2. Review on Chemistry of Coordination Compounds

3. Coordination Compounds Constitution [Co(NH 3 ) 5 Cl](NO 3 ) 2 Central atomCoordination number Inner sphere Coordination atomOuter sphere ions ligand

4. H 2 O, NH 3, Cl –, CN –, CO, SCN –, OH – CO 3 2-, NH 2 CH 2 CH 2 NH 2 (ethylenediamine, en), C 2 O 4 2– (oxalate ion) EDTA 4  (ethylenediaminetetraacetate ion) is a hexadentate ligand.

5. Polydentate ligands are also known as chelating agents

6. Naming Coordination Compounds The name of a complex is one word, with no space between the ligand names and no space between the names of the last ligand and the metal.

7. Naming Coordination Compounds a salt, name the cation first Name the ligands, in alphabetical order, before the metal.  Note: in anionic ligand endings from -ide to -o, and -ate to -ato. -ite to -ito  the names of the ligands differ slightly from their chemical names  in the chemical formula, the metal atom or ion is written before the ligands

8. Naming Coordination Compounds If the complex contains more than one ligand of a particular type  Use Greek prefixes (di-, tri-, tetra-, etc.), or bis- (2), tris-(3), tetrakis-(4), and so forth, and put the ligand name in parentheses for the later.  The ligands are listed in alphabetical order, and the prefixes are ignored in determining the order.

9. Naming Coordination Compounds A Roman numeral in parentheses follows the name of the metal to indicate the metal's oxidation state To name the metal  use the ending -ate if the metal is in an anionic complex, or the Latin names for some  -ium ending for Cationic coordination sphere, or same as the element

10. Isomers 异构体 Compounds with the same formula but a different arrangement of atoms are called isomers. Constitution isomers Stereoisomers Linkage isomers Ionization Isomer Diastereoisomers Enantiomers Isomers

11. Co NH 3 H3NH3N H3NH3N O O N Co NH 3 H3NH3N H3NH3N N O O nitro nitrito

12.  Coordination Isomerism [Co(NH 3 ) 6 ][Cr(CN) 6 ] and [Cr(NH 3 ) 6 ][Co(CN) 6 ] [Pt(NH 3 ) 4 ][PtCl 6 ] and [Pt(NH 3 ) 4 Cl 2 ][PtCl 4 ] [Pt(NH 3 ) 4 ][PtCl 4 ] and [Pt(NH 3 ) 4 Cl 2 ][PtCl 4 ]

13. Aquo Isomer  CrCl 3 ·6H 2 O [Cr(H 2 O) 5 Cl]Cl 2 ·H 2 O [Cr(H 2 O) 4 Cl 2 ]Cl·2H 2 O [Cr(H 2 O) 6 ]Cl 3 Green Violet

14. [Co(NH 3 ) 5 Br]SO 4 contains Co–Br bond, and the sulfate ion is free [Co(NH 3 ) 5 SO 4 ]Br contains Co–sulfate bond, and the bromide ion is free Ionization isomers 电离异构体

15. two kinds of stereoisomers: ddiastereoisomers 非对映异构体 eenantiomers 对映异构体 Stereoisomers 立体异构体 cistrans isomer

16. M X X X L L L M X X X L L L Facial Fac- Meridional Meri-

17. M O N N O O O

18. Optical Isomerism 光学异构现象  enantiomers chiral 手性的 achiral 非手性的  The [Co(en) 3 ] 3+ cation is chiral and exists in two nonidentical mirror-image forms.  properties identical except for their reactions with other chiral substances their effect on plane-polarized light: Optical Activity

19. Co N N N N N N N N N N N N

20.  d: dextrorotatory 右旋  l: levorotatory 左旋 used to indicate the direction of rotation. racemic 外消旋 A 50:50 mixture of both isomers produces no net optical rotation. The labels

21. Valence Bond Theory Vacant metal hybrid atomic orbital Coordinate covalent bond Occupied ligand hybrid atomic orbital

22. The hybrid orbitals used by the metal are determined by the geometry of the complex. The number of d electrons in the metal is determined by the oxidation state of the metal ion. The orbitals used to construct the hybrid orbitals for bonding must be vacant on the metal. Valence Bond Theory

23. For octahedral complexes it may be necessary to pair some electrons already in d orbitals to get vacant orbitals required for hybridization. This leads to a low spin complex. Contrast this with the use of higher energy vacant d orbitals. This leads to more unpaired d electrons and a high spin complex. Valence Bond Theory

24. Knowing whether a complex is paramagnetic or diamagnetic can help determine which d orbitals to use. It can also help determine whether a complex is square planar or tetrahedral Valence Bond Theory Spectrochemical series 光化学序列 Increasing  → Weak field ligands Strong field ligands I – < Br – < Cl – < F – < H 2 O < NH 3 < en < CN –

25. Hybrid Orbitals A unsuccessful example:  Cu(NH 3 ) 4 2+ A square planar complex Hybrid form: dsp 2 Cu 2+ [Ar] dsp 2

26. This explains the color and magnetic properties of the transition metal complexes. Bonding in complexes is viewed as entirely ionic and as arising from electrostatic interactions between the d electrons of the metal and the ligand electrons. Crystal Field Theory

27. It considers the effect of the ligand charges on the energies of the metal ion d orbitals. The d orbitals are raised in energy and are separated in energy based on the geometry of the complex. The energy separation is called the crystal field splitting, represented by the symbol Δ.

