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Chem 222 Problem Set 5 Answer key Intro to Inorganic Chemistry Summer 2011 Wednesday July 20, 2011 1. (H&S 20.10) (a) In the solid state, Fe(CO)5 possesses a trigonal bipyramidal structure. How many carbon environments are there? The trigonal bipyramidal structure places five chemically equivalent CO groups in two different spatial environments – two COs are in axial positions, and three CO’s occupy equatorial positions. CO axial equatorial OC Fe CO equatorial CO equatorial CO axial (b) Explain why only one signal is seen in the 13C NMR spectrum of solutions of Fe(CO)5, even at low temperature. This observation indicates that, in solution, this five-coordinate complex is undergoing Berry pseudorotation: rapid conversion between a trigonal bipyramidal geometry and a square-based pyramidal geometry. This has the effect of exchanging all five carbonyl groups between different environments around the iron, which gives an averaged 13C NMR shift. 2. (from H&S 20.11) Comment on the coordination number and the bond angle data for the iron complex shown below (what bond angle data would you normally expect for this geometry?). (Me3Si)3Si 137¡ 111¡ Fe Cl 112¡ (Me3Si)3Si This iron complex is three-coordinate, which is unusually low. It should have trigonal planar geometry: the sum of the bond angles is 360°, which is consistent with planarity at Fe, but the angles are distorted from the usual 120° we’d expect. This is because the two bulky “silyl” ligands, SiR3– (where R = Si(CH3)3) take up a lot of space in the coordination sphere. Thus the Si-FeSi angle is widened to 137° and the two Si-Fe-Cl angles are compressed. It is probably the size of these silyl ligands that stabilizes the low coordinate structure. 3. (from H&S 20.14) (a) The salt [Co(bpy)3]3+[Fe(CN)6]3– has an isomer of the formula [Co(CN)4(bpy)]–[Fe(CN)2(bpy)2]+ What type of isomerism do these two salts demonstrate? Coordination isomerism (arises from exchange of ligands between two complex ions in a salt). (b) What other types of isomerism exist for each of the complex ions in (a)? (Draw the structures and label the isomers appropriately.) [Co(bpy)3]3+ is chiral, and exists as the two stereoisomers shown below – they are enantiomers. N 3+ N N N Co N N Co N N 3+ N N N N [Co(CN)4(bpy)]– is limited to the cis diastereomer (a stereoisomer) because the bite angle of the bpy ligand is too small to allow the nitrogens to be trans to one another. Chem 222 Intro to Inorganic Chemistry CN N N Co NC CN trans CN N CN CN + N Fe CN N CN N N Fe N CN N + N N Fe Ğ N N + N Summer 2011 CN cis cis + [Fe(CN)2(bpy)2] exhibits both types of stereoisomerism. It has cis and trans isomers (diastereomers), and the cis isomer is chiral: it exists as a pair of enantiomers. 4. (H&S 20.20a) The conjugate base (i.e. deprotonated form) of the fluoro-substituted acetylacetone (aka pentanedione) shown below forms the complex [CoL3], which has mer and fac isomers. Draw the structures of these isomers, and explain why the labels mer and fac are used. H3 C CF3 The labels mer and fac indicate the relative arrangements of “like” ends of this unsymmetric chelating ligand. In the fac isomer, the three oxygens close to CH3 (and those close to CF3) occupy one face of the octahedron, while in the mer isomer, these “like” oxygens each trace out one edge of the octahedron. O O H3C F3C H3C F3C O O O O CF3 O Co H3C O O O F3C fac O CF3 Co CH3 H3C O O O CF3 H3C mer 5. (a) Draw the structure for an ionization isomer of [CoCl(NH3)5]SO4, indicating clearly features of the inner and outer coordination spheres. (b) Draw the structure for a hydration isomer of [Fe(OH2)6]Cl2, indicating clearly features of the inner and outer coordination spheres. Chem 222 Intro to Inorganic Chemistry Summer 2011 6. (a) (from H&S 21.1) Outline how you would apply crystal field theory to explain why the five dorbitals in an octahedral complex are not degenerate. (Include a labeled splitting diagram.) The CFT model sees ligands as negative point charges attracted to the positive charge of the metal ion, but repelled by the d-electrons. As the ligands are brought close to the metal, the d-orbitals are destabilized by this repulsion. Those d-orbitals with lobes pointing directly at the ligands in an octahedral field (i.e. six L approaching along x, y , and z axes) are most destabilized by this interaction: dx2-y2 and dz2. Those d-orbitals with lobes pointing between the approaching ligands are not destabilized as much: dxy, dxz, dyz (Relative to a spherical crystal field, these latter three are actually stabilized.) dz2, dx2-y2 eg Oh E dxy, dxz, dyz t2g free metal ion metal ion in spherical crystal field metal ion in octahedral crystal field (b) Based on the analogous splitting diagram, as derived from the MO diagram of a simple octahedral complex, explain why a strong -donor ligand will give a large Oh value relative to a weaker -donor. eg* -antibonding Oh E non-bonding t2g metal ion d-orbitals ligand donor orbitals -bonding This diagram shows only the d-orbitals interacting with ligand sigma-donor orbitals in an octahedral complex. (We’ve left out the s and p orbitals on the metal, and the four other sigma bonding and sigma antibonding sets above and below this diagram in energy.) Two of the d-orbitals are of appropriate symmetry to interact with two of the ligand orbitals in a sigma fashion. The interaction stabilizes two of the ligand orbitals (sigma bonding) and destabilizes two of the d-orbitals (sigma antibonding), and leaves three of the d-orbitals unchanged in energy (non-bonding). If the ligand is a strong sigma donor, the sigma bonding orbitals