Coursehero >> Washington >> Washington >> CSE 142

Course Hero has millions of student submitted documents similar to the one below including study guides, homework solutions, papers, exam answer keys and textbook solutions.

TWENTY CHAPTER TRANSITION METALS AND COORDINATION CHEMISTRY Transition Metals 6. Transition metal ions lose the s electrons before the d electrons. a. Ti: [Ar]4s23d2 Ti2+: [Ar]3d2 Ti4+: [Ar] or [Ne]3s23p6 7. b. Re: [Xe]6s24f145d5 Re2+: [Xe]4f145d5 Re3+: [Xe]4f145d4 c. Ir: [Xe]6s24f145d7 Ir2+: [Xe]4f145d7 Ir3+: [Xe]4f145d6 Cr and Cu are exceptions to the normal filling order of electrons. a. Cr: [Ar]4s13d5 Cr2+: [Ar]3d4 Cr3+: [Ar]3d3 b. Cu: [Ar]4s13d10 Cu+: [Ar]3d10 Cu2+: [Ar]3d9 c. V: [Ar]4s23d3 V2+: [Ar]3d3 V3+: [Ar]3d2 8. 9. Since transition metals form bonds to species which donate lone pairs of electrons, then transition metals are Lewis acids (electron pair acceptors). Most transition metals have unfilled d orbitals, which creates a large number of valence electrons that can be removed. Stable ions of the representative metals are determined by how many s and p valence electrons can be removed. In general, representative metals lose all of the s and p valence electrons to form their stable ions. Transition metals generally lose the s electron(s) to form +1 and +2 ions, but they can also lose some (or all) of the d electrons to form other oxidation states as well. The lanthanide elements are located just before the 5d transition metals. The lanthanide contraction is the steady decrease in the atomic radii of the lanthanide elements when going from left to right across the periodic table. As a result of the lanthanide contraction, the sizes of the 4d and 5d elements are very similar (see Exercise 12.85). This leads to a greater similarity in the chemistry of the 4d and 5d elements in a given vertical group. a. molybdenum(IV) sulfide; molybdenum(VI) oxide 560 10. 11. 12. b. MoS2, +4; MoO3, +6; (NH4)2Mo2O7, +6; (NH4)6Mo7O24 4 H2O, +6 a. 4 O atoms on faces 1/2 O/face = 2 O atoms, 2 O atoms inside body, Total: 4 O atoms 8 Ti atoms on corners 1/8 Ti/corner + 1 Ti atom/body center = 2 Ti atoms Formula of the unit cell is Ti2O4. The empirical formula is TiO2. b. +4 -2 0 0 +4 -1 +4 -2 +2 -2 2 TiO2 + 3 C + 4 Cl2 2 TiCl4 + CO2 + 2 CO; Cl is reduced and C is oxidized. Cl2 is the oxidizing agent and C is the reducing agent. +4 -1 0 +4 -2 0 TiCl4 + O2 TiO2 + 2 Cl2; O is reduced and Cl is oxidized. O2 is the oxidizing agent and TiCl4 is the reducing agent. 13. TiF4: ionic compound containing Ti4+ ions and F- ions. TiCl4, TiBr4, and TiI4: covalent compounds containing discrete, tetrahedral TiX4 molecules. As these covalent molecules get larger, the bp and mp increase because the London dispersion forces increase. TiF4 has the highest bp since the interparticle forces are usually stronger in ionic compounds as compared to covalent compounds. Coordination Compounds 14. 15. Fe2O3(s) + 6 H2C2O4(aq) 2 Fe(C2O4)33-(aq) + 3 H2O(l) + 6 H+(aq); The oxalate anion forms a soluble complex ion with iron in rust (Fe2O3), which allows rust stains to be removed. a. ligand: b. chelate: c. bidentate: d. complex ion: 16. Species that donates a pair of electrons to form a covalent bond to a metal ion. Ligands act as Lewis bases (electron pair donors). Ligand that can form more than one bond to a metal ion. Ligand that forms two bonds to a metal ion. Metal ion plus ligands. b. hexaaquacobalt(III) iodide d. potassium hexachloroplatinate(II) f. triamminetrinitrocobalt(III) a. hexaamminecobalt(II) chloride c. potassium tetrachloroplatinate(II) e. pentaamminechlorocobalt(III) chloride 