Answers to Selected Exercises

The answers listed here are from the Complete Solutions Guide, in which 109 mi/h 55. 100 ft 57. 18 g 59. 7 3 105 kg mercury 61. a. 22738C, rounding is carried out at each intermediate step in a calculation in order to 0 K; b. 240.8C, 233 K; c. 20.8C, 293 K; d. 4 3 107 8C, 4 3 107 K 63. a. 312.4 show the correct number of significant figures for that step. Therefore, an K; 102.68F; b. 248 K; 2138F; c. 0 K;24598F; d. 1074 K; 14708F 65. 160.8C answer given here may differ in the last digit from the result obtained by car- 5oX 5 320.8F 67. a. T 5 (T 1 10°C) ; b. 11.4°X; c. 152°C; 425 K; rying extra digits throughout the entire calculation and rounding at the end X C 14oC (the procedure you should follow). 306°F 69. It will float (density 5 0.80 g/cm3). 71. 1 3 106 g/cm3 73. a. 0.28 cm3; b. 49 carats 75. 3.8 g/cm3 77. a. Both are the same mass; Chapter 1 b. 1.0 mL mercury; c. Both are the same mass; d. 1.0 L benzene 79. a. 1.0 kg 19. A law summarizes what happens, e.g., law of conservation of mass in a feather; b. 100 g water; c. same 81. 2.77 cm 83. a. Picture iv represents a chemical reaction or the ideal gas law, PV 5 nRT. A theory (model) is an gaseous compound. Pictures ii and iii also contain a gaseous compound but attempt to explain why something happens. Dalton’s atomic theory explains have a gaseous element present. b. Picture vi represents a mixture of two why mass is conserved in a chemical reaction. The kinetic molecular theory gaseous elements. c. Picture v represents a solid element. d. Pictures ii and explains why pressure and volume are inversely related at constant tempera- iii both represent a mixture of a gaseous element and a gaseous compound. ture and moles of gas. 21. The fundamental steps are (1) making observa- 85. a. heterogeneous; b. homogeneous (hopefully); c. homogeneous; d. ho- tions; (2) formulating hypotheses; (3) performing experiments to test the mogeneous (hopefully); e. heterogeneous; f. heterogeneous 87. a. pure; hypotheses. The key to the scientific method is performing experiments to b. mixture; c. mixture; d. pure; e. mixture (copper and zinc); f. pure; g. mix- test hypotheses. If after the test of time, the hypotheses seem to account sat- ture; h. mixture; i. mixture. Iron and uranium are elements. Water is a com- isfactorily for some aspect of natural behavior, then the set of tested hypoth- pound. Table salt is usually a homogeneous mixture composed mostly of eses turns into a theory (model). However, scientists continue to perform sodium chloride (NaCl), but it usually will contain other substances that help experiments to refine or replace existing theories. 23. A qualitative obser- absorb water vapor (an anticaking agent). 89. Compound 91. a. physical; vation expresses what makes something what it is; it does not involve a num- b. chemical; c. physical; d. chemical 93. The sphere that floats has the ber; e.g., the air we breathe is a mixture of gases, ice is less dense than water, smaller density; this will be the object that has the same mass in a larger rotten milk stinks. The SI units are mass in grams, length in meters, and volume. So the object with the larger diameter will have the larger volume volume in the derived units of m3. The assumed uncertainty in a number is and will have the smaller density; the larger sphere will float. 95. 0.010 kg 61 in the last significant figure of the number. The precision of an instru- 97. 1.0 3 105 bags 99. 3.0 3 1017 m 101. a. 1.1 g; b. 4.9 g 103. 14 g ment is related to the number of significant figures associated with an ex- 105. 56.568C 107. a. Volume 3 density 5 mass; the orange block is more perimental reading on that instrument. Different instruments for measuring dense. Since mass (orange) . mass (blue) and volume (orange) , volume mass, length, or volume have varying degrees of precision. Some instru- (blue), the density of the orange block must be greater to account for the ments only give a few significant figures for a measurement while others will larger mass of the orange block. b. Which block is more dense cannot be give more significant figures. 25. Significant figuresare the digits we as- determined. Since mass (orange) . mass (blue) and volume (orange) . vol- sociate with a number. They contain all of the certain digits and the first un- ume (blue), the density of the orange block may or may not be larger than the certain digit (the first estimated digit). What follows is one thousand indi- blue block. If the blue block is more dense, then its density cannot be so large cated to varying numbers of significant figures: 1000 or 1 3 103 (1 S.F.); that the mass of the smaller blue block becomes larger than the orange block 1.0 3 103 (2 S.F.); 1.00 3 103 (3 S.F.); 1000. or 1.000 3 103 (4 S.F.). To mass. c. The blue block is more dense. Since mass (blue) 5 mass (orange) perform the calculation, the addition/subtraction significant rule is applied to and volume (blue) , volume (orange), the density of the blue block must be 1.5 2 1.0. The result of this is the one significant figure answer of 0.5. Next, larger to equate the masses. d. The blue block is more dense. Since mass the multiplication/division rule is applied to 0.5/0.50. A one significant num- (blue) . mass (orange) and the volumes are equal, the density of the blue block ber divided by a two significant number yields an answer with one significant must be larger to give the blue block the larger mass. 109. 8.5 6 0.5 g/cm3 4 6 figure (answer 5 1). 27. The slope of the TF vs. TC plot is 1.8 (5 9/5) and 111. 3.648 3 10 cable lengths; 6.671 3 10 m; 3648 nautical miles the y-intercept is 328F. The slope of TC vs. TK plot is 1 and the y-intercept 113. Phosphorus would be a liquid. 115. a. False; sugar is generally con-

is 22738C. 29. The density of a gas is much smaller than the density of a sidered to be the pure compound sucrose, C12H22O11. b. False; elements and solid or a liquid. The molecules in a solid and a liquid are very close together, compounds are pure substances. c. True; air is a mixture of mostly nitrogen whereas in the gas phase, the molecules are very far apart. 31. a. exact; and oxygen gases. d. False; gasoline has many additives, so it is a mixture. b. inexact; c. exact; d. inexact 33. a. 3; b. 4; c. 4; d. 1; e. 7; f. 1; g. 3; h. 3 e. True; compounds are broken down to elements by chemical change. 35. a. 3.42 3 1024; b. 1.034 3 104; c. 1.7992 3 101; d. 3.37 3 105 37. 2.85 117. In a subtraction, the result gets smaller, but the uncertainties add. If the 1 0.280 5 3.13 mL; the graduated cylinder on the left limits the precision of two numbers are very close together, the uncertainty may be larger than the the total volume; it is the least precise measuring device between the two result. For example, let us assume we want to take the difference of the fol- graduated cylinders. 39. a. 641.0; b. 1.327; c. 77.34; d. 3215; e. 0.420 lowing two measured quantities, 999,999 62 and 999,996 62. The differ- 41. a. 188.1; b. 12; c. 4 3 1027; d. 6.3 3 10226; e. 4.9; Uncertainty appears ence is 3 6 4. Because of the larger uncertainty, subtracting two similar

in the first decimal place. The average of several numbers can be only as numbers is bad practice. 119. a. 2%; b. 2.2%; c. 0.2% 121. dold 5 3 3 precise as the least precise number. Averages can be exceptions to the sig- 8.8 g/cm , dnew 5 7.17 g/cm ; dnew/dold 5 massnew/massold 5 0.81; The differ- nificant figure rules. f. 0.22 43. a. 84.3 mm; b. 2.41 m; c. 2.945 3 1025 cm; ence in mass is accounted for by the difference in the alloy used (if the as- 5 1 d. 14.45 km; e. 2.353 3 10 mm; f. 0.9033 mm 45. a. 8 lb and 9.9 oz; 204 in; sumptions are correct). 123. 7.0% 125. a. One possibility is that rope B b. 4.0 3 104 km, 4.0 3 107 m; c. 1.2 3 1022 m3, 12 L, 730 in3, 0.42 ft3 is not attached to anything and rope A and rope C are connected via a pair of 47. a. 4.00 3 102 rods; 10.0 furlongs; 2.01 3 103 m; 2.01 km; b. 8390.0 rods; pulleys and/or gears; b. Try to pull rope B out of the box. Measure the dis- 209.75 furlongs; 42,195 m; 42.195 km 49. a. 0.373 kg, 0.822 lb; b. 31.1 g, tance moved by C for a given movement of A. Hold either A or C firmly while 156 carats; c. 19.3 cm3 51. 24 capsules 53. 2.91 3 109 knots; 3.36 3 pulling on the other rope. 127. $160./person; 3.20 3 103 nickels/person;

A31

57404_Ans_A31-A64.indd 31 10/14/16 5:12 PM A32 Answers to Selected Exercises

85.6 £/person 129. 200.08F 5 93.338C; 2100.08F 5 273.338C; 93.338C tion, mass must not change. In this equation we have two oxygen atoms and 5 366.48 K; 273.338C 5 199.82 K; difference of temperatures in 8C 5 four hydrogen atoms both before and after the reaction occurs. 47. O, 7.94; 166.66; difference of temperatures in K 5 166.66; No, there is not a differ- Na, 22.8; Mg, 11.9; O and Mg are incorrect by a factor of < 2; correct for- 214 ence of 300 degrees in 8C or K. mulas are H2O, Na2O, and MgO. 49. Using r 5 5 3 10 cm, dnucleus 5 15 3 28 3 3 3 10 g/cm ; using r 5 1 3 10 cm, datom 5 0.4 g/cm 51. 37 Chapter 2 53. Sn, tin; Pt, platinum; Hg, mercury; Mg, magnesium; K, potassium; Ag, 19. A compound will always contain the same numbers (and types) of atoms. silver 55. a. 6; b. 6; c. 4; d. 7 57. a. Metals: Mg, Ti, Au, Bi, Ge, Eu, Am; A given amount of hydrogen will react only with a specific amount of oxy- nonmetals: Si, B, At, Rn, Br; b. metalloids: Si, Ge, B, At. The elements at the gen. Any excess oxygen will remain unreacted. 21. Law of conservation of boundary between the metals and the nonmetals are B, Si, Ge, As, Sb, Te, Po, mass: mass is neither created nor destroyed. The mass before a chemical re- and At. These elements are all considered metalloids. Aluminum has mostly action always equals the mass after a chemical reaction. Law of definite pro- properties of metals. 59. a. transition metals; b. alkaline earth metals; c. al- 17 37 60 portion: a given compound always contains exactly the same proportion of kali metals; d. noble gases; e. halogens 61. a. 8O; b. 17Cl; c. 27Co; 57 131 7 23 19 16 elements by mass. Water is always 1 g H for every 8 g oxygen. Law of mul- d. 26Fe; e. 53I; f. 3Li 63. a. 11Na b. 9F c. 8O 65. a. 35 p, 44 n, 35 e; tiple proportions: When two elements form a series of compounds, the ratios b. 35 p, 46 n, 35 e; c. 94 p, 145 n, 94 e; d. 55 p, 78 n, 55 e; e. 1 p, 2 n, 1 e; of the mass of the second element that combine with one gram of the first f. 26 p, 30 n, 26 e 67. a. 56 p; 54 e; b. 30 p; 28 e; c. 7 p; 10 e; d. 37 p; 36 e; 151 31 118 21 238 element can always be reduced to small whole numbers. For CO2 and CO e. 27 p; 24 e; f. 52 p; 54 e; g. 35 p; 36 e 69. 63Eu ; 50Sn 71. 92U, 40 21 51 31 discussed in Section 2.2, the mass ratios of oxygen that react with 1 g of 92 p, 146 n, 92 e, 0; 20Ca , 20 p, 20 n, 18 e, 21; 23V , 23 p, 28 n, 20 e, 31; 89 79 2 31 32 carbon in each compound are in a 2:1 ratio. 23. J. J. Thomson’s study of 39Y, 39 p, 50 n, 39 e, 0; 35Br , 35 p, 44 n, 36 e, 12; 15P , 15 p, 16 n, 18 e, cathode-ray tubes led him to postulate the existence of negatively charged 32 73. a. lose 2 e2 to form Ra21; b. lose 3 e2 to form In31; c. gain 3 e2 to particles, which we now call electrons. Ernest Rutherford and his alpha bom- form P32; d. gain 2 e2 to form Te22; e. gain 1 e2 to form Br2; f. lose 1 e2 to bardment of metal foil experiments led him to postulate the nuclear atom— form Rb1 75. a. sodium bromide; b. rubidium oxide; c. calcium sulfide; an atom with a tiny dense center of positive charge (the nucleus) with elec- d. aluminum iodide; e. SrF2; f. Al2Se3; g. K3N; h. Mg3P2 77. a. cesium trons moving about the nucleus at relatively large distances away; the fluoride; b. lithium nitride; c. silver sulfide (Silver forms only 11 ions so no distance is so large that an atom is mostly empty space. 25. The number Roman numerals are needed); d. manganese(IV) oxide; e. titanium(IV) ox- and arrangement of electrons in an atom determines how the atom will react ide; f. strontium phosphide 79. a. barium sulfite; b. sodium nitrite; c. po- with other atoms. The electrons determine the chemical properties of an tassium permanganate; d. potassium dichromate 81. a. dinitrogen tetrox- atom. The number of neutrons present determine the isotope identity and the ide; b. iodine trichloride; c. sulfur dioxide; d. diphosphorus pentasulfide mass number. 27. 2; the ratio increases 29. Metallic character increases 83. a. copper(I) iodide; b. copper(II) iodide; c. cobalt(II) iodide; d. sodium as one goes down a family and decreases from left to right across the periodic carbonate; e. sodium hydrogen carbonate or sodium bicarbonate; f. tetrasul- table. 31. In the paste, sodium chloride will dissolve to form separate Na1 fur tetranitride; g. selenium tetrachloride; h. sodium hypochlorite; i. bar- and Cl2 ions. With the ions present and able to move about, electrical im- ium chromate; j. ammonium nitrate 85. selenate; selenite; tellurate; tellu- pulses will be conducted. 33. a. This represents ionic bonding, which is rite 87. a. SF2; b. SF6; c. NaH2PO4; d. Li3N; e. Cr2(CO3)3; f. SnF2; the electrostatic attraction between anions and cations. b. This represents g. NH4C2H3O2; h. NH4HSO4; i. Co(NO3)3; j. Hg2Cl2; Mercury(I) exists as 21 covalent bonding, where electrons are shared between two atoms. This could Hg2 ions. k. KClO3; l. NaH 89. a. Na2O; b. Na2O2; c. KCN; d. Cu(NO3)2;

be the space-filling model for 2H O or SF2 or NO2, etc. 35. Statements a and e. SeBr4; f. HIO2; g. PbS2; h. CuCl; i. GaAs (from the positions in the peri- 31 32 b are true. Element 118 is a noble gas and will be a nonmetal. For statement odic table, Ga and As are the predicted ions); j. CdSe; k. ZnS; l. HNO2; c, hydrogen has mostly nonmetallic properties. For statement d, a family of m. P2O5 91. a. nitric acid, HNO3; b. perchloric acid, HClO4; c. acetic acid, elements is also known as a group of elements. For statement e, two items HC2H3O2; d. sulfuric acid, H2SO4; e. phosphoric acid, H3PO4 93. There are incorrect. When a metal reacts with a nonmetal, an ionic compound is should be no difference. The composition of insulin from both sources will

produced and the formula of the compound would be AX2 (since alkaline be the same and therefore it will have the same activity regardless of the earth metals for 12 ions and halogens form 21 ions in ionic compounds). source. 95. a. 53 protons, 78 neutrons; b. 81 protons, 120 neutrons The correct statement would be: When alkaline earth metal, A, reacts with 97. 26 p, 27 n, 24 e 99. Only statement a is true. For statement b, X has 34

a halogen, X, the formula of the ionic compound formed should be AX2. protons. For statement c, X has 45 neutrons. For statement d, X is selenium. 37. a. The composition of a substance depends on the number of atoms 101. a. lead(II) acetate; b. copper(II) sulfate; c. calcium oxide; d. magnesium of each element making up the compound (depends on the formula of sulfate; e. magnesium hydroxide; f. calcium sulfate; g. dinitrogen monoxide the compound) and not on the composition of the mixture from which or nitrous oxide (common) 103. X 5 Ra, 142 neutrons 105. a. Ca3N2; it was formed. b. Avogadro’s hypothesis implies that volume ratios calcium nitride; b. K2O; potassium oxide; c. RbF; rubidium fluoride; d. MgS; are equal to molecule ratios at constant temperature and pressure. magnesium sulfide; e. BaI2; barium iodide; f. Al2Se3; aluminum selenide;

H2 1g2 1 Cl2 1g2 n 2HCl1g2 ; from the balanced equation, the volume of g. Cs3P; cesium phosphide; h. InBr3; indium(III) bromide (In forms com- HCl produced will be twice the volume of H2 (or Cl2) reacted. 39. 299 g pounds with 11 and 13 ions. You would predict a 13 ion from the position 41. The ratio of the masses of carbon that combine with 1 g of oxygen is of In in the periodic table.) 107. a. P; 15 p; 16 n; b. I; 53 p; 74 n; c. K; 19 p; 0.751 2 20 n d. Yb; 70 p; 103 n 109. The law of multiple proportions does not in- 5 . Note that this supports the law of multiple proportions because 0.375 1 volve looking at the ratio of the mass of one element with the total mass of this carbon ratio is a small whole number. 43. For CO and CO2, it is easiest the compounds. To illustrate the law of multiple proportions, we compare the to concentrate on the mass of oxygen that combines with 1 g of carbon. From mass of carbon that combines with 1.0 g of oxygen in each compound: the formulas (two oxygen atoms per carbon atom in CO vs. one oxygen 27.2 g C 42.9 g C 2 compound 1: 5 0.374 g C/g O; compound 2: 5 0.751 g C/g O; atom per carbon atom in CO), CO2 will have twice the mass of oxygen that 72.8 g O 57.1 g O combines per gram of carbon as compared to CO. For CO and C O , it is 0.751 2 2 3 2 5 . Because the ratio is a small whole number, this supports the law easiest to concentrate on the mass of carbon that combines with 1 g of oxy- 0.374 1

gen. From the formulas (three carbon atoms per two oxygen atoms in C3O2 of multiple proportions. vs. one carbon atom per two oxygen atoms in CO2), C3O2 will have three 111. Number of Protons Number of Neutrons Symbol times the mass of carbon that combines per gram of oxygen as compared to 9 10 19F CO2. As expected, the mass ratios are whole numbers as predicted by the law 9 27 of multiple proportions. 45. Mass is conserved in a chemical reaction be- 13 14 13Al 127 cause atoms are conserved. Chemical reactions involve the reorganization of 53 74 53I 79 atoms, so formulas change in a chemical reaction, but the number and types 34 45 34Se 32 of atoms do not change. Because the atoms do not change in a chemical reac- 16 16 16S

57404_Ans_A31-A64.indd 32 10/14/16 5:12 PM Answers to Selected Exercises A33

113. Atom/Ion Protons Neutrons Electrons 1022 atoms N; b. 3.76 3 1022 atoms N; c. 4.78 3 1021 atoms N 65. 2.77 3 1019 molecules; 3.26 mg Cl 67. a. 0.9393 mol; b. 2.17 3 1024 mol; 120 50Sn 50 70 50 c. 2.5 3 1028 mol 69. a. 4.01 3 1022 atoms N; b. 5.97 3 1022 atoms N; 25 21 12Mg 12 13 10 c. 3.67 3 1022 atoms N; d. 6.54 3 1022 atoms N 71. 176.12 g/mol; 56 21 Fe 26 30 24 22 22 26 2.839 3 10 mol C6H8O6; 1.368 3 10 molecules 73. a. 165.39 g/mol; 79 34Se 34 45 34 b. 3.023 mol; c. 3.3 g; d. 5.5 3 1022 atoms; e. 1.6 g; f. 1.373 3 10219 g 35 17Cl 17 18 17 75. a. 50.00% C, 5.595% H, 44.41% O; b. 55.80% C, 7.025% H, 37.18% O; 63 29Cu 29 34 29 c. 67.90% C, 5.699% H, 26.40% N 77. NO 79. 6.54 3 104 g/mol 81. a. 39.99% C, 6.713% H, 53.30% O; b. 40.00% C, 6.714% H, 53.29% O; 115. carbon tetrabromide, CBr ; cobalt(II) phosphate, Co (PO ) ; magne- 4 3 4 2 c. 40.00% C, 6.714% H, 53.29% O (all the same except for rounding sium chloride, MgCl ; nickel(II) acetate, Ni(C H O ) ; calcium nitrate, 2 2 3 2 2 differences) 83. a. NO ; b. CH ; c. PO ; d. CH O 85. CH O Ca(NO ) 117. K will lose 1 e2 to form K1. Cs will lose 1 e2 to form Cs1. 2 2 2 5 2 3 6 2 3 2 87. compound I: HgO; compound II: Hg O 89. SN; S N 91. C H Cl; Br will gain 1 e2 to form Br2. Sulfur will gain 2 e2 to form S22. Se will gain 2 4 4 3 5 C H Cl CH93. O 95. C H , C H 97. a. C H O (s) 1 6O (g) n 2 e2 to form Se22. 119. Cu, Ag, and Au 121. C:H 5 8:18 or 4:9 6 10 2 2 3 4 9 12 6 12 6 2 6CO2(g) 1 6H2O(g); b. Fe2S3(s) 1 6HCl(g) n 2FeCl3(s) 1 3H2S(g); 123. The alchemists were incorrect. The solid residue must have come from MnO c. CS (l) 1 2NH (g) H S(g) 1 NH SCN(s) 99. 2H O (aq) 2 the flask. 125. a. The compounds are isomers of each other. Isomers are 2 3 n 2 4 2 2 h compounds with the same formula but the atoms are attached differently, 2H2O(l) 1 O2(g) 101. a. 3Ca(OH)2(aq) 1 2H3PO4(aq) n 6H2O(l) 1 resulting in different properties. b. When wood burns, most of the solid ma- Ca3(PO4)2(s); b. Al(OH)3(s) 1 3HCl(aq) n AlCl3(aq) 1 3H2O(l); terial is converted to gases, which escape. c. Atoms are not an indivisible c. 2AgNO3(aq) 1 H2SO4(aq) n Ag2SO4(s) 1 2HNO3(aq) 103. a. 2C6H6(l) particle. Atoms are composed of electrons, neutrons, and protons. d. The two 1 15O2(g) n 12CO2(g) 1 6H2O(g); b. 2C4H10(g) 1 13O2(g) n 8CO2(g) 1 hydride samples contain different isotopes of either hydrogen and/or lithium. 10H2O(g); c. C12H22O11(s) 1 12O2(g) n 12CO2(g) 1 11H2O(g); d. 4Fe(s) Isotopes may have different masses but have similar chemical properties. 1 3O2(g) n 2Fe2O3(s); e. 4FeO(s) 1 O2(g) n 2Fe2O3(s) 105. a. SiO2(s) 1 2C(s) n Si(s) 1 2CO(g); b. SiCl4(l) 1 2Mg(s) n Si(s) 1 2MgCl2(s); 127. CO2 129. tantalum(V) oxide; the formula would have the same sub- 41 99 c. Na2SiF6(s) 1 4Na(s) n Si(s) 1 6NaF(s) 107. CaF2 ? 3Ca3(PO4)2(s) 1 scripts, Ta2S5; 40 protons 131. Ge ; Tc 10H2SO4(aq) + 20H2O(l) n 6H3PO4(aq) 1 2HF(aq) + 10CaSO4 ? 2H2O(s) Chapter 3 109. 7.26 g Al; 21.5 g Fe2O3; 13.7 g Al2O3 111. 4.355 kg 113. 72.3 g 4 25. The atomic mass of any particular isotope is a relative mass to a specific 115. a. 0.076 g; b. 0.052 g 117. 32 kg 119. 7.2 3 10 g 121. 2NO(g) 1 3 standard. The standard is one atom of the carbon-12 isotope weighing exactly O2(g) n 2NO2(g); NO is limiting. 123. a. 1.22 3 10 g; 284 g H2 unre- 6 12.0000 u. One can determine from experiment how much heavier or lighter acted 125. 0.301 g H2O2; 1.75 g HCl in excess 127. 2.81 3 10 g HCN; 6 3 any specific isotope is than12 C. From this information, we assign an atomic 5.63 3 10 g H2O 129. 72.9% 131. 82.9% 133. 1.20 3 10 kg 5 12 1 16 mass value to that isotope. For example, experiment tells one that 16O is about 1.20 metric tons 135. C H2 O6 137. 139. 71.40% C, 8.689% H, 4/3 heavier than 12C, so a mass of 4/3(12.00) 5 16.00 u is assigned to 16O. The 5.648% F, 14.26% O 141. Sulfur 143. C8H11O3N 145. 1360 g/mol 22 22 atomic mass listed in the periodic table is also an average mass. Most elements 147. 9.25 3 10 H atoms 149. 4.30 3 10 mol; 2.50 g 151. 5 in nature occur as a mixture of isotopes. The atomic mass of an element is the 153. 42.8% 155. 81.1 g 157. 86.2% 159. C20H30O 161. 14.3% X, average mass of all the isotopes that make up a specific element, weighted by 85.7% Z 163. 1.242 mol Mg3(PO4)2; 1.846 mol Ca(NO3)2; 0.7291 mol abundance. 27. Each person would have 90 trillion dollars. 29. Only in b K2CrO4; 3.761 mol N2O5 165. C7H8O 167. a. false; this is what is done are the empirical formulas the same for both compounds illustrated. In b, gen- when equations are balanced; b. false; the coefficients give molecule ratios as well as mole ratios between reactants and products; c. false; the reactants eral formulas of X2Y4 and XY2 are illustrated, and both have XY2 for an empirical formula. 31. The mass percent of a compound is a constant no matter are on the left, with the products on the right; d. true; e. true; in order for what amount of substance is present. Compounds always have constant com- mass to be conserved, there must be the same number of atoms as well as the position. 33. c 35. The theoretical yield is the stoichiometric amount of same type of atoms on both sides. 169. 51.7 g Na2SO3; 7.39 g H2O product that should form if the limiting reactant is completely consumed and 171. I: NH3; II: N2H4; III: HN3. If we set the atomic mass of H equal to 1.008, the reaction has 100% yield. 37. The information needed is mostly the coef- the atomic mass for nitrogen is 14.01. 173. C7H5N3O6 175. 5.7 g FeO 6 ficients in the balanced equation and the molar masses of the reactants and and 22.4 g Fe2O3 177. 40.08% 179. N 4H6 181. 1.8 3 10 g Cu(NH3)4Cl2; 5 products. For percent yield, we would need the actual yield of the reaction and 5.9 3 10 g NH3 183. 207 u, Pb 185. Al2Se3 187. LaH2.90 contains 1 9 the amounts of reactants used. La21, or 10.% La21, and La31, or 90.% La31 189. 0.48 mol 4 10 10 a. mass of CB produced 5 1.00 3 10 molecules A2B2 3 191. a. 113 Fe atoms; b. mass 5 9.071 3 10220 g; c. 540 Ru atoms 1 mol A2B2 2 mol CB molar mass of CB 193. M 5 Y, X 5 Cl, yttrium(III) chloride; 1.84 g 23 3 3 6.022 3 10 molecules A2B2 1 mol A2B2 mol CB Chapter 4 4 2 atoms A b. atoms of A produced 5 1.00 3 10 molecules A2B2 3 17. a. Polarity is a term applied to covalent compounds. Polar covalent 1 molecule A2B2 4 c. mol of C reacted 5 1.00 3 10 molecules A2B2 3 compounds have an unequal sharing of electrons in bonds that results in an unequal charge distribution in the overall molecule. Polar molecules have a 1 mol A B 2 mol C 2 2 3 partial negative end and a partial positive end. These are not full charges as 6.022 3 1023 molecules A B 1 mol A B 2 2 2 2 in ionic compounds, but are charges much less in magnitude. Water is a polar actual mass molecule and dissolves other polar solutes readily. The oxygen end of water d. % yield 5 3 100; The theoretical mass of CB produced theoretical mass (the partial negative end of the polar water molecule) aligns with the partial was calculated in part a. If the actual mass of CB produced is given, then the positive end of the polar solute while the hydrogens of water (the partial percent yield can be determined for the reaction using the percent yield equa- positive end of the polar water molecule) align with the partial negative end tion. 39. 207.2 u, Pb 41. 185 u 43. 48% 151Eu and 52% 153Eu of the solute. These opposite charged attractions stabilize polar solutes in 45. There are three peaks in the mass spectrum, each 2 mass units apart. This water. This process is called hydration. Nonpolar solutes do not have perma- is consistent with two isotopes, differing in mass by two mass units. 47. 4.64 3 nent partial negative and partial positive ends; nonpolar solutes are not stabi- 10220 g Fe 49. 1.00 3 1022 atoms C 51. Prozac: 309.32 g/mol; Zoloft: lized in water and do not dissolve. b. KF is a soluble ionic compound so it is 306.22 g/mol 53. a. 17.03 g/mol; b. 32.05 g/mol; c. 252.08 g/mol a strong electrolyte. KF(aq) actually exists as separate hydrated K1 ions and 23 2 55. a. 0.0587 mol NH3; b. 0.0312 mol N2H4; c. 3.97 3 10 mol (NH4)2Cr2O7 hydrated F ions in solution: C6H12O6 is a polar covalent molecule that is a 57. a. 85.2 g NH3; b. 160. g N2H4; c. 1260 g (NH4)2Cr2O7 59. a. 70.1 g N; nonelectrolyte. C6H12O6 is hydrated as described in part a. c. RbCl is a solu- 22 22 1 b. 140. g N; c. 140. g N 61. a. 3.54 3 10 molecules NH3; b. 1.88 3 10 ble ionic compound so it exists as separate hydrated Rb ions and hydrated 21 2 molecules N2H4; c. 2.39 3 10 formula units (NH4)2Cr2O7 63. a. 3.54 3 Cl ions in solution. AgCl is an insoluble ionic compound so the ions stay

