PHOTOELECTRON SPECTROSCOPY PROBLEMS …ohschemistry.org/ap/03/03_04_hw.pdfPHOTOELECTRON SPECTROSCOPY...

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Page 1: PHOTOELECTRON SPECTROSCOPY PROBLEMS …ohschemistry.org/ap/03/03_04_hw.pdfPHOTOELECTRON SPECTROSCOPY PROBLEMS: 1. A hydrogen atom in its ground state is irradiated with 80 nm radiation.

PHOTOELECTRON SPECTROSCOPY PROBLEMS: 1. A hydrogen atom in its ground state is irradiated with 80 nm radiation. What is the kinetic energy (in eV) of the ejected photoelectron? Useful formulae: ν = c/λ E = hν 1 Joule = 6.2415×1018 eV E=-R/n2 R = 2.1799 × 10-18 J 2. When 1 nm radiation is used to obtain the photoelectron spectrum of neon, electrons are ejected with kinetic energies of 372, 1191, and 1218 eV. From what atomic orbitals do these electrons originate? What is the orbital energy of each of these orbitals? 3. The orbital energies of the atomic orbitals in argon are: -3206, -326, -249, -29, and -16 eV. Assuming that the intensity of the peaks in the photoelectron spectrum are proportional to the number of electrons contained in the atomic orbital, sketch on the axis below the photoelectron spectrum (as a function of electron kinetic energy, EKE) expected for argon that is subjected to X-rays with an energy of 3500 eV:

4. Two 3d transition metal elements have been formed into an alloy. Shown below is the XPS spectrum in the region corresponding to ejection of 3p electrons. Each peak shows an unresolved doubling, which is due to the difference in energy between 3p electrons that have the orbital angular momentum (ℓ 1 for p electrons) parallel to the spin, or antiparallel to the spin. Estimate the binding energy of the electrons from the graph, and use that information in conjunction with the table of core electron binding energies (next page) to deduce the two elements that are present in the alloy. Based on the relative intensities of the two peaks, estimate the percentage of each element in the alloy (a rough, eyeball estimate is good enough). | | | | |

Page 2: PHOTOELECTRON SPECTROSCOPY PROBLEMS …ohschemistry.org/ap/03/03_04_hw.pdfPHOTOELECTRON SPECTROSCOPY PROBLEMS: 1. A hydrogen atom in its ground state is irradiated with 80 nm radiation.

Table of Core Electron Binding Energies of the Elements (eV)

Z Element 1s 2s 2p1/2 2p3/2 3s 3p1/2 3p3/2 3d3/2 3d5/2 1 H 13.6 2 He 24.6 3 Li 54.7 4 Be 111.5 5 B 188 6 C 284.2 7 N 409.9 37.3 8 O 543.1 41.6 9 F 696.7 10 Ne 870.2 48.5 21.7 21.6 11 Na 1070.8 63.5 30.65 30.81 12 Mg 1303 88.7 49.78 49.5 13 Al 1559.6 117.8 72.95 72.55 14 Si 1839 149.7 99.82 99.42 15 P 2145.5 189 136 135 16 S 2472 230.9 163.6 162.5 17 Cl 2822.4 270 202 200 18 Ar 3205.9 326.3 250.6 248.4 29.3 15.9 15.7 19 K 3608.4 378.6 297.3 294.6 34.8 18.3 18.3 20 Ca 4038.5 438.4 349.7 346.2 44.3 25.4 25.4 21 S 4492 498 403.6 398.7 51.1 28.3 28.3 22 Ti 4966 560.9 460.2 453.8 58.7 32.6 32.6 23 V 5465 626.7 519.8 512.1 66.3 37.2 37.2 24 Cr 5989 696 583.8 574.1 74.1 42.2 42.2 25 Mn 6539 769.1 649.9 638.7 82.3 47.2 47.2 26 Fe 7112 844.6 719.9 706.8 91.3 52.7 52.7 27 Co 7709 925.1 793.2 778.1 101 58.9 59.9 28 Ni 8333 1008.6 870 852.7 110.8 68 66.2 29 Cu 8979 1096.7 952.3 932.7 122.5 77.3 75.1 30 Zn 9659 1196.2 1044.9 1021.8 139.8 91.4 88.6 10.2 10.1 31 Ga 10367 1299 1143.2 1116.4 159.5 103.5 100 18.7 18.7 32 Ge 11103 1414.6 1248.1 1217 180.1 124.9 120.8 29.8 29.2 33 As 11867 1527 1359.1 1323.6 204.7 146.2 141.2 41.7 41.7 34 Se 12658 1652 1474.3 1433.9 229.6 166.5 160.7 55.5 54.6 35 Br 13474 1782 1596 1550 257 189 182 70 69 36 Kr 14326 1921 1730.9 1678.4 292.8 222.2 214.4 95 93.8

Page 3: PHOTOELECTRON SPECTROSCOPY PROBLEMS …ohschemistry.org/ap/03/03_04_hw.pdfPHOTOELECTRON SPECTROSCOPY PROBLEMS: 1. A hydrogen atom in its ground state is irradiated with 80 nm radiation.

