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```Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 23
The Transition Elements and
Their Coordination Compounds
23-1
The Transition Elements and Their Coordination Compounds
23.1 Properties of the Transition Elements
23.2 The Inner Transition Elements
23.3 Highlights of Selected Transition Metals
23.4 Coordination Compounds
23.5 Theoretical Basis for the Bonding and Properties of Complexes
23-2
Figure 23.1
23-3
The transition elements (d block) and inner transition
elements (f block) in the periodic table.
Figure 23.2
23-4
The Period 4 transition metals.
23-5
Sample Problem 23.1
PROBLEM:
Writing Electron Configurations of Transition
Metal Atoms and Ions
Write condensed electron configurations for the following: (a) Zr;
(b) V3+; (c) Mo3+. (Assume that elements in higher periods
behave like those in Period 4.)
PLAN: The general configuration is [noble gas] ns2(n-1)dx. Recall that in ions
the ns electrons are lost first.
SOLUTION:
(a) Zr is the second element in the 4d series: [Kr]5s24d2.
(b) V is the thired element in the 3d series: [Ar]4s23d3. In forming V3+,
three electrons are lost (two 4s and one 3d), so V3+ is a d2 ion: [Ar]3d10.
(c) Mo lies below Cr in Group 6B(6), so we expect the same except in
configuration as for Cr. Thus, Mo is [Kr]5s14d5. In forming the ion, Mo
loses the one 5s and two of the 4d electrons to become a 4d3 ion: [Kr]4d3.
23-6
Figure 23.3
23-7
Horizontal trends in key atomic properties of the
Period 4 elements.
Figure 23.4
Vertical trends in key properties within the transition elements.
23-8
Figure 23.5 Aqueous oxoanions of transition elements.
One of the most characteristic
chemical properties of these
elements is the occurrence of
multiple oxidation states.
Mn(II)
Mn(VI)
Mn(VII)
Mn(VII)
Cr(VI)
V(V)
23-9
23-10
23-11
Figure 23.6
Colors of representative compounds of the Period 4 transition metals.
sodium chromate
titanium oxide
scandium oxide
dihydrate
23-12
nickel(II) nitrate
hexahydrate
potassium
ferricyanide
manganese(II)
chloride
tetrahydrate
cobalt(II)
chloride
hexahydrate
zinc sulfate
heptahydrate
copper(II)
sulfate
pentahydrate
23-13
Sample Problem 23.2
PROBLEM:
PLAN:
Finding the Number of Unpaired Electrons
The alloy SmCo5 forms a permanent magent because both
samarium and cobalt have unpaired electrons. How many
unpaired electrons are in the Sm atom (Z = 62)?
Write the condensed configuration of Sm and, using Hund’s
rule and the aufbau principle, place electrons into a partial
orbital diagram.
SOLUTION: Sm is the eighth element after Xe. Two electrons go into the 6s
sublevel and the remaining six electrons into the 4f (which fills
before the 5d).
Sm is [Xe]6s24f6
6s
4f
There are 6 unpaired e- in Sm.
23-14
5d
Figure 23.7
23-15
The bright colors of chromium (VI) compounds.
Orbital Occupancy
*Most common states in bold face.
23-16
Figure 23.8
23-17
Steps in producing a black-and-white negative.
Figure 23.9 Components of a coordination compound.
models
23-18
wedge diagrams
chemical formulas
Structures of Complex Ions:
Coordination Numbers, Geometries, and Ligands
•Coordination Number - the number of ligand atoms that are bonded
directly to the central metal ion. The coordination number is specific for
a given metal ion in a particular oxidation state and compound.
•Geometry - the geometry (shape) of a complex ion depends on the
coordination number and nature of the metal ion.
•Donor atoms per ligand - molecules and/or anions with one or more
donor atoms that each donate a lone pair of electrons to the metal ion to
form a covalent bond.
23-19
23-20
23-21
23-22
Formulas and Names of Coordination Compounds
Rules for writing formulas:
1. The cation is written before the anion.
2. The charge of the cation(s) is balanced by the charge
of the anion(s).
3. In the complex ion, neutral ligands are written before
anionic ligands, and the formula for the whole ion is
placed in brackets.
23-23
Formulas and Names of Coordination Compounds
Rules for naming complexes:
continued
1. The cation is named before the anion.
2. Within the complex ion, the ligands are named, in alphabetical
order, before the metal ion.
3. Neutral ligands generally have the molecule name, but there
are a few exceptions. Anionic ligands drop the -ide and add
-o after the root name.
4. A numerical prefix indicates the number of ligands of a
particular type.
5. The oxidation state of the central metal ion is given by a
Roman numeral (in parentheses).
6. If the complex ion is an anion we drop the ending of the metal
23-24
Sample Problem 23.3
Writing Names and Formulas of Coordination
Compounds
PROBLEM: (a) What is the systematic name of Na3[AlF6]?
(b) What is the systematic name of [Co(en)2Cl2]NO3?
(c) What is the formula of tetraaminebromochloroplatinum(IV)
chloride?
(d) What is the formula of hexaaminecobalt(III) tetrachloroferrate(III)?
PLAN:
Use the rules presented -
SOLUTION:
and
.
(a) The complex ion is
3-.
[AlF6] Six
(hexa-) fluorines (fluoro-) are the ligands - hexafluoro
Aluminum is the central metal atom - aluminate
Aluminum has only the +3 ion so we don’t need Roman
numerals.
sodium hexafluoroaluminate
23-25
Sample Problem 23.3
Writing Names and Formulas of Coordination
Compounds
continued
(b) There are two ligands, chlorine and ethylenediamine dichloro, bis(ethylenediamine)
The complex is the cation and we have to use Roman numerals for
the cobalt oxidation state since it has more than one - (III)
The anion, nitrate, is named last.
dichlorobis(ethylenediamine)cobalt(III) nitrate
(c)
Cl
Pt4+
Cltetraaminebromochloroplatinum(IV) chloride
4 NH3
Br-
[Pt(NH3)4BrCl]Cl2
(d)
6 NH3
Co3+
4 Cl-
Fe3+
hexaaminecobalt(III) tetrachloro-ferrate(III)
[Co(NH3)6][Cl4Fe]3
23-26
23-27
Figure 23.10
Important types of isomerism in coordination compounds.
ISOMERS
Same chemical formula, but different properties
Constitutional (structural) isomers
Stereoisomers
Atoms connected differently
Different spatial arrangement
Coordination
Coordination
isomers
isomers
isomers
isomers
Ligand
Ligand and
and
counter-ion
counter-ion
exchange
exchange
Different
Different donor
donor
atom
atom
23-28
Geometric
Optical
Optical isomers
isomers
Geometric (cis(cistrans)
(enantiomers)
(enantiomers)
trans) isomers
isomers
(diastereomers)
(diastereomers) Nonsuperimposable
Nonsuperimposable
Different
mirror
Different
mirror images
images
arrangement
arrangement
around
around metal
metal ion
ion
23-29
Figure 23.11
23-30
Geometric (cis-trans) isomerism.
Figure 23.12
Optical isomerism in an
octahedral complex ion.
23-31
Sample Problem 23.4
Determining the Type of Stereoisomerism
PROBLEM: Draw all stereoisomers for each of the following and state the type
of isomerism:
(b) [Cr(en)3]3+ (en = H2NCH2CH2NH2)
(a) [Pt(NH3)2Br2]
PLAN:
Determine the geometry around each metal ion and the nature of
the ligands. Place the ligands in as many different positions as
possible. Look for cis-trans and optical isomers.
SOLUTION: (a) Pt(II) forms a square planar complex and there are two pair
of monodentate ligands - NH3 and Br.
Br
NH3
H3N
Pt
H3N
23-32
Pt
Br
trans
Br
H3 N
Br
cis
These are geometric isomers;
they are not optical isomers
since they are superimposable
on their mirror images.
