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COORDINATION CHEMISTRY

COORDINATION CHEMISTRY. COORDINATION COMPLEX – Any ion or neutral species with a metal atom or ion bonded to 2 or more molecules or ions. Zn(NH 3 ) 4 2+. Al(OH) 4 -. Charged coordination complexes are called COMPLEX IONS.

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COORDINATION CHEMISTRY

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  1. COORDINATION CHEMISTRY COORDINATION COMPLEX – Any ion or neutral species with a metal atom or ion bonded to 2 or more molecules or ions Zn(NH3)42+ Al(OH)4- Charged coordination complexes are called COMPLEX IONS LIGAND – An ion or molecule that bonds to a central metal atom to form a coordination complex NH3 OH- 4A-1 (of 20)

  2. PROPERTIES OF LIGANDS MONODENTATE LIGAND – A ligand with one lone pair that can form one bond to a metal ion H2O, NH3, CN-, NO2-, SCN-, OH-, X- CHELATE – A ligand with more than one atom with a lone pair that can be used to bond to a metal ion Bidentate ligands: oxalate (ox) – C2O42- ethylenediamine (en) – H2N(CH2)2NH2 Polydentate ligands: diethylenetriamine (dien) – H2N(CH2CH2)NH(CH2CH2)NH2 ethylenediaminetetraacetate (EDTA) – (O2CCH2)2N(CH2CH2)N(CH2CO2)24- 4A-2 (of 20)

  3. COORDINATION NUMBER – The number of bonds formed between the metal ion and the ligands Coordination Number Shape of Complex Ag(NH3)2+ 2 linear Zn(OH)42- 4 square planar or tetrahedral SnCl62- 6 octahedral Fe(C2O4)33- 6 octahedral 4A-3 (of 20)

  4. COORDINATION COMPOUND – Any compound containing a complex ion and a counterion [Cu(NH3)4]Cl2 This is a 2+ charged complex ion, requiring 2 Cl-counterions to produce a neutral compound 4A-4 (of 20)

  5. NOMENCLATURE FOR COMPLEX IONS 1 – Name the ligands before the metal ion 2 – In naming ligands, molecules use their molecular names (with 4 common exceptions), and anions have their name end in -o H2O – aqua NH3 – ammine CO – carbonyl NO – nitrosyl ligands ending in –ide, drop the –ide Cl-– chloroF-– fluoroOH-– hydroxoCN-– cyano ligands ending in –ate or –ite, drop the –e SO42-– sulfatoNO3-– nitratoNO2-– nitritoC2O42-– oxalato 3 – Different ligands are named alphabetically 4A-5 (of 20)

  6. NOMENCLATURE FOR COMPLEX IONS 4 – Prefixes are used if a complex ion has more than one particular ligand 5 – Prefixes for polydentate ligands (or ligands whose names contain prefixes) are bis-, tris-, etc., with the ligand in parenthesis 6 – The charge of the metal is given as a roman numeral in parenthesis 7 – If the complex ion has a negative charge, the suffix –ate is added to the name of the metal 4A-6 (of 20)

  7. Co(NH3)63+ hexaamminecobalt(III) CoCl63- hexachlorocobaltate(III) Co(NH3)5Cl2+ pentaamminechlorocobalt(III) Fe(CN)63- not hexacyanoironate(III) , its hexacyanoferrate(III) Fe – ferrateCu – cupratePb– plumbate Sn – stannatePt – platinateMn - manganate Fe(C2O4)33- tris(oxalato)ferrate(III) Ni(CO)4 tetracarbonylnickel(0) 4A-7 (of 20)

  8. NOMENCLATURE FOR COMPOUNDS CONTAINING COMPLEX IONS (COORDINATION COMPOUNDS) When naming coordination compounds, name the cation, then name the anion [Cu(NH3)4]Cl2 This is a 2+ charged complex tetraamminecopper(II) chloride 4A-8 (of 20)

