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Transcript
COORDINATION CHEMISTRY
COORDINATION COMPLEX – Any ion or neutral species with a metal
atom or ion bonded to 2 or more molecules or ions
Ag(NH3)2+
Cu(NH3)42+
Zn(NH3)42+
Al(OH)4-
Zn(OH)42-
Charged coordination complexes are called COMPLEX IONS
LIGAND – An ion or molecule that bonds to a central metal atom to form
a complex ion
4A-1 (of 20)
PROPERTIES OF COORDINATION COMPLEXES (with transition metals)
(1) COLOR
Co(NO2)63-
Co(CN)63-
Co(H2O)63+
Co(CO3)33-
Color depends upon the chemical groups attached to the transition
metal
4A-2 (of 20)
(2) MAGNETISM
Some are DIAMAGNETIC (no unpaired electrons), and some are
PARAMAGNETIC (1 or more unpaired electrons)
4A-3 (of 20)
(3) GEOMETRY
COORDINATION NUMBER – The number of bonds formed between
the metal ion and the ligands
Coordination Number
2
4A-4 (of 20)
Geometry
linear
4
square planar or tetrahedral
6
octahedral
(a) PROPERTIES OF LIGANDS
MONODENTATE LIGAND – A ligand with one lone pair that can form one
bond to a metal ion
H2O, NH3, CN-, NO2-, SCN-, OH-, XCHELATE – A ligand with more than one atom with a lone pair that can be
used to bond to a metal ion
Bidentate ligands:
oxalate (ox) – C2O42ethylenediamine (en) – H2N(CH2)2NH2
Polydentate ligands:
diethylenetriamine (dien) – H2N(CH2CH2)NH(CH2CH2)NH2
ethylenediaminetetraacetate (EDTA) – (O2CCH2)2N(CH2CH2)N(CH2CO2)24-
4A-5 (of 20)
(b) NOMENCLATURE FOR COMPLEX IONS
1 – Name the ligands before the metal ion
2 – In naming ligands, molecules use their molecular names (with 4
common exceptions), and anions have their name end in -o
H2O – aqua
NH3 – ammine
CO – carbonyl
NO – nitrosyl
Cl- – chloro
F- – fluoro
OH- – hydroxo
CN- – cyano
NO2- – nitrito
C2O42- – oxalato
SO42- – sulfato NO3- – nitrato
3 – Different ligands are named alphabetically
4 – Prefixes are used if a complex ion has more than one particular
ligand
5 – Prefixes for polydentate ligands (or ligands whose names contain
prefixes) are bis-, tris-, tetrakis-, etc., with the ligand in parenthesis
6 – The charge of the metal is given as a roman numeral in parenthesis
7 – If the complex ion has a negative charge, the suffix –ate is added to
the name of the metal
4A-6 (of 20)
Co(NH3)63+
hexaamminecobalt(III)
CoCl63-
hexachlorocobaltate(III)
Co(NH3)5Cl2+
pentaamminechlorocobalt(III)
Fe(CN)63-
not hexancyanoironate(III) , its hexacyanoferrate(III)
Fe – ferrate
Sn – stannate
Cu – cuprate
Pt – platinate
Fe(C2O4)33-
tris(oxalato)ferrate(III)
Ni(CO)4
tetracarbonylnickel(0)
4A-7 (of 20)
Pb – plumbate
Mn - manganate
(c) NOMENCLATURE FOR COMPOUNDS CONTAINING COMPLEX IONS
COORDINATION COMPOUND – Any compound containing a complex ion
and a counterion
[Cu(NH3)4]Cl2
This is a 2+ charged complex ion, requiring 2 Cl- counterions to produce
a neutral compound
When naming coordination compounds, name the cation, then name
the anion
tetraamminecopper(II) chloride
4A-8 (of 20)
[Cr(NH3)6]Cl3
This is a 3+ charged complex ion
hexaamminechromium(III) chloride
[Pt(NH3)3Cl3]Cl
This is a 1+ charged complex ion
triamminetrichloroplatinum(IV) chloride
Mn(en)2Cl2
This is a neutral complex