Sample Problem 23.4
Determining the Type of Stereoisomerism
(b) Ethylenediamine is a bidentate ligand. Cr3+ is
hexacoordinated and will form an octahedral geometry.
continued
Since all of the ligands are identical, there will be no geometric isomerism
possible.
3+
3+
N
N
N
N
N
N
Cr
Cr
N
The mirror images are
nonsuperimposable
and are therefore
optical isomers.
N
N
N
N
N
rotate
3+
N
N
N
Cr
N
N
N
23-33
Figure 23.13
Hybrid orbitals and bonding in the octahedral [Cr(NH3)6]3+ ion.
23-34
Figure 23.14
Hybrid orbitals and bonding in the square planar [Ni(CN)4]2- ion.
23-35
Figure 23.15
Hybrid orbitals and bonding in the tetrahedral [Zn(OH)4]2- ion.
23-36
Figure 23.16
23-37
An artist’s wheel.
23-38
Figure 23.17
23-39
The five d-orbitals in an octahedral field of ligands.
Figure 23.18
Splitting of d-orbital energies by an octahedral field of
ligands.
Δ is the splitting energy
23-40
Figure 23.19
23-41
The effect of ligand on splitting energy.
Figure 23.20
23-42
The color of [Ti(H2O)6]3+.
Figure 23.21
Effects of the metal oxidation state and of ligand identity on color.
[V(H2O)6]3+
[V(H2O)6]2+
[Cr(NH3)6]3+
23-43
[Cr(NH3)5Cl ]2+
Figure 23.22
The spectrochemical series.
•For a given ligand, the color depends on the oxidation state of the metal ion.
•For a given metal ion, the color depends on the ligand.
I- < Cl- < F- < OH- < H2O < SCN- < NH3 < en < NO2- < CN- < CO
WEAKER FIELD
23-44
STRONGER FIELD
SMALLER Δ
LARGER Δ
LONGER λ
SHORTER λ
Sample Problem 23.5
Ranking Crystal Field Splitting Energies for
Complex Ions of a Given Metal
PROBLEM: Rank the ions [Ti(H2O)6]3+, [Ti(NH3)6]3+, and [Ti(CN)6]3- in terms of
the relative value of Δ and of the energy of visible light absorbed.
PLAN: The oxidation state of Ti is 3+ in all of the complexes so we are
looking at the crystal field strength of the ligands. The stronger the
ligand the greater the splitting and the higher the energy of the light
absorbed.
SOLUTION:
The field strength according to
is CN- > NH3 > H2O. So the
relative values of Δ and energy of light absorbed will be
[Ti(CN)6]3- > [Ti(NH3)6]3+ > [Ti(H2O)6]3+
23-45
Figure 23.23
23-46
High-spin and low-spin complex ions of Mn2+.
Figure 23.24 Orbital occupancy for high- and low-spin complexes
of d4 through d7 metal ions.
high spin:
weak-field
ligand
23-47
low spin:
strong-field
ligand
high spin:
weak-field
ligand
low spin:
strong-field
ligand
Sample Problem 23.6
Identifying Complex Ions as High Spin or Low Spin
PROBLEM: Iron (II) forms an essential complex in hemoglobin. For each of the
two octahedral complex ions [Fe(H2O)6]2+ and [Fe(CN)6]4-, draw an
orbital splitting diagram, predict the number of unpaired electrons,
and identify the ion as low or high spin.
PLAN:
The electron configuration of Fe2+ gives us information that the
iron has 6d electrons. The two ligands have field strengths shown
in
.
potential energy
Draw the orbital box diagrams, splitting the d orbitals into eg and
t2g. Add the electrons noting that a weak-field ligand gives the
maximum number of unpaired electrons and a high-spin complex
and vice-versa.
[Fe(CN)6]4[Fe(H2O)6]2+
SOLUTION:
4 unpaired e-eg
(high spin)
eg
no unpaired e-(low spin)
t2g
t2g
23-48
Figure 23.25
Splitting of d-orbital energies by a tetrahedral field
and a square planar field of ligands.
tetrahedral
square planar
23-49