5. Shown below is a the photoelectron spectrum of CO, plotted as a function of the electron binding energy (EBE=hν – EKE). The peaks in the spectrum fall at the following electron binding energies:

Peak # Energy (eV) Intensity 1 14.01 strong 2 14.28 weak 3 16.54 weak 5 16.73 medium 6 16.92 medium

7 17.11 medium 8 17.29 medium 9 17.46 weak 10 17.64 tiny 11 17.81 tiny 12 17.97 miniscule 13 19.67 weak 14 19.88 tiny 15 20.08 miniscule

Shown below is the CO molecular orbital diagram that is found in many textbooks:

Now, for the actual questions: (a). From which molecular orbital is the electron ejected in the 14.01 eV EBE peak? (b). From which molecular orbital is the electron ejected in the 16.54 eV EBE peak? (c). From which molecular orbital is the electron ejected in the 19.67 eV EBE peak? (d). What is the ionization potential of CO? (e). What is the vibrational frequency (in units of eV and cm-1) of the ground state of CO+? (f). What is the vibrational frequency (in units of eV and cm-1) of the first excited state of CO+? (g). What is the orbital energy of the 5σ orbital of CO, in eV? (h). What is the orbital energy of the 1π orbital of CO, in eV? (i). What is the orbital energy of the 4σ orbital of CO, in eV?

Page 4: PHOTOELECTRON SPECTROSCOPY PROBLEMS …ohschemistry.org/ap/03/03_04_hw.pdfPHOTOELECTRON SPECTROSCOPY PROBLEMS: 1. A hydrogen atom in its ground state is irradiated with 80 nm radiation.

6. The gold anion (also called auride) is rarely encountered, but salts with this anion have been created. It can be produced in the gas phase and probed by photoelectron spectroscopy, however. (a). What do you expect will be the ground electronic configuration of Au−, which has 80 electrons? (b). Which orbital lies highest in energy in the Au− anion? (c). Shown below is the photoelectron spectrum of Au−. Based on the spectrum, what would you estimate to be the electron affinity of gold? For comparison, the electron affinities of some nonmetallic elements are: C 1.263 eV N -0.07 eV O 1.461 eV F 3.399 eV Si 1.385 eV P 0.746 eV S 2.077 eV Cl 3.617 eV

As 0.81 eV Se 2.021 eV Br 3.365 eV Sb 1.07 eV Te 1.971 eV I 3.059 eV

Note: Nitrogen has a negative electron affinity because the N− anion lies higher in energy than an N atom plus an electron (in the gas phase); thus, the gas-phase N− anion will spontaneously eject an electron very, very quickly. The triply charged N3− nitride anion is only stable when it is surrounded by cations. The photoelectron spectrum explains why, in some situations, gold can behave as a pseudohalogen. It has a high electron affinity (higher than all of the typical nonmetals except for the halogens).

Page 5: PHOTOELECTRON SPECTROSCOPY PROBLEMS …ohschemistry.org/ap/03/03_04_hw.pdfPHOTOELECTRON SPECTROSCOPY PROBLEMS: 1. A hydrogen atom in its ground state is irradiated with 80 nm radiation.

7. A portion of the X-ray photoelectron spectrum (XPS) of ethylchloroformate is given below, along with the structure of the compound:

The structure of the ethylchloroformate molecule is given below: CH3-CH2-C-Cl ║ O You will notice that the 1s electrons on carbon have a binding energy of roughly 290 eV, while the 2s electrons on chlorine have a binding energy of roughly 273 eV. However, there are three distinct carbon 1s peaks, which correspond to carbons in chemically different environments. The three carbons in the molecule are the methyl carbon (CH3), the methylene carbon (CH2), and the carbonyl carbon (C=O).

(a). Which carbon atom gives the peak labeled (a) in the spectrum? (b). Which carbon atom gives the peak labeled (b) in the spectrum? (c). Which carbon atom gives the peak labeled (c) in the spectrum? (d). Explain how you have assigned the peaks in this manner.