Physical Chemistry Lab II Name: KEY CHEM 4644 Spring 2011 Final Exam 5 questions at 3 points each equals 15 total points possible. Constants: c = 3.00 10 8 m/s h = 6.63 10-34 J s 1 Hartree = 4.36 10-18 J = 27.21 ev N A = 6.02 10 23 /mol m e = 9.11 10-31 kg
Question 1 (atomic energy levels and emission) Score: /15 One might suppose, based on the lowest-energy valence electron configuration, that the emission spectrum of sodium would be similar to the emission spectrum of the hydrogen atom. The most intense visible line in the emission spectrum of hydrogen comes from a transition from a 3p orbital to a 2s orbital. What is the analogous orbital transition for the sodium atom? Answer: 5p to 4s,because Na is two rows below H in the periodic table. Recall that the energy levels of the hydrogen atom obey the formula E n = Z2, in units of 2 2n Hartrees. The transition suggested above is observed at λ=589 nm. Calculate the value of Z that would make the E n formula give the observed wavelength. Answer: I need to find Z so that hc = Z2 2 1 5 2 1 4 2 = 0. 01125Z2 hc/λ = 3.375 10-19 J 3.375 10-19 J / 4.36 10-18 J = 0.0774 Hartrees Z 2 = 0.0774 / 0.01125 = 6.88 Z = 2.62 (This implies shielding of 11-2.62 = 8.4 by the other electrons. Slater's rules suggest shielding of 8 0.85 + 2 = 8.8.) Actually, there are two lines, the famous sodium doublet, one at 589.16 and the other at 589.76 nm. One comes from the 2 P 1/2 term, the other from the 2 P 3/2 term. What is the term symbol for the final state, the state that corresponds to the ground 1s 2 2s 2 2p 6 3s 1 configuration?
Question 2 Emission spectra of a difluoroboron compound attached to a polymer were reported by Guoqing Zhang, et al., JACS, 129, 8942 (2007). Spectra taken under nitrogen are shown at left. The "Fluorescence" spectrum is a normal fluorescence spectrum. The spectrum labeled "Phosphorescence" shows both room-temperature phosphorescence and delayed fluorescence. The two spectra have been rescaled to fit conveniently on the same graph. Suppose that the diagram at right is completely adequate to explain energetics of absorption and emission. Neglect any complicating factors. By correlating the spectra above with the diagram, calculate the following three energies: a) ΔE 1,2 b) ΔE 1,3 c) ΔE s,t, the singlet-triplet energy splitting the excited state Units of Joules will be fine. If you need wavelengths, read them from the spectra. Answers: a) ΔE 1,2 represents phosphorescence. I'll use the the phosphorescence wavelength at maximum intensity to calculate ΔE 1,2. λ max = 510 nm. ΔE 1,2 = hc/λ = 6.63 10-34 J s * 3.00 10 8 m/s / 510 10-9 m = 3.90 10-19 Joules b) ΔE 1,3 represents fluorescence. I'll use the the fluorescence wavelength at maximum intensity to calculate ΔE 1,3. λ max = 440 nm. ΔE 1,3 = hc/λ = 6.63 10-34 J s * 3.00 10 8 m/s / 440 10-9 m = 4.52 10-19 Joules c) ΔE s,t = ΔE 1,3 - ΔE 1,2 = (4.52-3.90) 10-19 J = 6 10-20 J
Question 3 FTIR spectra of carbon monoxide gas are shown above. Which spectrum is the fundamental and which is the overtone? Note that the center of the right spectrum, 4260 cm -1, is similar to, but not exactly equal to twice the center of the other spectrum, 2 2142 cm -1 = 4284 cm -1. Please explain both points, why 4260 is approximately 4284, and why it is not exactly 4284. Answers: The right spectrum is the first overtone and corresponds to quantum number change Δn = 0 to 2. The left spectrum is the fundamental, corresponding the Δn = 0 to 1. For a harmonic oscillator, ν 02 = 2ν 01 = 2ν e. It is because CO is approximately a harmonic oscillator that 4260 is similar to 4284. Anharmonicity reduces the frequency of overtones. It is because of anharmonicity that the overtone center is at 4260, less than two times the fundamental spectrum's center.
Question 4. Absorption spectra of three cyanine dyes, and the dyes' structures, are shown below. These were published by Heinz Mustroph, et al., ChemPhysChem, 10, 835-840, 2009. Suppose I insist on using the one-dimensional particle-in-a-box model to interpret these spectra, and further insist on considering just the four π electrons from two double bonds. Under those assumptions, the spectra differ because of different effective box lengths, L. Recall that the energy of a particle in a box is E n = n 2 h 2 /(8m e L 2 ), where n is a quantum number and L is the length of the box. For dye 2, calculate the value of L that makes the particle-in-a-box theory predict that wavelength. Show your calculation. Give L in picometers. Answer: λ=521 nm. ΔE=hc/λ = 3.82 10-19 J. The HOMO quantum number is 2, because four π electrons occupy two molecular orbitals. The LUMO quantum number is 3. Assume that λ comes from the HOMO-to-LUMO transition. ΔE = E 3 - E 2 = (3 2-2 2 ) h 2 /(8 m e L 2 ) = 5h 2 /(8 m e L 2 ). Set the box ΔE equal to the spectroscopic ΔE. 3.82 10-19 J = 5h 2 /(8 m e L 2 ). L 2 = 5 (6.63 10-34 J s) 2 /(8 9.11 10-31 kg 3.82 10-19 J) = 7.89 10-19 m 2 L = 8.85 10-10 m = 885 pm. (For comparison 5 139pm = 695pm.)
Question 5 Recall the formula for the n th energy level of an anharmonic oscillator, in wave numbers: G n = e n 1 2 e x e n 1 2 2 The C-H bond strength in chloroform can be obtained by measuring vibrational frequencies in infrared and near-infrared spectra. Consider the possibility of measuring the C-Br bond strength in bromobenzene, Br-C 6 H 5. The fundamental Br-C stretching vibration frequency is e = 316 cm 1 and the anharmonicity x e =0.00279. Calculate the wavenumbers at which the fundamental and the first two overtones may be observed in IR spectra, assuming the spectrometer's range and sensitivity permit. Answers: x e e = 0.00279 316=0.88 cm 1 The spectrum will show absorptions at fundamental: 0 =G 1 G 0 = e 2x e e = 316 2 = 314 cm 1 first overtone: G 2 G 0 = 2 e 6 x e e = 632 5 = 627 cm 1 second overtone: G 3 G 0 = 3 e 12 x e e = 948 11 = 937 cm 1 (The range of our FTIR includes the first and second overtones.)