COORDINATION CHEMISTRY COORDINATION COMPLEX – Any ion or neutral species with a metal atom or ion bonded to 2 or more molecules or ions Ag(NH 3 ) 2 + Cu(NH 3 ) 4 2+ Zn(NH 3 ) 4 2+ Al(OH) 4 - Zn(OH) 4 2- 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)
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COORDINATION CHEMISTRY COORDINATION COMPLEX – Any ion or neutral species with a metal atom or ion bonded to 2 or more molecules or ions Ag(NH 3 ) 2 + Cu(NH.
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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)4
2+ 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)
Color depends upon the chemical groups attached to the transition metal
(1) COLOR
Co(H2O)63+Co(CN)6
3- Co(CO3)33-Co(NO2)6
3-
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)
COORDINATION NUMBER – The number of bonds formed between the metal ion and the ligands
Coordination Number Geometry
2
4
linear
square planar or tetrahedral
6 octahedral
4A-4 (of 20)
(3) GEOMETRY
MONODENTATE LIGAND – A ligand with one lone pair that can form one bond to a metal ion
(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)
opposite – trans
adjacent – cis
Cl
Co
Cl
NH3
NH3
NH3
NH3
Cl
Co
NH3
NH3
NH3
NH3
Cl
trans-tetraamminedichlorocobalt(II)
cis-tetraamminedichlorocobalt(II)
4A-15 (of 20)
How many geometrical isomers are there for diamminedichloroplatinum(II) if it has square planar geometry?
trans-diamminedichloroplatinum(II)
cis-diamminedichloroplatinum(II)
Pt
ClNH3
NH3
Cl
Pt
NH3
ClNH3
Cl
4A-16 (of 20)
How many geometrical isomers are there for diamminedichloroplatinum(II) if it has tetrahedral geometry?
NH3
Pt
ClCl NH3
Cl’s all always adjacent
only one
4A-17 (of 20)
How many geometrical isomers are there for triamminetrichlorocobalt(III) if it has octahedral geometry?
Cl
Co
Cl
NH3
NH3
NH3
Cl
Cl
Co
NH3
NH3
NH3
ClCl
each pair adjacent – fac
adjacent and opposite – mer
fac-triamminetrichlorocobalt(III)
mer-triamminetrichlorocobalt(III)
4A-18 (of 20)
O
Fe
O
OO
OO
(b) OPTICAL ISOMERS – Spatial isomers that ARE mirror images of each other, and they are nonsuperimposable
tris(oxalato)ferrate(III)
O
Fe
O
OO
OO
O
Fe
O
OO
OO
180º
These are nonsuperimposable
molecules
the compound tris(oxalato)ferrate(III) has 2 optical isomers
4A-19 (of 20)
Optical isomers are called ENANTIOMERS
anteater enantiomer
4A-20 (of 20)
Ag+ (aq) + 2NH3 (aq) → Ag(NH3)2+ (aq)
FORMATION CONSTANT (Kf) – The equilibrium constant for the complete formation of a complex ion
Kf = [Ag(NH3)2+]
_______________
[Ag+][NH3]2
4B-1 (of 15)
diamminesilver(I)
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)
= 0.100 M __________________________
(0.250 M)(1.00 x105)
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].
Co2+ (aq) + 6NH3 (aq) ⇆ Co(NH3)62+ (aq)
Kf = [Co(NH3)62+]
________________
[Co2+][NH3]6
[NH3]6 = [Co(NH3)62+]
________________
[Co2+]Kf
[NH3] = [Co(NH3)62+]
________________
[Co2+]Kf
6 = 0.126 M
4B-3 (of 15)
Write the reaction for the complete formation of hexaamminecobalt(II), and its Kf expression
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’sChange in M’sEquilibrium M’s
0.100 0.500 0
- x - 2x + x
0.100 - x x0.500 - 2x
4B-4 (of 15)
The reaction is going in the forward direction and has a large equilibrium constant, x will be a large number
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
0.100 0.500 0
- 0.100 - 0.200 + 0.100
0 0.1000.300
+ x + 2x - x
x 0.100 - x0.300 + 2x
Kf = [Ag(NH3)2+]
________________
[Ag+][NH3]2
1.00 x 106 = (0.100 – x)
___________________
(x)(0.300 + 2x)2
1.00 x 106 = (0.100)
____________
(x)(0.300)2
x = 1.11 x 10-6 = [Ag+]
4B-5 (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.
