1. Transition Metals and Coordination Compounds
Chapter 24
Transition Metals and Coordination Compounds

2. Contents The Colors of Rubies and Emeralds
Properties of Transition Metals
Coordination Compounds
Structure and Isomerization
Bonding in Coordination Compounds
Applications of Coordination Compounds

3. The Colors of Rubies and Emeralds
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Rubies: About 1% of the Al3+ ions in Al2O3 are replaced by Cr3+.
Emeralds: About 1% of the Al3+ ions in Be3Al2(SiO3)6 are replaced by Cr3+.
Gemstones: 寶石
Rubies: 紅寶石
Emeralds: 綠寶石

4. mole compound (conductivity exp)
Werner’s Theory of Coordination Chemistry
Old formula
mole ions/
mole compound (conductivity exp)
mole AgCl ppt by AgNO3/
per mole cpd
Correct
Werner formula
CoCl3‧6NH3 4 3
[Co(NH3)6]Cl3
CoCl3‧5NH3 2
[Co(NH3)5Cl]Cl2
CoCl3‧4NH3 1
[Co(NH3)4Cl2]Cl
CoCl3‧3NH3
[Co(NH3)3Cl3]
The two compounds have very similar formulas but are very different in appearance because of the different chemical structures.

5. Properties of Transition Metals
General energy ordering of orbitals for multielectron Atoms: 5

6. First-Row Transition Metal Orbital Occupancy
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ns and (n - 1)d sublevels are close in energy
Cr is 4s13d5, associated with a half-filled stability
Cu is 4s13d10, associated with a completely stability

7. Properties of the First-Row Transition Metals

8. Electron configurations for transition metals
[noble gas]ns2(n-1)dx
[noble gas]ns2(n-2)f14(n-1)dx
x: 1~10
Electron configurations for transition metals’ ion
losing electrons from the ns orbital before losing electrons from the (n - 1)d orbitals.

9. Example 24.1
Write the ground state electron configuration for Zr.
Solution
[Kr] 5s24d2
Example 24.2
*****
Write the ground state electron configuration for Co3+.
Solution
For Co [Ar] 3d7 4s2
For Co3+ [Ar] 3d6

10. Atomic Size
The third transition series atoms are about the same size as the second because of the lanthanide contraction.
The atomic radii of all the transition metals are very similar.
Small increase in size down a column

11. Lanthanide contraction
The decrease in expected atomic size for the third transition series atoms that come after the lanthanides.
14 between the second and third series go into 4f orbitals.
Electrons in f orbitals are not as good at shielding the valence electrons.
The result is a greater effective nuclear charge increase and therefore a stronger pull on the valence electrons—the lanthanide contraction.

12. Ionization Energy
The first IE of the transition metals slowly increases across a series.
The first IE of the third transition series is generally higher than the first and second series
– lanthanide contraction

13. Electronegativity
The electronegativity of the transition metals slowly increases across a series. Except for last element in the series.
Electronegativity slightly increases between first and second series, but the third transition series atoms are about the same as the second. Trend opposite to main group elements

14. Oxidation States
Unlike main group metals, transition metals often exhibit multiple oxidation states.

15. Coordination Compounds
Terminology
A complex consists of a central atom, which is usually a metal atom or ion, and attached groups (anions or neutral molecule) called ligands.
If a complex carries a net electric charge, it is called a complex ion.
When a complex ion combines with counterions to make a neutral compound, it is called a coordination compound.
The total number of points at which a central atom or ion attaches ligands, called coordination number.

16. Bonding in Complex
Coordinate covalent bond (dative bond): The covalent bonding between two atoms in which both electrons come from only one of the atoms (of the ligand).
Central metal atom (ion): Lewis acid, electron pair acceptor
Ligand: Lewis base, electron pair donor
Monodentate : Ligands that donate only one electron pair to the central metal, for example: H2O NH3, Cl−.
Chelating agent (chelator): polydentate ligand, for example:
Ethylenediamine (en): bidentate
Oxalato (ox): bidentate
Ethylenediaminetetraacetato (EDTA): hexadentate
Chelate: A complex ion that contains either a bidentate or polydentate.

17. Common Ligands

18. Four Common Structures of Complex Ions

19. Bidentate Ligands Coordinated to Co3+

20. Hexadentate Ligands Coordinated to Co3+

21. Example
What are the coordination number and the oxidation number of the central atom in (a) [CoCl4(NH3)2]– and (b) [Ni(CO)4]?
Solution:
Co: center atom, 6 ligands attached (4 Cl– and 2 NH3)
coordination number: 6.
Charge calculation:
x – = –1, x = +3
central ion is Co3+, oxidation number is +3.
(b) coordination number: 4
central atom is Ni, oxidation number is 0.
Werner’s definition:
The primary valence is the oxidation number of the metal.
The secondary valence is the number of ligands bonded to the metal (coordination number).

22. Naming Coordination Compounds
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Identify the cation and anion, either may be complex ion or uncomplex ion.
Naming complex cation and/or complex anions.
Write the compound name as the name of cation followed by the name of the anion.

23. Names and Formulas of Common Ligands:
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24. Naming complex cations:
Ligand first, metal (with oxidation number written in Roman numerals) after.
Name the ligands in alphabetical order (ignoring Greek numeric prefixes).
Designate the number of ligands in a complex with a Greek numeric prefix: di = 2, tri = 3, tetra = 4, hexa = 6.

25. Example
Name [CrCl2(NH3)4]+
Solution:
a complex cation
4 NH3: tetraammine, 2 Cl–: dichloro
alphabetical order (ignoring Greek numeric prefixes):
tetraamminedichloro
complex cation: unmodified name for central metal.
Cr oxidation number (x – = +1)
Ans: tetraamminedichlorochromium(III) ion.

26. For metal in complex anion: Replace the ending from –um to –ate.
Naming complex anion:
Ligand first, metal (with oxidation number written in Roman numerals) after.
Name the ligands in alphabetical order (ignoring Greek numeric prefixes).
Designate the number of ligands in a complex with a Greek numeric prefix: di = 2, tri = 3, tetra = 4, hexa = 6.
For metal in complex anion:
Replace the ending from –um to –ate.
Certain metals in complex anions, use the Latin-based names: Copper Cuprate
Gold Aurate
Iron Ferrate
Lead Plumbate
Silver Argentate
Tin Stannate

27. Example 22.4
Name [PtBrCl2NH3]-
Solution:
a complex anion
1 NH3: ammine, 1 Br–: bromo, 2 Cl–: dichloro
alphabetical order: amminebromodichloro
complex anion ending –ate
Pt oxidation number: (x – 1 – = –1)
Ans: amminebromodichloroplatinate(II).

28. Examples of Naming Coordination Compounds *****
Name [Cr(H2O)5Cl]Cl2
Name K3[Fe(CN)6]
Identify the cation and anion, and the name of the uncomplex ion.
[Cr(H2O)5Cl]2+ is a complex cation;
Cl− is chloride.
K+ is potassium;
[Fe(CN)6]3− is a complex anion.
Give each ligand a name and list them in alphabetical order.
H2O is aqua;
Cl− is chloro.
CN− is cyano.
Name the metal ion.
Cr3+ is chromium(III).
Fe3+ is ferrate(III) because the complex ion is anionic.
Name the complex ion by adding prefixes to indicate the number of each ligand followed by the name of each ligand followed by the name of the metal ion.
[Cr(H2O)5Cl]2+ is pentaquochloro-chromium(III).
[Fe(CN)6]3− is hexacyanoferrate(III).
Name the compound by writing the name of the cation before the anion. The only space is between ion names.
[Cr(H2O)5Cl]Cl2 is pentaquochloro-chromium(III) chloride.
K3[Fe(CN)6] is potassium hexacyanoferrate(III).

29. When writing a complex formula from name:
Cation first, anion after
For the complex
Center metal first, ligands after
Place the ligands in alphabetical order (ignoring Greek numeric prefixes)

30. Example
*****
Write the formula for sodium hexanitrocobaltate(III).
Solution:
Coordination compound: made up of Na+ cations and a complex anion.
Complex anion charge: –3 (1 Co3+: +3, 6NO2–: –6)
Cation first, anion after:
Ans: Na3[Co(NO2)6].

31. 4. Structure and Isomerization
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32. Structural isomers
Coordination isomers: the structural isomers occur when coordinated ligand exchanges places with the uncoordinated counterion, for example:
[Co(NH3)5Br]Cl pentaamminebromocobalt(II) chloride
[Co(NH3)5Cl]Br pentaamminechlorocobalt(II) bromide
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33. ii) Linkage isomers: the structural isomers that have ligands attached to the central cation through different ends of the ligand structure, for example:
[Co(NH3)5(NO2)]Cl [Co(NH3)5(ONO)]Cl2

34. Continued
Yellow = pentaamminenitrocobalt(III) chloride
[Co(NH3)5(NO2)]Cl2
Red = pentaamminenitritocobalt(III) chloride
[Co(NH3)5(ONO)]Cl2

35. a) Cis-trans isomers (MA2B2 and MA4B2 type)
2) Stereoisomers
Geometric isomers
a) Cis-trans isomers (MA2B2 and MA4B2 type)
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36. b) Fac-mer isomers (MA3B3 type)

37. Example 24.5
Drawing Geometric Isomers of [Co(en)2Cl2]+.
Solution
MA4B2 type, cis–trans isomers.

38. Optical isomers (enantiomers):
Molecules that are nonsuperimposable (not identical) mirror images of one another, like right and left hands.
Each enantiomer rotates polarized light in opposite directions.
For example:
Enantiomer: 鏡像異構物、對掌體異構物
Mirror image of each other
Two nonsuperimposable (not identical) structures
Ans: two optical isomers

39. Example 24.7a Determine whether the cis isomer of [Co(en)2Cl2]+ is optically active.
Solution
the two structures are not superimposable, so the cis isomer does exhibit optical activity.

40. Example 24.7b Determine whether the trans isomer of [Co(en)2Cl2]+ is optically active.
Solution
In this case the two are identical, so there is no optical activity.

41. 5. Bonding in Coordination Compounds
Valence Bond Theory/Hybridization of Atomic Orbitals

42. electron configuration of Ag: 4d105s1, Ag+: 4d10 for [Ag(NH3)2]+
Continued
electron configuration of Ag: 4d105s1, Ag+: 4d10
for [Ag(NH3)2]+
sp hybridization 4d 5s 5p
electron configuration of Zn: 3d104s2 , Zn2+: 3d10
for [Zn(NH3)4]2+
sp3 hybridization 3d 4s 4p

43. electron configuration of Pd: 4d10, Pd2+: 4d8 for [PdCl4]2-
Continued
electron configuration of Pd: 4d10, Pd2+: 4d8
for [PdCl4]2-
dsp2 hybridization 4d 5s 5p
electron configuration of Fe: 3d64s2, Fe3+: 3d5
for [Fe(H2O)6]3+
d2sp3 hybridization 3d 4s 4p

44. Crystal Field Theory *****
Assume the attractions between a central atom (or ion) and its ligands are largely electrostatic.
Ligands distort the d-orbitals of the central atom, leading to a splitting of energy levels of those orbitals.
Splitting energy (Δ): The splitting of energy levels of d-orbitals which are caused by ligands distoration of those orbitals.
The spectrochemical series shows the relative abilities of ligands (Δ) to split the d-orbital energy levels:
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* Increases the charge on the metal cation also increases the splitting energy (Δ), for example:
Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+

45. Schematic representation of d-level splitting:
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Schematic representation of d-level splitting: Δ Δ
Usually
High Spin
Using
Spectroche-mical series
Usually
Low Spin

46. *****
Crystal field theory can predict:
Magnetism
Paramagnetic, with spin (unpaired electron)
Diamagnetic, without spin
Magnetic strength
High spin, more unpaired electron (Δ small)
Low spin, less unpaired electron, (Δ large)
Complex color: According to the energy level transition.

47. *****
Example: Predict the magnetism for [Fe(H2O)6]2+ and [Fe(CN)6]4-.
Solution:
Both are octahedral complexes
Fe: [Ar]3d64s2, Fe2+: [Ar]3d6 (six 3d electrons)
Δ: CN– (large) > H2O (small)
Diamagnetic
Low-spin complex
Paramagnetic
High-spin complex
* From d1 through d10 metal ion in octahedral complexes, only electron d4, d5, d6, or d7 can have low and high spin possibilities.

48. *****
Example
How many unpaired electrons would you expect for the octahedral complex ion [CoF6]3–?
Solution:
Electron configuration:
Co: [Ar]3d74s2
Co3+: [Ar]3d6 (six 3d electrons)
F– ligand: Δ small
Ans: 4 unpaired electrons

49. How many unpaired electrons would you expect for the octahedral complex ion [Co(NH3)5NO2]2+?
Solution:
Electron configuration:
Co: [Ar]3d74s2
Co3+: [Ar]3d6 (six 3d electrons)
NH3 and NO2-: Δ large
Ans: No unpaired electrons

50. Example
How many unpaired electrons would you expect to find in the tetrahedral complex ion [NiCl4]2–?
Solution:
Electron configuration:
Ni: [Ar]3d84s2
Ni2+: [Ar]3d8
(Tetrahedral, usually Δ small, high spin)
Ans: 2 unpaired electrons

51. *****
Example
How many unpaired electrons would you expect to find in the
square planar complex ion [PtCl4]2–?
Solution:
Electron configuration:
Pt: [Xe]5d96s1
Pt2+: [Xe]5d8
(Square planar, usually Δ large, low spin)
Ans: no unpaired electrons
a diamagnetic species

52. Color In Complex Ions And Coordination Compounds
Many complex ions are colored because the energy differences between d orbitals match the energies of components of visible light.
Crystal field theory helps to explain the colors of complex ions.
Ions having the following electron configurations have no electron transitions in the energy range of visible light (colorless):
No electron in d orbital (d0-complex), e.g., Sc3+, Y3+, La3+ (noble-gas electron configuration) are colorless.
Electrons completely filled in d orbital (d10-complex), e.g., Zn2+, Cd2+, Hg2+, Cu+, and Ag+ are colorless.

53. The color of the transmitted light is the complementary color of the absorbed light.
The Color Wheel: Colors across from one another on the color wheel are said to be complementary.

54. Continued
Large Δ
High ν
Short λ
Small Δ
Low ν
Long λ
The Color of [Ti(H2O)6]3+ solution
The absorption pectrum [Ti(H2O)6]3+ solution

55. Complex Ion Color and Crystal Field Strength
The colors of complex ions are due to electronic transitions between the split d sublevel orbitals.
The wavelength of maximum absorbance can be used to determine the size of the energy gap between the split d sublevel orbitals.
Ephoton = hn = hc/l = D

56. *****
Example 24.8
The complex ion [Cu(NH3)6]2+ is blue in aqueous solution. Estimate the crystal field splitting energy (in kJ/mol) for this ion.
Solution
The color orange ranges from 580 to 650 nm, so you can estimate the average wavelength as 615 nm. E = hc/λ.
Convert J/ion into kJ/mol.

57. 6. Applications of Coordination Compounds
Extraction of metals from ores
Silver and gold as cyanide complexes
Nickel as Ni(CO)4(g)
Use of chelating agents in heavy metal poisoning
EDTA for Pb poisoning
Chemical analysis
Qualitative analysis for metal ions
Blue = CoSCN+, Red = FeSCN2+
Ni2+ and Pd2+ form insoluble colored precipitates with dimethylglyoxime.

59. *****
In hemoglobin, the iron complex is octahedral, with the four nitrogen atoms of the porphyrin in a square planar arrangement around the metal. A nitrogen atom from a nearby amino acid of the protein occupies the fifth coordination site, and either O2 or H2O occupies the last coordination site.
Hemoglobin: 血紅素蛋白
Heme: 血紅素
Porphyrin: 紫質

60. End of Chapter 24