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Chapter 24 Transition Metals and Coordination Compounds TMHsiung ©2015 Chapter 24 Slide 1 of 60 Contents 1. 2. 3. 4. 5. 6. The Colors of Rubies and Emeralds Properties of Transition Metals Coordination Compounds Structure and Isomerization Bonding in Coordination Compounds Applications of Coordination Compounds TMHsiung ©2015 Chapter 24 Slide 2 of 60 1. The Colors of Rubies and Emeralds ***** Rubies: About 1% of the Al3+ ions in Al2O3 are replaced by Cr3+. Emeralds: About 1% of the Al3+ ions in Be3Al2(SiO3)6 are replaced by Cr3+. TMHsiung ©2015 Chapter 24 Slide 3 of 60 Werner’s Theory of Coordination Chemistry Old formula mole ions/ mole compound (conductivity exp) mole AgCl ppt by AgNO3/ per mole cpd Correct Werner formula CoCl3‧6NH3 4 3 [Co(NH3)6]Cl3 CoCl3‧5NH3 3 2 [Co(NH3)5Cl]Cl2 CoCl3‧4NH3 2 1 [Co(NH3)4Cl2]Cl CoCl3‧3NH3 0 0 [Co(NH3)3Cl3] The two compounds have very similar formulas but are very different in appearance because of the different chemical structures. TMHsiung ©2015 Chapter 24 Slide 4 of 60 2. Properties of Transition Metals General energy ordering of orbitals for multielectron Atoms: TMHsiung ©2015 Chapter 24 Slide 5 of 60 First-Row Transition Metal Orbital Occupancy ***** TMHsiung ©2015 Chapter 24 Slide 6 of 60 Properties of the First-Row Transition Metals TMHsiung ©2015 Chapter 24 Slide 7 of 60 Electron configurations for transition metals [noble gas]ns2(n-1)dx [noble gas]ns2(n-2)f14(n-1)dx x: 1~10 Electron configurations for transition metals’ ion losing electrons from the ns orbital before losing electrons from the (n - 1)d orbitals. TMHsiung ©2015 Chapter 24 Slide 8 of 60 Example 24.1 Write the ground state electron configuration for Zr. Solution [Kr] 5s24d2 Example 24.2 Write the ground state electron configuration for Co3+. Solution ***** For Co [Ar] 3d7 4s2 For Co3+ [Ar] 3d6 TMHsiung ©2015 Chapter 24 Slide 9 of 60 Atomic Size • The third transition series atoms are about the same size as the second because of the lanthanide contraction. TMHsiung ©2015 Chapter 24 Slide 10 of 60 Lanthanide contraction The decrease in expected atomic size for the third transition series atoms that come after the lanthanides. • 14 between the second and third series go into 4f orbitals. • Electrons in f orbitals are not as good at shielding the valence electrons. • The result is a greater effective nuclear charge increase and therefore a stronger pull on the valence electrons—the lanthanide contraction. TMHsiung ©2015 Chapter 24 Slide 11 of 60 Ionization Energy • The first IE of the transition metals slowly increases across a series. • The first IE of the third transition series is generally higher than the first and second series – lanthanide contraction TMHsiung ©2015 Chapter 24 Slide 12 of 60 Electronegativity • The electronegativity of the transition metals slowly increases across a series. Except for last element in the series. • Electronegativity slightly increases between first and second series, but the third transition series atoms are about the same as the second. Trend opposite to main group elements TMHsiung ©2015 Chapter 24 Slide 13 of 60 Oxidation States Unlike main group metals, transition metals often exhibit multiple oxidation states. TMHsiung ©2015 Chapter 24 Slide 14 of 60 3. Coordination Compounds Terminology A complex consists of a central atom, which is usually a metal atom or ion, and attached groups (anions or neutral molecule) called ligands. If a complex carries a net electric charge, it is called a complex ion. When a complex ion combines with counterions to make a neutral compound, it is called a coordination compound. The total number of points at which a central atom or ion attaches ligands, called coordination number. TMHsiung ©2015 Chapter 24 Slide 15 of 60 Bonding in Complex Coordinate covalent bond (dative bond): The