Name CHM 4610/5620 Fall 2017 December 14 FINAL EXAMINATION SOLUTIONS Part I, from the Literature Reports I II III IV V VI VII VIII IX X Total This exam consists of several problems. Rough point values are given to help you judge the value of problems. The total will be scaled to 175 points after the exams are marked. Glance over the entire exam, and then attempt the problems in the order of your choice. You must show your work or explain your reasoning to receive credit for an answer. For calculations, draw a box around your final answer and be sure to include the correct units. Some additional information is provided as a separate handout. I. (25 points) A study of the photophysical properties of a series of gold(iii) corrole complexes was recently reported (Inorg. Chem. 2017, 56, 10991 10997). These complexes are of interest as potential probes for determining the concentration of oxygen in biological tissue. The structure of a representative complex is shown on the right in Part A. A. The charge of the gold ion is 3+, and the coordination geometry is square planar. Draw the d-orbital splitting diagram and populate it with the correct number of electrons. Au 3+ => d 8 d x2-y2 d xy d z2 d xz,d yz B. Why you might expect this complex to be square planar? d 8 complexes are often square planar because the eight electrons can occupy four d orbitals at very low energies relative to the d x2-y2 orbital. C. If you did not know that the charge of the gold ion is 3+, how could you figure it out from the structure above? Three nitrogen donor atoms in the ring only have two single bonds, so they must also have two lone pairs, making the formal charge on those nitrogens 5 (4 + 2) = -1. The complex is overall neutral, so the charge on gold must be 3+. D. A different family of square planar gold(iii) complexes was reported a few years ago, prepared according to the reaction shown on the right (Inorg. Chem. 2013, 52, 12713 12725). The C C stretching frequency is a bit lower than one might expect for a carbon-carbon triple bond. Provide an explanation based on the bonding in this complex. A complete explanation will include drawings. Backbonding from the gold(iii) ion to the empty π* orbital would reduce the C C stretching frequency.
CHM 4610, Final Exam (Exam Three), Fall 2017 page 2 II. (25 points) Cesium lead halide perovskite ( x Pb y X z, X = Cl, Br, and I) quantum dots are of great interest for their potential use in high-performance solar cells and light-emitting diodes (LEDs); J. Am. Chem. Soc. 2017, 139, 11443 11450. A. Determine the number of cesium ions in the unit cell. (Be sure to show your work.) 8 corner x 1/(8 corner) = 1 ion B. Determine the number of halide (X) ions in the unit cell. (Be sure to show your work.) 6 face X x 1/(2 face) = 3 X ion C. Determine the number of lead ions in the unit cell. (Be sure to show your work.) 1 interior Pb ion X D. What is the charge of the lead ions? Explain your reasoning. + + Pb n+ + 3(Cl - ) = 0 (zero) So n = 2 and the lead ion is Pb 2+ X E. Given that the unit cell is cubic and one side is 7.9 nm, calculate the density of this perovskite in g/cm 3. X = Cl, so the mass of the unit cell is [132.9 g/mol + 207.2 g/mol + 3(35.45 g/mol)] / (6.022 x 10 23 mol -1 ) = 7.414 x 10-22 g (7.9 nm) 3 x (100 cm / 10 9 nm) 3 = 4.930 x 10-19 cm 3 density = mass/volume = (7.414 x 10-22 g)/( 4.930 x 10-19 cm 3 ) =???problem??? III. (20 points) It was recently reported that the chromium(ii) coordination environment in LaCrAsO changes from tetrahedral to octahedral when hydride is introduced to the lattice at high pressure (Inorg. Chem. 2017, 56, 13642 13645). Draw the d-orbital splitting diagrams for both geometries and populate with the correct number of electrons. d xy, d xz, d yz d x2-y2, d z2 Cr 2+ => d 4 tetrahedral d x2-y2, d z2 octahedral d xy, d xz, d yz If hydride (H ) is a stronger field ligand than As, how would you expect the d-orbital splitting diagram for the octahedral coordination environment to change? The d orbitals aligned with the z axis should be higher in energy. Why might As be the stronger field ligand? The low-lying empty d orbitals might allow As to act as a π acid.
