1. Transition Elements and Coordination Chemistry

2. Chem 1A Review Why Study Coordinated Complexes of Transition metals?
These compounds are used as catalyst in oxidation of organic
compounds and pharmaceutical applications.
Chem 1A Review
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3. Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
< 4f< 5d< 6p< 7s< 5f< 6d< 7p
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4. Chapter 20

5. Electron Configuration and the Periodic Table
ns2np6
Electron Configuration and the Periodic Table
ns1
ns2np1
ns2np2
ns2np3
ns2np4
ns2np5
ns2
d10 d1 d5 4f 5f
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6. Using periodic table write Noble gas notation
for the following elements:
[Ne]3s23p4
a)S
[Ar] 4s23d6
b)Fe
[Ar] 4s23d104p4
c)Se
[Xe]6s24f75d1
d)Gd
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
< 4f< 5d< 6p< 7s< 5f< 6d< 7p

7. Chapter 20

8. Electron Configurations
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9. Why Study Coordinated Complexes of Transition metals?
Are used as catalyst in oxidation of organic compounds.
Medicinal applications.
“Cisplatin” - a cancer chemotherapy agent
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10. Coordination Compounds
Pt(NH3)2Cl2
“Cisplatin” - a cancer chemotherapy agent
Co(H2O)62+
Cu(NH3)42+
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11. Coordination Compounds of Ni2+
[Ni(NH3)6]+2
[Ni(NH2C2H4NH2)3]+2
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12. Electron Configurations
Sc (Z = 21): [Ar] 3d1 4s2
Zn (Z = 30): [Ar] 3d10 4s2
Remember that it’s common to list the s electrons last since they are removed first when forming ions.
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13. Coordination Compounds
Omit
Coordination Compounds
Many coordination compound consists of a complex ion.
A complex ion contains a central metal cation bonded to one or more molecules or ions.
The molecules or ions that surround the metal in a complex ion are called ligands.
A ligand has at least one unshared pair of valence electrons • H N H O • • Cl - • C O
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14. Coordination Compounds
The atom in a ligand that is bound directly to the metal atom is the donor atom. • H N H O •
Ligands with:
one donor atom
monodentate
H2O, NH3, Cl-
two donor atoms
bidentate
ethylenediamine
three or more donor atoms
polydentate
EDTA
The number of donor atoms surrounding the central metal atom in a complex ion is the coordination number.
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15. Coordination Compounds
Coordination Number: The number of ligand donor atoms that surround a central metal ion or atom.
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16. Ligands 02
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17. Ligands 03
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18. Poly DentateLigands
EDTA4– is often used to treat heavy metal poisoning such as Hg2+, Pb2+, and Cd2+.
EDTA4– bonds to Pb2+, which is excreted by the kidneys as [Pb(EDTA)]2–.
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19. Coordination Compounds
H2N CH2 CH NH2 •
bidentate ligand
polydentate ligand
(EDTA)
Bidentate and polydentate ligands are called chelating agents
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20. What is the Oxidation Numbers of Cu?
Naming coordinated complex compounds
What is the Oxidation Numbers of Cu? +2
Knowing the charge on a complex ion and the charge on each ligand, one can determine the oxidation number for the metal.
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21. Oxidation Number Rules
Applies to
Statement 1
Elements
The oxidation number of an atom in an element is zero. 2
Monatomic ions
The oxidation number of an atom in a monatomic ion equals the charge of the ion. 3
Oxygen
The oxidation number of oxygen is –2 in most of its compounds. (An exception is O in H2O2 and other peroxides, where the oxidation number is –1.)
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22. Oxidation Number Rules
Applies to
Statement 4
Hydrogen
+1, it will be -1 when hydrogen comes with metal. NaH 5
Halogens
Fluorine is –1 in all its compounds. Each of the other halogens is –1 in binary compounds unless the other element is oxygen. 6
Compounds and ions
The sum of the oxidation numbers of the atoms in a compound is zero. The sum in a polyatomic ion equals the charge on the ion.
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23. What is the charge on the following Complex, If the Oxidation number of Cr is +3?
Or, knowing the oxidation number on the metal and the charges on the ligands, one can calculate the charge on the complex ion.
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24. What are the oxidation numbers of the metals in K[Au(OH)4] and [Cr(NH3)6](NO3)3 ?
OH- has charge of -1
K+ has charge of +1
? Au x(-1) = 0
NO3- has charge of -1
Au = +3
NH3 has no charge
? Cr + 6x(0) + 3x(-1) = 0
Cr = +3
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25. Oxidation States of the 1st Row Transition Metals
(most stable oxidation numbers are shown in red)
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26. Learning Check
A complex ion contains a Cr3+ bound to four H2O molecules and two Cl– ions. Write its formula. +1
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27. N [Ag(NH3)2]2 SO4 Coordination Sphere Coordinate bond: N
Coordination Sphere: is the central metal and surrounding ligands. The square brackets separate the complex from counter ions such as SO42–. N N N
[Ag(NH3)2]2 SO4
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28. Geometry of Coordination Compounds
Coordination number
Structure 2
Linear 4
Tetrahedral or Square
planar (mostly d8 ) 6
Octahedral
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29. Coordination Number of 7&8
Geometry
Pentagonal bipyramid
Hexagonal bipyramid
Coordination Number of 7
Coordination Number of 8
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30. Coordination Compounds
Geometries:
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31. Nomenclature Co(H2O)62+ Hexaaquacobalt(II) H2O as a ligand is aqua
Cu(NH3)42+
Tetraamminecopper(II)
Pt(NH3)2Cl2
diamminedichloroplatinum(II)
NH3 as a ligand is ammine
Systematic naming specifies the type and number of ligands,
the metal, and its oxidation state.
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32. Ligand’s Names 01
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33. Chapter 20

34. Nomenclature
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35. Nomenclature Systematic naming follows IUPAC rules:
If compound is a salt, name cation first and then the anion, just as in naming simple salts.
In naming a complex ion or neutral complex, name ligands first and then the metal.
If the complex contains more than one ligand of a particular type, indicate the number with the appropriate Greek prefix: di–, tri–, tetra–, penta–, hexa–.
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36. Nomenclature
If the name of a ligand itself contains a Greek prefix, (ethylenediamine or triphenylphosphine) put the ligand name in parentheses and use: bis (2), tris (3), or tetrakis (4).
Use a Roman numeral in parentheses, immediately following the name of the metal, to indicate the metal’s oxidation state.
In naming the metal, use the ending –ate if metal is in an anionic complex.
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37. Name the following Complexes:
Tris(ethylenediamine)nickel(II)
Pt(
[Ni(NH2C2H4NH2)3]2+
IrCl(CO)(PPh3)2
Carbonylchlorobis(triphenylphosphine)iridium(I)
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38. What is the systematic name of [Cr(H2O)4Cl2]Cl ?
tetraaquadichlorochromium(III) chloride
Write the formula of
tris(ethylenediamine)cobalt(II) sulfate
[Co(en)3]SO4
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39. Constitutional Isomerism
1. Constitutional Isomers: Have different bonds among their constituent atoms.
Ionization Isomers :
[Co(NH3)5Br]SO4 (violet compound with Co–Br bond), [Co(NH3)5 SO4]Br (red compound with Co–SO4 bond).
Linkage Isomers form when a ligand can bond through two different donor atoms. Consider [Co(NH3)5NO2]2+ which is yellow with the Co–NO2 bond and red with the Co–ONO bond.
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40. Such a transformation could be used as an energy storage device.
Linkage Isomerism
sunlight
Such a transformation could be used as an energy storage device.
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41. 2.Stereoisomers
Geometric Isomers of Pt(NH3)2Cl2: In the cis isomer, atoms are on the same side. In the trans isomer, atoms are on opposite sides.
DNA-damaging antitumor agents
Inactive

42. 2.Stereoisomers
Geometric Isomers have the same connections among atoms but different spatial orientations of the metal–ligand bonds.
cis isomers have identical ligands in adjacent corners of a square.
trans isomers have identical ligands across the corners from each other.
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43. Isomers
Geometric Isomers of [Co(NH3)4Cl2]Cl:
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44. Enantiomers
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45. Enantiomers Enantiomers are stereoisomers of molecules or ions
that are nonidentical mirror images of each other.
Objects that have “handedness” are said to be chiral,
and objects that lack “handedness” are said to be achiral.
An object or compound is achiral if it has a symmetry plane cutting through the middle.
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46. Enantiomers
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47. © 2003 John Wiley and Sons Publishers
Unpolarized light.
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48. Plane-polarized light.
© 2003 John Wiley and Sons Publishers
Plane-polarized light.
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49. Reflected glare is plane-polarized light.
© 2003 John Wiley and Sons Publishers
Reflected glare is plane-polarized light.
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50. Polarizing sunglasses versus glare.
© 2003 John Wiley and Sons Publishers
Polarizing sunglasses versus glare.
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51. The effect of polarizing lenses on unpolarized light.
© 2003 John Wiley and Sons Publishers
Courtesy Andy Washnik
The effect of polarizing lenses on unpolarized light.
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52. plane-polarized light
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53. Enantiomers and Molecular Handedness
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54. Enantiomers Enantiomers have identical properties except for their
reaction with other chiral substances and their effect
on plane-polarized light.
Enantiomers are often called optical isomers; their
effect on plane-polarized light can be measured with
a polarimeter.
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55. Enantiomers
Plane-polarized light is obtained by passing ordinary light through a polarizing filter.
In a polarimeter the plane-polarized light is passed through a chiral solution and the polarization plane measured with an analyzing filter.
If the plane rotates to the right it is dextrorotatory.
If the plane rotates to the left it is levorotatory.
Equal amounts of each are racemic.
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56. Isomers 01
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57. sunlight See next slide for Diastereoisomers [Co(NH3)5Br]SO4 (violet),
[Co(NH3)5 SO4]Br (red ).
sunlight
See next slide for Diastereoisomers

