chem1612 - pharmacy week 8 : complexes i
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CHEM1612 - Pharmacy Week 8: Complexes I
Dr. Siegbert SchmidSchool of Chemistry, Rm 223Phone: 9351 4196E-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 Chapter 13 and Sections 10.4, 11.8 Biologically important metal-complexes Complex ions Kstab
Coordination compounds Chelates Geometry of complexes Solubility and complexes Nomenclature Isomerism in complexes
Co(EDTA)-
Lecture 22-4
Metal Ions as Lewis Acids
M2+
H2O(l)
[M(H2O)4]2+
adduct
º M2+(aq)(Hydrated M2+ ion)
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).
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 overall 2+ charge.
Blackman Figure 13.12
Lecture 22-6
They consist of:• Complex ion (metal ion with attached ligands)• Counter ions (additional anions/cations needed for zero net charge)
Eg. [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 behave like electrolytes: the complex ion exists as the cation and the 3 Cl- ions are separate.
Note: the counter ion may also be a complex ion.
Lecture 22-7
CoordinationCompound
ComplexIon
CounterIons
Ligands within the coordination sphere remain bound to the metal ion
Coordination compoundsFigure from Silberberg, “Chemistry”, McGraw Hill, 2006.
Lecture 22-8
e.g. Ag+(aq) + 2 NH3 Ag(NH3)2
+(aq)
Ligands must have a lone pair to donate to the metal. The ‘donation’ of the electron pair is sometimes referred to as a
“dative” bond.
Complex Ions
Lecture 22-9
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 22-10
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
TRE
NG
TH
Metal Ion HydrolysisEach hydrated metal ion that transfers a proton to water has a characteristic Ka value.
Lecture 22-11
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+ 4,6Zn2+ 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 22-12
Coordination Number and GeometryRemember Valence Shell Electron Pair Repulsion Theory (VSEPR)?
F
SbF
F F
FN C O S
FF
F
:
: :
::
: ::
:F: ::
F:
:
:
F: ::
Blackman Chapter 5
Lecture 22-13
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 22-14
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 bondse.g. ethylenediamine
Polydentate ligands – can form more than 2 bondse.g. EDTA - (hexadentate, can form 6 bonds)
Ligands
Lecture 22-15
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 22-16
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 22-17
Examples of ligands
Table from Silberberg, “Chemistry”, McGraw Hill, 2006.
Lecture 22-18
OH2Fe
OH2H2O OH2H2O OH2
3+NH2Fe
NH2H2N NH2
H2N NH23+
O
Fe
NN OO O
OO
OO
-
[Fe(H2O)6]3+ [Fe(en)3]3+ [Fe(EDTA)]-
monodentateligands
bidentateligands
hexadentateligands
The charge of a complex ion is the charge of the metal ion plus the charge of its ligands:
e.g. [Ni(H2O)6]2+ charge of complex ion is that of the Ni2+ ion.
eg [NiCl4]2- Ni2+ ion coordinated to four chloride (Cl-) ions giving overall (2-) charge.
Examples of ligands
Lecture 22-19
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 22-20
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 22-21
Metal ions gain ligands one at a time. Each step characterised by “stepwise stability constant” aka “stepwise
formation 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 =
Kstab = K1 x K2 = [Ag(NH3)2+] = 1.7 · 107
[Ag+] [NH3]2
Stepwise stability constant
Lecture 22-22
Demo: Nickel complexes
Ni2+ forms three complexes with ethylenediamine:
1. Mix [Ni(H2O)6]2+ and en in ratio 3:1 → some [Ni(en)(H2O)4]2+and [Ni(H2O)6]2
Green blue-green
2. Mix [Ni(H2O)6]2+ and en in ratio 1:1 → mostly [Ni(en)(H2O)4]2+
light blue
3. Mix [Ni(H2O)6]2+ and en in ratio 1:3 → mostly [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)en 2+
Lecture 22-23
Biologically Important Complexes Many biomolecules contain metal ions that act as Lewis acids.
Give some examples of naturally occurring complexes.
Heme
Chlorophyll
Vitamin B12
Enzyme Carbonic anhydrase
Lecture 22-24
Heme
Heme is a square planar complex of Fe2+ and the tetradentate ring ligand porphyrin (bonds to 4 donor N atoms). Present in hemoglobin, which carries oxygen in blood, and myoglobin, which stores oxygen in muscle.
Porphyrin ring
O2 bound to Fe2+
Myoglobin protein
Blackman Figure 13.37
Lecture 22-25
Chlorophyll
Chlorophyll is a photosynthetic pigment, that gives leaves the characteristic green colour.It is a complex of Mg2+ and a porphyrin ring system (four N atoms are the chelae).
Figure from S
ilberberg, “Chem
istry”, McG
raw H
ill, 2006.
Lecture 22-26
Dorothy Crowfoot HodgkinThe Nobel Prize in Chemistry 1964
Nobelprize.org
Vitamin B12
Image download from Wikipedia
Lecture 22-27
CO2(g) + 2H2O(l) H3O+(aq) + HCO3
- (aq)
Carbonic anhydraseTetrahedral complex of Zn2+.
Catalyses reaction between water and carbon dioxide during respiration. Coordinated to 3 N, fourth site left free to interact with molecule whose reaction is being catalysed (here with water).
By withdrawing electron density, makes water acidic to lose proton and OH- attacks partial positive C of CO2 much more vigorously. Cd2+ is toxic because it competes with zinc for this spot.
Figure downloaded from Wikipedia
Lecture 22-28
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 22-29
Metal complex formation can influence the solubility of a compound.
e.g. AgCl(s) + 2 NH3 [Ag(NH3)2]+ + Cl-
This occurs in 2 stages:
AgCl(s) Ag+ + Cl- (1)Ag+ + 2 NH3 [Ag(NH3)2]+ (2)
Complex formation removes the free Ag+ from solution and so drives the dissolution of AgCl forward.
Complex Formation and solubility
Lecture 22-30
Example: AgBr(s) Ag+(aq) + Br-
(aq)
Calculate the solubility of AgBr in:a) waterb) 1.0 M sodium thiosulfate (Na2S2O3)c) 1.0 M NH3
(Ksp (AgBr)= 5.0·10-13, Kstab ([Ag(S2O3)2]3- )= 4.7·1013; Kstab(Ag(NH3)2+)= 1.7·107)
Complex ion formation affects solubility
AgBr(s) Ag+(aq) + Br-
(aq)
Ksp = x2 = 5.0·10-13 x = 7.1 ·10-7 M
Ksp = [Ag+][Br-]
x x
a) Solubility of AgBr in water
Lecture 22-31
AgBr(s) Ag+(aq) + Br-
(aq)
Koverall = Ksp x Kstab = = 5.0·10-13 x 4.7·1013 = 24
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-2x
1.0 -2x
0+xx
0+xx
Substitute: Koverall = x2/(1.0 - 2x)2 = 24 x = 0.45
Solubility of AgBr in thiosulfate is 0.45 M (c.f. in water 7.1 x 10-7 M)
1.0 M Na2S2O3
Lecture 22-32
AgBr(s) Ag+(aq) + Br-
(aq)
Koverall = Ksp x Kstab = = 5.0·10-13 x 1.7·107 = 8.5·10-6
1.0 M NH3
Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+
(aq)
AgBr(s) + 2NH3(aq) [AgNH3]+(aq) + Br-
(aq)
(1)
(2)
(1)+(2)
c) Solubility of AgBr in ammonia
[Ag(NH3)2+][Br-]
[NH3]
Initial Conc.ChangeEquilibrium Conc.
