15. Atoms have an approximately spherical shape (on average). It is impossible to pack spheres together without some empty space among the spheres.

Transcription

1 CAPTER 0 LIQUIDS AND SOLIDS Questions. Chalk is composed of the ionic compound calcium carbonate (CaCO ). The electrostatic forces in ionic compounds are much stronger than the intermolecular forces in covalent compounds. Therefore, CaCO should have a much higher boiling point than the covalent compounds found in motor oil and in O. Motor oil is composed of nonpolar C C and C! bonds. The intermolecular forces in motor oil are therefore London dispersion forces. We generally consider these forces to be weak. owever, with compounds that have large molar masses, these London dispersion forces add up significantly and can overtake the relatively strong hydrogen-bonding interactions in water.. In the vapor phase, gas molecules are relatively far apart from each other, so far apart that gas molecules are assumed to exhibit no intermolecular forces. 4. ydrogen bonding occurs when hydrogen atoms are covalently bonded to highly electronegative atoms such as oxygen, nitrogen, or fluorine. Because the electronegativity difference between hydrogen and these highly electronegative atoms is relatively large, the N, O, and F bonds are very polar covalent bonds. This leads to strong dipole forces. Also, the small size of the hydrogen atom allows the dipoles to approach each other more closely than can occur between most polar molecules. Both of these factors make hydrogen bonding a special type of dipole interaction. 5. Atoms have an approximately spherical shape (on average). It is impossible to pack spheres together without some empty space among the spheres. 6. 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. 6

2 CAPTER 0 LIQUIDS AND SOLIDS 7 The kinetic energy distribution changes as one raises the temperature (T 4 > T c > T > T > 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 various temperature distributions shown in the plot are not to scale. The area under each temperature distribution should be equal to each other (area total number of molecules). 7. 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.. 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. 9. 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. 0. 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.. 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.. O(l) O(g) 44 kj/mol; heat must be absorbed by water in order to evaporate. This heat comes from the surroundings. ence, as water evaporates, the surroundings (which includes earth) get cooler.. Sublimation will occur, allowing water to escape as O(g). 4. Water boils when the vapor pressure equals the external pressure. Because the external pressure is significantly lower at high altitudes, a lower temperature is required to equalize the vapor pressure of water to the external pressure. Thus food cooked in boiling water at high elevations cooks at a lower temperature, so it takes longer. 5. 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 among 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

3 CAPTER 0 LIQUIDS AND SOLIDS 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 that hydrogen bonds will have the highest boiling point, whereas the polar compound with no hydrogen bonding will boil at a temperature in the middle of the other compounds. 6. 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 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 π electron system. b. Silica is a network solid having an empirical formula of SiO. The silicon atoms are singly bonded to four oxygens. Each silicon atom is at the center of a tetrahedral arrangement of oxygen atoms that 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 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 :, silicates have O/Si ratios greater than : 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. 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 CO, B O, and K O, with each compound serving a specific purpose relating to the properties of glass. 7. a. Both CO and 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 is nonpolar and only exhibits London dispersion forces. O exhibits the relatively strong hydrogen-bonding interactions. The differences in strength is evidenced by the solid-phase changes that occur at atm. CO (s) sublimes at a relatively low temperature of!7ec. In sublimation, all of the intermolecular forces are broken. owever, O(s) doesn t have a phase change until 0EC, 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 O(s) as compared to CO (s). Related to the intermolecular forces are the relative densities of the solid and liquid phases for these two compounds. CO (s) is denser than CO (l), whereas O(s) is less dense than O(l). For CO (s) and for most solids, the molecules pack together as close as possible; hence solids are usually more dense than the liquid phase. O is an

4 CAPTER 0 LIQUIDS AND SOLIDS 9 exception to this. Water molecules are particularly well suited for hydrogen bonding interaction with each other because each molecule has two polar O bonds and two lone pairs on the oxygen. This can lead to the association of four hydrogen atoms with each oxygen atom: two by covalent bonds and two by dipoles. To keep this arrangement (which maximizes the hydrogen-bonding interactions), the O(s) molecules occupy positions that create empty space in the lattice. This translates into a smaller density for O(s) as compared to 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.. 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. 9. If TiO conducts electricity as a liquid, then it is an ionic solid; if not, then TiO is a network solid. 0. The interparticle forces in ionic solids (the ionic bonds) are much stronger than the interparticle forces in molecular solids (dipole forces, London forces, etc.). The difference in intermolecular forces is most clearly shown in the huge difference in melting points between ionic and molecular solids. Table salt and ordinary sugar are both crystalline solids at room temperature that look very similar to each other. owever, sugar can be melted easily in a saucepan during the making of candy, whereas the full heat of a stove will not melt salt. When a substance melts, some interparticle forces must be broken. Ionic solids (salt) require a much larger amount of energy to break the interparticle forces as compared to the relatively weak forces in molecular solids (sugar).. The mathematical equation that relates the vapor pressure of a substance to temperature is: 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 versus /T with temperature in Kelvin, the slope of the straight line is! vap /R. Because vap is always positive, the slope of the straight line will be negative.. 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 would look like this:

5 40 CAPTER 0 LIQUIDS AND SOLIDS P 90 torr s l g Exercises 5 o C T 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. Ionic compounds have ionic forces. Covalent compounds all have London dispersion (LD) forces, whereas polar covalent compounds have dipole forces and/or hydrogen bonding forces. For hydrogen-bonding (-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 is a nonpolar covalent compound.) f. dipole, LD g. ionic 4. a. ionic b. dipole, LD (LD London dispersion) c. LD only d. LD only (For all practical purposes, a C bond can be considered as a nonpolar bond.) e. ionic f. LD only g. -bonding, LD 5. a. OCS; OCS is polar and has dipole-dipole forces in addition to London dispersion (LD) forces. All polar molecules have dipole forces. CO is nonpolar and only has LD forces. To predict polarity, draw the Lewis structure and deduce whether the individual bond dipoles cancel. b. SeO ; both SeO and SO are polar compounds, so they both have dipole forces as well as LD forces. owever, SeO is a larger molecule, so it would have stronger LD forces. c. NC C N ; more extensive hydrogen bonding (-bonding) is possible because two N groups are present.

6 CAPTER 0 LIQUIDS AND SOLIDS 4 d. CO; CO is polar, whereas C C is nonpolar. CO has dipole forces in addition to LD forces. C C only has LD forces. e. C O; C O can form relatively strong -bonding interactions, unlike CO. 6. Ar exists as individual atoms that are held together in the condensed phases by London dispersion forces. The molecule that 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 (70.90 g/mol) and F (.00 g/mol) are nonpolar. Because F has a molar mass closest to that of Ar, one would expect the boiling point of F to be close to that of Ar. 7. 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.. Ethanol, C 6 O, has (4) + 6() valence electrons. C C O Exhibits -bonding and London dispersion forces. Dimethyl ether, C 6 O, also has 0 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. Propane, C, has (4) + () 0 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,

7 4 CAPTER 0 LIQUIDS AND SOLIDS 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, 7.5EC; dimethyl ether,!ec; propane,!4.ec 9. Boiling points and freezing points are assumed directly related to the strength of the intermolecular forces, whereas vapor pressure is inversely related to the strength of the intermolecular forces. a. Br; Br is polar, whereas Kr and Cl are nonpolar. Br has dipole forces unlike Kr and Cl. b. NaCl; ionic forces are much stronger than the intermolecular forces for molecular substances. c. I ; all are nonpolar, so the largest molecule (I ) will have the strongest LD forces and the lowest vapor pressure. d. N ; nonpolar and smallest, so it has the weakest intermolecular forces. e. C 4 ; smallest, nonpolar molecule, so it has the weakest LD forces. f. F; F can form relatively strong -bonding interactions, unlike the others. g. C C C O; -bonding, unlike the others, so it has strongest intermolecular forces. 40. a. CBr 4 ; largest of these nonpolar molecules, so it has strongest LD forces. b. F ; ionic forces in LiF are much stronger than the molecular forces in F and Cl. Cl has dipole forces, whereas the nonpolar F does not exhibit these. So F has the weakest intermolecular forces and the lowest freezing point. c. C C O; can form -bonding interactions, unlike the other covalent compounds. d. O ; O-O structure produces stronger -bonding interactions than F, so it has the greatest viscosity. e. CO; CO is polar, so it has dipole forces, unlike the other nonpolar covalent compounds. f. I ; I has only LD forces, whereas CsBr and CaO have much stronger ionic forces. I has the weakest intermolecular forces, so it has smallest fusion. Properties of Liquids 4. The attraction of O for glass is stronger than the O O attraction. The miniscus is concave to increase the area of contact between glass and O. The g g attraction is greater than the g glass attraction. The miniscus is convex to minimize the g glass contact.

8 CAPTER 0 LIQUIDS AND SOLIDS 4 4. Water is a polar substance, and wax is a nonpolar substance; they are not attracted to each other. A molecule at the surface of a drop of water is subject to attractions only by water molecules below it and to each side. The effect of this uneven pull on the surface water molecules tends to draw them into the body of the water and causes the droplet to assume the shape that has the minimum surface area, a sphere. 4. The structure of O is O O, which produces greater hydrogen bonding than in water. Thus the intermolecular forces are stronger in O than in O, resulting in a higher normal boiling point for O and a lower vapor pressure. 44. CO is a gas at room temperature. As melting point and boiling point increase, the strength of the intermolecular forces also increases. Therefore, the strength of forces is CO < CS < CSe. From a structural standpoint, this is expected. All three are linear, nonpolar molecules. Thus only London dispersion forces are present. Because the molecules increase in size from CO < CS < CSe, the strength of the intermolecular forces will increase in the same order. Structures and Properties of Solids 45. nλ d sin θ, d nλ sin θ 54 pm sin 4. pm. 0 0 m 46. d nλ sin θ 54 pm sin.0 40 pm m 47. λ dsin θ n m sin 5.0 o m D 4. nλ d sin θ, d nλ sin θ.6 D 4.9 D m 49 pm sin 5.55 sin θ n λ.6 D d 4.9D 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 corner / atom + 6 faces 4 atoms face 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:

9 44 CAPTER 0 LIQUIDS AND SOLIDS 4r l l + l (4r) l 6r l r l l r 97 0 m 5.57 Volume of a unit cell l ( m cm 0 cm).7 0 cm Mass of a unit cell 4 Ca atoms mol Ca atoms 40.0 g Ca mol Ca.66 0 g Ca Density mass volume.66 0 g.54 g/cm.7 0 cm 50. There are four Ni atoms in each unit cell. For a unit cell: mass density volume Solving: l g/cm 4 Ni atoms 0 cm cube edge length mol Ni atoms l 5.69 g Ni mol Ni 4r l For a face-centered cube: (4r) l + l l r l, r l / r.5 0 cm / r.6 0 cm 6 pm l 5. The unit cell for cubic closest packing is the face-centered unit cell. The volume of a unit cell is: V l 0 (49 0 cm).9 0 cm There are four 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:

10 CAPTER 0 LIQUIDS AND SOLIDS 45 mass 4 Pb atoms mol Pb atoms 07. g Pb mol Pb. 0 g Density mass volume. 0 g.6 g/cm.9 0 cm From Exercise 49, the relationship between the cube edge length l and the radius r of an atom in a face-centered unit cell is l r. r l 49 pm 74 pm m 5. The volume of a unit cell is: V l (. 0 0 cm) cm There are four Ir 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: mol Ir 9. g Ir mass 4 Ir atoms.77 0 g atoms mol Ir Density mass volume.77 0 g.6 g/cm cm 5. A face-centered cubic unit cell contains four atoms. For a unit cell: mass of X volume density ( cm) 0.5 g/cm 7. 0 g mol X 4 atoms X mol X atoms mol X Molar mass g X mol X 0 g/mol; the metal is silver (Ag). 54. For a face-centered unit cell, the radius r of an atom is related to the length of a cube edge l by the equation l r (see Exercise 49). Radius r l / 9 0 m / m.9 0 cm The volume of a unit cell is l, so the mass of the unknown metal (X) in a unit cell is: volume density (.9 0 cm).45 g X cm.9 0 g X

11 46 CAPTER 0 LIQUIDS AND SOLIDS Because each face-centered unit cell contains four atoms of X: mol X in unit cell 4 atoms X mol X atoms X mol X Therefore, each unit cell contains.9 0 g X, which is equal to mol X. The molar mass of X is: g X mol X 94 g/mol From the periodic table, the best choice for the metal is platinum. / Ti 55. For a body-centered unit cell, corners corner + Ti at body center Ti atoms. All body-centered unit cells have two atoms per unit cell. For a unit cell where l cube edge length: mol Ti 47. g Ti atoms Ti density 4.50 g/cm atoms mol Ti l Solving: l edge length of unit cell. 0 cm pm Assume Ti atoms just touch along the body diagonal of the cube, so body diagonal 4 radius of atoms 4r. The triangle we need to solve is: l 4r -. x 0 cm l - (. x 0 cm) (4r) (. 0 cm) + [(. 0 cm) ], r.4 0 cm 4 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/. 56. From Exercise 55:

12 CAPTER 0 LIQUIDS AND SOLIDS 47 4r l l 6r l + l l 4r/ (.09)r l.09( pm) 5 pm 5. 0 cm In a body-centered cubit unit cell, there are two atoms per unit cell. For a unit cell: density mass volume atoms Ba mol Ba (5. 0 atoms cm) 7. g Ba mol Ba. g cm 57. If gold has a face-centered cubic structure, then there are four atoms per unit cell, and from Exercise 49: l 6r 4r l l r (44 pm) 407 pm l m cm l mol Au 4 atoms Au atoms Density ( cm) 97.0 g Au mol Au 9.4 g/cm If gold has a body-centered cubic structure, then there are two atoms per unit cell, and from Exercise 55: 6r l + l 4r l l l 4r/ pm 0 m l 0 0 cm. 0 cm Density atoms Au mol Au atoms (. 0 cm) 97.0 g Au mol Au 7.7 g/cm The measured density of gold is consistent with a face-centered cubic unit cell.

13 4 CAPTER 0 LIQUIDS AND SOLIDS 5. If face-centered cubic: l r (7 pm) 7 pm.7 0 cm density 4 atomsw mol (.7 0 atoms cm).9 g W mol. g/cm If body-centered cubic: l 4r 4 7 pm 6 pm.6 0 cm density atoms W mol.9 g W atoms mol (.6 0 cm) 9.4 g/cm The measured density of tungsten is consistent with a body-centered unit cell. 59. In a face-centered unit cell (ccp structure), the atoms touch along the face diagonal: 4r l (4r) l + l l r V cube l (r ) (.6)r There are four atoms in a face-centered cubic cell (see Exercise 49). Each atom has a volume of (4/)πr volume of a sphere. V atoms 4 4 πr (6.76)r So V V atoms. atoms cube (6.76)r , or 74.06% of the volume of each unit cell is occupied by (.6)r In a simple cubic unit cell, the atoms touch along the cube edge l: (radius) r l r l V cube l (r) r

