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(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 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) Color depends upon the chemical groups attached to the transition metal (1) COLOR Co(H 2 O) 6 3+ Co(CN) 6 3- Co(CO 3 ) 3 3- Co(NO 2 ) 6 3- 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. COORDINATION NUMBER – The number of bonds formed between the metal ion and the ligands Coordination NumberGeometry 2 4 linear square planar or tetrahedral 6octahedral 4A-4 (of 20) (3)GEOMETRY

5. MONODENTATE LIGAND – A ligand with one lone pair that can form one bond to a metal ion H 2 O, NH 3, CN -, NO 2 -, SCN -, OH -, X - Bidentate ligands: oxalate (ox) – C 2 O 4 2- ethylenediamine (en) – H 2 N(CH 2 ) 2 NH 2 Polydentate ligands: diethylenetriamine (dien) – H 2 N(CH 2 CH 2 )NH(CH 2 CH 2 )NH 2 ethylenediaminetetraacetate (EDTA) – (O 2 CCH 2 ) 2 N(CH 2 CH 2 )N(CH 2 CO 2 ) 2 4- CHELATE – A ligand with more than one atom with a lone pair that can be used to bond to a metal ion 4A-5 (of 20) (a)PROPERTIES OF LIGANDS

6. 4A-6 (of 20) (b)NOMENCLATURE FOR COMPLEX IONS 1 – Name the ligands before the metal ion 4 – Prefixes are used if a complex ion has more than one particular ligand 2 – In naming ligands, molecules use their molecular names (with 4 common exceptions), and anions have their name end in -o H 2 O – aquaNH 3 – ammineCO – carbonylNO – nitrosyl 3 – Different ligands are named alphabetically 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 Cl - – chloroF - – fluoro OH - – hydroxo CN - – cyano 7 – If the complex ion has a negative charge, the suffix –ate is added to the name of the metal SO 4 2- – sulfato NO 3 - – nitrato NO 2 - – nitrito C 2 O 4 2- – oxalato

7. Co(NH 3 ) 6 3+ hexaamminecobalt(III) Co(NH 3 ) 5 Cl 2+ pentaamminechlorocobalt(III) 4A-7 (of 20) CoCl 6 3- hexachlorocobaltate(III) Fe(CN) 6 3- not hexancyanoironate(III) Fe – ferrateCu – cupratePb – plumbate Sn – stannate Pt – platinate Mn - manganate Fe(C 2 O 4 ) 3 3- tris(oxalato)ferrate(III) Ni(CO) 4 tetracarbonylnickel(0), its hexacyanoferrate(III)

8. COORDINATION COMPOUND – Any compound containing a complex ion and a counterion [Cu(NH 3 ) 4 ]Cl 2 This is a 2+ charged complex ion, requiring 2 Cl - counterions to produce a neutral compound 4A-8 (of 20) (c)NOMENCLATURE FOR COMPOUNDS CONTAINING COMPLEX IONS When naming coordination compounds, name the cation, then name the anion tetraamminecopper(II) chloride

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

10. K[PtNH 3 Cl 5 ] This is a 1- charged complex ion potassium amminepentachloroplatinate(IV) 4A-10 (of 20) [Fe(en) 2 (NO 2 ) 2 ] 2 SO 4 This is a 1+ complex ion bis(ethylenediamine)dinitritoiron(III) sulfate K 4 Fe(CN) 6 This is a 4- charged complex ion potassium hexacyanoferrate(II)

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 4A-11 (of 20) Structural isomers have different names

12. HHHHHCCCCHHHHHHHHHHCCCCHHHHH C 4 H 10 HHCHHHHCCCHHHHHHCHHHHCCCHHHHH 4A-12 (of 20) butane methyl propane

13. [Pt(H 2 O) 4 (OH) 2 ]Cl 2 Cl - H 2 O Pt H 2 O H2OH2O H2OH2O OH 2+ OH - H 2 O Pt H 2 O H2OH2O H2OH2O Cl 2+ tetraaquadihydroxoplatinum(IV) chloride [Pt(H 2 O) 4 Cl 2 ](OH) 2 tetraaquadichloroplatinum(IV) hydroxide 4A-13 (of 20) Pt(H 2 O) 4 (OH) 2 Cl 2

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. CoCl 2 (NH 3 ) 4 tetraamminedichlorocobalt(II) opposite – trans adjacent – cis Cl Co Cl NH 3 Cl Co NH 3 Cl trans-tetraamminedichlorocobalt(II) cis-tetraamminedichlorocobalt(II) 4A-15 (of 20)