(a) Calculate molar solubility of zinc hydroxide in pure water.
x 2x
Zn(OH)2 (s) ⇆ Zn2+ (aq) + 2OH- (aq)
Initial M’sChange in M’sEquilibrium M’s
0 0
+ x + 2x
Ksp = [Zn2+][OH-]2 = (x)(2x)2
= 4x3
x = molar solubility of Zn(OH)2
4.5 x 10-17 = 4x3
= molar solubility of Zn(OH)2
2.2 x 10-6 M = x
4B-6 (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) ⇆ Zn2+ (aq) + 2OH- (aq)
Zn2+ (aq) + 4OH- (aq) ⇆ Zn(OH)42- (aq)
Zn(OH)2 (s) + 2OH- (aq) ⇆ Zn(OH)4
2- (aq)
Ksp = 4.5 x 10-17
Kf = 5.0 x 1014
K = 2.25 x 10-2
4B-7 (of 15)
Zn(OH)2 (s) ⇆ Zn2+ (aq) + 2OH- (aq)
Initial M’sChange in M’sEquilibrium M’s
0 0.10
No, because Zn2+ forms a complex ion with OH-
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)4
2- (aq)
Initial M’sChange in M’sEquilibrium M’s 0.10 - 2x x
0.10 0
- 2x + x
2.25 x 10-2 = x ______________
(0.10 – 2x)2
2.3 x 10-4 = x
K = [Zn(OH)42-]
______________
[OH-]2
= molar solubility of Zn(OH)2
4B-8 (of 15)
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)
E
3d
SPLITTING ENERGY (Δo) – The energy difference between the d orbitals in a ligand field
dxy dxz dyz
dx2-y2 dz2
4B-13 (of 15)
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
dxy dxz dyz
dx2-y2 dz2
4B-14 (of 15)
With a transition metal that has 6 d electrons:
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
↑ ↑
dxy dxz dyz
dx2-y2 dz2
4B-15 (of 15)
With a transition metal that has 6 d electrons:
dx2-y2 dz2
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
dxy dxz dyz
dx2-y2 dz2
4C-1 (of 21)
SPLITTING ENERGY
(Δo)
(2) The ligands attached to the metal
The ligands can be either strong-field ligands or weak field ligands
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:
BACK BONDING – A coordinate covalent pi bond formed between a d orbital of a metal and an empty antibonding orbital of a ligand
sp spsp
2p z
sp2p y
2py
2pz
sp sp
2p z
2p y
4C-2 (of 21)
π bonding MOπ antibonding MO
..
dxy AO
E
The increased stability of the dxy, dxz, and dyz increases the splitting energy
dxy dxz dyz
dx2-y2 dz2
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)
.. ..
E
The decreased stability of the dxy, dxz, and dyz reduces the splitting energy
dxy dxz dyz
dx2-y2 dz2
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
(c) Draw the splitting pattern for the cobalt
E
↑↓ ↑ ↑
↑ ↑
(b) Identify the ligands as strong-field or weak-field
(d) Identify the complex as high-spin or low-spin
weak-field
high-spin
4C-6 (of 21)
dxy dxz dyz
dx2-y2 dz2
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
(c) Draw the splitting pattern for the iron
E
↑↓
(b) Identify the ligands as strong-field or weak-field
(d) Identify the complex as high-spin or low-spin
strong-field
low-spin
↑↓ ↑↓
4C-7 (of 21)
dxy dxz dyz
dx2-y2 dz2
CoF63- Fe(CO)6
2+
↑↓ ↑ ↑
↑ ↑
E
↑↓ ↑↓ ↑↓
Complexes will absorb EM radiation to promote electrons from the low-energy d orbitals to the high-energy d orbitals
c = λν c = ν __
λ
E = hν E = hc ____
λ
If photons of visible light are absorbed, the complex will be colored
4C-8 (of 21)
dxy dxz dyz
dx2-y2 dz2
dxy dxz dyz
dx2-y2 dz2
CoF63- Fe(CO)6
2+
↑↓ ↑ ↑
↑ ↑
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)6
2+
↑↓ ↑ ↑
↑ ↑
E
↑↓ ↑↓ ↑↓
Calculate the splitting energy (Δo) of each
E = hc ____
λ
= (6.626 x 10-34 Js)(2.9979 x 108 ms-1) _____________________________________________
(6.5 x 10-7 m)
= 3.1 x 10-19 J
= (6.626 x 10-34 Js)(2.9979 x 108 ms-1) _____________________________________________
(4.5 x 10-7 m)
= 4.4 x 10-19 JE = hc ____
λ
4C-10 (of 21)
CoF63- Fe(CO)6
2+
↑↓ ↑ ↑
↑ ↑
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)
↑
(a) Coordination Number of 2 : Linear
CRYSTAL FIELD THEORY FOR OTHER GEOMETRIES
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)
E
3d
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+
dxy dx2-y2
dz2
dxz dyz
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)
E
3d
With 1 very unstable orbital ( 4 stable orbitals), this arrangement favors metal ions with d8 electron configurations
Pt2+ - PtCl42-
Au3+ - AuCl4-
dxz dyz
dxy
dz2
dx2-y2
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)
E
3d
This arrangement favors metal ions with a d7 electron configurations
Co2+ - CoCl42-
This 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-
dz2 dx2-y2
dxy dxz dyz
4C-17 (of 21)
: Br N Br :
: Br :
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 :
Coordinate covalent bonding is an example of a LEWIS ACID-BASE REACTION
4C-18 (of 21)
LIGAND FIELD THEORY – Assumes coordinate covalent bonding between the ligands and the metal using molecular orbital theory
LEWIS ACID – An electron pair acceptor
LEWIS BASE – An electron pair donor
BF3
NBr3
4C-19 (of 21)
A substance with an incomplete outershell or empty valence orbitals
A substance with a lone pair available for bonding