17. a. pentaamminechlororuthenium(III) ion c. tris(ethylenediamine)manganese(II) ion b. hexacyanoferrate(II) ion d. pentaamminenitrocobalt(III) ion b. sodium hexacyanocobaltate(III) 18. a. pentaaquabromochromium(III) bromide 561 562 CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY 19. c. bis(ethylenediamine)dinitroiron(III) chloride d. tetraamminediiodoplatinum(IV) tetraiodoplatinate(II) a. K2[CoCl4] b. [Pt(H2O)(CO)3]Br2 c. Na3[Fe(CN)2(C2O4)2] d. [Cr(NH3)3Cl(NH2CH2CH2NH2)]I2 b. [Ru(NH3)5H2O]3+ d. [Pt(NH3)Cl3]- 20. a. FeCl4c. [Cr(CO)4(OH)2]+ 21. BaCl2 gives no precipitate so SO42- must be in the coordination sphere. A precipitate with AgNO3 means that the Cl- is not in the coordination sphere. Since there are only four ammonia molecules in the coordination sphere, then the SO42- must be acting as a bidentate ligand (assuming an octahedral complex ion). The structure is: + NH3 O O H 3N Co H 3N NH3 O S O Cl - 22. CN- is a weak base so OH- ions are present. When the acid H2S is added, OH- and CN- ions are removed as H2O and HCN. The two reactions are: Ni2+(aq) + 2 OH-(aq) Ni(OH)2(s); The precipitate is Ni(OH)2(s). Ni(OH)2(s) + 4 CN-(aq) Ni(CN)42-(aq) + 2 OH- (aq); Ni(CN)42- is a soluble species. 23. a. isomers: Species with the same formulas but different properties. See text for examples of the following types of isomers. b. structural isomers: Isomers that have one or more bonds that are different. c. stereoisomers: Isomers that contain the same bonds but differ in how the atoms are arranged in space. d. coordination isomers: Structural isomers that differ in the atoms that make up the complex ion. e. linkage isomers:Structural isomers that differ in how one or more ligands are attached to the transition metal. f. geometric isomers: (cis - trans isomerism) Stereoisomers that differ in the positions of atoms with respect to a rigid ring, bond, or each other. CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY 563 24. g. optical isomers: Stereoisomers that are nonsuperimposable mirror images of each other; that is, they are different in the same way that our left and right hands are different. a. 2; Forms bonds through the lone pairs on the two oxygen atoms. b. 3; Forms bonds through the lone pairs on the three nitrogen atoms. c. 4; Forms bonds through the two nitrogen atoms and the two oxygen atoms. d. 4; Forms bonds through the four nitrogen atoms. 25. a. cis trans Note: C2O42- is a bidentate ligand. Bidentate ligands bond to the metal at two positions which are 90 apart from each other in octahedral complexes. Bidentate ligands do not bond to the metal at positions 180 apart. b. cis c. trans 564 CHAPTER 20 cis TRANSITION METALS AND COORDINATION CHEMISTRY trans CHAPTER 20 d. TRANSITION METALS AND COORDINATION CHEMISTRY 565 + N H 3N Cr I I NH 3 H 3N N I Cr I NH 3 N N + N I Cr H 3N I NH 3 N + 26. Note: N a. H 2C N is an abbreviation for the bidentate ligand ethylenediamine (NH2CH2CH2NH2). b. H 2N Pt Cl Cl H 2C H 2N Cl Co NH 2 CH 2 H 2C H 2N H 2C H 2N Cl NH 2 CH 2 c. d. e. 27. 566 CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY M = transition metal ion CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY 567 28. 29. Linkage isomers differ in the way that the ligand bonds to the metal. SCN- can bond through the sulfur or through the nitrogen atom. NO2- can bond through the nitrogen or through the oxygen atom. OCN- can bond through the oxygen or through the nitrogen atom. N3-, NH2CH2CH2NH2 and I- are not capable of linkage isomerism. 30. 