chem1612 - pharmacy week 8: complexes i dr. siegbert schmid school of chemistry, rm 223 phone: 9351...
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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
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