Name CHM 4610/5620 Fall 2016 December 15 FINAL EXAMINATION SOLUTIONS I. (80 points) From the literature... A. The synthesis and properties of copper(ii) complexes with ligands containing phenanthroline and histidine were recently reported (Inorg. Chem. 2016, 55, 11801 11814). These complexes are five-coordinate with the donor atoms distributed in a square pyramidal geometry. N N Cl Cu N N 1. The d-orbital splitting diagram for square pyramidal is given on the right. Populate it with the correct number of d electrons for this complex. d 9 => 2. In class we deduced the d-orbital splitting diagram for square planar by starting with an octahedral complex, and then slowly pulling away the ligands along the z axis and shifting the orbital energies accordingly. We obtained a very similar diagram as the one shown here for square pyramidal, except the d z2 and d xy are switched in energy. Explain this difference. A complete answer will include a drawing showing how the orbital energies change as the ligands are pulled away from an octahedral complex along the z axis. Figure 11.12 from Huheey 4e (shown on the right) shows the effect pulling away the ligands along the z-axis has on the splitting of the d orbitals. Removing only one ligand as shown below L L L L L would reduce the electric field along the z axis less than removing both ligands, more like stretching but not removing both ligands, hence the d z2 and d xy orbitals would not cross.
CHM 4610/5620, Exam Three, Fall 2016 page 2. A. Ru(II)-Mo(VI) bimetallic complex has been prepared as a model for the biological catalyst molybdenum oxotransferase (Inorg. Chem. 2016). The cartoon below illustrates the chemistry, but let s focus on the metal complexes. oth are essentially octahedral. 1. Draw d-orbital splitting diagrams for both complexes, and populate with the correct number of d electrons. Ru(II) complex (d 6 ) Mo(VI) complex (d 0 ) 2. Which, if any, of these complexes can be either high spin or low spin? Ru(II) complex 3. For any of complexes that can be either high spin or low spin, explain why you picked the spin state you used to answer Part1. (1) 2nd row transition metal => Δ o large (2) ligands are acceptors => strong field 4. oth complexes can exhibit charge transfer electronic transitions. Describe the type of charge transfer you would expect for two complexes and explain your reasoning. the Ru(II) complex the Mo(IV) complex
CHM 4610/5620, Exam Three, Fall 2016 page 3 D. A theoretical investigation of the substitution reactions of cobalt(iii) pentaammine looked at the energies of potential intermediates in order to better understand the substitution mechanisms (Inorg. Chem. 2016). In class stated that crystal field energies could be used to estimate a crystal field activation energy to explain the kinetic stability of transition metal complexes. The d-orbital energies for various geometries are given in the table below. 1. Draw a d-orbital splitting diagram for an octahedral cobalt(iii) complex assuming that it is low spin, and calculate the total energy of the electrons using the values in the table. 2. Assuming a square pyramid geometry for the cobalt(iii) intermediate cobalt(iii), draw a d-orbital splitting diagram and calculate the total energy of the electrons using the values in the table. 3. Calculate the crystal field activation energy by subtracting your answer to Part 1 from your answer to Part 2. 4. Repeat parts 1-3 above, only this time assume high spin for both the initial complex and the intermediate. ased on the calculations above, which do you expect to undergo faster substitution reactions: high spin cobalt (II) or low spin cobalt(ii)?
II. (45 points) Titanium dioxide. Part One. CHM 4610/5620, Exam Three, Fall 2016 page 4 A novel technique for producing small metal-oxide clusters in water was reported this month in the online journal Chem. (2016, 1, 887 901). They can be produced in a precise and consistent manner, giving spherical clusters of about 100 atoms, hopefully leading to new ways to exploit the chemistry of metal oxides. One important metal oxide is titanium dioxide, TiO 2 (s), which is used as a photocatalyst and as a component of numerous materials. A. Draw a thermodynamic cycle that would allow you to calculate the enthalpy of formation for TiO 2 from the lattice energy along with the other information provided in the info handout. IE 4 (Ti) Ti 4+ ( ) IE 3 (Ti) Ti 3+ ( ) Ti 2+ ( ) 2O 2- ( ) IE 2 (Ti) -2EA 2 (O) Ti + ( ) 2O - ( ) IE 1 (Ti) -2EA 1 (O) 0[TiO 2 ( )] Ti( ) 2O( ) o atom(ti) o E (O=O) Ti( ) + O 2 (g) o f [TiO 2 ( )] TiO 2 ( ) ΔH f Ti(s) + O 2 (g) TiO 2 (s). Using your cycle from Part A, write an equation that would allow you to calculate ΔH f for TiO 2 (s). ΔH f [TiO 2 (s)] = ΔH atom (Ti) + IE 1 (Ti) + IE 2 (Ti) + IE 3 (Ti) + IE 4 (Ti) + ΔH E (O=O) + 2[-EA 1 (O)] + 2[-EA 2 (O)] + U 0 [TiO 2 (s)] U 0 [TiO 2 (s)] = ΔH f [TiS 2 (s)] - ΔH atom (Ti) - IE 1 (Ti) - IE 2 (Ti) - IE 3 (Ti) - IE 4 (Ti) - ΔH E (O=O) - 2[-EA 1 (S)] - 2[-EA 2 (S)] C. Using your equation from Part, the determine ΔH f for TiO 2 (s). The lattice energy of TiO 2 (s) is 12,493 kj/mol.
