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CHEM1612 - Pharmacy Week 8: Complexes I Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: 9351 4196

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Presentation on theme: "CHEM1612 - Pharmacy Week 8: Complexes I Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: 9351 4196"— Presentation transcript:

1 CHEM Pharmacy Week 8: Complexes I Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone:

2 Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, John Wiley & Sons Australia, Ltd ISBN:

3 Lecture 22-3 Complexes Blackman, Bottle, Schmid, Mocerino & Wille Chapters 13,10.4, 11.8 Complex ions Coordination compounds Geometry of complexes Chelates K stab Solubility and complexes Nomenclature Isomerism in complexes Biologically important metal-complexes Co(EDTA) -

4 Lecture 22-4 Whenever a metal ion enters water, a complex ion forms with water as the ligand. Metal ions act as Lewis acid (accepts electron pair). Water is the Lewis base (donates electron pair). Metal Ions as Lewis Acids M 2+ H 2 O(l) [M(H 2 O) 4 ] 2+ adduct +

5 Lecture 22-5 Complex Ions Definition: A central metal ion covalently bound to two or more anions or molecules, called ligands. Neutral ligands, e.g., water, CO, NH 3 Ionic ligands, e.g., OH -, Cl -, CN - [Ni(H 2 O) 6 ] 2+, a typical complex ion: Ni 2+ is the central metal ion Six H 2 O molecules are the ligands O are the donor atoms overall 2+ charge.

6 Lecture 22-6 They consist of: Complex ion (metal ion with attached ligands) Counter ions (additional anions/cations needed for zero net charge) e.g. [Co(NH 3 ) 6 ]Cl 3 (s) [Co(NH 3 ) 6 ] 3+ (aq) + 3 Cl - (aq) Coordination Compounds Complex ionCounter ions e.g. [Co(H 2 O) 6 ][CoCl 4 ] 3 (s) [Co(H 2 O) 6 ] 3+ (aq) + 3 [CoCl 4 ] - (aq) In water coordination compounds dissociate into the complex ion (cation in this example) and the counterions (3 Cl - ions here). Note: the counter ion may also be a complex ion.

7 Lecture 22-7 A small and multiply-charged metal ion acts as an acid in water, i.e. the hydrated metal ion transfers an H + ion to water. 6 bound H 2 O molecules 5 bound H 2 O molecules 1 bound OH - (overall charge reduced by 1) Acidity of Aqueous Transition Metal Ions Acidic solution Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.

8 Lecture 22-8 Free IonHydrated IonKaKa Fe 3+ Fe(H 2 O) 6 3+ (aq) 6 x Cr 3+ Cr(H 2 O) 6 3+ (aq)1 x Al 3+ Al(H 2 O) 6 3+ (aq)1 x Be 2+ Be(H 2 O) 4 2+ (aq)4 x Cu 2+ Cu(H 2 O) 6 2+ (aq)3 x Fe 2+ Fe(H 2 O) 6 2+ (aq) 4 x Pb 2+ Pb(H 2 O) 6 2+ (aq)3 x Zn 2+ Zn(H 2 O) 6 2+ (aq)1 x Co 2+ Co(H 2 O) 6 2+ (aq)2 x Ni 2+ Ni(H 2 O) 6 2+ (aq) 1 x ACID STRENGTH Metal Ion Hydrolysis Each hydrated metal ion that transfers a proton to water has a characteristic K a value.

9 Lecture 22-9 M + Coord no.M 2+ Coord no.M 3+ Coord no. M + Coord no.M 2+ Coord no.M 3+ Coord no. Cu + 2,4Mn 2+ 4,6Sc 3+ 6 Ag + 2Fe 2+ 6Cr 3+ 6 Au + 2,4Co 2+ 4,6Co 3+ 6 Ni 2+ 4,6Au 3+ 4 Cu 2+ 4,6 Zn 2+ 4,6 The number of ligand atoms attached to the metal ion is called the coordination number.  varies from 2 to 8 and depends on the size, charge, and electron configuration of the metal ion. Typical coordination numbers for some metal ions are: Coordination Number

10 Lecture Coordination number Coordination geometry 2linear 4square planar 4tetrahedral 6octahedral Examples [Ag(NH 3 ) 2 ] + [AuCl 2 ] - [Pd(NH 3 ) 4 ] 2+ [PtCl 4 ] 2- [Zn(NH 3 ) 4 ] 2+ [CuCl 4 ] 2- [Co(NH 3 ) 6 ] 3+ [FeCl 6 ] 3- Coordination Number and Geometry

11 Lecture Ligands must have a lone pair to donate to the metal. The covalent bond formed is sometimes referred to as a “dative” bond. Ligands that can form 1 bond with the metal ion are called monodentate (denta – tooth) e.g. H 2 O, NH 3, Cl - (a single donor atom). Some ligands have more than one atom with lone pairs that can be bonded to the metal ion – these are called CHELATES (greek: claw). n Bidentate ligands can form 2 bonds e.g. Ethylenediamine n Polydentate ligands – can form more than 2 bonds For a list of ligands see the recommended textbook. Ligands

12 Lecture Ethylenediamine (en) has two N atoms that can form a bond with the metal ion, giving a five-membered ring. Bidentate Chelate Ligands M X+ (en) Blackman, Bottle, Schmid, Mocerino & Wille, Figure 13.10

