Review on Chemistry of Coordination Compounds Coordination Compounds Constitution [Co(NH 3 ) 5...

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Transcript of Review on Chemistry of Coordination Compounds Coordination Compounds Constitution [Co(NH 3 ) 5...

Review on Chemistry of Coordination Compounds

Coordination Compounds• Constitution

[Co(NH3)5Cl](NO3)2

Central atom Coordination number

Inner sphere

Coordination atom Outer sphere ions

ligand

H2O, NH3, Cl–, CN–, CO, SCN–, OH–

CO32-, NH2CH2CH2NH2 (ethylenediami

ne,

en), C2O42– (oxalate ion)

EDTA4 (ethylenediaminetetraacetate ion) is a hexadentate ligand.

Polydentate ligands are also known as chelating agents

Naming Coordination Compounds

• The name of a complex is one word, with no space between the ligand names and no space between the names of the last ligand and the metal.

Naming Coordination Compounds

• a salt, name the cation first • Name the ligands, in alphabetical ord

er, before the metal. Note: in anionic ligand endings from

-ide to -o , and -ate to -ato. -ite to -ito the names of the ligands differ slight

ly from their chemical namesin the chemical formula, the metal at

om or ion is written before the ligands

Naming Coordination Compounds• If the complex contains more than o

ne ligand of a particular typeUse Greek prefixes (di-, tri-, tetra-, et

c.), or bis- (2), tris-(3), tetrakis-(4), and so forth, and put the ligand name in parentheses for the later.

The ligands are listed in alphabetical order, and the prefixes are ignored in determining the order.

Naming Coordination Compounds• A Roman numeral in parentheses

follows the name of the metal to indicate the metal's oxidation state

• To name the metaluse the ending -ate if the metal is

in an anionic complex, or the Latin names for some

-ium ending for Cationic coordination sphere, or same as the element

Isomers 异构体

Compounds with the same formula but a different arrangement of atoms are called isomers.

Constitution isomers Stereoisomers

Linkage isomers

Ionization Isomer

Diastereoisomers

Enantiomers

Isom

ers

Co

NH3

NH3

NH3H3N

H3N

O

ON

Co

NH3

NH3H3N

H3N

NO O

NH3

nitro nitrito

Coordination Isomerism

[Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]

[Pt(NH3)4][PtCl6] and

[Pt(NH3)4Cl2][PtCl4]

[Pt(NH3)4][PtCl4] and

[Pt(NH3)4Cl2][PtCl4]

Aquo Isomer

CrCl3·6H2O

[Cr(H2O)5Cl]Cl2·H2O

[Cr(H2O)4Cl2]Cl·2H2O

[Cr(H2O)6]Cl3

Green

Green

Violet

[Co(NH3)5Br]SO4 contains Co–Br

bond, and the sulfate ion is free

[Co(NH3)5SO4]Br contains Co–sulfate

bond, and the bromide ion is free

Ionization isomers 电 离 异 构体

two kinds of stereoisomers:

diastereoisomers 非对映异构体 enantiomers 对映异构体

Stereoisomers 立 体 异 构体

cis trans isomer

M

X

X

X

L

L L

M

X

X

X

L

L

L

FacialFac-

MeridionalMeri-

M

O

N

N

O

O

O

•Optical Isomerism 光学异构现象enantiomerschiral 手性的 achiral 非手性的The [Co(en)3]3+ cation is chiral and exists in two nonidentical mirror-image forms. properties identical except fortheir reactions with other chiral substancestheir effect on plane-polarized light: Optical Activity

Co

NN

NN

N

N

Co

NN

NN

N

N

d: dextrorotatory 右旋l: levorotatory 左旋used to indicate the direction of rotation.

•racemic 外消旋 A 50:50 mixture of both isomers

produces no net optical rotation.

•The labels

Valence Bond Theory

Vacant metalhybrid atomic orbital

Coordinate covalent bond

Occupied ligandhybrid atomic orbital

•The hybrid orbitals used by the metal are determined by the geometry of the complex.

•The number of d electrons in the metal is determined by the oxidation state of the metal ion.

•The orbitals used to construct the hybrid orbitals for bonding must be vacant on the metal.

Valence Bond Theory

•For octahedral complexes it may be necessary to pair some electrons already in d orbitals to get vacant orbitals required for hybridization. This leads to a low spin complex .

•Contrast this with the use of higher energy vacant d orbitals. This leads to more unpaired d electrons and a high spin complex.

Valence Bond Theory

•Knowing whether a complex is paramagnetic or diamagnetic can help determine which d orbitals to use. It can also help determine whether a complex is square planar or tetrahedral

Valence Bond Theory

Spectrochemical series 光化学序列

Increasing    →

Weak field ligands Strong field ligands

I– <   Br– <   Cl– <   F– < H2O     < NH3  <  en < CN–

Hybrid Orbitals

• A unsuccessful example:Cu(NH3)4

2+

A square planar complexHybrid form: dsp2

Cu2+ [Ar]

dsp2

•This explains the color and magnetic properties of the transition metal complexes.