31. Similar to the molecules discussed in Figures 20.16 and 20.17 of the text, Cr(acac)3 and cis-Cr(acac)2(H2O)2 are optically active. The mirror images of these two complexes are nonsuperimposable. There is a plane of symmetry in trans-Cr(acac)2(H2O)2, so it is not optically active. A molecule with a plane of symmetry is never optically active as the mirror images are 568 CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY always superimposable. A plane of symmetry is a plane through a molecule where one side reflects on the other side of the molecule. 32. There are five geometrical isomers (labeled i-v). Only isomer v where the CN-, Br- and H2O ligands are cis to each other is optically active. The nonsuperimposable mirror image is shown for isomer v. Bonding, Color, and Magnetism in Coordination Compounds 33. a. Ligand that will give complex ions with the maximum number of unpaired electrons. b. Ligand that will give complex ions with the minimum number of unpaired electrons. c. Complex with a minimum number of unpaired electrons (low-spin = strong-field). d. Complex with a maximum number of unpaired electrons (high-spin = weak-field). 34. Cu2+: [Ar]3d9; Cu+: [Ar]3d10; Cu(II) is d9 and Cu(I) is d10. Color is a result of the electron transfer between split d orbitals. This cannot occur for the filled d orbitals in Cu(I). Cd2+, like Cu+, is also d10. We would not expect Cd(NH3)4Cl2 to be colored since the d orbitals are filled in this Cd2+ complex. Sc3+ has no electrons in d orbitals. Ti3+ and V3+ have d electrons present. The color of transition metal complexes results from electron transfer between split d orbitals. If no d electrons are present, no electron transfer can occur and the compounds are not colored. All these complex ions contain Co3+ bound to different ligands so the difference in d-orbital 35. 36. CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY 569 splitting for each complex ion is due to the difference in ligands. The spectrochemical series indicates that CN- is a stronger field ligand than NH3 which is a stronger field ligand than F-. Therefore, Co(CN)63- will have the largest d-orbital splitting and will absorb the lowest wavelength electromagnetic radiation ( = 290 nm) since energy and wavelength are inversely related ( = hc/E). Co(NH3)63+ will absorb 440 nm electromagnetic radiation while CoF63- will absorb the longest wavelength electromagnetic radiation ( = 770 nm) since F- is the weakest field ligand present. 37. a. Fe2+: [Ar]3d6 High spin, small [Ar]3d 8 Low spin, large b. Fe3+: [Ar]3d5 c. Ni2+: High spin, small d. Zn2+: [Ar]3d10 e. Co2+: [Ar]3d7 High spin, small 38. Low spin, large NH3 and H2O are neutral ligands so the oxidation states of the metals are Co3+ and Fe2+. Both have six d electrons ([Ar]3d6). To explain the magnetic properties, we must have a strong-field for Co(NH3)63+ and a weak-field for Fe(H2O)62+. 570 Co3+: [Ar]3d6 CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY Fe2+: [Ar]3d6 large small Only this splitting of d-orbitals gives a diamagnetic Co(NH3)63+ (no unpaired electrons) and a paramagnetic Fe(H2O)62+ (unpaired electrons present). 39. To determine the crystal field diagrams, you need to determine the oxidation state of the transition metal which can only be determined if you know the charges of the ligands (see Table 20.13). The electron configurations and the crystal field diagrams follow. a. Ru2+: [Kr]4d6, no unpaired eb. Ni2+: [Ar]3d8, 2 unpaired e Low spin, large c. V3+: [Ar]3d2, 2 unpaired e- Note: Ni2+ must have 2 unpaired electrons, whether high-spin or low-spin, and V3+ must have 2 unpaired electrons, whether high-spin or low-spin. 40. In both compounds, iron is in the +3 oxidation state