1 / 59

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. Ag(NH 3 ) 2 +. Cu(NH 3 ) 4 2+. Zn(NH 3 ) 4 2+. Al(OH) 4 -. Zn(OH) 4 2-. Charged coordination complexes are called COMPLEX IONS.

susan
Télécharger la présentation

COORDINATION CHEMISTRY

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. 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 4A-2 (of 20)

  3. (2) MAGNETISM Some are DIAMAGNETIC (no unpaired electrons), and some are PARAMAGNETIC (1 or more unpaired electrons) 4A-3 (of 20)

  4. (3) GEOMETRY COORDINATION NUMBER – The number of bonds formed between the metal ion and the ligands Coordination Number Geometry 2 linear 4 square planar or tetrahedral 6 octahedral 4A-4 (of 20)

  5. (a) 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-5 (of 20)

  6. (b) 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 Cl-– chloroF-– fluoroOH-– hydroxoCN-– cyano SO42-– sulfatoNO3-– nitratoNO2-– nitritoC2O42-– oxalato 3 – Different ligands are named alphabetically 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-, tetrakis-, 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 hexancyanoironate(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. (c) NOMENCLATURE FOR COMPOUNDS CONTAINING COMPLEX IONS 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 When naming coordination compounds, name the cation, then name the anion 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. (4) 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+ Cl- Cl- H2O Pt H2O H2O OH H2O OH [Pt(H2O)4Cl2](OH)2 tetraaquadichloroplatinum(IV) hydroxide H2O Pt H2O 2+ H2O Cl 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, but with the atoms bonded in different spatial orientations (a) GEOMETRICAL ISOMERS – Spatial isomers that ARE NOT mirror images of each other 4A-14 (of 20)

  15. CoCl2(NH3)4 tetraamminedichlorocobalt(II) opposite – trans Cl Co Cl trans-tetraamminedichlorocobalt(II) NH3 NH3 NH3 NH3 Cl Co NH3 adjacent – cis NH3 NH3 NH3 Cl cis-tetraamminedichlorocobalt(II) 4A-15 (of 20)

  16. How many geometrical isomers are there for diamminedichloroplatinum(II) if it has square planar geometry? Pt Pt NH3 Cl NH3 NH3 NH3 Cl Cl Cl trans-diamminedichloroplatinum(II) cis-diamminedichloroplatinum(II) 4A-16 (of 20)

  17. NH3 Pt Cl NH3 Cl How many geometrical isomers are there for diamminedichloroplatinum(II) if it has tetrahedral geometry? only one Cl’sall always adjacent 4A-17 (of 20)

  18. How many geometrical isomers are there for triamminetrichlorocobalt(III) if it has octahedral geometry? each pair adjacent – fac Cl Co Cl Cl Co NH3 NH3 NH3 NH3 Cl fac-triamminetrichlorocobalt(III) NH3 NH3 Cl Cl adjacent and opposite – mer mer-triamminetrichlorocobalt(III) 4A-18 (of 20)

  19. (b) OPTICAL ISOMERS – Spatial isomers that ARE mirror images of each other, and they are nonsuperimposable tris(oxalato)ferrate(III) 180º O Fe O O Fe O O Fe O O O O O O O O O O O O O These are nonsuperimposable molecules  the compound tris(oxalato)ferrate(III) has 2 optical isomers 4A-19 (of 20)

  20. Optical isomers are called ENANTIOMERS anteater enantiomer 4A-20 (of 20)

  21. To form a complex ion: Ag+ (aq) + NH3 (aq) → Ag(NH3)+ (aq) Ag(NH3)+ (aq) + NH3 (aq) → Ag(NH3)2+ (aq) Ag+ (aq) + 2NH3 (aq) → Ag(NH3)2+ (aq) This is the reaction for the complete formation of the complex ion FORMATION CONSTANT (Kf) – The equilibrium constant for the complete formation of a complex ion Kf= [Ag(NH3)2+] _______________ [Ag+][NH3]2 4B-1 (of 15)

  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 15)

  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 15)

  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 15)

  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 15)

  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 15)

  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 15)

  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) 0.10 0 Initial M’s Change in M’s Equilibrium M’s - 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 15)

  29. BONDING IN COORDINATION COMPLEXES Theories attempt to explain (a) geometries (shapes) (b) magnetism (paired or unpaired electrons) (c) color (electronic energy level differences) 4B-9 (of 15)

  30. 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-10 (of 15)

  31. Coordination Number of 6 : Octahedral When 6 ligands surround a metal atom, they arrange octahedrally to minimize repulsion (VSEPR theory) 4B-11 (of 15)

  32. E 4B-12 (of 15)

  33. dx2-y2 dz2 dxy dxz dyz 3d E SPLITTING ENERGY (Δo) – The energy difference between the d orbitals in a ligand field 4B-13 (of 15)

  34. With a transition metal that has 6 d electrons: dx2-y2 dz2 ↑↓ ↑↓ ↑↓ dxy dxz dyz 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-14 (of 15)

  35. With a transition metal that has 6 d electrons: dx2-y2 dz2 ↑ ↑ dx2-y2 dz2 ↑↓ ↑ ↑ dxy dxz dyz 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-15 (of 15)

  36. 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 21)

  37. 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 .. 2pz 2pz sp sp sp sp sp sp 2py 2py 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 21)

  38. dx2-y2 dz2 dxy dxz dyz E The increased stability of the dxy, dxz, and dyz increases the splitting energy 4C-3 (of 21)

  39. 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: .. .. Both the d orbital of the metal and the p orbital of the ligand contain electrons, and repel 4C-4 (of 21)

  40. dx2-y2 dz2 dxy dxz dyz E The decreased stability of the dxy, dxz, and dyz reduces the splitting energy 4C-5 (of 21)

  41. 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 21)

  42. 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 21)

  43. 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 21)

  44. 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 21)

  45. CoF63- Fe(CO)62+ ↑ ↑ ↑↓ ↑ ↑ ↑↓ ↑↓ ↑↓ E Calculate the splitting energy (Δo)of each E = hc ____ λ = (6.626 x 10-34 Js)(2.9979 x 108 ms-1) _____________________________________________ (6.5 x 10-7 m) = 3.1 x 10-19 J E = hc ____ λ = (6.626 x 10-34 Js)(2.9979 x 108 ms-1) _____________________________________________ (4.5 x 10-7 m) = 4.4 x 10-19 J 4C-10 (of 21)

  46. CoF63- Fe(CO)62+ ↑ ↑ ↑ ↑↓ ↑ ↑ ↑↓ ↑↓ ↑↓ E Because of the 3 stable d orbitals, this arrangement favors metal ions with d3 or d6 electron configurations d3 - Cr3+, Mn4+ d6– Co3+, Fe2+ Metal ions with a d5 electron configuration are very stable as a high-spin octahedral complex d5 - Fe3+, Mn2+ 4C-11 (of 21)

  47. CRYSTAL FIELD THEORY FOR OTHER GEOMETRIES (a) Coordination Number of 2 : Linear Ligands pointing along the z-axis make the dz2 the most unstable d orbitals on the xy plane will be the most stable 4C-12 (of 21)

  48. dz2 dxz dyz dxy dx2-y2 3d E With only 2 ligands, energies do not increase as much as with 6 ligands  with 5 stable orbitals, this arrangement favors metal ions with d10electron configurations Ag+ - Ag(NH3)2+ 4C-13 (of 21)

More Related