•Bonding in complexes is viewed as entirely ionic and as arising from electrostatic interactions between the d electrons of the metal and the ligand electrons.

Crystal Field Theory

Crystal Field Theory• It considers the effect of the liga

nd charges on the energies of the metal ion d orbitals.

•The d orbitals are raised in energy and are separated in energy based on the geometry of the complex.

•The energy separation is called the crystal field splitting, represented by the symbol Δ.

• In octahedral complexes the dx2 - y2 and the dz2 orbitals are higher tin energy than the dxy, the dxz, and the dyz because the negative charge of the electrons from the ligands point directly at the negative charges of the electrons in the d orbitals that lie on the x,y, and z axes.

Crystal Field Theory

•The color of the complexes is due to electronic transitions from one set of d orbitals to another.

•Visible light can supply enough energy to promote an electron from the lower enegy to the higher energy orbitals.

•Light at a particular wavelength is absorbed and the complementary color is seen.

Crystal Field Theory

The angular distribution of d orbitals

Crystal Field Splitting

o = 10 Dq

Spheric field

Free atom

+6 Dq

4 Dq

o

eg

t2g

dxy, dyz, dxz

dx2-y2, dz2

Octahedral field

CFSE: Crystal Field Stabilization Energy 晶体场稳定化能

Pairing Energy 成对能 P

Tetrahedral Field

Splitting in Tetrahedral Field

t = 4/9o = 10 Dq

Spheric field

Free atom

t

e

t2

dxy, dyz, dxz

dx2-y2, dz2

Tetrahedral field

Color• Complementary colors:

• R-G

• O-B

• Y-V

•An absorbance spectrum It plots the absorbance (amount of light absorbed by a substance) as a function of wavelength

[Ti(H2O)6]3+

Spectrochemical series 光化学序列

Increasing    →

Weak field ligands Strong field ligands

I– <   Br– <   Cl– <   F– < H2O     < NH3  <  en < CN–

Back-bonding 反馈键

Magnetism

Paramagnetic: unpaired electronDiamagnetic: no unpaired electron

[Co(NH3)6]3+:

Co3+: d6

diamagnetic

t2g4 eg

2paramagnetic

t2g6 eg

0

[CoF6]3

Oh t2g eg

Jahn-Teller EffectsFor a non-linear molecule that is in an electronically degenerate state, distortion must occur to lower the symmetry, remove the degeneracy, and lower the energy.Jahn-Teller effects do not predict which distortion will occur other than that the center of symmetry will remain.The distortion by the unsymmetrical distribution of electrons in eg orbital is stronger than that of t2g.

Stability Constant

Cu2+ + 4NH3 [Cu(NH3)4]2+

Overall Stability ConstantStepwise stability constants: K1, K2, …, K4

Overall stability constants: 1, 2, 3, 4

Kstability = [Cu(NH3)4

2+ ][Cu2+ ] [NH3]4

1 = K1

2 = K1 K2

3 = K1 K2 K3

4 = K1 K2 K3K4

Factors That Determine The Stability Of Coordination Compounds

1. metal ions: charge, radius and electronic configuration;2. ligand: basicity, chelate effects.

Chelate Effects 螯合效应 : entropy effect.

Entropy-driven reaction (process)

[Cu(H2O)4]2+ (aq) + 2 NH3 (aq) → [Cu(NH3)2(H2O)2]2+ (aq)lg2 = 7.65

[Cu(H2O)4]2+ (aq) + en (aq) → [Cu(en)(H2O)2]2+ (aq)lg1 = 10.64

Example 1

Solution:

Ksp = [Ag+][Cl] = 1.6 × 10

Step 2:

Step 1:

Ag+ + 2NH3 = Ag(NH3)2+

AgCl(s) = Ag+ + Cl

2 =[Ag(NH3)2

+]

[Ag+][NH3]2= 1.5 × 10

Calculate the molar solubility of AgCl in a 6 M NH3 solution

Example 1 (continue)

AgCl(s) + 2NH3 = Ag(NH3)2++ Cl

Overall reaction

K =[Ag(NH3)2

+] [Cl]

[NH3]2

= (1.6 × 10)= Ksp2(1.5 × 10)

= 2.4 × 10

AgCl(s) + 2NH3 = Ag(NH3)2++ Cl

Initial (M): 6.0 0.0 0.0Change (M): -2s +s +sEqui (M): 6 - 2s s s

K = (s)(s)(6 – 2s)

s = 0.016 M

Step 3

0.1

0.045

Example 2

Calculate the equilibrium concentration of every relevant species of a 0.1 M Ag(NH3)2

+ solution.

K1 = 2.2 × 10=

K2 = 5.1 × 10=

[Ag(NH3)+]

[Ag+][NH3]

[Ag(NH3)2+]

[Ag(NH3)+][NH3]

2 =[Ag(NH3)2

+]

[Ag+][NH3]2= 1.5 × 10

With only a few exceptions, there is generally a slowly descending progression in the values of the Ki’s in any particular system.