with an electron configuration of [Ar]3d5. Fe3+ complexes have one unpaired electron when a strong-field case and five unpaired electrons when a weak-field case. Fe(CN)62- is a strong-field case and Fe(SCN)63- is a weak-field case. Therefore, cyanide, CN-, is a stronger field ligand than thiocyanate, SCN-. Coordination compounds exhibit the color complementary to that absorbed. From Table 20.16, [Cr(H2O)6]Cl3 absorbs yellow-green light (~570 nm), [Cr(H2O)4Cl2]Cl absorbs red light (~650 nm) and [Cr(NH3)6]Cl3 absorbs blue light (~450 nm). [Cr(NH3)6]Cl3 absorbs the shorter wavelength light so , the d-orbital splitting, is largest for [Cr(NH3)6]Cl3 since energy and wavelength of light are inversely related (E = hc/). From the wavelengths of light absorbed by the various compounds, NH3 is a stronger field ligand than H2O, which is a stronger field ligand than Cl-. 41. CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY 571 This is consistent with the spectrochemical series. 42. Octahedral Cr2+ complexes should be used. Cr2+: [Ar]3d4; High-spin (weak-field) Cr2+ complexes have 4 unpaired electrons and low-spin (strong-field) Cr2+ complexes have 2 unpaired electrons. Ni2+: [Ar]3d8; Octahedral Ni2+ complexes will always have 2 unpaired electrons, whether high or low-spin. Therefore, Ni2+ complexes cannot be used to distinguish weak from strong-field ligands by examining magnetic properties. Alternatively, the ligand field strengths can be measured using visible spectra. Either Cr2+ or Ni2+ complexes can be used for this method. a. Ru(phen)32+ exhibits optical isomerism [similar to Co(en)33+ in Figure 20.16 of the text]. b. Ru2+: [Kr]4d6; Since there are no unpaired electrons, then Ru2+ is a strong-field (low-spin) case. 43. 44. Co2+: [Ar]3d7; The corresponding d-orbital splitting diagram for tetrahedral Co2+ complexes is: All tetrahedral complexes high are spin since the d-orbital splitting is small. Ions with 2 or 7 d electrons should give the most stable tetrahedral complexes since they have the greatest number of electrons in the lower energy orbitals as compared to the number of electrons in the higher energy orbitals. 45. CoBr64- has an octahedral structure and CoBr42- has a tetrahedral structure (as do most Co2+ complexes with four ligands). Coordination complexes absorb electromagnetic radiation (EMR) of energy equal to the energy difference between the split d-orbitals. Since the tetra-hedral d-orbital splitting is less than one-half of the octahedral d-orbital splitting, then tetrahedral complexes will absorb lower energy EMR which corresponds to longer wavelength EMR (E = hc/). Therefore, CoBr62- will absorb EMR having a wavelength shorter than 3.4 10-6 m. Pd is in the +2 oxidation state in PdCl42-; Pd2+: [Kr]4d8. If PdCl42- were a tetrahedral complex, then it would have 2 unpaired electrons and would be paramagnetic (see diagram below). Instead, PdCl42- has a square planar molecular structure with a d-orbital splitting diagram shown below. Note that all electrons are paired in the square planar diagram which explains the diamagnetic properties of PdCl42-. 46. 572 CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY tetrahedral d8 square planar d8 Additional Exercises 47. 