16. How many geometrical isomers are there for diamminedichloroplatinum(II) if it has square planar geometry? trans-diamminedichloroplatinum(II) cis-diamminedichloroplatinum(II) Pt Cl NH 3 Cl Pt NH 3 Cl NH 3 Cl 4A-16 (of 20)

17. How many geometrical isomers are there for diamminedichloroplatinum(II) if it has tetrahedral geometry? NH 3 Pt Cl NH 3 Cl’s all always adjacent only one 4A-17 (of 20)

18. How many geometrical isomers are there for triamminetrichlorocobalt(III) if it has octahedral geometry? Cl Co Cl NH 3 Cl Co NH 3 Cl each pair adjacent – fac adjacent and opposite – mer fac-triamminetrichlorocobalt(III) mer-triamminetrichlorocobalt(III) 4A-18 (of 20)

19. O Fe O O O O O (b)OPTICAL ISOMERS – Spatial isomers that ARE mirror images of each other, and they are nonsuperimposable tris(oxalato)ferrate(III) O Fe O O O O O Fe O O O O O 180º 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)

22. Ag + (aq) + 2NH 3 (aq) → Ag(NH 3 ) 2 + (aq) FORMATION CONSTANT (K f ) – The equilibrium constant for the complete formation of a complex ion K f = [Ag(NH 3 ) 2 + ] _______________ [Ag + ][NH 3 ] 2 4B-1 (of 15) diamminesilver(I)

23. Write the reaction for the complete formation of hexaamminecobalt(II), and its K f expression Co 2+ (aq) + 6NH 3 (aq) ⇆ Co(NH 3 ) 6 2+ (aq) K f = [Co(NH 3 ) 6 2+ ] ________________ [Co 2+ ][NH 3 ] 6 4B-2 (of 15)

24. = 0.100 M __________________________ (0.250 M)(1.00 x10 5 ) 6 If a solution is 0.250 M Co 2+ and 0.100 M Co(NH 3 ) 6 2+ at equilibrium, and the formation constant is 1.00 x 10 5, calculate [NH 3 ]. Co 2+ (aq) + 6NH 3 (aq) ⇆ Co(NH 3 ) 6 2+ (aq) K f = [Co(NH 3 ) 6 2+ ] ________________ [Co 2+ ][NH 3 ] 6 [NH 3 ] 6 = [Co(NH 3 ) 6 2+ ] ________________ [Co 2+ ]K f [NH 3 ]= [Co(NH 3 ) 6 2+ ] ________________ [Co 2+ ]K f 6 = 0.126 M 4B-3 (of 15) Write the reaction for the complete formation of hexaamminecobalt(II), and its K f expression

25. The formation constant for diamminesilver(I) is 1.00 x 10 6. Calculate [Ag + ] in a solution that was originally 0.100 M Ag + and 0.500 M NH 3. Ag + (aq) + 2NH 3 (aq) ⇆ Ag(NH 3 ) 2 + (aq) Initial M’s Change in M’s Equilibrium M’s 0.1000.5000 - x- 2x+ x 0.100 - xx0.500 - 2x 4B-4 (of 15) The reaction is going in the forward direction and has a large equilibrium constant,  x will be a large number

26. The formation constant for diamminesilver(I) is 1.00 x 10 6. Calculate [Ag + ] in a solution that was originally 0.100 M Ag + and 0.500 M NH 3. Ag + (aq) + 2NH 3 (aq) ⇆ Ag(NH 3 ) 2 + (aq) Initial M’s Shift M’s New Initial M’s Change M’s Equilibrium M’s 0.1000.5000 - 0.100- 0.200+ 0.100 00.1000.300 + x+ 2x- x x0.100 - x0.300 + 2x K f = [Ag(NH 3 ) 2 + ] ________________ [Ag + ][NH 3 ] 2 1.00 x 10 6 = (0.100 – x) ___________________ (x)(0.300 + 2x) 2 1.00 x 10 6 = (0.100) ____________ (x)(0.300) 2 x = 1.11 x 10 -6 = [Ag + ] 4B-5 (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 10 14. (a)Calculate molar solubility of zinc hydroxide in pure water. x2x Zn(OH) 2 (s) ⇆ Zn 2+ (aq) + 2OH - (aq) Initial M’s Change in M’s Equilibrium M’s 0 0 + x+ 2x K sp = [Zn 2+ ][OH - ] 2 = (x)(2x) 2 = 4x 3 x = molar solubility of Zn(OH) 2 4.5 x 10 -17 = 4x 3 = molar solubility of Zn(OH) 2 2.2 x 10 -6 M = x 4B-6 (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 10 14. (b)Calculate molar solubility of zinc hydroxide in 0.10 M NaOH. Zn(OH) 2 (s) ⇆ Zn 2+ (aq) + 2OH - (aq) Zn 2+ (aq) + 4OH - (aq) ⇆ Zn(OH) 4 2- (aq) Zn(OH) 2 (s) + 2OH - (aq) ⇆ Zn(OH) 4 2- (aq) K sp = 4.5 x 10 -17 K f = 5.0 x 10 14 K = 2.25 x 10 -2 4B-7 (of 15) Zn(OH) 2 (s) ⇆ Zn 2+ (aq) + 2OH - (aq) Initial M’s Change in M’s Equilibrium M’s 00.10 No, because Zn 2+ forms a complex ion with OH -