  9. [Cr(NH3)6]Cl3 This is a 3+ charged complex ion hexaamminechromium(III) chloride [Pt(NH3)3Cl3]Cl This is a 1+ charged complex ion triamminetrichloroplatinum(IV) chloride Mn(en)2Cl2 This is a neutral complex dichlorobis(ethylenediamine)manganese(II) 4A-9 (of 20)

  10. K[PtNH3Cl5] This is a 1- charged complex ion potassium amminepentachloroplatinate(IV) K4Fe(CN)6 This is a 4- charged complex ion potassium hexacyanoferrate(II) [Fe(en)2(NO2)2]2SO4 This is a 1+ complex ion bis(ethylenediamine)dinitritoiron(III) sulfate 4A-10 (of 20)

  11. ISOMERISM ISOMERS – Compounds with the same chemical formula, but with different properties (1)STRUCTURAL ISOMERS – Compounds with the same chemical formula, but with the atoms bonded in different orders Structural isomers have different names 4A-11 (of 20)

  12. C4H10 H H C H H H H C C C H H H H H H H H H C C C C H H H H H butane methyl propane 4A-12 (of 20)

  13. Pt(H2O)4(OH)2Cl2 [Pt(H2O)4(OH)2]Cl2 tetraaquadihydroxoplatinum(IV) chloride 2+ H2O Pt H2O Cl- Cl- OH H2O H2O OH [Pt(H2O)4Cl2](OH)2 tetraaquadichloroplatinum(IV) hydroxide H2O Pt H2O H2O Cl 2+ H2O Cl OH- OH- 4A-13 (of 20)

  14. (2) SPATIAL ISOMERS – Compounds with the same chemical formula and with the atoms bonded in the same order (so not structural isomers), but with the atoms bonded in different spatialorientations (a) GEOMETRICAL ISOMERS – Spatial isomers that ARE NOT mirror images of each other These are not the same, and they are not mirror images – they are geometrical isomers 4A-14 (of 20)

  15. Find the number of geometrical isomers for CoCl2(NH3)4 trans – the 2 common ligands are across from each other cis – the 2 common ligands are adjacent to each other Cl- T C Cl Co Cl trans-tetraamminedichlorocobalt(II) NH3 NH3 NH3 NH3 Cl Co NH3 NH3 NH3 NH3 Cl cis-tetraamminedichlorocobalt(II) 4A-15 (of 20)

  16. Find the number of geometrical isomers for PtCl2(NH3)4 if it has square planar geometry Cl- NH3 T T T C C T C C trans-diamminedichloroplatinum(II) Pt Pt Cl NH3 NH3 NH3 Cl NH3 Cl Cl cis-diamminedichloroplatinum(II) 4A-16 (of 20)

  17. NH3 Pt Cl NH3 Cl Find the number of geometrical isomers for PtCl2(NH3)4 if it has tetrahedral geometry Cl- NH3 C C only one Cl’s all always adjacent 4A-17 (of 20)

  18. Find the number of geometrical isomers for CoCl3(NH3)3 if it has octahedral geometry fac – all 3 common ligands are adjacent mer – 2 of the 3 common ligands are opposite Cl Co Cl Cl Co NH3 NH3 Cl- F F T C C T C C NH3 NH3 NH3 Cl fac-triamminetrichlorocobalt(III) NH3 NH3 Cl Cl mer-triamminetrichlorocobalt(III) 4A-18 (of 20)

  19. (b) OPTICAL ISOMERS – Spatial isomers that ARE mirror images of each other, and they are nonsuperimposable (not the same) These mirror images are the same – they are not geometrical isomers or optical isomers These mirror images are not the same (nonsuperimposable) – they are optical isomers, or ENANTIOMERS This is an anteater This is its enantiomer 4A-19 (of 20)

