chapter 17 the colorful chemistry of transition metals

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 IonsComplex 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

Coordinationnumber

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

Kf = Cu(NH3)

42 +

Cu2 +

NH3 4

= 5.0 x 1013

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

KKspsp values of some common salts values of some common salts

HgS(s)HgS(s) KKspsp = [Hg = [Hg2+2+][S][S2-2-] = 4.0 x 10] = 4.0 x 10-53-53

Fe(OH)Fe(OH)33(s)(s) KKspsp = [Fe = [Fe3+3+][OH][OH--]]33 = 2.8 x 10 = 2.8 x 10-39-39

AgI(s)AgI(s) KKspsp = [Ag = [Ag1+1+][I][I1-1-] = 8.5 x 10] = 8.5 x 10-17-17

CaCOCaCO33(s)(s) KKspsp = [Ca = [Ca2+2+][CO][CO332-2-] = 9.8 x 10] = 9.8 x 10-9-9

CaSOCaSO44(s)(s) KKspsp = [Ca = [Ca2+2+][SO][SO442-2-] = 4.9 x 10] = 4.9 x 10-5-5

AgAg22SOSO33(s)(s)KKspsp = [Ag = [Ag1+1+]]22[SO[SO332-2-] = 1.2 x 10] = 1.2 x 10-5-5

NaCl(s)NaCl(s) KKspsp = [Na = [Na1+1+][Cl][Cl1-1-] = 6.2] = 6.2

Solubility Problem17.58. 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

Ethylenediamine- tetraacetic 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 d-orbital 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)6

2+ --> 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

Fig. 17.15: The mineral turquoise, CuAl6(PO4)4(OH)84H2O, has Cu2+ ions occupying irregular shaped octahedral holes. Consequently, the crystal field spitting produces four energy levels among the d-orbitals.

How many d-electrons does Cu2+ have?

Tetrahedral Crystal Field Splitting

Magnetism and Spin States

Magnetism and Spin StatesBoth the high- and low-spin 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.

ΔE = geμBB0 (i.e. the splitting is proportional to the magnetic field strength (B0)

An unpaired electron can move between the two energy levels by either absorbing or emitting electromagnetic radiation of energy ε = hν

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]3-

Aquapentacyanoferrate(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)2Cl2cis-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.

Oxidizing Strength of Al3+, Ni, and Cu2

+

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) C) Cu2

+(aq)

Oxidizing Strength of Al3+, Ni, and Cu2

+

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).

Ag/Cu Electrochemical Cell

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 B) No C) Depends

Ag/Cu Electrochemical Cell

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.

Oxidation of Fe2+ by Ag+

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) B) C)

Oxidation of Fe2+ by Ag+

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.

pH of an Electrochemical Cell

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  = 0.80 V

2 H+ + 2 e- → H2  = 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 B) pH = 0 C) pH < 0

redEredE

pH of an Electrochemical Cell

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.

Cu/Cu2+ Concentration 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/Cu2+ 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.