29. The solubility product constant for zinc hydroxide is 4.5 x 10 -17, and the formation constant for tetrahydroxozincate(II) is 5.0 x 10 14. (b)Calculate molar solubility of zinc hydroxide in 0.10 M NaOH. Zn(OH) 2 (s) + 2OH - (aq) ⇆ Zn(OH) 4 2- (aq) Initial M’s Change in M’s Equilibrium M’s 0.10 - 2x x 0.100 - 2x + x 2.25 x 10 -2 = x ______________ (0.10 – 2x) 2 2.3 x 10 -4 = x K = [Zn(OH) 4 2- ] ______________ [OH - ] 2 = molar solubility of Zn(OH) 2 4B-8 (of 15)

30. 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)

31. 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)

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

33. E 4B-12 (of 15)

34. E 3d SPLITTING ENERGY (Δ o ) – The energy difference between the d orbitals in a ligand field d xy d xz d yz d x 2 -y 2 d z 2 4B-13 (of 15)

35. 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 d xy d xz d yz d x 2 -y 2 d z 2 4B-14 (of 15) With a transition metal that has 6 d electrons:

36. E ↑↓↑↑ HIGH-SPIN COMPLEX – A complex with a small splitting energy, resulting in electrons distributing into all of the d orbitals, and producing a high number of unpaired electrons Because of the small splitting energy, the d electrons remain unpaired as long as possible, causing the complex to be paramagnetic ↑↑ d xy d xz d yz d x 2 -y 2 d z 2 4B-15 (of 15) With a transition metal that has 6 d electrons: d x 2 -y 2 d z 2

38. 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 d xy d xz d yz d x 2 -y 2 d z 2 4C-1 (of 21) SPLITTING ENERGY (Δo)(Δo) (2) The ligands attached to the metal The ligands can be either strong-field ligands or weak field ligands

39. STRONG-FIELD LIGANDS – Ligands that produce a strong electrostatic field for the d orbitals, causing the splitting energy to be large CN -, CO, and NO 2 - are strong-field ligands :C O: BACK BONDING – A coordinate covalent pi bond formed between a d orbital of a metal and an empty antibonding orbital of a ligand sp 2p z sp 2p y 2p z sp 2p z 2p y 4C-2 (of 21) π bonding MO π antibonding MO.. d xy AO

40. E The increased stability of the d xy, d xz, and d yz increases the splitting energy d xy d xz d yz d x 2 -y 2 d z 2 4C-3 (of 21)

41. 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)..

42. E The decreased stability of the d xy, d xz, and d yz reduces the splitting energy d xy d xz d yz d x 2 -y 2 d z 2 4C-5 (of 21)

43. Hexafluorocobaltate(III) is found to be a paramagnetic complex (a)Give the electron configuration of the cobalt ion Co atom: [Ar]4s 2 3d 7 Co 3+ ion: [Ar]3d 6 (c)Draw the splitting pattern for the cobalt E ↑↓↑↑ ↑↑ (b)Identify the ligands as strong-field or weak-field (d)Identify the complex as high-spin or low-spin weak-field high-spin 4C-6 (of 21) d xy d xz d yz d x 2 -y 2 d z 2

44. Hexacarbonyliron(II) is found to be a diamagnetic complex (a)Give the electron configuration of the iron ion Fe atom: [Ar]4s 2 3d 6 Fe 2+ ion: [Ar]3d 6 (c)Draw the splitting pattern for the iron E ↑↓ (b)Identify the ligands as strong-field or weak-field (d)Identify the complex as high-spin or low-spin strong-field low-spin ↑↓ 4C-7 (of 21) d xy d xz d yz d x 2 -y 2 d z 2