14 CAPTER 0 LIQUIDS AND SOLIDS 49 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 (4.9)r So 0.56, or 5.6% of the volume of each unit cell is occupied by V cube r atoms. A cubic closest packed structure (face-centered cubic unit cell) packs the atoms much more efficiently than a simple cubic structure. 60. From Exercise 55, a body-centered unit cell contains two 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 (4r/ ) (.)r Volume of atoms in unit cell 4 πr (.7)r So V V atoms cube (.7)r % occupied (.)r To determine the radius of the Fe atoms, we need to determine the cube edge length l. Volume of unit cell Fe atoms Volume l.6 mol Fe atoms 0 cm, l.7 0 cm 55.5 g Fe cm mol Fe 7.6 g.6 0 cm l 4r/, r l / cm / cm 6. 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). 6. A rectifier is a device that produces a current that flows in one direction from an alternating current that 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.

15 50 CAPTER 0 LIQUIDS AND SOLIDS 6. In has fewer valence electrons than Se. Thus Se doped with In would be a p-type semiconductor. 64. 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. 65. E gap.5 ev J/eV J; we want E gap E light hc/λ, so: hc λ E ( J s)( J m / s) m nm 66. E hc λ ( J s)( m m / s) J energy of band gap / Cl 67. Sodium chloride structure: corners corner + 6 faces / Cl face 4 Cl ions / 4 Na edges + Na at body center 4 Na ions; NaCl is the formula. edge / Cl Cesium chloride structure: Cs ion at body center; corners Cl ion; CsCl is the formula. corner Zinc sulfide structure: There are four Zn ions inside the cube. / S corners corner / S + 6 faces face 4 S ions; ZnS is the formula. Titanium oxide structure: / Ti corners corner + Ti at body center Ti ions 4 faces / O face + O inside cube 4 O ions; TiO is the formula. / Ni 6. Both As ions are inside the unit cell. corners corner + 4 edges / 4 Ni Ni ions edge The unit cell contains ions of Ni and ions of As, which gives a formula of NiAs. 69. There is one octahedral hole per closest packed anion in a closest packed structure. If onehalf of the octahedral holes are filled, then there is a : ratio of fluoride ions to cobalt ions in the crystal. The formula is CoF.

16 CAPTER 0 LIQUIDS AND SOLIDS There are two tetrahedral holes per closest packed anion. Let f fraction of tetrahedral holes filled by the cations. f Na O: Cation-to-anion ratio, f ; all the tetrahedral holes are filled by Na + cations. f CdS: Cation-to-anion ratio, f ; one-half the tetrahedral holes are filled by Cd + cations. f ZrI 4 : Cation-to-anion ratio, f ; one-eighth the tetrahedral holes are filled 4 by Zr 4+ cations. 7. 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 four S ions, eight tetrahedral holes, and four octahedral holes. In this structure we have /() Zn + ion and /(4) Al + ions present, along with the 4 S ions. The formula is ZnAl S A repeating pattern in the two-dimensional structure is: Assuming the anions A are the larger circles, there are four anions completely in this repeating square. The corner cations (smaller circles) are shared by four different repeating squares. Therefore, there is one cation in the middle of the square plus /4 (4) net cation from the corners. Each repeating square has two cations and four anions. The empirical formula is MA. 7. F ions at corners / F /corner F ion per unit cell; Because there is one cubic hole per cubic unit cell, there is a : ratio of F ions to metal ions in the crystal if only half of the body centers are filled with the metal ions. The formula is MF, where M + is the metal ion. 74. Mn ions at corners: (/) Mn ion; F ions at edges: (/4) F ions; the formula is MnF. Assuming fluoride is charged, then the charge on Mn is From Figure 0.7, MgO has the NaCl structure containing 4 Mg + ions and 4 O ions per face-centered unit cell. 4 MgO formula units mol MgO atoms 40.g MgO mol MgO.67 0 g MgO

17 5 CAPTER 0 LIQUIDS AND SOLIDS 76. Volume of unit cell.67 0 g MgO cm.5 g Volume of unit cell l, l cube edge length; l ( cm For a face-centered unit cell, the O ions touch along the face diagonal: 4. 0 cm l 4r, r.49 O O 4 0 cm ) / 4. 0 cm 0 cm The cube edge length goes through two radii of the O anions and the diameter of the Mg + cation, so: l r + +, 4. 0 cm (.49 0 cm) + r, r 6.5 O r Mg Mg + Mg cm Assuming K + and Cl just touch along the cube edge l: l (4 pm) 6 pm 6. 0 cm Volume of unit cell l The unit cell contains four K + and four Cl ions. For a unit cell: density 4 KCl formula units mol KCl formula units (6. 0 cm) g KCl mol KCl.00 g/cm 77. 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 + + r l, l length of cube edge Cs Cl In each unit cell: mol CsCl mass CsCl formula unit formula units volume l.796 l g CsCl cm.97 g CsCl g CsCl mol CsCl 0 cm 0 cm, l 4. 0 cm 4 pm length of cube edge g