28. In octahedral complexes the dx2 - y2 and the dz2 orbitals are higher tin energy than the dxy, the dxz, and the dyz because the negative charge of the electrons from the ligands point directly at the negative charges of the electrons in the d orbitals that lie on the x,y, and z axes. Crystal Field Theory

29. The color of the complexes is due to electronic transitions from one set of d orbitals to another. Visible light can supply enough energy to promote an electron from the lower enegy to the higher energy orbitals. Light at a particular wavelength is absorbed and the complementary color is seen. Crystal Field Theory

30. The angular distribution of d orbitals

31. Crystal Field Splitting  o = 10 Dq Spheric field Free atom +6 Dq  4 Dq oo egeg t 2g d xy, d yz, d xz d x2-y2, d z2 Octahedral field

32. CFSE: Crystal Field Stabilization Energy 晶体场稳定化能 Pairing Energy 成对能 P

34. Tetrahedral Field

35. Splitting in Tetrahedral Field  t = 4/9  o = 10 Dq Spheric field Free atom tt e t2t2 d xy, d yz, d xz d x2-y2, d z2 Tetrahedral field

37. Color Complementary colors: R-G O-B Y-V

38. An absorbance spectrum  It plots the absorbance (amount of light absorbed by a substance) as a function of wavelength [Ti(H 2 O) 6 ] 3+

39. Spectrochemical series 光化学序列 Increasing  → Weak field ligands Strong field ligands I – < Br – < Cl – < F – < H 2 O < NH 3 < en < CN –

40. Back-bonding 反馈键

41. Magnetism  Paramagnetic: unpaired electron  Diamagnetic: no unpaired electron [Co(NH 3 ) 6 ] 3+ : Co 3+ : d6d6 diamagnetic t 2g 4 e g 2 paramagnetic t 2g 6 e g 0 [CoF 6 ] 3  OhOh t 2g e g

42. Jahn-Teller Effects  For a non-linear molecule that is in an electronically degenerate state, distortion must occur to lower the symmetry, remove the degeneracy, and lower the energy.  Jahn-Teller effects do not predict which distortion will occur other than that the center of symmetry will remain. The distortion by the unsymmetrical distribution of electrons in e g orbital is stronger than that of t 2g.

43. Stability Constant Cu 2+ + 4NH 3  [Cu(NH 3 ) 4 ] 2+ Overall Stability Constant Stepwise stability constants: K 1, K 2, …, K 4 Overall stability constants:  1,  2,  3,  4 K stability = [Cu(NH 3 ) 4 2+ ] [Cu 2+ ] [NH 3 ] 4  1 = K 1  2 = K 1 K 2  3 = K 1 K 2 K 3  4 = K 1 K 2 K 3 K 4

44. Factors That Determine The Stability Of Coordination Compounds 1. metal ions: charge, radius and electronic configuration; 2. ligand: basicity, chelate effects. Chelate Effects 螯合效应 : entropy effect. Entropy-driven reaction (process) [Cu(H 2 O) 4 ] 2+ (aq) + 2 NH 3 (aq) → [Cu(NH 3 ) 2 (H 2 O) 2 ] 2+ (aq) lg  2 = 7.65 [Cu(H 2 O) 4 ] 2+ (aq) + en (aq) → [Cu(en)(H 2 O) 2 ] 2+ (aq) lg  1 = 10.64

45. Example 1 Solution: K sp = [Ag + ][Cl  ] = 1.6 × 10  Step 2: Step 1: Ag + + 2NH 3 = Ag(NH 3 ) 2 + AgCl(s) = Ag + + Cl  2 =2 = [Ag(NH 3 ) 2 + ] [Ag + ][NH 3 ] 2 = 1.5 × 10  Calculate the molar solubility of AgCl in a 6 M NH 3 solution

46. Example 1 (continue) AgCl(s) + 2NH 3 = Ag(NH 3 ) 2 + + Cl  Overall reaction K = [Ag(NH 3 ) 2 + ] [Cl  ] [NH 3 ] 2 = (1.6 × 10  ) = K sp  2 (1.5 × 10  ) = 2.4 × 10  AgCl(s) + 2NH 3 = Ag(NH 3 ) 2 + + Cl  Initial (M): 6.0 0.0 0.0 Change (M): -2s +s +s Equi (M): 6 - 2s s s K = (s)(s) (6 – 2s) s = 0.016 M Step 3 0.1 0.045

47. Example 2 Calculate the equilibrium concentration of every relevant species of a 0.1 M Ag(NH 3 ) 2 + solution. K 1 = 2.2 × 10  = K 2 = 5.1 × 10  = [Ag(NH 3 ) + ] [Ag + ][NH 3 ] [Ag(NH 3 ) 2 + ] [Ag(NH 3 ) + ][NH 3 ] 2 =2 = [Ag(NH 3 ) 2 + ] [Ag + ][NH 3 ] 2 = 1.5 × 10  With only a few exceptions, there is generally a slowly descending progression in the values of the K i ’s in any particular system.