CHM 4610/5620, Exam Three, Fall 2016 page 5 II. Titanium dioxide. Part Two. C. The unit cell for TiO 2 (s) is shown on the right. (Yes, I know, but it really fit in nicely with this problem. Let s call it my Christmas present to you.) 1. How many Ti ions are in one unit cell? (show your work) 8 corners x 1/8 + 1 interior = 2 Ti 4+ ions 2. How many O ions are in one unit cell? (show your work) 4 face x 1/2 + 2 interior = 4 O 2- ions 3. Calculate the mass of one unit cell in grams. (show your work) [(2 x 47.88 g/mol) + (4 x 16.00 g/mol)] / 6.022 x 10 23 mol -1 = 2.653 x 10-22 g 4. The density of titanium dioxide is 4.23 g/cm 3. Determine the volume of one unit cell in pm 3. (show your work) (2.653 x 10-22 g) / 4.23 g/cm 3 = 6.27 x 10-23 cm 3 x (10 12 pm / 100 cm) 3 = 6.27 x 10 7 pm 3 D. The use of titanium dioxide as a sunscreen and photocatalyst is based on its ability to absorb photons to undergo an electron transition. Would the electronic transition be best described as a d-d transition, a metal-to-ligand charge transfer, or a ligand to metal charge transfer? Explain your reasoning. E. Titanium dioxide can be doped with nitrogen to make it a better conductor, as shown in the cartoon on the right. Is this n doping or p doping? n doping Explain how you determined the type of doping. e - e - e - The band gap shown by the arrow in the drawing for N-TiO 2 indicates n doping, where the electrons provided by the dopant must be excited to the conduction band.
CHM 4610/5620, Exam Three, Fall 2016 page 6 III. (20 points) Consider the hypothetical molecule A 2. A. Group molecular orbitals constructed from the p orbitals on for the hypothetical bent molecule A 2 is shown on the below. Determine the symmetry of these group orbitals. In other words, what is the Mulliken symbol for the irreducible representation corresponding to these orbitals? The character table for C 2v was included with this exam. A 1 2 Explain or show how you made these assignments. E C 2 xz yz E C 2 xz yz 1 1 1 1 => A 1 1-1 -1 1 => 2. Which, if any, atomic orbitals (s, p x, p y, p z, d x2-y2, d z2, d xy, d xz, or d yz ) on A could mix with the group orbital on the right? s, p z, d z2, d x2-y2 Sketch the bonding and antibonding molecular orbitals that would result from mixing these orbitals with the group orbital. For example, mixing with d x2-y2 gives... C. Which, if any, atomic orbitals (s, p x, p y, p z, d x2-y2, d z2, d xy, d xz, or d yz ) on A could mix with the group orbital on the right? p y, d yz Sketch the bonding and antibonding molecular orbitals that would result from mixing these orbitals with the group orbital.
CHM 4610/5620, Exam Three, Fall 2016 page 7 IV. (20 points) For transition metal ML 6 complexes we generally assume the geometry is octahedral unless observations indicate otherwise. When considering other possible geometries for any kind of molecule or polyatomic ion, a D 6h planar A 6 geometry would be highly controversial and require solid supporting evidence. A. Generate the reducible representation that would correspond to all vibrations for D 6h A 6 compound, but do not resolve the representation. Let C 2 and v go through the A- bonds, and C 2 and d go between the A- bonds. A D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 3 2S 6 h 3 d 3 v. Generate the reducible representation for just the A stretches. D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 3 2S 6 h 3 d 3 v 6 0 0 0 2 0 0 0 0 6 0 2 C. How many of the A stretches are IR active? Give the symmetry of these stretches. Vibrations must have A 2u (z) or E 1u (x,y) symmetry to be IR active. D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 3 2S 6 h 3 d 3 v A 2u 1 1 1 1-1 -1-1 -1-1 -1 1 1 6 0 0 0 2 0 0 0 0 6 0 2 1(1)6 0 0 0 3(-1)2 0 0 0 0 1(-1)6 0 3(1)2 6 + (-6) + (-6) + 6 = 0 A 2u D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 3 2S 6 h 3 d 3 v E 1u 2 1-1 -2 0 0-2 -1-1 2 0 0 6 0 0 0 2 0 0 0 0 6 0 2 1(2)6 0 0 0 0 0 0 0 0 1(2)6 0 0 12 + 12 = 24/24 = 1 E 1u D. An octahedral A 6 molecule would be expected to show one A stretch in the IR. Could IR spectroscopy be used to distinguish between the octahedral and hexagonal planar geometries? Explain. No. oth geometries give only one A stretch in the IR.
CHM 4610/5620, Exam Three, Fall 2016 page 8 V. (extra credit) Olefins bond sideways to transitional metals as shown. Showing the orbitals involved... A. sketch the -bonding interaction.. sketch the -backbonding interaction. C. Why might such complexes be useful as catalysts? ackbonding electron density into the * orbital weakens the C=C bond, making it more reactive.
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