45. CoF 6 3- Fe(CO) 6 2+ ↑↓↑↑ ↑↑ E Complexes will absorb EM radiation to promote electrons from the low- energy d orbitals to the high-energy d orbitals c = λνc = ν __ λ E = hνE = hc ____ λ If photons of visible light are absorbed, the complex will be colored 4C-8 (of 21) d xy d xz d yz d x 2 -y 2 d z 2 d xy d xz d yz d x 2 -y 2 d z 2

46. CoF 6 3- Fe(CO) 6 2+ ↑↓↑↑ ↑↑ E CoF 6 3- absorbs EM radiation with a wavelength of 6.5 x 10 -7 m, while Fe(CO) 6 2+ absorbs EM radiation with a wavelength of 4.5 x 10 -7 m. 4C-9 (of 21)

47. CoF 6 3- Fe(CO) 6 2+ ↑↓↑↑ ↑↑ E Calculate the splitting energy (Δ o ) of each E = hc ____ λ = (6.626 x 10 -34 Js)(2.9979 x 10 8 ms -1 ) _____________________________________________ (6.5 x 10 -7 m) = 3.1 x 10 -19 J = (6.626 x 10 -34 Js)(2.9979 x 10 8 ms -1 ) _____________________________________________ (4.5 x 10 -7 m) = 4.4 x 10 -19 JE = hc ____ λ 4C-10 (of 21)

48. CoF 6 3- Fe(CO) 6 2+ ↑↓↑↑ ↑↑ E Because of the 3 stable d orbitals, this arrangement favors metal ions with d 3 or d 6 electron configurations d 3 - Cr 3+, Mn 4+ d 6 – Co 3+, Fe 2+ Metal ions with a d 5 electron configuration are very stable as a high-spin octahedral complex d 5 - Fe 3+, Mn 2+ 4C-11 (of 21) ↑

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

50. E 3d With only 2 ligands, energies do not increase as much as with 6 ligands  with 5 stable orbitals, this arrangement favors metal ions with d 10 electron configurations Ag + - Ag(NH 3 ) 2 + d xy d x 2 -y 2 dz2dz2 d xz d yz 4C-13 (of 21)

51. (b)Coordination Number of 4 : Square Planar Ligands on the xy plane make the d x 2 -y 2 the most unstable The d xy will be the next most unstable The d z 2 will be the next most unstable because of the doughnut 4C-14 (of 21)

52. E 3d With 1 very unstable orbital (  4 stable orbitals), this arrangement favors metal ions with d 8 electron configurations Pt 2+ - PtCl 4 2- Au 3+ - AuCl 4 - d xz d yz d xy dz2dz2 d x 2 -y 2 4C-15 (of 21)

53. (c)Coordination Number of 4 : Tetrahedral The d xz, d xz, and d yz point closest to the ligands The d x 2 - y 2 and d z 2 will be the most stable 4C-16 (of 21)

54. E 3d This arrangement favors metal ions with a d 7 electron configurations Co 2+ - CoCl 4 2- This arrangement also favors metal ions with a d 4 electron configuration, but it is not a stable arrangement – it is a strong reducing agent, producing the more stable d 3 electron configuration and an octahedral complex Cr 2+ → Cr 3+ + e - d z 2 d x 2 - y 2 d xy d xz d yz 4C-17 (of 21)

55. : Br N Br : : Br : Ligands bond to metal atoms with lone pairs COORDINATE COVALENT BOND – A covalent bond in which the 2 shared electrons come from the same atom : F : : F B F : Coordinate covalent bonding is an example of a LEWIS ACID-BASE REACTION 4C-18 (of 21) LIGAND FIELD THEORY – Assumes coordinate covalent bonding between the ligands and the metal using molecular orbital theory

56. LEWIS ACID – An electron pair acceptor LEWIS BASE – An electron pair donor BF 3 NBr 3 4C-19 (of 21) A substance with an incomplete outershell or empty valence orbitals A substance with a lone pair available for bonding Na +, Mg 2+, Al 3+ Cl -, S 2-, NO 2 -, CO 3 2-

57. Fe(CN) 6 4- hexacyanoferrate(II)Fe 2+ : [Ar]3d 6 4C-20 (of 21)

58. FeCl 6 4- hexachloroferrate(II)Fe 2+ : [Ar]3d 6 4C-21 (of 21)