  20. (b) OPTICAL ISOMERS – Spatial isomers that ARE mirror images of each other, and they are nonsuperimposable (not the same) Fe(C2O4)3- O Fe O 180º O Fe O O Fe O O O O O O O O O O O O O These are not the same (nonsuperimposable) • the compound tris(oxalato)ferrate(III) has 2 optical isomers Ʌ-tris(oxalato)ferrate(III) Δ-tris(oxalato)ferrate(III) 4A-20 (of 20)

  21. FORMATION CONSTANT (Kf) – The equilibrium constant for the complete formation of a complex ion diamminesilver(I) Ag+ (aq) + 2NH3 (aq) → Ag(NH3)2+ (aq) Kf= [Ag(NH3)2+] _______________ [Ag+][NH3]2 4B-1 (of 19)

  22. Write the reaction for the complete formation of hexaamminecobalt(II), and its Kf expression Co2+ (aq) + 6NH3 (aq) ⇆Co(NH3)62+ (aq) Kf= [Co(NH3)62+] ________________ [Co2+][NH3]6 4B-2 (of 19)

  23. Write the reaction for the complete formation of hexaamminecobalt(II), and its Kf expression Co2+ (aq) + 6NH3 (aq) ⇆Co(NH3)62+ (aq) Kf= [Co(NH3)62+] ________________ [Co2+][NH3]6 If a solution is 0.250 M Co2+ and 0.100 M Co(NH3)62+ at equilibrium, and the formation constant is 1.00 x 105, calculate [NH3]. [NH3]6= [Co(NH3)62+] ________________ [Co2+]Kf = 0.100 M _________________________ (0.250 M)(1.00 x105) [NH3]= [Co(NH3)62+] ________________ [Co2+]Kf 6 6 = 0.126 M 4B-3 (of 19)

  24. The formation constant for diamminesilver(I) is 1.00 x 106. Calculate [Ag+] in a solution that was originally 0.100 M Ag+ and 0.500 M NH3. Ag+(aq) + 2NH3(aq)⇆ Ag(NH3)2+(aq) Initial M’s Change in M’s Equilibrium M’s 0.100 0.500 0 - x - 2x + x 0.100 - x 0.500 - 2x x The reaction is going in the forward direction and has a large equilibrium constant, xwill be a large number 4B-4 (of 19)

  25. The formation constant for diamminesilver(I) is 1.00 x 106. Calculate [Ag+] in a solution that was originally 0.100 M Ag+ and 0.500 M NH3. Ag+(aq) + 2NH3(aq)⇆ Ag(NH3)2+(aq) Initial M’s Shift M’s New Initial M’s Change M’s Equilibrium M’s 0.100 0.500 0 - 0.100 - 0.200 + 0.100 0 0.300 0.100 + x + 2x - x x 0.300 + 2x 0.100 - x Kf=[Ag(NH3)2+] _______________ [Ag+][NH3]2 1.00 x 106= (0.100 – x) ___________________(x)(0.300 + 2x)2 1.00 x 106= (0.100) _____________ (x)(0.300)2 x = 1.11 x 10-6 = [Ag+] 4B-5 (of 19)

  26. The solubility product constant for zinc hydroxide is 4.5 x 10-17, and the formation constant for tetrahydroxozincate(II) is 5.0 x 1014. (a) Calculate molar solubility of zinc hydroxide in pure water. Zn(OH)2(s) ⇆Zn2+(aq) + 2OH-(aq) Initial M’s Change in M’s Equilibrium M’s 0 0 + x + 2x x 2x • Ksp= [Zn2+][OH-]2 • = (x)(2x)2 • = 4x3 x = molar solubility of Zn(OH)2 • 4.5 x 10-17 = 4x3 • 2.2 x 10-6 M = x • = molar solubility of Zn(OH)2 4B-6 (of 19)