48. Ni(CO)4 is composed of 4 CO molecules and Ni. Thus, nickel has an oxidation state of zero. CN- and CO form much stronger complexes with Fe(II) than O2. Thus, O2 cannot be transported by hemoglobin in the presence of CN- or CO. 49. i. 0.0203 g CrO3 0.0106 g 52.00 g Cr = 0.0106 g Cr; % Cr = 100 = 10.1% Cr 0.105 g 100.0 g CrO3 0.100 mmol HCl 1 mmol NH3 17.03 mg NH3 _ _ = 56.1 mg NH3 mL mmol HCl mmol ii. 32.93 mL HCl % NH3 = 56.1 mg 100 = 16.5% NH3 341 mg iii. 73.53% + 16.5% + 10.1% = 100.1%; The compound must be composed of only Cr, NH3, and I. Out of 100.00 of compound: 10.1 g Cr 1 mol = 0.194 52.00 g 1 mol = 0.969 17.03 g 0.194 = 1.00 0.194 0.969 = 4.99 0.194 16.5 g NH3 CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY 573 73.53 g I 1 mol = 0.5794 126.9 g 0.5794 = 2.99 0.194 Cr(NH3)5I3 is the empirical formula. Cr(III) forms octahedral complexes. So compound A is made of the octahedral [Cr(NH3)5I]2+ complex ion and two I- ions as counter ions; the formula is [Cr(NH3)5I]I2. Lets check this proposed formula using the freezing point data. iv. Tf = iKfm; For [Cr(NH3)5I]I2, i = 3.0 (assuming complete dissociation). molality = m = 0.601 g complex 1 mol complex _ = 0.116 mol/kg -2 1.000 _ 10 kg H2 O 517.9 g complex Tf = 3.0 1.86 C kg/mol 0.116 mol/kg = 0.65 C Since Tf is close to the measured value, then this is consistent with the formula [Cr(NH3)5I]I2. 50. a. Copper is both oxidized and reduced in this reaction, so yes this reaction is an oxidationreduction reaction. The oxidation state of copper in [Cu(NH3)4]Cl2 is +2, the oxidation state of copper in Cu is zero and the oxidation state of copper in [Cu(NH3)4]Cl is +1. b. Total mass of copper used: 10,000 boards (8.0 cm _ 16.0 cm _ 0.060 cm) 8.96 g _ = 6.9 105 g Cu61 6 ff1 3 board cm Amount of Cu to be recovered = 0.80 6.9 105 g = 5.5 105 g Cu 5.5 105 g Cu 1 mol [Cu( NH3 )4] Cl2 202.59 g [Cu( NH3 )4] Cl2 1 mol Cu _ _ 63.55 g Cu mol Cu mol [Cu( NH3 )4] Cl2 = 1.8 106 g [Cu(NH3)4]Cl2 5.5 105 g Cu 1 mol Cu 4 mol NH3 17.03 g NH3 _ _ = 5.9 105 g NH3 63.55 g Cu mol Cu mol NH3 51. M = metal ion 574 CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY 52. a. The optical isomers of this compound are similar to the ones discussed in Figure 20.16 of the text. In the following structures, we omitted the 4 NH3 ligands coordinated to the outside cobalt atoms. mirror b. All are Co(III). The three "ligands" each contain 2 OH- and 4 NH3 groups. If each cobalt is in the +3 oxidation state, then each ligand has a +1 overall charge. The +3 charge from the three ligands along with the +3 charge of the central cobalt atom gives the overall complex a +6 charge. This is balanced by the -6 charge of the six Cl- ions. c. Co3+: [Ar]3d6; There are zero unpaired electrons if a low-spin (strong-field) case. large CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY 575 53. No; In all three cases, six bonds are formed between Ni 2+ and nitrogen, so H values should be similar. S for formation of the complex ion is most 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 become 1, so S is least negative for this reaction as compared to the other reactions. Thus, the chelate effect occurs because the more bonds a chelating agent can form to the metal, the more favorable S is for the formation of the complex ion and the larger the formation constant. CrCl3 6H2O contains nine possible ligands; only six of which are used to form the octahedral complex ion. The three species not present in the complex ion will either be counter ions to balance the charge of the complex ion and/or waters of hydration. The number of counter ions for each compound can be determined from the silver chloride precipitate data and the number of waters of hydration can be determined from the dehydration data. In all experiments, the ligands in the complex ion do not react. Compound I: mol CrCl3 6H2O = 0.27 g 54. 