dichlorobis(ethylenediamine)manganese(II)
4A-9 (of 20)
K[PtNH3Cl5]
This is a 1- charged complex ion
potassium amminepentachloroplatinate(IV)
K4Fe(CN)6
This is a 4- charged complex ion
potassium hexacyanoferrate(II)
[Fe(en)2(NO2)2]2SO4
This is a 1+ complex ion
bis(ethylenediamine)dinitritoiron(III) sulfate
4A-10 (of 20)
(4) ISOMERISM
ISOMERS – Compounds with the same chemical formula, but with
different properties
(1) STRUCTURAL ISOMERS – Compounds with the same chemical
formula, but with the atoms bonded in different orders
Structural isomers have different names
4A-11 (of 20)
C4H10
H
H
H
H
H
H
H
C
C
C
C
H
H
H
H
butane
4A-12 (of 20)
C
H
H
H
H
H
C
C
C
H
H
H
H
methyl propane
Pt(H2O)4(OH)2Cl2
[Pt(H2O)4(OH)2]Cl2
H2O
H2O
H2O
Pt
tetraaquadihydroxoplatinum(IV) chloride
2+
OH
OH
Cl- Cl-
H2O
[Pt(H2O)4Cl2](OH)2
H2O
H2O
H2O
Pt
H2O
4A-13 (of 20)
tetraaquadichloroplatinum(IV) hydroxide
2+
Cl
Cl
OH- OH-
(2) SPATIAL ISOMERS – Compounds with the same chemical formula
and with the atoms bonded in the same order, but with the atoms
bonded in different spatial orientations
(a) GEOMETRICAL ISOMERS – Spatial isomers that ARE NOT
mirror images of each other
4A-14 (of 20)
CoCl2(NH3)4
tetraamminedichlorocobalt(II)
Cl
NH3
NH3
Co
NH3
NH3
opposite – trans
Cl
trans-tetraamminedichlorocobalt(II)
Cl
NH3
NH3
Co
NH3
Cl
NH3
cis-tetraamminedichlorocobalt(II)
4A-15 (of 20)
adjacent – cis
How many geometrical isomers are there for diamminedichloroplatinum(II)
if it has square planar geometry?
Cl
NH3
Pt
NH3
Cl
trans-diamminedichloroplatinum(II)
NH3
Cl
Pt
NH3
Cl
cis-diamminedichloroplatinum(II)
4A-16 (of 20)
How many geometrical isomers are there for diamminedichloroplatinum(II)
if it has tetrahedral geometry?
NH3
Cl
Pt
Cl
NH3
only one
Cl’s all always adjacent
4A-17 (of 20)
How many geometrical isomers are there for triamminetrichlorocobalt(III)
if it has octahedral geometry?
Cl
NH3
NH3
Co
Cl
Cl
each pair adjacent – fac
NH3
fac-triamminetrichlorocobalt(III)
Cl
NH3
NH3
Co
NH3
Cl
Cl
mer-triamminetrichlorocobalt(III)
4A-18 (of 20)
adjacent and opposite – mer
(b) OPTICAL ISOMERS – Spatial isomers that ARE mirror images
of each other, and they are nonsuperimposable
tris(oxalato)ferrate(III)
O
O
O
Fe
O
O
O
O
O
O
O
Fe
O
O
O
These are
nonsuperimposable
molecules
 the compound
tris(oxalato)ferrate(III)
has 2 optical isomers
4A-19 (of 20)
180º
O
O
Fe
O
O
O
Optical isomers are called ENANTIOMERS
anteater
4A-20 (of 20)
enantiomer
FORMATION CONSTANT (Kf) – The equilibrium constant for the complete
formation of a complex ion
diamminesilver(I)
Ag+ (aq) + 2NH3 (aq) → Ag(NH3)2+ (aq)
Kf =
[Ag(NH3)2+]
_______________
[Ag+][NH3]2
4B-1 (of 15)
Write the reaction for the complete formation of hexaamminecobalt(II),
and its Kf expression
Co2+ (aq) + 6NH3 (aq) ⇆ Co(NH3)62+ (aq)
Kf =
[Co(NH3)62+]
________________
[Co2+][NH3]6
4B-2 (of 15)
Write the reaction for the complete formation of hexaamminecobalt(II),
and its Kf expression
Co2+ (aq) + 6NH3 (aq) ⇆ Co(NH3)62+ (aq)
Kf =
[Co(NH3)62+]
________________
[Co2+][NH3]6
If a solution is 0.250 M Co2+ and 0.100 M Co(NH3)62+ at equilibrium, and
the formation constant is 1.00 x 105, calculate [NH3].