will be lower in energy, because a stronger bonding interaction will occur. Accordingly, the eg* antibonding orbitals will be raised in energy, while the t2g non-bonding orbitals’ energy will remain the same. This increases the separation between the t2g and the eg* sets of d-orbitals (Oh). (c) Why does additional π-donation from a ligand such as a halide ion cause the Oh value to diminish? As shown in the diagram below, filled p orbitals on halide ligands of the appropriate symmetry can overlap with empty non-bonding orbitals on the metal in a π-fashion. This interaction stabilizes the ligands’ orbitals (π) and destabilizes the metal d-orbitals (π*). This gives the t2g set (of d-orbitals) slight antibonding character – and raises them in energy relative to the eg* set, which makes Oh smaller. Chem 222 Intro to Inorganic Chemistry e g* Summer 2011 -antibonding Oh ¹ -antibonding t2g* E filled porbitals ¹ -bonding metal ion d-orbitals ligand donor orbitals -bonding 7. (a) (RC&O 19.33) The complex [Co(NH3)6]Cl3 is yellow-orange, while [Co(OH2)3F3] is blue. Suggest an explanation for the difference in colour. The yellow-orange complex must be absorbing higher energy, blue-violet light, while the blue complex is absorbing lower energy orange light. This is consistent with colours arising from electronic transitions between the t2g and eg sets of d-orbitals in these complexes. The hexaammine complex has a larger Oh (ligand field or crystal field splitting) than the tris(aqua)trifluoro complex, because NH3 is a reasonably strong field ligand ( a strong sigma donor), while H2O is not quite as strong a sigma donor and F– is a π-donor ligand – both are weaker-field ligands than NH3. Because the energy gap is larger, a higher energy light is required to prompt the electronic transition for the hexaammine complex. (b) One of the above complexes is diamagnetic, while the other is paramagnetic, with a µ eff of approximately 5.0 µB at room temperature. Which is which, and why? Both complexes contain Co3+, which has a d6 configuration. If the complex is diamagnetic, it has no unpaired electrons, which corresponds to a low spin d6 arrangement. The µeff value for the paramagnetic complex is close to that predicted for four unpaired electrons by the spin-only formula µeff = (n(n+2))1/2, 4.90µB. This corresponds to a high spin d6 arrangement. The compound with the larger Oh value is most likely to be low spin – hence the hexaammine complex is diamagnetic, while the aqua/fluoro complex, with its smaller Oh must be the high spin complex with four unpaired electrons. eg low spin Oh Oh eg high spin t2g t2g 8. (RC&O 19.17) For which member of the following pairs of complex ions would Oh be greater? Explain your reasoning. (a) [MnF6]2– and [ReF6]2– Mn and Re are both in group 7. For metals in the same triad with analogous ligand sets and oxidation states, the ligand field splitting will be larger for the heavier element(s). (This is because M-L bonding tends to strengthen going down the triad – see 7(b) above.) (b) [Fe(CN)6]4– and [Fe(CN)6]3– The first complex has Fe2+ and the second complex contains Fe3+. For metal complexes with identical M and ligand set, the complex with the metal in the higher oxidation state will have larger ligand field splitting. (This is because oxidation of the metal lowers the d-orbitals in energy, which puts the t2g d-orbital set lower relative to the eg set.) 9. Explain the following: magnetic measurements indicate that [CoCl4]2– has three unpaired electrons, [NiBr4]2– has two unpaired electrons, and [Ni(CN)4]2– is diamagnetic. Include coordination geometries and labeled splitting diagrams in your answer. What µeff values would you expect? Chem 222 Intro to Inorganic Chemistry Summer 2011 2– 2+ 7 [CoCl4] contains Co , which is d . It is probably tetrahedral. Placing seven electrons into the Td dorbital splitting diagram does give an arrangement with three unpaired electrons. It should have a µeff of around 3.87 µB (the spin-only estimated value). recall: dxy, dxz, dyz t2 E Td e dx2-y2, dz2 Both nickel complexes contain four-coordinate Ni2+, which is d8. We might expect them both to have Sq. Pl. geometry, since that is favoured for d8 complexes, but the fact that they have different numbers of unpaired electrons (two, for which the spin-only µeff is calculated to be 2.83µeff, and zero, for which µeff = 0 µB) indicates that they must have different geometries, and hence different d-orbital splitting. Two unpaired electrons is consistent with a tetrahedral d8 arrangement for the chloro complex, while no unpaired electrons is consistent with a square planar arrangement for the cyano complex. (For the nickel chloride complex, the ligand field splitting is small enough that having four electrons in the t2 set is not too destabilizing. For the nickel cyano complex, the ligand field splitting becomes large enough that the square planar geometry is preferred, since it keeps the electrons at lower energy.) recall: dx2-y2 Oh t2 E Td e Td d8 Sq Pl d8 Definitions/Concepts: inner/outer coordination sphere, coordination geometry, Kepert model, electroneutrality principle, isomerism, stereoisomerism, diastereomer, enantiomer, optical isomers, chirality, structural isomerism, ionization isomers, hydration isomers, linkage isomers, coordination isomers, MO diagram for simple ML6 complex with only -bonding, d-orbital splitting, Oh, Td, -donor ligand, π-donor ligand, π-acceptor ligand, weak field ligand, strong field ligand, spectrochemical series, high spin, low spin, crystal field theory for Oh, Td, Sq Pl geometries, max, CO, effective magnetic moment, diamagnetic, paramagnetic, pairing energy.