covalent bonding between two atoms in which both electrons come from only one of the atoms (of the ligand). Central metal atom (ion): Lewis acid, electron pair acceptor Ligand: Lewis base, electron pair donor Monodentate : Ligands that donate only one electron pair to the central metal, for example: H2O NH3, Cl−. Chelating agent (chelator): polydentate ligand, for example: • Ethylenediamine (en): bidentate • Oxalato (ox): bidentate • Ethylenediaminetetraacetato (EDTA): hexadentate Chelate: A complex ion that contains either a bidentate or polydentate. TMHsiung ©2015 Chapter 24 Slide 16 of 60 Common Ligands TMHsiung ©2015 Chapter 24 Slide 17 of 60 Four Common Structures of Complex Ions TMHsiung ©2015 Chapter 24 Slide 18 of 60 Bidentate Ligands Coordinated to Co3+ TMHsiung ©2015 Chapter 24 Slide 19 of 60 Hexadentate Ligands Coordinated to Co3+ TMHsiung ©2015 Chapter 24 Slide 20 of 60 Example What are the coordination number and the oxidation number of the central atom in (a) [CoCl4(NH3)2]– and (b) [Ni(CO)4]? Solution: (a) Co: center atom, 6 ligands attached (4 Cl– and 2 NH3) coordination number: 6. Charge calculation: x – 4 + 0 = –1, x = +3 central ion is Co3+, oxidation number is +3. (b) coordination number: 4 central atom is Ni, oxidation number is 0. Werner’s definition: • The primary valence is the oxidation number of the metal. • The secondary valence is the number of ligands bonded to the metal (coordination number). TMHsiung ©2015 Chapter 24 Slide 21 of 60 Naming Coordination Compounds ***** 1. Identify the cation and anion, either may be complex ion or uncomplex ion. 2. Naming complex cation and/or complex anions. 3. Write the compound name as the name of cation followed by the name of the anion. TMHsiung ©2015 Chapter 24 Slide 22 of 60 Names and Formulas of Common Ligands: TMHsiung ©2015 Chapter 24 ***** Slide 23 of 60 Naming complex cations: • Ligand first, metal (with oxidation number written in Roman numerals) after. • Name the ligands in alphabetical order (ignoring Greek numeric prefixes). • Designate the number of ligands in a complex with a Greek numeric prefix: di = 2, tri = 3, tetra = 4, hexa = 6. TMHsiung ©2015 Chapter 24 Slide 24 of 60 Example Name [CrCl2(NH3)4]+ Solution: a complex cation 4 NH3: tetraammine, 2 Cl–: dichloro alphabetical order (ignoring Greek numeric prefixes): tetraamminedichloro complex cation: unmodified name for central metal. Cr oxidation number +3 (x – 2 + 0 = +1) Ans: tetraamminedichlorochromium(III) ion. TMHsiung ©2015 Chapter 24 Slide 25 of 60 Naming complex anion: • Ligand first, metal (with oxidation number written in Roman numerals) after. • Name the ligands in alphabetical order (ignoring Greek numeric prefixes). • Designate the number of ligands in a complex with a Greek numeric prefix: di = 2, tri = 3, tetra = 4, hexa = 6. • For metal in complex anion: Replace the ending from –um to –ate. Certain metals in complex anions, use the Latinbased names: Copper Cuprate Gold Iron Lead Silver Tin TMHsiung ©2015 Chapter 24 Aurate Ferrate Plumbate Argentate Stannate Slide 26 of 60 Example 22.4 Name [PtBrCl2NH3]Solution: a complex anion 1 NH3: ammine, 1 Br–: bromo, 2 Cl–: dichloro alphabetical order: amminebromodichloro complex anion ending –ate Pt oxidation number: +2 (x – 1 – 2 + 0 = –1) Ans: amminebromodichloroplatinate(II). TMHsiung ©2015 Chapter 24 Slide 27 of 60 Examples of Naming Coordination Compounds Identify the cation and anion, and the name of the uncomplex ion. Give each ligand a name and list them in alphabetical order. 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. © 2014 Pearson Education, Inc. TMHsiung ©2014 ***** Name [Cr(H2O)5Cl]Cl2 Name K3[Fe(CN)6] [Cr(H2O)5Cl]2+ is a complex cation; Cl− is chloride. K+ is potassium; [Fe(CN)6]3− is a complex anion. 