[NH3]6 =
[Co(NH3)62+]
________________
[Co2+]Kf
[NH3] =
6
[Co(NH3)62+]
________________
[Co2+]Kf
4B-3 (of 15)
=
6
0.100 M
__________________________
(0.250 M)(1.00 x105)
= 0.126 M
The formation constant for diamminesilver(I) is 1.00 x 106. Calculate [Ag+]
in a solution that was originally 0.100 M Ag+ and 0.500 M NH3.
Ag+ (aq) + 2NH3 (aq) ⇆ Ag(NH3)2+ (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.100
-x
0.100 - x
0.500
- 2x
0.500 - 2x
0
+x
x
The reaction is going in the forward direction and has a large equilibrium
constant,  x will be a large number
4B-4 (of 15)
The formation constant for diamminesilver(I) is 1.00 x 106. Calculate [Ag+]
in a solution that was originally 0.100 M Ag+ and 0.500 M NH3.
Ag+ (aq) + 2NH3 (aq) ⇆ Ag(NH3)2+ (aq)
Initial M’s
Shift M’s
New Initial M’s
Change M’s
Equilibrium M’s
Kf =
[Ag(NH3)2+]
________________
0.100
- 0.100
0
+x
x
1.00 x 106 =
[Ag+][NH3]2
x = 1.11 x 10-6 = [Ag+]
4B-5 (of 15)
0.500
- 0.200
0.300
+ 2x
0.300 + 2x
0
+ 0.100
0.100
-x
0.100 - x
(0.100 – x)
___________________
(x)(0.300 + 2x)2
1.00 x 106 =
(0.100)
____________
(x)(0.300)2
The solubility product constant for zinc hydroxide is 4.5 x 10-17, and the
formation constant for tetrahydroxozincate(II) is 5.0 x 1014.
(a) Calculate molar solubility of zinc hydroxide in pure water.
Zn(OH)2 (s)
⇆
Initial M’s
Change in M’s
Equilibrium M’s
Zn2+ (aq)
0
+x
x
Ksp = [Zn2+][OH-]2 = (x)(2x)2 = 4x3
x = molar solubility of Zn(OH)2
4.5 x 10-17 = 4x3
2.2 x 10-6 M = x = molar solubility of Zn(OH)2
4B-6 (of 15)
+
2OH- (aq)
0
+ 2x
2x
The solubility product constant for zinc hydroxide is 4.5 x 10-17, and the
formation constant for tetrahydroxozincate(II) is 5.0 x 1014.
(b) Calculate molar solubility of zinc hydroxide in 0.10 M NaOH.
Zn(OH)2 (s)
Initial M’s
Change in M’s
Equilibrium M’s
⇆
Zn2+ (aq)
0
+
2OH- (aq)
0.10
No, because Zn2+ forms a complex ion with OHZn(OH)2 (s) ⇆ Zn2+ (aq) + 2OH- (aq)
Ksp = 4.5 x 10-17
Zn2+ (aq) + 4OH- (aq) ⇆ Zn(OH)42- (aq)
Kf
= 5.0 x 1014
Zn(OH)2 (s) + 2OH- (aq) ⇆ Zn(OH)42- (aq)
K
= 2.25 x 10-2
4B-7 (of 15)
The solubility product constant for zinc hydroxide is 4.5 x 10-17, and the
formation constant for tetrahydroxozincate(II) is 5.0 x 1014.
(b) Calculate molar solubility of zinc hydroxide in 0.10 M NaOH.
