Transcript Solubility and complexes

CHEM1612 - Pharmacy
Week 8: Complexes I
Dr. Siegbert Schmid
School of Chemistry, Rm 223
Phone: 9351 4196
E-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
Complexes
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Blackman Chapter 13 and Sections 10.4, 11.8
Biologically important metal-complexes
Complex ions
Kstab
Co(EDTA)Coordination compounds
Chelates
Geometry of complexes
Solubility and complexes
Nomenclature
Isomerism in complexes
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
 M2+(aq)
(Hydrated M2+ ion)
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
overall 2+ charge.
Blackman Figure 13.12
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)
Eg. [Co(NH3)6]Cl3 (s)
Complex ion
[Co(NH3)6]3+(aq) + 3 Cl-(aq)
Counter ions
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.
e.g.
[Co(H2O)6][CoCl4]3 (s)
[Co(H2O)6]3+(aq) + 3 [CoCl4]-(aq)
Lecture 22-6
Coordination compounds
Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.
Ligands within the
coordination sphere
remain bound to the
metal ion
Coordination
Compound
Complex
Ion
Counter
Ions
Lecture 22-7
Complex Ions
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.
Lecture 22-8
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 22-9
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 22-10
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 22-11
Coordination Number and Geometry
:
Remember Valence Shell Electron Pair Repulsion Theory (VSEPR)?
O
Blackman Chapter 5
F
Sb
F
F
: F:
:
S
: F:
: F:
:
C
:
:
F
N
: F:
: F: : F :
:
F
Lecture 22-12
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 22-13
Ligands
 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
e.g. EDTA - (hexadentate, can form 6 bonds)
Lecture 22-14
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 22-15
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 22-16
Examples of ligands
Table from Silberberg, “Chemistry”, McGraw Hill, 2006.
Lecture 22-17
Examples of ligands
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.
O
OH2
H2O
OH2
Fe
H2O
OH2
OH2
[Fe(H2O)6]3+
monodentate
ligands
3+
H2N
H2N
NH2 NH
2
Fe
NH2
NH2
[Fe(en)3]3+
bidentate
ligands
3+
O
-
O
O
O
Fe
N
O
N
O
O
[Fe(EDTA)]hexadentate
ligands
Lecture 22-18
Lewis bases: water and ammonia
The stepwise exchange of NH3 for H2O in M(H2O)42+.
NH3
3
more
steps
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 22-19
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)

Kstab
[Ag(NH3 )2 ]

[Ag][NH3 ]2
Lecture 22-20
Stepwise stability constant



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) +
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)
Kstab = K1 x K2 = [Ag(NH3)2+] = 1.7 · 107
[Ag+] [NH3]2
Lecture 22-21
Demo: Nickel complexes
Ni2+ forms three complexes with ethylenediamine:
1.
2.
3.
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
Mix [Ni(H2O)6]2+ and en in ratio 1:1 → mostly [Ni(en)(H2O)4]2+
light blue
Mix [Ni(H2O)6]2+ and en in ratio 1:3 → mostly [Ni(en)3]2+
purple
Lecture 22-22
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 22-23
Heme
O2 bound
to Fe2+
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
Myoglobin protein
Blackman Figure 13.37
Lecture 22-24
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 Silberberg, “Chemistry”, McGraw Hill, 2006.
Lecture 22-25
Vitamin B12
Dorothy Crowfoot Hodgkin
The Nobel Prize in Chemistry 1964
Image download from Wikipedia
Nobelprize.org
Lecture 22-26
Carbonic anhydrase
Tetrahedral 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).
Figure downloaded from Wikipedia
CO2(g) + 2H2O(l)
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.
H3O+(aq) + HCO3- (aq)
Lecture 22-27
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 22-28
Complex Formation and solubility

Metal complex formation can influence the solubility of a compound.
e.g.
This
AgCl(s) + 2 NH3
[Ag(NH3)2]+ + Cl-
occurs in 2 stages:
AgCl(s)
Ag+ + ClAg+ + 2 NH3
[Ag(NH3)2]+
(1)
(2)
Complex formation removes the free Ag+ from solution and so drives
the dissolution of AgCl forward.

Lecture 22-29
Complex ion formation affects solubility
Ag+(aq) + Br-(aq)

Example: AgBr(s)

Calculate the solubility of AgBr in:
a) water
b) 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) of AgBr in water
a) Solubility
AgBr(s)
Ag+(aq) + Br-(aq)
x
Ksp = x2 = 5.0·10-13
Ksp = [Ag+][Br-]
x
x = 7.1 ·10-7 M
Lecture 22-30
b) Solubility of AgBr in sodium thiosulfate
1.0 M Na2S2O3
AgBr(s)
Ag+(aq) + 2S2O32-(aq)
AgBr(s) + 2S2O32-(aq)
Initial Conc.
Change
Equilibrium Conc.
(1)
Ag+(aq) + Br-(aq)
[Ag(S2O3)2]3-(aq)
[Ag(S2O3)2]3-(aq) + Br-(aq)
0
+x
x
1.0 M
-2x
1.0 -2x
Koverall = Ksp x Kstab =
(2)
[Ag(S2O3)23-][Br-]
[S2O32-]2
Substitute: Koverall = x2/(1.0 - 2x)2 = 24
(1)+(2)
0
+x
x
= 5.0·10-13 x 4.7·1013 = 24
x = 0.45
Solubility of AgBr in thiosulfate is 0.45 M (c.f. in water 7.1 x 10-7 M)
Lecture 22-31
c) Solubility of AgBr in ammonia
1.0 M NH3
AgBr(s)
(1)
Ag+(aq) + Br-(aq)
Ag+(aq) + 2NH3(aq)
[Ag(NH3)2]+(aq)
AgBr(s) + 2NH3(aq)
[AgNH3]+(aq) + Br-(aq)
Initial Conc.
Change
Equilibrium Conc.
0
+x
x
1.0 M
-2x
1.0 - 2x
Koverall = Ksp x Kstab =
[Ag(NH3)2+][Br-]
(2)
(1)+(2)
0
+x
x
= 5.0·10-13 x 1.7·107 = 8.5·10-6
[NH3]
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 22-32
The One Pot Reaction
Start with a AgNO3 aqueous solution. Add sequentially :
+ NaOH
Ag+ + OH-  AgOH(s) (brown)
Ksp = 10-7.70 M2
+ Na2HPO4 2 AgOH(s) + HPO42-  Ag3PO4(s) (yellow)Ksp = 10-16 M3
+ HNO3
Ag3PO4(s) + HNO3  3Ag+ + NO3- + HPO42-
+ NaCl
Ag+ + Cl-  AgCl (s) (white)
Ksp = 1.8 x 10-10 M2
+ NH3
AgCl(s) + 2NH3  [Ag(NH3)2]+ + Cl-
Kstab = 1.7 x 107 M-2
+ KBr
[Ag(NH3)2]+ + Br-  AgBr (s)(green/white)Ksp = 5 x 10-13 M2
+ Na2S2O3 AgBr(s) + 2S2O32-  [Ag(S2O3)2]3- + Br-
Kstab = 2.5 x 1013 M-2
+ KI
[Ag(S2O3)2]3- + I-  AgI (s) (yellow)
Ksp = 8.3 x 10-17 M2
+ KCN
AgI(s) + 2CN-  [Ag(CN)2]- + I-
Kstab = 6.3 x 1019 M-2
+ Na2S
2 Ag(CN)2- + S2-  Ag2S + CN-(black)
Ksp = 8 x 10-51 M3
Lecture 22-33
Nomenclature
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
Lecture 22-34
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:
New IUPAC Nomenclature:
all anions ending in – ‘ide’ become -‘ido’.
(Please modify accordingly
pp.518-519 of your book)
Fluorido
Chlorido
Bromido
Iodido
Hydroxido
Cyanido
Lecture 22-35
Nomenclature of Ligands



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:
Not Ironate
Not Copperate
Not Leadate
Not Silverate
Not Goldate
Not Tinnate
Lecture 22-36
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 22-37
Assigning oxidation numbers

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
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 22-38
About naming complexes
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.

