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Chapter 24
Transition Metals and
Coordination Compounds
TMHsiung ©2015
Chapter 24
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Contents
1.
2.
3.
4.
5.
6.
The Colors of Rubies and Emeralds
Properties of Transition Metals
Coordination Compounds
Structure and Isomerization
Bonding in Coordination Compounds
Applications of Coordination Compounds
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1.
The Colors of Rubies and Emeralds
*****
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+.
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 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
3
2
[Co(NH3)5Cl]Cl2
CoCl3‧4NH3
2
1
[Co(NH3)4Cl2]Cl
CoCl3‧3NH3
0
0
[Co(NH3)3Cl3]
The two compounds have very
similar formulas but are very
different in appearance because of
the different chemical structures.
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2. Properties of Transition Metals
 General energy ordering of orbitals for multielectron
Atoms:
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
First-Row Transition Metal Orbital Occupancy
*****
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 Properties of the First-Row Transition Metals
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
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.
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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
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 Atomic Size
• The third transition
series atoms are
about the same size as
the second because of
the lanthanide
contraction.
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 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.
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 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
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 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
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 Oxidation States
Unlike main group metals, transition metals often exhibit
multiple oxidation states.
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3. 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.
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






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.
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 Common Ligands
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 Four Common Structures of Complex Ions
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 Bidentate Ligands Coordinated to Co3+
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 Hexadentate Ligands Coordinated to Co3+
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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:
(a) Co: center atom, 6 ligands attached (4 Cl– and 2 NH3)
coordination number: 6.
Charge calculation:
x – 4 + 0 = –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).
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 Naming Coordination Compounds
*****
1. Identify the cation and anion, either may be complex ion or
uncomplex ion.
2. Naming complex cation and/or complex anions.
3. Write the compound name as the name of cation followed by
the name of the anion.
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
Names and Formulas of Common Ligands:
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
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.
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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 +3
(x – 2 + 0 = +1)
Ans: tetraamminedichlorochromium(III) ion.
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
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 Latinbased names:
Copper
Cuprate
Gold
Iron
Lead
Silver
Tin
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Chapter 24
Aurate
Ferrate
Plumbate
Argentate
Stannate
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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: +2
(x – 1 – 2 + 0 = –1)
Ans: amminebromodichloroplatinate(II).
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Examples of Naming Coordination Compounds
Identify the cation and anion, and
the name of the uncomplex ion.
Give each ligand a name and
list them in alphabetical order.
Name the metal ion.
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.
Name the compound by writing
the name of the cation before the
anion. The only space is between
ion names.
© 2014 Pearson Education, Inc.
TMHsiung ©2014
*****
Name [Cr(H2O)5Cl]Cl2
Name K3[Fe(CN)6]
[Cr(H2O)5Cl]2+ is a
complex cation;
Cl− is chloride.
K+ is potassium;
[Fe(CN)6]3− is a
complex anion.
H2O is aqua;
Cl− is chloro.
CN− is cyano.
Cr3+ is chromium(III).
Fe3+ is ferrate(III)
because the complex
ion is anionic.
[Cr(H2O)5Cl]2+ is
pentaquochlorochromium(III).
[Fe(CN)6]3− is
hexacyanoferrate(III).
[Cr(H2O)5Cl]Cl2 is
pentaquochlorochromium(III) chloride.
K3[Fe(CN)6] is
potassium
hexacyanoferrate(III).
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 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)
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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].
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4. Structure and Isomerization
*****
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1) Structural isomers
*****
i) 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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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)]Cl2
[Co(NH3)5(ONO)]Cl2
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Continued
Yellow =
pentaamminenitrocobalt(III)
chloride
[Co(NH3)5(NO2)]Cl2
Red =
pentaamminenitritocobalt(III)
chloride
[Co(NH3)5(ONO)]Cl2
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2) Stereoisomers
i) Geometric isomers
a) Cis-trans isomers (MA2B2 and MA4B2 type)
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*****
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b) Fac-mer isomers (MA3B3 type)
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Example 24.5
Drawing Geometric Isomers of [Co(en)2Cl2]+.
Solution
MA4B2 type, cis–trans isomers.
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ii) 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:
Mirror image of each other
Two nonsuperimposable (not identical) structures
Ans: two optical isomers
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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.
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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.
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5. Bonding in Coordination Compounds
 Valence Bond Theory/Hybridization of Atomic
Orbitals
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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
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4s
Chapter 24
4p
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Continued
electron configuration of Pd: 4d10, Pd2+: 4d8
for [PdCl4]2dsp2 hybridization
4d
5s
5p
electron configuration of Fe: 3d64s2, Fe3+: 3d5
for [Fe(H2O)6]3+
d2sp3 hybridization
3d
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4s
Chapter 24
4p
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 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 dorbitals 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:
*****
*
Increases the charge on the metal cation also increases the
splitting energy (Δ), for example:
Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+
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
*****
Schematic representation of d-level splitting:
Δ
Δ
Usually Using
High Spin Spectrochemical series
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Usually
Low Spin
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
*****
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.
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]2+
Example: Predict the magnetism for [Fe(H2O)6
[Fe(CN)6]4-.
Solution:
Both are octahedral complexes
Fe: [Ar]3d64s2, Fe2+: [Ar]3d6 (six 3d electrons)
Δ: CN– (large) > H2O (small)
and
*****
Diamagnetic
Paramagnetic
Low-spin complex
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.
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*****
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
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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
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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
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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
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 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.
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
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.
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Continued
Large Δ
High ν
Short λ
The Color of
[Ti(H2O)6]3+ solution
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Chapter 24
Small Δ
Low ν
Long λ
The absorption
pectrum [Ti(H2O)6]3+
solution
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
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

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*****
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.
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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.
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*****
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.
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End of Chapter 24
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