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). Chapter 24 Slide 28 of 42 When writing a complex formula from name: • Cation first, anion after • For the complex Center metal first, ligands after Place the ligands in alphabetical order (ignoring Greek numeric prefixes) TMHsiung ©2015 Chapter 24 Slide 29 of 60 Example Write the formula for sodium hexanitrocobaltate(III). ***** Solution: Coordination compound: made up of Na+ cations and a complex anion. Complex anion charge: –3 (1 Co3+: +3, 6NO2–: –6) Cation first, anion after: Ans: Na3[Co(NO2)6]. TMHsiung ©2015 Chapter 24 Slide 30 of 60 4. Structure and Isomerization ***** TMHsiung ©2015 Chapter 24 Slide 31 of 60 1) Structural isomers ***** i) Coordination isomers: the structural isomers occur when coordinated ligand exchanges places with the uncoordinated counterion, for example: [Co(NH3)5Br]Cl pentaamminebromocobalt(II) chloride [Co(NH3)5Cl]Br pentaamminechlorocobalt(II) bromide TMHsiung ©2015 Chapter 24 Slide 32 of 60 ii) Linkage isomers: the structural isomers that have ligands attached to the central cation through different ends of the ligand structure, for example: [Co(NH3)5(NO2)]Cl2 [Co(NH3)5(ONO)]Cl2 TMHsiung ©2015 Chapter 24 Slide 33 of 60 Continued Yellow = pentaamminenitrocobalt(III) chloride [Co(NH3)5(NO2)]Cl2 Red = pentaamminenitritocobalt(III) chloride [Co(NH3)5(ONO)]Cl2 TMHsiung ©2015 Chapter 24 Slide 34 of 60 2) Stereoisomers i) Geometric isomers a) Cis-trans isomers (MA2B2 and MA4B2 type) TMHsiung ©2015 Chapter 24 ***** Slide 35 of 60 b) Fac-mer isomers (MA3B3 type) TMHsiung ©2015 Chapter 24 Slide 36 of 60 Example 24.5 Drawing Geometric Isomers of [Co(en)2Cl2]+. Solution MA4B2 type, cis–trans isomers. TMHsiung ©2015 Chapter 24 Slide 37 of 60 ii) Optical isomers (enantiomers): Molecules that are nonsuperimposable (not identical) mirror images of one another, like right and left hands. Each enantiomer rotates polarized light in opposite directions. For example: Mirror image of each other Two nonsuperimposable (not identical) structures Ans: two optical isomers TMHsiung ©2015 Chapter 24 Slide 38 of 60 Example 24.7a Determine whether the cis isomer of [Co(en)2Cl2]+ is optically active. Solution the two structures are not superimposable, so the cis isomer does exhibit optical activity. TMHsiung ©2015 Chapter 24 Slide 39 of 60 Example 24.7b Determine whether the trans isomer of [Co(en)2Cl2]+ is optically active. Solution In this case the two are identical, so there is no optical activity. TMHsiung ©2015 Chapter 24 Slide 40 of 60 5. Bonding in Coordination Compounds Valence Bond Theory/Hybridization of Atomic Orbitals TMHsiung ©2015 Chapter 24 Slide 41 of 60 Continued electron configuration of Ag: 4d105s1, Ag+: 4d10 for [Ag(NH3)2]+ sp hybridization 4d 5s 5p electron configuration of Zn: 3d104s2 , Zn2+: 3d10 for [Zn(NH3)4]2+ sp3 hybridization 3d TMHsiung ©2015 4s Chapter 24 4p Slide 42 of 60 Continued electron configuration of Pd: 4d10, Pd2+: 4d8 for [PdCl4]2dsp2 hybridization 4d 5s 5p electron configuration of Fe: 3d64s2, Fe3+: 3d5 for [Fe(H2O)6]3+ d2sp3 hybridization 3d TMHsiung ©2015 4s Chapter 24 4p Slide 43 of 60 Crystal Field Theory ***** Assume the attractions between a central atom (or ion) and its ligands are largely electrostatic. Ligands distort the d-orbitals of the central atom, leading to a splitting of energy levels of those orbitals. • Splitting energy (Δ): The splitting of energy levels of dorbitals which are caused by ligands distoration of those orbitals. • The spectrochemical series shows the relative abilities of ligands (Δ) to split the d-orbital energy levels: ***** * Increases the charge on the metal cation also increases the splitting energy (Δ), for example: Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+ TMHsiung ©2015 Chapter 24 Slide 44 of 60 ***** Schematic representation of d-level splitting: Δ Δ Usually Using High Spin Spectrochemical series TMHsiung ©2015 Chapter 24 Usually Low Spin