chapter 19 transition metals and coordination chemistry

1 - Que signifient les lettres c-m-u ?2 - Quels sont les avantages qu’elle propose ? ( au moins unexemple pour illustrer ).3 - En quoi cette mesure tente-t-elle de diminuer les inégalités?Affiche pour la CMU3 pts1 - De quand le RMI date-t-il ? ( justifie )2 - Que signifient les lettres R-M-I ? ( le R signifie revenu )3 - Qui peut bénéficier de cette mesure, et à quoi sert-elle ?4 - Que signifie le dialogue entre les trois personnes ?Caricature sur le RMI, 20024 ptsTg

734 CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY+H 3 NNH 3OOCoSCl -H 3 NNH 3OO28. CN − is a weak base, so OH − ions are present. When the acid H 2 S is added, OH − and CN − ionsare removed as H 2 O and HCN. The hydrated Ni 2+ complex ion forms after the OH − and CN −ions are removed by addition of H 2 S. The two reactions are:Ni 2+ (aq) + 2 OH − (aq) → Ni(OH) 2 (s); the precipitate is Ni(OH) 2 (s).Ni(OH) 2 (s) + 4 CN − (aq) → Ni(CN) 4 2− (aq) + 2 OH − (aq); Ni(CN) 4 2− is a soluble species.Ni(CN) 4 2− (aq) + 4 H 2 S(aq) + 6 H 2 O(l) → Ni(H 2 O) 6 2+ (aq) + 4 HCN(aq) + 4 HS − (aq)29. Because each compound contains an octahedral complex ion, the formulas for the compoundsare [Co(NH 3 ) 6 ]I 3 , [Pt(NH 3 ) 4 I 2 ]I 2 , Na 2 [PtI 6 ], and [Cr(NH 3 ) 4 I 2 ]I. Note that in some cases the I −ions are ligands bound to the transition metal ion as required for a coordination number of 6,whereas in other cases the I − ions are counter ions required to balance the charge of thecomplex ion. The AgNO 3 solution will only precipitate the I − counter ions and will notprecipitate the I − ligands. Therefore, 3 mol AgI will precipitate per mole of [Co(NH 3 ) 6 ]I 3 , 2mol AgI will precipitate per mole of [Pt(NH 3 ) 4 I 2 ]I 2 , 0 mol AgI will precipitate per mole ofNa 2 [PtI 6 ], and l mol AgI will precipitate per mole of [Cr(NH 3 ) 4 I 2 ]I.30. Test tube 1: Added Cl − reacts with Ag + to form a silver chloride precipitate. The net ionicequation is Ag + (aq) + Cl − (aq) → AgCl(s). Test tube 2: Added NH 3 reacts with Ag + ions toform a soluble complex ion Ag(NH 3 ) + 2 . As this complex ion forms, Ag + is removed fromsolution, which causes the AgCl(s) to dissolve. When enough NH 3 is added, all the silverchloride precipitate will dissolve. The equation is AgCl(s) + 2 NH 3 (aq) → Ag(NH 3 ) + 2 (aq) +Cl − (aq). Test tube 3: Added H + reacts with the weak base NH 3 to form NH + 4 . As NH 3 isremoved from the Ag(NH 3 ) + 2 complex ion, Ag + ions are released to solution and can thenreact with Cl − to reform AgCl(s). The equations are Ag(NH 3 ) + 2 (aq) + 2 H + (aq) → Ag + (aq)+ 2 NH + 4 (aq) and Ag + (aq) + Cl − (aq) → AgCl(s).31. a. Isomers: species with the same formulas but different properties; they are different compounds.See the text for examples of the following types of isomers.b. Structural isomers: isomers that have one or more bonds that are different.c. Steroisomers: isomers that contain the same bonds but differ in how the atoms arearranged in space.d. Coordination isomers: structural isomers that differ in the atoms that make up thecomplex ion.

CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 735e. Linkage isomers: structural isomers that differ in how one or more ligands are attached tothe transition metal.f. Geometric isomers: (cis-trans isomerism); steroisomers that differ in the positions ofatoms with respect to a rigid ring, bond, or each other.g. Optical isomers: steroisomers 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.32. 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.33. a.d. 4; forms bonds through the four nitrogen atoms.OH 2 OH 2 OCOCoOCCOOCOOOOCCOOOH 2OCoOOH 2CCOOOcistransNote: C 2 O 4 2- is a bidentate ligand. Bidentate ligands bond to the metal at two positions thatare 90° apart from each other in octahedral complexes. Bidentate ligands do not bond to themetal at positions 180° apart from each other.b.H 3 NH 3 NIPtNH 32+2+IIH 3 N NH 3PtNH 3 H 3 N NH 3Icistrans

CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 737e.2+H 2 CH 2 NOH 2NH 2CH 2CuH 2 CH 2 NOH 2NH 2CH 235.ClClClH 3 NH 3 NCoClNH 3NH 3H 3 NH 3 NCoNH 3Cl ClNH 3CoNH 3 NH 3 NH 3NH 3mirrortrans(mirror image issuperimposable)cisThe mirror image of the cisisomer is also superimposable.No; both the trans and the cis forms of Co(NH 3 ) 4 Cl 2 + have mirror images that are superimposable.For the cis form, the mirror image only needs a 90E rotation to produce the originalstructure. Hence neither the trans nor cis form is optically active.36.H 3 NH 3 NCl2+NH 3CoNH 3NH 3To form the trans isomer, Cl − would replace the NH 3 ligand that is bold in the structureabove. If any of the other four NH 3 molecules are replaced by Cl − , the cis isomer results.Therefore, the expected ratio of the cis:trans isomer in the product is 4:1.37.MOH 2 NCOCH 2M = transition metal ion

738 CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRYOOOCONH 2 CH 2COOCCuandCuH 2 CH 2 NOCH 2 CH 2 NNH 2CH 2O38.monodentatebidentatebridgingM O COOMOOCOMMOOCO39. Linkage isomers differ in the way that the ligand bonds to the metal. SCN − can bond throughthe sulfur or through the nitrogen atom. NO 2 − can bond through the nitrogen or through theoxygen atom. OCN − can bond through the oxygen or through the nitrogen atom. N 3 − ,NH 2 CH 2 CH 2 NH 2 and I − are not capable of linkage isomerism.40.H 3 NH 3 NNO 2CoNO 2NH 3NH 3O 2 NH 3 NNO 2CoNH 3NH 3NH 3H 3 NH 3 NONONH 3CoNH 3ONOONOH 3 NONOCoNH 3NH 3NH 3H 3 NH 3 NONOCoNH 3NO 2NH 3O 2 NH 3 NONOCoNH 3NH 3NH 341. Similar to the molecules discussed in Figures 19.16 and 19.17 of the text, Cr(acac) 3 and cis-Cr(acac) 2 (H 2 O) 2 are optically active. The mirror images of these two complexes arenonsuperimposable. There is a plane of symmetry in trans-Cr(acac) 2 (H 2 O) 2 , so it is notoptically active. A molecule with a plane of symmetry is never optically active because themirror images are always superimposable. A plane of symmetry is a plane through amolecule where one side exactly reflects the other side of the molecule.

CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 73942. There are five geometric isomers (labeled i-v). Only isomer v, where the CN − , Br - , and H 2 Oligands are cis to each other, is optically active. The nonsuperimposable mirror image isshown for isomer v.i CNii OH 2iiiBr OH 2Br CNPtPtBr OH 2Br CNCNOH 2H 2 OH 2 OBrPtBrCNCNivBrNCOH 2PtOH 2CNBrvCNBr CNPtBr OH 2OH 2opticallyactivemirrorNCNC BrPtH 2 O BrH 2 Omirror image of v(nonsuperimposable)Bonding, Color, and Magnetism in Coordination Compounds43. 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).44. Cu 2+ : [Ar]3d 9 ; Cu + : [Ar]3d 10 ; Cu(II) is d 9 and Cu(I) is d 10 . Color is a result of the electrontransfer between split d orbitals. This cannot occur for the filled d orbitals in Cu(I). Cd 2+ ,like Cu + , is also d 10 . We would not expect Cd(NH 3 ) 4 Cl 2 to be colored since the d orbitals arefilled in this Cd 2+ complex.45. Sc 3+ has no electrons in d orbitals. Ti 3+ and V 3+ have d electrons present. The color oftransition metal complexes results from electron transfer between split d orbitals. If no delectrons are present, no electron transfer can occur, and the compounds are not colored.46. All these complex ions contain Co 3+ bound to different ligands, so the difference in d-orbitalsplitting for each complex ion is due to the difference in ligands. The spectrochemical seriesindicates that CN − is a stronger field ligand than NH 3 , which is a stronger field ligand than F − .3−Therefore, Co(CN) 6 will have the largest d-orbital splitting and will absorb the lowestwavelengthelectromagnetic radiation (λ = 290 nm) because energy and wavelength areinversely related (λ = hc/E). Co(NH 3 ) 3+6 will absorb 440-nm electromagnetic radiation,whereas CoF 3− 6 will absorb the longest-wavelength electromagnetic radiation (λ = 770 nm)because F − is the weakest field ligand present.

740 CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY47. a. Fe 2+ : [Ar]3d 6↑↑↑↓↑↑↑↓↑↓↑↓High spin, small ΔLow spin, large Δb. Fe 3+ : [Ar]3d 5 c. Ni 2+ : [Ar]3d 8↑↑↑↑↑↑↑↑↓↑↓↑↓High spin, small Δd. Zn 2+ : [Ar]3d 10↑↓↑↓↑↓↑↓↑↓e. Co 2+ : [Ar]3d 7↑↑↑↑↓↑↓↑↑↓ ↑↓ ↑↓High spin, small ΔLow spin, large Δ48. NH 3 and H 2 O are neutral ligands, so the oxidation states of the metals are Co 3+ and Fe 2+ .Both have six d electrons ([Ar]3d 6 ). To explain the magnetic properties, we must have astrong field for Co(NH 3 ) 6 3+ and a weak field for Fe(H 2 O) 6 2+ .Co 3+ : [Ar]3d 6 Fe 2+ : [Ar]3d 6↑↓↑↓↑↓large Δ↑↓↑↑↑↑small ΔOnly this splitting of d orbitals gives a diamagnetic Co(NH 3 ) 6 3+ complex ion (no unpairedelectrons) and a paramagnetic Fe(H 2 O) 6 2+ complex ion (unpaired electrons present).

CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 74149. To determine the crystal field diagrams, you need to determine the oxidation state of thetransition metal, which can only be determined if you know the charges of the ligands (seeTable 19.13). The electron configurations and the crystal field diagrams follow.a. Ru 2+ : [Kr]4d 6 , no unpaired e - b. Ni 2+ : [Ar]3d 8 , 2 unpaired e -↑↑↑↓↑↓↑↓↑↓↑↓↑↓Low spin, large Δc. V 3+ : [Ar]3d 2 , 2 unpaired e -↑↑Note: Ni 2+ must have two unpaired electrons, whether high spin or low spin, and V 3+must have two unpaired electrons, whether high spin or low spin.50. All have octahedral Co 3+ ions, so the difference in d-orbital splitting and the wavelength oflight absorbed only depends on the ligands. From the spectrochemical series, the order of theligands from strongest to weakest field is CN − > en > H 2 O > I − . The strongest-field ligandproduces the greatest d-orbital splitting (largest Δ) and will absorb light having the smallestwave-length. The weakest-field ligand produces the smallest Δ and absorbs light having thelongest wavelength. The order is:Co(CN) 3− 6 < Co(en) 3+ 3 < Co(H 2 O) 3+ 63−< CoI 6shortest λlongest λabsorbedabsorbed51. Replacement of water ligands by ammonia ligands resulted in shorter wavelengths of lightbeing absorbed. Energy and wavelength are inversely related, so the presence of the NH 3ligands resulted in a larger d-orbital splitting (larger Δ). Therefore, NH 3 is a stronger-fieldligand than H 2 O.52. In both compounds, iron is in the +3 oxidation state with an electron configuration of [Ar]3d 5 .Fe 3+ complexes have one unpaired electron when a strong-field case and five unpairedelectrons when a weak-field case. Fe(CN) 2− 6 is a strong-field case, and Fe(SCN) 3− 6 is a weakfieldcase. Therefore, cyanide (CN − ) is a stronger-field ligand than thiocyanate (SCN − ).53. From Table 19.16 of the text, the violet complex ion absorbs yellow-green light (λ ≈ 570nm), the yellow complex ion absorbs blue light (λ ≈ 450 nm), and the green complex ionabsorbs red light (λ ≈ 650 nm). The spectrochemical series shows that NH 3 is a stronger-fieldligand than H 2 O, which is a stronger-field ligand than Cl − . Therefore, Cr(NH 3 ) 6 3+ will have

742 CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRYthe largest d-orbital splitting and will absorb the lowest wavelength electromagnetic radiation(λ ≈ 450 nm) because energy and wavelength are inversely related (λ = hc/E). Thus theyellow solution contains the Cr(NH 3 ) 3+6 complex ion. Similarly, we would expect theCr(H 2 O) 4 Cl + 2 complex ion to have the smallest d-orbital splitting because it contains theweakest-field ligands. The green solution with the longest wavelength of absorbed light+contains the Cr(H 2 O) 4 Cl 2 complex ion. This leaves the violet solution, which contains theCr(H 2 O) 3+ 6 complex ion. This makes sense because we would expect Cr(H 2 O) 3+ 6 to absorblight of a wavelength between that of Cr(NH 3 ) 3+ 6 and Cr(H 2 O) 4 Cl + 2 .54. Octahedral Cr 2+ complexes should be used. Cr 2+ : [Ar]3d 4 ; High spin (weak field) Cr 2+complexes have four unpaired electrons, and low spin (strong field) Cr 2+ complexes have twounpaired electrons. Ni 2+ : [Ar]3d 8 ; octahedral Ni 2+ complexes will always have two unpairedelectrons, whether high or low spin. Therefore, Ni 2+ complexes cannot be used to distinguishweak- from strong-field ligands by examining magnetic properties. Alternatively, the ligandfield strengths can be measured using visible spectra. Either Cr 2+ or Ni 2+ complexes can beused for this method.55. a. Ru(phen) 3 2+ exhibits optical isomerism [similar to Co(en) 3 3+ in Figure 19.16 of the text].b. Ru 2+ : [Kr]4d 6 ; because there are no unpaired electrons, Ru 2+ is a strong-field (low-spin)case.↑↓ ↑↓ ↑↓large Δ56. Co 2+ : [Ar]3d 7 ; the corresponding d-orbital splitting diagram for tetrahedral Co 2+ complexesis:↑ ↑ ↑↑↓↑↓All tetrahedral complexes are high spin since the d-orbital splitting is small. Ions with two orseven d electrons should give the most stable tetrahedral complexes because they have thegreatest number of electrons in the lower-energy orbitals as compared with the number ofelectrons in the higher-energy orbitals.57. The crystal field diagrams are different because the geometries of where the ligands point aredifferent. The tetrahedrally oriented ligands point differently in relationship to the d orbitalsthan do the octahedrally oriented ligands. Also, we have more ligands in an octahedralcomplex.See Figure 19.28 for the tetrahedral crystal field diagram. Notice that the orbitals are reverseof that in the octahedral crystal field diagram. The degenerate d and d are at a lower22zx − y2energy than the degenerate d xy , d xz , and d yz orbitals. Again, the reason for this is that

CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 743tetrahedral ligands are oriented differently than octahedral field ligands, so the interactionswith specifically oriented d orbitals are different. Also notice that the difference in magnitudeof the d-orbital splitting for the two geometries. The d-orbital splitting in tetrahedralcomplexes is less than one-half the d-orbital splitting in octahedral complexes. There are noknown ligands powerful enough to produce the strong-field case; hence all tetrahedralcomplexes are weak field or high spin.58. From Table 19.16, the red octahedral Co(H 2 O) 6 2+ complex ion absorbs blue-green light (λ ≈490 mm), whereas the blue tetrahedral CoCl 42−complex ion absorbs orange light (λ ≈ 600nm). Because tetrahedral complexes have a d-orbital splitting much less than octahedralcomplexes, one would expect the tetrahedral complex to have a smaller energy differencebetween split d orbitals. This translates into longer-wavelength light absorbed (E = hc/λ) fortetrahedral complex ions compared to octahedral complex ions. Information from Table19.16 confirms this.59. CoBr 6 4− has an octahedral structure, and CoBr 4 2− has a tetrahedral structure (as do most Co 2+complexes with four ligands). Coordination complexes absorb electromagnetic radiation(EMR) of energy equal to the energy difference between the split d orbitals. Because thetetrahedral d-orbital splitting is less than one-half the octahedral d-orbital splitting, tetrahedralcomplexes will absorb lower-energy EMR, which corresponds to longer-wavelengthEMR (E = hc/λ). Therefore, CoBr 6 4− will absorb EMR having a wavelength shorter than 3.4× 10 −6 m.60. Pd is in the +2 oxidation state in PdCl 4 2− ; Pd 2+ : [Kr]4d 8 . If PdCl 4 2− were a tetrahedralcomplex, then it would have two unpaired electrons and would be paramagnetic (see diagrambelow). Instead, PdCl 4 2− has a square planar molecular structure with a d-orbital splittingdiagram shown below. Note that all electrons are paired in the square planar diagram, whichexplains the diamagnetic properties of PdCl 4 2− .↑↓ ↑ ↑↑↓↑↓ ↑↓ ↑↓↑↓↑↓tetrahedral d 8 square planar d 8Additional Exercises61. Hg 2+ (aq) + 2 I − (aq) → HgI 2 (s), orange precipitateHgI 2 (s) + 2 I − (aq) → HgI 4 2− (aq), soluble complex ion

744 CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRYHg 2+ is a d 10 ion. Color is the result of electron transfer between split d orbitals. This cannotoccur for the filled d orbitals in Hg 2+ . Therefore, we would not expect Hg 2+ complex ions toform colored solutions.62. The transition metal ion must form octahedral complex ions; only with the octahedralgeometry are two different arrangements of d electrons possible in the split d orbitals. Thesetwo arrangements depend on whether a weak field or a strong field is present. For fourunpaired electrons in the first row, the two possible weak-field cases are for transition metalions having either a 3d 4 or a 3d 6 electron configuration:small Δsmall Δd 4 d 6Of these two, only d 6 ions have no unpaired electron in the strong-field case.large ΔTherefore, the transition metal ion has a 3d 6 valence electron configuration. Fe 2+ and Co 3+ aretwo possible metal ions that are 3d 6 . Thus one of these ions is probably present in the fourcoordination compounds, with each complex ion having a coordination number of 6 due tothe octahedral geometry.The colors of the compounds are related to the magnitude of Δ (the d-orbital splitting value).The weak-field compounds will have the smallest Δ, so the wavelength of light absorbed willbe longest. Using Table 19.16, the green solution (absorbs 650-nm light) and the bluesolution (absorbs 600-nm light) absorb the longest-wavelength light; these solutions containthe complex ions that are the weak-field cases with four unpaired electrons. The red solution(absorbs 490-nm light) and yellow solution (absorbs 450-nm light) contain the two strongfieldcase complex ions because they absorb the shortest-wavelength (highest-energy) light.These complex ions are diamagnetic.63. Ni(CO) 4 is composed of 4 CO molecules and Ni. Thus nickel has an oxidation state of zero.64. CN - and CO form much stronger complexes with Fe 2+ than O 2 . Thus O 2 cannot betransported by hemoglobin in the presence of CN − or CO because the binding sites prefer thetoxic CN − and CO ligands.65. i. 0.0203 g CrO 3 ×52.00 g Cr100.0 g CrO 3= 0.0106 g Cr; % Cr =0.0106 g× 100 = 10.1% Cr0.105 gii. 32.93 mL HCl ×.100 mmol HClmL1 mmol NH 17.03 mg NH× 3 ×= 56.1 mg NH 3mmol HCl mmol0 3

CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 745% NH 3 =56.1 mg× 100 = 16.5% NH 3341 mgiii. 73.53% + 16.5% + 10.1% = 100.1%; the compound must be composed of only Cr, NH 3 ,and I.Out of 100.00 g of compound:10.1 g Cr ×1 mol52.00 g= 0.194 mol;0.1940.194= 1.0016.5 g NH 3 ×1 mol17.03 g= 0.969 mol;0.9690.194= 4.9973.53 g I ×1 mol126.9 g= 0.5794 mol;0.57940.194= 2.99Cr(NH 3 ) 5 I 3 is the empirical formula. Cr 3+ forms octahedral complexes. So compound Ais made of the octahedral [Cr(NH 3 ) 5 I] 2+ complex ion and two I − ions as counter ions; theformula is [Cr(NH 3 ) 5 I]I 2 . Lets check this proposed formula using the freezing-point data.iv. ΔT f = iK f m; for [Cr(NH 3 ) 5 I]I 2 , i = 3.0 (assuming complete dissociation).Molality = m =0.601g complex 1 mol complex×= 0.116 mol/kg− 21.000 × 10 kg H O 517.9 g complex2ΔT f = 3.0 × 1.86 °C kg/mol × 0.116 mol/kg = 0.65°CBecause ΔT f is close to the measured value, this is consistent with the formula[Cr(NH 3 ) 5 I]I 2 .66. a. Copper is both oxidized and reduced in this reaction, so, yes, this reaction is an oxidationreductionreaction. The oxidation state of copper in [Cu(NH 3 ) 4 ]Cl 2 is +2, the oxidationstate of copper in Cu is zero, and the oxidation state of copper in [Cu(NH 3 ) 4 ]Cl is +1.b. Total mass of copper used:(8.0 cm × 16.0 cm × 0.060cm) 8.96 g10,000 boards × ×3boardcm= 6.9 × 10 5 g CuAmount of Cu to be recovered = 0.80(6.9 × 10 5 g) = 5.5 × 10 5 g Cu

746 CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY5.5 × 10 5 g Cu ×mol Cu 1 mol[Cu(NH3)4]Cl×63.55 g Cu mol Cu1 2×202.59 g [Cu(NH3)4]Clmol[Cu(NH ) ]Cl3 422= 1.8 × 10 6 g [Cu(NH 3 ) 4 ]Cl 267. M = metal ion5.5 × 10 5 1 mol Cu 4 mol NH317.03 g NH3g Cu × ××= 5.9 × 10 5 g NH 363.55 g Cu mol Cu mol NH3CH 2 OHCH 2 SHHSCHHSCHHSCH 2MHS CH 2MHO CH 2MHOCH 2CHSH68. a. The optical isomers of this compound are similar to the ones discussed in Figure 19.16 ofthe text. In the following structures we omitted the 4 NH 3 ligands coordinated to theoutside cobalt atoms.CoHOOHHO6+HHOOCoOH6+CoCoCoCoHOCoOHOHOHOHOHComirrorb. All are Co(III). The three “ligands” each contain 2 OH − and 4 NH 3 groups. If eachcobalt is in the +3 oxidation state, then each ligand has a +1 overall charge. The +3charge from the three ligands, along with the +3 charge of the central cobalt atom, givesthe overall complex a +6 charge. This is balanced by the -6 charge of the six Cl − ions.

CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 747c. Co 3+ : [Ar]3d 6 ; There are zero unpaired electrons if a low-spin (strong-field) case.large Δ↑↓↑↓↑↓69. No; in all three cases, six bonds are formed between Ni 2+ and nitrogen, so ΔH values shouldbe similar. ΔS° for formation of the complex ion is most negative for 6 NH 3 moleculesreacting with a metal ion (seven independent species become one). For penten reacting witha metal ion, two independent species become one, so ΔS° is least negative for this reactioncompared to the other reactions. Thus the chelate effect occurs because the more bonds achelating agent can form to the metal, the more favorable ΔS° is for the formation of thecomplex ion, and the larger is the formation constant.70. CrCl 3 •6H 2 O contains nine possible ligands, only six of which are used to form the octahedralcomplex ion. The three species not present in the complex ion will either be counter ions tobalance the charge of the complex ion and/or waters of hydration. The number of counterions for each compound can be determined from the silver chloride precipitate data, and thenumber of waters of hydration can be determined from the dehydration data. In allexperiments, the ligands in the complex ion do not react.Compound I:mol CrCl 3 •6H 2 O = 0.27 g ×1 mol266.5 g= 1.0 × 10 −3 mol CrCl 3 •6H 2 Omol waters of hydration = 0.036 g H 2 O ×1 mol18.02 g= 2.0 × 10 −3 mol H 2 Omol waters of hydrationmol compound2.0 × 10=−31.0 × 10−3molmol=2.0In compound I, two of the H 2 O molecules are waters of hydration, so the other four watermolecules are present in the complex ion. Therefore, the formula for compound I must be[Cr(H 2 O) 4 Cl 2 ]Cl•2H 2 O. Two of the Cl − ions are present as ligands in the octahedral complexion, and one Cl − ion is present as a counter ion. The AgCl precipitate data that refer to thiscompound are the one that produces 1430 mg AgCl:mol Cl − 0.100 mol [Cr(H O) 4Cl2]Cl2H2Ofrom compound I = 0.1000 L ×L−1 mol Cl×= 0.0100 mol Cl −mol[Cr(H O) Cl ]Cl 2H O2 •2 4 2 •2