CHM 4610, Final Exam (Exam Three), Fall 2017 page 3 IV. (40 points) Prussian blue has been known and used for centuries, but it is still actively investigated for new applications (Inorg. Chem., 2017, 56, 14373 14382). The structure of Prussian blue is a 3-dimensional lattice of alternating Fe 2+ and Fe 3+ ions bridged by cyanate ions, CN. A. Write the dot structure for CN. - :C N: B. Predict which end of the CN ions is bonded to Fe 2+ and which end is bonded to Fe 3+ in the Prussian blue lattice. Explain your reasoning. Expect Fe 2+ to bond to the carbon. Hard-soft acid-base theory. Also, the highest energy nonbonding (lone) pair is on carbon. C. Okay, you saw this coming... sketch the MO diagram for CN and populate it with the correct number of electrons. Similar to MO diagram for B 2, except... 1. 2s and 2p orbitals higher for carbon than nitrogen 2. difference in energy between 2s and 2p less for carbon than nitrogen 3. HOMO should be localized more on the carbon side * * 2p nb 2s 2p nb 2s C CN N How well does your MO diagram agree with your dot structure?
CHM 4610, Final Exam (Exam Three), Fall 2017 page 4 V. (24 points) Molecules of the type AB 3 can be trigonal planar (D 3h ), trigonal pyramidal (C 3v ), or T-shaped (C 2v ). A. Generate and resolve the reducible representation corresponding to all stretching vibrations of trigonal planar AB 3. Which of these stretches are IR active? Which of these stretches are Raman active? B. Generate and resolve the reducible representation corresponding to all stretching vibrations of trigonal pyramidal AB 3. Which of these stretches are IR active? Which of these stretches are Raman active? C. Generate and resolve the reducible representation corresponding to all stretching vibrations of T-shaped AB 3. Which of these stretches are IR active? Which of these stretches are Raman active? D. Can you use the A B stretches to distinguish between these geometries?
VII. (35 points) Answer the following questions and explain your reasoning. CHM 4610, Final Exam (Exam Three), Fall 2017 page 5 A. Which compound do you expect to be more acidic, cyclobutane (CH 2 ) 4 or cyclohexane (CH 2 ) 6? cyclobutane, (CH 2 ) 4 Cyclobutane has smaller CCC bond angles, so the C C bonds have more p character; therefore, the C H bonds will have more s character. Great s character lower energy greater effective EN more acidic. B. If you deprotonated cyclohexene, from which carbon would you remove H +? The olefinic hydrogens are more acidic. sp 2 hybridization great s character lower energy greater effective EN more acidic. C. Can you deprotonate cyclohexene in water? No. It is easier to deprotonate water than it is to deprotonate cyclohexene (water is more acidic than cyclohexene), so any base strong enough to deprotonate cyclohexene will deprotonate water instead; the base would be leveled to OH in water. D. Which of the following is formed when Cl 2 O 5 is added to water: HCl, HClO, HClO 2, HClO 3, or HClO 4? Cl 2 O 5 + H 2 O 2 HClO 3 The oxidation number of Cl is +5 for both Cl 2 O 5 and HClO 3. E. Predict the product of the solid-state reaction Cl 2 O 7 + CaO Ca(ClO 4 ) 2 Cl 2 O 7 is an acidic oxide (nonmetal oxides are acidic, see table of Lux-Flood parameters) and CaO is a basic oxide (metal oxides are basic, see table of Lux-Flood parameters). According to Lux-Flood acid-base theory, acids are oxide acceptors and bases are oxide donors, so CaO will transfer an oxide to Cl 2 O 7 to form an anion with a 2:8 calcium-to-chlorine ratio, consistent with the product being Ca(ClO 4 ) 2. F. Is the reaction AgF + LiI AgI + LiF favored towards reactants or products? Towards products because the softer acid Ag + prefers the softer base I, and the harder acid Li + prefers the harder base F.
CHM 4610, Final Exam (Exam Three), Fall 2017 page 6 VIII. (15 points) In HW08 we also predicted that InSb would be a better conductor than AgI or CdTe. Indeed, InSb is one of the major semiconductor materials used in transmitters and receivers such as those found in remote control devices. The band gap for InSb is 2.6 10 19 J. It turns out that band gap is too large for some applications, so InSb is often doped with Cd. Would this be n-type or p-type doping? p-type On the right, sketch a band gap diagram for Cd-doped InSb. Label the valence band, conduction band, and ΔE corresponding to the band gap energy required to achieve conduction. Would the wavelength of light required for photoconduction in the Cd-doped InSb be greater than, less than, or about the same as the wavelength required for pure InSb? the wavelength of light required for photoconduction in the Cddoped InSb be greater than pure InSb Explain your reasoning. smaller energy gap lower frequency longer wavelength Extra credit (4 points). Calculate the wavelength of light corresponding to the InSb band gap given in Question VIII. The band gap for InSb is 2.6 10 19 J = ΔE/h = 2.6 x 10 19 J / 6.626 x 10-34 J s = 3.9 x 10 14 s 1 c = = c/ = (3.00 x 10 8 m/s)/(3.9 x 10 14 s 1 ) = 7.6 x 10-7 m = 760 nm