CHEM1612 - Pharmacy Week 8: Complexes I

Dr. Siegbert Schmid

School of Chemistry, Rm 223

Phone: 9351 4196

E-mail: siegbert.schmid@sydney.edu.au

Unless otherwise stated, all images in this file have been reproduced from:

Blackman, Bottle, Schmid, Mocerino and Wille,     Chemistry, John Wiley & Sons Australia, Ltd. 2008

     ISBN: 9 78047081 0866

Lecture 22-3

Complexes

Blackman, Bottle, Schmid, Mocerino & Wille Chapters 13,10.4, 11.8

Complex ions Coordination compounds Geometry of complexes Chelates Kstab

Solubility and complexes Nomenclature Isomerism in complexes Biologically important metal-complexes

Co(EDTA)-

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

M2+

H2O(l)

[M(H2O)4]2+

adduct

+

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, NH3

Ionic ligands, e.g., OH-, Cl-, CN-

[Ni(H2O)6]2+, a typical complex ion: Ni2+ is the central metal ion Six H2O molecules are the ligands O are the donor atoms overall 2+ charge.

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(NH3)6]Cl3 (s) [Co(NH3)6]3+(aq) + 3 Cl-(aq)

Coordination Compounds

Complex ion Counter ions

e.g. [Co(H2O)6][CoCl4]3 (s) [Co(H2O)6]3+(aq) + 3 [CoCl4]-

(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.

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 H2O molecules5 bound H2O molecules

1 bound OH- (overall charge reduced by 1)

Acidity of Aqueous Transition Metal Ions

Acidicsolution

Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.

Lecture 22-8

Free Ion Hydrated Ion Ka

Fe3+ Fe(H2O)63+(aq) 6 x 10-3

Cr3+ Cr(H2O)63+(aq) 1 x 10-4

Al3+ Al(H2O)63+(aq) 1 x 10-5

Be2+ Be(H2O)42+(aq) 4 x 10-6

Cu2+ Cu(H2O)62+(aq) 3 x 10-8

Fe2+ Fe(H2O)62+(aq) 4 x 10-9

Pb2+ Pb(H2O)62+(aq) 3 x 10-9

Zn2+ Zn(H2O)62+(aq) 1 x 10-9

Co2+ Co(H2O)62+(aq) 2 x 10-10

Ni2+ Ni(H2O)62+(aq) 1 x 10-10

AC

ID S

TR

EN

GT

H

Metal Ion HydrolysisEach hydrated metal ion that transfers a proton to water has a characteristic Ka value.

Lecture 22-9

M+ Coord no. M2+ Coord no. M3+ Coord no.Cu+ 2,4 Mn2+ 4,6 Sc3+ 6Ag+ 2 Fe2+ 6 Cr3+ 6Au+ 2,4 Co2+ 4,6 Co3+ 6

Ni2+ 4,6 Au3+ 4Cu2+ 4,6

Zn2+ 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

Lecture 22-10

Coordination number

Coordination

geometry

2 linear

4 square planar

4 tetrahedral

6 octahedral

Examples

[Ag(NH3)2]+

[AuCl2]-

[Pd(NH3)4]2

+

[PtCl4]2-

[Zn(NH3)4]2

+

[CuCl4]2-

[Co(NH3)6]3

+

[FeCl6]3-

Coordination Number and Geometry

Lecture 22-11

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. H2O, NH3, 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).

Bidentate ligands can form 2 bonds

e.g. Ethylenediamine

Polydentate ligands – can form more than 2 bonds

For a list of ligands see the recommended textbook.

Ligands

Lecture 22-12

H2C CH2

NH2

Mx+

H2N

Ethylenediamine (en) has two N atoms that can form a bond with the metal ion, giving a five-membered ring.

Bidentate Chelate LigandsMX+(en)

Blackman, Bottle, Schmid, Mocerino & Wille, Figure 13.10

Lecture 22-13

Demo: Nickel Complexes

Ni2+ forms three complexes with ethylenediamine:

1. Mix [Ni(H2O)6]2+ and en in ratio 1:1 → [Ni(en)(H2O)4]2+

green blue-green

2. Mix [Ni(H2O)6]2+ and en in ratio 1:2 → [Ni(en)2(H2O)2]2+

light blue

3. Mix [Ni(H2O)6]2+ and en in ratio 1:3 → [Ni(en)3]2+ purple

Ni2+

H O2

H O2

H O2

H O2

NH2

NH2

CH2CH2

Ni2+

H O2

H O2

H O2

H O2

en Ni( ) (aq)en2+

Lecture 22-14

Ethylenediaminetetraacetate tetraanion (EDTA4-)

N N

O O

OOO

O O

O

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=blueO=red

Lecture 22-15

M(H2O)42+

M(H2O)3(NH3)2+

M(NH3)42+

NH3

The stepwise exchange of NH3 for H2O in M(H2O)42+.