18 CAPTER 0 LIQUIDS AND SOLIDS 5 r + + r l (4 pm) 75 pm Cs Cl The distance between ion centers r r 75 pm/ 5 pm Cs + + Cl From ionic radius: r + 69 pm and r pm; r + + Cs Cl Cs r Cl pm The distance calculated from the density is pm (.%) greater than that calculated from tables of ionic radii. 7. a. The NaCl unit cell has a face-centered cubic arrangement of the anions with cations in the octahedral holes. There are four NaCl formula units per unit cell, and since there is a : ratio of cations to anions in MnO, then there would be four MnO formula units per unit cell, assuming an NaCl-type structure. The CsCl unit cell has a simple cubic structure of anions with the cations in the cubic holes. There is one CsCl formula unit per unit cell, so there would be one MnO formula unit per unit cell if a CsCl structure is observed. Formula units of MnO Unit cell ( cm) 5. g MnO cm mol MnO g MnO formula units MnO 4.00 formula units MnO mol MnO From the calculation, MnO crystallizes in the NaCl type structure. b. From the NaCl structure and assuming the ions touch each other, then l cube edge length r. + + r Mn O l cm + (.40 0 cm), + r.5 0 cm 4 pm + r Mn 79. a. CO : molecular b. SiO : network c. Si: atomic, network d. C 4 : molecular e. Ru: atomic, metallic f. I : molecular g. KBr: ionic h. O: molecular i. NaO: ionic j. U: atomic, metallic k. CaCO : ionic l. P : molecular 0. a. diamond: atomic, network b. P : molecular c. : molecular d. Mg: atomic, metallic e. KCl: ionic f. quartz: network g. N 4 NO : ionic h. SF : molecular i. Ar: atomic, group A j. Cu: atomic, metallic k. C 6 O 6 : molecular. 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. 4 Mn

19 54 CAPTER 0 LIQUIDS AND SOLIDS b. / atom from corners + atom at body center; empirical formula NiTi c. Both have a coordination number of (both are surrounded by atoms). / Al. Al: corners corner Al; Ni: 6 face centers / Ni face center Ni The composition of the specific phase of the superalloy is AlNi.. Structure Structure / Ca / Ti corners Ca atom corners Ti atom corner corner 6 faces / O face / 4 O O atoms 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. 4. With a cubic closest packed array of oxygen ions, we have 4 O ions per unit cell. We need to balance the total! charge of the anions with a + charge from the Al + and Mg + cations. The only combination of ions that gives a + charge is Al + ions and Mg + ion. The formula is Al 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 Al + and Mg + 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 + ions. 5. a. Y: Y in center; Ba: Ba in center / Cu Cu: corners corner Cu, edges / 4 Cu edge Cu, total Cu atoms / 4 O O: 0 edges edge 5 oxygen, faces / O face 4 oxygen, total 9 O atoms Formula: YBa Cu O 9 b. The structure of this superconductor material follows the second perovskite structure described in Exercise. The YBa 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.

20 CAPTER 0 LIQUIDS AND SOLIDS 55 / 4 O edges edge O, faces / O face 4 O, total 7 O atoms Superconductor formula is YBa Cu O a. Structure (a): / Tl Ba: Ba inside unit cell; Tl: corners corner Tl Cu: 4 edges O: 6 faces Structure (b): / 4 Cu edge / O face Cu / 4 O + edges edge 5 O; Formula TlBa CuO 5. Tl and Ba are the same as in structure (a). / 4 Cu Ca: Ca inside unit cell; Cu: edges Cu edge O: 0 faces Structure (c): / O face / 4 O + edges edge 7 O; Formula TlBa CaCu O 7. Tl and Ba are the same, and two Ca are located inside the unit cell. Cu: edges / 4 Cu edge Cu; O: 4 faces / O face / 4 O + edges edge 9 O Formula TlBa Ca Cu O 9. Structure (d): Following similar calculations, formula TlBa 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 CuO 5 : + () + x + 5( ) 0, x + Only Cu + is present in each formula unit. TlBa CaCu O 7 : + () + + (x) + 7( ) 0, x +5/ Each formula unit contains Cu + and Cu +. TlBa Ca Cu O 9 : + () + () + (x) + 9( ) 0, x +7/ Each formula unit contains Cu + and Cu +. TlBa Ca Cu 4 O : + () + () + 4(x) + ( ) 0, x +9/4 Each formula unit contains Cu + and Cu +.

