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Chapter 17
The Colorful Chemistry of
Transition Metals
Lewis Acids and Bases
• A Lewis base is a substance that donates a
pair of electrons in a chemical reaction.
• A Lewis acid is a substance that accepts a
pair of electrons in a chemical reaction.
Figure 17.1
Another Lewis Example
Complex Ions
• A complex ion is an ionic species consisting
of a metal ion bonded to one or more Lewis
bases.
• A coordinate bond forms when one anion or
molecule donates a pair of electrons to
another ion or molecule to form a covalent
bond.
• A ligand is a Lewis base bonded to the
central metal ion of a complex ion.
• The inner coordination sphere of a metal
consists of the ligands that are bound directly
to the metal via coordinate covalent bonds.
Complex Ions
Dissolved metal ions are Lewis Acids, and form aqueous
complexes with Lewis Bases
More Metal Complexes
Terms
• Counter ions are the ions that balance the
electrical charges of complex ions in
coordination compounds (e.g. in basic
solution OH- ions could act as counter ions
Zn(NH3)4(OH)2
• Coordination compounds are made up of
one or more complex ions; typically no
more than 6 ligands surround the metal.
• The coordination number (CN) of a metal
ion identifies the number of electron pairs
surrounding it in a complex.
Which ions are the counter ions in the
following:
Na2[Zn(CN)4]
[Co(NH3)4Cl2]NO3
Bonding in Zn(NH3)42+ Ion
Common Coordination Numbers and Shapes for
Complex Ions
Coordinationnu
mber
Shape
Hybridization
Example
6
Octahedral
d2sp3
Fe(H2O)63+
4
Tetrahedral
sp3
Zn(H2O)42+
4
Square Planar
dsp2
Pt(NH3)42+
2
Linear
sp
Ag(NH3)2+
Are any of
these NOT
Lewis bases?
Would you
expect H2O
or NH3 to be
the stronger
Lewis Base?
Complex-Ion Equilibria
Cu2+(aq) + 4 NH3(aq) <==> Cu(NH3)42+(aq)
Figure 17.5

2 +
Cu(NH
)
3 4 



13
Kf =
=
5.0
x
10
4
 2 +
Cu  NH 3 



17.26.
When a strong base is added to
a solution of CuSO4, which is pale blue, a
precipitate forms and the solution above the
precipitate is colorless. When ammonia is
added, the precipitate dissolves and the
solution turns a deep navy blue. Use
appropriate chemical equations to explain
why the observed changes occurred.
Formation Reactions
The equilibrium constants associated with complexation are
called formation constants
Kf = [Cu(NH3)42+]/[Cu2+][NH3]4
= 5.0E+13
Metal ions as Lewis Acid also promote hydrolysis of water, and
the formation of H3O+
Metal cations (Fe3+, Cr3+ and Al3+)with large positive charges
are more likely to cause hydrolysis.
Ka Values of Hydrated Metal Ions
Ion
Ka
Fe3+(aq)
3 x 10-3
Cr3+(aq)
1 x 10-4
Al3+(aq)
1 x 10-5
Cu2+(aq)
3 x 10-8
Pb2+(aq)
3 x 10-8
Zn2+(aq)
1 x 10-9
Co2+(aq)
2 x 10-10
Ni2+(aq)
1 x 10-10
17.45.
Sketch the titration curve (pH versus
volume of 0.50 M NaOH) for a 25 mL sample
of 0.5 M FeCl3 (Ka=3E-3)
17.5. Solubility of Ionic Compounds
Minerals in contact with ground water will dissolve to some
extent.
CaCO3(s) ↔ Ca2+(aq) + CO32-(aq)
Write the equilibrium expression for this dissolution.
Solubility Equilibria
CaCO3(s) ↔ Ca2+(aq) + CO32-(aq)
K = [Ca2+(aq)][CO32-(aq)] = Ksp
The equilibrium expression is called the solubility product
(sp), because it involves only products of the concentrations
of the dissolved species and NOT the solid.
If Ksp is known, the solubility (at equilibrium) of the solid can
be calculated
Solubilities of Ionic Compounds
The solubility-product constant (Ksp) is
an equilibrium constant that describes
the formation of a saturated solution of
a slightly soluble salt.
Mg(OH)2(s) <==> Mg2+(aq) + 2 OH-(aq)
Ksp = [Mg2+][OH-]2 = 5.6 x 10-12
Ksp values of some common salts
HgS(s)
Fe(OH)3(s)
AgI(s)
Ksp = [Hg2+][S2-] = 4.0 x 10-53
Ksp = [Fe3+][OH-]3 = 2.8 x 10-39
Ksp = [Ag1+][I1-] = 8.5 x 10-17
CaCO3(s)
CaSO4(s)
Ag2SO3(s)
Ksp = [Ca2+][CO32-] = 9.8 x 10-9
Ksp = [Ca2+][SO42-] = 4.9 x 10-5
Ksp = [Ag1+]2[SO32-] = 1.2 x 10-5
NaCl(s)
Ksp = [Na1+][Cl1-] = 6.2
Solubility Problem
17.58. What are the equilibrium concentrations of Pb2+
and F– in a saturated solution of lead fluoride if the Ksp
value of PbF2 is 3.2 × 10–8?
Solubility Problem
2.75 grams of BaF2 (FW=175.3) is placed in enough water to
make 1.00 L at 25°C. After equilibrium has been
established…the F- concentration equal 0.0150 M, what is
the Ksp for BaF2.
Solubility Problem
50 mg of PbSO4 (FW=303.3) is placed in 250 mL of pure
water; What percentage of the solid dissolves?
Ksp(PbSO4)=1.8E-8 (Table A5.4)
Solubility Problem
Calculate the pH of a saturated solution of zinc hydroxide,
Ksp = 4.0E-17
Solubility Problem 120
Calculate the solubility of silver chloride in seawater with a
chloride concentration of 0.547 M. Ksp(AgCl) = 1.8E-10
Terms
• A monodentate ligand is a species that
forms only a single coordinate bond to
a metal ion in a complex.
• A polydentate ligand is a species that
can form more than one coordinate
bond per molecule.
• Chelation is the interaction of a metal
with a polydentate ligand (chelating
agent). (e.g. Google “chelation therapy”)
Chelation-Examples
(a) bidentate chelation Ni2+(aq) by
ethylenediamine (b) 3 ethylenediamime
molecules bind to Ni2+
tridentate chelation by
diethylenetriamine.
A Hexadentate Ligand
Ethylenediaminetetraacetic acid (EDTA)
forms 6 stable,
coordinate bonds with
many metals
Ligands and Complex Colors
NiCl2
[Ni(NH3)]Cl2
Complex Ions of Nickel
• Ni(H2O)62+ is equivalent to Ni2+(aq) and is green
in solution.
Ni2+(aq) + 6NH3(aq) <==> Ni(NH3)62+(aq)
Kf = 5 x 108
Ni2+(aq) + 3en <==> Ni(en)32+(aq)
Kf = 1.1 x 1018
en=ethylenediamine
• The chelate effect is the greater affinity of
metal ions for polydentate ligands compared
to the corresponding monodentate ligands.
Crystal Field Theory…
why transition metal complexes are colored
• Crystal field splitting is the separation
of a set of d-orbitals into subsets with
different energies as a result of
interactions between electrons in those
orbitals and pairs of electrons in ligands
surrounding the orbitals.
• Crystal field splitting energy () is the
difference in energy between subsets of
d-orbitals split by interactions in a
crystal field.
d-orbitals in an octahedral field
Light (photons) Can Promote
Electrons in a Complex Ion’s Orbitals
Fig. 17.11: Crystal field spitting of dorbital energies that results from the
interaction with p-orbitals in an
octahedral geometry.
The difference in energy is called the
crystal field splitting energy ().
An example of this is the substitution
of Cr3+ for Al3+ in the octahedral
holes of the silicate structure of beryl.
The Be2+ sits in tetrahedral holes.
Be3CrxAl2-xSi6O18
Review: See chapter 7 for
shapes of atomic orbitals.
The Color of Compounds
• Our eyes perceive
the transmitted
colors of complex
ions.
• [Cu(NH3)4]2+
absorbs much of the
yellow, orange, and
red wavelengths of
light, so we see the
complex as being
navy blue.
Color Wheel
Ni(H2O)62+ --> Ni(NH3)62+ --> Ni(en)32+
Green
Blue
Violet
• Ni(H2O)62+ is green because it absorbs the
color red on the opposite side of the color
wheel.
• Ni(NH3)62+ is blue because it absorbs the
color opposite blue, which is orange.
• Ni(en)32+ is violet because it absorbs colors
centered on the complement of violet, which
is yellow (en is the ethlyenediamine ligand).
Crystal Field Splitting Energy
  must change for the three nickel compounds.
• H2O as a ligand must yield the smallest  because
the red light, which is lowest in energy, is absorbed
by the nickel complex.
• NH3 must cause a little larger  than water
because a higher energy of orange light is
absorbed.
• The ligand en must cause the largest , because
its complex absorbs yellow light, which is highest in
energy of the three examined.
The Spectrochemical Series
A list of ligands
ordered by their
abilities to increase ,
the crystal field
splitting energy of the
d-orbitals.
Square Planar Crystal Field Splitting
How many delectrons does
Cu2+ have?
Fig. 17.15: The mineral turquoise,
CuAl6(PO4)4(OH)84H2O, has Cu2+ ions occupying
irregular shaped octahedral holes. Consequently,
the crystal field spitting produces four energy
levels among the d-orbitals.
Tetrahedral Crystal Field Splitting
Magnetism and Spin States
Magnetism and Spin States
Both the high- and lowspin species of Fe(III)
complexes are
paramagnetic.
High-spin species would
have a greater
interaction with a
magnetic field.
High-spin complexes
result from weak field
ligands (H2O); low-spin
complexes result from
strong-field ligands (CN-)
The Spectrochemical Series
A list of ligands
ordered by their
abilities to increase ,
the crystal field
splitting energy of the
d-orbitals.
Electron Paramagnetic Resonance spectroscopy is a
technique for measuring chemical species with one or more
unpaired elections.
An unpaired electron can
move between the two
energy levels by either
absorbing or emitting
electromagnetic radiation
of energy ε = hν
ΔE = geμBB0 (i.e. the splitting
is proportional to the
magnetic field strength (B0)
Naming Positively Charged
Complex Ions
• Start with the name of ligand(s) (see Table
17.3 for common ligand names).
• Use the usual prefixes to indicate the number
of each type of ligand (di(2), tri(3), etc.).
• Next write the name of the central metal ion
using a Roman numeral to indicate the
oxidation state of the transition metal ion.
• Ligands are named alphabetically (excluding
prefixes).
Examples
Ni(H2O)62+
Hexaaquanickel(II)
Co(NH3)63+
Hexaamminecobalt(III)
[Cu(NH3)4(H2O)2]2+
Tetraamminediaquacopper(II)
Naming Negatively Charged
Complex Ions
• Follow the steps for a positively charged
complex ion.
• Use an -ate ending on the name of the
central metal ion to indicate the charge
of the complex ion.
Examples
• Fe(CN)63-
Hexacyanoferrate(III)
• [Fe(H2O)(CN)5]3Aquapentacyanoferrate(II)
• [Al(H2O)2(OH)4]Diaquotetrahydroxoaluminate
Coordination Compounds
• [Ni(H2O)6]Cl2
Hexaaquanickel(II) Chloride
• K3[Fe(CN)6]
Potassium Hexacyanoferrate(III)
Geometric Isomers
cis-Pt(NH3)2Cl2
trans-Pt(NH3)2Cl2
cis-platinum is a widely in chemotherapy to treat cancer,
while the trans-isomer is ineffective.
Geometric Isomers of octahedral
complexes
Enantiomers
• Enantiomers are geometric isomers that are
nonsuperimposable mirror images of each
other.
• Enantiomeric ions and molecules are called
chiral.
Metal Complexes in Biomolecules
Heme
Cytochrome c Protein
Metalloporphyrins
ChemTour: Crystal Field Splitting
Click to launch animation
PC | Mac
This animation illustrates how transition metals can fill
tetrahedral and octahedral holes in crystal lattices, resulting
in the splitting of the energies of the d orbitals and the
brilliant color of gemstones.