3NH3

3moresteps

Lewis bases: water and ammonia

Ammonia is a stronger Lewis base than waterFigure from Silberberg, “Chemistry”, McGraw Hill, 2006.

Lecture 22-16

Equilibrium Constant Kstab

The complex formation equilibrium is characterised by a stability constant, Kstab (also called formation constant):

Ag+(aq) + 2 NH3 Ag(NH3)2

+(aq)

23

23stab ][NH [Ag]

])[Ag(NH K

Metal Ion + nLigand Complex

The larger Kstab, the more stable the complex, e.g.

nstab [Ligand] [Metal]

[Complex] K

Lecture 22-17

Metal ions gain ligands one at a time. Each step characterised by a specific stability constant. Overall formation constant: Kstab = K1 x K2…x Kn

Example:

Ag+(aq) + NH3(aq) [Ag(NH3)]+

(aq) K1 = 2.1 · 103

[Ag(NH3)]+(aq) + NH3(aq) [Ag(NH3)2]+

(aq) K2 = 8.2 · 103

Ag+(aq) + 2 NH3(aq) [Ag(NH3)2]+

(aq) Kstab = ?

Stepwise Stability Constant

Lecture 22-18

Example: AgBr(s) Ag+(aq) + Br-

(aq)

Calculate the solubility of AgBr in:

a) water

b) 1.0 M sodium thiosulfate (Na2S2O3)

(Ksp (AgBr)= 5.0·10-13, Kstab ([Ag(S2O3)2]3- )= 4.7·1013 )

Complex Formation and Solubility

AgBr(s) Ag+(aq) + Br-

(aq)Ksp = [Ag+][Br-]

a) Solubility of AgBr in water

Lecture 22-19

AgBr(s) Ag+(aq) + Br-

(aq)

Koverall = Ksp x Kstab = =

Ag+(aq) + 2S2O3

2-(aq) [Ag(S2O3)2]3-

(aq)

AgBr(s) + 2S2O32-

(aq) [Ag(S2O3)2]3-(aq) + Br-

(aq)

(1)

(2)

(1)+(2)

b) Solubility of AgBr in sodium thiosulfate

[Ag(S2O3)23-][Br-]

[S2O32-]2

Initial Conc.ChangeEquilibrium Conc.

1.0 M Na2S2O3

Lecture 22-20

Ag+(aq) + OH-(aq) AgOH(s) brown Ksp =

AgOH(s) + H2PO4-(aq) Ag3PO4(s) yellow Ksp =

Ag3PO4(s) + HNO3(aq) Ag+(aq) + H3PO4(aq)

Ag+(aq) + Cl-(aq) AgCl(s) white Ksp =

AgCl(s) + 2 NH3(aq) [Ag(NH3)2]+(aq) + Cl-(aq) Kstab =

[Ag(NH3)2]+(aq) + Br-(aq) AgBr(s) (green/white) Ksp =

AgBr(s) + 2 S2O32-(aq) [Ag(S2O3)2]3-(aq)+Br-(aq) Kstab =

[Ag(S2O3)2]3-(aq) + I-(aq) AgI(s) (yellow) Ksp =

AgI(s) + 2 CN-(aq) [Ag(CN)2]-(aq) + I-(aq) Kstab =

[Ag(CN)2]-(aq) + S2-(aq) Ag2S(s) (black) Ksp =

The One Pot Reaction

* Note: Not all reaction equations are balanced

Lecture 22-21

Additional Exercise

M104(0.48)10

0.01

][CN10

][Ag(CN)][Ag

10]][CN[Ag

][Ag(CN)

22220.02-20.0

2

20.02-

2

stabK

0.01 moles of AgNO3 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 Kstab [Ag(CN)2]– =1020 M–2.

(careful with the direction of the equation represented by Kstab!)

Ag+ + 2CN– [Ag(CN)2]–

initial /M 0.01 0.500 0 change ~ -0.01 -0.02 0.01equilibrium /M x 0.480 0.01

Lecture 22-22

AgBr(s) Ag+(aq) + Br-

(aq)

1.0 M NH3

Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+

(aq)

AgBr(s) + 2NH3(aq) [AgNH3]+(aq) + Br-

(aq)

(1)

(2)

(1)+(2)

Solubility of AgBr in Ammonia

Initial Conc.ChangeEquilibrium Conc.

Kstab(Ag(NH3)2+)= 1.7·107