  27. The solubility product constant for zinc hydroxide is 4.5 x 10-17, and the formation constant for tetrahydroxozincate(II) is 5.0 x 1014. (b) Calculate molar solubility of zinc hydroxide in 0.10 M NaOH. Zn(OH)2(s) ⇆Zn2+(aq) + 2OH-(aq) Initial M’s Change in M’s Equilibrium M’s 0 0.10 No, because Zn2+ forms a complex ion with OH- Zn(OH)2(s) ⇆ Zn2+(aq) + 2OH-(aq) Ksp= 4.5 x 10-17 Zn2+ (aq) + 4OH- (aq) ⇆ Zn(OH)42-(aq) Kf = 5.0 x 1014 Zn(OH)2(s) + 2OH- (aq) ⇆ Zn(OH)42-(aq) K = 2.25 x 10-2 4B-7 (of 19)

  28. The solubility product constant for zinc hydroxide is 4.5 x 10-17, and the formation constant for tetrahydroxozincate(II) is 5.0 x 1014. (b) Calculate molar solubility of zinc hydroxide in 0.10 M NaOH. Zn(OH)2(s) + 2OH- (aq) ⇆Zn(OH)42-(aq) Initial M’s Change in M’s Equilibrium M’s 0.10 0 - 2x + x 0.10 - 2x x • K= [Zn(OH)42-] • ______________ • [OH-]2 2.25 x 10-2 = x ______________ (0.10 – 2x)2 2.3 x 10-4 = x = molar solubility of Zn(OH)2 4B-8 (of 19)

  29. PROPERTIES OF COORDINATION COMPLEXES (with transition metals) (1) COLOR Co(NO2)63- Co(CN)63- Co(H2O)63+ Co(CO3)33- Color depends upon the chemical groups attached to the transition metal 4B-9 (of 19)

  30. (2) MAGNETISM Some are DIAMAGNETIC (no unpaired electrons), and some are PARAMAGNETIC (1 or more unpaired electrons) 4B-10 (of 19)

  31. BONDING IN COORDINATION COMPLEXES Theories attempt to explain (a) geometries (shapes) (b) color (electronic energy level differences) (c) magnetism (paired or unpaired electrons) CRYSTAL FIELD THEORY – Assumes ionic bonding between the ligands and the metal The ligand’s lone pairs affect the energies of the metal’s d orbitals 4B-12 (of 19)

  32. 2px 2py 2pz 2nd EL 2s 1st EL 1s E 4B-13 (of 19)

  33. 3dxy 3dxz 3dyz 3dx2-y2 3dz2 3rd EL 3px 3py 3pz 3s E 4B-14 (of 19)

  34. Coordination Number of 6 : Octahedral When 6 ligands surround a metal atom, they arrange octahedrally to minimize repulsion (VSEPR theory) 4B-15 (of 19)

  35. E 4B-16 (of 19)

  36. dx2-y2 dz2 (Δo) dxy dxz dyz 3d E OCTAHEDRAL SPLITTING ENERGY (Δo) – The energy difference between the d orbitals in an octahedral ligand field 4B-17 (of 19)

  37. With a transition metal that has 6 d electrons: dx2-y2 dz2 ↑↓ ↑↓ ↑↓ dxydxzdyz E LOW-SPIN COMPLEX – A complex with a large splitting energy, resulting in electrons remaining in the lower energy d orbitals, and producing a low number of unpaired electrons Because of the large splitting energy, the d electrons are all paired in the 3 stable d orbitals, causing the complex to be diamagnetic 4B-18 (of 19)

  38. With a transition metal that has 6 d electrons: dx2-y2 dz2 ↑ ↑ dx2-y2 dz2 ↑↓ ↑ ↑ dxydxzdyz E HIGH-SPIN COMPLEX – A complex with a small splitting energy, resulting in electrons distributing into all of the d orbitals, and producing a highnumber of unpaired electrons Because of the small splitting energy, the d electrons remain unpaired as long as possible, causing the complex to be paramagnetic 4B-19 (of 19)