• Note to instructors: The following in-class
review questions are also applicable to
chapter 18, Electrochemistry and Electric
Vehicles.
• Per the note in chapter 10, Forces between
Ions and Molecules and Colligative
Properties, the review question posted there
is also applicable to this chapter.
Under standard conditions, copper will
plate out onto a nickel rod immersed in a
solution containing Cu2+ ions, but
aluminum will not plate onto a nickel rod
immersed in a solution containing Al3+
ions. Which of the following is the
strongest oxidizing agent?
A) Ni(s)
B) Al3+(aq)
Oxidizing Strength of Al +, Ni, and Cu
+
C) Cu2+(aq)
Consider the following arguments for each answer
and vote again:
A. Ni(s), a metallic conductor, can accept and release
electrons most readily.
B. Al3+(aq) is the strongest oxidizing agent because it
has the greatest positive charge.
C. Cu2+(aq) can oxidize Ni(s) and Ni2+(aq) can oxidize
Al(s), so Cu2+(aq) can oxidize Al(s).
Oxidizing Strength of Al +, Ni, and Cu
+
For the reaction
Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq),
G° = 88.3 kJ/mole. Under standard
conditions, will the Cu/Ag cell pictured to
the left produce a current?
A) Yes
Ag/Cu Electrochemical Cell
B) No
C) Depends
Consider the following arguments for each answer
and vote again:
A. Under standard conditions, Ag+ can oxidize Cu to
form Cu2+. Therefore, electrons will flow from the
Cu electrode to the Ag electrode.
B. ΔG° is positive, and so the reaction is not
spontaneous and no electrons will flow between the
two electrodes.
C. Whether a current will flow in the electrochemical
cell depends on which metal, Cu or Ag, is the anode
and which is the cathode.
Ag/Cu Electrochemical Cell
For the oxidation of Fe2+ by Ag+,
Fe2+(aq) + Ag+(aq)
Fe3+(aq) + Ag(s),
H° and S° are both negative. Which of the
following plots shows the correct relationship between
the electromotive force, E°, (y-axis) and the
temperature (x-axis)?
A)
Oxidation of Fe + by Ag+
B)
C)
Consider the following arguments for each answer
and vote again:
A. At lower temperatures, the negative enthalpy
dominates, so the reaction is more spontaneous.
B. All electrochemical reactions have an increased E°
at higher temperatures due to the increase in kinetic
energy provided to the transferred electrons.
C. E° depends only on the concentrations of the
species in solution and is independent of the
temperature.
Oxidation of Fe + by Ag+
For the cell pictured to the left, the voltage,
E, is measured to be 0.79 V. Given the
following standard reduction potentials,
Ag+ + e→ Ag
2 H+ + 2 e- → H2

E red = 0.80 V

E red = 0.00 V
what can be said of the pH in the right half
of the cell if the partial pressure of H2(g) is
1.0 atm?
A) pH > 0
pH of an Electrochemical Cell
B) pH = 0 C) pH < 0
Consider the following arguments for each answer
and vote again:
A. A lower H3O+ concentration in the right half of the
cell would result in a decrease in the efficiency of the
cell, causing the cell voltage to drop.
B. This electrochemical reaction is spontaneous only
under standard conditions, where the concentration of
H3O+ is 1.0 M and the pH is 0.
C. A pH less than zero favors the formation of H2(g) and
thus would decrease the overall voltage of the cell.
pH of an Electrochemical Cell
For the electrochemical cell
Cu | Cu2+(1.0 M) || Cu2+(0.1 M) | Cu,
what will happen to the color in the
darker solution (1.0 M Cu2+) once the
circuit is completed?
A) It gets darker.
B) It gets lighter. C) It stays the same.
Cu/Cu + Concentration Cell
Consider the following arguments for each answer
and vote again:
A. Formation of Cu2+(aq) through the oxidation of Cu
occurs in both cells, so both solutions should get
darker.
B. During the approach to equilibrium, which reduces
the cell potential, Cu will plate onto the electrode in
the darker solution.
C. Both half-cells contain the same type of ions and the
electrons do not know in which direction to flow.
Cu/Cu + Concentration Cell