Download - CHEM1612 - Pharmacy Week 8: Complexes II
CHEM1612 - Pharmacy Week 8: Complexes II
Dr. Siegbert SchmidSchool of Chemistry, Rm 223Phone: 9351 4196E-mail: [email protected]
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
Example: AgBr(s) Ag+(aq) + Br-
(aq)
Calculate the solubility of AgBr in:a) waterb) 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 = x2 = 5.0·10-13 x = 7.1 ·10-7 M
Ksp = [Ag+][Br-]
x x
a) Solubility of AgBr in water
Lecture 22-5
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 22-6
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 I
Lecture 22-7
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:
(Please modify accordingly pp.518-519 of your book)
Ligand Name FormulaFluorido F-
Chlorido Cl-Bromido Br-
Iodido I-
Cyanido CN-
Hydroxido OH-
Ligand Name FormulaAmmine NH3
Aqua H2OCarbonylCONitrosyl NO
Lecture 22-8
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 22-9
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 22-10
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 22-11
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 22-12
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 22-13
Linkage isomers Occur when a ligand has two alternative donor atoms.
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 22-14
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 22-15
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 22-16
[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 22-17
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 cases are possible, e.g.
1. Complex with four ligands of two types.
Lecture 22-18
[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 22-19
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 22-20
Octahedral complex - stereoisomerism
rotation of I by 270° gives III = II
Mirror
image
Trans-
Dichlorido
Bis(ethylendiamine)cobalt(III) ion
Figure from S
ilberberg, “Chem
istry”, McG
raw H
ill, 2006.
Lecture 22-21
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 22-22
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 22-23
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 22-24
Dorothy Crowfoot HodgkinThe Nobel Prize in Chemistry 1964
Nobelprize.org
Vitamin B12
Image download from Wikipedia
Lecture 22-25
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 22-26
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
Lecture 22-27
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 22-28
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 22-29
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)
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)
Kstab(Ag(NH3)2+)= 1.7·107)