COORDINATION CHEMISTRY

COORDINATION COMPLEX – Any ion or neutral species with a metal atom or ion bonded to 2 or more molecules or ions

Zn(NH3)42+ Al(OH)4

-

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)

NH3 OH-

MONODENTATE LIGAND – A ligand with one lone pair that can form one bond to a metal ion

H2O, NH3, CN-, NO2-, SCN-, OH-, X-

Bidentate ligands:

oxalate (ox) – C2O42-

ethylenediamine (en) – H2N(CH2)2NH2

Polydentate ligands:

diethylenetriamine (dien) – H2N(CH2CH2)NH(CH2CH2)NH2

ethylenediaminetetraacetate (EDTA) – (O2CCH2)2N(CH2CH2)N(CH2CO2)24-

CHELATE – A ligand with more than one atom with a lone pair that can be used to bond to a metal ion

4A-2 (of 20)

PROPERTIES OF LIGANDS

COORDINATION NUMBER – The number of bonds formed between the metal ion and the ligands

Coordination Number

24

linearsquare planar or tetrahedral

6 octahedral

4A-3 (of 20)

Shape of Complex

Ag(NH3)2+

Zn(OH)42-

SnCl62-

Fe(C2O4)33- 6 octahedral

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

4A-4 (of 20)

1893 ALFRED WERNERIdentified 2 types of bonding that exists in coordination compounds

PRIMARY BONDS – Ionic bonds attracting the charged complex ion and the counter ions

[Cu(NH3)4]Cl2

SECONDARY BONDS – Covalent bonds attracting the metal of the complex and the ligands

Ionic bonding between the Cu(NH3)42+ and each Cl-

Covalent between the Cu2+ and each NH3

4A-5 (of 20)

4A-6 (of 20)

NOMENCLATURE FOR COMPLEX IONS

1 – Name the ligands before the metal ion

4 – Prefixes are used if a complex ion has more than one particular ligand

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

3 – Different ligands are named alphabetically

5 – Prefixes for polydentate ligands (or ligands whose names contain prefixes) are bis-, tris-, etc., with the ligand in parenthesis

6 – The charge of the metal is given as a roman numeral in parenthesis

Cl- – chloro F- – fluoro OH- – hydroxo CN- – cyano

7 – If the complex ion has a negative charge, the suffix –ate is added to the name of the metal

SO42- – sulfato NO3

- – nitrato NO2- – nitrito C2O4

2- – oxalato

Co(NH3)63+ hexaamminecobalt(III)

Co(NH3)5Cl2+ pentaamminechlorocobalt(III)

4A-7 (of 20)

CoCl63- hexachlorocobaltate(III)

Fe(CN)63- not hexancyanoironate(III)

Fe – ferrate Cu – cuprate Pb – plumbateSn – stannate Pt – platinate Mn - manganate

Fe(C2O4)33- tris(oxalato)ferrate(III)

Ni(CO)4 tetracarbonylnickel(0)

, its hexacyanoferrate(III)

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

4A-8 (of 20)

NOMENCLATURE FOR COMPOUNDS CONTAINING COMPLEX IONS

When naming coordination compounds, name the cation, then name the anion

tetraamminecopper(II) chloride

[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

4A-9 (of 20)

Mn(en)2Cl2

This is a neutral complex

dichlorobis(ethylenediamine)manganese(II)

K[PtNH3Cl5]

This is a 1- charged complex ion

potassium amminepentachloroplatinate(IV)

4A-10 (of 20)

[Fe(en)2(NO2)2]2SO4

This is a 1+ complex ion

bis(ethylenediamine)dinitritoiron(III) sulfate

K4Fe(CN)6

This is a 4- charged complex ion

potassium hexacyanoferrate(II)

ISOMERISMISOMERS – 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

4A-11 (of 20)

Structural isomers have different names

H H H H

H C C C C H

H H H H

C4H10 HH C HH H

H C C C H

H H H

4A-12 (of 20)

butane methyl propane

[Pt(H2O)4(OH)2]Cl2

Cl- Cl-

H2O

PtH2O

H2OH2O

OHOH

2+

OH- OH-

H2O

PtH2O

H2OH2O

ClCl

2+

tetraaquadihydroxoplatinum(IV) chloride

[Pt(H2O)4Cl2](OH)2 tetraaquadichloroplatinum(IV) hydroxide

4A-13 (of 20)

Pt(H2O)4(OH)2Cl2

(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

NH3NH3

NH3NH3

Cl

Co

NH3

NH3NH3

NH3Cl

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

NH3Cl

Pt

NH3 Cl

NH3Cl

4A-16 (of 20)

How many geometrical isomers are there for diamminedichloroplatinum(II) if it has tetrahedral geometry?

NH3

Pt Cl

Cl NH3

Cl’s all always adjacentonly one

4A-17 (of 20)

How many geometrical isomers are there for triamminetrichlorocobalt(III) if it has octahedral geometry?