  39. SPLITTING ENERGY dx2-y2 dz2 (Δo) ↑↓ ↑↓ ↑↓ dxy dxz dyz E The splitting energy depends upon: (1) The charge of the metal The greater the charge of the metal ion, the larger the splitting energy (2) The ligands attached to the metal The ligands can be either strong-field ligands or weak field ligands 4C-1 (of 24)

  40. STRONG-FIELD LIGANDS – Ligands that produce a strong electrostatic field for the d orbitals, causing the splitting energy to be large CN-, CO, and NO2- are strong-field ligands :C O: 2pz .. sp sp 2py πantibonding MO π bonding MO dxy AO BACK BONDING – A coordinate covalent pi bond formed between a d orbital of a metal and an empty antibonding orbital of a ligand 4C-2 (of 24)

  41. dx2-y2 dz2 dxy dxz dyz E The increased stability of the dxy, dxz, and dyz increases the splitting energy 4C-3 (of 24)

  42. WEAK-FIELD LIGANDS – Ligands that produce a weak electrostatic field for the d orbitals, causing the splitting energy to be small I-, Br-, Cl-, and F- are weak-field ligands - :Cl: .. .. p AO dxy AO Both the d orbital of the metal and the p orbital of the ligand contain electrons, and repel 4C-4 (of 24)

  43. dx2-y2 dz2 dxy dxz dyz E The decreased stability of the dxy, dxz, and dyz reduces the splitting energy 4C-5 (of 24)

  44. Hexafluorocobaltate(III) is found to be a paramagnetic complex (a) Give the electron configuration of the cobalt ion Co atom: [Ar]4s23d7 Co3+ ion: [Ar]3d6 (b) Identify the ligands as strong-field or weak-field weak-field (c) Draw the splitting pattern for the cobalt ↑ ↑ dx2-y2 dz2 ↑↓ ↑ ↑ dxy dxz dyz E (d) Identify the complex as high-spin or low-spin high-spin 4C-6 (of 24)

  45. Hexacarbonyliron(II) is found to be a diamagnetic complex (a) Give the electron configuration of the iron ion Fe atom: [Ar]4s23d6 Fe2+ ion: [Ar]3d6 (b) Identify the ligands as strong-field or weak-field strong-field (c) Draw the splitting pattern for the iron dx2-y2 dz2 ↑↓ ↑↓ ↑↓ dxy dxz dyz E (d) Identify the complex as high-spin or low-spin low-spin 4C-7 (of 24)

  46. CoF63- Fe(CO)62+ dx2-y2 dz2 ↑ ↑ dx2-y2 dz2 ↑↓ ↑ ↑ ↑↓ ↑↓ ↑↓ dxydxzdyz dxy dxz dyz E Complexes will absorb EM radiation to promote electrons from the low-energy d orbitals to the high-energy d orbitals E = hν c = λν c = ν __ λ E = hc ____ λ If photons of visible light are absorbed, the complex will be colored 4C-8 (of 24)

  47. CoF63- Fe(CO)62+ ↑ ↑ ↑↓ ↑ ↑ ↑↓ ↑↓ ↑↓ E CoF63- absorbs EM radiation with a wavelength of 6.5 x 10-7 m, while Fe(CO)62+ absorbs EM radiation with a wavelength of 4.5 x 10-7 m. 4C-9 (of 24)

  48. CoF63- Fe(CO)62+ ↑ ↑ ↑↓ ↑ ↑ ↑↓ ↑↓ ↑↓ E Calculate the splitting energy (Δo)of each E = hc ____ λ = (6.626 x 10-34Js)(2.9979 x 108 ms-1) _____________________________________________ (6.5 x 10-7 m) = 3.1 x 10-19 J E = hc ____ λ = (6.626 x 10-34Js)(2.9979 x 108 ms-1) _____________________________________________ (4.5 x 10-7 m) = 4.4 x 10-19 J 4C-10 (of 24)

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