1.0 M-2x
1.0 - 2x
0+xx
0+xx
Substitute: Koverall = x2/(1.0-2x)2 = 8.5·10-6 x = 2.9·10-3 M
Solubility of AgBr in NH3 is 2.9·10-3 M (c.f. in thiosulfate 0.45 M)
Lecture 22-33
The One Pot ReactionStart with a AgNO3 aqueous solution. Add sequentially :
Ag+ + OH- AgOH(s) (brown)
2 AgOH(s) + HPO42- Ag3PO4(s) (yellow)
Ag3PO4(s) + HNO3 3Ag+ + NO3- + HPO4
2-
Ag+ + Cl- AgCl (s) (white)
AgCl(s) + 2NH3 [Ag(NH3)2]+ + Cl-
[Ag(NH3)2]+ + Br- AgBr (s)(green/white)
AgBr(s) + 2S2O32- [Ag(S2O3)2]3- + Br-
[Ag(S2O3)2]3- + I- AgI (s) (yellow)
AgI(s) + 2CN- [Ag(CN)2]- + I-
2 Ag(CN)2- + S2- Ag2S + CN-(black)
+ NaOH
+ Na2HPO4
+ HNO3
+ NaCl
+ NH3
+ KBr
+ Na2S2O3
+ KI
+ KCN
+ Na2S
Ksp = 10-7.70 M2
Ksp = 10-16 M3
Ksp = 1.8 x 10-10 M2
Kstab = 1.7 x 107 M-2
Ksp = 5 x 10-13 M2
Kstab = 2.5 x 1013 M-2
Ksp = 8.3 x 10-17 M2
Kstab = 6.3 x 1019 M-2
Ksp = 8 x 10-51 M3
Lecture 22-34
Rules for nomenclature of coordination compounds:
Name cation, then anion, as separate words.Examples:
[Pt(NH3)4Cl2](NO2)2 tetraamminedichloridoplatinum(IV) nitrite
[Pt(NH3)4(NO2)2]Cl2 tetraamminedinitritoplatinum(IV) chloride
Name the ligands then the metal, all in same word.
Number of ligands as Greek prefixes (di-, tri-, tetra-, penta-, hexa-), except ligands that already have numerical prefixes which use Latin prefixes (bis, tris, tetrakis…) e.g. bis(ethylenediamine) for (en)2
Nomenclature
Lecture 22-35
Nomenclature II Oxidation state in Roman numeral in parentheses after name of metal
e.g. [Ag(NH3)2]NO3 diamminesilver(I) nitrate
Anionic ligands end in '-ido';
Neutral ligands named as molecule, except those listed here:
FluoridoChloridoBromidoIodidoHydroxidoCyanido
New IUPAC Nomenclature: all anions ending in – ‘ide’ become -‘ido’.(Please modify accordingly pp.518-519 of your book)
Lecture 22-36
Ligands named in alphabetical order (but prefixes do not affect the order) e.g. [Co(NH3)5Cl]SO4 pentaamminechloridocobalt(III) sulfate
Anionic complexes end in ‘-ate’ e.g. K3[CrCl6] potassium hexachloridochromate(III)
Some metals in anionic complexes use Latin -ate names:
Nomenclature of Ligands
Not IronateNot CopperateNot LeadateNot SilverateNot GoldateNot Tinnate
Lecture 22-37
Nomenclature - Exercises [Co(H2O)6]CO3
hexaaquacobalt(II) carbonate
[Cu(NH3)4]SO4
tetraamminecopper(II) sulfate
(NH4)3[FeF6]ammonium hexafluoridoferrate(III)
K4[Mn(CN)6]potassium hexacyanidomanganate(II)
Lecture 22-38
Example 1:Find O.N. of Co in : [Co(NH3)5Cl]SO4 pentaamminechloridocobalt(?) sulfate
[Co(NH3)5Cl]2+ ammine is neutral, chloride is -1
O.N. -1 = +2 (sum of O.N.s = overall charge)
O.N. = +3
Assigning oxidation numbers
Example 2:Find O.N. of Mn in :K4[Mn(CN)6] potassium hexacyanidomanganate(?)
[Mn(CN)6]4- (CN) is -1 overall
O.N. + 6x(-1) = -4 (sum of O.N.s = overall charge)
ON = +2
Lecture 22-39
You won’t be asked to draw formulae of complicated biological complexes.
You should be able to use the naming rules to write formulae from names and names from formulae.
About naming complexes
Lecture 22-40
Isomerism in ComplexesComplexes can have several types of isomers:
Structural Isomers: different atom connectivities
1. Coordination sphere isomerism2. Linkage isomerism
Stereoisomers: same atom connectivities but different arrangement of atoms in space
3. Geometric isomerism4. Optical isomerism
Lecture 22-41
CoordinationCompound
ComplexIon
CounterIons
Ligands within the coordination sphere remain bound to the metal ion
Coordination compoundsFigure from Silberberg, “Chemistry”, McGraw Hill, 2006.