Slide 45 of 60 ***** Crystal field theory can predict: Magnetism Paramagnetic, with spin (unpaired electron) Diamagnetic, without spin Magnetic strength High spin, more unpaired electron (Δ small) Low spin, less unpaired electron, (Δ large) Complex color: According to the energy level transition. TMHsiung ©2015 Chapter 24 Slide 46 of 60 ]2+ Example: Predict the magnetism for [Fe(H2O)6 [Fe(CN)6]4-. Solution: Both are octahedral complexes Fe: [Ar]3d64s2, Fe2+: [Ar]3d6 (six 3d electrons) Δ: CN– (large) > H2O (small) and ***** Diamagnetic Paramagnetic Low-spin complex High-spin complex * From d1 through d10 metal ion in octahedral complexes, only electron d4, d5, d6, or d7 can have low and high spin possibilities. TMHsiung ©2015 Chapter 24 Slide 47 of 60 ***** Example How many unpaired electrons would you expect for the octahedral complex ion [CoF6]3–? Solution: Electron configuration: Co: [Ar]3d74s2 Co3+: [Ar]3d6 (six 3d electrons) F– ligand: Δ small Ans: 4 unpaired electrons TMHsiung ©2015 Chapter 24 Slide 48 of 60 How many unpaired electrons would you expect for the octahedral complex ion [Co(NH3)5NO2]2+? Solution: Electron configuration: Co: [Ar]3d74s2 Co3+: [Ar]3d6 (six 3d electrons) NH3 and NO2-: Δ large Ans: No unpaired electrons TMHsiung ©2015 Chapter 24 Slide 49 of 60 Example How many unpaired electrons would you expect to find in the tetrahedral complex ion [NiCl4]2–? Solution: Electron configuration: Ni: [Ar]3d84s2 Ni2+: [Ar]3d8 (Tetrahedral, usually Δ small, high spin) Ans: 2 unpaired electrons TMHsiung ©2015 Chapter 24 Slide 50 of 60 Example How many unpaired electrons would you expect to find in the square planar complex ion [PtCl4]2–? Solution: Electron configuration: Pt: [Xe]5d96s1 Pt2+: [Xe]5d8 (Square planar, usually Δ large, low spin) ***** Ans: no unpaired electrons a diamagnetic species TMHsiung ©2015 Chapter 24 Slide 51 of 60 Color In Complex Ions And Coordination Compounds Many complex ions are colored because the energy differences between d orbitals match the energies of components of visible light. Crystal field theory helps to explain the colors of complex ions. Ions having the following electron configurations have no electron transitions in the energy range of visible light (colorless): No electron in d orbital (d0-complex), e.g., Sc3+, Y3+, La3+ (noble-gas electron configuration) are colorless. Electrons completely filled in d orbital (d10-complex), e.g., Zn2+, Cd2+, Hg2+, Cu+, and Ag+ are colorless. TMHsiung ©2015 Chapter 24 Slide 52 of 60 The color of the transmitted light is the complementary color of the absorbed light. The Color Wheel: Colors across from one another on the color wheel are said to be complementary. TMHsiung ©2015 Chapter 24 Slide 53 of 60 Continued Large Δ High ν Short λ The Color of [Ti(H2O)6]3+ solution TMHsiung ©2015 Chapter 24 Small Δ Low ν Long λ The absorption pectrum [Ti(H2O)6]3+ solution Slide 54 of 60 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 TMHsiung ©2015 Chapter 24 Slide 55 of 60 ***** Example 24.8 The complex ion [Cu(NH3)6]2+ is blue in aqueous solution. Estimate the crystal field splitting energy (in kJ/mol) for this ion. Solution The color orange ranges from 580 to 650 nm, so you can estimate the average wavelength as 615 nm. E = hc/λ. Convert J/ion into kJ/mol. TMHsiung ©2015 Chapter 24 Slide 56 of 60 6. 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. TMHsiung ©2015 Chapter 24 Slide 57 of 60 TMHsiung ©2015 Chapter 24 Slide 58 of 60 ***** In hemoglobin, the iron complex is octahedral, with the four nitrogen atoms of the porphyrin in a square planar arrangement around the metal. A nitrogen atom from a nearby amino acid of the protein occupies the fifth coordination site, and either O2 or H2O occupies the last coordination site. TMHsiung ©2015 Chapter 24 Slide 59 of 60 End of Chapter 24 TMHsiung ©2015 Chapter 24 Slide 60 of 60