1 mol = 1.0 10-3 mol CrCl3 6H2O 266.5 mol waters of hydration = 0.036 g H2O 1 mol = 2.0 10-3 mol H2O 18.02 g 2.0 _ 10-3 mol mol waters of hydration = 2.0 = 1.0 _ 10-3 mol mol compound In compound I, two of the H2O molecules are waters of hydration so the other four water molecules are present in the complex ion. Therefore, the formula for compound I must be [Cr(H2O)4Cl2]Cl 2H2O. Two of the Cl- ions are present as ligands in the octahedral complex ion and one Cl- ion is present as a counter ion. The AgCl precipitate data that refers to this compound is the one that produces 1430 mg AgCl: mol Cl- from compound I = 0.1000 L 0.100 mol [Cr(H 2 O )4 Cl2] Cl _ 2 H 2 O L 1 mol Cl= 0.0100 mol Clmol [Cr(H 2 O )4 Cl2] Cl _ 2 H 2 O 576 CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY mass AgCl produced = 0.0100 mol Cl- AgCl Compound II: 1 mol AgCl 143.4 g AgCl _ = 1.43 g =1430 mg mol Cl mol AgCl 1 mol mol waters of hydration 18.02 g = = 1.0 -3 mol compound 1.0 _ 10 mol compound 0.018 g H 2 O _ The formula for compound II must be [Cr(H2O)5Cl]Cl2 H2O. The 2870 mg AgCl precipitate data refers to this compound. For 0.0100 mol of compound II, 0.0200 mol Cl- are present as counter ions: mass AgCl produced = 0.0200 mol Cl- AgCl Compound III: This compound has no mass loss on dehydration so there are no waters of hydration present. The formula for compound III must be [Cr(H 2 O )6] Cl3 . 0.0100 mol of this compound produces 4300 mg of AgCl(s) when treated with AgNO3. 0.0300 mol Cl- 1 mol AgCl 143.4 g _ = 2.87 g = 2870 mg mol Cl mol 1 mol AgCl 143.4 g AgCl _ = 4.30 g = 4.30 103 mg AgCl mol Cl mol AgCl The structural formulas for the compounds are: Compound I CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY 577 Compound II Compound III From Table 20.16 of the text, the violet compound will be the one that absorbs light with the shortest wavelength (highest energy). This should be compound III. H 2 O is a stronger field ligand than Cl- ; compound III with the most coordinated H 2 O molecules will have the largest d-orbital splitting and will absorb the higher energy light. 55. (H2O)5Cr - Cl - Co(NH3)5 (H2O)5Cr - Cl - Co(NH3)5 Cr(H2O)5Cl2+ + Co(II) complex Yes, this is consistent. After the oxidation, the ligands on Cr(III) won't exchange. Since Cl- is in the coordination sphere, then it must have formed a bond to Cr(II) before the electron transfer occurred (as proposed through the formation of the intermediate). II III III II 56. a. Be(tfa)2 exhibits optical isomerism. A representation for the tetrahedral optical isomers are: 578 CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY Note: The dotted line indicates a bond pointing into the plane of the paper and the wedge indicates a bond pointing out of the plane of the paper. b. Square planar Cu(tfa)2 molecules exhibit geometric isomerism. In one geometric isomer, the CF3 groups are cis to each other and in the other isomer, the CF3 groups are trans. Challenge Problems 57. d x2 - y2 , d xy d z2 d xz , d yz CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY 579 The d x 2- y2 and dxy orbitals are in the plane of the three ligands and should be destabilized the most. The amount of destabilization should be about equal when all the possible interactions are considered. The d z2 orbital has some electron density in the xy plane (the doughnut) and should be destabilized a lesser amount as compared to the d x 2- y2 and dxy orbitals. The dxz and dyz orbitals have no electron density in the plane and should be lowest in energy. 