57404_Ans_A31-A64.indd 33 10/14/16 5:12 PM A34 Answers to Selected Exercises

21 22 together in solution and fall to the bottom of the container as a precipitate. Ca (aq) 1 SO4 (aq) n CaSO4(s); d. K2S(aq) 1 Ni(NO3)2(aq) n 1 1 22 21 2 d. HNO3 is a strong acid and exists as separate hydrated H ions and hy- 2KNO3(aq) 1 NiS(s); 2K (aq) 1 S (aq) 1 Ni (aq) 1 2NO3 (aq) n 2 1 2 21 22 drated NO3 ions in solution. CO is a polar covalent molecule and is hy- 2K (aq) 1 2NO3 (aq) 1 NiS(s); Ni (aq) 1 S (aq) n NiS(s) 21 22 drated as explained in part a. 19. Only statement b is true. A concentrated 55. a. CuSO4(aq) 1 Na2S(aq) n CuS(s) 1 Na2SO4(aq); Cu (aq) 1 S (aq) solution can also contain a nonelectrolyte dissolved in water, e.g., concen- n CuS(s); the gray spheres are the Na1 spectator ions and the blue-green 22 trated sugar water. Acids are either strong or weak electrolytes. Some ionic spheres are the SO4 spectator ions; b. CoCl2(aq) 1 2NaOH(aq) n 21 2 compounds are not soluble in water, so they are not labeled as a specific type Co(OH)2(s) 1 2NaCl(aq); Co (aq) 1 2OH (aq) n Co(OH)2(s); the gray 1 2 of electrolyte. 21. Bromides: NaBr, KBr, and NH4Br (and others) would be spheres are the Na spectator ions and the green spheres are the Cl specta- 1 2 soluble and AgBr, PbBr2, and Hg2Br2 would be insoluble. Sulfates: Na2SO4, tor ions; c. AgNO3(aq) 1 KI(aq) n AgI(s) 1 KNO3(aq); Ag (aq) 1 I (aq) 1 K2SO4, and (NH4)2SO4 (and others) would be soluble and BaSO4, CaSO4, n AgI(s); the red spheres are the K spectator ions and the blue spheres are 2 21 22 and PbSO4 (or Hg2SO4) would be insoluble. Hydroxides: NaOH, KOH, and the NO3 spectator ions 57. a. Ba (aq) 1 SO4 (aq) n BaSO4(s); 21 2 Ca(OH)2 (and others) would be soluble and Al(OH)3, Fe(OH)3, and Cu(OH)2 b. Pb (aq) 1 2Cl (aq) n PbCl2(s); c. no reaction; d. no reaction; 21 2 21 21 21 (and others) would be insoluble. Phosphates: Na3PO4, K3PO4, and (NH4)3PO4 e. Cu (aq) 1 2OH (aq) n Cu(OH)2( s) 59. Ca , Sr , or Ba could all (and others) would be soluble and Ag3PO4, Ca3(PO4)2, and FePO4 (and oth- be present. 61. 0.607 g 63. 0.668 g 65. 2.82 g 67. a. 2KOH(aq) 1

ers) would be insoluble. Lead: PbCl2, PbBr2, PbI2, Pb(OH)2, PbSO4, and PbS Mg(NO3)2(aq) n Mg(OH)2(s) 1 2KNO3(aq); b. magnesium hydroxide; 21 2 22 21 1 2 (and others) would be insoluble. Pb(NO3)2 would be a soluble Pb salt. c. 0.583 g; d. 0 M OH , 5.00 3 10 M Mg , 0.100 M K , 0.200 M NO3 23. The Brønsted–Lowry definitions are best for our purposes. An acid is a 69. 23 u; Na 71. a. 2HClO4(aq) 1 Mg(OH)2(s) n Mg(ClO4)2(aq) 1 1 1 2 21 2 proton donor and a base is a proton acceptor. A proton is an H ion. Neutral 2H2O(l); 2H (aq) 1 2ClO4 (aq) 1 Mg(OH)2(s) n Mg (aq) 1 2ClO4 (aq) 1 1 21 hydrogen has 1 electron and 1 proton, so an H ion is just a proton. An acid– 1 2H2O(l); 2H (aq) 1 Mg(OH)2(s) n Mg (aq) 1 2H2O(l); b. HCN(aq) 1 1 1 2 base reaction is the transfer of an H ion (a proton) from an acid to a base. NaOH(aq) n NaCN(aq) 1 H2O(l); HCN(aq) 1 Na (aq) 1 OH (aq) n 1 2 2 2 25. a. The species reduced is the element that gains electrons. The reducing Na (aq) 1 CN (aq) 1 H2O(l); HCN(aq) 1 OH (aq) n H2O(l) 1 CN (aq); 1 2 agent causes reduction to occur by itself being oxidized. The reducing agent c. HCl(aq) 1 NaOH(aq) n NaCl(aq) 1 H2O(l); H (aq) 1 Cl (aq) 1 1 2 1 2 1 2 is generally listed as the entire formula of the compound/ion that contains the Na (aq) 1 OH (aq) n Na (aq) 1 Cl (aq) 1 H2O(l); H (aq) 1 OH (aq) element oxidized. b. The species oxidized is the element that loses electrons. n H2O(l) 73. a. KOH(aq) 1 HNO3(aq) n H2O(l) 1 KNO3(aq); The oxidizing agent causes oxidation to occur by itself being reduced. The b. Ba(OH)2(aq) 1 2HCl(aq) n 2H2O(l) 1 BaCl2(aq); c. 3HClO4(aq) oxidizing agent is generally listed as the entire formula of the compound/ion 1 Fe(OH)3(s) n 3H2O(l) 1 Fe(ClO4)3(aq); d. AgOH(s) 1 HBr(aq) n that contains the element reduced. c. For simple binary ionic compounds, the AgBr(s) 1 H2O(l); e. Sr(OH)2(aq) 1 2HI(aq) n 2H2O(l) 1 SrI2(aq) actual charges on the ions are the oxidation states. For covalent compounds, 75. a. 100. mL; b. 66.7 mL; c. 50.0 mL 77. 2.0 3 1022 M excess OH2 nonzero oxidation states are imaginary charges the elements would have if 79. 0.102 M 81. 43.8 mL 83. 0.4178 g 85. a. K, 11; O, 22; Mn, 17; they were held together by ionic bonds (assuming the bond is between two b. Ni, 14; O, 22; c. Na, 11; Fe, 12; O, 22; H, 11 d. H, 11; O, 22; N, 23; 8 different nonmetals). Nonzero oxidation states for elements in covalent com- P, 15; e. P, 13; O, 22; f. O, 22; Fe, 13 ; g. O, 22; F, 21; Xe, 16; h. S, 14; pounds are not actual charges. Oxidation states for covalent compounds are F, 21; i. C, 12; O, 22; j. C, 0; H, 11; O, 22 87. a. 23; b. 23; c. 22; a bookkeeping method to keep track of electrons in a reaction. d. 12; e. 11; f. 14; g. 13; h. 15; i. 0 27. 89. Oxidizing Reducing Substance Substance Redox? Agent Agent Oxidized Reduced – – + Br– Cl Cl Na Mg2+ a. Yes Ag1 Cu Cu Ag1 + Mg2+ Br– Na – b. No — — — — Cl Cl– – + Cl c. No — — — — Na – 2+ Cl– Br Mg d. Yes SiCl4 Mg Mg SiCl4 (Si) e. No — — — — a. b. c. For answers c–i, we will describe what should be in each solution. For c, In b, c, and e, no oxidation numbers change from reactants to products. 2 31 2 1 2 2 2 the drawing should have three times as many NO3 anions as Al cations. 91. a. 3I (aq) 1 2H (aq) 1 ClO (aq) n I3 (aq) 1 Cl (aq) 1 H2O(l); 1 22 1 2 d. The drawing should have twice as many NH4 cations as SO4 anions. b. 7H 2O(l) 1 4H (aq) 1 3As2O3(s) 1 4NO3 (aq) n 4NO(g) 1 6H3AsO4(aq) 1 2 2 21 e. The drawing should have equal numbers of Na1 cations and OH2 anions. c. 16H (aq) 1 2MnO4 (aq) 1 10Br (aq) n 5Br2(l) 1 2Mn (aq) 1 21 22 1 22 31 f. The drawing should have equal numbers of Fe cations and SO4 anions. 8H2O(l) d. 8H (aq) 1 3CH3OH(aq) 1 Cr2O7 (aq) n 2Cr (aq) 1 1 2 2 g. The drawing should have equal numbers of K cations and MnO4 anions. 3CH2O(aq) 1 7H2O(l) 93. a. 2H2O(l) 1 Al(s) 1 MnO4 (aq) n 1 2 2 2 2 2 h. The drawing should have equal numbers of H cations and ClO4 anions. Al(OH)4 (aq) 1 MnO2(s); b. 2OH (aq) 1 Cl2(g) n Cl (aq) 1 OCl (aq) 1 1 2 2 2 2 i. The drawing should have equal numbers of NH4 cations and C2H3O2 H2O(l); c. OH (aq) 1 H2O(l) 1 NO2 (aq) 1 2Al(s) n NH3(g) 1 2AlO2 (aq) 21 2 anions. 29. CaCl2(s) n Ca (aq) 1 2Cl (aq) 31. a. 0.2677 M; b. 1.255 95. 4NaCl(aq) 1 2H2SO4(aq) 1 MnO2(s) n 2Na2SO4(aq) 1 MnCl2(aq) 1 23 23 22 22 21 2 Cl2(g) 1 2H2O( l) 97. 1.622 3 10 M 99. a. 4.46 3 10 M; b. 24.8 mL 3 10 M; c. 8.065 3 10 M 33. a. MCa 5 1.00 M, MNO3 5 2.00 M;

1 22 1 2 1 101. We should weigh out between 4.24 and 4.32 g of KIO3. We should weigh b. MNa 5 4.0 M, MSO4 5 2.0 M; c. MNH4 5 MCl 5 0.187 M; d. MK 5

32 it to the nearest milligram or 0.1 mg. Dissolve the KIO3 in water, and dilute 0.0564 M, MPO4 5 0.0188 M 35. 100.0 mL of 0.30 M AlCl3 37. 4.00 g 39. 0.0472 g 41. a. Place 20.0 g NaOH in a 2-L volumetric flask; add water to the mark in a 1-L volumetric flask. This will produce a solution whose to dissolve the NaOH and fill to the mark. b. Add 500. mL of the 1.00 M NaOH concentration is within the limits and is known to at least the fourth decimal 21 22 stock solution to a 2-L volumetric flask; fill to the mark with water. c. As in a, place. 103. increase by 0.019 M 105. Ca (aq) 1 C2O4 (aq) n

instead using 38.8 g K2CrO4. d. As in b, instead using 114 mL of 1.75 M CaC2O4(s) 107. a. AgNO3, Pb(NO3)2, and Hg2(NO3)2 would form precipi- 2 1 2 21 2 1 22 tates with the Cl ion; Ag (aq) 1 Cl (aq) n AgCl(s); Pb (aq) 1 2Cl (aq) K2CrO4 stock solution. 43. MNH4 5 0.272 M, MSO4 5 0.136 M 45. 4.5 M 21 2 47. 5.94 3 1028 M 49. Aluminum nitrate, magnesium chloride, and ru- n PbCl2(s); Hg2 (aq) 1 2Cl (aq) n Hg2Cl2(s); b. Na2SO4, Na2CO3, and 21 21 22 bidium sulfate are soluble. 51. a. No precipitate forms; b. Al(OH)3(s); Na3PO4 would form precipitates with the Ca ion; Ca (aq) 1 SO4 (aq) 21 22 21 32 c. CaSO4(s); d. NiS(s) 53. a. No reaction occurs because all possible prod- n CaSO4(s); Ca (aq) 1 CO3 (aq) n CaCO3(s); 3Ca (aq) 1 2PO4 (aq) ucts are soluble salts. b. 2Al(NO3)3(aq) 1 3Ba(OH)2(aq) n 2Al(OH)3(s) 1 n Ca3(PO4)2(s); c. NaOH, Na2S, and Na2CO3 would form precipitates with 31 2 21 2 31 31 2 31 22 3Ba(NO3)2(aq); 2Al (aq) 1 6NO3 (aq) 1 3Ba (aq) 1 6OH (aq) n the Fe ion; Fe (aq) 1 3OH (aq) n Fe(OH)3(s); 2Fe (aq) 1 3S (aq) 21 2 31 2 31 22 2Al(OH)3(s) 1 3Ba (aq) 1 6NO3 (aq); Al (aq) 1 3OH (aq) n n Fe2S3(s); 2Fe (aq) 1 3CO3 (aq) n Fe2(CO3)3(s); d. BaCl2, Pb(NO3)2, 21 22 21 Al(OH)3(s); c. CaCl2(aq) 1 Na2SO4(aq) n CaSO4(s) 1 2NaCl(aq); Ca (aq) and Ca(NO3)2 would form precipitates with the SO4 ion; Ba (aq) 1 2 1 22 1 2 22 21 22 21 1 2Cl (aq) 1 2Na (aq) 1 SO4 (aq) n CaSO4(s) 1 2Na (aq) 1 2Cl (aq); SO4 (aq) n BaSO4(s); Pb (aq) 1 SO4 (aq) n PbSO4(s); Ca (aq) 1 22 SO4 (aq) n CaSO4(s); e. Na2SO4, NaCl, and NaI would form precipitates

57404_Ans_A31-A64.indd 34 10/14/16 5:12 PM Answers to Selected Exercises A35

21 21 22 21 with the Hg2 ion; Hg2 (aq) 1 SO4 (aq) n Hg2SO4(s); Hg2 (aq) 1 pressure of NH3. The partial pressure of H2 will be three times the partial 2 21 2 2Cl (aq) n Hg2Cl2(s); Hg2 (aq) 1 2I (aq) n Hg2I2(s); f. NaBr, Na2CrO4, pressure of N2.CO 37. 2 39. From the plot, Ne appears to behave most 1 1 2 3 3 5 and Na3PO4 would form precipitates with the Ag ion; Ag (aq) 1 Br (aq) n ideally. 41. a. 3.6 3 10 mm Hg; b. 3.6 3 10 torr; c. 4.9 3 10 Pa; d. 71 psi 1 22 1 32 3 22 AgBr(s); 2Ag (aq) 1 CrO4 (aq) n Ag2CrO4(s); 3Ag (aq) 1 PO4 (aq) n 43. 65 torr, 8.7 3 10 Pa, 8.6 3 10 atm 45. a. 642 torr, 0.845 atm; 4 5 Ag3PO4( s) 109. Ba 111. 1.465% 113. 2.00 M 115. 180.2 g/mol 8.56 3 10 Pa; b. 975 torr; 1.28 atm; 1.30 3 10 Pa; c. 517 torr; 850. torr 1 22 117. C6H8O6 119. 4.05% 121. 63.74% 123. 4H (aq) 1 Mn(s) 1 47. 0.972 atm 49. 8.01 L 51. a. 14.0 L; b. 4.72 3 10 mol; c. 678 K; 2 21 21 2NO3 (aq) n Mn (aq) 1 2NO2(g) 1 2H2O(l); 3H2O(l) 1 2Mn (aq) 1 d. 133 atm 53. 88.9 g He; 44.8 g H2 55. 0.0449 mol 57. a. 69.6 K; 2 2 2 1 23 22 5IO4 (aq) n 2MnO4 (aq) 1 5IO3 (aq) 1 6H (aq) 125. 2.979 3 10 M b. 32.3 atm 59. 1.1 3 10 air particles 61. 1.27 mol 63. PB 5 2PA 4 127. 747 mL 129. 0.052 M 131. 0.328 M 133. MgSO4: S has an oxida- 65. a. 12.8 atm; b. 161 K; c. 620. K 67. 5.1 3 10 torr 69. The volume of tion state of 16. PbSO4: Lead has an oxidation state of 12. O2: O has an the balloon increases from 1.00 L to 2.82 L, so the change in volume is 7 7 oxidation state of zero. Ag: Ag has an oxidation state of zero. CuCl2: Copper 1.82 L. 71. 3.21 g Al 73. 135 g NaN3 75. 1.5 3 10 g Fe, 2.6 3 10 g 28 29 has an oxidation state of 12. 135. a. 2.5 3 10 M; b. 8.4 3 10 M; 98% H2SO4 77. 33.2 g 79. 2.47 mol H2O 81. a. 2CH4(g) 1 2NH3(g) 1 24 27 c. 1.33 3 10 M; d. 2.8 3 10 M 137. a. 24.8% Co, 29.7% Cl, 5.09% H, 3O2(g) n 2HCN(g) 1 6H2O(g); b. 13.3 L 83. Cl2 85. 12.6 g/L 87. PHe 5

40.4% O; b. CoCl2 ? 6H2O; c. CoCl2 ? 6H2O(aq) 1 2AgNO3(aq) n 2AgCl(s) 0.50 atm, PNe 5 0.30 atm, PAr 5 0.20 atm 89. PHe 5 18.4 atm; PO2 5

1 Co(NO3)2(aq) 1 6H2O(l), CoCl2 ? 6H2O(aq) 1 2NaOH(aq) n Co(OH)2(s) 4.63 atm; PTOTAL 5 23.0 atm 91. PH2 5 317 torr, PN2 5 50.7 torr, PTOTAL 5 1 2NaCl(aq) 1 6H2O(l), 4Co(OH)2(s) 1 O2(g) n 2Co2O3(s) 1 4H2O(l) 368 torr 93. PHe 5 50.0 torr; PNe 5 76.0 torr; PAr 5 90.0 torr; PTOTAL 5 22 22 x x 139. 14.6 g Zn and 14.4 g Ag 141. 0.123 g SO4 , 60.0% SO4 ; 216.0 torr 95. 0.230 97. a. CH4 5 0.412, O2 5 0.588; b. 0.161 mol; 61% K2SO4 and 39% Na2SO4 143. 4.90 M 145. Y, 2.06 mL/min; Z, c. 1.06 g CH4, 3.03 g O2 99. 0.990 atm; 0.625 g 101. 18.0% 103. 46.5% 22 4.20 mL/min 147. 57.6 mL 149. 4.7 3 10 M 151. Citric acid has 105. Ptotal 5 6.0 atm; PH2 5 1.5 atm; PN2 5 4.5 atm 107. PN2 5 0.98 atm; three acidic hydrogens per citric acid molecule. 153. 0.07849 6 0.00016 M PTOTAL 5 2.9 atm 109. Both CH4(g) and N2(g) have the same average kinetic 31 22 221 or 0.0785 6 0.0002 M 155. a. 7H2O(l) 1 2Cr (aq) 1 3S2O8 (aq) n energy at the various temperatures. 273 K, 5.65 3 10 J/molecule; 546 K, 22 1 22 1 21 220 Cr2O7 (aq) 1 14H (aq) 1 6SO4 (aq); 14H (aq) 1 6Fe (aq) 1 1.13 3 10 J/molecule 111. CH4: 652 m/s (273 K); 921 m/s (546 K); N2: 22 31 31 24 Cr2O7 (aq) n 2Cr (aq) 1 6Fe (aq) 1 7H2O(l); b. 3.00 3 10 cm 493 m/s (273 K); 697 m/s (546 K) 113. The number of gas particles is con- 157. 3(NH4)2CrO4(aq) 1 2Cr(NO2)3(aq) n 6NH4NO2 (aq) 1 Cr2(CrO4)3(s); stant, so at constant moles of gas, either a temperature change or a pressure 7.34 g 159. X 5 Se; H2Se is hydroselenic acid; 0.252 g change results in the smaller volume. If the temperature is constant, an increase in the external pressure would cause the volume to decrease. Gases are mostly empty space, so gases are easily compressible. If the pressure is con- Chapter 5 stant, a decrease in temperature would cause the volume to decrease. As the 23. higher than; 13.6 times taller; When the pressure of the column of liquid temperature is lowered, the gas particles move with a slower average velocity standing on the surface of the liquid is equal to the pressure exerted by air on and don’t collide with the container walls as frequently and as forcefully. As the rest of the surface of the liquid, then the height of the column of liquid a result, the internal pressure decreases. In order to keep the pressure constant, is a measure of atmospheric pressure. Because water is 13.6 times less the volume of the container must decrease in order to increase the gas particle dense than mercury, the column of water must be 13.6 times longer than collisions per unit area. that of mercury to match the force exerted by the columns of liquid standing on the surface. 25. The P versus 1yV plot is incorrect. The plot should be 115. Wall-Collision linear with positive slope and a y-intercept of zero. PV 5 k so P 5 k (1/V), Average KE Average Velocity Frequency which is in the form of the straight-line equation y 5 mx 1 b. 27. d 5 a. Increase Increase Increase (molar mass)PyRT; density is directly proportional to the molar mass of the b. Decrease Decrease Decrease gas. Helium, which has the smallest molar mass of all the noble gases, will c. Same Same Increase have the smallest density. 29. At STP (T 5 273.2 K and P 5 1.000 atm), d. Same Same Increase the volume of 1.000 mole of gas is:

0.08206 L atm 117. a. All the same; b. Flask C 119. NO 121. The relative rates of 1.000 mol 3 3 273.2 K effusion of 12C16O, 12C17O, and 12C18O are 1.04, 1.02, and 1.00. Advantage: nRT K mol V 5 5 5 22.42 L CO isn’t as toxic as CO. Disadvantage: Can get a mixture of oxygen iso- P 1.000 atm 2 topes in CO2, so some species would effuse at about the same rate. The volume of 1.000 mole of any gas is 22.42 L, assuming the gas behaves 123. a. 12.24 atm; b. 12.13 atm; c. The ideal gas law is high by 0.91%. ideally. Therefore, the molar volume of He(g) and N (g) at STP both equal 26 14 3 2 125. 5.2 3 10 atm; 1.3 3 10 atoms He/cm S(s127.) 1 O 2(g) n 22.42 L/mol. If the temperature increases to 25.0°C (298.2 K), the molar SO2(g); 2SO2(g) 1 O2(g) n 2SO3(g); SO3(g) 1 H2O(l) n H2SO4(aq) volume of a gas will be larger than 22.42 L/mol because volume and tem- 129. a. 0.19 torr; b. 6.6 3 1021 molecules CO/m3; c. 6.6 3 1015 molecules perature are directly related at constant pressure. If 1.000 mole of a gas is CO/cm3 collected over water at a total pressure of 1.000 atm, the partial pressure of 131. a. b. the collected gas will be less than 1.000 atm because water vapor is present (P 5 P 1 P ). At some partial pressure below 1.000 atm, the molar total gas H2O P volume of a gas will be larger than 22.42 L/mol because volume and pressure PV are inversely related at constant temperature. 31. a. As temperature increases, the average kinetic energy will increase; b. As temperature increases, the average velocity of the gas molecules will increase; c. At constant T temperature, the lighter the gas molecules (the smaller the molar mass), the faster the average velocity. 33. No; At any nonzero Kelvin temperature, there is a distribution of kinetic energies. Similarly, there is a distribution of V velocities at any nonzero Kelvin temperature. 35. 2NH3(g) n N2(g) 1 c. d. 3H2(g); At constant P and T, volume is directly proportional to the moles of gas present. In the reaction, the moles of gas double as reactants are converted to products, so the volume of the container should double. At constant T P V and T, P is directly proportional to the moles of gas present. As the moles

of gas double, the pressure will double. The partial pressure of N2 will be 1y2 the initial pressure of NH and the partial pressure of H will be 3/2 the initial 3 2 V V

57404_Ans_A31-A64.indd 35 10/14/16 5:12 PM A36 Answers to Selected Exercises

e. f. 33. 16 kJ 35. 70. J 37. 375 J heat transferred to the system 39. 213.2 kJ 41. 11.0 L 43. q 5 30.9 kJ, w 5 212.4 kJ, DE 5 18.5 kJ 45. This is an P endothermic reaction, so heat must be absorbed to convert reactants into PV products. The high-temperature environment of internal combustion engines T provides the heat. 47. a. endothermic; b. exothermic; c. exothermic; d. endothermic 49. a. 1650 kJ released; b. 826 kJ released; c. 7.39 kJ released; 1/V d. 34.4 kJ released 51. 4400 g C3H8 53. When a liquid is converted into a gas, there is an increase in volume. The 2.5 kJ/mol quantity can be considered as the work done by the vaporization process in pushing back the atmo- 3 P sphere. 55. H2O(l); 2.30 3 10 J; Hg(l); 1408C 57. Al(s) 59. 311 K 61. 23.78C 63. 0.25 J/8C ? g 65. 266 kJ/mol 67. 170 J/g; 20. kJ/mol 133. 0.772 atm ? L; in Example 5.3, 1.0 mole of gas was present at 08C. The 69. 39.28C 71. –25 kJ/g; –2700 kJ/mol 73. a. 31.5 kJ/8C; b. 21.10 3 moles of gas and/or the temperature must have been different for Boyle’s 103 kJ/mol 75. 2220.8 kJ 77. 1268 kJ; No, because this reaction is very data. 135. P 5 582 torr, P 5 18 torr 137. 13.3% 139. 1490 He Xe endothermic, it would not be a practical way of making ammonia due to the 141. 24 torr 143. 4.1 3 106 L air; 7.42 3 105 L H 145. C H N; C H N 2 2 3 2 3 high energy costs. 79. –202.6 kJ 81. 2713 kJ 83. The enthalpy change 147. C H NO; C H N O 149. Processes b, c, and d will all result in a 12 21 24 42 2 2 for the formation of one mole of a compound from its elements, with all doubling of the pressure. Process b doubles the pressure because the absolute substances in their standard states. Na(s) 1 1 Cl (g) n NaCl(s); H (g) 1 temperature is doubled (from 200. K to 400. K). Process c has the effect of 2 2 2 1 O (g) n H O(l); 6C (s) 1 6H (g) 1 3O (g) n C H O (s); Pb(s) 1 halving the volume, which would double the pressure by Boyle’s law. Pro- 2 2 2 graphite 2 2 6 12 6 S (s) 1 2O (g) n PbSO (s) 85. a. 2940. kJ; b. 2265 kJ; c. 2176 kJ cess d doubles the pressure because the moles of gas are doubled (28 g N is rhombic 2 4 2 87. a. 2908 kJ, 2112 kJ, 2140. kJ; b. 12NH (g) 1 21O (g) n 8HNO (aq) 1 mol of N and 32 g O is 1 mol of O ). Process a won’t double the pressure 3 2 3 2 2 2 1 4NO(g) 1 14H O(g), exothermic 89. 22677 kJ 91. 2169 kJ/mol because 28 g O is less than 1 mol of gas, and process e won’t double the 2 2 93. 132 kJ 95. 229.67 kJ/g 97. For C H (g), 250.37 kJ/g vs. 247.7 kJ/g temperature since the absolute temperature is not doubled (goes from 303 K 3 8 for octane. Because of the low boiling point of propane, there are extra safety to 333 K). 151. DV 5 142 L 153. 51.4 g 155. All the gases have the hazards associated with storing the propane in high-pressure compressed gas same average kinetic energy since they are all at the same temperature. The tanks. 99. 1.05 3 105 L 101. 4900 g 103. a. 2SO (g) 1 O (g) n 2SO (g); average velocity order from slowest to fastest is F , N , He. 157. 13.4% 2 2 3 2 2 w . 0; b. COCl (g) n CO(g) 1 Cl (g); w , 0; c. N (g) 1 O (g) n 2NO(g); CaO, 86.6% BaO 159. C H 161. a. 8.7 3 103 L air/min; b. x 5 0.0017, 2 2 2 2 2 6 CO w 5 0; Compare the sum of the coefficients of all the product gases in the x 5 0.032, x 5 0.13, x 5 0.77, x 5 0.067 163. a. 1.01 3 104 g; CO2 O2 N2 H2O balanced equation to the sum of the coefficients of all the reactant gases. b. 6.65 3 104 g; c. 8.7 3 103 g 165. a. Due to air’s larger average molar When a balanced reaction has more moles of product gases than moles of mass, a given volume of air at a given set of conditions has a higher density reactant gases, then the reaction will expand in volume (DV is positive) and than helium. We need to heat the air to greater than 258C to lower the air the system does work on the surroundings (w , 0). When a balanced reac- density (by driving air out of the hot-air balloon) until the density is the tion has a decrease in the moles of gas from reactants to products, then the same as that for helium (at 258C and 1.00 atm). b. 2150 K 167. C H is the 3 8 reaction will contract in volume (DV is negative) and the surroundings does most likely identity. 169. row 7 171. 0.023 mol 173. 4.8 g/L; UF will 3 compression work on the system (w . 0). When there is no change in the effuse 1.02 times faster. moles of gas from reactants to products, then DV 5 0 and w 5 0. 105. a. C H O (s) 1 12O (g) n 12CO (g) 1 11H O(l); b. 25630 kJ/mol Chapter 6 12 22 11 2 2 2 107. 3.0 miles 109. 4.2 kJ heat released 111. The calculated DH value 13. Path-dependent functions for a trip from Chicago to Denver are those will be less positive (smaller) than it should be. 113. 25.918C quantities that depend on the route taken. One can fly directly from Chicago 115. 1C 1 F n A 1 B 1 D; 47.0 kJ 117. a. 632 kJ; b. C H ( g) 119. 28 g to Denver or one could fly from Chicago to Atlanta to Los Angeles and then 2 2 2 121. a. 2361 kJ; b. 2199 kJ; c. 2227 kJ; d. 2112 kJ 123. Work is done to Denver. Some path-dependent quantities are miles traveled, fuel consump- by the surroundings on the system when there is a compression. This tion of the airplane, time traveling, airplane snacks eaten, etc. State functions will occur when the moles of gas decrease when going from reactants are path-independent; they only depend on the initial and final states. Some to products. This will occur for reactions a and c. The other reactions have state functions for an airplane trip from Chicago to Denver would be longi- an increase in the moles of gas as reactants are converted to products. tude change, latitude change, elevation change, and overall time zone In these reactions, the system does PV work on the surroundings. change. 15. Both q and w are negative. 17. a. 446 kJ released; b. 446 kJ 125. 3.97 3 103 kJ heat is released 127. 51.5°C 129. An element in its released 19. exothermic, evolved standard state has a standard enthalpy of formation equal to zero. At 25°C 21. and 1 atm, chlorine is found as Cl (g) and hydrogen is found as H (g). So H O1l2 1 1CO 1g2 1CH 1g2 1 O 1g2 DH 5 21 12891 kJ2 2 2 2 2 2 S 2 4 2 1 2 these two elements (a and b) have enthalpies of formation equal to zero. The 1CH 1g2 1 O 1g2 1CO 1g2 1 H O1g2 DH 5 1 12803 kJ2 2 4 2 S 2 2 2 2 2 other two choices (c and d) do not have the elements in their standard state.

H2O1l2 S H2O1g2 DH 5 DH1 1 DH2 5 44 kJ The standard state for nitrogen is N2(g) and the standard state for chlorine is 2 4 23. The zero points for DHf8 values are elements in their standard state. All Cl2(g). 131. w 5 22940 J; DE 5 27.8 kJ 133. 25 m 135. 1 3 10 steps substances are measured in relationship to this zero point. 25. No matter 137. 56.9 kJ 139. q 5 DH 5 28.51 kJ, w 5 1.83 kJ, DE 5 26.68 kJ how insulated your thermos bottle, some heat will always escape into the 141. 1.74 kJ 143. 3.3 cm surroundings. If the temperature of the thermos bottle (the surroundings) is high, less heat initially will escape from the coffee (the system); as a result Chapter 7 your coffee will stay hotter for a longer period of time. 27. Fossil fuels 23. The equations relating the terms are vl 5 c, E 5 hv, and E 5 hcyl. From contain carbon; the incomplete combustion of fossil fuels produces CO(g) the equations, wavelength and frequency are inversely related, photon energy instead of CO2(g). This occurs when the amount of oxygen reacting is not and frequency are directly related, and photon energy and wavelength are in- sufficient to convert all of the carbon in fossil fuels to CO2. Carbon monox- versely related. The unit of 1 joule (J) 5 1 kg m2/s2. This is why you must ide is a poisonous gas to humans. 29. 150 J 31. a. Potential energy is change mass units to kg when using the de Broglie equation. 25. The photo energy due to position. Initially, ball A has a higher potential energy than ball electric effect refers to the phenomenon in which electrons are emitted from B because the position of ball A is higher than that of ball B. In the final the surface of a metal when light strikes it. The light must have a certain mini- position, ball B has the higher position, so ball B has the higher potential mum frequency (energy) in order to remove electrons from the surface of a energy. b. As ball A rolled down the hill, some of the potential energy lost by metal. Light having a frequency below the minimum results in no electrons A was converted to random motion of the components of the hill (frictional being emitted, whereas light at or higher than the minimum frequency does heating). The remainder of the lost potential energy was added to B to ini- cause electrons to be emitted. For light having a frequency higher than the tially increase its kinetic energy and then to increase its potential energy. minimum frequency, the excess energy is transferred into kinetic energy for the

57404_Ans_A31-A64.indd 36 10/14/16 5:12 PM Answers to Selected Exercises A37

emitted electron. Albert Einstein explained the photoelectric effect by applying 89. a. or quantum theory. 27. Example 7.3 calculates the de Broglie wavelength of a ball and of an electron. The ball has a wave length on the order of 10234 m. This 1s 2s 2p 3s 3s is incredibly short and, as far as the wave-particle duality is concerned, the b. wave properties of large objects are insignificant. The electron, with its tiny mass, also has a short wavelength: on the order of 10210 m. However, this 1s 2s 2p 3s 3p wavelength is significant as it is on the same order as the spacing between atoms in a typical crystal. For very tiny objects like electrons, the wave proper- or ties are important. The wave properties must be considered, along with the particle properties, when hypothesizing about the electron motion in an atom. 4s 3d 3d 29. The Bohr model was an important step in the development of the current c. quantum mechanical model of the atom. The idea that electrons can occupy only certain, allowed energy levels is illustrated nicely (and relatively easily). 1s 2s 2p 3s 3p We talk about the Bohr model to present the idea of quantized energy levels. 31. When the p and d orbital functions are evaluated at various points in space, the results sometimes have positive values and sometimes have negative values. The term phase is often associated with the 1 and 2 signs. For example, 4s 3d 4p a sine wave has alternating positive and negative phases. This is analogous to 91. Si: 1s22s22p63s23p2 or [Ne]3s23p2; Ga: 1s22s22p63s23p64s23d104p1 or 2 10 1 2 10 3 2 10 2 2 1 the positive and negative values (phases) in the p and d orbitals. 33. a. mℓ; [Ar]4s 3d 4p ; As: [Ar]4s 3d 4p ; Ge: [Ar]4s 3d 4p ; Al: [Ne]3s 3p ; Cd: 2 10 2 4 2 10 4 b. n; c. ms; d. ℓ 35. If one more electron is added to a half-filled subshell, [Kr]5s 4d ; S: [Ne]3s 3p ; Se: [Ar]4s 3d 4p 93. a. 2; b. 10; c. 10; d. 0 electron–electron repulsions will increase because two electrons must now oc- 95. a. I: [Kr]5s24d105p5; b. element 120: [Rn]7s25f 146d107p68s2; c. Rn: cupy the same atomic orbital. This may slightly decrease the stability of the [Xe]6s24f 145d106p6; d. Cr: [Ar]4s13d5 97. a. S; b. N; c. Se2 99. a. 18; atom. 37. The valence electrons are strongly attracted to the nucleus for ele- b. 30; c. 8; d. 40 ments with large ionization energies. One would expect these species to read- 101. B: 1s22s22p1 N: 1s22s22p3 ily accept another electron and have very exothermic electron affinities. The n , m, ms n , m, ms noble gases are an exception; they have a large ionization energy but an endo- 1 1 thermic electron affinity. Noble gases have a filled valence shell. The added 1s 1 0 0 12 1s 1 0 0 12 1 1 electron must go into a higher n value atomic orbital, which would have a sig- 1s 1 0 0 22 1s 1 0 0 22 nificantly higher energy. This is unfavorable. 39. For hydrogen and hydrogen- 2s 2 0 0 11 2s 2 0 0 11 like (one electron) ions, all orbitals with the same value of n have the same 2 2 1 1 energy. For polyatomic atoms/ions, the energy of the orbitals also depends 2s 2 0 0 22 2s 2 0 0 22 on . Because there are more nondegenerate energy levels for polyatomic atoms/ 1 1 , 2p 2 1 21 12 2p 2 1 21 12 ions as compared with hydrogen, there are many more possible electronic tran- 2p 2 1 0 11 sitions resulting in more complicated line spectra. 41. Yes, the maximum 2 1 number of unpaired electrons in any configuration corresponds to a minimum 2p 2 1 11 12 14 –1 in electron–electron repulsions. 43. a. n; b. n and ℓ 45. 4.5 3 10 s For boron, there are six possibilities for the 2p electron. For nitrogen, all the 47. 3.0 3 1010 s21, 2.0 3 10223 J/photon, 12 J/mol 49. 9.4 3 1014 s21 to 1 1 1 2 2p electrons could have ms 5 22 . 103. 1A: 1, ns , Li, 2s ; 2A: 2, ns , Ra, 15 21 24 1.1 3 10 s 51. Wave a has the longer wavelength (4.0 3 10 m). Wave b 7s2; 3A: 3, ns2np1, Ga, 4s24p1; 4A: 4, ns2np2, Si, 3s23p2; 5A: 5, ns2np3, Sb, 12 21 has the higher frequency (1.5 3 10 s ) and larger photon energy (9.9 3 5s25p3; 6A: 6, ns2np4, Po, 6s26p4; 7A: 7, ns2np5, Ts, 7s27p5; 8A: 8, ns2np6, Ne, 222 10 J). Since both of these waves represent electromagnetic radiation, they 2s22p6 105. none; an excited state; energy released 107. C, O, Si, S, Ti, both should travel at the same speed, c, the speed of light. Both waves repre- Ni, Ge, Se 109. Li (1 unpaired electron), N (3 unpaired electrons), Ni 23 sent infrared radiation. 53. 1.50 3 10 atoms 55. 427.7 nm 57. 276 nm (2 unpaired electrons), and Te (2 unpaired electrons) are all expected to be 211 234 227 59. a. 2.4 3 10 m; b. 3.4 3 10 m 61. 1.6 3 10 kg 63. a. 656.7 nm paramagnetic because they have unpaired electrons. 111. a. S , Se , Te; (visible); b. 486.4 nm (visible); c. 121.6 nm (ultraviolet) b. Br , Ni , K; c. F , Si , Ba 113. a. Te , Se , S; b. K , Ni , Br; 65. 4 c. Ba , Si , F 115. a. He; b. Cl; c. element 116; d. Si; e. Na1 3 2 14 4 22 ab 117. a. [Rn]7s 5f 6d ; b. W; c. SgO3 and SgO4 probably would form 2 (similar to Cr). 119. Se is an exception to the general ionization trend. There are extra electron–electron repulsions in Se because two electrons are E in the same 4p orbital, resulting in a lower ionization energy than expected. c 121. a. As we remove succeeding electrons, the electron being removed is closer to the nucleus, and there are fewer electrons left repelling it. The remaining electrons are more strongly attracted to the nucleus, and it takes 2 2 6 2 1 1 more energy to remove these electrons. b. Al: 1s 2s 2p 3s 3p . For I4, we 67. a. 6; b. n 5 4 to n 5 3; c. n 5 4 to n 5 1 69. n 5 1 n n 5 5, begin removing an electron with n 5 2. For I3, we remove an electron with l 5 95.00 nm; n 5 2 n n 5 6, l 5 410.4 nm; visible light has sufficient n 5 3 (the last valence electron). In going from n 5 3 to n 5 2, there is a energy for the n 5 2 n n 5 6 transition but does not have sufficient energy big jump in ionization energy because the n 5 2 electrons are closer to for the n 5 1 n n 5 5 transition. 71. n 5 1, 91.20 nm; n 5 2, 364.8 nm the nucleus on average than the n 5 3 electrons. Since the n 5 2 electrons 73. n 5 2 75. a. 5.79 3 1024 m; b. 3.64 3 10233 m; c. The diameter of an are closer, on average, to the nucleus, they are held more tightly and require H atom is roughly 1.0 3 1028 cm. The uncertainty in position is much larger a much larger amount of energy to remove compared to the n 5 3 electrons. than the size of the atom. d. The uncertainty is insignificant compared to the In general, valence electrons are much easier to remove than inner-core elec-

size of a baseball. 77. n 5 1, 2, 3, . . . ; , 5 0, 1, 2, . . . (n 2 1); m, 5 2,, trons. 123. a. C, Br; b. N, Ar; c. C, Br 125. Al (244), Si (2120), P (274), . . . , 22, 21, 0, 1, 2, . . . , 1,. 79. b. For , 5 3, m, can range from 23 to S (2200.4), Cl (2348.7); Based on the increasing nuclear charge, we would 13; thus 14 is not allowed. c. n cannot equal zero. d. , cannot be a negative expect the electron affinity (EA) values to become more exothermic as we go number. 81. c 2 gives the probability of finding the electron at that point. from left to right in the period. Phosphorus is out of line. The reaction for the 83. He: 1s2; Ne: 1s22s22p6; Ar: 1s22s22p63s23p6; each peak in the diagram EA of P is corresponds to a subshell with different values of n. Corresponding subshells P1g2 1 e2 P2 1g2 are closer to the nucleus for heavier elements because of the increased nu- S clear charge. 85. 3; 1; 5; 25; 16 87. a. 32; b. 8; c. 25; d. 10; e. 6 3Ne43s23p3 3 Ne43s23p4

