Problems 14, 15, 16, 19, 22, 24, and 25 in DeKock and Gray, pages 411–414. 14. “All octahedral complexes of V3+ have the same number of unpaired electrons, no matter what the nature of the ligand. Why is this so?” V3+ has a d2 electron configuration; in an octahedral ligand environment, regardless of the type of ligand interactions, the lowest energy orbitals are the triply degenerate 2 t2g orbitals. A d octahedral metal complex will always have 2 unpaired electrons. For the same reason, octahedral d1, d3, d8, d9, and d10 also only have one possible electronic configuration. 15. “How does the molecular-orbital theory account for the order of ligands in the spectrochemical series.” The spectrochemical series (courtesy of Wikipedia): I− < Br− < S2− < SCN− < Cl− < − − − − 2− − NO3 < N3 < F < OH < C2O4 ≈ H2O < NCS < CH3CN < pyridine < NH3 < − − ethylenediamine < 2,2'-bipyridine < 1,10-phenanthroline < NO2 < PPh3 < CN ≈ CO. This series can be explained by the type of π interactions on the ligands. The ligands with the smallest octahedral splitting will be π-donors (i.e. I-, Br-, etc) who will bond with the t2g orbitals on the metal, raising it in energy, thus decreasing the - t2g eg splitting. Similarly, π-acceptors (i.e. CN , CO, etc) will have the largest octahedral splitting energy as the t2g orbitals on the metal now become bonding and decrease in energy. The ligands in the middle of the series have little π-interaction with the metal and therefore the t2g remain nonbonding. 3- 4- 16. “Explain why Co(CN)6 is extremely stable but Co(CN)6 is not.” 3- 6 6 4- 7 6 1 Co(CN)6 is low spin d , therefore t2g , while Co(CN)6 is low spin d , t2g eg* . The trivalent cobalt species only has electrons in bonding orbitals while the population of an antibonding orbital in the divalent cobalt species significantly weakens the stability of the complex by decreasing the metal-ligand bond order. Population of the σ* orbital will also increase the metal–carbon bond length, reducing orbital overlap for the π bonds, and further reducing the stability of the complex. 19. “ Carbon monoxide is a strong-field ligand that stabilizes transition metals in - unusually low oxidation states. For example, V(CO)6 and V(CO)6 are both stable complexes. What are the ground-state electronic configuration of these two complexes in the ligand-field levels t2g and eg*? Which member of the series V(CO)6, Cr(CO)6, and Mn(CO)6 would you expect to be the most stable? Which would be least stable? Why?” From the back of the book: 5 V(CO)6: t2g - 6 V(CO)6 : t2g 6 Predicted stability: Cr(CO)6 > V(CO)6 >> Mn(CO)6 because Mn(CO)6 would be t2g 1 eg* . In a bit more detail, carbonyl complexes are strong field (low spin), thus the t2g orbitals are π-bonding in character. The more electrons in t2g, the more stable the - 6 complex; Cr(CO)6 is isoelectronic with V(CO)6 (t2g ) thus it is more stable than the neutral vanadium species. Manganese hexacarbonyl has an electron in an antibonding orbital and therefore is significantly destabilized. In fact, the manganese complex readily loses a CO and dimerizes to the staggered Mn2(CO)10 binuclear complex. Unlike for many coordination complexes, the 18e− rule works quite well for low spin complexes. 22. a) “Give the ligand-field electronic configuration of the ground state of W(CO)6. Is the complex diamagnetic or paramagnetic? The first electronic absorption band occurs at about 31,000 cm-1. Assign this band to an electronic transition in the complex. Is the photochemical dissociation of W(CO)6 to W(CO)5+CO, described in Problem 21, reasonable? Explain. 6 W(CO)6 has the diamagnetic electronic configuration t26 . -1 Cotton assumes that the transition at 31,000 cm is a t2g to eg* transition. Apart from being both spin and Laporte forbidden, making the transition a fairly weak one, this absorption occurs in the UV and the compound is colorless. This transition weakens the W-CO bond, making the dissociation reaction described in Problem 21 more 0 likely; in fact, many reactions that use W(CO)6 as a W source must be irradiated with UV light to proceed at an appreciable rate. In reality, this transition is actually a MLCT band (because W is in the third row and forms very strong M-L bonds), and the dissociation pathway is a little more complex. (e.g. See Phys. Chem. Chem. Phys., 2010, 12, 13197) b) “The W(CO)5 molecule has a square-pyramidal structure. Assume a reference coordinate system in which only one CO ligand is along the Z axis and predict the -1 ligand-field splitting for this complex. W(CO)5 absorbs light strongly at 25,000 cm . Assign the band to an electronic transition in W(CO)6 and explain why the absorption is at lower energy than in W(CO)6. With a loss of a CO ligand along the z-axis from W(CO)6, the t2g orbitals lose degeneracy as both the dxz and dyz orbitals become slightly less bonding than before. 