13 Lecture Demo: Nickel Complexes Ni 2+ forms three complexes with ethylenediamine: 1. Mix [Ni(H 2 O) 6 ] 2+ and en in ratio 1:1 → [Ni(en)(H 2 O) 4 ] 2+ green blue-green 2. Mix [Ni(H 2 O) 6 ] 2+ and en in ratio 1:2 → [Ni(en) 2 (H 2 O) 2 ] 2+ light blue 3. Mix [Ni(H 2 O) 6 ] 2+ and en in ratio 1:3 → [Ni(en) 3 ] 2+ purple

14 Lecture Ethylenediaminetetraacetate tetraanion (EDTA 4- ) EDTA forms very stable complexes with many metal ions. EDTA is used for treating heavy-metal poisoning, because it removes lead and other heavy metal ions from the blood and other bodily fluids. Hexadentate ligand: EDTA Co(III) [Co(EDTA)] - N=blue O=red

15 Lecture M(H 2 O) 4 2+ M(H 2 O) 3 (NH 3 ) 2+ M(NH 3 ) 4 2+ NH 3 The stepwise exchange of NH 3 for H 2 O in M(H 2 O) NH 3 3 more steps Lewis bases: water and ammonia Ammonia is a stronger Lewis base than water Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.

16 Lecture Equilibrium Constant K stab The complex formation equilibrium is characterised by a stability constant, K stab (also called formation constant): Ag + (aq) + 2 NH 3 Ag(NH 3 ) 2 + (aq) Metal Ion + nLigand Complex The larger K stab, the more stable the complex, e.g.

17 Lecture Metal ions gain ligands one at a time. Each step characterised by a specific stability constant. Overall formation constant: K stab = K 1 x K 2 …x K n Example: Ag + (aq) + NH 3(aq) [Ag(NH 3 )] + (aq) K 1 = 2.1 · 10 3 [Ag(NH 3 )] + (aq) + NH 3(aq) [Ag(NH 3 ) 2 ] + (aq) K 2 = 8.2 · 10 3 Ag + (aq) + 2 NH 3(aq) [Ag(NH 3 ) 2 ] + (aq) K stab = ? Stepwise Stability Constant

18 Lecture Example: AgBr (s) Ag + (aq) + Br - (aq) Calculate the solubility of AgBr in: a) water b) 1.0 M sodium thiosulfate (Na 2 S 2 O 3 ) (K sp (AgBr)= 5.0·10 -13, K stab ([Ag(S 2 O 3 ) 2 ] 3- )= 4.7·10 13 ) Complex Formation and Solubility AgBr (s) Ag + (aq) + Br - (aq) K sp = [Ag + ][Br - ] a) Solubility of AgBr in water

19 Lecture AgBr (s) Ag + (aq) + Br - (aq) K overall = K sp x K stab = = Ag + (aq) + 2S 2 O 3 2- (aq) [Ag(S 2 O 3 ) 2 ] 3- (aq) AgBr (s) + 2S 2 O 3 2- (aq) [Ag(S 2 O 3 ) 2 ] 3- (aq) + Br - (aq) (1) (2) (1)+(2) b) Solubility of AgBr in sodium thiosulfate [Ag(S 2 O 3 ) 2 3- ][Br - ] [S 2 O 3 2- ] 2 Initial Conc. Change Equilibrium Conc. 1.0 M Na 2 S 2 O 3

20 Lecture Ag + (aq) + OH - (aq) AgOH(s) brown K sp = AgOH(s) + H 2 PO 4 - (aq) Ag 3 PO 4 (s) yellow K sp = Ag 3 PO 4 (s) + HNO 3 (aq) Ag + (aq) + H 3 PO 4 (aq) Ag + (aq) + Cl - (aq) AgCl(s) white K sp = AgCl(s) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq) + Cl - (aq) K stab = [Ag(NH 3 ) 2 ] + (aq) + Br - (aq) AgBr(s) (green/white) K sp = AgBr(s) + 2 S 2 O 3 2- (aq) [Ag(S 2 O 3 ) 2 ] 3- (aq)+Br - (aq) K stab = [Ag(S 2 O 3 ) 2 ] 3- (aq) + I - (aq) AgI(s) (yellow) K sp = AgI(s) + 2 CN - (aq) [Ag(CN) 2 ] - (aq) + I - (aq) K stab = [Ag(CN) 2 ] - (aq) + S 2- (aq) Ag 2 S(s) (black) K sp = The One Pot Reaction * Note: Not all reaction equations are balanced

21 Lecture Additional Exercise 0.01 moles of AgNO 3 are added to a 500 mL of a 1.00 M solution of KCN. Then enough water is added to make 1.00 L of solution. Calculate the equilibrium [Ag + ] given K stab [Ag(CN) 2 ] – =10 20 M –2. (careful with the direction of the equation represented by K stab !) Ag + + 2CN – [Ag(CN) 2 ] – initial /M change ~ equilibrium /M x

22 Lecture AgBr (s) Ag + (aq) + Br - (aq) 1.0 M NH 3 Ag + (aq) + 2NH 3(aq) [Ag(NH 3 ) 2 ] + (aq) AgBr (s) + 2NH 3(aq) [AgNH 3 ] + (aq) + Br - (aq) (1) (2) (1)+(2) Solubility of AgBr in Ammonia Initial Conc. Change Equilibrium Conc. K stab (Ag(NH 3 ) 2 + )= 1.7·10 7


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