57404_Ans_A31-A64.indd 37 10/14/16 5:12 PM A38 Answers to Selected Exercises

The additional electron in P2 will have to go into an orbital that already has 189. a. Fr: [Rn]7s1, Fr1: [Rn] 5 [Xe]6s24f145d106p6; b. 7.7 3 1022 Fr atoms; one electron. There will be greater repulsions between the paired electrons in c. 2.27790 3 10222 g P2, causing the EA of P to be less favorable than predicted based solely on attractions to the nucleus. 127. a. Se , S; b. I , Br , F , Cl 129. Electron– Chapter 8 electron repulsions are much greater in O2 than in S2 because the electron goes into a smaller 2p orbital versus the larger 3p orbital in sulfur. This re- 19. a. This diagram represents a polar covalent bond as in HCl. In a polar sults in a more favorable (more exothermic) EA for sulfur. 131. a. Se31(g) covalent bond, there is an electron-rich region (indicated by the red color) and n Se41(g) 1 e2; b. S2(g) 1 e2 n S22(g); c. Fe31(g) 1 e2 n Fe21(g); an electron-poor region (indicated by the blue color). In HCl, the more elec- d. Mg(g) n Mg1(g) 1 e2 133. C: 1s22s22p2; in the ground state, carbon tronegative Cl atom has a slightly greater ability to attract the bonding elec- has three different orbitals that hold electrons. Each orbital will have a sepa- trons, which in turn produces a dipole moment. b. This diagram represents an rate peak on the PES spectrum, so we should have three different peaks in the ionic bond as in NaCl. Here, the electronegativity differences between the Na spectrum. The peak at the lowest binding energy corresponds to the peak and Cl are so great that the valence electron of sodium is transferred to the with the easiest electrons to remove. This will be the 2p electrons. The peak chlorine atom. This results in the formation of a cation, an anion, and an ionic at the highest energy corresponds to the 1s orbital electrons since they will bond. c. This diagram represents a pure covalent bond as in H2. Both atoms have the largest binding energy. The area under each peak is directly related attract the bonding electrons equally, so there is no bond dipole formed. This to the number of electrons in each orbital. Since all orbitals in carbon have is illustrated in the electrostatic potential diagram as there is no one region two electrons, each peak area will be equal. This is shown on the PES spec- that is red nor one region that is blue (there is no specific partial negative end trum for carbon: three peaks with equal area. 135. potassium peroxide, and no specific partial positive end). 21. (NH4)2SO4 and Ca3(PO4)2 are com- 21 14 21 219 pounds with both ionic and covalent bonds. 23. Electronegativity increases K2O2; K unstable 137. 6.582 3 10 s ; 4.361 3 10 J 139. Yes; the ionization energy general trend decreases down a group and the atomic ra- left to right across the periodic table and decreases from top to bottom. dius trend increases down a group. The data in Table 7.8 confirm both of Hydrogen has an electronegativity value between B and C in the second row, these general trends. 141. No; lithium metal is very reactive. It will react and identical to P in the third row. Going further down the periodic table, somewhat violently with water, making it completely unsuitable for human H has an electronegativity value between As and Se (row 4) and identical to consumption. Lithium has a small first ionization energy, so it is more likely Te (row 5). It is important to know where hydrogen fits into the electronega- that the lithium prescribed will be in the form of a soluble lithium salt (a tivity trend, especially for rows 2 and 3. If you know where H fits into the 1 trend, then you can predict bond dipole directions for nonmetals bonded to soluble ionic compound with Li as the cation). 143. a. 6Li(s) 1 N2(g) n hydrogen. 25. For ions, concentrate on the number of protons and the number 2Li3N(s); b. 2Rb(s) 1 S(s) n Rb2S(s) 145. a. 6; b. O, S, Se, and Te; c. K2X would be the predicted formula, where X is the unknown nonmetal. of electrons present. The species whose nucleus holds the electrons most tightly d. smaller; e. smaller 147. 386 nm 149. 200 s 151. l 5 4.104 3 will be smallest. For example, compare the size of an anion to the neutral atom. 1025 cm so violet light is emitted. 153. a. true for H only; b. true for all The anion has more electrons held by the same number of protons in the atoms; c. true for all atoms 155. 1p: n 5 1,  5 1 is not possible; 3f: n 5 3, nucleus. These electrons will not be held as tightly, resulting in a bigger size for  5 3 is not possible; 2d: n 5 2,  5 2 is not possible. In all three incorrect the anion as compared to the neutral atom. For isoelectronic ions, the same cases, n 5 . The maximum value  can have is n 2 1, not n. 157. smallest number of electrons are held by different numbers of protons in the various ions. size: O; largest size: K; smallest 1st ionization energy: K; largest 1st ioniza- The ion with the most protons holds the electrons tightest and has smallest size. tion energy: N 159. Valence electrons are easier to remove than inner-core 27. Fossil fuels contain a lot of carbon and hydrogen atoms. Combustion of fossil fuels (reaction with O2) produces CO2 and H2O. Both these compounds electrons. The large difference in energy between I2 and I3 indicates that this element has two valence electrons. The element is most likely an alkaline have very strong bonds. Because much stronger product bonds are formed than earth metal because alkaline earth metals have two valence electrons. reactant bonds broken, combustion reactions are very exothermic. 161. a. 146 kJ; b. 407 kJ; c. 1117 kJ; d. 1524 kJ 163. 251 nm 165. 15 29. qOC 167. The number of electrons with  5 1 is 15. The number of electrons with Carbon: formal charge 5 21; oxygen: formal charge 5 11 m 5 0 is 15. The number of electrons with m 5 1 is 7. 169. a. Te; b. Ge; c. F 171. a. true; b. false; cations are smaller than the parent atom; c. false; Electronegativity predicts the opposite polarization with the partial negative all ionization energies are endothermic; d. true; e. true; this is hard to predict end of the molecule around the more electronegative oxygen atom. The two because the general atomic radius trends go against each other when going opposing effects seem to partially cancel to give a much less polar molecule from Li to Al. Figure 7.37 of the text was used to answer this question. than expected. 173. S 175. a. line A, n 5 6 to n 5 3; line B, n 5 5 to n 5 3; b. 121.6 nm 31. a. C , N , O; b. Se , S , Cl; c. Sn , Ge , Si; d. Tl , Ge , S 51 2 33. a. GeOF; b. POCl; c. SOF; d. TiOCl 35. Order of electronegativity 177. X 5 C, m 5 5; C is the ion. 179. For r 5 a0 and u 5 08, c 5 2.46 28 2 from Fig. 8.3: a. C (2.5) , N (3.0) , O (3.5), same; b. Se (2.4) , S (2.5) , Cl 3 10 . For r 5 a0 and u 5 908, c 5 0. As expected, the xy plane is a node (3.0), same; c. Si 5 Ge 5 Sn (1.8), different; d. Tl (1.8) 5 Ge (1.8) , S (2.5), for the 2pz atomic orbital. different. Most polar bonds using actual electronegativity values: a. SiOF and 181. a. 1 2 GeOF equal polarity (GeOF predicted); b. POCl (same as predicted); 3 4 c. SOF (same as predicted); d. TiOCl (same as predicted) 37. Incorrect: d2 d1 d2 d1 5 6 7 8 9 10 11 12 b, d, e; b. ClOI ; d. nonpolar bond so no dipole moment; e. OOP 39. a. ionic; b. covalent; c. polar covalent; d. ionic; e. polar covalent; f. cova- 13 14 15 16 17 18 19 20 lent 41. FOH . OOH . NOH . COH . POH 43. a. Has a permanent b. 2, 4, 12, and 20; c. There are many possibilities. One example of each for- dipole; b. has no permanent dipole; c. has no permanent dipole; d. has a permanent dipole; e. has no permanent dipole; f. has no permanent dipole. mula is XY 5 1 1 11, XY2 5 6 1 11, X2Y 5 1 1 10, XY3 5 7 1 11, and 45. Al31: [He]2s22p6; Ba21: [Kr]5s24d105p6; Se22: [Ar]4s23d104p6; I2: X2Y3 5 7 1 10; d. 6; e. 0; f. 18 183. The ratios for Mg, Si, P, Cl, and Ar are 2 10 6 1 32 about the same. However, the ratios for Na, Al, and S are higher. For Na, the [Kr]5s 4d 5p 47. a. Li and N are the expected ions. The formula of the 31 22 second IE is extremely high because the electron is taken from n 5 2 (the first compound would be Li3N (lithium nitride). b. Ga and O ; Ga2O3, 1 2 electron is taken from n 5 3). For Al, the first electron requires a bit less energy gallium(III) oxide or gallium oxide; c. Rb and Cl ; RbCl, rubidium chloride; 21 22 21 31 2 2 6 1 than expected by the trend due to the fact it is a 3p electron. For S, the first d. Ba and S ; BaS, barium sulfide 49. a. Mg and Al : 1s 2s 2p ; K : 2 2 6 2 6 32 22 2 2 2 6 22 2 10 6 electron requires a bit less energy than expected by the trend due to electrons 1s 2s 2p 3s 3p ; b. N , O , and F : 1s 2s 2p ; Te : [Kr]5s 4d 5p 51. a. Sc31; b. Te22; c. Ce41, Ti41; d. Ba21 53. a. F2, O22, or N32; b. Cl2, being paired in one of the p orbitals. 185. a. Zeff 5 43.33; b. Silver is element S22, or P32; c. F2, O22, or N32; d. Br2, Se22, or As32 55. Se22, Br2, Rb1, 47, so Z 5 47 for silver. Our calculated Zeff value is less than 47. Electrons in 21 31 41 other orbitals can penetrate the 1s orbital. Thus a 1s electron can be slightly Sr , Y , and Zr are some ions that are isoelectronic with Kr. The size trend is shielded from the nucleus, giving a Zeff close to but less than Z. 187. Solving for the molar mass of the element gives 40.2 g/mol; this is calcium. Zr41 , Y31 , Sr21 , Rb1 , Br2 , Se22.

57404_Ans_A31-A64.indd 38 10/14/16 5:12 PM Answers to Selected Exercises A39

57. a. Cu . Cu1 . Cu21; b. Pt21 . Pd21 . Ni21; c. O22 . O2 . O; d. La31 – – 93. a. – ššš mn š š š . Eu31 . Gd31 . Yb31; e. Te22 . I2 . Cs1 . Ba21 . La31 59. a. NaCl, NO2 P NO O Oð ðO O N P O Na1 smaller than K1; b. LiF, F2 smaller than Cl2; c. MgO, O22 greater charge

ð – than OH2; d. Fe(OH) , Fe31 greater charge than Fe21; e. Na O, O22 greater – – – ð 3 2 NO3 Oðð ðOð Oðð charge than Cl2; f. MgO, Mg21 smaller than Ba21, and O22 smaller than S22. B A A 61. 2411 kJ/mol 63. The lattice energy for Mg21O22 will be much more N mn N mn N D G J G MD exothermic than for Mg1O2. 65. 161 kJ/mol 67. Ca21 has greater charge ýOOý ý½OOÂ Â½ ý ýOOÂ Â½ ý Â½ Â than Na1, and Se22 is smaller than Te22. Charge differences affect lattice energy values more than size differences, and we expect the trend from most N O exothermic to least exothermic to be: 2 4 O O O O NN NN CaSe . CaTe . Na Se . Na Te 2 2 O O O O (22862) (22721) (22130) (22095) 69. a. 2183 kJ; b. 2109 kJ 71. 242 kJ 73. 21282 kJ 75. 21228 kJ 77. 485 kJ/mol 79. a. Using standard enthalpies of formation, DH8 5 2184 kJ vs. 2183 kJ from bond energies; b. Using standard enthalpies of O O O O formation, DH 5 292 kJ vs. 2109 kJ from bond energies. Bond energies NN NN give a reasonably good estimate for DH, especially when all reactants and O O O O products are gases. 81. a. Using SF4 data: DSF 5 342.5 kJ/mol. Using SF6 data: DSF 5 327.0 kJ/mol. b. The SOF bond energy in the table is 327 kJ/mol. The value in the table was based on the SOF bond in SF . c. S(g) and F(g) – – – 6 b. OCN– ðš O CO q Nð š P CO P Nš ð qCO ONš ð are not the most stable forms of the elements at 258C. The most stable forms are S (s) and F (g); DH8 5 0 for these two species. 83. 158.4 kJ/mol vs. 8 2 f – – – 160. kJ/mol in Table 8.5 – š š š š SCN ð O CS q Nð P CS P N ð qCS ON ð 85. a. FOF b. OPO c. CqO – – – – š š š š H N3 ð O NN q Nð P NN P N ð qNN ON ð A d. HCO OH e. HONOH f. HOOOH g. HOF A A 95. H H H A A H H H H H HEENC HEKHC C C C C 87. a. Cl b. ClOON Cl B A A B A A EC CH EC CH ClOOC Cl Cl H CKH H H CEN H A A A Cl H H 97. With resonance all carbon–carbon bonds are equivalent (we indicate this c. ClOOSe Cl d. IOCl with a circle in the ring), giving three different structures: 89. H—Be—H H Cl Cl Cl B Cl HH Cl 91. F F Cl PF5 P F F Localized double bonds would give four unique structures. F 99. H H ýk H C NNCPP O mn H C POCOO mn F F H H SF4 S F H F H C NOCqO

F H 22 ClF 101. Longest n shortest COO bond: CH3OH . CO3 . CO2 . CO; 3 Cl F 22 weakest n strongest COO bond: CH3OH , CO3 , CO2 , CO F 103. F +1 F 0 B A – _ B –1 B 0 Br Br O Br O Br D G D G 3 0 F F 0 0 F F 0

Row 3 and heavier nonmetals can have more than 8 electrons around them The first Lewis structure obeys the octet rule but has a 11 formal charge on the when they have to. Row 3 and heavier elements have empty d orbitals that most electronegative element there is, fluorine, and a negative formal charge are close in energy to valence s and p orbitals. These empty d orbitals can on a much less electronegative element, boron. This is just the opposite of what accept extra electrons. we expect: negative formal charge on F and positive formal charge on B. The other Lewis structure does not obey the octet rule for B but has a zero formal

57404_Ans_A31-A64.indd 39 10/14/16 5:12 PM A40 Answers to Selected Exercises

charge on each element in BF . Since structures generally want to minimize for- 3 ðšFð mal charge, BF3 with only single bonds is best from a formal charge point of view. A 105. a–f and h all have similar Lewis structures: c. CF Tetrahedral, nonpolar 4 EHC ðšFFA šð ðFð ðšYð A ðš XOO YY šð A SeF ð OOSeF Fð See-saw, polar š š 4 DG ðY ð FFðð ðð š

– FF – g. ClO3 ðOCO lOOð A KrF4 Kr Square planar, nonpolar Oðð FF

Formal charges: a. 11; b. 12; c. 13; d. 11; e. 12; f. 14; g. 12; h. 11 ðšFð ðšF šFð D AD 107. F O O F d. IF5 I Square pyramid, polar G G ðšF šFð Formal charge: 0 0 0 0 Oxidation number: Ϫ1 ϩ1 ϩ1 Ϫ1 ððF ðF Oxidation numbers are more useful. We are forced to assign 11 as the oxida- G A tion number for oxygen. Oxygen is very electronegative and 11 is not a AsF5 AsO Fð Trigonal bipyramid, nonpolar D A stable oxidation state for this element. ðF Fðð 109. Cl S S Cl 129. Element E is a halogen (F, Cl, Br, or I); trigonal pyramid; ,109.58

111. H qNC H 131. GðFð Â½ Â

H C C CP O C H P NPN NPN DG D Â O H ðFFð Fðð 113. [87] a. tetrahedral, 109.58; b. trigonal pyramid, ,109.58; c. V-shaped, Polar Nonpolar 2 1 2 2 1 1 ,109.58; d. linear, no bond angle in diatomic molecules. [93] a. NO2 : 133. a. radius: N , N , N ; I.E.: N , N , N ; b. radius: Cl , Cl , Se 2 2 2 1 21 1 2 V‑shaped, ,1208; NO3 : trigonal planar, 1208; N2O4: trigonal planar about both , Se ; I.E.: Se , Se , Cl , Cl ; c. radius: Sr , Rb , Br ; I.E.: 2 2 1 21 N atoms, 1208; b. all are linear, 1808 115. Br3 : linear; ClF3: T-shaped; SF4: Br , Rb , Sr ; 135. a. 1549 kJ; b. 1390. kJ c. 1312 kJ; d. 1599 kJ see-saw 117. a. V-shaped or bent; b. see-saw; c. trigonal pyramid; d. trigo- 137. a. NaBr: In NaBr2, the sodium ion would have a 12 charge assuming nal bipyramid; e. tetrahedral 119. a. trigonal planar; 1208; b. V-shaped; each bromine has a 21 charge. Sodium doesn’t form stable Na21 com- 2 ,1208 121. a. linear; 1808; b. T-shaped; ,908; c. see-saw; ,908 and ,1208; pounds. b. ClO4 : ClO4 has 31 valence electrons so it is impossible to satisfy d. trigonal bipyramid; 908 and 1208 123. SeO2 (bond dipoles do not cancel the octet rule for all atoms in ClO4. The extra electron from the 21 charge in 2 each other out in SeO2) 125. ICl3 and TeF4 (bond dipoles do not cancel each ClO4 allows for complete octets for all atoms. c. XeO4: We can’t draw a other out in ICl3 and TeF4) Lewis structure that obeys the octet rule for SO4 (30 electrons), unlike XeO4 (32 electrons). d. SeF4: Both compounds require the central atom to expand 127. a. OCl V-shaped, polar 2 O its octet. O is too small and doesn’t have low-energy d orbitals to expand its DG octet (which is true for all row 2 elements). 139. a. Both have one or more Cl Cl 1808 bond angles; both are made up entirely of Xe and Cl; both have the individual bond dipoles arranged so they cancel each other (both are nonpolar); KrF š ýk š Linear, nonpolar 2 ðFFOOKr ð both have lone pairs on the central Xe atom; both have a central Xe atom that has more than 8 electrons around it. b. All have lone pairs on the central

BeH2 HOBeOH Linear, nonpolar atom; all have a net dipole moment (all are polar). 141. 25681 kJ 143. Yes, each structure has the same number of effective pairs around the central SO šS V-shaped, polar atom. (We count a multiple bond as a single group of electrons.) 2 JG ý k – OOÂ½Â 145. F F F

(one other resonance structure possible) GDA GDTe ððO F F B b. SO S Trigonal planar, nonpolar 3 The lone pair of electrons around Te exerts a stronger repulsion than the ý DG k ½ÂOO½ Â bonding pairs. The stronger repulsion pushes the four square planar F atoms (two other resonance structures possible) away from the lone pair, reducing the bond angles between the axial F atom and the square planar F atoms. 147. a. nonpolar covalent; b. ionic; c. ionic; d. polar covalent; e. polar covalent; f. polar covalent; g. polar covalent; NF3 N Trigonal pyramid, polar EH h. ionic 149. a. O22 . O2 . O; b. Fe21 . Ni21 . Zn21; c. Cl2 . K1 ðšFFA šð 21 Fðð . Ca 151. 2303 kJ 153. CO2: carbon dioxide, linear; NH3: ammonia (nitrogen trihydride), trigonal pyramid; SO3: sulfur trioxide, trigonal planar; H O: water (dihydrogen monoxide), V-shaped or bent; ClO 2: perchlorate F 2 4 ion, tetrahedral 155. a. 25 3 10218 J; b. 26 3 10218 J 157. 17 kJ/mol IF3 IF T-shaped, polar 159. See Fig. 8.11 to see the data supporting MgO as an ionic compound. F Note that the lattice energy is large enough to overcome all of the other pro-

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cesses (removing 2 electrons from Mg, and so on). The bond energy for O2 Note: OH, CH, H2C, CH2, and CH3 are short-hand notation. In this notation, (247 kJ/mol) and electron affinity (737 kJ/mol) are the same when making single bonds to the various H atoms are assumed. Most of the bond angles in CO. However, the energy needed to ionize carbon to form a C21 ion must be the molecule are predicted to be 109.5 degrees because most of the carbon too large. Figure 7.32 shows that the first ionization energy for carbon is atoms exhibit tetrahedral geometry. The exceptions are the two carbons with about 350 kJ/mol greater than the first IE for magnesium. If all other num- the *. These two carbons exhibit trigonal planar geometry and are predicted bers were equal, the overall energy change would be 250 kJ/mol (see Fig. to have 120-degree bond angles. Because most of the carbon atoms exhibit 8.11). It is not unreasonable to assume that the second ionization energy for tetrahedral geometry, the molecule is not planar. 171. X is iodine; square carbon is more than 250 kJ/mol greater than the second ionization energy for pyramid 173. [Ar]4s13d5 5 Cr; [Ne]3s23p3 5 P; [Ar]4s23d104p3 5 As; magnesium. This would result in a positive DH value for the formation of CO [Ne]3s23p5 5 Cl; Cr , As , P , Cl as an ionic compound. One wouldn’t expect CO to be ionic if the energetics were unfavorable. 161. As the halogen atoms get larger, it becomes more Chapter 9 difficult to fit three halogen atoms around the small nitrogen atom, and the 11. In hybrid orbital theory, some or all of the valence atomic orbitals of the NX3 molecule becomes less stable. 163. reaction i: 22636 kJ; reaction ii: central atom in a molecule are mixed together to form hybrid orbitals; these 23471 kJ; reaction iii: 23543 kJ; Reaction iii yields the most energy per kg hybrid orbitals point to where the bonded atoms and lone pairs are oriented. (28085 kJ/kg) The s bonds are formed from the hybrid orbitals overlapping head to head 165. HCOON with an appropriate orbital on the bonded atom. The p bonds in hybrid orbital M CHO theory are formed from unhybridized p atomic orbitals. The p orbitals overlap D side to side to form the p bond where the p electrons occupy the space above CONOH A and below a line joining the atoms p the internuclear axis). Assuming the HCO OH z-axis is the internuclear axis, then the pz atomic orbital will always be hybrid- A ized whether the hybridization is sp, sp2, sp3, dsp3 or d2sp3. For sp hybridiza- HOON COCOOOH A A B tion, the px and py atomic orbitals are unhybridized; they are used to form two 2 H H O p bonds to the bonded atom(s). For sp hybridization, either the px or py We would expect 1208 bond angles about the carbon atom labeled 1 and atomic orbital is hybridized (along with the s and pz orbitals); the other p or- 3 ,109.58 bond angles about the nitrogen atom labeled 2. The nitrogen bond bital is used to form a p bond to a bonded atom. For sp hybridization, the angles should be slightly smaller than 109.58 due to the lone pair of electrons s and all of the p orbitals are hybridized; no unhybridized p atomic orbitals are 3 3 on nitrogen. present, so typical p bonds do not form with sp hybridization. For dsp and d2sp3 hybridization, we just mix in one or two d orbitals into the hybridization 167. a. Two possible structures exist; each has a T-shaped molecular structure: process. Which specific d orbitals are used is not important to our discussion. F I 13. We use d orbitals when we have to; i.e., we use d orbitals when the central atom on a molecule has more than eight electrons around it. The d orbitals are

Br I Br F necessary to accommodate the electrons over eight. Row 2 elements never I I have more than eight electrons around them so they never hybridize d orbitals. We rationalize this by saying there are no d orbitals close in energy to the 90° bond angle between I atoms 180° bond angle between I atoms valence 2s and 2p orbitals (2d orbitals are forbidden energy levels). However, b. Three possible structures exist; each has a see-saw molecular structure: for row 3 and heavier elements, there are 3d, 4d, 5d, etc. orbitals that will be close in energy to the valence s and p orbitals. It is row 3 and heavier nonmet- O O F O F O als that hybridize d orbitals when they have to. For sulfur, the valence elec- Xe Xe Xe trons are in 3s and 3p orbitals. Therefore, 3d orbitals are closest in energy and F F O are available for hybridization. Arsenic would hybridize 4d orbitals to go with F O F the valence 4s and 4p orbitals while iodine would hybridize 5d orbitals since 90° bond angle 180° bond angle 120° bond angle the valence electrons are in n 5 5. between O atoms between O atoms between O atoms š š 15. P CO P O c. Three possible structures exist; each has a square pyramid molecular struc- sp ture: The darker green orbitals about carbon are hybrid orbitals. The lighter green orbitals about each oxygen are sp2 hybrid orbitals, and the gold orbit- – – – F Cl Cl als about all of the atoms are unhybridized p atomic orbitals. In each double Cl Cl Cl F F Cl bond in CO2, one s and one p bond exists. The two carbon—oxygen s Te Te Te bonds are formed from overlap of sp hybrid orbitals from carbon with a sp2 Cl F Cl F Cl F hybrid orbital from each oxygen. The two carbon—oxygen p bonds are formed from side-to-side overlap of the unhybridized p atomic orbitals from One F atom is 180° Both F atoms are 90° Both F atoms are 90° carbon with an unhybridized p atomic orbital from each oxygen. These two from lone pair. from lone pair and 90° from lone pair and 180° p bonds are oriented perpendicular to each other as illustrated in the figure. from each other. from each other. 17. Bonding and antibonding molecular orbitals are both solutions to the

169. CH3 CH2 CH2 CH3 quantum mechanical treatment of the molecule. Bonding orbitals form when in-phase orbitals combine to give constructive interference. This results in CH CH2 CH H H H enhanced electron probability located between the two nuclei. The end result CH3 CH3 is that a bonding MO is lower in energy than the atomic orbitals of which it C C is composed. Antibonding orbitals form when out-of-phase orbitals com- H C C CH 2 2 bine. The mismatched phases produce destructive interference leading H H H CH3 H to a node in the electron probability between the two nuclei. With electron C C C CH2 distribution pushed to the outside, the energy of an antibonding orbital is H2C C C H higher than the energy of the atomic orbitals of which it is composed. H 19. The localized electron model does not deal effectively with molecules con- C *C CH taining unpaired electrons. We can draw all of the possible resonance structures * 2 HO C C for NO, but still not have a good feel for whether the bond in NO is weaker or H H stronger than the bond in NO2. MO theory can handle odd electron species H without any modifications. In addition, hybrid orbital theory does not predict that NO2 is paramagnetic. The MO theory correctly makes this prediction.

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ð ð 3 21. H2Oð O f. f a <1208, see-saw, dsp , polar

D D bððF ðF f HH Hf A b <908

a f Te ð H O has a tetrahedral arrangement of the electron pairs about the O atom that E A 2 ðF requires sp3 hybridization. Two of the sp3 hybrid orbitals are used to form ððF bonds to the two hydrogen atoms, and the other two sp3 hybrid orbitals hold the two lone pairs on oxygen.

g. Fðð a 5 908, trigonal bipyramid, dsp3, nonpolar ðF f f a G A b 5 1208

23. H2COð ððO f b f AsOFð B D A C ðF D D ððF HH The central carbon atom has a trigonal planar arrangement of the electron h. FOýkKrO F linear, 1808, dsp3, nonpolar pairs that requires sp2 hybridization. Two of the sp2 hybrid orbitals are used ð ð to form the two bonds to hydrogen. The other sp2 hybrid orbital forms the s bond to oxygen. The unchanged (unhybridized) p orbital on carbon is used to i. Fðð square planar, 908, d2sp3, nonpolar form the p bond between carbon and oxygen. D ðOKrF OFð 25. Ethane: H H D Fðð H C C H j. octahedral, 908, d2sp3, nonpolar H H ððF Fð AD The carbon atoms are sp3 hybridized. The six C H bonds are formed from ðFFSe OOOO ð O the sp3 hybrid orbitals on C with the 1s atomic orbitals from the hydrogen D A ðF Fðð atoms. The carbon–carbon bond is formed from an sp3 hybrid orbital on each C atom. k. F l. F Ethanol: H H F F I IF H C C O H F F F H H The two C atoms and the O atom are all sp3 hybridized. All bonds are formed square pyramid, <908, T-shaped, <908, 2 3 3 from these sp3 hybrid orbitals. The COH and OOH bonds form from sp3 d sp , polar dsp , polar hybrid orbitals and the 1s atomic orbitals from the hydrogen atom. The 35. The p bond forces all six atoms into the same plane. 37. a. There are 2 COC and COO bonds are formed from sp3 hybrid orbitals on each atom. 33 s and 9 p bonds. b. All C atoms are sp hybridized since all have a trigo- 3 2 2 2 2 nal planar arrangement of the electrons. 39. Biacetyl: 27. [87] All are sp hybridized. [93] a. NO2 , sp ; NO3 , sp ; N2O4, both N atoms are sp2 hybridized; b. All are sp hybridized. 29. All exhibit dsp3 ýkO

2 sp2

B h hybridization. 31. The molecules in Exercise 119 all exhibit sp hybridi H h š zation about the central atom; the molecules in Exercise 120 all exhibit sp3 AEHCOK ð

HOCCh 3h hybridization about the central atom. 33. a. tetrahedral, 109.58, sp3, non A sp A polar H HOCOH A H Fðð A All CCO angles are 1208. The six atoms are not forced to lie in the same EHC plane. 11 s and 2 p bonds; Acetoin: ðFFA ð F 2

ðð sp ýkO Bf f

3 H H a

b. trigonal pyramid, ,109.58, sp , polar E H

A A f C f f E E OOCH f CCh A Af b A E ð ONF sp3 H DG H ðOð H FF G ðð ð sp3 H 5 8 5 8 s p c. V-shaped, ,109.58, sp3, polar angle a 120 , angle b 109.5 , 13 bonds and 1 bond 41. a. H ýkO N DG A a e x ðFFk C HON D k ý D f g CON c O H2CPC M MJ D d 2 O b NOC COO OCH3 d. trigonal planar, 1208, sp , nonpolar D B h NH 2 O Fðð Azodicarbonamide Methyl cyanoacrylate A B Note: NH2, CH2, and CH3 are shorthand for carbon atoms singly bonded to DG Fk Fk hydrogen atoms. b. In azodicarbonamide, the two carbon atoms are sp2 hybrid- ý kk ý ized. The two nitrogen atoms with hydrogens attached are sp3 hybridized and e. linear, 1808, sp, nonpolar the other two nitrogens are sp2 hybridized. In methyl cyanoacrylate, the 3 CH3 carbon is sp hybridized, the carbon with the triple bond is sp hybridized, HOBeOH and the other three carbons are sp2 hybridized. c. Azodicarbonamide contains