748 CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRYmass AgCl produced = 0.0100 mol Cl − ×Compound II:1 mol AgCl−mol Cl×143.4 g AgClmol AgCl= 1.43 g= 1430 mg AgClmol waters of hydrationmol compound0.018 g=−1.0 × 101 molH2O×18.02 g3mol compound= 1.0The formula for compound II must be [Cr(H 2 O) 5 Cl]Cl 2 •H 2 O. The 2870-mg AgCl precipitatedata refer to this compound. For 0.0100 mol of compound II, 0.0200 mol Cl − is present ascounter ions:mass AgCl produced = 0.0200 mol Cl − ×1 mol AgCl 143.4 g× = 2.87 g−mol Cl mol= 2870 mg AgClCompound III:This compound has no mass loss on dehydration, so there are no waters of hydrationpresent. The formula for compound III must be [Cr(H 2 O) 6 ]Cl 3 . 0.0100 mol of thiscompound produces 4300 mg of AgCl(s) when treated with AgNO 3 .0.0300 mol Cl − 1 mol AgCl 143.4 g AgCl× ×= 4.30 g = 4.30 × 10 3 mg AgCl−mol Cl mol AgClThe structural formulas for the compounds are:Compound IH 2 OH 2 OClCrClOH 2+OH 2Cl• 2 H 2 OorH 2 OH 2 OClCrOH 2ClOH 2+Cl• 2 H 2 OCompound IICompound IIIH 2 OH 2 OClCrOH 2OH 22+OH 2Cl 2•H 2 OH 2 OH 2 OOH 2CrOH 2OH 23+OH 2Cl 3

CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 749From Table 19.16 of the text, the violet compound will be the one that absorbs light with theshortest wavelength (highest energy). This should be compound III. H 2 O is a stronger fieldligand than Cl − ; compound III with the most coordinated H 2 O molecules will have the largestd-orbital splitting and will absorb the higher-energy light.II III III II71. (H 2 O) 5 Cr‒Cl‒Co(NH 3 ) 5 → (H 2 O) 5 Cr‒Cl‒Co(NH 3 ) 5 → Cr(H 2 O) 5 Cl 2+ + Co(II) complexYes; this is consistent. After the oxidation, the ligands on Cr(III) won't exchange. BecauseCl − is in the coordination sphere, it must have formed a bond to Cr(II) before the electrontransfer occurred (as proposed through the formation of the intermediate).72. a. Be(tfa) 2 exhibits optical isomerism. Representations for the tetrahedral optical isomersare:COOCHCBeCHCCH 3CH 3OOCF 3C OCF 3 CH 3COOCHCBeCHCOCCF 3CH 3CF 3mirrorNote: The dotted line indicates a bond pointing into the plane of the paper, and thewedge indicates a bond pointing out of the plane of the paper. Also note that theplacement of the double bonds in the tfa isomer is not important. The two double bondscan resonate between adjacent carbon or oxygen atoms. Therefore, multiple resonancestructures can be drawn. Resonance structures are not different isomers.b. Square planar Cu(tfa) 2 molecules exhibit geometric isomerism. In one geometric isomer,the CF 3 groups are cis to each other, and in the other isomer, the CF 3 groups are trans.CF 3CF 3CF 3CH 3COOCCOOCHCCuCHHCCuCHCOOCCOOCCH 3CH 3CH 3CF 3cistrans

750 CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY73. a. Fe(H 2 O) 6 3+ + H 2 O ⇌ Fe(H 2 O) 5 (OH) 2+ + H 3 O +Initial 0.10 M 0 ~0Equil. 0.10 ! x x xK a =[Fe(H2O)5[Fe(H(OH )2O)2+3+6][H]3O+]= 6.0 × 10 −3 =2x0.10 − x≈2x0.10x = 2.4 ×210 − ; assumption is poor (x is 24% of 0.10). Using successive approximations:2x0.10 − 0.0243= 6.0 × 10 − , x = 0.0212x0.10 − 0.0213= 6.0 × 10 − , x = 0.022;2x0.10 − 0.0223= 6.0 × 10 − , x = 0.022x = [H + ] = 0.022 M; pH = 1.66b. Because of the lower charge, Fe 2+ (aq) will not be as strong an acid as Fe 3+ (aq). Asolution of iron(II) nitrate will be less acidic (have a higher pH) than a solution with thesame concentration of iron(III) nitrate.74. We need to calculate the Pb 2+ concentration in equilibrium with EDTA 4- . Because K is largefor the formation of PbEDTA 2- , let the reaction go to completion; then solve an equilibriumproblem to get the Pb 2+ concentration.Pb 2+ + EDTA 4− ⇌ PbEDTA 2− K = 1.1 × 10 18Before 0.010 M 0.050 M 00.010 mol/L Pb 2+ reacts completely (large K)Change !0.010 !0.010 → +0.010 Reacts completelyAfter 0 0.040 0.010 New initial conditionx mol/L PbEDTA 2− dissociates to reach equilibriumEquil. x 0.040 + x 0.010 - x1.1 × 10 18 =(0.010 − x )( x)(0.040+ x)≈(0.010)x(0.040), x = [Pb 2+ ] = 2.3 ×1910 − M; assumptions good.Now calculate the solubility quotient for Pb(OH) 2 to see if precipitation occurs. Theconcentration of OH − is 0.10 M because we have a solution buffered at pH = 13.00.Q = [Pb2+] [OH0− 2] 0= (2.3 ×1910 − )(0.10) 2 = 2.3 ×Pb(OH) 2 (s) will not form because Q is less than K sp .2110 − < K sp (1.2 ×1510 − )75. a. In the lungs, there is a lot of O 2 , and the equilibrium favors Hb(O 2 ) 4 . In the cells, there isa deficiency of O 2 , and the equilibrium favors HbH 4 4+ .

CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 751b. CO 2 is a weak acid in water; CO 2 + H 2 O ⇌ HCO − 3 + H + . Removing CO 2 essentiallydecreases H + . Hb(O 2 ) 4 is then favored, and O 2 is not released by hemoglobin in thecells. Breathing into a paper bag increases [CO 2 ] in the blood, thus increasing [H + ],which shifts the reaction left.c. CO 2 builds up in the blood, and it becomes too acidic, driving the equilibrium to the left.Hemoglobin can't bind O 2 as strongly in the lungs. Bicarbonate ion acts as a base inwater and neutralizes the excess acidity.76. HbO 2 → Hb + O 2 ΔG° = −(−70 kJ)Hb + CO → HbCOΔG° = −80 kJ____________________________________________________________________HbO 2 + CO → HbCO + O 2ΔG° = −10 kJΔG° = −RT ln K, K =⎛ − ΔG⎞⎜o 3⎡ − ( −10× 10 J) ⎤⎟= exp ⎢−1−1⎥⎝ RT ⎠ ⎣(8.3145 J K mol )(298 kJ) ⎦= 60Challenge Problems77.LLMLzaxis(pointing outof the planeof the paper)____ ____ d2, d xy____ dz 2x 2 − y____ ____ d xz,d yzThe dx 2 − y 2and d xy orbitals are in the plane of the three ligands and should be destabilizedthe most. The amount of destabilization should be about equal when all the possibleinteractions are considered. The d orbital has some electron density in the xy plane (thez 2doughnut) and should be destabilized a lesser amount than the dx 2 − y 2and d xy orbitals. Thed xz and d yz orbitals have no electron density in the plane and should be lowest in energy.78. Ni 2+ = d 8 ; if ligands A and B produced very similar crystal fields, the trans-[NiA 2 B 4 ] 2+complex ion would give the following octahedral crystal field diagram for a d 8 ion:This is paramagnetic.

752 CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY79.Because it is given that the complex ion is diamagnetic, the A and B ligands must producedifferent crystal fields, giving a unique d-orbital splitting diagram that would result in adiamagnetic species.LLLMLLzaxis(pointing outof the planeof the paper)____ dz 2____ ____ d2, d xyx 2 − y____ ____ d xz,d yzThe d orbital will be destabilized much more than in the trigonal planar case (see Exercisez 219.77). The d orbital has electron density on the z axis directed at the two axial ligands.z 2The dx 2 − y 2and d xy orbitals are in the plane of the three trigonal planar ligands and should bedestabilized a lesser amount than the d orbital; only a portion of the electron density in thez 2dx 2 −and dy 2 xy orbitals is directed at the ligands. The d xz and d yz orbitals will be destabilizedthe least since the electron density is directed between the ligands.80. Ni 2+ : [Ar]3d 8 ; the coordinate system for trans-[Ni(NH 3 ) 2 (CN) 4 ] 2− is shown below. BecauseCN − produces a much stronger crystal field, it will dominate the d-orbital splitting. From thecoordinate system, the CN − ligands are in a square planar arrangement. Therefore, thediagram will most resemble the square planar diagram given in Figure 19.29. Note that therelative position of d orbital is hard to predict; it could switch positions with the dz 2xy orbital.NCNCNH 3NiNH 3CNCNyxdx 2 − y 2dz 2d xyd xzd yzz81. a. Consider the following electrochemical cell:Co 3+ + e − → Co 2+E o c = 1.82 VCo(en) 2+ 3 → Co(en) 3+ 3 + e − − E o a = ?_________________________________________________________________________________________Co 3+ + Co(en) 3 2+ → Co 2+ + Co(en) 33+oE cell = 1.82 !oE a

CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 753The equilibrium constant for this overall reaction is:Co 3+ 3++ 3 en → Co(en) 3 K 1 = 2.0 × 10 47Co(en) 2+ 3 → Co 2+ + 3 en K 2 = 1/1.5 × 10 12__________________________________________________________________________Co 3+ + Co(en) 2+ 3 → Co(en) 3+ 3 + Co 2+ 2.0 × 10K = K 1 K 2 =121.5 × 10From the Nernst equation for the overall reaction:47= 1.3 × 10 35o 0.05910.059135E cell = log K = log(1.3 × 10 ),n1Eocell= 2.08 VoE cell = 1.82 !oao aE = 2.08 V, − E = 2.08 V − 1.82 V = 0.26 V, sooE c = −0.26 Vb. The stronger oxidizing agent will be the more easily reduced species and will have themore positive standard reduction potential. From the reduction potentials, Co 3+ (E° =1.82 V) is a much stronger oxidizing agent than Co(en) 3 3+ (E° = !0.26 V).c. In aqueous solution, Co 3+ forms the hydrated transition metal complex Co(H 2 O) 6 3+ . Inboth complexes, Co(H 2 O) 6 3+ and Co(en) 3 3+ , cobalt exists as Co 3+ , which has six d electrons.Assuming a strong-field case for each complex ion, the d-orbital splitting diagramfor each is:e g↑↓ ↑↓ ↑↓t 2gWhen each complex gains an electron, the electron enters a higher-energy e g orbital.Because en is a stronger-field ligand than H 2 O, the d-orbital splitting is larger forCo(en) 3 3+ , and it takes more energy to add an electron to Co(en) 3 3+ than to Co(H 2 O) 6 3+ .Therefore, it is more favorable for Co(H 2 O) 6 3+ to gain an electron than for Co(en) 3 3+ togain an electron.82. a. Cr 3+ : [Ar]3d 3 ; we will assume the A ligands lie on the z axis.dz 2d x 2 y 2 −smallsplittingd xz d yzd xy

754 CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRYb.4p4seg*3dt 2gc. The major difference in the two diagrams involves the d xy , d xz , and d yz orbitals. These aredegenerate in the MO diagram (only σ bonding assumed) but are not degenerate in thecrystal field diagram. If the d orbitals were involved in π bonding, this might reconcilethe differences between the two diagrams. One expects involvement of d xy to be differentfrom d xz and d yz because of the symmetry of the complex. This would remove thedegeneracy of these orbitals in the MO picture.83. a. AgBr(s) ⇌ Ag + + Br − K sp = [Ag + ][Br - ] = 5.0 × 10 −13Initial s = solubility (mol/L) 0 0Equil. s sK sp = 5.0 × 10 −13 = s 2 , s = 7.1 × 10 −7 mol/Lb. AgBr(s) ⇌ Ag + + Br - K sp = 5.0 × 10 −13Ag + ++ 2 NH 3 ⇌ Ag(NH 3 ) 2 K f = 1.7 × 10 7____________________________________________________________________________________________________AgBr(s) + 2 NH 3 (aq) ⇌ Ag(NH 3 ) 2 + (aq) + Br - (aq)K = K sp × K f = 8.5 × 10 −6AgBr(s) + 2 NH 3 ⇌ Ag(NH 3 ) 2++ Br -Initial 3.0 M 0 0s mol/L of AgBr(s) dissolves to reach equilibrium = molar solubilityEquil. 3.0 − 2s s s

CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 755[Ag(NH3)2 ][Br ]K =2[NH ]Assumption good.3+−s2=2( 3.0 − 2s), 8.5 × 10 −6 s≈2(3.0)2, s = 8.7 × 10 −3 mol/Lc. The presence of NH 3 increases the solubility of AgBr. Added NH 3 removes Ag + fromsolution by forming the complex ion Ag(NH 3 ) 2 + . As Ag + is removed, more AgBr(s) willdissolve to replenish the Ag + concentration.d. Mass AgBr = 0.2500 L ×−38.7 × 10 mol AgBrL×187.8 g AgBrmol AgBr= 0.41 g AgBre. Added HNO 3 will have no effect on the AgBr(s) solubility in pure water. Neither H + norNO 3 − reacts with Ag + or Br - ions. Br - is the conjugate base of the strong acid HBr, so it isa terrible base. Added H + will not react with Br - to any great extent. However, addedHNO 3 will reduce the solubility of AgBr(s) in the ammonia solution. NH 3 is a weak base(K b = 1.8 × 10 −5 ). Added H + will react with NH 3 to form NH 4 + . As NH 3 is removed, asmaller amount of the Ag(NH 3 ) 2 + complex ion will form, resulting in a smaller amount ofAgBr(s) that will dissolve.84. For a linear complex ion with ligands on the x axis, the dx 2 − y 2will be destabilized the most,with the lobes pointing directly at the ligands. The d yz orbital has the fewest interactions withx-axis ligands, so it is destabilized the least. The d xy and d xz orbitals will have similar destabilizationbut will have more interactions with x-axis ligands than the d yz orbital. Finally, thed 2 orbital with the doughnut of electron density in the xy plane will probably be destabilizedzmore than the d xy and d xz orbitals but will have nowhere near the amount of destabilizationthat occurs with the dx 2 − y 2orbital. The only difference that would occur in the diagram ifthe ligands were on the y axis is the relative positions of the d xy , d xz , and d yz orbitals. The d xzwill have the smallest destabilization of all these orbitals, whereas the d xy and d yz orbitals willbe degenerate since we expect both to be destabilized equivalently from y-axis ligands. Thed-orbital splitting diagrams are:a. b.dx 2 − y 2dz 2d xy,d xzd yzlinear x-axis ligandslinear y-axis ligandsdx 2 − y 2dz 2d xy,d yzd xz