chapter 23 transition metals and coordination...
TRANSCRIPT
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Chapter 23
Transition Metals and Coordination
Chemistry
Yonsei University
Lecture Presentation
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23.1 The Transition Metals
Minerals• Most metals, including transition metals, are found in solid
inorganic compounds known as minerals.
• Minerals are named by common, not chemical, names.
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Metallurgy
Metallurgy: the science and technology of extracting metals from natural sources and preparing them for practical use.
• mining (getting the ore out of the ground)
• concentrating (preparing it for further treatment) – Differences in the chemical and physical properties of the mineral of
interest and the undesired material, called gangue, are used to separate these components.
– Example: Iron can be separated from gangue in finely ground magnetite by using a magnet to attract the iron.
• reduction (to obtain the free metal in the 0 oxidation state)
• purifying or refining (to obtain the pure metal)
• mixing with other metals (to form an alloy)– Alloys are metallic materials composed of two or more elements.
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Atomic RadiiAs one goes from left to right across a row, we see a decrease, then an increase in the radius of transition metals.
• In general, increasing effective nuclear charge tends to make atoms smaller.
• On the other hand, the strongest (and, therefore, shortest) metallic bonds are found in the center of the transition metals.
• The lanthanide contractionbalances the increase in size we anticipate between Hf (4B) and Zr.
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Traits of Transition Metals
• Because most transition metals have only partially occupied d subshells, the metals and/or their compounds often– Have more than one
oxidation state, commonly +1, +2 or +3
– Are pigmented
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Traits of Transition Metals
• Because most transition metals have only partially occupied d subshells, the metals and/or their compounds often– Have more than one
oxidation state,
– Are pigmented,
– Have magnetic properties.
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Paramagnetism
• Paramagnetism, as you recall from Chapter 9, results from an atom having unpaired electrons.
• Such atoms, then, show attraction to a magnet placed close by.
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Ferromagnetism
• In ferromagnetic substances, the unpaired spins influence each other to align in the same direction, thereby exhibiting strong attractions to an external magnetic field.
• Such species are permanent magnets.– SmCo5, Nd5Fe14B
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Antiferromagnetism
• Antiferromagnetic substances have unpaired spins on adjacent atoms that align in opposing directions.
• These magnetic fields tend to cancel each other.
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Ferrimagnetism
• Ferrimagnetic substances have spins that align opposite each other, but the spins are not equal, so there is a net magnetic field.
• Examples are NiMnO3, Y3Fe5O12, and Fe3O4.
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23.2 Transition-Metal Complexes
• Commonly, transition metals can have molecules or ions that bond to them.
• These give rise to complex ions or coordination compounds.
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Ligands
The molecules or ions that bind to the central metal are called ligands (from the Latin ligare, meaning “to bind”).
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Coordination
One of the properties that has led to the fascination with complexes and transition metals is the wide range of stunning colors found in them.
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Werner Theory
• Alfred Werner (Swiss) deduced that there was a difference between the oxidation number of a metal and the number of ligands it took on, which he called the coordination number.
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• Thus, although the first two complexes in the table above each have 3 chlorines, in the first all three serve as anions, while in the second one of the chlorines is tightly bound to the cobalt and the other two are counterions.
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Sample Exercise 23.1 Identifying the Coordination Sphere of a Complex
Palladium(II) tends to form complexes with coordination number 4. A compound has the composition PdCl2 · 3 NH3. (a) Write the formula for this compound that best shows the coordination structure. (b) When an aqueous solution of the compound is treated with excess AgNO3(aq), how many moles of AgCl(s) are formed per mole of PdCl2 · 3 NH3?
Solution(a)[Pd(NH3)3Cl]Cl(b)[Pd(NH3)3Cl]Cl(aq) + AgNO3(aq) → [Pd(NH3)3Cl]NO3(aq) + AgCl(s)
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Sample Exercise 23.2 Determining the Oxidation Number of a Metal in a Complex
What is the oxidation number of the metal in [Rh(NH3)5Cl](NO3)2?
Solutionx + 5(0) + (–1) + 2(–1) = 0
[Rh(NH3)5Cl](NO3)2
The oxidation number of rhodium, x, must therefore be +3.
Practice ExerciseWhat is the charge of the complex formed by a platinum(II) metal ion surrounded by two ammonia molecules and two bromide ions?
Answer: zero
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The Metal–Ligand Bond
• The reaction between a metal and a ligand is a reaction between a Lewis acid (the metal) and Lewis base (the ligand).
• The new complex has distinct physical and chemical properties.– Ag+(aq) + e– Ag(s) E° = +0.799 V
– [Ag(CN)2]–(aq) + e– Ag(s) + 2CN– (aq) E° = –0.31 V
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Coordination Numbers
• The coordination number of a metal depends upon the size of the metal and the size of the ligands.
• CN=4 or 6– CN=6 for Cr3+ and Co3+
– CN=4 for d8 metal ions such as Pt2+ and Au3+.