57404_Ans_A31-A64.indd 42 10/14/16 5:12 PM Answers to Selected Exercises A43

3 p bonds and methyl cyanoacrylate contains 4 p bonds. d. a: 109.5°, b: 1208, c. Square pyramid; d2sp3 c: 120°, d: 1208, e: 1808, f: 1208, g: 109.5°, h: 1208 43. To complete the O F Lewis structure, add lone pairs to complete octets for each atom. a. 6; b. 4; F F F F c. The center N in ONPNPN group; d. 33 s; e. 5 p; f. 1808; g. <109.58; Xe or Xe h. sp3 45. a. The bonding molecular orbital is on the right and the antibond- F F F O ing molecular orbital is on the left. The bonding MO has the greatest electron probability between the nuclei, while the antibonding MO has greatest elec- d. T-shaped; dsp3 tron probability on either side of the nuclei. b. The bonding MO is lower in F O energy. Because the electrons in the bonding MO have the greatest probability of lying between the two nuclei, these electrons are attracted to two XeO or F Xe different nuclei, resulting in a lower energy. 47. a. H 1, H , H 2; b. He 21 2 2 2 2 F F 1 2 and He2 49. a. (s2s) ; B.O. 5 1; diamagnetic (0 unpaired electrons); 2 2 4 2 b. (s2s) (s2s*) (p2p) ; B.O. 5 2; diamagnetic (0 unpaired electrons); c. (s3s) e. Trigonal bipyramid; dsp3 2 2 4 2 (s3s*) (s3p) (p3p) (p3p*) ; B.O. 5 2; paramagnetic (2 unpaired electrons) 1 2 F F O 51. N2 and N2 both have bond orders of 2.5. 53. The peroxide ion has a O O F bond order of 1 and the superoxide ion has a bond order of 1.5. The superoxide Xe O or Xe F or Xe F ion with the larger bond order should have the shorter (and stronger) bond. O O O 2 2 4 2 2 2 4 F O O 55. a. (s2s) (s2s*) (p2p) (s2p) ; B.O. 5 3; diamagnetic; b. (s2s) (s2s*) (p2p) 1 2 2 4 (s2p) ; B.O. 5 2.5; paramagnetic; c. (s2s) (s2s*) (p2p) ; B.O. 5 2; diamagnetic; 69. a. No, some atoms are attached differently; b. Structure 1: All N atoms 1 21 21 1 bond length: CO , CO , CO ; bond energy: CO , CO , CO 57. H2; are sp3 hybridized, all C atoms are sp2 hybridized; structure 2: All C and N 22 B2; C2 atoms are sp2 hybridized; c. The first structure with the carbon–oxygen dou- 59. ble bonds is slightly more stable. 71. ϩ ϩ O O C H C H Ϫ ϩ ϩ Ϫ H H H H OCCOH OCCOH Ϫ Ϫ H C C H C C H mn H CH C C CH C C N H N H H H 61. a. The electrons would be closer to F on the average. The F atom is more electronegative than the H atom, and the 2p orbital of F is lower in energy a. 21 s bonds; 4 p bonds (The electrons in the 3 p bonds in the ring are de- than the 1s orbital of H; b. The bonding MO would have more fluorine 2p localized.) b. angles a, c, and g: <109.58; angles b, d, e, and f: <1208; character because it is closer in energy to the fluorine 2p orbital; c. The anti- c. 6 sp2 carbons; d. 4 sp3 atoms; e. Yes, the p electrons in the ring are delocal- bonding MO would place more electron density closer to H and would have ized. The atoms in the ring are all sp2 hybridized. This leaves a p orbital a greater contribution from the higher-energy hydrogen 1s atomic orbital. perpendicular to the plane of the ring from each atom. Overlap of all six of 63. [:C { C:]22 sp hybrid orbitals form the s bond and the two unhybridized these p orbitals results in a p molecular orbital system where the electrons p atomic orbitals from each carbon form the two p bonds. are delocalized above and below the plane of the ring (similar to benzene in Fig. 9.47 of the text). 73. 267 kJ/mol; this amount of energy must be sup- MO: (s )2(s *)2(p )4(s )2, B.O. 5 (8 2 2)/2 5 3 2s 2s 2p 2p plied to break the p bond. Both give the same picture, a triple bond composed of a s bond and two p 75. a. H Trigonal planar, nonpolar, 1208, sp2 bonds. Both predict the ion to be diamagnetic. Lewis structures deal well with diamagnetic (all electrons paired) species. The Lewis model cannot really pre- B 2 dict magnetic properties. 65. O3 and NO2 have identical Lewis structures, HH so we need to discuss only one of them. The Lewis structures for O3 are b. N2F2

O O Â K H ½ mn ½ H K Â GðFð OÂ ½OÂ Â O ½ Â O Â½ Â NPN NPNÂ Localized electron model: The central oxygen atom is sp2 hybridized, which DG D ðFFð Fðð is used to form the two s bonds and hold the lone pair of electrons. An unchanged (unhybridized) p atomic orbital forms the p bond with the neigh- Polar Nonpolar boring oxygen atoms. The actual structure of O3 is an average of the two V-shaped about both N atoms, <1208, sp2 resonance structures. Molecular orbital model: There are two localized s bonds and a p bond that is delocalized over the entire surface of the mole- These are distinctly different molecules. cule. The delocalized p bond results from overlap of a p atomic oribtal on c. H Trigonal planar about all C atoms, each oxygen atom in O . 3 H H nonpolar, 1208, sp2 67. a. Trigonal pyramid; sp3 C C C C H H Xe H O O O 77. a. The NNO structure is correct. From the Lewis structures we would predict both NNO and NON to be linear, but NON would be nonpolar. NNO 3 b. Tetrahedral; sp is polar. O b. ý ý ½NNPPO ½ mn ðNNq OmOð n ð NN O qO ð Xe –1 +1 0 0+1–1 –2 +1 +1 Formal O O charges O

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The central N is sp hybridized. We can probably ignore the third resonance All of the carbon atoms in the rings have a trigonal planar arrangement of structure on the basis of formal charge. c. sp hybrid orbitals on the center N electron pairs, so these five carbon atoms are sp2 hybridized. The other overlap with atomic orbitals (or hybrid orbitals) on the other two atoms to three carbon atoms are sp3 hybridized because they have a tetrahedral ar- form two s bonds. The remaining p orbitals on the center N overlap with rangement of electron pairs. The three nitrogen atoms that are bonded to a 3 p orbitals on the other N to form two p bonds. 79. In O2, an antibonding OCH3 group all have the CON bond formed from overlap of an sp hybrid electron is removed, which will increase the bond order to 2.5 [5 (8 2 3)/2]. orbital from carbon with an sp3 hybrid orbital from nitrogen. Note that each The bond order increases as an electron is removed, so the bond strengthens. of these nitrogen atoms has a tetrahedral arrangement of electron pairs, 3 In N2, a bonding electron is removed, which decreases the bond order to hence they are sp hybridized. There are eight lone pairs of electrons, and 2.5 5 [(7 2 2)/2]. So the bond strength weakens as an electron is removed the molecule has four p bonds (each double bond contains one p bond). 2 2 2 4 4 2 2 4 2 1 2 2 4 1 2 2 from N2. 81. F2: (s2s) (s2s*) (s2p) (p2p) (p2p*) ; F2 should have a lower 91. N2: (s2s) (s2s*) (p2p) (s2p) ; N2 : (s2s) (s2s*) (p2p) (s2p) ; N2 : (s2s) 2 4 2 1 ionization energy than F. The electron removed from F2 is in a p2p* antibond- (s2s*) (p2p) (s2p) (p2p*) ing molecular orbital, which is higher in energy than the 2p atomic orbitals 93. H from which the electron in atomic fluorine is removed. Since the electron re- A O moved from F2 is higher in energy than the electron removed from F, it should H H HCO OH be easier to remove an electron from F than from F. 83. p molecular orbital A 2 C C N N C 85. šSe šSe H CHO JG JG ýOOk mn ýOOk C C Â½Â ½Â ½ O M N N A HCO OH P A EHA š H ðClš Clð Clðð 6 C and N atoms exhibit 120° bond angles; 6 C and N atoms are sp3 hybridized; Two other resonance NNO 0 C and N atoms are sp hybridized; PP structures are possible. 25 σ bonds and 4 π bonds

Two other resonance 95. a. NCN22: SCPPO structures are possible. 2– 2– 2– ý NCP PmN ý n ðNCq OmNð n ð NCO q N ð ½ ½ P EH H +100 –1 H 00 ðšFFA šð G G Fðð H NCN: NPCPmNk n NOCqNð 2 D Â D Â H H All of these compounds are polar. In each compound, the bond dipoles do not

cancel each other out, leading to a polar compound. SeO2 is the only com- Favored by formal charge pound exhibiting ,120-degree bond angles. PCl3 and PF3 both have a tetra- Dicyandiamide: hedral arrangement of electron pairs, so these are the compounds that have 3 H H central atoms that are sp hybridized. NNO and COS both have a linear mo- Â 0 D Â 0 D N 000 N lecular structure. –1 0+10D G 0 ý D G + ½NPCNP PmC H n ðNqCNO PC H Favored 87. F O G G 0 F 0 NHO N by formal ½ ð charge Cl Cl A D G H HH F F O O F 3 3 Melamine: see-saw, dsp tetrahedral, sp H H Note: Similar to Exercise 67 c, d, and e, F3ClO has one additional Lewis A A structure that is possible, and F ClO (below) has two additional Lewis struc- 3 2 HOONNN H HON N NOH tures that are possible. The predicted hybridization is unaffected. G E N D A GDN E A C C H C C H B A mn A B O N N N N ½ H K Â ½ K H Â F trigonal bipyramid C C F Cl A A dsp3 F H N HONð O A H H 89. A Lewis structure for this compound is: b. NCN22: C is sp hybridized. Each resonance structure predicts a different H hybridization for the N atoms. For the remaining compounds, we will predict A O HCO OH hybrids for the favored resonance structures only. H H A C C Â N H NH2 N C G ½ G m

H CHO sp3 m NCm N NCN C m O q ð ð q m O P C C D m m m D m M N H O N sp3 sp sp sp2 ½NH2 A 3 2 HCO OH Melamine: N in NH2 groups are all sp hybridized. Atoms in ring are all sp A hybridized; c. NCN22: 2 s and 2 p bonds; H NCN: 4 s and 2 p bonds; di- H 2 cyandiamide: 9 s and 3 p bonds; melamine: 15 s and 3 p bonds; d. The p

57404_Ans_A31-A64.indd 44 10/14/16 5:12 PM Answers to Selected Exercises A45

system forces the ring to be planar just as the benzene ring is planar. e. The decreases. b. increase T: increase rate; c. increase surface area: increase rate

structure 25. C2H5OH1l2 nC2H5OH1g2 is an endothermic process. Heat is absorbed ½ when liquid ethanol vaporizes; the internal heat from the body provides this ðNq ONC M heat, which results in the cooling of the body. 27. Sublimation will occur, CONOH allowing water to escape as H O( g). 29. The strength of intermolecular D A 2 ðN forces determines relative boiling points. The types of intermolecular forces DG H HH for covalent compounds are London dispersion forces, dipole forces, and hydrogen bonding. Because the three compounds are assumed to have similar best agrees with experiment because it has three different CN bonds. This molar mass and shape, the strength of the London dispersion forces will be structure is also favored on the basis of formal charge. 97. a. 25-nm light about equal among the three compounds. One of the compounds will be non- has sufficient energy to ionize N and N2 and to break the triple bond in N2. polar so it only has London dispersion forces. The other two compounds will 1 1 Thus, N2, N2 , N, and N will all be present, assuming excess N2. b. 85.33 nm be polar so they have additional dipole forces and will boil at a higher tem- , l # 127 nm; c. The ionization energy of a substance is the energy it takes perature than the nonpolar compound. One of the polar compounds will have s 2 s 2 p 4 s 2 to completely remove an electron. N2: ( 2s) ( 2s*) ( 2p) ( 2p) ; the electron an H covalently bonded to either N, O, or F. This gives rise to the strongest s removed from N2 is in the 2p molecular orbital, which is lower in energy than type of covalent intermolecular force, hydrogen bonding. This compound ex- p the 2 atomic orbital from which the electron in atomic nitrogen is removed. hibiting hydrogen bonding will have the highest boiling point while the polar Since the electron removed from N2 is lower in energy than the electron re- compound with no hydrogen bonding will boil at an intermediate tempera- moved in N, then the ionization energy of N2 is greater than the ionization ture. 31. a. Both CO2 and H2O are molecular solids. Both have an ordered energy of N. 99. Both reactions apparently involve only the breaking of the array of the individual molecules, with the molecular units occupying the 1 NOCl bond. However, in the reaction ONCl n NO Cl some energy is re- lattice points. A difference within each solid lattice is the strength of the inter- leased in forming the stronger NO bond, lowering the value of DH. Therefore, molecular forces. CO2 is nonpolar and only exhibits London dispersion the apparent NOCl bond energy is artificially low for this reaction. The first forces. H2O exhibits the relatively strong hydrogen bonding interactions. The reaction involves only the breaking of the NOCl bond. 101. The ground difference in strength is evidenced by the solid phase change that occurs at 2 2 state MO electron configuration for He2 is (s1s) (s1s*) giving a bond order of 1 atm. CO2 sublimes at a relatively low temperature of 2788C. In sublima- 0. Therefore, He2 molecules are not predicted to be stable (and are not stable) tion, all of the intermolecular forces are broken. However, H2O doesn’t have in the lowest-energy ground state. However, in a high-energy environment, a solid phase change until 08C, and in this phase change from ice to water, the electron(s) from the antibonding orbitals in He2 can be promoted into only a fraction of the intermolecular forces are broken. The higher tempera- higher-energy bonding orbitals, thus giving a nonzero bond order and a “rea- ture and the fact that only a portion of the intermolecular forces are broken are son” to form. For example, a possible excited-state MO electron configuration attributed to the strength of the intermolecular forces in H O as compared to 2 1 1 2 for He2 would be (s1s) (s1s*) (s2s) , giving a bond order of (3 2 1)/2 5 1. CO2. Related to the intermolecular forces are the relative densities of the solid Thus excited He2 molecules can form, but they spontaneously break apart as and liquid phases for these two compounds. CO (s) is denser than CO (l) the electron(s) fall back to the ground state, where the bond order equals zero. 2 2 while H2O(s) is less dense than H2O(l). For CO2(s) and for most solids, the 103. The species with the smallest ionization energy has the highest energy molecules pack together as close as possible, which is why solids are usually 22 2 1 electron. O2, N2 , N2 , and O2 all have at least one electron in the high- more dense than the liquid phase. Water is an exception to this. Water mole- p 22 energy 2p* orbitals. Because N2 has the highest ratio of electrons to pro- cules are particularly well suited to interact with each other because each tons, the p * electrons are least attracted to the nuclei and easiest to remove, 2p molecule has two polar OOH bonds and two lone pairs on each oxygen. This translating into the smallest ionization energy. 105. a. The CO bond is polar, can lead to the association of four hydrogen atoms with each oxygen: two by with the negative end at the more electronegative oxygen atom. We would covalent bonds and two by dipoles. To keep this symmetric arrangement expect metal cations to be attracted to and to bond to the oxygen end of CO (which maximizes the hydrogen bonding interactions), the H O(s) molecules 2 2 on the basis of electronegativity. b. The formal charge on C is 1, and the occupy positions that create empty space in the lattice. This translates into a formal charge on O is 11. From formal charge, we would expect metal cat- smaller density for H2O(s) (less mass per unit volume). b. Both NaCl and ions to bond to the carbon (with the negative formal charge). c. In molecular CsCl are ionic compounds with the anions at the lattice points of the unit cell orbital theory, only orbitals with proper symmetry overlap to form bonding and the cations occupying the empty spaces created by the anions (called orbitals. The metals that form bonds to CO are usually transition metals, all of holes). In NaCl, the Cl2 anions occupy the lattice points of a face-centered d which have outer electrons in the orbitals. The only molecular orbitals of unit cell with the Na1 cations occupying the octahedral holes. Octahedral d p CO that have proper symmetry to overlap with orbitals are the 2p* orbitals, holes are the empty spaces created by six Cl2 ions. CsCl has the Cl2 ions at d whose shape is similar to that of the orbitals. Since the antibonding molecu- the lattice points of a simple cubic unit cell with the Cs1 cations occupying lar orbitals have more carbon character, one would expect the bond to form the middle of the cube. 33. Chalk is composed of the ionic compound cal- 5 5 through carbon. 107. a. Li2 bond order 1; B2 bond order 1; b. 4 elec- cium carbonate (CaCO ). The electrostatic forces in ionic compounds are 3 5 dsp3 3 trons must be removed; c. 4.5 10 kJ 109. T-shaped and hybridized much stronger than the intermolecular forces in covalent compounds. There- fore, CaCO should have a much higher boiling point than the covalent com- Chapter 10 3 pounds found in motor oil and in H O. Motor oil is composed of nonpolar 15. Intermolecular forces are the forces between molecules that hold the sub- 2 COC and COH bonds. The intermolecular forces in motor oil are therefore stances together in the solid and liquid phases. Hydrogen bonding is a specific London dispersion forces. We generally consider these forces to be weak. type of intermolecular force. In this figure, the dotted lines represent the hy- However, with compounds that have large molar masses, these London dis- drogen bonding interactions that hold individual H2O molecules together in persion forces add up significantly and can overtake the relatively strong the solid and liquid phases (answer a is correct). The solid lines represent the hydrogen bonding interactions in water. 35. In the ln(Pvap) versus 1yT plot, OOH covalent bonds. 17. N2 because it will have the weaker intermolecu- the slope of the straight line is equal to 2DHvap/R. Because DHvap is always lar forces. 19. Atoms have an approximately spherical shape. It is impossi- positive, the slope of the line will always be negative. 37. a. LD (London ble to pack spheres together without some empty space between the spheres. dispersion); b. dipole, LD; c. hydrogen bonding, LD; d. ionic; e. LD; f. di- 21. An alloy is a substance that contains a mixture of elements and has metal- pole, LD; g. ionic 39. a. OCS; b. SeO ; c. H NCH CH NH ; d. H CO; lic properties. In a substitutional alloy, some of the host metal atoms are re- 2 2 2 2 2 2 e. CH3OH 41. a. The compound with the stronger intermolecular forces placed by other metal atoms of similar size (e.g., in brass, pewter, plumber’s will boil at the higher temperature. Both HCl and Ar have about the same solder). An interstitial alloy is formed when some of the interstices (holes) in molar mass, so the London dispersion forces are approximately equivalent. the closest packed metal structure are occupied by smaller atoms (e.g., in car- However, HCl is a polar molecule leading to additional dipole forces that the bon steels). 23. a. As intermolecular forces increase, the rate of evaporation

57404_Ans_A31-A64.indd 45 10/14/16 5:12 PM A46 Answers to Selected Exercises

nonpolar Ar does not have. HCl has the stronger intermolecular forces, and it 101. 1680 kJ 103. 1490 g 105. A: solid; B: liquid; C: vapor; D: solid 1 boils at the higher temperature. b. HF is capable of hydrogen bonding; HCl is vapor; E: solid 1 liquid 1 vapor (triple point); F: liquid 1 vapor; G: liquid not. c. LiCl is ionic, and HCl is a molecular solid with only dipole forces and 1 vapor (critical point); H: vapor; the first dashed line (at the lower tempera- London dispersion forces. Ionic forces are much stronger than the forces for ture) is the normal freezing point, and the second dashed line is the normal molecular solids. d. n-Hexane is a larger molecule, so it has stronger London boiling point. The solid phase is denser. 107. a. two; b. higher pressure dispersion forces. 43. a. HBr has dipole forces in addition to LD forces; triple point: graphite, diamond, and liquid; lower pressure triple point: b. NaCl, stronger ionic forces; c. I2, larger molecule so stronger LD forces; graphite, liquid and vapor; c. It is converted to diamond (the more dense d. N2, smallest nonpolar compound present, has weakest LD forces; e. CH4, solid form); d. Diamond is more dense, which is why graphite can be con- smallest nonpolar compound present, has weakest LD forces; f. HF, can form verted to diamond by applying pressure. 109. Because the density of the relatively strong hydrogen bonding interactions, unlike the other compounds; liquid phase is greater than the density of the solid phase, the slope of the

g. CH3CH2CH2OH, unlike others, has relatively strong hydrogen bonding. solid/liquid boundary line is negative (as in H2O). With a negative slope, 45. H2O is attracted to glass while Hg is not. 47. The structure of H2O2 pro- the melting points increase with a decrease in pressure so the normal melting duces greater hydrogen bonding than water. 49. 313 pm 51. 0.704 Å point of X should be greater than 2258C. 111. The covalent bonds within a 53. 1.54 g/cm3 55. 174 pm; 11.6 g/cm3 g 57. A 59. edge, 328 pm; radius, molecule are much stronger than the intermolecular forces between mole- 142 pm 61. face-centered cubic unit cell 63. For a cubic closest packed cules. Consider water, which exhibits a very strong type of dipole force structure, 74.06% of the volume of each unit cell is occupied by atoms; in a called hydrogen bonding. When water boils at 1008C, hydrogen bonding in-

simple cubic unit cell structure, 52.36% is occupied. The cubic closest packed termolecular forces that hold the H2O molecules together in the liquid phase structure provides the more efficient means for packing atoms. 65. Doping are broken, resulting in H2O(l) converting to H2O(g). At 1008C, the covalent silicon with phosphorus produces an n-type semiconductor. The phosphorus bonds within each water molecule are not broken; if they were to be broken,

adds electrons at energies near the conduction band of silicon. Electrons do then H2(g) and O2(g) would be the produced. It takes a lot more energy than not need as much energy to move from filled to unfilled energy levels so con- what is provided at 1008C to break the covalent bonds within a water mole- duction increases. Doping silicon with gallium produces a p-type semicon- cule; the intramolecular forces within a covalent compound are much stron- ductor. Because gallium has fewer valence electrons than silicon, holes (un- ger than the intermolecular forces between molecules. 113. As the physical

filled energy levels) at energies in the previously filled molecularorbitals are properties indicate, the intermolecular forces are slightly stronger in D2O created, which induces greater electron movement (greater conductivity). than in H2O. 115. A: CH4; B: SiH4; C: NH3 117. 4.67 min 119. XeF2 2 1 2 1 2 67. p-type 69. 5.0 3 10 nm 71. NaCl: 4Na , 4Cl ; CsCl: 1Cs , 1Cl ; 121. B2H6, molecular; SiO2, network; CsI, ionic; W, metallic 123. 57.8 torr 21 22 41 22 ZnS: 4Zn , 4S ; TiO2: 2Ti , 4O 73. CoF2 75. ZnAl2S4 MF77. 2 125. A mixture of H2O(s) and H2O(l) will be present at 0°C when 215 kJ of 28 29 79. NaCl structure; rO22 5 1.49 3 10 cm; rMg21 5 6.15 3 10 cm 81. The heat are removed from the gas sample. 127. If we extend the liquid–vapor calculated distance between ion centers is 358 pm. Ionic radii give a distance line of the water phase diagram to below the freezing point, we find that su- of 350. pm. The distance calculated from the density is 8 pm (2.3%) greater percooled water will have a higher vapor pressure than ice at 210°C (see than that calculated from the table of ionic radii. 83. a. CO2: molecular; Fig. 10.44). To achieve equilibrium, there must be a constant vapor pressure. b. SiO2: network; c. Si: atomic, network; d. CH4: molecular; e. Ru: atomic, Over time, supercooled water will be transformed through the vapor into ice metallic; f. I2: molecular; g. KBr: ionic; h. H2O: molecular; i. NaOH: ionic; in an attempt to equilibrate the vapor pressure. Eventually there will only be

j. U: atomic, metallic; k. CaCO3: ionic; l. PH3: molecular 85. a. The unit ice at 210°C along with H2O(g) at the vapor pressure given by the solid– cell consists of Ni at the cube corners and Ti at the body center or Ti at the vapor line in the phase diagram. 129. 4.65 kg/h 131. CO2 and XeF4 are cube corners and Ni at the body center. b. NiTi; c. Both have a coordination both nonpolar covalent compounds, so they only exhibit London dispersion

number of 8. 87. CaTiO3; each structure has six oxygens around each Ti. forces. CH3CH2OH and HF are covalent compounds that contain either an 89. a. YBa2Cu3O9; b. The structure of this superconductor material is based NOH, OOH, or FOH bond, so they can form hydrogen bonding forces. SF4 on the second perovskite structure. The YBa2Cu3O9 structure is three of these and ICl5 are polar covalent compounds, so they can exhibit dipole–dipole cubic perovskite unit cells stacked on top of each other. The oxygens are in forces. 133. a. False; the ionic forces in LiF are much stronger than the

the same places, Cu takes the place of Ti, two Ca are replaced by two Ba, and molecular intermolecular forces found in H2S. b. True; HF is capable of one Ca is replaced by Y. c. YBa2Cu3O7 91. Li, 158 kJ/mol; Mg, 139 kJ/mol. H bonding; HBr is not. c. True; the larger Cl2 molecule will have the stronger Bonding is stronger in Li. 93. 898C 95. 28.6 kJ/mol London dispersion forces. d. True; HCl is a polar compound while CCl4 is a 97. nonpolar compound. e. False; the ionic forces in MgO are much stronger

than the molecular intermolecular forces found in CH3CH2OH. 135. 41 137. a. Kr: group 8A; b. SO2: molecular; c. Ni: metallic; d. SiO2: network; 5 e. NH3: molecular; f. Pt: metallic 139. 207 torr 141. DE 5 27.86 kJ/mol; DH 5 30.79 kJ/mol 143. 46.7 kJ/mol; 90.%

80 4 145. H H H H H C C O H H C O C H 60 and H H H H 40 Liquid (H-bonding) Gas (no H-bonding)

C) 3 Slope 5 > slope 3 > slope 1 ° 20 Time 4 = 4 × time 2 The first structure with the —OH bond is capable of forming hydrogen bonding; the other structure is not. Therefore, the liquid (which has the stronger Temp ( 0 intermolecular forces) is the first structure, and the gas is the second structure. 2 147. The solids with high melting points (NaCl, MgCl , NaF, MgF , AlF ) –20 2 2 3 1 are all ionic solids. SiCl4, SiF4, Cl2, F2, PF5, and SF6 are nonpolar covalent –40 molecules with LD forces. PCl3 and SCl2 are polar molecules with LD and dipole forces. In these 8 molecular substances the intermolecular forces are –60 weak and the melting points low. AlCl3 is intermediate. The melting point Time indicates there are stronger forces present than in the nonmetal halides, 99. a. Much more energy is required to break the intermolecular forces when but not as strong as for an ionic solid. AlCl3 illustrates a gradual transition going from a liquid to a gas than is required to go from a solid to a liquid. from ionic to covalent bonding; from an ionic solid to discrete molecules. 21 31 149. TiO1.182 or Ti0.8462O; 63.7% Ti , 36.3% Ti (a 1.75;1 ion ratio) Hence DHvap is much larger than DHfus; b. 113 J; c. 4220 J; d. 4220 J released 151. 6.58 g/cm3

57404_Ans_A31-A64.indd 46 10/14/16 5:12 PM Answers to Selected Exercises A47

153. that of the cells, the cells shrivel as there is a net transfer of water out of the a cells. This is called crenation. Hemolysis occurs when the red blood cells are bathed in a solution having lower osmotic pressure than that inside the cell. Here, the cells rupture as there is a net transfer of water to inside the red

blood cells. 35. 1.9% 37. 1.06 g/mL; 0.0180 mole fraction H3PO4, 0.9820 mole fraction H O; 0.981 mol/L; 1.02 mol/kg 39. HCl: 12 M, 17 m, P cb 2 0.23; HNO3: 16 M, 37 m, 0.39; H2SO4: 18 M, 200 m, 0.76; HC2H3O2: 17 M, 2000 m, 0.96; NH3: 15 M, 23 m, 0.29 41. 35%; 0.39; 7.3 m; 3.1 M 43. 10.1% by mass; 2.45 mol/kg 45. 23.9%; 1.6 m, 0.028, 4.11 N 1 2 47. NaI1s2 S Na 1aq2 1 I 1aq2 DHsoln 5 28 kJ/mol 49. The attraction of water molecules for Al31 and OH2 cannot overcome the larger lattice energy T of Al(OH)3. 51. a. CCl4; b. H2O; c. H2O; d. CCl4; e. H2O; f. H2O; g. CCl4 As P is lowered, we go from a to b on the phase diagram. The water boils. 53. a. NH3; b. CH3CN; c. CH3COOH 55. As the length of the hydrocarbon The boiling of water is endothermic and the water is cooled 1b S c2, forming chain increases, the solubility decreases because the nonpolar hydrocarbon some ice. If the pump is left on, the ice will sublime until none is left. This is chain interacts poorly with the polar water molecules. 57. 1.04 3 the basis of freeze drying. 1023 mol/L ? atm; 1.14 3 1023 mol/L 59. 136 torr 61. 0.918 4 63. a. 290 torr; b. 0.69 65. xmethanol 5 xpropanol 5 0.500 67. solution c 155. The volume of the hole is p3 1"3 2 12r43 157. CdS; n-type 3 69. a. These two molecules are named acetone (CH3COCH3) and water. As 159. 253 torr; 6.38 3 1022 atoms discussed in Section 11.4 on nonideal solutions, acetone–water solutions exhibit negative deviations from Raoult’s law. Acetone and water have the abil- Chapter 11 ity to hydrogen bond with each other, which gives the solution stronger inter- 13. 9.74 M 15. 160 mL 17. 4.5 M 19. As the temperature increases, the molecular forces compared to the pure states of both solute and solvent. In gas molecules will have a greater average kinetic energy. A greater fraction the pure state, acetone cannot hydrogen bond with itself, so the middle dia- of the gas molecules in solution will have kinetic energy greater than the at- gram illustrating negative deviations from Raoult’s law is the correct choice tractive forces between the gas molecules and the solvent molecules. More for acetone–water solutions. b. These two molecules are named ethanol gas molecules will escape to the vapor phase, and the solubility of the gas (CH3CH2OH) and water. Ethanol–water solutions show positive deviations will decrease. 21. The levels of the liquids in each beaker will become from Raoult’s law. Both substances can hydrogen bond in the pure state, and constant when the concentration of solute is the same in both beakers. Be- they can continue this in solution. However, the solute–solvent interactions cause the solute is less volatile, the beaker on the right will have a larger are weaker for ethanol–water solutions due to the significant nonpolar part of volume when the concentrations become equal. There will be a larger net ethanol (CH3OCH2 is the nonpolar part of ethanol). This nonpolar part of transfer of water molecules into the the right beaker than the net transfer of ethanol weakens the intermolecular forces in solution, so the first diagram

solute molecules to the left beaker. Eventually the rate that solute and H2O illustrating positive deviations from Raoult’s law is the correct choice for leave and return to each beaker will become equal when the concentrations ethanol–water solutions. c. These two molecules are named heptane (C7H16)

become equal. 23. No. For an ideal solution, DHsoln 5 0. 25. Normality is and hexane (C6H14). Heptane and hexane are very similar nonpolar sub- the number of equivalents per liter of solution. For an acid or a base, an stances; both are composed entirely of nonpolar COC bonds and relatively equivalent is the mass of acid or base that can furnish 1 mole of protons nonpolar COH bonds, and both have a similar size and shape. Solutions of (if an acid) or accept 1 mole of protons (if a base). A proton is an H1 ion. heptane and hexane should be ideal, so the third diagram illustrating no de- Molarity is defined as the moles of solute per liter of solution. When viation from Raoult’s law is the correct choice for heptane–hexane solutions. the number of equivalents equals the number of moles of solute, then d. These two molecules are named heptane (C7H16) and water. The interac- normality 5 molarity. This is true for acids that only have one acidic proton tions between the nonpolar heptane molecules and the polar water molecules in them and for bases that accept only one proton per formula unit. Examples will certainly be weaker in solution compared to the pure solvent and pure

of acids where equivalents 5 moles solute are HCl, HNO3, HF, and HC2H3O2. solute interactions. This results in positive deviations from Raoult’s law (the Examples of bases where equivalents 5 moles solute are NaOH, KOH, and first diagram). 71. 101.58C 73. 14.8 g C3H8O3 75. Tf 5 229.98C, Tb 5 8 3 22 NH3. When equivalents 2 moles solute, then normality 2 molarity. This is 108.2 C 77. 6.6 10 mol/kg; 590 g/mol (610 g/mol if no rounding of D 5 3 258 p 5 true for acids that donate more than one proton (H2SO4, H3PO4, H2CO3, etc.) numbers) 79. a. T 2.0 10 C, 0.20 torr; b. Osmotic pressure is and for bases that react with more than one proton per formula unit [Ca(OH)2, better for determining the molar mass of large molecules. A temperature 258 Ba(OH)2, Sr(OH)2, etc.]. 27. Only statement b is true. A substance freezes change of 10 C is very difficult to measure. A change in height of a column when the vapor pressures of the liquid and solid phases are the same. When of mercury by 0.2 mm is not as hard to measure precisely. a solute is added to water, the vapor pressure of the solution at 08C is less 81. 2.51 3 105 g/mol 83. Dissolve 210 g sucrose in some water and dilute than the vapor pressure of the solid; the net result is for any ice present to to 1.0 L in a volumetric flask. To get 0.62 6 0.01 mol/L, we need 212 6 convert to liquid in order to try to equalize the vapor pressures (which never 3 g sucrose. 85. a. 0.010 m Na3PO4 and 0.020 m KCl; b. 0.020 m HF; can occur at 08C). A lower temperature is needed to equalize the vapor pres- c. 0.020 m CaBr2 87. a. Tf 5 2138C; Tb 5 103.58C; b. Tf 5 24.78C; Tb 5 sures of water and ice, hence the freezing point is depressed. For statement 101.38C 89. 1.67 91. a. Tf 5 20.288C; Tb 5 100.0778C; b. Tf 5 a, the vapor pressure of a solution is directly related to the mole fraction of 20.378C; Tb 5 100.108C 93. 2.63 (0.0225 m), 2.60 (0.0910 m), 2.57 solvent (not solute) by Raoult’s law. For statement c, colligative properties (0.278 m); iaverage 5 2.60 95. a. yes; b. no 97. Benzoic acid is capable of depend on the number of solute particles present and not on the identity of hydrogen bonding, but a significant part of benzoic acid is the nonpolar ben-

the solute. For statement d, the boiling point of water is increased because zene ring. In benzene, a hydrogen bonded dimer forms.

the sugar solute decreases the vapor pressure of the water; a higher tempera- O O H D

ture is required for the vapor pressure of the solution to equal the external M

pressure so boiling can occur. 29. Adding a solute to a solvent increases the CCM D O boiling point and decreases the freezing point of the solvent. Thus, the sol- O H vent is a liquid over a wider range of temperatures when a solute is dissolved. 31. The vapor pressure, boiling point, and osmotic pressure will be the same The dimer is relatively nonpolar and thus more soluble in benzene than in between the two different solutions. 33. Isotonic solutions are those that water. Because benzoic acid forms dimers in benzene, the effective solute have identical osmotic pressures. Crenation and hemolysis refer to phenom- particle concentration will be less than 1.0 molal. Therefore, the freezing- ena that occur when red blood cells are bathed in solutions having a mis- point depression would be less than 5.12°C (DTf 5 Kf m). 99. The main match in osmotic pressure between the inside and the outside of the cell. factor for stabilization seems to be electrostatic repulsion. The center of a When red blood cells are in a solution having a higher osmotic pressure than colloid particle is surrounded by a layer of same charged ions, with oppo-