Zn(OH)2 (s) + 2OH- (aq) ⇆ Zn(OH)42- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
K =
[Zn(OH)42-]
______________
0.10
- 2x
0.10 - 2x
2.25 x 10-2 =
x
______________
(0.10 – 2x)2
[OH-]2
2.3 x 10-4 =
4B-8 (of 15)
0
+x
x
x = molar solubility of Zn(OH)2
BONDING IN COORDINATION COMPLEXES
Theories attempt to explain
(a) geometries (shapes)
(b) magnetism (paired or unpaired electrons)
(c) color (electronic energy level differences)
4B-9 (of 15)
CRYSTAL FIELD THEORY – Assumes ionic bonding between the ligands
and the metal
The ligand’s lone pairs affect the energies of the metal’s d orbitals
4B-10 (of 15)
Coordination Number of 6 : Octahedral
When 6 ligands surround a metal atom, they arrange octahedrally to
minimize repulsion (VSEPR theory)
4B-11 (of 15)
E
4B-12 (of 15)
dx2-y2 dz2
dxy dxz dyz
3d
E
SPLITTING ENERGY (Δo) – The energy difference between the d orbitals
in a ligand field
4B-13 (of 15)
With a transition metal that has 6 d electrons:
dx2-y2 dz2
↑↓ ↑↓
↑↓
dxy dxz dyz
E
LOW-SPIN COMPLEX – A complex with a large splitting energy, resulting
in electrons remaining in the lower energy d orbitals, and producing a
low number of unpaired electrons
Because of the large splitting energy, the d electrons are all paired in the
3 stable d orbitals, causing the complex to be diamagnetic
4B-14 (of 15)
With a transition metal that has 6 d electrons:
↑
↑↓ ↑
dx2-y2 dz2
↑
dx2-y2 dz2
↑
dxy dxz dyz
E
HIGH-SPIN COMPLEX – A complex with a small splitting energy, resulting
in electrons distributing into all of the d orbitals, and producing a high
number of unpaired electrons
Because of the small splitting energy, the d electrons remain unpaired as
long as possible, causing the complex to be paramagnetic
4B-15 (of 15)
SPLITTING ENERGY
dx2-y2 dz2
(Δo)
↑↓ ↑↓
↑↓
dxy dxz dyz
E
The splitting energy depends upon:
(1) The charge of the metal
The greater the charge of the metal ion, the larger the splitting energy
(2) The ligands attached to the metal
The ligands can be either strong-field ligands or weak field ligands
4C-1 (of 21)
STRONG-FIELD LIGANDS – Ligands that produce a strong electrostatic
field for the d orbitals, causing the splitting energy to be large
CN-, CO, and NO2- are strong-field ligands
:C
O:
2pz
..
2pz
sp
sp
2pz
sp
sp
sp
sp
2py
dxy AO
ππ
antibonding
MO
bonding MO
BACK BONDING – A coordinate covalent pi bond formed between a d
orbital of a metal and an empty antibonding orbital of a ligand
4C-2 (of 21)
dx2-y2 dz2
dxy dxz dyz
E
The increased stability of the dxy, dxz, and dyz increases the splitting energy
4C-3 (of 21)
WEAK-FIELD LIGANDS – Ligands that produce a weak electrostatic field
for the d orbitals, causing the splitting energy to be small
I-, Br-, Cl-, and F- are weak-field ligands
: Cl :
-
..
..
Both the d orbital of the metal and the p orbital of the ligand contain
electrons, and repel
4C-4 (of 21)
dx2-y2 dz2
dxy dxz dyz
E
The decreased stability of the dxy, dxz, and dyz reduces the splitting energy
4C-5 (of 21)
Hexafluorocobaltate(III) is found to be a paramagnetic complex
(a) Give the electron configuration of the cobalt ion
Co atom: [Ar]4s23d7
Co3+ ion: [Ar]3d6
(b) Identify the ligands as strong-field or weak-field
weak-field
(c) Draw the splitting pattern for the cobalt
↑
↑↓ ↑
↑
dx2-y2 dz2
↑
dxy dxz dyz
E
(d) Identify the complex as high-spin or low-spin
4C-6 (of 21)
high-spin
Hexacarbonyliron(II) is found to be a diamagnetic complex
(a) Give the electron configuration of the iron ion
Fe atom: [Ar]4s23d6
Fe2+ ion: [Ar]3d6
(b) Identify the ligands as strong-field or weak-field
strong-field
(c) Draw the splitting pattern for the iron
dx2-y2 dz2
↑↓
↑↓
↑↓
dxy dxz dyz
E
(d) Identify the complex as high-spin or low-spin
4C-7 (of 21)
low-spin
CoF63-
dx2-y2 dz2
dxy dxz dyz
↑
↑↓ ↑
Fe(CO)62+
dx2-y2 dz2
↑
↑
↑↓
↑↓ ↑↓
dxy dxz dyz
E
Complexes will absorb EM radiation to promote electrons from the lowenergy d orbitals to the high-energy d orbitals
E = hν
c = λν
c = ν
__
λ
E = hc
____
λ
If photons of visible light are absorbed, the complex will be colored
4C-8 (of 21)
CoF63↑
↑↓ ↑
Fe(CO)62+
↑
↑
↑↓
↑↓ ↑↓
E
CoF63- absorbs EM radiation with a wavelength of 6.5 x 10-7 m, while
Fe(CO)62+ absorbs EM radiation with a wavelength of 4.5 x 10-7 m.