58. d z2 d x2 - y2 , d xy d xz , d yz The d z2 orbital will be destabilized much more than in the trigonal planar case (see Exercise 20.57). The d z2 orbital has electron density on the z-axis directed at the two axial ligands. The d x2- y2 and dxy orbitals are in the plane of the three trigonal planar ligands and should be destabilized a lesser amount as compared to the d z2 orbital; only a portion of the electron density in the d x 2- y2 and dxy orbitals is directed at the ligands. The dxz and dyz orbitals will be destabilized the least since the electron density is directed between the ligands. 59. a. Consider the following electrochemical cell: Co3+ + e- Co2+ Co(en)32+ Co(en)33+ + eCo3+ + Co(en)32+ Co2+ + Co(en)33+ Ec = 1.82 V - Ea = ? Ecell = 1.82 - Ea The equilibrium constant for this overall reaction is: Co3+ + 3 en Co(en)33+ Co(en)32+ Co2+ + 3 en K1 = 2.0 1047 12 K2 = 1/1.5 10 Co3+ + Co(en)32+ Co(en)33+ + Co2+ K = K1K2 = 2.0 _ 1047 = 1.3 1035 12 1.5 _ 10 From the Nernst equation for the overall reaction: Ecell = 0.0591 0.0591 log K = log (1.3 _ 1035), Ecell = 2.08 V n 1 580 CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY Ecell = 1.82 - Ea = 2.08 V, - Ea = 2.08 V -1.82 V = 0.26 V so Ec = -0.26 V b. The strongest oxidizing agent is the species most easily reduced and will have the most positive standard reduction potential. From the reduction potentials, Co3+ (E = 1.82 V) is a much stronger oxidizing agent than Co(en)33+ (E = -0.26 V). c. In aqueous solution, Co3+ forms the hydrated transition metal complex, Co(H2O)63+. In both complexes, Co(H2O)63+ and Co(en)33+, cobalt exists as Co3+ which has 6 d electrons. Assuming a strong-field case for each complex ion, then the d-orbital splitting diagram for each is: When each complex gains an electron, the electron enters the higher energy eg orbitals. Since en is a stronger field ligand than H2O, then the d-orbital splitting is larger for Co(en)33+ and it takes more energy to add an electron to Co(en)33+ than to Co(H2O)63+. Therefore, it is more favorable for Co(H2O)63+ to gain an electron than for Co(en)33+ to gain an electron. 60. a. Initial Equil. AgBr(s) _ Ag+ + 0 s Br- Ksp = [Ag+][Br-] = 5.0 10-13 0 s s = solubility (mol/L) Ksp = 5.0 10-13 = s2, s = 7.1 10-7 mol/L b. AgBr(s) _ Ag+ + BrAg+ + 2 NH3 _ Ag(NH3)2+ Ksp = 5.0 10-13 Kf = 1.7 107 AgBr(s) + 2 NH3(aq) _ Ag(NH3)2+(aq) + Br-(aq) K = Ksp Kf = 8.5 10-6 AgBr(s) Initial Equil. K= + 2 NH3 _ Ag(NH3)2+ + Br- 3.0 M 0 0 s mol/L of AgBr(s) dissolves to reach equilibrium = molar solubility 3.0 - 2s s s + 2 2 [Ag( NH3 )2 ][Br -] s s = 8.5 10-6 , s = 8.7 10-3 mol/L = 2 2 2 [ NH3 ] (3.0 ) (3.0 - 2 s ) Assumption good. CHAPTER 20 TRANSITION METALS AND COORDINATION CHEMISTRY 581 c. The presence of NH3 increases the solubility of AgBr. Added NH3 removes Ag+ from solution by forming the complex ion Ag(NH3)2+. As Ag+ is removed, more AgBr(s) will dissolve to replenish the Ag+ concentration. d. mass AgBr = 0.2500 L 8.7 _ 10-3 mol AgBr 187.8 g AgBr = 0.41 g AgBr _ L mol AgBr e. Added HNO3 will have no effect on the AgBr(s) solubility in pure water. Neither H+ nor NO3react with Ag+ or Br- ions. Br- is the conjugate base of the strong acid HBr, so it is a terrible base. Added H+ will not react with Br- to any great extent. However, added HNO3 will reduce the solubility of AgBr(s) in the ammonia solution. NH3 is a weak base (Kb = 1.8 10-5). Added H+ will react with NH3 to form NH4+. As NH3 is removed, a smaller amount of the Ag(NH3)2+ complex ion will form resulting in a smaller amount of AgBr(s) which will dissolve.

Find millions of documents here - Study Guides, Homework Solutions, Papers, Exam Answer Keys and more. Course Hero has millions of course related materials that will enable you to learn better, faster and get an A in all your courses.
Below is a small sample set of documents: