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Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Chapter 24
Transition
Metals and
Coordination
Compounds
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright  2011 Pearson Education, Inc.
Gemstones
• The colors of rubies and emeralds are both due
to the presence of Cr3+ ions – the difference
lies in the crystal hosting the ion
Some Al3+
ions in
Be3Al2(SiO3)6
are replaced
by Cr3+
Some Al3+
ions in Al2O3
are replaced
by Cr3+
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Properties and Electron
Configuration of Transition Metals
• The properties of the transition metals are
similar to each other
and very different to the properties of the main
group metals
high melting points, high densities, moderate to
very hard, and very good electrical conductors
• The similarities in properties come from
similarities in valence electron configuration
– they generally have 2 valence electrons
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Electron Configuration
• For 1st & 2nd transition series = ns2(n−1)dx
1st = [Ar]4s23dx; 2nd = [Kr]5s24dx
• For 3rd & 4th transition series = ns2(n−2)f14(n−1)dx
• Some individuals deviate from the general pattern
by “promoting” one or more s electrons into the
underlying d to complete the subshell
Group 1B
• Form ions by losing the ns electrons first, then
the (n – 1)d
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Example 24.1: Write the ground state
electron configuration of Zr
identify the noble gas that precedes the
[Kr]
element and write it in square brackets
Count down the periods to determine the
outer principal quantum level—this is the
quantum level for the s orbital. Subtract one
to obtain the quantum level for the d orbital.
[Kr]5s4d
If the element is in the third or
fourth transition series, include (n − 2)f14
electrons in the configuration.
count across the row to see how many
Zr has 4 more
electrons are in the neutral atom and fill the electrons than Kr
orbitals accordingly
[Kr]5s24d2
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Example 24.2: Write the ground state
electron configuration of Co3+
identify the noble gas that precedes the
element and write it in square brackets
Count down the periods to determine the
outer principal quantum level—this is the
quantum level for the s orbital. Subtract one
to obtain the quantum level for the d orbital.
[Ar]
[Ar]4s3d
count across the row to see how many
electrons are in the neutral atom and fill the
orbitals accordingly
Co has 9 more
electrons than Ar
[Ar]4s23d7
for an ion, remove the required number of
electrons, first from the s and then from the
d orbitals.
lost 3 e−
[Ar]3d6
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Practice – Use the Periodic Table to write the short
electron configuration and short orbital diagram for
each of the following
[Ar]4s23d8
• Ni (at. no. 28)
5s
4d
[Kr]4d5
• Tc2+ (at. no. 43)
5s
[Xe]6s24f145d4
• W (at. no. 74)
6s
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4d
4f
7
5d
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Irregular Electron Configurations
• We know that because of sublevel splitting, the
4s sublevel is lower in energy than the 3d; and
therefore the 4s fills before the 3d
• But the difference in energy is not large
• Some of the transition metals have irregular
electron configurations in which the ns only
partially fills before the (n−1)d or doesn’t fill at
all
• Therefore, their electron configuration must be
found experimentally
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Irregular Electron Configurations
•
•
•
•
•
•
•
•
•
•
•
•
Expected
Cr = [Ar]4s23d4
Cu = [Ar]4s23d9
Mo = [Kr]5s24d4
Ru = [Kr]5s24d6
Pd = [Kr]5s24d8
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Found Experimentally
Cr = [Ar]4s13d5
Cu = [Ar]4s13d10
Mo = [Kr]5s14d5
Ru = [Kr]5s14d7
Pd = [Kr]5s04d10
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Atomic Size
• The atomic radii of all the
transition metals are very
similar
 small increase in size down
a column
• The 3rd transition series
atoms are about the same
size as the 2nd
 the lanthanide contraction
is the decrease in expected
atomic size for the 3rd
transition series atoms that
come after the lanthanides
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Why Aren’t the 3rd Transition Series
Atoms Bigger?
• 14 of the added 32 electrons between the 2nd
•
•
and 3rd series go into 4f orbitals
Electrons in f orbitals are not as good at
shielding the valence electrons from the pull of
the nucleus
The result is a greater effective nuclear charge
increase and therefore stronger pull on the
valence electrons – the lanthanide contraction
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Ionization Energy
• The 1st IE of the transition
•
metals slowly increases
across a series
The 1st IE of the 3rd
transition series is
generally higher than
the 1st and 2nd series
indicating the valence
electrons are held more
tightly – why?
trend opposite to main
group elements
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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 1st
and 2nd series, but the 3rd
transition series atoms
are about the same as
the 2nd
 trend opposite to main
group elements
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Oxidation States
• Unlike main group metals, transition metals often
•
•
exhibit multiple oxidation states
Vary by 1
Highest oxidation state is the same as the group
number for Groups 3B to 7B
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Complex Ions
• When a monatomic cation combines with
•
•
multiple monatomic anions or neutral
molecules it makes a complex ion
The attached anions or neutral molecules are
called ligands
The charge on the complex ion can then be
positive or negative, depending on the
numbers and types of ligands attached
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Coordination Compounds
• When a complex ion combines with counter-ions
•
•
to make a neutral compound it is called a
coordination compound
The primary valence is the oxidation number of
the metal
The secondary valence is the number of ligands
bonded to the metal
 coordination number
• Coordination numbers range from 2 to 12, with
the most common being 6 and 4
CoCl36H2O = [Co(H2O)6]Cl3
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Coordination Compound
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Complex Ion Formation
• Complex ion formation is a type of Lewis acid–
base reaction
• A bond that forms when the pair of electrons is
donated by one atom is called a coordinate
covalent bond
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Ligands with Extra Teeth
• Some ligands can form more than one
coordinate covalent bond with the metal atom
lone pairs on different atoms that are separated
enough so that both can bond to the metal
• A chelate is a complex ion containing a
multidentate ligand
the ligand is called
the chelating agent
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EDTA
a Polydentate Ligand
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Complex Ions with Polydentate Ligands
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Geometries in Complex Ions
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Naming Coordination Compounds
1. Determine the name of the noncomplex ion
2. Determine the ligand names and list them in
alphabetical order
3. Determine the name of the metal cation
4. Name the complex ion b:
a) naming each ligand alphabetically, adding a prefix in
front of each ligand to indicate the number found in the
complex ion
b) following with the name of the metal cation
5. Write the name of the cation followed by the
name of the anion
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Common Ligands
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Common Metals found in
Anionic Complex Ions
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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 simple ion
[Cr(H2O)5Cl]2+ is a
complex cation,
Cl− is chloride
K+ is potassium,
[Fe(CN)6]3− is a
complex ion
Give each ligand a name and list
them in alphabetical order
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)
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.
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Practice – Name each of the following
K2CuI4
potassium tetraiodocuprate(II)
[Co(NH3)2(CO)4](NO3)3
diamminotetracarbonylcobalt(III) nitrate
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Isomers
• Structural isomers are molecules that have
•
the same number and type of atoms, but they
are attached in a different order
Stereoisomers are molecules that have the
same number and type of atoms, and that are
attached in the same order, but the atoms or
groups of atoms point in a different spatial
direction
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Linkage Isomers
• Linkage isomers are structural isomers that
have ligands attached to the central cation
through different ends of the ligand structure
yellow complex =
pentamminonitrocobalt(III)
red complex =
pentamminonitritocobalt(III)
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Geometric Isomers
• In fac–mer isomerism three identical ligands
• •In
cis–trans
isomerism
identical
ligands
inGeometric
an
octahedral
complex
either
are adjacent
isomers
aretwo
stereoisomers
that
are
either
adjacent
toorientation
eachface
other
to
each
making
one
(fac)
oror
form an
differ
inother
the
spatial
of(cis)
ligands
arc
around
the other
center(trans)
(mer) in
in the
the structure
structure
opposite
each
• cis–trans
isomerismininoctahedral
octahedral
complexes
MA34B
B32 2B2
fac–mer isomerism
square-planar
complexes
complexes
MA
MA
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Example 24.5: Draw the structures and label
the type for all isomers of [Co(en)2Cl2]+
the ethylenediamine ligand (en = H2NCH2CH2NH2) is
bidentate
each Cl ligand is monodentate
octahedral
MA4B2
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Practice – Identify each of the following
stereoisomers as cis, trans, fac, or mer
trans
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[Co(en)3]3+
Optical Isomers
• Optical isomers are
stereoisomers that are
nonsuperimposable mirror
images of each other
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Ex 24.7 – Determine if the cis-trans isomers of
[Co(en)2Cl2]+ are optically active
• Draw the mirror
•
image of the
given isomer
Check to see if
the two are
superimposable
cis isomer
trans isomer
mirror image
identical
is to
nonsuperimposable
its mirror image
nooptical
opticalisomers
isomerism
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Practice – Determine if the following is
optically active
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Practice – Determine if the following is
optically active
mirror images are superimposable, therefore not optically active
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Bonding in Coordination Compounds:
Valence Bond Theory
• Bonding takes place when the filled atomic
•
orbital on the ligand overlaps an empty atomic
orbital on the metal ion
Explains geometries well, but doesn’t explain
color or magnetic properties
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Bonding in Coordination Compounds:
Crystal Field Theory
• Bonds form due to the attraction of the electrons
•
•
•
on the ligand for the charge on the metal cation
Electrons on the ligands repel electrons in the
unhybridized d orbitals of the metal ion
The result is the energies of the d orbitals are split
The difference in energy depends on the complex
formed and the kinds of ligands
 crystal field splitting energy
 strong field splitting and weak field splitting
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Crystal Field Splitting
The repulsions
other
ligands
d orbitals
in an lie
between
electron
octahedral
thecomplex
axes pairs
and
in
the
ligands
and
have
are
located
nodes
in
directly
the any
on
potential
electrons
in
the axes,
same
space
which
as the
results
the
d orbitals
in less
lobes
ofrepulsion
the result
and
andin
an
increase
in the
lower
energies
for
orbitals
energies
of these
these three
orbitals
orbitals
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Splitting of d Orbital Energies Due to
Ligands in an Octahedral Complex
The size of the crystal field splitting energy, D,
depends on the kinds of ligands and their relative
positions on the complex ion, as well as the kind of
metal ion and its oxidation state
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Color and Complex Ions
• Transition metal ions show many intense colors
in host crystals or solution
• The color of light absorbed by the complexed ion
is related to electronic energy changes in the
structure of the complex
the electron “leaping” from
a lower energy state to
a higher energy state
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Complex Ion Color
• The observed color is the complementary color
of the one that is absorbed
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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: Estimate the Crystal Field Splitting
Energy for a blue solution of [Cu(NH3)6]2+
use the color wheel to
determine approximately what
wavelength is being absorbed
Because the solution is blue it is
absorbing orange. The midpoint
of orange is 615 nm
Calculate the energy of a
photon with this wavelength. It
is the same as the energy gap
between the d orbitals, D.