4C-9 (of 21)
CoF63↑
Fe(CO)62+
↑
↑↓ ↑
↑
↑↓
↑↓ ↑↓
E
Calculate the splitting energy (Δo) of each
E = hc = (6.626 x 10-34 Js)(2.9979 x 108 ms-1)
____
_____________________________________________
λ
(6.5 x 10-7 m)
E = hc = (6.626 x 10-34 Js)(2.9979 x 108 ms-1)
____
= 3.1 x 10-19 J
_____________________________________________
λ
4C-10 (of 21)
(4.5 x 10-7 m)
= 4.4 x 10-19 J
CoF63↑
↑ ↑
↑↓
Fe(CO)62+
↑
↑
↑↓
↑↓ ↑↓
E
Because of the 3 stable d orbitals, this arrangement favors metal ions
with d3 or d6 electron configurations
d3 - Cr3+, Mn4+
d6 – Co3+, Fe2+
Metal ions with a d5 electron configuration are very stable as a high-spin
octahedral complex
d5 - Fe3+, Mn2+
4C-11 (of 21)
CRYSTAL FIELD THEORY FOR OTHER GEOMETRIES
(a) Coordination Number of 2 : Linear
Ligands pointing along the z-axis make the dz2 the most unstable
d orbitals on the xy plane will be the most stable
4C-12 (of 21)
d z2
dxz dyz
dxy dx2-y2
3d
E
With only 2 ligands, energies do not increase as much as with 6 ligands
 with 5 stable orbitals, this arrangement favors metal ions with d10
electron configurations
Ag+ - Ag(NH3)2+
4C-13 (of 21)
(b) Coordination Number of 4 : Square Planar
Ligands on the xy plane make the dx2-y2 the most unstable
The dxy will be the next most unstable
The dz2 will be the next most unstable because of the doughnut
4C-14 (of 21)
dx2-y2
3d
dxy
d z2
dxz dyz
E
With 1 very unstable orbital ( 4 stable orbitals), this arrangement favors
metal ions with d8 electron configurations
Pt2+ - PtCl42Au3+ - AuCl4-
4C-15 (of 21)
(c) Coordination Number of 4 : Tetrahedral
The dxz, dxz, and dyz point closest to the ligands
The dx2-y2 and dz2 will be the most stable
4C-16 (of 21)
dxy dxz dyz
dz2 dx2-y2
3d
E
This arrangement favors metal ions with a d7 electron configurations
Co2+ - CoCl42This arrangement also favors metal ions with a d4 electron configuration,
but it is not a stable arrangement – it is a strong reducing agent, producing
the more stable d3 electron configuration and an octahedral complex
Cr2+ → Cr3+ + e-
4C-17 (of 21)
LIGAND FIELD THEORY – Assumes coordinate covalent bonding
between the ligands and the metal using molecular orbital theory
Ligands bond to metal atoms with lone pairs
COORDINATE COVALENT BOND – A covalent bond in which the 2
shared electrons come from the same atom
:F:
:F
B
F:
: Br
N
Br :
: Br :
Coordinate covalent bonding is an example of a LEWIS ACID-BASE
REACTION
4C-18 (of 21)
LEWIS ACID – An electron pair acceptor
BF3
A substance with an incomplete outershell or empty valence orbitals
Na+, Mg2+, Al3+
LEWIS BASE – An electron pair donor
NBr3
A substance with a lone pair available for bonding
Cl-, S2-, NO2-, CO32-
4C-19 (of 21)
Fe(CN)64-
4C-20 (of 21)
hexacyanoferrate(II)
Fe2+: [Ar]3d6
FeCl64-
4C-21 (of 21)
hexachloroferrate(II)
Fe2+: [Ar]3d6