21 56 CAPTER 0 LIQUIDS AND SOLIDS 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 in part c. The superconductor material in Exercise 5 achieves variable copper oxidation states by omitting oxygen at various sites in the lattice. Phase Changes and Phase Diagrams 7. If we graph ln P vap versus /T with temperature in Kelvin, the slope of the resulting straight line will be vap /R. P vap ln P vap T (Li) /T T (Mg) /T torr 0 0 K K 9 K. 0 4 K For Li: We get the slope by taking two points (x, y) that are on the line we draw. For a line, slope 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 vap /R, vap slope R K.45 J/K mol vap J/mol 5 kj/mol

22 CAPTER 0 LIQUIDS AND SOLIDS 57 For Mg: The equation of the line is: ln P vap (/T) +.7, slope K vap slope R K.45 J/K mol vap J/mol 9 kj/mol The bonding is stronger in Li because vap is larger for Li.. We graph ln P vap vs /T. The slope of the line equals! vap /R. T(K) 0 /T (K ) P vap (torr) ln P vap Slope ( ) K!4600 K K R vap vap, vap,000 J/mol kj/mol.45 J/K mol To determine the normal boiling point, we can use the following formula: P vap ln P R T T At the normal boiling point, the vapor pressure equals.00 atm or 760. torr. At 7 K, the vapor pressure is 4.4. torr (from data in the problem). 4.4,000 J/mol ln, J/K mol T 7 K! (/T!.66 0 )! /T.0 0, T 57 K normal boiling point 9. At 00. C (7 K), the vapor pressure of O is.00 atm 760. torr. For water, vap 40.7 kj/mol. P ln P R vap T T or P ln P R vap T T

23 5 CAPTER 0 LIQUIDS AND SOLIDS 50. torr J/mol ln, 760. torr.45 J/K mol 7 K T! K! ! T, T.76 0, T.76 0 T 6 K or 9 C 90. At 00. C (7 K), the vapor pressure of O is.00 atm. For water, vap 40.7 kj/mol. P ln P R vap T T or P ln P R vap T T P ln.00 atm J/mol,.45 J/K mol 7 K K ln P 0.5, P e atm.50 ln J/mol.45 J/K mol 7 K T, K T T, T.4 0, T K or 40. C 9. P vap 6 torr vap ln, ln P R T T torr.45 J/K mol K 5 K Solving: vap. 0 4 J/mol; for the normal boiling point, P.00 atm 760. torr torr ln torr 4. 0 J/mol.45 J/K mol K T, T T 50. K 77EC; the normal boiling point of CCl 4 is 77EC. 9. P ln P R vap T T P 760. torr, T 56.5 C K; P 60. torr, T? J/mol ln, J/K mol T T , T.0 0, T 4.5 K 5. C T

24 CAPTER 0 LIQUIDS AND SOLIDS torr.0 0 J/mol ln, P.45 J/K mol ln 60.! ln P.05 ln P 5.40, P e 5.40 torr X(g, 00. C) X(g, 75 C), T -5 C.0 J q s gas m T 50. g (!5 C)!600 J 6. kj g C X(g, 75 C) X(l, 75 C), q 50. g mol 0. kj!67 kj 75.0 g mol.5 J X(l, 75 C) X(l, 5 C), q 50. g ( 90. C) 56,000 J 56 kj g C X(l, 5 C) X(s,!5 C), q g mol 5.0 kj!7 kj 75.0 g mol.0 J X(s, 5 C) X(s,!50. C), q g (!5 C)!6,000 J!6 kj g C q total q + q + q + q 4 + q kj 95. a. Many more intermolecular forces must be broken to convert a liquid to a gas as compared with converting a solid to a liquid. Because more intermolecular forces must be broken, much more energy is required to vaporize a liquid than is required to melt a solid. Therefore, vap is much larger than fus. b..00 g Na mol Na.60 kj 0. kj J to melt.00 g Na.99 g mol Na

25 60 CAPTER 0 LIQUIDS AND SOLIDS c..00 g Na mol Na.99 g 97.0 kj 4. kj 40 J to vaporize.00 g Na mol Na d. This is the reverse process of that described in part c, so the energy change is the same quantity but opposite in sign. Therefore, q 40 J; i.e., 40 of heat will be released. 96. Melt:.5 g C 6 6 mol C6 7. g kj.05 kj mol C 6 6 Vaporize:.5 g C 6 6 mol C6 7. g kj.4 kj mol C 6 6 As is typical, the energy required to vaporize a certain quantity of substance is much larger than the energy required to melt the same quantity of substance. A lot more intermolecular forces must be broken to vaporize a substance as compared to melting a substance. 97. To calculate q total, break up the heating process into five steps. O(s, 0. C) O(s, 0 C), T 0. C.0 J q s ice m T g 0. C J 0. kj o g C O(s, 0 C) O(l, 0 C), q mol 6.0 kj g O 67 kj.0 g mol O(l, 0 C) O(l, 00. C), q 4. J g 00. C. 0 5 J 0 kj o g C O(l, 00. C) O(g, 00. C), q g mol.0 g 40.7 kj 0 kj mol.0 J O(g, 00. C) O(g, 50. C), q g 50. C J o g C 50 kj q total q + q + q + q 4 + q kj 9. O(g, 5 C) O(g, 00. C), q.0 J/g C C 75.0 g (!5 C) 00 J!. kj O(g, 00. C) O(l, 00. C), q 75.0 g mol 40.7 kj 69 kj.0 g mol O(l, 00. C) O(l, 0 C), q 4. J/g C C 75.0 g ( 00. C),000 J kj To convert O(g) at 5 C to O(l) at 0 C requires (!. kj! 69 kj! kj )!05 kj of heat removed. To convert from O(l) at 0 C to O(s) at 0 C requires:

26 CAPTER 0 LIQUIDS AND SOLIDS 6 q g mol 6.0 kj!5. kj.0 g mol This amount of energy puts us over the!5 kj limit (!05 kj! 5. kj!0. kj). Therefore, a mixture of O(s) and O(l) will be present at 0 C when 5 kj of heat is removed from the gas sample. 99. Total mass O cubes 0.0 g cube eat removed to produce ice at -5.0 C: mol O 540. g; 540. g O 0.0 mol O.0 g. J g C 4 o 540. g.0 C mol J 0.0 mol.0 J o g 5.0 C g C J J J J J g CF Cl.49 0 g CF Cl must be vaporized. 5 J 00. eat released 0.50 g Na mol.99 g 6 kj.00 kj mol To melt 50.0 g of ice requires: 50.0 g ice mol O 6.0 kj 6.7 kj.0 g mol The reaction doesn't release enough heat to melt all of the ice. The temperature will remain at 0 C. 0. 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 following plot). Normal boiling point: Temperature at which liquid-vapor line is at.0 atm (see following plot )..0 atm nfp Because the solid-liquid line equilibrium has a positive slope, the solid phase is denser than the liquid phase. nbp

27 6 CAPTER 0 LIQUIDS AND SOLIDS 0. a. b. Triple point at 95. C: rhombic, monoclinic, gas Triple point at 5. C: monoclinic, liquid, gas Triple point at 5 C: rhombic, monoclinic, liquid c. From the phase diagram, the monoclinic solid phase is stable at T 00. C and P atm. d. Normal melting point 5. C; normal boiling point C; the normal melting and boiling points occur at P.0 atm. e. Rhombic is the densest phase because the rhombic-monoclinic equilibrium line has a positive slope, and because the solid-liquid equilibrium lines also have positive slopes. 5 f. No; P.0 0 atm is at a pressure somewhere between the 95. and 5. C triple points. At this pressure, the rhombic and gas phases are never in equilibrium with each 5 other, so rhombic sulfur cannot sublime at P.0 0 atm. owever, monoclinic sulfur can sublime at this pressure. 0. a. two g. From the phase diagram, we would start off with gaseous sulfur. At 00. C and ~ 0 5 atm, S(g) would convert to the solid monoclinic form of sulfur. Finally at 00. C and some large pressure less than 40 atm, S(s, monoclinic) would convert to the solid rhombic form of sulfur. Summarizing, the phase changes are S(g) S(monoclinic) S(rhombic). b. igher-pressure triple point: graphite, diamond and liquid; lower-pressure triple point: graphite, liquid and vapor c. It is converted to diamond (the more dense solid form). d. Diamond is more dense, which is why graphite can be converted to diamond by applying pressure. 04. The following sketch of the Br phase diagram is not to scale. Because the triple point of Br is at a temperature below the freezing point of Br, the slope of the solid-liquid line is positive. 00 Pressure (atm) 0.05 Solid Liquid Gas Temperature ( o C)

28 CAPTER 0 LIQUIDS AND SOLIDS 6 The positive slopes of all the lines indicate that Br (s) is more dense than Br (l), which is more dense than Br (g). At room temperature (~ C) and atm, Br (l) is the stable phase. Br (l) cannot exist at a temperature below the triple-point temperature of!7. C or at a temperature above the critical-point temperature of 0 C. The phase changes that occur as temperature is increased at 0.0 atm are solid liquid gas. 05. Because the density of the liquid phase is greater than the density of the solid phase, the slope of the solid-liquid boundary line is negative (as in O). With a negative slope, the melting points increase with a decrease in pressure, so the normal melting point of X should be greater than 5EC. 06. P (torr) 760 s l 0 g T( C) From the three points given, the slope of the solid-liquid boundary line is positive, so Xe(s) is more dense than Xe(l). Also, the positive slope of this line tells us that the melting point of Xe increases as pressure increases. The same direct relationship exists for the boiling point of Xe because the liquid-gas boundary line also has a positive slope. Connecting to Biochemistry 07. The structures of the two C 6 O compounds (0 valence e ) are:

29 64 CAPTER 0 LIQUIDS AND SOLIDS C C O exhibits relatively strong hydrogen bonding C O C does not exhibit hydrogen bonding The liquid will have the stronger intermolecular forces. Therefore, the first compound above (ethanol) is the liquid, whereas the second compound (dimethyl ether) with the weaker intermolecular forces is the gas. 0. Benzene Naphthalene LD forces only LD forces only Note: London dispersion forces in molecules such as benzene and naphthalene are fairly large. The molecules are flat, and there is efficient surface-area contact between molecules. Large surface-area contact leads to stronger London dispersion forces. Cl Carbon tetrachloride (CCl 4 ) has polar bonds but is a nonpolar molecule. CCl 4 only has LD forces. Cl C Cl Cl In terms of size and shape CCl 4 < C 6 6 < C 0. The strengths of the LD forces are proportional to size and are related to shape. Although the size of CCl 4 is fairly large, the overall spherical shape gives rise to relatively weak LD forces as compared to flat molecules such as benzene and naphthalene. The physical properties given in the problem are consistent with the order listed above. Each of the physical properties will increase with an increase in intermolecular forces.

30 CAPTER 0 LIQUIDS AND SOLIDS 65 Acetone Acetic acid O C C C O C C O LD, dipole LD, dipole, -bonding Benzoic acid O C O LD, dipole, -bonding We would predict the strength of interparticle forces of the last three molecules to be: acetone < acetic acid < benzoic acid polar -bonding -bonding, but large LD forces because of greater size and shape This ordering is consistent with the values given for boiling point, melting point, and vap. The overall order of the strengths of intermolecular forces based on physical properties are: acetone < CCl 4 < C 6 6 < acetic acid < naphthalene < benzoic acid The order seems reasonable except for acetone and naphthalene. Because acetone is polar, we would not expect it to boil at the lowest temperature. owever, in terms of size and shape, acetone is the smallest molecule, and the LD forces in acetone must be very small compared to the other molecules. Naphthalene must have very strong LD forces because of its size and flat shape. 09. Both molecules are capable of -bonding. owever, in oil of wintergreen the hydrogen bonding is intramolecular (within each molecule): O C O C O In methyl-4-hydroxybenzoate, the -bonding is intermolecular (between molecules), resulting in stronger forces between molecules and a higher melting point. 0. With the charged ends, this tripeptide exhibits ionic forces. Between the charged ends, there are nonpolar C groups, polar CO groups, and polar N groups. The nonpolar C groups will interact with C groups on other tripeptides through London dispersion forces. The polar CO group allows the tripeptides to interact with each other through dipole forces. And finally, the N bonds allow for hydrogen-bonding interactions with other tripeptides.

31 66 CAPTER 0 LIQUIDS AND SOLIDS. C 5 O(l) C 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.. O(l) O(g) 44 kj/mol; the vaporization of water is an endothermic process. In order to evaporate, water must absorb heat from the surroundings. In this case, part of the surroundings is our body. So, as water evaporates, our body supplies heat, and as a result, our body temperature can cool down. From Le Chatelier s principle, the less water vapor in the air, the more favorable the evaporation process. Thus the less humid the surroundings, the more favorably water converts into vapor, and the more heat that is lost by our bodies.. The phase change O(g) O(l) releases heat that can cause additional damage. Also, steam can be at a temperature greater than 00 C. 4. P vap 760 torr vap ln, ln P R T T 400. torr.45 J/K mol 9. K 07. K Solving: vap. 0 4 J/mol. kj/mol Additional Exercises 5. As the physical properties indicate, the intermolecular forces are slightly stronger in D O than in O. 6. C CO : -bonding + dipole forces + LD forces C ClCO : C CO C : -bonding + larger electronegative atom replacing (greater dipole) + LD forces Dipole forces (no -bonding) + LD forces From the intermolecular forces listed above, we predict C CO C to have the weakest intermolecular forces and C ClCO to have the strongest. The boiling points are consistent with this view. 7. At any temperature, the plot tells us that substance A has a higher vapor pressure than substance B, with substance C having the lowest vapor pressure. Therefore, the substance with the weakest intermolecular forces is A, and the substance with the strongest intermolecular forces is C. N can form hydrogen-bonding interactions, whereas the others cannot. Substance C is N. The other two are nonpolar compounds with only London dispersion forces. Because C 4 is smaller than Si 4, C 4 will have weaker LD forces and is substance A. Therefore, substance B is Si 4.

32 CAPTER 0 LIQUIDS AND SOLIDS 67. As the electronegativity of the atoms covalently bonded to increases, the strength of the hydrogen-bonding interaction increases. N N < N O < O O < O F < F F weakest / Xe 9. corners corner / 4 F + Xe inside cell Xe; edges edge strongest + F inside cell 4 F Empirical formula is XeF. This is also the molecular formula. 0. One B atom and one N atom together have the same number of electrons as two C atoms. The description of physical properties sounds a lot like the properties of graphite and diamond, the two solid forms of carbon. The two forms of BN have structures similar to graphite and diamond.. B 6 : This compound contains only nonmetals, so it is probably a molecular solid with covalent bonding. The low boiling point confirms this. SiO : This is the empirical formula for quartz, which is a network solid. CsI: This is a metal bonded to a nonmetal, which generally form ionic solids. The electrical conductivity in aqueous solution confirms this. W: Tungsten is a metallic solid as the conductivity data confirm.. Ar is cubic closest packed. There are four Ar atoms per unit cell, and with a face-centered unit cell, the atoms touch along the face diagonal. Let l length of cube edge. Face diagonal 4r l, l 4(90. pm)/ 57 pm cm mol 9.95 g 4 atoms mass atoms mol Density.7 g/cm volume (5.7 0 cm). 4.7 g C 6 6 P C 6 6 mol 7. g 0.6 mol C L atm 0.6 mol 9. K nrt K mol atm, or 57. torr V 00.0 L 4. In order to set up an equation, we need to know what phase exists at the final temperature. To heat 0.0 g of ice from!0.0 to 0.0 C requires: q.0 J g C 0.0 g 0.0EC 406 J