDS AND SOLIDS mol 6.02kJ q 4 = 75.0 g = 25 kj.02g mol This amount of energy puts us over the -25 kj limit (-205 kj - 25 kj = -20. kj). Therefore, a mixture of 2 O(s) and 2 O(l) will be present at 0 C when 25 kj of heat are removed from the gas sample. 9. Total mass 2 O = cubes eat removed to produce ice at -5.0 C: 0.0 g = 540. g; 540. g 2 O cube mol2 O = 0.0 mol 2 O.02g 4.J 540. g 22.0 C + g C mol J 0.0 mol + 2.0J 540. g 5.0 C g C J J J = J J eat released = g Na g CF2 Cl =.49 0 g CF 2 Cl 2 must be vaporized. 5J mol 22.99g 6kJ 2 mol = 2.00 kj To melt 50.0 g of ice requires: 50.0 g ice mol2o 6.02kJ = 6.7 kj.02g mol The reaction doesn't release enough heat to melt all of the ice. The temperature will remain at 0 C. 9. A: solid B: liquid C: vapor D: solid + vapor E: solid + liquid + vapor F: liquid + vapor G: liquid + vapor : vapor triple point: E critical point: G normal freezing point: temperature at which solid - liquid line is at.0 atm (see plot below). normal boiling point: temperature at which liquid - vapor line is at.0 atm (see plot below)..0 atm nfp Since the solid-liquid line has a positive slope, the solid phase is denser than the liquid phase. nbp