58. Diasteromers
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59. Bonding in Complexes 01
Bonding Theories attempt to account for the color and magnetic properties of transition metal complexes.
Ni Cu2+
Co2+
Zn2+
Solutions of [Ni(H2O)6]2+, [Ni(NH3)6]2+, & [Ni(en)3]2+
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60. Color of Transition Metal Complexes
DE = E2 - E1 = hn = l hc or
Using c = ln. DE hc
l =
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61. Color of Transition Metal Complexes
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62. Color of Transition Metal Complexes
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63. Bonding in Complexes: Valence Bond Theory
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64. Bonding in Complexes: Which empty orbital is metal using for bonding S, p, d or f ?
The complex is paramagnetic since there are 1 or more unpaired electrons.
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65. Hybridization and sp3 Hybrid Orbitals
How can the bonding in CH4 be explained?
4 valence electrons
2 unpaired electrons
How can carbon have 4 bonds if there are only 2 unpaired valence electrons?
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66. Hybridization and sp3 Hybrid Orbitals
How can the bonding in CH4 be explained?
4 valence electrons
4 unpaired electrons
Promotion of 1 electron allows for 4 unpaired valence electrons.
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67. Hybridization and sp3 Hybrid Orbitals
How can the bonding in CH4 be explained?
4 nonequivalent orbitals
The 4 bonds in CH4 must be equivalent (tetrahedral distribution).
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68. Hybridization and sp3 Hybrid Orbitals
How can the bonding in CH4 be explained?
4 equivalent orbitals
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69. Hybridization and sp3 Hybrid Orbitals
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70. Hybridization and sp3 Hybrid Orbitals
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71. Other Kinds of Hybrid Orbitals
Summary.
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72. Hybrid Orbitals in Coordinated Complexes
We should look at magnetic property of the complex
to see if they are high or low spin. Then we could
decide whether they are using d2sp3 or sp3d2 hybrid

73. The octahedral d2sp3 and sp3d2
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74. Square Planar geometry of four dsp2
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75. Bonding in Complexes: Valence Bond Theory
The complex is paramagnetic since there are 1 or more unpaired electrons.
Experimental results: High Spin
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76. Bonding in Complexes: Valence Bond Theory
The complex is diamagnetic since there are 1 or more unpaired electrons.
Experimental results: Low Spin
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77. High- and Low-Spin Complexes
Experimental results : High Spin
[CoF6]3-
[Co(CN)6]3-
High spin: Maxium number of unpaired electron, Paramagnetic
Low spin: Minimum number of unpaired electron
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78. Crystal Field Theory Crystal Field Theory:
Effect of charges of ligand on transition metal
d-electrons.
A model that views the bonding in complexes as
arising from electrostatic interactions and considers
the effect of the ligand charges on the energies of
the metal ion d orbitals.
Only electrostatic interactions- no covalent bonds, no shared electrons, and no hybrid orbitals.
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79. Directed between ligands
Crystal Field Theory
Octahedral Complexes
Directed between ligands
Directed at ligands
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80. Crystal Field Theory
Octahedral Complexes
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81. [Ni(X)6]2+ X=H2O, NH3, and ethylenediamine (en)
Crystal Field Theory
Octahedral Complexes
[Ni(X)6]2+ X=H2O, NH3, and ethylenediamine (en)
[Ti(H2O)6]3+ see p845 fig 20.25
[Ni(H2O6]2+ see p844 fig 20.24
(red-violet)
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82. [Ti(H2O)6]3+
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83. Crystal Field Theory Octahedral Complexes
The crystal field splitting changes depending on nature of the legand.
[Ni(X)6]2+ X=H2O, NH3, and ethylenediamine (en)
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84. The absorption maximum for the complex ion [Co(NH3)6]3+ occurs at 470 nm. What is the color of the complex and what is the crystal field splitting in kJ/mol? hc l =
DE = hn
= 4.23 x J
DE (kJ/mol) ?
= 255 kJ/mol
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