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
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
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Co(EDTA)-
Lecture 22-3
Metal Ions as Lewis Acids

Whenever a metal ion enters water, a complex ion forms with water
as the ligand.
+
[M(H2O)4]2+
M2+
H2O(l)


adduct
Metal ions act as Lewis acid (accepts electron pair).
Water is the Lewis base (donates electron pair).
Lecture 22-4
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 22-5
Coordination Compounds
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)
Complex ion
[Co(NH3)6]3+(aq) + 3 Cl-(aq)
Counter ions
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.
e.g.
[Co(H2O)6][CoCl4]3 (s)
[Co(H2O)6]3+(aq) + 3 [CoCl4]-(aq)
Lecture 22-6
Acidity of Aqueous Transition Metal Ions
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 molecules
5 bound H2O molecules
1 bound OH(overall charge reduced by 1)
Acidic
solution
Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.
Lecture 22-7
Metal Ion Hydrolysis
Each hydrated metal ion that transfers a proton to water has a
characteristic Ka value.
Free Ion
Hydrated Ion
Ka
Fe3+
6 x 10-3
Cr3+
Fe(H2O)63+(aq)
Cr(H2O)63+(aq)
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+
4 x 10-9
Pb2+
Fe(H2O)62+(aq)
Pb(H2O)62+(aq)
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
3 x 10-9
ACID STRENGTH
1 x 10-4
Lecture 22-8
Coordination Number
 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:
M+
Cu+
Ag+
Au+
Coord no.
2,4
2
2,4
M2+
Mn2+
Fe2+
Co2+
Ni2+
Cu2+
Coord no.
4,6
6
4,6
4,6
4,6
Zn2+
4,6
M3+ Coord no.
Sc3+
6
Cr3+
6
Co3+
6
Au3+
4
Lecture 22-9
Coordination Number and Geometry
Coordination
number
2
4
4
6
Coordination
geometry
Examples
linear
[Ag(NH3)2]+
[AuCl2]-
square planar
[Pd(NH3)4]2+
[PtCl4]2-
tetrahedral
[Zn(NH3)4]2+
[CuCl4]2-
octahedral
[Co(NH3)6]3+
[FeCl6]3-
Lecture 22-10
Ligands
 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.
Lecture 22-11
Bidentate Chelate Ligands
MX+(en)
Mx+
Ethylenediamine (en) has two N atoms that
can form a bond with the metal ion, giving a
five-membered ring.
H2N
H 2C
NH2
CH2
Blackman, Bottle, Schmid, Mocerino & Wille, Figure 13.10
Lecture 22-12
Demo: Nickel Complexes
Ni2+ forms three complexes with ethylenediamine:
Mix [Ni(H2O)6]2+ and en in ratio 1:1 →
green
[Ni(en)(H2O)4]2+
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
1.
Lecture 22-13
Hexadentate ligand: EDTA
Ethylenediaminetetraacetate tetraanion (EDTA4-)
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.
Co(III)
O
N=blue
O=red
O
O
O
N
N
O
O
O
O
[Co(EDTA)]Lecture 22-14
Lewis bases: water and ammonia
The stepwise exchange of NH3 for H2O in M(H2O)42+.
3
more
steps
NH3
3NH3
M(H2O)42+
M(H2O)3(NH3)2+
Ammonia is a stronger Lewis base than water
Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.
M(NH3)42+
Lecture 22-15
Equilibrium Constant Kstab
Metal Ion + nLigand
Complex
The complex formation equilibrium is characterised by a stability
constant, Kstab (also called formation constant):
K stab 
The
[Complex]
[Metal] [Ligand]n
larger Kstab, the more stable the complex, e.g.
Ag+(aq) + 2 NH3
Ag(NH3)2+(aq)

[Ag(NH 3 )2 ]
K stab 
[Ag] [NH 3 ]2
Lecture 22-16
Stepwise Stability Constant




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) +
[Ag(NH3)]+(aq)
K1 = 2.1 · 103
[Ag(NH3)]+(aq) + NH3(aq)
[Ag(NH3)2]+(aq)
K2 = 8.2 · 103
Ag+(aq) +
[Ag(NH3)2]+(aq)
Kstab = ?
NH3(aq)
2 NH3(aq)
Lecture 22-17
Complex Formation and Solubility
Ag+(aq) + Br-(aq)

Example: AgBr(s)

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 )
a) Solubility of AgBr in water
AgBr(s)
Ag+(aq) + Br-(aq)
Ksp = [Ag+][Br-]
Lecture 22-18
b) Solubility of AgBr in sodium thiosulfate
1.0 M Na2S2O3
AgBr(s)
(1)
Ag+(aq) + Br-(aq)
Ag+(aq) + 2S2O32-(aq)
AgBr(s) + 2S2O32-(aq)
[Ag(S2O3)2]3-(aq)
[Ag(S2O3)2]3-(aq) + Br-(aq)
(2)
(1)+(2)
Initial Conc.
Change
Equilibrium Conc.
Koverall = Ksp x Kstab =
[Ag(S2O3)23-][Br-]
[S2O3
2-]2
=
Lecture 22-19
The One Pot Reaction
Ag+(aq) + OH-(aq)
AgOH(s) + H2PO4-(aq)
Ag3PO4(s) + HNO3(aq)
Ag+(aq) + Cl-(aq)
AgCl(s) + 2 NH3(aq)
[Ag(NH3)2]+(aq) + Br-(aq)
AgBr(s) + 2 S2O32-(aq)
AgOH(s) brown
Ag3PO4(s) yellow
Ksp =
Ksp =
Ag+(aq) + H3PO4(aq)
AgCl(s) white
[Ag(NH3)2]+(aq) + Cl-(aq)
AgBr(s) (green/white)
[Ag(S2O3)2]3-(aq)+Br-(aq)
Ksp =
Kstab =
Ksp =
Kstab =
[Ag(S2O3)2]3-(aq) + I-(aq)
AgI(s) (yellow)
AgI(s) + 2 CN-(aq)
[Ag(CN)2]-(aq) + I-(aq) Kstab =
[Ag(CN)2]-(aq) + S2-(aq)
Ag2S(s) (black)
Ksp =
Ksp =
* Note: Not all reaction equations are balanced
Lecture 22-20
Additional Exercise
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+ +
initial /M
0.01
change
~ -0.01
equilibrium /M
x
2CN–
0.500
-0.02
0.480
[Ag(CN)2]–
0
0.01
0.01

K stab
[Ag(CN) 2 ]
20.0


10
[Ag ][CN- ]2

[Ag(CN) 2 ]
0.01
 22
[Ag ]  20.0


4

10
M
- 2
20.0
2
10 [CN ]
10 (0.48)
Lecture 22-21
Solubility of AgBr in Ammonia
1.0 M NH3
AgBr(s)
Kstab(Ag(NH3)2+)= 1.7·107
Ag+(aq) + Br-(aq)
Ag+(aq) + 2NH3(aq)
[Ag(NH3)2]+(aq)
AgBr(s) + 2NH3(aq)
[AgNH3]+(aq) + Br-(aq)
(1)
(2)
(1)+(2)
Initial Conc.
Change
Equilibrium Conc.
Lecture 22-22