Lecture 22-42
Coordination Isomers Ligands and counter-ions exchange place:Example:
[Pt(NH3)4Cl2](NO2)2 tetraamminedichloridoplatinum(IV) nitrite
[Pt(NH3)4(NO2)2]Cl2 tetraamminedinitritoplatinum(IV) chloride
Two sets of ligands are reversed:[Cr(NH3)6][Co(CN)6] NH3 is a ligand for Cr3+
[Co(NH3)6][Cr(CN)6] NH3 is a ligand for Co3+
ligands counterions
Lecture 22-43
Linkage isomers Occur when a ligand has two alternative donor atoms. Example 1:
NCSThiocyanate ion
H3N Co
NH3
NH3
NNH3
H3NC S
2+
H3N Co
NH3
NH3
SNH3
H3NC N
2+
and
cyanate ion NCOcyanato NCO:→ isocyanato OCN:→
Thiocyanato NCS:→ Isothiocyanato SCN:→
Pentaammineisothiocyanatocobalt(III) pentaamminethiocyanatocobalt (III)
Lecture 22-44
NO2- nitro O2N:→
nitrito ONO:→
Linkage Isomers Example 2:
N N
O
OOO
[Co(NH3)5(NO2)]Cl2
Pentaamminenitrocobalt(III) chloride
[Co(NH3)5(ONO)]Cl2
Pentaamminenitritocobalt(III) chloride
Figure from S
ilberberg, “Chem
istry”, McG
raw H
ill, 2006.
Lecture 22-45
Isomerism in ComplexesComplexes can have several types of isomers:
Structural Isomers: different atom connectivities
1. Coordination sphere isomerism2. Linkage isomerism
Stereoisomers: same atom connectivities but different arrangement of atoms in space
3. Geometric isomerism4. Optical isomerism
Lecture 22-46
Square planar complex. Four coordinate: cis- and trans-[Pt(NH3)2Cl2]
Stereoisomers: Geometric Isomers
No
anti-tumour
effect
cisplatin –
highly effective
anti-tumour agent
Figure from S
ilberberg, “Chem
istry”, McG
raw H
ill, 2006.
Lecture 22-47
Stereoisomers: Geometric Isomers
2 Cl next to each other
Octahedral complex. Six coordinate: cis- and trans- [Co(NH3)4Cl2]+
violet
green2 Cl axial to each other
Lecture 22-48
[NiClBrFI]2-
Stereoisomers: Optical Isomers
When a molecule is non-superimposable with its mirror image. Example: four different substituents about tetrahedral centre. Same physical properties, except direction in which they rotate the
plane of polarized light.
Lecture 22-49
cis-[Co(NH3)4Cl2]+ cis-[Co(en)2Cl2]+
ClCo
NH3
NH3H3N
Cl
NH3
ClCo
NH2
NH2H2N
Cl
NH2
+ +Has no optical
isomers
Has optical
isomers
Stereoisomers: Optical isomers Metal atoms with tetrahedral or octahedral geometries (but not
square planar) may be chiral due to having different ligands. For the octahedral case, several chiralities are possible, e.g.
1. Complex with four ligands of two types.
Lecture 22-50
[M(en)3]n+ complexes have optical isomers:
Notsuperimposable
H2N CoNH2
NH2H2N
NH2
NH2
NH2Co
NH2
NH2NH2
H2N
H2N
3+ 3+
Mirrorplane
Stereoisomers: Optical isomers
2. Having three bidentate ligands of only one type - gives a propeller-type structure.
www.pt-boat.com
Lecture 22-51
Octahedral complex - stereoisomerism
rotation of I by 180° gives III ≠ II
Mirror
image
Cis-
Dichlorido
Bis(ethylendiamine)cobalt(III) ion
Figure from S
ilberberg, “Chem
istry”, McG
raw H
ill, 2006.
Lecture 22-52
Octahedral complex - stereoisomerism
rotation of I by 90° gives III = II
Mirror
image
Trans-
Dichlorido
Bis(ethylendiamine)cobalt(III) ion
Figure from S
ilberberg, “Chem
istry”, McG
raw H
ill, 2006.
Lecture 22-53
Question Does the square planar complex ion [Pt(NH3)(N3)BrCl]- have optical
isomers?
BrPt
N=N=N
NH3
Cl
Br
Pt
NH3
ClN=N=N
This complex has no optical isomers because it can be superimposed
on its mirror image.
Lecture 22-54
Summary
Concepts: Complex formation Stability constant and stepwise stability constant Acidity of some metal ions in solution Coordination compounds and geometry Nomenclature of coordination compounds Isomerism in Complexes
Calculations Complex Formation Equilibria in solution: complex formation + solubility
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