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Practice – Estimate the crystal field splitting
energy for a red solution of [Fe(CN)6]3−
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Practice – Estimate the crystal field splitting
energy for a red solution of [Fe(CN)6]3−
use the color wheel to
determine approximately what
wavelength is being absorbed
Because the solution is red it is
absorbing green. The midpoint of
green is 525 nm
Calculate the energy of a
photon with this wavelength. It
is the same as the energy gap
between the d orbitals, D.
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Ligands and
Crystal Field Strength
• The size of the energy gap depends on what
kind of ligands are attached
strong field ligands include CN─ > NO2─ > en > NH3
weak field ligands include H2O > OH─ > F─ > Cl─ >
Br─ > I─
• The size of the energy gap also depends on the
type of cation
increases as the charge on the metal cation
increases
Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+
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Magnetic Properties and
Crystal Field Strength
• The electron configuration of the metal ion with
•
•
split d orbitals depends on the strength of the
crystal field
The 4th and 5th electrons will go into the higher
energy
if the field is weak and the
energy gap is small – leading to unpaired
electrons and a paramagnetic complex
The 4th thru 6th electrons will pair the electrons in
the dxy, dyz, and dxz if the field is strong and the
energy gap is large – leading to paired electrons
and a diamagnetic complex
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Low Spin & High Spin Complexes
diamagnetic
paramagnetic
low-spin complex
high-spin complex
only electron configurations d4, d5, d6, or d7
can have low or high spin
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Example 24.9: How many unpaired electrons
are in [CoF6]3−
determine the charge and
number of d electrons on the
metal cation
because F is −1, Co is 3+
Co is atomic number 27, so Co3+ is
[Ar]4s03d6
determine if the ligand is strong
field or weak field
F− is weak field
decide if the complex ion is low
spin or high spin
weak field ligands have small D
and yield high spin configurations
draw the energy diagram and
count the unpaired electrons
there are four unpaired electrons
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Practice – How many unpaired electrons
are in [Fe(CN)6]3−
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Practice – How many unpaired electrons are
in [Fe(CN)6]3−
because CN is −1, Fe is 3+
Fe is atomic number 26, so Fe3+ is
[Ar]4s03d5
determine if the ligand is strong
field or weak field
CN− is strong field
decide if the complex ion is low
spin or high spin
strong field ligands have large D
and yield low spin configurations
draw the energy diagram and
count the unpaired electrons
shere is one unpaired electron
Energy
determine the charge and
number of d electrons on the
metal cation
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Tetrahedral Geometry and
Crystal Field Splitting
• Because the ligands interact more strongly
•
•
with the planar orbitals in the tetrahedral
geometry, their energies are raised
This reverses the order of energies compared
to the octahedral geometry
Almost all tetrahedral complexes are high spin
because of reduced metal orbital – ligand
interaction
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Crystal Field Splitting in the
Tetrahedral Geometry
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Square Planar Geometry and
Crystal Field Splitting
• d8 metals
Pt2+, Pd2+, Ir+, Au3+
• The most complex splitting pattern
• Almost all are low-spin complexes
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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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Applications of
Coordination Compounds
• Commercial coloring agents
Prussian blue = mixture of hexacyanoFe(II) and
Fe(III)
inks, blueprinting, cosmetics, paints
• Biomolecules
porphyrin ring
cytochrome C
hemoglobin
chlorophyll
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Applications of
Coordination Compounds
• Carbonic anhydrase
catalyzes the reaction between water and CO2
contains tetrahedrally complexed Zn2+
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Applications of
Coordination Compounds
• Drugs and therapeutic agents
cisplatin
anticancer drug
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