57404_Ans_A31-A64.indd 47 10/14/16 5:12 PM A48 Answers to Selected Exercises

sitely charged ions forming another charged layer on the outside. Overall, the rate by affecting the value of the rate constant k. The value of the rate there are equal numbers of charged and oppositely charged ions, so the col- constant is dependent on temperature. It also depends on the activation en- loidal particles are electrically neutral. However, since the outer layers are ergy. A catalyst will change the value of k because the activation energy

the same charge, the particles repel each other and do not easily aggregate for changes. Increasing the concentration (partial pressure) of either O2 or NO precipitation to occur. Heating increases the velocities of the colloidal parti- does not affect the value of k, but it does increase the rate of the reaction cles. This causes the particles to collide with enough energy to break the ion because both concentrations appear in the rate law. 15. In a unimolecular barriers, allowing the colloids to aggregate and eventually precipitate out. reaction, a single reactant molecule decomposes to products. In a bimolecu- Adding an electrolyte neutralizes the adsorbed ion layers, which allows col- lar reaction, two molecules collide to give products. The probability of the loidal particles to aggregate and then precipitate out. 101. No, the solution simultaneous collision of three molecules with enough energy and orienta- is not behaving ideally. If the solution were behaving ideally, the freezing tion is very small, making termolecular steps very unlikely. 17. When the point would be 21.9°C. 103. a. Water boils when the vapor pressure rate doubles as the concentration quadruples, the order is 1y2. For a reactant equals the pressure above the water. In an open pan, Patm < 1.0 atm. In a that has an order of 21, the rate will decrease by a factor of 1y2 when the pressure cooker, Pinside . 1.0 atm and water boils at a higher temperature. concentrations are doubled. 19. Two reasons are: 1) the collision must in- The higher the cooking temperature, the faster the cooking time. b. When volve enough energy to produce the reaction; i.e., the collision energy must water freezes from a solution, it freezes as pure water, leaving behind a more equal or exceed the activation energy. 2) the relative orientation of the reac-

concentrated salt solution. c. On the CO2 phase diagram, the triple point is tants must allow formation of any new bonds necessary to produce prod- above 1 atm and CO2(g) is the stable phase at 1 atm and room temperature. ucts. 21. Enzymes are very efficient catalysts. As is true for all catalysts, CO2(l) can’t exist at normal atmospheric pressures, which explains why dry enzymes speed up a reaction by providing an alternative pathway for reactants ice sublimes rather than boils. In a fire extinguisher, P . 1 atm and CO2(l) to convert to products. This alternative pathway has a smaller activation en- can exist. When CO2 is released from the fire extinguisher, CO2(g) forms as ergy and, hence, a faster rate. Also true is that catalysts are not used up in the predicted from the phase diagram. 105. 0.600 107. Pideal 5 188.6 torr; overall chemical reaction. Once an enzyme comes in contact with the correct xacetone 5 0.512, xmethanol 5 0.488; the actual vapor pressure of the solution is reagent, the chemical reaction quickly occurs, and the enzyme is then free to smaller than the ideal vapor pressure, so this solution exhibits a negative catalyze another reaction. Because of the efficiency of the reaction step, only deviation from Raoult’s law. This occurs when solute–solvent attractions are a relatively small amount of enzyme is needed to catalyze a specific reaction,

stronger than for the pure substances. 109. 776 g/mol 111. C2H4O3; no matter how complex the reaction. 23. The rate of a reaction is indepen- 151 g/mol (exp.); 152.10 g/mol (calc.); C4H8O6 113. 1.97% NaCl dent of the enthalpy change for the reaction. The combustion of carbohy- 115. a. 100.778C; b. 23.1 mm Hg; c. Assume an ideal solution; assume no drates has a lower activation energy, which is why it is the faster process. 24 23 ions form (i 5 1); assume the solute is nonvolatile. 117. 639 g/mol; 25. P4: 6.0 3 10 mol/L ? s; H2: 3.6 3 10 mol/L ? s 27. a. average rate 25 33.7 torr 119. 3 kJ/mol 121. 0.732 123. 100.25°C 125. 43% NaNO3 of decomposition of H2O2 5 2.31 3 10 mol/L ? s, rate of production of 25 and 57% Mg(NO3)2 by mass O2 5 1.16 3 10 mol/L ? s; b. average rate of decomposition of H2O2 5 1.16 3 1025 mol/L ? s, rate of production of O 5 5.80 3 1026 mol/L ? s 0.30y 2 ? ? 21 ? 2 2 ? 5 127. 30.% A: xA 5 , xB 5 1 2 xA; 29. a. mol/L s; b. mol/L s; c. s ; d. L/mol s; e. L /mol s 31. a. rate 0.70x 1 0.30y 2 2 2 2 2 k[NO] [Cl2]; b. 1.8 3 10 L /mol ? min 33. a. rate 5 k[NOCl] ; b. 6.6 3 y y 229 3 28 2 2 50.% A: xA 5 , xB 5 1 2 ; 10 cm /molecules ? s; c. 4.0 3 10 L/mol ? s 35. a. rate 5 k[I ][OCl ]; x 1 y x 1 y b. 3.7 L/mol ? s; c. 0.083 mol/L ? s 37. a. first order in Hb and first order in 0.80y 80.% A: x 5 , x 5 1 2 x ; CO; b. rate 5 k[Hb][CO]; c. 0.280 L/mmol ? s; d. 2.26 mmol/L ? s 39. rate A 0.20x 1 0.80y B A 5 k[H2O2]; ln[H2O2] 5 2kt 1 ln[H2O2]0; V 0.30x V 0.30x 30.% A: xA 5 , xB 5 1 2 ; 24 21 2 1 1 0.30x 1 0.70y 0.30x 1 0.70y k 5 8.3 3 10 s ; 0.037 M 41. rate 5 k[NO2] ; 5 kt 1 ; 3NO2 4 3NO2 40 V x V V 50.% A: x 5 , x 5 1 2 x ; 24 A x 1 y B A k 5 2.08 3 10 L/mol ? s; 0.131 M 43. a. rate 5 k; [C2H5OH] 5 2kt 1 [C H OH] ; because slope 5 2k, k 5 4.00 3 1025 mol/L ? s; b. 156 s; V 0.80x V V 2 5 0 80.% A: xA 5 , xB 5 1 2 xA 0.80x 1 0.20y 2 1 1 22 c. 313 s 45. rate 5 k[C4H6] ; 5 kt 1 ; k 5 1.4 3 10 3C4H6 4 3C4H6 40 129. 72.5% sucrose and 27.5% NaCl by mass; 0.313 131. 0.050 133. 44% L/mol ? s 47. second order; 0.1 M 49. a. [A] 52kt 1 [A] ; b. 1.0 3 naphthalene, 56% anthracene 135. 20.208C, 100.0568C 137. a. 46 L; 0 1022 s; c. 2.5 3 1024 M 51. 9.2 3 1023 s21; 75 s 53. 150. s b. No; a reverse osmosis system that applies 8.0 atm can only purify water 55. 1.0 3 102 min 57. a. 1.1 3 1022 M; b. 0.025 M 59. 1.6 L/mol ? s with solute concentrations less than 0.32 mol/L. Salt water has a solute con- 61. a. rate 5 k[CH3NC]; b. rate 5 k[O3][NO]; c. rate 5 k[O3]; d. rate 5 centration of 2(0.60 M) 5 1.2 M ions. The solute concentration of salt water 2 1 k[O3][O] 63. Rate 5 k[C4H9Br]; C4H9Br 1 2H2O n C4H9OH 1 Br 1 H3O ; is much too high for this reverse osmosis unit to work. 139. a. 26.6 kJ/mol; 1 1 the intermediates are C4H9 and C4H9OH2 . 65. The rate law is Rate 5 b. 2657 kJ/mol 141. i 5 3.00; CdCl2 2 k[NO] [Cl2]. If we assume the first step is rate determining, we would expect 5 k Chapter 12 the rate law to be Rate 1[NO][Cl2]. This isn’t correct. However, if we assume the second step to be rate determining, then Rate 5 k [NOCl ][NO]. To 11. One experimental method to determine rate laws is the method of initial 2 2 see if this agrees with experiment, we must substitute for the intermediate rates. Several experiments are carried out using different initial concentra- NOCl concentration. Assuming a fast-equilibrium first step (rate reverse 5 tions of reactants, and the initial rate is determined for each experiment. The 2 rate forward): results are then compared to see how the initial rate depends on the initial k concentrations. This allows the orders in the rate law to be determined. The k [NOCl ] 5 k [NO][Cl ], [NOCl ] 5 1 [NO][Cl ]; 21 2 1 2 2 k 2 value of the rate constant is determined from the experiments once the orders 21 substituting into the rate equation: are known. The second experimental method utilizes the fact that the inte-

grated rate laws can be put in the form of a straight-line equation. Concentra- k2k1 2 2 k2k1 Rate 5 [NO] [Cl2] 5 k[NO] [Cl2] where k 5 tion versus time data are collected for a reactant as a reaction is run. These k21 k21 data are then manipulated and plotted to see which manipulation gives a This is a possible mechanism with the second step the rate-determining straight line. From the straight-line plot we get the order of the reactant, and step because the derived rate law agrees with the experimentally deter- the slope of the line is mathematically related to k, the rate constant. 13. All mined rate law. of these choices would affect the rate of the reaction, but only b and c affect

57404_Ans_A31-A64.indd 48 10/14/16 5:12 PM Answers to Selected Exercises A49

67. Note that the heights of the second and third humps must be lower than the first-step activation energy. However, the height of the third hump could be higher than the second hump. One cannot determine this absolutely from the information in the problem.

b. F2 S 2F slow E Ea P F 1 H2 S HF 1 H fast ∆E H 1 F S HF fast

R F2 1 H2 S 2HF

c. F2 was the limiting reactant. 119. a. [B] .. [A], so [B] can be consid- RC ered constant over the experiments. This gives us a pseudo-order rate-law 2 2 2 69. 341 kJ/mol 71. The graph of ln(k) versus 1yT is linear with slope 5 equation. b. Rate 5 k[A] [B], k 5 0.050 L /mol ? s. c. i. This mechanism 4 2 25 gives the wrong stoichiometry, so it can’t be correct. ii. Rate 5 k[E][A]; 2EayR 5 21.2 3 10 K; Ea 5 1.0 3 10 kJ/mol 73. 9.5 3 10 L/mol ? s 1 2 75. 518C 77. H O (aq) 1 OH (aq) 2H O(l) should have the faster rate. k1 3A4 3B4 kk1 2 3 n 2 k [A][B] 5 k [E]; [E] 5 ; Rate 5 3A4 3B4. This mechanism 1 2 41 1 21 H3O and OH will be electrostatically attracted to each other; Ce and k21 k21 21 gives the correct stoichiometry and gives the correct rate law when it is de- Hg2 will repel each other (so Ea is much larger). 79. a. NO; b. NO2; rived from the mechanism. This is a possible mechanism for this reaction. c. 2.3 81. CH2DOCH2D should be the product. If the mechanism is pos- iii. Rate 5 k[A]2. This mechanism gives the wrong derived rate law, so it sible, then the reaction must be C2H4 1 D2 n CH2DCH2D. If we got this 2 product, then we could conclude that this is a possible mechanism. If we got can’t be correct. Only mechanism ii is possible. 121. Rate 5 k[A][B] , k 5 1.4 3 1022 L2/mol2 ? s 123. 2.20 3 1025 s21; 5.99 3 1021 molecules some other product, e.g., CH3CHD2, then we would conclude that the mecha- 25 21 nism is wrong. Even though this mechanism correctly predicts the products of 125. 1.3 3 10 s ; 112 torr the reaction, we cannot say conclusively that this is the correct mechanism; we might be able to conceive of other mechanisms that would give the same Chapter 13 product as our proposed one. 83. The rate depends on the number of reac- 13. No, equilibrium is a dynamic process. Both the forward and reverse reactant molecules adsorbed on the surface of the catalyst. This quantity is pro- tions are occurring at equilibrium, just at equal rates. Thus the forward and portional to the concentration of reactant. However, when all the catalyst 14 reverse reactions will distribute C atoms between CO and CO2. 15. A surface sites are occupied, the rate becomes independent of the concentration large value for K (K @ 1) indicates there are relatively large concentrations of of reactant. 85. 2158C 87. a. 20 min; b. 30 min; c. 15 min 89. 427 s product gases and/or solutes as compared with the concentrations of reactant 2 91. 1.0 3 10 kJ/mol 93. a. second order; b. 20 s; c. 190 s; d. 847 s; gases and/or solutes at equilibrium. A reaction with a large K value is a good e. 31 kJ/mol. 95. 6.58 3 1026 mol/L ? s 97. a. 25 kJ/mol; b. 12 s; source of products. 17. 4 molecules H2O, 2 molecules CO, 4 molecules H2, c. T Interval 54 2 2(Intervals) and 4 molecules CO2 are present at equilibrium. 19. K and Kp are equilibrium constants as determined by the law of mass action. For K, the units used

21.08C 16.3 s 218C for concentrations are mol/L, and for Kp, partial pressures in units of atm are 27.88C 13.0 s 288C used (generally). Q is called the reaction quotient. Q has the exact same form 30.08C 12 s 30.8C as K or Kp, but instead of equilibrium concentrations, initial concentrations are used to calculate the Q value. Q is of use when it is compared to the K This rule of thumb gives excellent agreement to two significant figures. value. When Q 5 K (or when Qp 5 Kp), the reaction is at equilibrium. When 3 3 ? 5 3 25 M 5 99. 12.5 s 101. a. 115 L /mol s; b. 87.0 s; c. [A] 1.27 10 , [B] Q 2 K, the reaction is not at equilibrium and one can determine what has to 1.00 M 103. The reaction is first order with respect to both A and B; rate 5 be the net change for the system to get to equilibrium. 21. We always try to k[A][B] where k 5 1.6 L/mol · s. 105. 35.4 s 107. a. False; from the half- make good assumptions that simplify the math. In some problems, we can set life expression for a zero-order reaction in Table 12.6 of the text, there is a up the problem so that the net change, x, that must occur to reach equilibrium direct relationship between concentration and the half-life value. As the reac- is a small number. This comes in handy when you have expressions like tion proceeds for a zero-order reaction, the half-life value will decrease be- 0.12 2 x or 0.727 1 2x. When x is small, we assume that it makes little dif- cause concentration decreases. b. True. c. False; from the half-life expression ference when subtracted from or added to some relatively big number. When for a first-order reaction in Table 12.6, the half-life does not depend on con- this is true, 0.12 2 x < 0.12 and 0.727 1 2x < 0.727. If the assumption holds centration. d. True; the half-life for a second-order reaction increases with by the 5% rule, then the assumption is assumed valid. The 5% rule refers to x time because there is an inverse relationship between concentration and the (or 2x or 3x, etc.) that was assumed small compared to some number. If x (or half-life value. 109. 53.9 kJ/mol; the chirping rate will be about 45 chirps 2x or 3x, etc.) is less than 5% of the number the assumption was made against, k I2 OCl2 3 4 3 4 1 21 then the assumption will be assumed valid. If the 5% rule fails to work, one per minute at 7.5°C. 111. rate 5 2 ; k 5 6.0 3 10 s 3OH 4 can generally use a math procedure called the method of successive approxi- 113. a. first order with respect to both reactants; b. rate 5 k[NO][O3]; mations to solve the quadratic or cubic equation. 23. There will be a net in- 21 21 214 3 c. k9 5 1.8 s ; k0 5 3.6 s ; d. k 5 1.8 3 10 cm /molecules ? s crease in the amount of N2O present once equilibrium is reestablished. As k k 1/2 2 1 3/2 3/2 N2O(g) is added, the reaction will shift right to use up some of the added 115. a. Rate 5 k3 a b [CO][Cl2] 5 k[CO][Cl2] . b. Cl and COCl k22 k21 N2O. However, only some of the added N2O will react, not all. The net effect are intermediates. 117. a. For a three-step reaction with the first step limiting, is for the amount of N2O to increase once equilibrium is reestablished. As far the energy-level diagram could be as the K value is concerned, as long as the temperature didn’t change, K will remain a constant value. 2 2 2 3NO4 3NO2 4 3SiCl4 4 3H2 4 25. a. K 5 ; b. K 5 ; c. K 5 2; 3N2 4 3O2 4 3N2O4 4 3SiH4 4 3Cl2 4 E 2 3 3PCl3 4 3Br2 4 24 6 d. K 5 2 3 27. a. 0.11; b. 77; c. 8.8; d. 1.7 3 10 29. 4.0 3 10 R 3PBr3 4 3Cl2 4 P 31. 1.7 3 1025 33. 6.3 3 10213 35. 4.6 3 103 Reaction coordinate 3H O4 PH O 37. a. K 5 2 , K 5 2 ; b. K 5 [N ][Br ]3, 2 p 2 3 2 2 3NH3 4 3CO2 4 PNH3 PCO2 3H O4 PH O 5 3 3 5 3 5 3 5 2 5 2 Kp PN2 PBr2 ; c. K [O2] , Kp PO2 ; d. K , Kp 3H2 4 PH2

57404_Ans_A31-A64.indd 49 10/14/16 5:12 PM A50 Answers to Selected Exercises

–9 2 39. only reaction d 41. 1.2 3 10 43. 4.07 45. a. not at equilibrium; donates a proton. NH2 is the conjugate base of the NH3 acid. In the reverse 2 Q . K, shift left; b. at equilibrium; c. not at equilibrium; Q . K, shift left reaction, NH2 accepts a proton. Conjugate acid–base pairs differ only by an 47. a. decrease; b. no change; c. no change; d. increase 49. 0.16 mol H1 in the formula. 29. a. These would be 0.10 M solutions of strong acids

51. 3.4 53. 0.056 55. [N2]0 5 10.0 M, [H2]0 5 11.0 M 57. [SO3] 5 like HCl, HBr, HI, HNO3, H2SO4, or HClO4. b. These are salts of the conju- 22 [NO] 5 1.06 M; [SO2] 5 [NO2] 5 0.54 M 59. 7.8 3 10 atm gate acids of the bases in Table 14.3. These conjugate acids are all weak ac- 5 5 5 5 61. PSO2 0.38 atm; PO2 0.44 atm; PSO3 0.12 atm 63. a. [NO] ids. Three examples would be 0.10 M solutions of NH4Cl, CH3NH3NO3, and 0.032 M, [Cl2] 5 0.016 M, [NOCl] 5 1.0 M; b. [NO] 5 [NOCl] 5 1.0 M, C2H5NH3Br. Note that the anions used to form these salts are conjugate bases 25 23 [Cl2] 5 1.6 3 10 M; c. [NO] 5 8.0 3 10 M, [Cl2] 5 1.0 M, [NOCl] 5 of strong acids; this is because they have no acidic or basic properties in 5 5 5 3 25 2 2.0 M 65. PCOCl2 1.0 atm; PCO PCl2 8.2 10 atm 67. 0.27 atm water (with the exception of HSO4 , which has weak acid properties). 5 5 69. PCO2 3.39 atm, PCO 2.61 atm 71. a. no effect; b. shifts left; c. shifts c. These would be 0.10 M solutions of strong bases like LiOH, NaOH, KOH, right 73. a. no change; b. products increase; c. reactants increase RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. d. These are salts of the 75. a. right; b. right; c. no effect; d. left; e. no effect 77. a. left; b. right; conjugate bases of the neutrally charged weak acids in Table 14.2. The con- c. left; d. no effect; e. no effect; f. right 79. increase 81. 2.6 3 1081 jugate bases of weak acids are weak bases themselves. Three examples 26 31 24 83. 6.74 3 10 85. a. 0.379 atm; b. 0.786 87. [Fe ] 5 2 3 10 M, would be 0.10 M solutions of NaClO2, KC2H3O2, and CaF2. The cations used [SCN2] 5 0.08 M, [FeSCN21] 5 0.020 M 89. 1.43 3 1022 atm 91. pink to form these salts are Li1, Na1, K1, Rb1, Cs1, Ca21, Sr21, and Ba21 since 2 1 1 93. Added OH reacts with H to produce H2O. As H is removed, the reac- these cations have no acidic or basic properties in water. Notice that these are 1 22 22 tion shifts right to produce more H and CrO4 . Because more CrO4 the cations of the strong bases that you should memorize. e. There are two is produced, the solution turns yellow. 95. 9.0 3 1023 M 97. 0.50 ways to make a neutral salt. The easiest way is to combine a conjugate base 25 2 99. [H2CO] 5 [H2] 5 2.1 3 10 M; [CH3OH] 5 1.24 M; as formaldehyde of a strong acid (except for HSO4 ) with one of the cations from the strong is removed from the equilibrium by forming some other substance, the equi- bases. These ions have no acidic/basic properties in water so salts of these

librium shifts right to produce more formaldehyde. Hence the concentration ions are neutral. Three examples would be 0.10 M solutions of NaCl, KNO3, of methanol (a reactant) decreases as formaldehyde (a product) reacts to and SrI2. Another type of strong electrolyte that can produce neutral solu- form formic acid. 101. 6.5 3 1026 M 103. 0.81 tions are salts that contain an ion with weak acid properties combined with

105. a. NOCl NO Cl2 an ion of opposite charge having weak base properties. If the Ka for the weak acid ion is equal to the Kb for the weak base ion, then the salt will produce a Initial 2.6 0 0 neutral solution. The most common example of this type of salt is ammo- 2 x 1 x 1x Change 2 2 nium acetate, NH C H O . For this salt, K for NH 1 5 K for C H O 2 5 2 x x x 4 2 3 2 a 4 b 2 3 2 Equilibrium 2.6 2 2 5.6 3 10210. This salt, at any concentration, produces a neutral solution. b. [NO] 5 0.060 M; [Cl2] 5 0.030 M; [NOCl] 5 2.5 M 107. Add N2(g): [N2] 1 2 31. a. H2O1l2 1 H2O1l2 m H3O 1aq2 1 OH 1aq2 or will increase overall, [H2] will decrease, and [NH3] will increase. Remove 1 2 1 2 H2O1l2 m H 1aq2 1 OH 1aq2 K 5 Kw 5 3H 4 3OH 4 H2(g): [N2] will increase, [H2] will decrease overall, and [NH3] will decrease. b. HF aq 1 H O l F2 aq 1 H O1 aq or Add NH3(g): [N2] and [H2] will increase, while [NH3] will also increase over- 1 2 2 1 2 m 1 2 3 1 2 1 2 all. Add Ne(g) at constant volume: no change for [N2], [H2], or [NH3]. In- 1 2 3H 4 3F 4 HF1aq2 m H 1aq2 1 F 1aq2 K 5 Ka 5 crease temperature (add heat): [N2] and [H2] will increase, while [NH3] will 3HF4 1 2 decrease. Decrease volume: [N2] and [H2] will decrease, while [NH3] will in- c. C5H5N1aq2 1 H2O1l2 m C5H5NH 1aq2 1 OH 1aq2 crease. Add a catalyst: no change for [N2], [H2], or [NH3]. 109. PCO 5 1 2 3C5H5NH 4 3OH 4 0.58 atm, P 5 1.65 atm 111. [NOCl] 5 2.0 M, [NO] 5 0.050 M, [Cl2] 5 K 5 Kb 5 CO2 3C H N4 0.025 M 113. 2.1 3 1023 atm 115. P 5 0.704 atm, P 5 0.12 atm 5 5 NO2 N2O4 33. a. H SO ; b. HClO ; c. H PO ; NaOH and KOH are ionic compounds 117. a. Assuming mol/L units for concentrations, K 5 8.16 3 1047; b. 33.9 L 2 3 3 3 3 composed of either Na1 or K1 cations and OH2 anions. When soluble ionic 119. 0.63 121. 0.240 atm 123. a. 2.33 3 1024; b. The argon gas will in- compounds dissolve in water, they form the ions from which they are formed. crease the volume of the container. This is because the container is a constant- The acids in this problem are all covalent compounds. When these acids dis- pressure system, and if the number of moles increases at constant T and P, solve in water, the covalent bond between oxygen and hydrogen breaks to form the volume must increase. An increase in volume will dilute the concentra- H1 ions. 35. a. This expression holds true for solutions of monoprotic tions of all gaseous reactants and gaseous products. Because there are more strong acids having a concentration greater than 1.0 3 1026 M. For example, moles of product gases than reactant gases (3 moles vs. 2 moles), the dilution 0.10 M HCl, 7.8 M HNO , and 3.6 3 1024 M HClO are solutions where this will decrease the numerator of K more than the denominator will decrease. 3 4 expression holds true. b. This expression holds true for solutions of weak This causes Q , K and the reaction shifts right to get back to equilibrium. acids where the two normal assumptions hold. The two assumptions are that Because temperature was unchanged, the value of K will not change. K is a the contribution of H1 from water is negligible and that the acid is less than constant as long as temperature is constant. 125. 192 g; 1.3 atm 127. 33 5% dissociated in water (from the assumption that x is small compared to some number). This expression will generally hold true for solutions of weak Chapter 14 acids having a K value less than 1 3 1024, as long as there is a significant 21. as an acid: HCO 2(aq) 1 H O(l) 34 CO 22(aq) 1 H O1(aq); as a a 3 2 3 3 amount of weak acid present. Three ex ample solutions are 1.5 M HC H O , base: HCO 2(aq) 1 H O(l) 34 H CO (aq) 1 OH2(aq); as an acid: 2 3 2 3 2 2 3 0.10 M HOCl, and 0.72 M HCN. c. This expression holds true for strong H PO 2(aq) 1 H O(l) 34 HPO 22(aq) 1 H O1(aq); as a base: H PO 2(aq) 2 4 2 4 3 2 4 bases that donate 2 OH2 ions per formula unit. As long as the concentration 1 H O(l) 34 H PO (aq) 1 OH2( aq). 23. b, c, and d 25. 10.78 (4 sig- 2 3 4 of the base is above 5 3 1027 M, this expression will hold true. Three ex- nificant figures); 6.78 (3 significant figures); 0.78 (2 significant figures); apH amples are 5.0 3 1023 M Ca(OH) , 2.1 3 1024 M Sr(OH) , and 9.1 3 value is a logarithm. The numbers to the left of the decimal place identify the 2 2 1025 M Ba(OH) . d. This expression holds true for solutions of weak bases power of 10 to which [H1] is expressed in scientific notation—for example, 2 where the two normal assumptions hold. The assumptions are that the 10211, 1027, 1021. The number of decimal places in a pH value identifies the OH2 contribution from water is negligible and that the base is less than number of significant figures in 1[H ]. In all three pH values, the [H1] should 5% ionized in water. For the 5% rule to hold, you generally need bases with be expressed only to two significant figures since these pH values have only K , 1 3 1024 and concentrations of weak base greater than 0.10 M. Three two decimal places. b examples are 0.10 M NH , 0.54 M C H NH , and 1.1 M C H N. 37. One 27. NH 1 NH NH 2 1 NH 1 3 6 5 2 5 5 3 3 m 2 4 reason HF is a weak acid is that the H—F bond is unusually strong and thus, Acid Base Conjugate Conjugate is difficult to break. This contributes to the reluctance of the HF molecules to Base Acid dissociate in water. 39. a. HClO (aq) 1 H O(l) n H O1(aq) 1 ClO 2(aq) One of the NH molecules acts as a base and accepts a proton to form NH 1. 4 2 3 4 3 4 or HClO (aq) n H1(aq) 1 ClO 2(aq); water is commonly omitted from The other NH molecule acts as an acid and donates a proton to form NH 2. 4 4 3 2 34 1 2 1 1 Ka reactions. b. CH3CH2CO2H(aq) H (aq) 1 CH3CH2CO2 (aq); NH4 is the conjugate acid of the NH3 base. In the reverse reaction, NH4 1 1 c. NH4 (aq) 34 H (aq) 1 NH3(aq) 41. a. H2O, base; H2CO3, acid;

57404_Ans_A31-A64.indd 50 10/14/16 5:12 PM Answers to Selected Exercises A51

1 2 1 H3O , conjugate acid; HCO3 , conjugate base; b. C5H5NH , acid; H2O, base; blood. As the CO2 concentration in the blood increases, this drives the equi- 1 2 1 1 C5H5N, conjugate base; H3O , conjugate acid; c. HCO3 , base; C5H5NH , librium to the right, which results in an increased production of H and a 41 acid; H2CO3, conjugate acid; C5H5N, conjugate base 43. a. HClO4, strong lowering of the blood pH. 149. a. Hb(O2)4 in lungs, HbH4 in cells; b. De- 1 acid; b. HOCl, weak acid; c. H2SO4, strong acid; d. H2SO3, weak acid creasing [CO2] will decrease [H ], favoring Hb(O2)4 formation. Breathing 1 45. HClO4 . HClO2 . NH4 . H2O 47. a. HCl; b. HNO2; c. HCN since into a bag raises [CO2]. c. NaHCO3 lowers the acidity from accumulated 2 27 22 it has a larger Ka value. 49. a. [OH ] 5 1.0 3 10 M; the solution is CO2. 151. 4.2 3 10 M 153. a. 2.62; b. 2.4%; c. 8.48 155. a. HA is a neutral. pH 5 7.00; pOH 5 7.00; b. [OH2] 5 12 M; the solution is basic. weak acid. Most of the acid is present as HA molecules; only one set of H1 pH 5 15.08; pOH 5 21.08; c. [OH2] 5 8.3 3 10216 M; the solution is and A2 ions is present. In a strong acid, all of the acid would be dissociated acidic. pH 5 –1.08; pOH 5 15.08; d. [OH2] 5 1.9 3 10210 M; the solution into H1 and A2 ions. b. 2.2 × 1023 157. 7.5 g 159. a. 1.66; b. Fe21 ions is acidic. pH 5 4.27; pOH 5 9.73 51. a. endothermic; b. [H1] 5 [OH2] 5 will produce a less acidic solution (higher pH) due to the lower charge on 2.34 3 1027 M 53. a. [H1] 5 4.0 3 1028 M; [OH2] 5 2.5 3 1027 M; Fe21 as compared with Fe31. As the charge on a metal ion increases, acid 1 216 2 1 2 22 basic; b. [H ] 5 5 3 10 M; [OH ] 5 20 M; basic; c. [H ] 5 10 M; strength of the hydrated ion increases. 161. acidic; HSO4 m SO4 1 [OH2] 5 1 3 10215 M; acidic; d. [H1] 5 6.3 3 1024 M; [OH2] 5 H1; 1.54 163. a. 2.80; b. 1.1 3 1023 M 165. 2.5 3 1023 167. 2.4 211 2 25 1 29 1 2 1.6 3 10 M; acidic; e. [OH ] 5 1 3 10 M; [H ] 5 1 3 10 M; 169. HNO3: [H ] 5 0.0070 M; pH 5 2.15; pOH 5 11.85; [OH ] 5 1.4 3 basic; f. [OH2] 5 2.5 3 10210 M; [H1] 5 4.0 3 1025 M; acidic 10212 M; KOH: [H1] 5 3.3 3 10215 M; pH 5 14.48; pOH 5 –0.48; [OH2] 5 1 23 2 212 2 55. pOH 5 11.9, [H ] 5 8 3 10 M, [OH ] 5 1 3 10 M, acidic 3.0 M OH 171. a. C2H5NH2 and H2O; b. 12.28 173. NaCl: neutral; 1 2 1 2 57. a. H , ClO4 , H2O; 0.602; b. H , NO3 , H2O; 0.602 59. a. 1.00; RbOCl: basic; KI: neutral; Ba(ClO4)2: neutral; NH4NO3: acidic 175. AlCl3 25 b. 20.70; c. 7.00 61. 5.6 3 10 M 63. Add 4.2 mL of 12 M HCl to water , CsClO4 , NaF , NaCN , KOH 177. 7.20. 179. 4540 mL 181. 4.17 32 with mixing; add enough water to bring the solution volume to 1600 mL. 183. 0.022 M 185. 917 mL of water evaporated 187. PO4 , Kb 5 0.021; 1 2 22 27 2 212 65. a. HNO2 and H2O, 2.00; b. HC2H3O2 and H2O, 2.68 67. [H ] 5 [F ] 5 HPO4 , Kb 5 1.6 3 10 ; H2PO4 , Kb 5 1.3 3 10 ; from the Kb values, 23 2 212 32 3.5 3 10 M, [OH ] 5 2.9 3 10 M, [HF] 5 0.017 M, 2.46 PO4 is the strongest base. 189. a. basic; b. acidic; c. basic; d. acidic; 2 1 23 23 24 23 69. [HC3H5O2] 5 0.099 M, [C3H5O2 ] 5 [H ] 5 1.1 3 10 M, e. acidic 191. 1.0 3 10 193. 5.4 3 10 195. 2.5 3 10 [OH2] 5 9.1 3 10212, pH 5 2.96, 1.1% dissociated. 71. 2.68 73. 1.57; 5.9 3 1029 M 75. a. 0.60%; b. 1.9%; c. 5.8%; d. Dilution shifts equilib- Chapter 15 rium to the side with the greater number of particles (% dissociation 11. When an acid dissociates, ions are produced. A common ion is when one increases). e. [H1] also depends on initial concentration of weak acid. of the product ions in a particular equilibrium is added from an outside 77. 1.4 3 1024 79. 0.16 81. 0.024 M 83. 6 3 1023 source. For a weak acid dissociating to its conjugate base and H1, the com- 1 2 mon ion would be the conjugate base; this would be added by dissolving a 1 2 3NH4 43OH 4 85. a. NH3 1aq2 1 H2O1l2 mNH4 1aq2 1 OH 1aq2 Kb 5 ; soluble salt of the conjugate base into the acid solution. The presence of the 3NH3 4 1 2 b. C5H5N1aq2 1 H2O1l2 m C5H5NH 1aq2 1 OH 1aq2 conjugate base from an outside source shifts the equilibrium to the left so 1 2 3C5H5NH 4 3OH 4 less acid dissociates. 13. The more weak acid and conjugate base present, Kb 5 the more H1 and/or OH2 that can be absorbed by the buffer without signifi- 3C5H5N4 cant pH change. When the concentrations of weak acid and conjugate base 87. NH . C H N . H O . NO 2; HNO . C H NH1 . NH 1 . H O 3 5 5 2 3 3 5 5 4 2 are equal (so that pH 5 pK ), the buffer system is equally efficient at absorb- 89. a. 13.00; b. 7.00; c. 14.30 91. a. K1, OH2, and H O, 0.015 M, 12.18; a 2 ing either H1 or OH2. If the buffer is overloaded with weak acid or with b. Ba21, OH2, and H O, 0.030 M, 12.48 93. 0.16 g 95. NH and H O, 2 3 2 conjugate base, then the buffer is not equally efficient at absorbing either 1H 1.6 3 1023 M, 11.20 97. a. [OH2] 5 1.2 3 1025 M; [H1] 5 8.3 3 10210 M; or OH2. 15. a. Let’s call the acid HA, which is a weak acid. When HA is pH 5 9.08; b. [OH2] 5 1.3 3 1022 M; [H1] 5 7.7 3 10213 M; pH 5 12.11 present in the beakers, it exists in the undissociated form, making it a weak 99. 12.00 101. a. 1.3%; b. 4.2%; c. 6.4% 103. 1.0 3 1029 acid. A strong acid would exist as separate H1 and A2 ions. b. beaker c → 105. H SO aq HSO 2 aq 1 H1 aq K reaction 2 3 1 2 m 3 1 2 1 2 a1 beaker a → beaker e → beaker b → beaker d c. pH 5 pKa when a buffer 2 22 1 1 HSO3 1aq2 m SO3 1aq2 H 1aq2 Ka2 reaction solution is present that has equal concentrations of the weak acid and conju- 219 107. 3.00 109. 4.00; 1.0 3 10 M 111. 20.30 113. HCl . NH4Cl . gate base. This is beaker e. d. The equivalence point is when just enough 2 2 1 27 2 KNO3 . KCN . KOH 115. OCl 117. a. [OH ] 5 [H ] 5 1.0 3 10 OH has been added to exactly react with all of the acid present initially. M, pH 5 7.00; b. [OH2] 5 2.4 3 1025 M, [H1] 5 4.2 3 10210 M, pH 5 9.38 This is beaker b. e. Past the equivalence point, the pH is dictated by the con- 2 26 1 2 119. a. 5.82; b. 10.95 121. [HN3] 5 [OH ] 5 2.3 3 10 M, [Na ] 5 centration of excess OH added from the strong base. We can ignore the 2 1 29 0.010 M, [N3 ] 5 0.010 M, [H ] 5 4.3 3 10 M 123. NaF amount of hydroxide added by the weak conjugate base that is also present. 2 125. 3.66 127. 3.08 129. a. neutral; b. basic; NO2 1 H2O m HNO2 This is beaker d. 17. The three key points to emphasize in your sketch are 2 1 1 1 1 OH ; c. acidic; C5H5NH m C5H5N 1 H ; d. acidic because NH4 is the initial pH, the pH at the halfway point to equivalence, and the pH at the 2 1 1 2 a stronger acid than NO2 is a base; NH4 34 NH3 1 H ; NO2 1 H2O equivalence point. For the two weak bases titrated, pH 5 pKa at the halfway 2 2 2 34 HNO2 1 OH ; e. basic; OCl 1 H2O 34 HOCl 1 OH ; f. basic point to equivalence (50.0 mL HCl added) because [weak base] 5 [conju- 2 1 2 because OCl is a stronger base than NH4 is an acid; OCl 1 H2O 34 gate acid] at this point. For the initial pH, the strong base has the highest pH 2 1 1 HOCl 1 OH , NH4 34 NH3 1 H 131. a. HIO3 , HBrO3; as the elec- (most basic), while the weakest base has the lowest pH (least basic). At the tronegativity of the central atom increases, acid strength increases. b. HNO2 equivalence point, the strong base titration has pH 5 7.0. The weak bases , HNO3; as the number of oxygen atoms attached to the central atom in- titrated have acidic pHs at the equivalence point because the conjugate acids creases, acid strength increases. c. HOI , HOCl; same reasoning as in part of the weak bases titrated are the major species present. The weakest base

a. d. H3PO3 , H3PO4; same reasoning as in part b. 133. a. H2O , H2S , has the strongest conjugate acid so its pH will be lowest (most acidic) at the H2Se; acid strength increases as bond energy decreases. b. CH3CO2H , equivalence point. FCH2CO2H , F2CHCO2H , F3CCO2H; as the electronegativity of the 1 1 neighboring atoms increases, acid strength increases. c. NH4 , HONH3 ; 1 1 Strong base same reasoning as in part b. d. NH4 , PH4 ; same reasoning as in part a. 135. a. basic; CaO1s2 1 H2O1l2 S Ca1OH2 2 1aq2 ; b. acidic; SO2(g) 1 H2O(l) n H2SO3(aq); c. acidic; Cl2O1g2 1 H2O1l2 S 2HOCl1aq2 137. a. B(OH)3, 1 2 –5 acid; H2O, base; b. Ag , acid; NH3, base; c. BF3, acid; F , base 7.0 Kb = 10 1 31 2 pH 139. Al(OH)3(s) 1 3H (aq) n Al (aq) 1 3H2O(l); Al(OH)3(s) 1 OH (aq) 2 31 n Al(OH)4 ( aq) 141. Fe ; because it is smaller with a greater positive 31 –10 charge, Fe will be more strongly attracted to a lone pair of electrons from Kb = 10 a Lewis base. 143. 990 mL H2ONH 145. 4Cl 147. If breathing is slow or very weak, CO2 is not exhaled normally, resulting in a CO2 buildup in the Volume HCl added (mL)