• While iron(III) can bind to 6 fluorides, it can only accommodate 4 of the larger chlorides.– [FeF6]3– vs. [FeCl4] –
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23.3 Common Ligands in Coordination Chemistry
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Monodentateligands coordinate to one site on the metal, bidentate to two, and so forth.
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chelating agents
Bi and polydentate ligands are also called chelating agents.
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Chelate effect
• [Ni(H2O)6]2+(aq) + 6NH3 [Ni(NH3)6]2+(aq) + 6H2O(l)
Kf = 1.2 109
• [Ni(H2O)6]2+(aq) + 3en [Ni(en)3]2+(aq) + 6H2O(l)
Kf = 6.8 1017
Cf) sequestering agents are used to selectively remove toxic metal ions (e.g., Hg2+ and Pb2+) while leaving biologically important metals.
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Chelates in Biological Systems
• There are many transition metals that are vital to human life.– V, Cr, Mn, Fe, Co, Cu, Zn,
Mo, Cd, and Ni
• Several of these are bound to chelating agents.– porphine
– porphyrin : a metal complex derived from porphine.
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Myoglobin
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Chelates in Biological Systems
• For instance, the iron in hemoglobin carries O2and CO2 through the blood.
• CO and CN- are poisonous because they will bind more tightly to the iron than will oxygen.– Kf (-CO) > 210 Kf(-O2)
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• porphyrins that contain Mg(II)
• Chlorophyll a
• Mg2+ is in the center
• conjugated double bonds
Chlorophylls6CO2 + 12H2O C6H12O6 + 6O2 + 6H2O
Photosynthesis: cholophyll-containing pigments in the plant
leaves absorbs photons glucose formation
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23.4 Nomenclature and Isomerism in Coordination Chemistry
Nomenclature
1. In naming complexes that are salts, the name of the cation is given before the name of the anion: cation + anion
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Nomenclature in Coordination Chemistry
2. In naming complex ions or molecules, the ligands are named before the metal. Ligands are listed in alphabetical order, regardless of their charges. :
Ligand (alphabetical order)+ metal
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Nomenclature in Coordination Chemistry
3. The names of anionic ligands end in the letter o, but electrically neutral ligands ordinarily bear the name of the molecules.
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Nomenclature in Coordination Chemistry
4. Greek prefixes (di-, tri-, tetra-, etc.) are used to indicate the number of each kind of ligand when more than one is present. If the ligand contains a Greek prefix or is polydentate, the prefixes bis-, tris-, tetrakis-, etc. are used and the ligand name is placed in parentheses.
– [Co(en)3]Cl3 tris(ethylenediamine)cobalt(III) chloride.
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Nomenclature in Coordination Chemistry
5. If the complex is an anion, its name ends in -ate.
6. The oxidation number of the metal is given in parentheses in Roman numerals following the name of the metal.
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Sample Exercise 23.4 Naming Coordination Compounds
Name the compounds (a) [Cr(H2O)4Cl2]Cl, (b) K4[Ni(CN)4].
+3 + 4(0) + 2(–1) + (–1) + 0
[Cr(H2O)4Cl2]Cl
tetraaquadichlorochromium(III) chloride
4(+1) + 0 + 4(–1) = 0
K4[Ni(CN)4]
potassium tetracyanonickelate(0)
Solution
Practice ExerciseName the compounds (a) [Mo(NH3)3Br3]NO3, (b) (NH4)2[CuBr4]. (c) Write the formula for sodium diaquabis(oxalato)ruthenate(III).Answer: (a) triamminetribromomolybdenum(IV) nitrate, (b) ammonium tetrabromocuprate(II) (c) Na[Ru(H2O)2(C2O4)2]
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Practice ExerciseName the compounds (a) [Mo(NH3)3Br3]NO3, (b) (NH4)2[CuBr4]. (c) Write the formula for sodium diaquabis(oxalato)ruthenate(III).Answer: (a) triamminetribromomolybdenum(IV) nitrate, (b) ammonium tetrabromocuprate(II) (c) Na[Ru(H2O)2(C2O4)2]
Sample Exercise 23.4 Naming Coordination CompoundsContinued
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Isomers
• Isomers have the same molecular formula but a different arrangement of atoms.
• There are two main subgroupings: structural isomers and stereoisomers.
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Linkage Isomers
In linkage isomers the ligand is bound to the metal by a different atom.
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Coordination Sphere Isomers
• Coordination sphere isomers differ in what ligands are bound to the metal and which fall outside the coordination sphere.
• For example, CrCl3(H2O)6 exists as– [Cr(H2O)6]Cl3,
– [Cr(H2O)5Cl]Cl2H2O, or
– [Cr(H2O)4Cl2]Cl2H2O.
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Geometric Isomers
• In geometric isomers, the ligands have a different spatial relationship.
• In the complexes above, the chlorines can be adjacent to each other (cis) or opposite each other (trans).
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The Lewis structure indicates that the CO molecule has two lone pairs of electrons. When CO binds to a transition-metal atom, it nearly always does so by using the C lone pair. How many geometric isomers are there for tetracarbonyldichloroiron(II)?