Cl

Co

Cl

NH3NH3

NH3Cl

Cl

Co

NH3

NH3NH3

ClCl

all 3 are adjacent – fac

2 of the 3 are 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 This anteater is an 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 19)

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 19)

= 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 19)

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 19)

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’sShift M’sNew Initial M’sChange M’sEquilibrium 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 19)

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 19)

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)4

2- (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 19)

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 19)

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-

4B-9 (of 19)

(2) MAGNETISMSome are DIAMAGNETIC (no unpaired electrons), and some are PARAMAGNETIC (1 or more unpaired electrons)

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BONDING IN COORDINATION COMPLEXES

Theories attempt to explain(a) geometries (shapes)(b) color (electronic energy level differences)(c) magnetism (paired or unpaired electrons)

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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

E1s

2s

2px 2py 2pz

1st EL

2nd EL

4B-13 (of 19)

E

3s

3px 3py 3pz

3dxy 3dxz 3dyz 3dx2-y2

3rd EL

3dz2

4B-14 (of 19)

4B-15 (of 19)

Coordination Number of 6 : OctahedralWhen 6 ligands surround a metal atom, they arrange octahedrally to minimize repulsion (VSEPR theory)

E

4B-16 (of 19)

E3d

SPLITTING ENERGY (Δo) – The energy difference between the d orbitals in a ligand field

dxy dxz dyz

dx2-y2 dz2

4B-17 (of 19)

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-18 (of 19)

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-19 (of 19)

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

SPLITTING ENERGY

(Δo)

(2) The ligands attached to the metalThe ligands can be either strong-field ligands or weak field ligands

4C-1 (of 24)

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

sp

sp

2pz

sp 2py

2py

2pz

sp sp

2pz

2py

4C-2 (of 24)

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 24)

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 24)

.. ..

E

The decreased stability of the dxy, dxz, and dyz reduces the splitting energy

dxy dxz dyz

dx2-y2 dz2

4C-5 (of 24)

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 24)

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 24)

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 ____

λIf photons of visible light are absorbed, the complex will be colored

4C-8 (of 24)

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 24)

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 24)

CoF63- Fe(CO)6

2+

↑↓ ↑ ↑

↑ ↑

E↑↓ ↑↓ ↑↓

Because of the 3 stable d orbitals, this arrangement favors metal ions with d3 or d6 electron configurationsd3 - Cr3+, Mn4+ d6 – Co3+, Fe2+ Metal ions with a d5 electron configuration are very stable as a high-spin octahedral complexd5 - Fe3+, Mn2+

4C-11 (of 24)

↑

(a) Coordination Number of 2 : Linear

CRYSTAL FIELD THEORY FOR OTHER GEOMETRIES

Ligands pointing along the z-axis make the dz2 the most unstabled orbitals on the xy plane will be the most stable

4C-12 (of 24)

E3d

With the repulsion of 6 ligands, energies of the d orbitals are increased significantlyWith the repulsion of only 2 ligands, energies do not increase as much with 5 stable orbitals, this arrangement favors metal ions with d10

electron configurationsAg+ - Ag(NH3)2

+

dxy dx2-y2

dz2

dxz dyz

4C-13 (of 24)

dxy dxz dyz

dx2-y2 dz2

(b) Coordination Number of 4 : Square Planar

Ligands on the xy plane make the dx2-y2 the most unstableThe dxy will be the next most unstableThe dz2 will be the next most unstable because of the doughnut

4C-14 (of 24)

E3d

With 1 very unstable orbital ( 4 stable orbitals), this arrangement favors metal ions with d8 electron configurationsPt2+ - PtCl4

2-

Au3+ - AuCl4-

dxz dyz

dxy

dz2

dx2-y2

4C-15 (of 24)

(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 24)

E3d

This arrangement favors metal ions with a d7 electron configurationsCo2+ - CoCl4

2-

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 complexCr2+ → Cr3+ + e-

dz2 dx2-y2

dxy dxz dyz

4C-17 (of 24)

4C-18 (of 24)

LIGAND FIELD THEORY – Assumes coordinate covalent bonding between the ligands and the metal using molecular orbital theory

COORDINATE COVALENT BONDS – Covalent bond in which the 2 shared electrons come from the same atomFound between the ligand and the metal of the complex

4C-19 (of 24)

..

: F :

Coordinate covalent bonding is an example of a LEWIS ACID-BASE REACTION

LEWIS ACID – An electron pair acceptor

LEWIS BASE – An electron pair donor

: Br N Br :

: Br :

: F :: F B F :

Fe3+

Lewis acid – BF3

Lewis base – NBr3

Lewis acid – Fe3+

Lewis base – F-

(uses its electron pair to make a bond)

4C-20 (of 24)

-

A Lewis acid is a substance with an incomplete outershell or empty valence orbitals

A Lewis base is a substance with a lone pair available to make bonds

BF3, Na+, Mg2+, Al3+

NBr3, Cl-, S2-, NO2-, CO3

2-

4C-21 (of 24)

O C O . .H – O : H

+ →

HO

O C O H

Lewis acid Lewis base

4C-22 (of 24)

Fe(CN)64- hexacyanoferrate(II) Fe2+: [Ar]3d6

4C-23 (of 24)

FeCl64- hexachloroferrate(II) Fe2+: [Ar]3d6

4C-24 (of 24)