Lecture 22-39
Isomerism in Complexes
Complexes can have several types of isomers:

Structural Isomers: different atom connectivities
1.
Coordination sphere isomerism
Linkage isomerism
2.

3.
4.
Stereoisomers: same atom connectivities but different
arrangement of atoms in space
Geometric isomerism
Optical isomerism
Lecture 22-40
Coordination compounds
Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.
Ligands within the
coordination sphere
remain bound to the
metal ion
Coordination
Compound
Complex
Ion
Counter
Ions
Lecture 22-41
Coordination Isomers
Ligands and counter-ions exchange place:
Example:


[Pt(NH3)4Cl2](NO2)2 tetraamminedichloridoplatinum(IV) nitrite
ligands
counterions

[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+
Lecture 22-42
Linkage isomers


Occur when a ligand has two alternative donor atoms.
Example 1:
S
Thiocyanate ion
cyanate ion
C
O
C
N
N
Thiocyanato NCS:→
Isothiocyanato SCN:→
cyanato NCO:→
isocyanato OCN:→
2+
2+
NH3
H3N
H3N
Co
NH3
NH3
N
C
S
and
NH3
Pentaammineisothiocyanatocobalt(III)
H3N
H3N
Co
NH3
S
C
N
NH3
pentaamminethiocyanatocobalt (III)
Lecture 22-43
Linkage Isomers

Example 2:
NO2-
nitro O2N:→
nitrito ONO:→
O
O
Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.
O
N
N
O
[Co(NH3)5(NO2)]Cl2
[Co(NH3)5(ONO)]Cl2
Pentaamminenitrocobalt(III) chloride
Pentaamminenitritocobalt(III) chloride
Lecture 22-44
Isomerism in Complexes
Complexes can have several types of isomers:

Structural Isomers: different atom connectivities
1.
Coordination sphere isomerism
Linkage isomerism
2.

3.
4.
Stereoisomers: same atom connectivities but different
arrangement of atoms in space
Geometric isomerism
Optical isomerism
Lecture 22-45
Stereoisomers: Geometric Isomers
cisplatin –
highly effective
anti-tumour agent
No
anti-tumour
effect
Lecture 22-46
Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.
Square planar complex. Four coordinate: cis- and trans-[Pt(NH3)2Cl2]
Stereoisomers: Geometric Isomers
Octahedral complex. Six coordinate: cis- and trans- [Co(NH3)4Cl2]+
2 Cl next to
each other
2 Cl axial to
each other
violet
green
Lecture 22-47
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.
[NiClBrFI]2-
Lecture 22-48
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.

cis-[Co(NH3)4Cl2]+
Has no optical Cl
isomers
H3N
Cl
+ NH3
Co
NH3
NH3
cis-[Co(en)2Cl2]+
Cl
H2N
Cl
+ NH2
Co
NH2
NH2
Has optical
isomers
Lecture 22-49
Stereoisomers: Optical isomers
Having three bidentate ligands of only
one type - gives a propeller-type structure.
2.
[M(en)3]n+ complexes have optical isomers:
H2N
NH2
3+ NH2
Co
NH2
NH2
H2N
H2N
H2N
www.pt-boat.com
NH2
3+ NH2
Co
NH2
NH2
Not
superimposable
Mirror
plane
Lecture 22-50
Octahedral complex - stereoisomerism
CisFigure from Silberberg, “Chemistry”, McGraw Hill, 2006.
Dichlorido
Bis(ethylendiamine)cobalt(III) ion
Mirror
image
rotation of I by 180° gives III ≠ II
Lecture 22-51
Octahedral complex - stereoisomerism
Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.
TransDichlorido
Bis(ethylendiamine)cobalt(III) ion
Mirror
image
rotation of I by 90° gives III = II
Lecture 22-52
Question

Does the square planar complex ion [Pt(NH3)(N3)BrCl]- have optical
isomers?
NH 3
Br
H 3N
Pt
Cl
Br
Pt
N=N=N
N=N=N
Cl
This complex has no optical isomers because it can be superimposed
on its mirror image.
Lecture 22-53
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 22-54