57404_Ans_A31-A64.indd 51 10/14/16 5:12 PM A52 Answers to Selected Exercises

19. At 50.0 mL of NaOH added, this is the first halfway point to equivalence Volume (mL) pH where H AsO and H AsO – are the major species present (along with H O). 3 4 2 4 2 25.0 5.28 At this point, we have a buffer solution in which [H AsO ] 5 [H AsO –] and 3 4 2 4 25.1 3.7 pH 5 pK 5 –log(5.5 3 10–3) 5 2.26. At 150.0 mL of NaOH added, this is a1 26.0 2.71 the second halfway point to equivalence at which H AsO – and HAsO 2– are 2 4 4 28.0 2.24 the major species present (along with H O). At this point, we have a buffer 2 30.0 2.04 – 5 2– 5 5 3 solution in which [H2AsO4 ] [HAsO4 ] and pH pKa2 –log(1.7 –7 2– See Solutions Guide for pH plot. 10 ) 5 6.77. At the third halfway point to equivalence, [HAsO4 ] 5 3– 5 71. a. 4.19, 8.45; b. 10.74; 5.96; c. 0.89, 7.00 73. 2.1 3 1026 75. a. yel- [AsO4 ] and pH pKa3 . This third halfway point will be at 250.0 mL of NaOH added. 21. Only the third beaker represents a buffer solution. A low; b. 8.0; c. blue 77. phenolphthalein 79. Phenol red is one possible in- weak acid and its conjugate base both must be present in large quantities in dicator for the titration in Exercise 63. Phenolphthalein is one possible indica- order to have a buffer solution. This is only the case in the third beaker. The tor for the titration in Exercise 65. 81. Phenolphthalein is one possible first beaker represents a beaker full of strong acid that is 100% dissociated. indicator for Exercise 67. Bromcresol green is one possible indicator for The second beaker represents a weak acid solution. In a weak acid solution, Exercise 69. 83. The pH is between 5 and 8. 85. a. yellow; b. green; 2 only a small fraction of the acid is dissociated. In this representation, 1/10 of c. yellow; d. blue 87. a. 200.0 mL; b. i. H2A, H2O; ii. H2A, HA , H2O, 1 2 1 2 22 1 22 the weak acid has dissociated. The only B2 present in this beaker is from the Na ; iii. HA , H2O, Na ; iv. HA , A , H2O, Na ; v. A , 2 H O, Na1; vi. A22, H O, Na1, OH2; c. K 5 1 3 1024; K 5 1 3 1028 dissociation of the weak acid. A buffer solution has B added from another 2 2 a1 a2 source. 23. When strong acid or strong base is added to a sodium bicarbon- 3acid4 89. pOH 5 pK 1 log 91. a. The optimal buffer pH is about 8.1; ate/sodium carbonate buffer mixture, the strong acid/base is neutralized. The b 3base4 reaction goes to completion resulting in the strong acid/base being replaced b. 0.083 at pH 5 7.00; 8.3 at pH 5 9.00; c. 8.08; 7.95 93. a. potassium with a weak acid/base. This results in a new buffer solution. The reactions fluoride 1 HCl; b. benzoic acid 1 NaOH; c. acetic acid 1 sodium acetate; 1 22 2 2 2 are H (aq) 1 CO3 (aq) n HCO3 (aq); OH (aq) 1 HCO3 (aq) n d. HOCl 1 NaOH; e. ammonium chloride 1 NaOH 95. a. 1.1 < 1; b. A 22 CO3 (aq) 1 H2O(l) 25. a. 2.96; b. 8.94; c. 7.00; d. 4.89 27. 1.1% vs. 1.3 best buffer has approximately equal concentrations of weak acid and conju- 22 2 2 3 10 % dissociated; the presence of C3H5O2 in solution 25d greatly inhib- gate base, so pH < pKa for a best buffer. The pKa value for an H3PO4H2PO4 23 its the dissociation of HC3H5O2. This is called the common ion effect. buffer is 2log(7.5 3 10 ) 5 2.12. A pH of 7.15 is too high for an 2 29. a. 1.70; b. 5.49; c. 1.70; d. 4.71 31. a. 4.29; b. 12.30; c. 12.30; d. 5.07 H3PO4H2PO4 buffer to be effective. At this high of a pH, there would be so 33. solution d; solution d is a buffer solution that resists pH changes. 35. 3.40 little H3PO4 present that we could hardly consider it a buffer; this solution 37. 3.48; 3.22 39. a. 5.14; b. 4.34; c. 5.14; d. 4.34 41. 4.37 43. a. 7.97; would not be effective in resisting pH changes, especially when a strong base – b. 8.73; both solutions have an initial pH 5 8.77. The two solutions differ in is added. 97. Removal of NaHCO3 (HCO3 ) causes the buffer equilibrium their buffer capacity. Solution b with the larger concentrations has the greater to shift right to produce more H1. This causes the blood to become more 2– – capacity to resist pH change. 45. The SO3 concentration is greater in this acidic, so pH decreases. Treatment involves the IV addition of HCO3 . buffer; 0.56 M 47. 15 g 49. a. 0.19; b. 0.59; c. 1.0; d. 1.9 99. a. 1.8 3 109; b. 5.6 3 104; c. 1.0 3 1014 101. 4.4 L 103. 65 mL 51. 1.3 3 1022 M 53. HOCl; there are many possibilities. One possibility is 105. 180. g/mol; 3.3 3 1024; assumed acetylsalicylic acid is a weak monopro- a solution with [HOCl] 5 1.0 M and [NaOCl] 5 0.35 M. 55. 8.18; add tic acid. 107. 0.210 M 109. 1.74 3 1028 111. 3.62 113. 0.92 M 0.20 mole NaOH 57. solution d 59. a. 1.0 mole; b. 0.30 mole; 115. a. False; the buffer with the larger concentration has the greater buffer c. 1.3 moles 61. a. ,22 mL base added; b. buffer region is from ,1 mL to capacity. b. True; when the base component of the buffer has a larger concen- ,21 mL base added. The maximum buffering region would be from ,5 mL tration than the acid component, then pH . pKa. c. True; as more of the acid to ,17 mL base added with the halfway point to equivalence (,11 mL) as component of the buffer is added, the pH will become more acidic (pH will 211 the best buffer point. c. ,11 mL base added; d. 0 mL base added; e. ,22 mL decrease). d. False; for this buffer, pKa 5 2log(2.3 3 10 ) 5 10.64. base added (the stoichiometric point); f. any point after the stoichiometric Here, we have more of the acid component of the buffer than the base com-

point (volume base added . ,22 mL) 63. a. 0.699; b. 0.854; c. 1.301; ponent. When this occurs, pH , pKa. So we can say that for this situation, d. 7.00; e. 12.15 65. a. 2.72; b. 4.26; c. 4.74; d. 5.22; e. 8.79; f. 12.15 pH , 10.64, not that pH , 3.36. e. True; when [weak acid] 5 [conjugate

base] in a buffer, pH 5 pKa. 117. a. 5.10; b. 9.21; c. 9.69; d. 10.95; 67. Volume (mL) pH e. 12.04 119. a. The correct order of increasing pH at the halfway point is 0.0 2.43 ii , iv , iii , i. b. The correct order of increasing pH at the equivalence 4.0 3.14 point is i , ii , iv , iii. c. All require the same volume of titrant to reach the 8.0 3.53 equivalence point. 121. 49 mL 123. 3.9 L 125. 4.8 g cacodylic acid and 12.5 3.86 14 g sodium cacodylate 127. a. The major species present at the various 1 22 1 22 20.0 4.46 points after H reacts completely are A: CO3 , H2O, Na ; B: CO3 , 2 2 1 2 2 1 2 24.0 5.24 HCO3 , H2O, Cl , Na ; C: HCO3 , H2O, Cl , Na ; D: HCO3 , CO2 2 1 2 1 1 24.5 5.6 (H2CO3), H2O, Cl , Na ; E: CO2 (H2CO3), H2O, Cl , Na ; F: H , 2 1 24.9 6.3 CO2 (H2CO3), H2O, Cl , Na ; b. B: pH 5 10.25; D: pH 5 6.37 210 25.0 8.28 129. pH < 5.0; Ka < 1 3 10 131. 3.00 133. 2.78 25.1 10.3 26.0 11.30 Chapter 16 28.0 11.75 11. K values can only be compared to determine relative solubilities when 30.0 11.96 sp the salts produce the same number of ions. Here, Ag S and CuS do not pro- See Solutions Guide for pH plot. 2 duce the same number of ions when they dissolve, so each has a different

69. Volume (mL) pH mathematical relationship between the Ksp value and the molar solubility. To 0.0 11.11 determine which salt has the larger molar solubility, you must do the actual 4.0 9.97 calculations and compare the two molar solubility values. 13. i. This is the 8.0 9.58 result when you have a salt that breaks up into two ions. Examples of these 12.5 9.25 salts would be AgCl, SrSO4, BaCrO4, and ZnCO3. ii. This is the result when 20.0 8.65 you have a salt that breaks up into three ions, either two cations and one an- 24.0 7.87 ion or one cation and two anions. Some examples are SrF2, Hg2I2, and 24.5 7.6 Ag2SO4. iii. This is the result when you have a salt that breaks up into four 24.9 6.9 ions, either three cations and one anion (Ag3PO4) or one cation and three

57404_Ans_A31-A64.indd 52 10/14/16 5:12 PM Answers to Selected Exercises A53

anions (ignoring the hydroxides, there are no examples of this type of salt in 31 22 1 b. V 1 C2O4 m VC2O4 Table 16.1). iv. This is the result when you have a salt that breaks up into five 1 22 2 VC2O4 1 C2O4 m V1C2O42 2 ions, either three cations and two anions [Sr3(PO4)2] or two cations and three V1C O 2 2 1 C O 22 m V1C O 2 32 anions (no examples of this type of salt are in Table 16.1). 15. For the Ksp 2 4 2 2 4 2 4 3 reaction of a salt dissolving into its respective ions, the common ion would 31 22 32 V 1aq2 1 3C2O4 1aq2 m V1C2O42 3 1aq2 be if one of the ions in the salt was added from an outside source. When a 67. 1.0 3 1042 69. Hg21 aq 1 2I2 aq HgI s (orange precipitate); common ion is present, the Ksp equilibrium shifts to the left resulting in less 1 2 1 2 m 2 1 2 2 22 232 of the salt dissolving into its ions. 17. Some people would automatically HgI2(s) 1 2I (aq) m HgI4 (aq) (soluble complex ion) 71. 3.3 3 10 M think that an increase in temperature would increase the solubility of a salt. 73. a. 1.0 3 1023 M; b. 2.0 3 1027 M; c. 8.0 3 10215 M 75. a. 1.2 3 1028 24 This is not always the case because some salts show a decrease in solubility mol/L; b. 1.5 3 10 mol/L; c. The presence of NH3 increases the solubility 1 as temperature increases. The two major methods used to increase solubility of AgI. Added NH3 removes Ag from solution by forming the complex ion 1 1 1 of a salt both involve removing one of the ions in the salt by reaction. If the Ag(NH3)2 . As Ag is removed, more AgI will dissolve to replenish the Ag salt has an ion with basic properties, adding H1 will increase the solubility concentration. 77. 4.7 3 1022 mol/L 79. Test tube 1: added Cl2 reacts of the salt because the added H1 will react with the basic ion, thus removing with Ag1 to form the silver chloride precipitate. The net ionic equation is 1 2 1 it from solution. More salt dissolves in order to make up for the lost ion. Ag 1aq2 1 Cl 1aq2 S AgCl1s2 . Test tube 2: added NH3 reacts with Ag ions to form the soluble complex ion Ag(NH ) 1. As this complex ion forms, Some examples of salts with basic ions are AgF, CaCO3, and Al(OH)3. The 3 2 other way to remove an ion is to form a complex ion. For example, the Ag1 Ag1 is removed from solution, which causes AgCl(s) to dissolve. When 1 enough NH is added, then all of the silver chloride precipitate will dissolve. ion in silver salts forms the complex ion Ag(NH3)2 as ammonia is added. 3 1 The equation is AgCl s 1 2NH aq Ag NH 1 aq 1 Cl2 aq . Test Silver salts increase their solubility as NH3 is added because the Ag ion is 1 2 3 1 2 S 1 32 2 1 2 1 2 1 1 removed through complex ion formation. 19. For conjugate acid–base tube 3: added H reacts with the weak base NH3 to form NH4 . As NH3 is pairs, the weaker the acid, the stronger is the conjugate base. Because HX is removed, Ag1 ions are released to solution, which can then react with Cl2 to 2 re-form AgCl(s). The equations are Ag(NH ) 1(aq) 1 2H1(aq) Ag1(aq) a stronger acid (has a larger Ka value) than HY, Y will be a stronger base 3 2 n 1 1 2 than X2. In acidic solution, Y2 will have a greater affinity for the 1H ions. 1 2NH4 (aq) and Ag 1aq2 1 Cl 1aq2 S AgCl1s2 . 81. 1.7 g AgCl; 5 3 Therefore, AgY(s) will be more soluble in acidic solution because more Y2 1029 mol/L 83. 2.7 3 1025 mol/L; the solubility of hydroxyapatite will will be removed through reaction with H1, which will cause more AgY(s) increase as a solution gets more acidic, since both phosphate and hydroxide 1 can react with H1. 6 3 1028 mol/L; the hydroxyapatite in the tooth enamel to dissolve. 21. In 2.0 M NH3, the soluble complex ion Ag(NH3)2 forms, which increases the solubility of AgCl(s). The reaction is is converted to the less soluble fluorapatite by fluoride-treated water. The 1 2 less soluble fluorapatite will then be more difficult to dissolve, making teeth AgCl(s) 1 2NH3(aq) m Ag(NH3)2 (aq) 1 Cl (aq). In 2.0 M NH4NO3, 1 less susceptible to decay. 85. Precipitation of Al(OH) (s) will begin when NH3 is only formed by the dissociation of the weak acid NH4 . There is not 3 pH . 3.7. 87. 7.0 3 1028 89. AgOH, 10.15; Cd(OH) , 9.34; Pb(OH) , enough NH3 produced by this reaction to dissolve AgCl(s) by the formation 2 2 1 2 9.11 91. 6.2 3 105 93. a. 6.7 3 1026 mol/L; b. 1.2 3 10213 mol/L; of the complex ion. 23. a. AgC2H3O2 1s2 m Ag 1aq2 1 C2H3O2 1aq2; 1 2 31 2 c. Pb(OH) (s) will not form since Q , K . 95. a. 1.6 3 1026; b. 0.056 Ksp 5 3Ag 4 3C2H3O2 4; b. Al1OH2 3 1s2 m Al 1aq2 1 3OH 1aq2 ; Ksp 5 2 sp 31 2 3 21 32 mol/L 97. 7.4 3 1024 99. 1.1 3 10232 101. 6.1 3 1026 mol/L [Al ][OH ] ; c. Ca3(PO4)2(s) m 3Ca (aq) 1 2PO4 (aq); Ksp 5 21 3 32 2 29 219 28 103. 1.08 3 1026 mol/L 105. a. 0.33 mol/L; b. 0.33 M; c. 4.8 3 1023 M [Ca ] [PO4 ] 25. a. 2.3 3 10 ; b. 8.20 3 10 27. 1.4 3 10 27 23 29. 3.92 3 1025 31. a. 1.6 3 1025 mol/L; b. 9.3 3 1025 mol/L; c. 6.5 3 107. a. 7.1 3 10 mol/L; b. 8.7 3 10 mol/L; c. The presence of NH3 in- 1 1027 mol/L 33. 0.89 g 35. 1.1 3 1025 mol/L 37. 2 3 10211 mol/L creases the solubility of AgBr. Added NH3 removes Ag from solution by 217 211 forming the complex ion, Ag(NH ) 1. As Ag1 is removed, more AgBr(s) will 39. a. CaF2; b. FePO4 41. a. 4 3 10 mol/L; b. 4 3 10 mol/L; c. 4 3 3 2 1 10229 mol/L 43. a. 1.4 3 1022 mol/L; b. 1.2 3 1023 mol/L; c. 3.9 3 dissolve to replenish the Ag concentration. d. 0.41 g AgBr; e. Added HNO3 1023 mol/L 45. 2.3 3 10211 mol/L 47. 3.5 3 10210 49. If the anion in will have no effect on the AgBr(s) solubility in pure water. Neither H1 nor 2 1 2 2 the salt can act as a base in water, then the solubility of the salt will increase NO3 react with Ag or Br ions. Br is the conjugate base of the strong acid as the solution becomes more acidic. Added H1 will react with the base, HBr, so it is a terrible base. However, added HNO3 will reduce the solubility 25 forming the conjugate acid. As the basic anion is removed, more of the salt of AgBr(s) in the ammonia solution. NH3 is a weak base (Kb 5 1.8 3 10 ). 1 1 will dissolve to replenish the basic anion. The salts with basic anions are Added H will react with NH3 to form NH4 . As NH3 is removed, a smaller amount of the Ag(NH ) 1 complex ion will form, resulting in a smaller Ag3PO4, CaCO3, CdCO3, and Sr3(PO4)2. Hg2Cl2 and PbI2 do not have any pH 3 2 dependence because Cl2 and I2 are terrible bases (the conjugate bases of amount of AgBr(s) that will dissolve. 109. 5.7 3 1022 mol/L 111. 3 M strong acids). 113. 4.8 3 10211 M 115. a. 5.8 3 1024 mol/L; b. greater; F2 is a weak base 211 2 1 (K 5 1.4 3 10 ), so some of the F is removed by reaction with water. As 1 1 22 excess H b Ag3PO4 1s2 1 H 1aq2 h 3Ag 1aq2 1 HPO4 1aq2 8888n 2 23 F is removed, more SrF2 will dissolve; c. 3.5 3 10 mol/L 117. pH 5 1 3Ag 1aq2 1 H PO 1aq2 21 3 4 0.70; 0.20 M Ba ; 23 g BaSO4(s) 119. Precipitation of CaF2(s) will begin 1 26 1 21 2 excess H when just over 9.0 3 10 g Ca(NO3)2 has been added. CaCO3 1s2 1 H 1aq2 h Ca 1aq2 1 HCO3 1aq2 8888n 21 Ca 1aq2 1 H2CO3 1aq2 3H2O1l2 1 CO2 1g2 4 Chapter 17 1 21 2 excess H1 11. Living organisms need an external source of energy to carry out these CdCO3 1s2 1 H 1aq2 h Cd 1aq2 1 HCO3 1aq2 8888n 21 processes. For all processes combined, DSuniv must be greater than zero (the Cd 1aq2 1 H2CO3 1aq2 3H2O1l2 1 CO2 1g2 4 2nd law). 13. As any process occurs, DSuniv will increase; DSuniv cannot de- 1 21 22 excess H1 Sr3 1PO42 2 1s2 1 2H 1aq2 h 3Sr 1aq2 1 2HPO4 1aq2 8888n crease. Time also goes in one direction, just as DSuniv goes in one direction. 21 3Sr 1aq2 1 2H3PO4 1aq2 15. This reaction is kinetically slow but thermodynamically favorable (DG , 0). Thermodynamics only tells us if a reaction can occur. To answer 51. 1.5 3 10219 g 53. No precipitate forms. 55. PbF (s) will not form. 2 the question will the reaction occur, one also needs to consider the kinetics 57. [K1] 5 0.160 M, [C O 22] 5 3.3 3 1027 M, [Ba21] 5 0.0700 M, [Br2] 5 2 4 (speed of reaction). The ultraviolet light provides the activation energy for this 0.300 M 59. 8.0 3 1026 mol/L 61. [AgNO ] . 5.6 3 1025 M 3 slow reaction to occur. 17. DS 5 2DH/T; heat flow (DH) into or out of 63. PbS(s) will form first, followed by Pb (PO ) (s), and PbF (s) will form surr 3 4 2 2 the system dictates DS . If heat flows into the surroundings, the random mo- last. surr tions of the surroundings increase, and the entropy of the surroundings in- 65. a. Ni21 1 CN 2 m NiCN1 creases. The opposite is true when heat flows from the surroundings into the 1 2 NiCN 1 CN m Ni1CN2 2 system (an endothermic reaction). Although the driving force described here 2 2 Ni1CN2 2 1 CN m Ni1CN2 3 really results from the change in entropy of the surroundings, it is often de- 2 2 22 scribed in terms of energy. Nature tends to seek the lowest possible en- Ni1CN2 3 1 CN m Ni1CN2 4 ergy. 19. a. increases; b. no change; c. decreases 21. Note that these sub- 21 2 22 Ni 1aq2 1 4CN 1aq2 m Ni1CN2 4 1aq2

57404_Ans_A31-A64.indd 53 10/14/16 5:12 PM A54 Answers to Selected Exercises

stances are not in the solid state but are in the aqueous state; water molecules taneous at standard concentrations and 298 K. c. T > 630 K 69. CH4(g) 1 are also present. There is an apparent increase in ordering when these ions are CO2(g) n CH3CO2H(l), DH8 5 216 kJ, DS8 5 2240. J/K, DG8 5 56 kJ; placed in water. The hydrating water molecules must be in a highly ordered CH3OH(g) 1 CO(g) n CH3CO2H(l), DH8 5 2173 kJ, DS8 5 2278 J/K, state when surrounding these anions. 23. One can determine DS8 and DH8 DG8 5 290. kJ; the second reaction is preferred. It should be run at tem- for the reaction using the standard entropies and standard enthalpies of forma- peratures below 622 K. 71. 2188 kJ 73. a. shifts right; b. no shift since tion in Appendix 4, then use the equation DG8 5 DH8 2 TDS8. One can also the reaction is at equilibrium; c. shifts left 75. 2198 kJ; 5.07 3 1034 use the standard free energies of formation in Appendix 4. And finally, one 77. 8.72; 0.0789 79. 140 kJ 81. a. 2.22 3 105; b. 94.3 83. 271 kJ/mol can use Hess’s law to calculate DG8. Here, reactions having known DG8 val- 85. DH8 5 1.1 3 105 J/mol; DS8 5 330 J/K ? mol; the major difference in the ues are manipulated to determine DG8 for a different reaction. For tempera- plot is the slope of the line. An endothermic process has a negative slope tures other than 258C, DG8 is estimated using the DG8 5 DH8 2 TDS8 equa- for the ln(K) versus 1yT plot, whereas an exothermic process has a positive tion. The assumptions made are that the DH8 and DS8 values determined from slope. 87. All have positive signs at 258C. 89. 447 J/K ? mol 91. DSuniv Appendix 4 data are temperature independent. We use the same DH8 and DS8 must be positive. 93. a. DH8 5 2183 kJ; DS8 5 2100. J/K; DG8 5 2153 kJ;

values as determined when T 5 258C, then plug in the new temperature in b. DH8 5 2177 kJ; DS8 5 2129 J/K; DG°343 5 2133 kJ; c. DH8 5 21336 kJ; 24 kelvins into the equation to estimate DG8 at the new temperature. 25. The DS8 5 88 J/K; DG°973 5 21422 kJ 95. 43.7 K 97. 7.0 3 10 99. 60 light source for the first reaction is necessary for kinetic reasons. The first re- 101. DS is more favorable (less negative) for reaction 2 than for reaction 1,

action is just too slow to occur unless a light source is available. The kinetics resulting in K2 . K1. In reaction 1, seven particles in solution form one of a reaction are independent of the thermodynamics of a reaction. Even particle. In reaction 2, four particles form one, which results in a smaller though the first reaction is more favorable thermodynamically (assuming decrease in positional probability than for reaction 1. 103. 725 K standard conditions), it is unfavorable for kinetic reasons. The second reaction 105. A negative value for DS8 indicates a process that has a decrease in posi- has a negative DG8 value and is a fast reaction, so the second reaction occurs tional probability. Processes b, c, and d all are expected to have negative very quickly. When considering if a reaction will occur, thermodynamics and values for DS8. In these processes, the number of moles of gaseous molecules kinetics must both be considered. 27. Only statement e is true. 29. a, b, c decreases as reactants are converted to products. Whenever this occurs, posi- 31. Possible arrangements for one molecule: tional probability decreases. Processes a and e have positive DS8 values. In process a, the number of gaseous molecules increase as reactants are converted to products, so positional probability increases. In process e, the liquid state has a larger positional probability compared to the solid state.

1 way1 way 107. DS8 5 49 J/K; DSsurr = 2140 J/K 109. 74.4 kJ/mol 111. a. False; the Both are equally probable. moles of gaseous molecules decrease in this reaction, so DS8 is negative Possible arrangements for two molecules: (unfavorable). From the problem DG8 is negative, which can only occur if DH8 is negative (exothermic). b. True. c. False; heat is a product in this exothermic reaction. As heat is added (temperature increases), this reaction will shift left by Le Châtelier’s principle. This results in a decrease in the value 1 way2 ways 1 way of K. At equilibrium at the higher temperature, reactant concentrations will Most probable increase and the product concentration will decrease. Therefore, the ratio of [PCl5]/[PCl3] will decrease with an increase in temperature. d. False; Possible arrangement for three molecules: when the signs of DH8 and ΔS8 are both negative, the reaction will only be spontaneous at temperatures below some value (at low temperatures) where the favorable DH8 term dominates. e. True; when DG8 , 0, K must be greater than 1. 113. a. Vessel 1: At 08C, this system is at equilibrium, so 1 way3 ways 3 ways 1 way DSuniv 5 0 and DS 5 DSsurr. Because the vessel is perfectly insulated, q 5 0 so DSsurr 5 0 5 DSsys. b. Vessel 2: The presence of salt in water lowers the Equally most probable freezing point of water to a temperature below 08C. In vessel 2 the conversion of ice into water will be spontaneous at 08C, so DS > 0. Because 33. We draw all of the possible arrangements of the two particles in the three univ the vessel is perfectly insulated, DS 5 0. Therefore, DS must be positive levels. surr sys (DS > 0) in order for DSuniv to be positive. 115. DH8 5 286 kJ; DG8 5 326 kJ; x x xx 5 3 258 5 3 241 2 kJ — — — — — — K 7.22 10 ; PO3 3.3 10 atm; this partial pressure represents 1 kJ — —x — —xx —x — one molecule of ozone per 9.5 3 1017 L of air. Equilibrium is probably not 0 kJ —xx —x —x — — — maintained under the conditions because the concentration of ozone is not Total E 5 0 kJ 1 kJ 2 kJ 2 kJ 3 kJ 4 kJ large enough to maintain equilibrium 117. a. The most likely total energy is 2 kJ. 35. a. H2 at 1008C and 0.5 atm; b. N2 at 2Ea 21Ea 2 DG°2 kf 5 A exp a b and kr 5 A exp a b , STP; c. H2O(l) 37. a. negative; b. positive 39. Spontaneous (DG , 0) for RT RT b, c, d 41. 89.3 J/K ? mol 43. a. yes (DG , 0); b. 196 K 45. a. negative; b. negative; c. negative; d. positive 47. a. C (s); b. CH OH(g); kf 2Ea 1Ea 2 DG°2 2DG° graphite 2 5 5 exp a 1 b 5 exp a b c. CO2(g) 49. a. negative, 2186 J/K; b. positive, 187 J/K; c. hard to predict kr RT RT RT since Dn 5 0; 138 J/K 51. 262 J/K ? mol 53. a. DH and DS are both posi- 2DG° From DG8 5 2RT ln K, K also equals the same expression: K 5 exp a b, tive; b. Srhombic 55. a. DH and DS are both negative; b. low temperatures RT k 57. a. DH8 5 2803 kJ, DS8 5 24 J/K, DG8 5 2802 kJ; b. DH8 5 2802 kJ, so K 5 f . b. A catalyst increases the value of the rate constant (increases rate) DS8 5 2262 J/K, DG8 5 2880. kJ; c. DH8 5 2416 kJ, DS8 5 2209 J/K, kr DG8 5 2354 kJ; d. DH8 5 2176 kJ, DS8 5 2284 J/K, DG8 5 291 kJ by lowering the activation energy. For the equilibrium constant K to remain 59. a. DH8 5 292 kJ; DS8 5 2199 J/K; DG8 5 233 kJ; b. This reaction is constant, both kf and kr must increase by the same factor. Therefore, a catalyst spontaneous at standard conditions. c. At T , 460 K and standard pressures must increase the rate of both the forward and the reverse reactions. (1 atm), the favorable DH8 term dominates and the reaction is spontaneous 119. a. 0.333; b. PA 5 1.50 atm; PB 5 0.50 atm; c. DG 5 DG8 1 (DG8 , 0). 61. At 90.8C: DG8 5 1.0 kJ/mol; at 110.8C: DG8 5 21.1 kJ/mol RT ln(PByPA) 5 2722 J 2 2722 J 5 0 121. greater than 7.5 torr 123. 6 M 63. 2817 kJ 65. 2731 kJ/mol 67. a. 53 kJ; b. No, the reaction is not spon- 125. 61 kJ/mol 127. 0.14 atm 129. 24.1 kJ/mol