Sample Exercise 23.5 Determining the Number of Geometric Isomers
Solutiontwo isomers possible: one with the Cl– ligands across the metal from each other, trans-[Fe(CO)4Cl2], and one with the Cl– ligands adjacent to each other, cis-[Fe(CO)4Cl2].
Practice ExerciseHow many isomers exist for the square-planar molecule [Pt(NH3)2ClBr]?Answer: two
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Optical Isomers
Optical isomers, or enantiomers, are non-superimposable mirror images of one another.
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Detection of Optical Activity
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Sample Exercise 23.6 Predicting Whether a Complex Has Optical Isomers
Does either cis-[Co(en)2Cl2]+ or trans-[Co(en)2Cl2]– have optical isomers?
SolutionThe trans isomer of [Co(en)2Cl2]+ and its mirror image are:
The mirror image of the cis isomer cannot be superimposed on the original: Thus, the two cis structures are optical isomers (enantiomers). We say that cis-[Co(en)2Cl2]+ is a chiral complex.
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Sample Exercise 23.6 Predicting Whether a Complex Has Optical Isomers
Continued
Practice ExerciseDoes the square-planar complex ion [Pt(NH3)(N3)ClBr]– have optical isomers? Explain your answer.
Answer: no, because the complex is flat. This complex ion does, however, have geometric isomers (for example, the Cl and Br ligands could be cis or trans).
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Racemic mixtures
• Dextrorotatory solutions rotate the plane of polarized light to the right. – dextro or d isomer.
• Levorotatory solutions rotate the plane of polarized light to the left. – levo or l isomer.
• Racemic mixtures contain equal amounts of land d isomers.
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23.5 Color and Magnetism in Coordination Chemistry
• The complex [Ti(H2O)6]3+
at the right appears red-violet because those are the wavelengths of visible light not absorbed by the complex.
• Many complexes are pigmented because they absorb in the visible part of the spectrum.
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Crystal-Field Theory
• As was mentioned earlier, ligands are Lewis bases that are attracted to a Lewis acid (the metal).
• But d electrons on the metal would repelthe ligand.
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23.6 Crystal-Field Theory
• Therefore, the d orbitals on a metal in a complex would not be degenerate.
• Those that point toward ligands would be higher in energy than those that do not.
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Crystal-Field Theory
The energy gap between d orbitals often corresponds to the energy in a photon of visible light.
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Spectrochemical Series
The spectrochemical series ranks ligands in order of their ability to increase the energy gap between d orbitals.
• Weak-field ligands : low-∆ end
• Strong-field ligands : high-∆ end
• Example: [CrF6]3- [Cr(CN)6]3–
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Crystal-Field Theory
The stronger the crystal-field strength of the ligand, the larger the energy gap between d orbitals, and the shorter the wavelength of light absorbed by the complex.
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[Ti(H2O)6]3+, [Ti(en)3]3+, and [TiCl6]3– all absorb visible light. Which one absorbs at the shortest wavelength?
Sample Exercise 23.8 Using the Spectrochemical Series
Solution[Ti(en)3]3+.
Practice ExerciseThe absorption spectrum of a Ti(III) complex containing the ligand L, [TiL6]3–, shows a peak maximum at a wavelength intermediate between the wavelengths of the absorption maxima for [TiCl6]3– and [TiF6]3–.What can we conclude about the place of L in the spectrochemical series?
Answer: It lies between Cl– and F–.
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Electron Configurations in Octahedral Complexes
low-spin complexes
high-spin complexes
∆ vs. spin-pairing energy
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high-spin complexes vs.low-spin complexes
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Sample Exercise 23.9 Predicting the Number of Unpaired Electrons in an Octahedral Complex
Predict the number of unpaired electrons in high-spin and low-spin Fe3+
complexes that have a coordination number of 6.
SolutionFe3+ is a d5 ion. In a high-spin complex, all five electrons are unpaired, with three in the t2 orbitals and two in the e orbitals. In a low-spin complex, all five electrons reside in the t2 set, so there is one unpaired electron:
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Tetrahedral Complexes
• Because there are only four ligands, ∆Td <∆O
• This causes all tetrahedral complexes to be high-spin.
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Crystal Field Splitting
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Square Planar Complexes
• Most d8 metal ions form square-planar complexes.
• The majority of complexes are low-spin (i.e., diamagnetic).
• Examples: Pd2+, Pt2+, Ir+, and Au3+.
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Nickel(II) complexes in which the metal coordination number is 4 can have either squareplanar or tetrahedral geometry. [NiCl4]2– is paramagnetic, and [Ni(CN)4]2– is diamagnetic. One of these complexes is square planar, and the other is tetrahedral. Use the relevant crystalfield splitting diagrams in the text to determine which complex has which geometry.
Solution
Sample Exercise 23.10 Populating d Orbitals in Tetrahedral and Square-Planar Complexes
[NiCl4]2– is tetrahedral, and [Ni(CN)4]2– is square planar.
Practice ExerciseHow many unpaired electrons do you predict for the tetrahedral [CoCl4]2– ion?Answer: three
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Problems
• 4,12,20,28,32,38,40,56,60,70,72