2 Similarly the eg* orbitals are also not degenerate as the dz becomes less antibonding. As shown in the figure below, the new HOMO to LUMO transition, now that the molecule has C4v symmetry becomes the e to a1* transition. Both of these orbitals are brought closer together in energy from the octahedral complex, thus decreasing the transition energy. 24. “A common inorganic chemical compound, ferric ammonium sulfate, has the formula Fe2(SO4)3 24H2O. Large crystals of the compound are very pale violet due to weak absorption (ε values between 0.05 and 1) in the visible region of the spectrum. The 3+ 3+ Fe in the compound is present as the hexaaquo complex ion, Fe(H2O)6 . Using ligand-field theory formulate an explanation of the weak absorption bands. 3+ Being a weak π-donor, H2O is a weak field ligand, which leads to a high spin Fe 5 ( )3 2 (d ) octahedron ion with the electronic configuration t2g * eg* . This electronic configuration means that the HOMO-LUMO transition will be both spin forbidden and Laporte forbidden, which matches the very low ε values. 4- 25. “The ferrocyanide ion, Fe(CN)6 , does not exhibit an absorption band in the visible 3- -1 region, but ferricyanide Fe(CN)6 , absorbs strongly at approximately 25,000 cm , what type of electronic transition is responsible for the strong absorption that makes 3- Fe(CN)6 red?” -1 The actual absorption at 25,000 cm , however, is due to a LMCT from the t1u orbitals on the ligand to the t2g metal orbitals, which is both spin and Laporte allowed. Note that this absorption can only occur in the d5 complex where the hole allows for LMCT to occur. Both complexes show higher energy MLCT transitions to the ligand π* orbitals. In addition, answer the following questions that are added below: 1. In problem 22, the observed νCO is 2000 cm−1. In the free state νCO is 2143 cm−1. Explain the difference using molecular orbital language. The CO bond is weakened due to π-backbonding from the metal into an antibonding orbital on CO, thus lowering the vibrational energy. Additionally, the HOMO of CO is weakly bonding in character so σ donation from CO to the metal further weakens the C–O bond order. −1 4− −1 2. In problem 24, νCN is 2094 cm in [Fe(CN)6] in water and 2135 cm in 3− −1 [Fe(CN)6] in water; νCN is 2080 cm for NaCN in water. Explain the difference using molecular orbital language. The ferricyanide vs ferrocyanide change is quite easy to explain on the basis of π 4− 6 backbonding. The electron configuration of ferrocyanide [Fe(CN)6] is (t2g) while 3− 5 the configuration of ferricyanide [Fe(CN)6] is (t2g) . These orbitals are pi-bonding in nature, which means increased electron density will result in increased electron donation from the metal d orbitals to the ligand π* orbitals. Thus, the more reduced iron species (ferrocyanide) should display more backbonding, weakening the CN bond, and reducing the frequency of the CN stretching mode. Additionally, the d orbital energies for Fe2+ should align more closely with the relevant pi orbitals on cyanide than the d orbitals of Fe3+, further strengthening the back donation. When cyanide binds to iron, the CN stretching frequency increases relative to the “free” ion (aqueous sodium cyanide). This is true despite the presence of backbonding. This would suggest that σ-donation from the HOMO of cyanide strengthens the CN bond – this tells us that the HOMO of CN must be weakly antibonding in nature. We can rationalize the difference between CO and cyanide using s-p mixing. Because the orbitals of nitrogen align more closely (compared to oxygen) with those of carbon, the effects of s-p mixing should be increased. This raises the HOMO of cyanide, and causes it to switch from weakly bonding to weakly antibonding. 4− − 3. [Fe(CN)6] exchanges slowly with labeled CN in water, but irradiation with visible 3− 3− light forms [Fe(CN)5(H2O)] rapidly. In contrast, [Fe(CN)6] exchanges rapidly with labeled CN− in water. Explain. The explanation for this follows directly from the logic in problem 16 and problem 4− 6 22a. [Fe(CN)6] is low spin d and has all bonding orbitals filled.