57404_Ans_A31-A64.indd 54 10/14/16 5:12 PM Answers to Selected Exercises A55

contains the oxidation half-reaction compounds/ions, and the cathode com- Chapter 18 partment contains the reduction half-reaction compounds/ions. The electrons 17. Oxidation: increase in oxidation number, loss of electrons; reduction: flow from the anode to the cathode. For each of the following answers, all decrease in oxidation number, gain of electrons 19. Reactions a, b, and c 31 solutes are 1.0 M and all gases are at 1.0 atm. a. 7H2O(l) 1 2Cr (aq) 1 are oxidation–reduction reactions. 22 2 1 3Cl2(g) n Cr2O7 (aq) 1 6Cl (aq) 1 14H (aq); cathode: Pt electrode; 2 31 1 Oxidizing Reducing Substance Substance Cl2 bubbled into solution, Cl in solution; anode: Pt electrode; Cr , H , 22 21 21 Agent Agent Oxidized Reduced and Cr2O7 in solution; b. Cu (aq) 1 Mg(s) n Cu(s) 1 Mg (aq); cathode: Cu electrode; Cu21 in solution; anode: Mg electrode; Mg21 in solution

a. H2O CH4 CH4(C) H2O(H) 39. a. 0.03 V; b. 2.71 V 41. See Exercise 37 for a description of a galvanic b. AgNO3 Cu Cu AgNO3(Ag) cell. For each of the following answers, all solutes are 1.0 M and all gases are c. HCl Zn Zn HCl(H) at 1.0 atm. In the salt bridge, cations flow to the cathode and anions flow to the anode. a. Cl (g) 1 2Br2(aq) n Br (aq) 1 2Cl2(aq), ° 5 0.27 V; cathode: 21. a. 4H1(aq) 1 NO 2(aq) 1 Cr(s) → Cr31(aq) 1 NO(g) 1 2H O(l); 2 2 3 2 Pt electrode; Cl (g) bubbled in, Cl2 in solution; anode: Pt electrode; Br and b. H O(l) 1 CH OH(aq) 1 6Ce41(aq) → 6Ce31(aq) 1 CO (g) 1 6H1(aq); 2 2 2 3 2 Br2 in solution; b. 3H O(l) 1 5IO 2(aq) 1 2Mn21(aq) n 5IO 2(aq) 1 c. 5SO 22(aq) 1 6H1(aq) 1 2MnO 2(aq) → 2Mn21(aq) 1 3H O(l) 1 2 4 3 3 4 2 2MnO 2(aq) 1 6H1(aq), ° 5 0.09 V; cathode: Pt electrode; IO 2, IO 2, 5SO 22(aq) 23. Electrochemistry is the study of the interchange of chemical 4 4 3 4 and H SO (as a source of H1) in solution; anode: Pt electrode; Mn21, and electrical energy. A redox (oxidation–reduction) reaction is a reaction in 2 4 MnO 2, and H SO in solution 43. 37a. PtuCr31 (1.0 M), H1 (1.0 M), which one or more electrons are transferred. In a galvanic cell, a spontaneous 4 2 4 Cr O 22 (1.0 M)uuCl (1.0 atm)uCl2 (1.0 M)uPt; 37b. MguMg21 (1.0 M)uu redox reaction occurs which produces an electric current. In an electrolytic 2 7 2 Cu21 (1.0 M)uCu; 41a. PtuBr2 (1.0 M), Br (1.0 M)uuCl (1.0 atm)uCl2 (1.0 cell, electricity is used to force a nonspontaneous redox reaction to occur. 2 2 M)uPt; 41b. PtuMn21 (1.0 M), MnO 2 (1.0 M), H1 (1.0 M)uuIO 2 (1.0 M), 25. Magnesium is an alkaline earth metal; Mg will oxidize to Mg21. The 4 4 H1 (1.0 M), IO 2 (1.0 M) uPt 45. a. Au31(aq) 1 3Cu1(aq) n 3Cu21(aq) 1 oxidation state of hydrogen in HCl is 11. To be reduced, the oxidation state 3 Au(s), ° 5 1.34 V; b. 2VO 1(aq) 1 4H1(aq) 1 Cd(s) n Cd21(aq) 1 of H must decrease. The obvious choice for the hydrogen product is H (g) 2 2 2VO21(aq) 1 2H O(l), ° 5 1.40 V 47. a. 16H1(aq) 1 2MnO 2(aq) 1 where hydrogen has a zero oxidation state. The balanced reaction is: Mg(s) 1 2 4 10I2(aq) n 5I (aq) 1 2Mn21(aq) 1 8H O(I), ° 5 0.97 V, spontaneous; 2HCl(aq) n MgCl (aq) 1 H (g). Mg goes from the 0 to the 12 oxidation 2 2 cell 2 2 b. 16H1(aq) 1 2MnO 2(aq) 1 10F2(aq) n 5F (g) 1 2Mn21(aq) 1 state by losing two electrons. Each H atom goes from the 11 to the 0 oxida- 4 2 8H O(I), ° 5 21.36 V, not spontaneous 49. ° 5 0.41 V, DG8 5 279 kJ tion state by gaining one electron. Since there are two H atoms in the bal- 2 cell 51. 45a. 2388 kJ; 45b. 2270. kJ 53. 20.829 V; the two values agree to two anced equation, then a total of two electrons are gained by the H atoms. significant figures. 55. 1.24 V 57. Mg21 , Fe21 , Fe31 , Cr O 22 , Cl Hence, two electrons are transferred in the balanced reaction. When the elec- 2 7 2 , MnO 2 59. a. no; b. yes; c. yes; 61. a. Of the species available, Ag1 trons are transferred directly from Mg to H1, no work is obtained. In order to 4 would be the best oxidizing agent because it has the largest ° value. harness this reaction to do useful work, we must control the flow of electrons b. Of the species available, Zn would be the best reducing agent because it through a wire. This is accomplished by making a galvanic cell, which sepa- has the largest 2° value. c. SO 22(aq) can oxidize Pb(s) and Zn(s) at stan- rates the reduction reaction from the oxidation reaction in order to control the 4 dard conditions. When SO 22(aq) is coupled with these reagents, ° is flow of electrons through a wire to produce a voltage. 27. An extensive 4 cell positive. d. Al(s) can reduce Ag1(aq) and Zn21(aq) at standard conditions property is one that depends on how many times the reaction occurs. The free because ° . 0. 63. a. Cr O 22, O , MnO , IO 2; b. PbSO , Cd21, Fe21, energy change for a reaction depends on whether 1 mole of product is pro- cell 2 7 2 2 3 4 Cr31, Zn21, H O 65. a. larger; b. smaller 67. Electron flow is always duced or 2 moles of product is produced or 1 million moles of product is 2 from the anode to the cathode. For the cells with a nonzero cell potential, produced. This is not the case for cell potentials, which do not depend on we will identify the cathode, which means the other compartment is the how many times the reaction occurs. The equation that relates DG to E is anode. a. 0; b. 0.018 V; compartment with [Ag1] 5 2.0 M is cathode; DG 5 2nFE. It is the n term that converts the intensive property E into the c. 0.059 V; compartment with [Ag1] 5 1.0 M is cathode; d. 0.26 V; compart- extensive property DG. n is the number of moles of electrons transferred in ment with [Ag1] 5 1.0 M is cathode; e. 0 69. 2.12 V 71. 1.54 V the balanced reaction that DG is associated with. 29. A potential hazard 73. [37]. a. DG8 5 220 kJ; 1 3 103; b. DG8 5 2523 kJ; 5.12 3 1091; [41]. when jump-starting a car is that the electrolysis of H O(l) can occur. When 2 a. DG8 5 252 kJ; 1.4 3 109; b. DG8 5 290 kJ; 2 3 1015 75. a. Fe21(aq) 1 H O(l) is electrolyzed, the products are the explosive gas mixture of H (g) 2 2 Zn(s) n Zn21(aq) 1 Fe(s) ° 5 0.32 V; b. 262 kJ; 6.8 3 1010; c. 0.20 V and O (g). A spark produced during jump-starting a car could ignite any cell 2 77. a. 0.23 V; b. 1.2 3 1025 M 79. 0.16 V, copper is oxidized. 81. 1.7 3 10230 H (g) and O (g) produced. Grounding the jumper cable far from the battery 2 2 83. a. no reaction; b. Cl (g) 1 2I2(aq) n I (s) 1 2Cl2(aq), ° 5 0.82 V; DG8 minimizes the risk of a spark nearby the battery where H (g) and O (g) could 2 2 cell 2 2 5 2160 kJ; K 5 5.6 3 1027; c. no reaction; d. 4Fe21(aq) 1 4H1(aq) 1 O (g) be collecting. 31. You need to know the identity of the metal so you know 2 n 4Fe31(aq) 1 2H O(l), ° 5 0.46 V; DG8 5 2180 kJ; K 5 1.3 3 1031; which molar mass to use. You need to know the oxidation state of metal ion 2 cell 85. 0.151 V; 229.1 kJ 87. 3.9 3 10228 89. 20.14 V 91. a. 30. hours; in the salt so the moles of electrons transferred can be determined. And fi- b. 33 s; c. 1.3 hours 93. a. 16 g; b. 25 g; c. 71 g; d. 4.9 g 95. MgCl nally, you need to know the amount of current and the time the current was 2 97. 9.12 L F (anode), 29.2 g K (cathode) 99. 1 3 105 A 101. 1.14 3 passed through the electrolytic cell. If you know these four quantities, then 2 1022 M 103. Au followed by Ag followed by Ni followed by Cd the mass of metal plated out can be calculated. 33. Only statement e is true. 105. Cathode: 2H O 1 2e2 n H (g) 1 2OH2; anode: 2H O n O (g) 1 4H1 The attached metals that are more easily oxidized than iron are called sacri- 2 2 2 2 1 4e2 107. a. cathode: Ni21 1 2e2 n Ni; anode: 2Br2 n Br 1 2e2; ficial metals. For statement a, corrosion is a spontaneous process, like the 2 b. cathode: Al31 1 3e2 n Al; anode: 2F2 n F 1 2e2; c. cathode: Mn21 1 ones harnessed to make galvanic cells. For statement b, corrosion of steel is 2 2e2 n Mn; anode: 2I2 n I 1 2e2 109. a. cathode: Ni21 1 2e2 n Ni; the oxidation of iron coupled with the reduction of oxygen. For statement c, 2 anode: 2Br2 n Br 1 2e2; b. cathode: 2H O 1 2e2 n H 1 2OH2; anode: cars rust more easily in high-moisture areas (the humid areas) because water 2 2 2 2H O n O 1 4H1 1 4e2; c. cathode: 2H O 1 2e2 n H 1 2OH2; anode: is a reactant in the reduction half-reaction as well as providing a medium for 2 2 2 2 2I2 n I 1 2e2 111. a. 0.10 V, SCE is anode; b. 0.53 V, SCE is anode; ion migration (a salt bridge of sorts). For statement d, salting roads adds ions 2 c. 0.02 V, SCE is cathode; d. 1.90 V, SCE is cathode; e. 0.47 V, SCE is cath- to the corrosion process, which increases the conductivity of the aqueous ode 113. a. decrease; b. increase; c. decrease; d. decrease; e. same solution and, in turn, accelerates corrosion. 35. The reducing agent causes 115. a. DG8 5 2582 kJ; K 5 3.45 3 10102; ° 5 1.01 V; b. 20.65 V reduction to occur; it does this by containing the species that is oxidized. 117. Aluminum has the ability to form a durable oxide coating over its sur- Oxidation occurs at the anode, so the reducing agent will be in the anode face. Once the HCl dissolves this oxide coating, Al is exposed to H1 and is compartment. The oxidizing agent causes oxidation to occur; it does this by easily oxidized to Al31. Thus, the Al foil disappears after the oxide coating is containing the species that is reduced. Reduction occurs at the cathode, so dissolved. 119. H O as an oxidizing agent: H O 1 2H1 1 2e2 → 2H O the oxidizing agent will be in the cathode compartment. Electron flow is 2 2 2 2 2 ° 5 1.78 V; H O as a reducing agent: H O → O 1 2H1 1 2e2 ° 5 always from the anode compartment to the cathode compartment. 37. See red 2 2 2 2 2 ox 20.68 V; from the more positive potential, H O is a better oxidizing Fig. 18.3 of the text for a typical galvanic cell. The anode compartment 2 2

57404_Ans_A31-A64.indd 55 10/14/16 5:12 PM A56 Answers to Selected Exercises

agent than it is a reducing agent at standard conditions. 121. 1.14 V be ordered at a minimum. 37. 17.7% 39. The fraction that remains is

123. wmax 5 213,200 kJ; the work done can be no larger than the free 0.0041, or 0.41%. 41. 26 g 43. 2.3 counts per minute per gram of C. No; for energy change. If the process were reversible all of the free energy released a 10.-mg C sample, it would take roughly 40 min to see a single disintegration. would go into work, but this does not occur in any real process. Fuel cells are This is too long to wait, and the background radiation would probably be much more efficient in converting chemical energy to electrical energy; they are greater than the 14C activity. Thus 14C dating is not practical for very small sam- also less massive. Major disadvantage: They are expensive. 125. 0.98 V ples. 45. 3.8 3 109 yr 47. 4.3 3 106 kg/s 49. 232Pu, 21.715 3 1014 J/mol; 127. 7.44 3 104 A 129. To produce 1.0 kg Al by the Hall-Heroult process 231Pa, 21.714 3 1014 J/mol 51. 12C: 1.230 3 10212 J/nucleon; 235U: requires 5.4 3 104 kJ of energy. To melt 1.0 kg Al requires 4.0 3 102 kJ of 1.2154 3 10212 J/nucleon; since 56Fe is the most stable known nucleus, the energy. It is feasible to recycle Al by melting the metal because, in theory, it binding energy per nucleon for 56Fe would be larger than that of 12C or 235U. takes less than 1% of the energy required to produce the same amount of Al (See Fig. 19.9 of the text.) 53. 6.01513 u 55. 22.0 3 1010 J/g of hydro- by the Hall-Heroult process. 131. 13 133. a. 1.93 V; b. Fe21; c. The anode gen nuclei 57. The Geiger–Müller tube has a certain response time. After would be composed of a La electrode and 1.0 M La31; d. The electrons flow from the gas in the tube ionizes to produce a “count,” some time must elapse for the La/La31 compartment to the Fe21/Fe compartment; e. 6; f. 1.87 V the gas to return to an electrically neutral state. The response of the tube 135. a. 3.87 V; b. 0.40 M 137. 3.97 A 139. a. Co21 is the primary pro- levels off because, at high activities, radioactive particles are entering the duct assuming standard conditions. b. Even with concentrated nitric acid tube faster than the tube can respond to them. 59. Water is produced in this (16 M), the only oxidation that can occur is to oxidize Co to Co21. reaction by removing an OH group from one substance and an H from the TDS° DH° other substance. There are two ways to do this: 141. ° 5 2 ; if we graph ° versus T, we should get a straight nF nF i. O O line (y 5 mx 1 b). The slope of the line is equal to DS8ynF and the y-intercept B 18 B 18 is equal to 2DH8 nF. ° will have little temperature dependence for cell y CH3C OH+ H OCH3 8n CH3C OCH3 + HO H reactions with DS8 close to zero. 143. 9.8 3 1026 145. 2.39 3 1027 26 1 147. a. 60.02 pH units; 66 3 10 M H ; b. 60.001 V 149. a. 0.16 V; ii. O O 1 218 21 b. 8.6 mol 151. [Ag ] 5 4.6 3 10 M; [Ni ] 5 1.5 M 153. 0.64 V B 18 B 18 155. a. 5.77 3 1010; b. 212.2 kJ/mol 157. Osmium(IV) nitrate; [Ar]4s13d10 CH3CO H + H O CH3 8n CH3CO CH3 + H OH

Chapter 19 Because the water produced is not radioactive, methyl acetate forms by 1. The characteristic frequencies of energies emitted in a nuclear reaction the first reaction, where all of the oxygen-18 ends up in methyl acetate. suggest that discrete energy levels exist in the nucleus. The extra stability of 61. 2 neutrons; 4 b particles 63. Strontium. Xe is chemically unreactive certain numbers of nucleons and the predominance of nuclei with even num- and not readily incorporated into the body. Sr can be easily oxidized to Sr21. bers of nucleons suggest that the nuclear structure might be described by us- Strontium is in the same family as calcium and could be absorbed and con- ing quantum numbers. 3. Radiotracers generally have short half-lives. The centrated in the body in a fashion similar to Ca21. The chemical properties radioactivity from the radiotracers is monitored to study a specific area of the determine where radioactive material may be concentrated in the body or body. However, we don’t want long-lived radioactivity in order to minimize how easily it may be excreted. 65. a. unstable; beta production; b. stable; potential damage to healthy tissue by the radioactivity. 5. b-particle pro- c. unstable; positron production or electron capture; d. unstable, positron duction has the net effect of turning a neutron into a proton. Radioactive production, electron capture, or alpha production. 67. The products of 214 210 210 210 nuclei having too many neutrons typically undergo b-particle decay. Posi- the various steps are 84Po, 82Pb, 83Bi, and 84Po. 69. 3800 decays/s tron production has the net effect of turning a proton into a neutron. Nuclei 71. The third-life will be the time required for the number of nuclides to

having too many protons typically undergo positron decay. 7. The trans- reach one-third of the original value (N03). The third-life of this nuclide is uranium elements are the elements having more protons than uranium. They 49.8 years. 73. 1975 75. 900 g 235U 77. 7 3 105 m/s; 8 3 10216 J/nuclei;

are synthesized by bombarding heavier nuclei with neutrons and positive 79. All evolved O2(g) comes from water. 81. The equations for the 2 239 235 4 214 214 0 ions in a particle accelerator. 9. DE 5 Dmc ; The key difference is the mass nuclear reactions are 94Pu S 92U 1 2He; 82Pb S 83Bi 1 21e; 60 60 0 99 99 0 239 239 0 change when going from reactants to products. In chemical reactions, the 27Co S 28Ni 1 21e; 43Tc S 44Ru 1 21e; 93Np S 94Pu 1 21e mass change is indiscernible. In nuclear processes, the mass change is dis- 83. 15.8 yr 85. 3.177 3 10–11 J 87. 77% 238U and 23% 235U 89. Assum- cernible. It is the conversion of this discernible mass change into energy that ing that (1) the radionuclide is long lived enough that no significant decay results in the huge energies associated with nuclear processes. 11. The occurs during the time of the experiment, and (2) the total activity is 12 temperatures of fusion reactions are so high that all physical containers uniformly distributed only in the rat’s blood, V 5 10. mL. 91. a. 6C; would be destroyed. At these high temperatures, most of the electrons are b. 13N, 13C, 14N, 15O, and 15N; c. 25.950 3 1011 J/mol 1H 93. 4.3 3 10229 249 22 267 1 2 14 5 stripped from the atoms. A plasma of gaseous ions is formed that can be 95. 97Bk 1 10Ne n 107Bh 1 4 0n; 62.7 s; [Rn]7s 5f 6d controlled by magnetic fields. 13. Somatic damage is immediate damage to the organism itself, resulting in sickness or death. Genetic damage is damage Chapter 20

to the genetic material, which can be passed on to future generations. The 1. The gravity of the earth cannot keep the light H2 molecules in the atmo- organism will not feel immediate consequences from genetic damage, but its sphere. 3. Calcium is found in the structural material that makes up bones 3 3 0 8 8 0 offspring may be damaged. 15. a. 1H n 2He 1 21e; b. 3Li n 4Be 1 21e, and teeth. Magnesium plays a vital role in metabolism and in muscle func- 8 4 8 4 0 7 0 7 4Be n 2 2He; overall reaction: 3Li n 2 2He 1 21e; c. 4Be 1 21e n 3Li; tion. 5. For Groups 1A–3A, the small size of H (as compared to Li), Be (as 8 8 0 234 0 d. 5B n 4Be 1 11e 17. a. 90Th; this is a-particle production. b. 21e; this is compared to Mg), and B (as compared to Al) seems to be the reason why 68 0 68 62 0 62 b-particle production. 19. a. 31Ga 1 21e n 30Zn; b. 29Cu n 11e 1 28Ni; these elements have nonmetallic properties, while others in Groups 1A–3A 212 4 208 129 0 129 c. 87Fr n 2He 1 85At; d. 51Sb n 21e 1 52Te a 21. 7 particles; 4 b particles are strictly metallic. The small size of H, Be, and B also causes these species 241 4 237 209 23. a. 95Am n 2He 1 93Np; b. 83Bi; c. The intermediate radionuclides are to polarize the electron cloud in nonmetals, thus forcing a sharing of elec- 237 233 233 229 225 225 221 217 213 213 209 93Np, 91Pa, 92U, 90Th, 88Ra, 89Ac, 87Fr, 85At, 83Bi, 84Po, and 82Pb. trons when bonding occurs. For Groups 4A–6A, a major difference between 25. 8B and 9B contain too many protons or too few neutrons. Electron capture the first and second members of a group is the ability to form p bonds. The and positron production are both possible decay mechanisms that increase smaller elements form stable p bonds, while the larger elements do not ex- the neutron-to-proton ratio. Alpha-particle production also increases the hibit good overlap between parallel p orbitals and, in turn, do not form strong neutron-to-proton ratio, but it is not likely for these light nuclei. 12B and 13B p bonds. For Group 7A, the small size of F as compared to Cl is used to ex- contain too many neutrons or too few protons. Beta-particle production plain the low electron affinity of F and the weakness of the OF F bond. 7. In lowers the neutron-to-proton ratio, so we expect these to be beta-emitters. order to maximize hydrogen bonding interactions in the solid phase, ice is 249 18 263 1 259 27. a. 98Cf 1 8O n 106Sg 1 4 0n; b. 104Rf 29. For t1/2 5 12,000 years: forced into an open structure. This open structure is why H2O(s) is less dense 12 18 1.1 3 10 decays/s; for t1/2 5 12 h: 9.6 3 10 decays/s; for t1/2 5 12 s: than H2O( l). 9. Group 1A and 2A metals are all easily oxidized. They must 22 47 3.5 3 10 decays/s 31. 690 hours 33. 5.0 min 35. 15 mg CaCO3 should be produced in the absence of materials (H2O, O2) that are capable of oxidiz-

57404_Ans_A31-A64.indd 56 10/14/16 5:12 PM Answers to Selected Exercises A57

ing them. 11. The reaction N2(g) 1 3H2(g) m 2NH3(g) is exothermic. 51. N Cl Thus, the value of the equilibrium constant K decreases as the temperature HH increases. Lower temperatures are favored for maximum yield of ammonia. H Cl 3 However, at lower temperatures the rate is slow; without a catalyst the rate is Trigonal pyramid; sp Cl As too slow for the process to be feasible. The discovery of a catalyst increased Cl the rate of reaction at a lower temperature favored by thermodynamics. 13. Boranes are covalent compounds produced between boron and hydrogen. Cl The B—H bonds in boranes are relatively weak and are very reactive. Trigonal bipyramid; dsp3 When boranes are reacted with oxygen, the product bonds are much stron- – ger than the reactant bonds. When this is the case, very exothermic reactions F result, which are the bases for good rocket fuels. 15. a. DH8 5 207 kJ,

DS8 5 216 J/K; b. T . 958 K 17. 4Li(s) 1 O2(g) n 2Li2O(s); 2Li(s) 1 F F S(s) n Li2S(s); 2Li(s) 1 Cl2(g) n 2LiCl(s); 12Li(s) 1 P4(s) n 4Li3P(s); P 2Li(s) 1 H2(g) n 2LiH(s); 2Li(s) 1 2H2O(l) n 2LiOH(aq) 1 H2(g); F F 2Li(s) 1 2HCl(aq) n 2LiCl(aq) 1 H2( g) 19. When lithium reacts with 1 22 F excess oxygen, Li2O forms, which is composed of Li and O ions. This is called an oxide salt. When sodium reacts with oxygen, Na2O2 forms, which 1 22 2 3 is composed of Na and O2 ions. This is called a peroxide salt. When Octahedral; d sp potassium (or rubidium or cesium) reacts with oxygen, KO forms, which is 2 Nitrogen does not have low-energy d orbitals it can use to expand its octet. composed of K1 and O 2 ions. For your information, this is called a super- 2 Both NF and NCl 2 would require nitrogen to have more than 8 valence oxide salt. So the three types of alkali metal oxides that can form differ 5 6 electrons around it; this never happens. 53. 1 N (g) 1 1 O (g) n NO(g); in the oxygen anion part of the formula (O22 vs. O 22 vs. O 2). Each of 2 2 2 2 2 2 DG8 5 DG8 ; NO (and some other oxides of nitrogen) have weaker bond’s these anions have unique bonding arrangements and oxidation states. f(NO) as compared with the triple bond of N and the double bond of O . Because 21. The small size of the Li1 cation results in a much greater attraction to 2 2 of this, NO (and some other oxides of nitrogen) has a higher (positive) stan- water. The attraction to water is not so great for the other alkali metal ions. dard free energy of formation as compared to the relatively stable N and O Thus lithium salts tend to absorb water. 23. CaCO (s) 1 H SO (aq) n 2 2 3 2 4 molecules. 55. The pollution provides nitrogen and phosphorus nutrients CaSO (aq) 1 H O(l) 1 CO ( g) 25. 3.84 3 106 g Ba 27. Beryllium 4 2 2 so the algae can grow. The algae consume dissolved oxygen, causing fish to has a small size and a large electronegativity as compared with the other die. 57. The acidic hydrogens in the oxyacids of phosphorus all are bonded alkaline earth metals. The electronegativity of Be is so high that it does to oxygen. The hydrogens bonded directly to phosphorus are not acidic. not readily give up electrons to nonmetals, as is the case for the other H PO has three oxygen-bonded hydrogens, and it is a triprotic acid. H PO alkaline earth metals. Instead, Be has significant covalent character in its 3 4 3 3 has only two of the hydrogens bonded to oxygen, and it is a diprotic acid. The bonds; it prefers to share valence electrons rather than give them up to third oxyacid of phosphorus, H PO , has only one of the hydrogens bonded form ionic bonds. 29. 9.3 3 1025 mol/L 31. Nh: [Rn]7s25f 146d107p1; 3 2 to an oxygen; it is a monoprotic acid. 59. 821 nm 61. H SeO (aq) 1 Nh falls below Tl in the periodic table. We would expect Nh, like Tl, 2 4 3SO (g) n Se(s) 1 3SO (g) 1 H O( l) 63. In the upper atmosphere, O to form 11 and 13 oxidation states in its compounds. 2 3 2 3 acts as a filter for ultraviolet (UV) radiation: 33. B2H6(g) 1 3O2(g) n 2B(OH)3( s) 35. 2Ga(s) 1 3F2(g) n 2GaF3(s); hn 4Ga(s) 1 3O2(g) n 2Ga2O3(s); 2Ga(s) 1 3S(s) n Ga2S3(s); 2Ga(s) 1 O3 1g2 h O2 1g2 1 O1g2 6HCl(aq) n 2GaCl3(aq) 1 3H2( g) 37. An amphoteric substance is one that can behav e as either an acid or a base. Al2O3 dissolves in both acidic and O3 is also a powerful oxidizing agent. It irritates the lungs and eyes, and at 1 31 basic solutions. The reactions are Al2O3(s) 1 6H (aq) n 2Al (aq) 1 high concentration, it is toxic. The smell of a “spring thunderstorm” is O3 2 2 3H2O(l) and Al2O3(s) 1 2OH (aq) 1 3H2O(l) n 2Al(OH)4 (aq). formed during lightning discharges. Toxic materials don’t necessarily smell 39. Compounds containing SiOSi single and multiple bonds are rare, unlike bad. For example, HCN smells like almonds. 65. The M.O. electron con- compounds of carbon. The bond strengths of the SiOSi and COC single figuration of 2O has two unpaired electrons in the degenerate pi antibonding * bonds are similar. The difference in bonding properties must be for other (p2p) orbitals. A substance with unpaired electrons is paramagnetic. reasons. One reason is that silicon does not form strong pi bonds, unlike 67. From the following Lewis structure, each oxygen atom has a tetrahedral carbon. Another reason is that silicon forms particularly strong sigma bonds arrangement of electron pairs. Therefore, bond angles are <109.58, and each 3 to oxygen, resulting in compounds with SiOO bonds instead of SiOSi O is sp hybridized. bonds. 41. O C PP O The darker green orbitals about carbon are sp hybrid orbitals. The lighter green orbitals about each oxygen are sp2 hybrid F O O F orbitals, and the gold orbitals about all of the atoms are unhybridized Formal charge: 0 0 0 0 p atomic orbitals. In each double bond in CO2, one s bond and one p bond Oxidation state: –1 +1 +1 –1 exist. The two carbon–oxygen s bonds are formed from overlap of sp hybrid orbitals from carbon with an sp2 hybrid orbital from each oxygen. The two Oxidation states are more useful. We are forced to assign 11 as the oxidation carbon–oxygen p bonds are formed from side-to-side overlap of the unhy- state for oxygen. Oxygen is very electronegative, and 11 is not a stable oxi- bridized p atomic orbitals from carbon with an unhybridized p atomic orbital dation state for this element. from each oxygen, as illustrated in the figure. 43. a. SiO (s) 1 2C(s) n 2 69. S Si(s) 1 2CO(g); b. SiCl4(l) 1 2Mg(s) n Si(s) 1 2MgCl2(s); c. Na2SiF6(s) 1 F 4Na(s) n Si(s) 1 6NaF(s) 45. 2:1 47. Nitrogen’s small size does not F 32 provide room for all four oxygen atoms, making NO4 unstable. Phosphorus V-shaped; ,109.5° 32 2 is larger so PO4 is more stable. To form NO3 , a pi bond must form. Phos- phorus doesn’t form strong pi bonds as readily as nitrogen. 49. 2Bi2S3(s) 1 F F 9O (g) n 2Bi O (s) 1 6SO (g); 2Bi O (s) 1 3C(s) n 4Bi(s) 1 3CO (g); 2 2 3 2 2 3 2 F F F 2Sb S (s) 1 9O (g) n 2Sb O (s) 1 6SO (g); 2Sb O (s) 1 3C(s) n 4Sb(s) 1 2 3 2 2 3 2 2 3 S S 3CO2(g) F F F

F F See-saw; <120°, <90° Octahedral; 90°

57404_Ans_A31-A64.indd 57 10/14/16 5:12 PM A58 Answers to Selected Exercises

32 1 OF4 would not form. This compound would require oxygen to have more than 9. Fe2O3(s) 1 6H2C2O4(aq) n 2Fe(C2O4)3 (aq) 1 3H2O(l) 1 6H (aq); the 8 valence electrons around it. This never occurs for oxygen; oxygen does not oxalate anion forms a soluble complex ion with iron in rust (Fe2O3), which have low-energy d orbitals it can use to expand its octet. 71. The oxyacid allows rust stains to be removed.

strength increases as the number of oxygens in the formula increases. There- 22 11. a. CoCl4 is an example of a weak-field case having fore, the order of the oxyacids from weakest to strongest acid is HOCl , HClO2 2 2 2 2 2 three unpaired electrons. , HClO3 , HClO4. 73. a. IO4 ; b. IO3 ; c. IF2 ; d. IF4 ; e. IF6 O O O 75. Helium is unreactive and doesn’t combine with any other elements. It is a very light gas and would easily escape the earth’s gravitational pull as the planet O O was formed. 77. One would expect RnF2, RnF4, and maybe RnF6 to form in small Δ fashion similar to XeF2, XeF4, and XeF6. 79. N2H4(l) 1 O2(g) n N2(g) 1 OO 2 32 2H2O(g); DH 5 2590. kJ 81. If the compound contained Ga(II), it would be b. CN is a strong-field ligand so Co(CN)6 will be a low- paramagnetic and if the compound contained Ga(I) and Ga(III), it would be spin case having zero unpaired electrons. diamagnetic. Paramagnetic compounds have an apparent greater mass in a mag- O O O netic field. 83. NaH is an ionic compound composed of Na1 and H2 ions. large Δ 2 Acid–base: A proton is transferred from an acid, H2O, to a base, H , forming 21 2 2 13. From Table 21.16, the red octahedral Co(H2O)6 complex ion absorbs the conjugate base of water, OH , and the conjugate acid of H , H2. Oxidation– 22 blue-green light (l < 490 nm), whereas the blue tetrahedral CoCl4 complex reduction: The oxidation state of hydrogen is 21 in NaH, 11 in H2O, and zero ion absorbs orange light (l < 600 nm). Because tetrahedral complexes have a in H2. Hydrogen is oxidized when it goes from NaH to H2 (from 21 S 0) and d-orbital splitting much less than octahedral complexes, one would expect the hydrogen is reduced when it goes from H2O to H2 (from 11 S 0). In this reaction, an electron is transferred from the hydride ion to a hydrogen in tetrahedral complex to have a smaller energy difference between split d orbit- hv als. This translates into longer-wavelength light absorbed (E 5 hc/l) for tetra- water when forming H2. 85. a. AgCl(s) 8n Ag(s) 1 Cl; the reactive chlorine atom is trapped in the crystal. When light is removed, Cl reacts with silver at- hedral complex ions compared to octahedral complex ions. Information from 2 2 2 oms to re-form AgCl; i.e., the reverse reaction occurs. In pure AgCl, the Table 21.16 confirms this. 15. SCN , NO2 , and OCN can form linkage 31 Cl atoms escape, making the reverse reaction impossible. b. Over time, chlo- isomers; all are able to bond to the metal ion in two different ways. 17. Sc 31 31 rine is lost and the dark silver metal is permanent. 87. In solution, Tl31 can has no electrons in d orbitals. Ti and V have d electrons present. The color 2 2 of transition metal complexes results from electron transfer between split d oxidize I to I3 1°cell . 02 . Thus we expect TlI3 to be thallium(I) triiodide. 89. Both ∆H and ∆S will be positive. Since ∆S is positive for this process, the orbitals. If no d electrons are present, no electron transfer can occur, and the rhombic form of solid sulfur will have the more ordered form (smaller posi- compounds are not colored. 19. 0; Ni(CO)4 21. At high altitudes, the oxy- tional probability and smaller S). 91. Strontium and calcium are both alkaline gen content of air is lower, so less oxyhemoglobin is formed which diminishes earth metals, so both have similar chemical properties. Since milk is a good the transport of oxygen in the blood. A serious illness called high-altitude sick- source of calcium, strontium could replace some calcium in milk without much ness can result from the decrease of O2 in the blood. High-altitude acclimatiza- 22 tion is the phenomenon that occurs in the human body in response to the lower difficulty. 93. 16 oxidation state: SO4 , SO3, SF6; 14 oxidation state: 22 amounts of oxyhemoglobin in the blood. This response is to produce more he- SO3 , SO2, SF4; 12 oxidation state: SCl2; 0 oxidation state: S8 and all other moglobin, and, hence, increase the oxyhemoglobin in the blood. High-altitude elemental forms of sulfur; 22 oxidation state: H2S, Na2S 95. a. DH° 5 132 kJ; DS° 5 134 J/K; b. At T . 985 K and standard pressures, the favorable DS° acclimatization takes several weeks to take hold for people moving from lower 2 1 2 6 term dominates, and the reaction is spontaneous 1DG° . 02. 97. 0.013 mol/L altitudes to higher altitudes. 23. a. Sc: [Ar]4s 3d ; b. Ru: [Kr]5s 4d (ex- 1 7 2 14 7 2 5 99. a. sp; b. sp2; c. sp2; d. sp 101. a. d 2sp3; b. dsp3; c. sp3; d. d 2sp3 pected), [Kr]5s 4d (actual); c. Ir: [Xe]6s 4f 5d ; d. Mn: [Ar]4s 3d 103. a. 0.17 g; b. 5 3 1015 atoms 105. Ca; 12.698 107. I 109. For the reaction 25. Transition metal ions lose the s electrons before the d electrons. Also, Pt is an exception to the normal filling order of electrons (see Figure 7.29). O a. Co: [Ar]4s23d7, Co21: [Ar]3d7, Co31: [Ar]3d6; b. Pt: [Xe]6s14f 145d9, M Pt21: [Xe]4f 145d8, Pt41: [Xe]4f 145d6; c. Fe: [Ar]4s23d6, Fe21: [Ar]3d6, NNOPONmn O2 + NO D Fe31: [Ar]3d5 27. a. Fe31: [Ar]3d 5; b. Ag1: [Kr]4d10; c. Ni21: [Ar]3d 8; O d. Cr31: [Ar]3d3 29. a. molybdenum(IV) sulfide, molybdenum(VI) oxide;

the activation energy must in some way involve the breaking of a nitrogen– b. MoS2, 14; MoO3, 16; (NH4)2Mo2O7, 16; (NH4)6Mo7O24 ? 4H2O, 16 2 nitrogen single bond. For the reaction 31. NH3 is a weak base that produces OH ions in solution. The white pre- 21 2 cipitate is Cu(OH)2(s). Cu (aq) 1 2OH (aq) n Cu(OH)2(s); with excess 21 21 O NH3 present, Cu forms a soluble complex ion, Cu(NH3)4 . Cu(OH)2(s) 1 M 21 2 NNOPOOmn 2 + N2O 4NH3(aq) n Cu(NH3)4 (aq) 1 2OH (aq) 33. Only [Cr(NH3)6]Cl3 will D – O form a precipitate since only this compound will have Cl ions in solution. The Cl– ions in the other compounds are ligands and are bound to the central 31 – at some point nitrogen–oxygen bonds must be broken. NON single bonds Cr ion. The Cl ions in [Cr(NH3)6]Cl3 are counterions needed to produce a 31 (160 kJ/mol) are weaker than NOO single bonds (201 kJ/mol). In addition, neutral compound; the NH3 molecules are the ligands bound to Cr . resonance structures indicate that there is more double-bond character in the 35. a. hexacyanomanganate(II) ion; b. cis-tetraamminedichlorocobalt(III) ion; c. pentaamminechlorocobalt(II) ion 37. a. hexaamminecobalt(II) chloride; NOO bonds than in the NON bond. Thus NO2 and NO are preferred by kinetics because of the lower activation energy. 111. 5.89 113. 20. g b. hexaaquacobalt(III) iodide; c. potassium tetrachloroplatinate(II); d. potas- 115. 6.5 3 1027 atoms 117. a. 7.1 g; b. 979 nm; this electromagnetic radia- sium hexachloroplatinate(II); e. pentaamminechlorocobalt(III) chloride; tion is not visible to humans; it is in the infrared region of the electromag- f. triamminetrinitrocobalt(III) 39. a. K2[CoCl4]; b. [Pt(H2O)(CO)3]Br2; netic radiation spectrum; c. n-type 119. a. 16; b. 4.42 c. Na3[Fe(CN)2(C2O4)2]; d. [Cr(NH3)3Cl(H2NCH2CH2NH2)]I2

41. a. – –

Chapter 21 H

A K E O

5. The lanthanide elements are located just before the 5d transition metals.

O OH2 K The lanthanide contraction is the steady decrease in the atomic radii of the H O O OK O A O O 2 E A E E E E E lanthanide elements when going from left to right across the periodic table. Co A Co A KH H H H K H A H O O A O O

As a result of the lanthanide contraction, the sizes of the 4d and 5d elements H2O O OH

O H 2

are very similar. This leads to a greater similarity in the chemistry of the 4d E K and 5d elements in a given vertical group. 7. No; both the trans and the cis K O 1 O forms of Co(NH3)4Cl2 have mirror images that are superimposable. For the cis form, the mirror image only needs a 908 rotation to produce the original cis trans structure. Hence, neither the trans nor the cis form is optically active.

57404_Ans_A31-A64.indd 58 10/14/16 5:12 PM Answers to Selected Exercises A59

2 22 21 b. 2+ 2+ precipitate; HgI2(s) 1 2I (aq) n HgI4 (aq), soluble complex ion; Hg is I I 10 A I A NH3 a d ion. Color is the result of electron transfer between split d orbitals. This D D 21 3NOOPtH NH3 3NOOPtH NH3 cannot occur for the filled d orbitals in Hg . Therefore, we would not expect DA DA Hg21 complex ions to form colored solutions. 75. The molecular formula H N H N 3 NH3 3 I 2 2 is EuC15H21O6. Because each acac is C5H7O2 , an abbreviated molecular 22 cis trans formula is Eu(acac)3. 77. 0.66 V; 2130 kJ; 2.2 3 10 79. There are four geometrical isomers (labeled i–iv). Isomers iii and iv are optically active, and c. Cl Cl the nonsuperimposable mirror images are shown. A Cl A Cl D D + + H3NOOIr Cl H3NOOIr NH3 i. ii. Br NH3 D A D A H N NH CH Br NH CH H N H N 3 2 2 2 2 3 NH3 3 Cl Cr Cr NH CH NH CH H3N 2 2 Cl 2 2 Cl NH d. + + + 3 N N N H3N A N INA INA G G G G G G Cr Cr Cr D D A D D A D A D + + I NH3 H3N I H3N NH3 iii. I NH3 I Br Br Cl NH2 CH2 H2C H2N Cl Cr Cr en = N NNM H CH CH NH NH CH H N NH 2 2 2 2 H3N 2 2 H2C 2 3 NH3 H3N 43. monodentate bidentate bridging O O MOO M D D D optically active mirror mirror image of iii M OO CO M COP COP (nonsuperimposable) D D D D O O MOO iv. + + 45. a. 2; b. 3; c. 4; d. 4 Cl Cl Br NH2 CH2 H2C H2N Br 47. NO NO ONO Cr Cr 2 2 NH CH H N NH H3N NH3 O2N NH3 H3N NH3 H3N 2 2 H2C 2 3 Co Co Co NH H N NH NH NH 3 3 H3N 3 H3N 3 H3N 3 NO2 NH3 ONO optically active mirror mirror image of iv ONO ONO ONO (nonsuperimposable) ONO NH3 H3N NH3 O2N NH3 Co Co Co 81. Octahedral Cr21 complexes should be used. Cr21: [Ar]3d4; high- H N NH H N NH3 H N NH3 3 3 3 3 spin (weak-field) Cr21 complexes have four unpaired electrons and low- NH3 NO2 NH3 spin (strong-field) Cr21 complexes have two unpaired electrons. Ni21: 8 21 49. Cr(acac)3 and cis-Cr(acac)2(H2O)2 are optically active. 51. With five [Ar]3d ; octahedral Ni complexes will always have two unpaired elec- electrons each in a different orbital, diagram (a) is for the weak-field trons, whether high or low spin. Therefore, Ni21 complexes cannot be used 31 Fe(H2O)6 complex ion. With three electrons, diagram (b) is for the to distinguish weak- from strong-field ligands by examining magnetic prop- 21 Cr(NH3)5Cl complex ion. With six electrons all paired up, diagram (c) is erties. Alternatively, the ligand field strengths can be measured using visible 1 21 21 for the strong-field Co(NH3)4Br2 complex ion. spectra. Either Cr or Ni complexes can be used for this method. 83. a. [Co(C H N) ]Cl ; b. [Cr(NH ) I]I ; c. [Ni(NH CH CH NH ) ]Br ; 53. a. Fe21 __h_ __h_ ___ ___ 5 5 6 3 3 5 2 2 2 2 2 3 2 d. K2[Ni(CN)4]; e. [Pt(NH3)4Cl2][PtCl4] 85. [Co(NH3)6]I3: 3 moles of AgI; hg h h hg hg hg ___ ___ ___ ___ ___ ___ [Pt(NH3)4I2]I2: 2 moles of AgI; Na2[PtI6]: 0 moles of AgI; [Cr(NH3)4I2]I: High spin Low spin 1 mole of AgI 87. 60 89. a. [Ar]4s13d10 is correct for Cu; b. correct; c. correct; d. [Xe]6s25d1 is correct for La; e. correct 91. Zn21: [Ar]3d10; b. Fe31 __h_ __h_ c. Ni21 __h_ __h_ Cu21: [Ar]3d9; Mn31: [Ar]3d4; Ti41: [Ne]3s23p6; color is a result of the elec- __h_ __h_ __h_ _h_g_ _h_g_ _h_g_ tron transfer between split d orbitals. This cannot occur for the filled d orbit- High spin als in Zn21. This also cannot occur for Ti41, which has no d electrons. So Zn21 and Ti41 compounds/ions will not be colored. Cu21 and Mn31 do have 32 31 55. weak field 57. a. 0; b. 2; c. 2 59. Co(CN)6 , Co(en)3 , d orbitals that are partially filled, so we would expect compounds/ions of 31 32 Co(H2O)6 , CoI6 61. The violet complex ion absorbs yellow-green Cu21 and Mn31 to be colored. 93. a. Zn21 has a 3d10 electron configuration. light (l < 570 nm), the yellow complex ion absorbs blue light (l < 450 nm), This diagram is for a d10 octahedral complex ion, not a d10 tetrahedral com- and the green complex ion absorbs red light (l < 650 nm). The violet complex as is given. So this diagram is incorrect. b. This diagram is correct. In 31 31 plex ion is Cr(H2O)6 , the yellow complex ion is Cr(NH3)6 , and the green this complex ion, Mn has a 13 oxidation state and an [Ar]3d4 electron con- 1 22 complex ion is Cr(H2O)4Cl2 . 63. CoBr4 is a tetrahedral complex ion, figuration. This strong field diagram for a d4 ion is shown. c. This diagram is 42 while CoBr6 is an octahedral complex ion. Since tetrahedral d-orbital split- correct. In this complex ion, Ni21 is present, which has a 3d8 electron con- ting is less than one-half the octahedral d-orbital splitting, the octahedral figuration. The diagram shown is correct for a diamagnetic square planar 42 complex ion (CoBr6 ) will absorb higher-energy light, which will have a complex. 95. Ni21 5 d 8; if ligands A and B produced very similar crystal 26 21 shorter wavelength than 3.4 3 10 m (E 5 hcyl). 65. 5 67. a. 211 kJ; fields, the cis-[NiA2B4] complex ion would give the following octahedral b. DH8 5 172.5 kJ; DS8 5 176 J/K; T . 980. K 69. Fe2O3: iron has a 13 crystal field diagram for a d 8 ion: oxidation state; Fe3O4: iron has a 18/3 oxidation state. The three iron ions in h h Fe3O4 must have a total charge of 18. The only combination that works is to ___ ___ have two Fe31 ions and one Fe21 ion per formula unit. This makes sense from _h_g_ _h_g_ _h_g_ 21 the other formula for magnetite, FeO ? Fe2O3. FeO has an Fe ion and Fe2O3 31 2 This is paramagnetic. has two Fe ions. 71. 8CN (aq) 1 4Ag(s) 1 O2(g) 1 2H2O(l) n 2 2 21 2 Because it is given that the complex ion is diamagnetic, the A and B ligands 4Ag(CN)2 (aq) 1 4OH (aq) 73. Hg (aq) 1 2I (aq) n HgI2(s), orange must produce different crystal fields, giving a unique d-orbital splitting dia-

57404_Ans_A31-A64.indd 59 10/14/16 5:12 PM A60 Answers to Selected Exercises

gram that would result in a diamagnetic species. 97. a. 20.26 V; b. From e. 3-methylhexane f. 4-ethylpentane standard reduction potentials, Co31 (° 5 1.82 V) is a much stronger oxidiz- 31 31 CH3CH2CHCH2CH2CH3 CH3CH2CH2CHCH3 ing agent than Co(en)3 (° 5 20.26 V); c. In aqueous solution, Co forms A A 31 the hydrated transition metal complex, Co(H2O)6 . In both complexes, CH3 CH2CH3 31 31 31 Co(H2O)6 and Co(en)3 , cobalt exists as Co , which has 6 d electrons. If we assume a strong-field case for each complex ion, then the d-orbital split- 3-methylhexane is correct. All six of these are the same compound. They only differ from each other by ting diagram for each has the six electrons paired in the lower-energy t2g orbitals. When each complex ion gains an electron, the electron enters the rotations about one or more carbon–carbon single bonds. Only one isomer of C7H16 is present in all of these names: 3-methylhexane. higher-energy eg orbitals. Since en is a stronger-field ligand than 2H O, then 31 5. Hydrocarbons are nonpolar substances exhibiting only London disper- the d-orbital splitting is larger for Co(en)3 , and it takes more energy to add 31 31 sion forces. Size and shape are the two most important structural features an electron to Co(en)3 than to Co(H2O)6 . Therefore, it is more favorable 31 31 relating to the strength of London dispersion forces. For size, the bigger the for Co(H2O)6 to gain an electron than for Co(en)3 to gain an electron. 99. No, in all three cases six bonds are formed between Ni21 and nitrogen. molecule (the larger the molar mass), the stronger the London dispersion So DH values should be similar. DS8 for formation of the complex ion is most forces and the higher the boiling point. For shape, the more branching present in a compound, the weaker the London dispersion forces and the lower negative for 6 NH3 molecules reacting with a metal ion (7 independent species become 1). For penten reacting with a metal ion, 2 independent species the boiling point. 7. The amide functional group is: become 1, so DS8 is the least negative. Thus the chelate effect occurs because O H B A the more bonds a chelating agent can form to the metal, the less unfavorable OOC NO DS8 becomes for the formation of the complex ion, and the larger the forma- When the amine end of one amino acid reacts with the carboxylic acid end tion constant. of another amino acid, the two amino acids link together by forming an am- 101. ___ d 2 z ide functional group. A polypeptide has many amino acids linked together, ___ ___ d 2 2, d x 2 y xy with each linkage made by the formation of an amide functional group. Be- ___ ___ d , d xz yz cause all linkages result in the presence of the amide functional group, the 103. The coordinate system for the complex ion is shown below. From the resulting polymer is called a polyamide. For nylon, the monomers also link coordinate system, the CN2 ligands are in a square planar arrangement. together by forming the amide functional group (the amine end of one mono- Since CN2 produces a much stronger crystal field, the diagram will most mer reacts with a carboxylic acid end of another monomer to give the amide resemble that of a square planar complex: functional group linkage). Hence nylon is also a polyamide. The correct order of strength is polyhydrocarbon , polyester , polyamide. The differ- NH3 y NC CN ence in strength is related to the types of intermolecular forces present. All of Ni NC CN these polymers have London dispersion forces. However, polyhydrocarbons x only have London dispersion forces. The polar ester group in polyesters and NH3 the polar amide group in polyamides give rise to additional dipole forces. z The polyamide has the ability to form relatively strong hydrogen bonding interactions, hence why it would form the strongest fibers. dx2 _ y2 O 9. a. OH H dz2 O H+ A A d xy O CH2PCH2 + H2OCH2 O CH2 1° alcohol dxz O O dyz b. OH H With the NH3 ligands on the z axis, we will assume that the dz2 orbital is de- H+ A A stabilized more than the dxy orbital. This may or may not be the case. CH3CHP CH2 + H2OCH3CH O CH2 2° alcohol 105. No; if 0.10 mol of AgBr could dissolve, Q would be greater than the K sp major product value, and a precipitate of AgBr(s) would have to re-form. 107. [Cr(NH3)5I]I2; octahedral c. OH H H+ A A Chapter 22 CH3CCP H2 + H2O CH3C O CH2 3° alcohol A A 1. Carbon has the unusual ability to form bonds to itself, whether they are CH CH single, double, or triple bonds, as well as the ability to form long chains or 3 3 major product rings of carbon atoms. Carbon also forms strong bonds to other nonmetals such as hydrogen, oxygen, phosphorus, sulfur, and the halogens. Because of d. O this bonding ability, carbon can form millions of compounds, including bio- oxidation P molecules, which are the basis for the existence of life. CH3CH2OH CH3CH aldehyde 3. a. 1-sec-butylpropane b. 4-methylhexane e. OH O CH CH CH CH A A 2 2 3 A 3 oxidation P CH3CHCH3 CH3CCH3 ketone CH3CHCH2CH3 CH3CH2CH2CHCH2CH3 3-methylhexane is correct. 3-methylhexane is correct. f. O

c. 2-ethylpentane d. 1-ethyl-1-methylbutane oxidation P CH3CH3CH3OH CH3CH2COOH carboxylic acid CH CHCH CH CH CH2CH3 3A 2 2 3 A or CH2CH3 CHCH2CH2CH3 A O O

CH3 P oxidation P CH3CH2CH CH3CH2COOH 3-methylhexane is correct. 3-methylhexane is correct. g. O O P P CH3OH + HOCCH3 CH3OOOCCH3 + H2O ester

57404_Ans_A31-A64.indd 60 10/14/16 5:12 PM Answers to Selected Exercises A61

11. a. A polyester forms when an alcohol functional group reacts with a 23. a. 2,2,4-trimethylhexane; b. 5-methylnonane; c. 2,2,4,4-tetramethylpentane; carboxylic acid functional group. The monomer for a homopolymer polyes- d. 3-ethyl-3-methyloctane ter must have an alcohol functional group and a carboxylic acid functional 25. CH3 iCH2 iCH2 iCH3; H H group present in the structure. b. A polyamide forms when an amine func- A A HCOCO OH tional group reacts with a carboxylic acid functional group. For a copolymer A A polyamide, one monomer would have at least two amine functional groups HCOCO OH present and the other monomer would have at least two carboxylic acid func- A A tional groups present. For polymerization to occur, each monomer must have HH two reactive functional groups present. c. To form an addition polymer, a Each carbon is bonded to four other atoms. carbon–carbon double bond must be present. To form a polyester, the mono- 27. a. 1-butene; b. 4‑methyl‑2‑hexene; c. 2,5-dimethyl-3-heptene mer would need the alcohol and carboxylic acid functional groups present. 29. a. CH3CH2CHPCHCH2CH3; b. CH3CHPCHCHPCHCH2CH3; To form a polyamide, the monomer would need the amine and carboxylic c. CH3CHCHP CHCH2CH2CH2CH3 acid functional groups present. The two possibilities are for the monomer to A CH have a carbon–carbon double bond, an alcohol functional group, and a car- 3 boxylic acid functional group all present, or to have a carbon–carbon double CH bond, an amine functional group, and a carboxylic acid functional group 31. a. 3 b. CH3 CH3 A A present. CH2CH3 H3COCCOCH3 13. Denaturation is the breakdown of the three-dimensional structure of pro- A A tein. Both the secondary and the tertiary structures of the protein are dis- CH3 CH3 rupted in denaturation. The primary structure (the order in which the amino c. d. acids link to each other) is not affected by denaturation. CH2CH3 CH2CHP CHCH3

15. CH3 CH2 CH2 CH2 CH2 CH3 CH3 CH2CH3 CH3 CH CH2 CH2 CH3 CH3 33. a. 1,3-dichlorobutane; b. 1,1,1-trichlorobutane; c. 2,3-dichloro-2, 4‑dimethylhexane; d. 1,2-difluoroethane 35. CH2Cl i CH2Cl,1,2‑ CH3 CH2 CH CH2 CH3 dichloroethane: There is free rotation about the C i C single bond that doesn’t CH3 lead to different compounds. CHCl w CHCl, 1,2-dichloroethene: There is no CH3 C CH2 CH3 rotation about the C w C double bond. This creates the cis and trans isomers, which are different compounds. 37. [27], compounds b and c; [29], all com- CH3 pounds CH3 CH3 39. CH2PCHCH2CH2CH3 CH3CHPCHCH2CH3 CH3 CH CH CH3 CH2 P CCH2CH3 CH3CP CHCH3 A A 17. a. CH3 CH3 CH3 CH CHCH CH CH CH CH 3 2 2 2 2 3 CH3CHCHPCH2 2-methylheptane A CH3 CH3 41. Cl CH HCH CH3CH2CHCH2CH2CH2CH3 D D 3 D D 3 3-methylheptane C PC C PC G G G G Cl HH Cl H CH3 CH2CHP CH2 CH2P CCH3 CH3CH2CH2CHCH2CH2CH3 A A 4-methylheptane Cl Cl

43. FCH2CH3 H3C CH3 b. CH3 CH3 D D D D C PC C PC CH C C CH G G G G 3 3 H H F H CH3 CH3 F CH3 2,2,3,3-tetramethylbutane A A CH2 PCHCHCH3 CH2 PCCH2 19. a. CH3 b. CH A A A 3 F CH3CHCH3 CH3CHCH2CH3 FCH3 F D D A C PC c. CH3 d. CH3 CH PCCH CH A A G G 2 2 3 H3C H CH3CHCH2CH2CH3 CH3CHCH2CH2CH2CH3 F CH3 H2CF CH A A D D 3 21. a. CH3CH2 CH CH2CH2CH3 b. CH3 C PC CHP CCH3 G G CH CH CH CH C CH CH CH H H 2 3 3 2 3 CH CH CH H CH CH H CH 3 3 3 D D 2 3 D D 3 C PC C PC c. CH CH CH CH CH d. 4-ethyl-2-methylheptane; G G G G 3 2 2 3 F H H CF H 2,2,3-trimethylhexane 2 CH C CH 3 3 F A CH3 CH2 PCHCH2CH2

57404_Ans_A31-A64.indd 61 10/14/16 5:13 PM

A62 Answers to Selected Exercises

D D D 45. a. CH3 CH2CH2CH3 b. CH3 D H 67. a. CH3CH2CH2CH3 B CC CCB D D D D b. Cl Cl Cl Cl c. Cl HH HCH

3 + HCl

CH2CHCHCHCH

D c. CH3 D CH2CH3 CH CH CCB 3 3 D D Cl Cl d. C4H8(g) 1 6O2(g) 8n 4CO2(g) 1 4H2O(g)

47. There are many possible structural isomers for the formula C6H12. Two 69. For the iron-catalyzed reaction, one of the ortho or para hydrogens in structural isomers are: benzene is replaced by chlorine. When an iron catalyst is not present, the benzene hydrogens are unreactive. To substitute for an alkane hydrogen, HH H H light must be present. For toluene, the light-catalyzed reaction substitutes a chlorine for a hydrogen in the methyl group attached to the benzene ring. H H H H 71. a. O O

H H HCCH2CHCH3 + HO CCH2CHCH3 H H CH2 P CHCH2CH2CH2CH3 CH CH Cyclohexane 1-hexene 3 3

The structural isomer 2-hexene (plus others) exhibits geometrical isomerism. b. O

D D D D CH C CHCH H3CCH2CH2CH3 H CH2CH2CH3 3 3

CCB CCB CH D D D D 3 HH H C H 3 c. No reaction cis trans The structural isomer 3-methyl-1-pentene exhibits optical isomerism (the aster- d. O O isk marks the chiral carbon). CH+ COH

CH3 A * e. O CH2 P CHOOC CH2CH3 A H CH3 For this compound, the chiral carbon has four different groups bonded to it; the mirror image of this compound will be nonsuperimposable. Optical isomerism f. O OH O OH is also possible with some of the cyclobutane and cyclopropane structural iso- CH3 CH3 mers. CH+ C OH 49. a. 3 monochloro isomers of n-pentane; b. 4 monochloro isomers of 2- methylbutane; c. 3 monochloro isomers of 2,4-dimethylpentane; d. 4 mono O O

chloro isomers of methylcyclobutane 51. a. ketone; b. aldehyde; c. carboxylic acid; d. amine 73. a. CH3CHPCH2 1 Br2 n CH3CHBrCH2Br;

b. OH O

53. a. HH D D A B H H D D amine oxidation CH3 OOCH CH3 CH3 OOCH3;C HO NO CCB

D D

D B D amine ONC O CO O CO O C O O O H ketone B carboxylic c. CH3 CH3 C B C A + A D D HHO acid H O OH H CH2 POC CH3 + H2O CH2 OOCH3;C alcohol A A H OH

2 b. 5 carbons in ring and the carbon in OCO2H: sp ; the other two carbons: d. O sp3; c. 24 sigma bonds, 4 pi bonds 55. a. 3-chloro-1-butanol, primary; B KMnO4 b. 3-methyl-3-hexanol, tertiary; c. 2-methylcyclopentanol, secondary CH3CH2CH2OH CH3CH2COOH 57. 1-butanol, 2-butanol, 2-methyl-1-propanol, and 2-methyl-2-propanol are the possible alcohols having the C4H10O formula. There are three possible 75. a. O ethers with the formula C4H10O. 59. Butanal and 2-methylpropanal are the B CH COH HOCH (CH ) CH possible aldehydes, and 2-butanone is the only ketone having the C4H8O 3 O + 2 2 6 3 formula. 61. a. 4,5-dichloro-3-hexanone; b. 2,3-dimethylpentanal; c. 3‑ O methylbenzaldehyde or m-methylbenzaldehyde 63. a. 4-chlorobenzoic B acid or p-chlorobenzoic acid; b. 3-ethyl-2-methylhexanoic acid; c. metha- CH3COOOCH2(CH2)6CH3 + H2O; noic acid (common name 5 formic acid) 65. Only statement d is false. b. O B O CH3CH2COOH + HOCH2(CH2)4CH3 P 2-butenal: HCCHP CHCH3 . O B The formula of 2-butenal is C4H6O, while the alcohol has a formula of CH3CH2COOO CH2(CH2)4CH3 + H2O C4H8O. 77. CFClPCF2

57404_Ans_A31-A64.indd 62 10/14/16 5:13 PM Answers to Selected Exercises A63

79. CN CN 105. aldohexose: glucose, mannose, galactose; aldopentose: ribose, arabi- A A nose; ketohexose: fructose; ketopentose; ribulose 107. They differ in the CCOO H2 OO CCOOH2 orientation of a hydroxy group on a particular carbon. Starch is composed A A n from a-d-glucose, and cellulose is composed from b-d-glucose. CO OCH3 CO OCH3 B B 109. The chiral carbons are marked with asterisks. O O H H 81. O O A A * * B B H3CCOOC H2CH3 H3COOOHC OOHN NHC C A A n * * H2NCOOO2HC H2NCOOO2HC A A 83. HO2C CO2H H H

H2NNH2 and Isoleucine Threonine

HO2C CO2H 111. Cl A * 85. Divinylbenzene has two reactive double bonds that are used during for- BrOOC CHP CH2 mation of the polymer. The key is for the double bonds to insert themselves A H into two different polymer chains. When this occurs, the two chains are bonded together (are crosslinked). The chains cannot move past each other is optically active. The chiral carbon is marked with an asterisk. because of the crosslinks making the polymer more rigid. 87. a. The poly- 113. Aspartame has two chiral carbon atoms. mer from 1,2-diaminoethane and terephthalic acid is stronger because of the 115. C–C–A–G–A–T–A–T–G 117. Uracil will H-bond to adenine. possibility of hydrogen bonding between chains. b. The polymer of

A A B BOH, ONH NP

HO CO2H HN NH,,N ON A K is more rigid because the chains are stiffer due to the rigid benzene rings in O N N B N Sugar A the chains. c. Polyacetylene is nHCqCH n O( CHPCHO)n. Polyacetylene A Sugar O Adenine is more rigid because the double bonds in the chains make the chains stiffer. H 89. a. serine; tyrosine; threonine; b. aspartic acid; glutamic acid; c. histidine; Uracil lysine; arginine; tryptophan; d. glutamine; asparagine 91. a. aspartic acid 119. a. glu: CTT, CTC; val: CAA, CAG, CAT, CAC; met: TAC; trp: ACC; and phenylalanine; b. Aspartame contains the methyl ester of phenylalanine. phe: AAA, AAG; asp: CTA, CTG; b. ACC–CTT–AAA–TAC or ACC–CTC– This ester can hydrolyze to form methanol, RCO2CH3 1 H2O 34 RCO2H AAA–TAC or ACC–CTT–AAG–TAC or ACC–CTC–AAG–TAC; c. four 1 CH3OH. (see answer to part b); d. met–asp–phe; e. TAC–CTA–AAG; TAC–CTA– 93. O O AAA; TAC–CTG–AAA 121. a. 2,3,5,6-tetramethyloctane; b. 2,2,3,5- tetramethylheptane; c. 2,3,4-trimethylhexane; d. 3-methyl-1-pentyne H2NCHC NHCHCO2H H2NCHC NHCHCO2H 123. Cl O Cl CH2 CH3 CH3 CH2

OH OH Cl O Cl ser–ala ala–ser There are many possibilities for isomers. Any structure with four chlorines 95. a. Six tetrapeptides are possible. From NH2 to CO2H end: phe–phe–gly– replacing four hydrogens in any of the numbered positions would be an gly, gly–gly–phe–phe, gly–phe–phe–gly, phe–gly–gly–phe, phe–gly–phe– isomer; i.e., 1,2,3,4-tetrachloro-dibenzo-p-dioxin is a possible isomer. gly, gly–phe–gly–phe; b. Twelve tetrapeptides are possible. From NH2 to 125. CO2H end: phe–phe–gly–ala, phe–phe–ala–gly, phe–gly–phe–ala, phe–gly– ala–phe, phe–ala–phe–gly, phe–ala–gly–phe, gly–phe–phe–ala, gly–phe– ala–phe, gly–ala–phe–phe, ala–phe–phe–gly, ala–phe–gly–phe, ala–gly– phe–phe 97. The secondary structure of a protein describes the arrangement n-hexane 2-methylpentane 3-methylpentane in space of the protein’s polypeptide chain. The most common secondary structures are the a-helix and the pleated sheet. The a-helix secondary structure gives a protein elasticity; examples are wool, hair, and tendons. The pleated sheet secondary structure has hydrogen bonding interactions between protein chains. As several protein chains interact with each other through hydrogen bonding, fibers result that are very strong and are resistant to stretching. 2,3-dimethylbutane 2,2-dimethylbutane Examples of proteins having pleated sheet secondary structures are silk and muscle fibers. 99. Ionic: his, lys, or arg with asp or glu; hydrogen bonding: 127. 2238C: CH3OOOCH3; 78.58C: CH3OCH2OOH 129. Alcohols ser, glu, tyr, his, arg, asn, thr, asp, gln, or lys with any amino acid; covalent: cys consist of two parts, the polar OH group and the nonpolar hydrocarbon chain with cys; London dispersion: all amino acids with nonpolar R groups (gly, ala, attached to the OH group. As the length of the nonpolar hydrocarbon chain pro, phe, ile, trp, met, leu, val); dipole–dipole: tyr, thr, and ser with each other increases, the solubility of the alcohol decreases. In methyl alcohol (metha- 101. Glutamic acid has a polar R group and valine has a nonpolar R group. nol), the polar OH group can override the effect of the nonpolar CH3 group, The change in polarity of the R groups could affect the tertiary structure of and methyl alcohol is soluble in water. In stearyl alcohol, the molecule con- hemoglobin and affect the ability of hemoglobin to bond to oxygen. sists mostly of the long nonpolar hydrocarbon chain, so it is insoluble in water. 131. n-hexane, 698C; pentanal, 1038C; 1-pentanol, 1378C; butanoic 103. CH2OH CH2OH H acid, 1648C. 133. 2-propanone; propanoic acid 135. In nylon, hydrogen- O O H H bonding interactions occur due to the presence of NiH bonds in the poly- H mer. For a given polymer chain length, there are more NiH groups in HH OH HO nylon-46 as compared to nylon-6. Hence, nylon-46 forms a stronger polymer H OH HO OH compared to nylon-6 due to the increased hydrogen-bonding interactions. OHOH HH D-Ribose D-Mannose

57404_Ans_A31-A64.indd 63 10/14/16 5:13 PM A64 Answers to Selected Exercises

137. a. 159. 6.07 161. a. No; the mirror image is superimposable. H2NNH2 and HO2CCO2H b. OH OH OH OH A A A A e C OOO Ce e C OOO Ce O O H CO H HO C H B B 002 2 00 OHN NHC CO CO2HH H CO2H b. Repeating unit: 163. O n B HCqCCO q CHOOCCPPCCHOOHCPCH PCHHC O CH2 OOH The two polymers differ in the substitution pattern on the benzene rings. The 13 12 11 10 987 65 43 21 Kevlar chain is straighter, and there is more efficient hydrogen bonding be-

tween Kevlar chains than between Nomex chains. 139. a. The bond angles 165. a. CH3CHCH2CH3 in the ring are about 608. VSEPR predicts bond angles close to 1098. The bonding electrons are much closer together than they prefer, resulting in CH3 strong electron–electron repulsions. Thus ethylene oxide is unstable (reac- b. tive). b. The ring opens up during polymerization and the monomers link CH2 CCH3 together through the formation of OOC bonds. CH3 O OCO H2CH2 OOOOCH2CH2 OOO CH2CH2 c. n CH3 CH2 – + 141. H2NCO HO CO2H H2NorO CHO CO2 Na A A CH CH CO –Na+ CH CH CO H 2 2 2 2 2 2 The first structure is MSG, which is impossible for you to predict. 143. DG 5 d. CH2PCHCH3 DH 2 TDS; for the reaction, we break a PO and OH bond and form a PO CH3 and OH bond, so DH < 0 based on bond dissociation energies. DS for this A e. CH CH CH CH CH CH CHCH CH process is negative (unfavorable) because positional probability decreases. A 2 2 2 2 3 A 2 2 3 Thus, DG . 0 due to the unfavorable DS term, and the reaction is not ex- OH OH pected to be spontaneous. CH3 CH3

145. 3 OCH A A CH CHCH CH CH OCOCH + + 3 2 2 2 3 1.0 M H : H3N CH C OH; A AA

OH OH CH3 3 OCH − − 167. O O O O 1.0 M OH : H2N CH C O B B B B OOOCH CH OCNH NHCOCH CH OCNH NHC 147. Both DH and DS are positive values. 149. a. pentane or n-pentane; 2 2 2 2 n b. 3-ethyl-2,5-dimethylhexane; c. 4-ethyl-5-isopropyloctane 151. a. iodo- cyclopropane; b. 2-chloro-1,3,5-trimethylcyclohexane; c. 1-chloro-2-ethyl- cyclopentane; d. 1-ethyl-2-methylcyclopentene 153. a. 1-pentanol, 169. a. The temperature of the rubber band increases when it is stretched; 2-methyl-2,4-pentadiol; b. trans-1,2-cyclohexadiol, 4-methyl-1-penten-3-ol b. exothermic (heat is released); c. As the chains are stretched, they line up (does not exhibit cis–trans isomerism) 155. The necessary carboxylic acid more closely resulting in stronger London dispersion forces between the is 2-chloropropanoic acid. chains. Heat is released as the strength of the intermolecular forces increases. O d. DG is positive and DS is negative; e. The structure of the stretched poly- B mer chains is more ordered (has a smaller positional probability) than in 157. CH3(CH2)8OOCOO CH2CH3 unstretched rubber. Therefore, entropy decreases as the rubber band is stretched. 171. 0.11% 173. a. 37.50%; b. The hybridization changes from sp2 to sp3; c. 3,4-dimethyl-3-hexanol

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