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Practice Final Exam Answer Key

Practice Final Exam Answer Key - Chem 10181 Fall 2007(1...

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Chem 10181, Fall 2007 PRACTICE FINAL EXAM KEY (1) The element polonium, and particularly the isotope Po-210, has recently attracted international notoriety as an unwelcome seasoning found in sushi consumed by Russian dissidents. Although its scarcity and intense radioactivity have limited the studies of polonium, some of its chemistry has been explored. (a) (4 pts.) Give the number of protons and neutrons in an atom of 210 Po. # of protons = 84 # of neutrons = 126 (b) (6 pts.) Give the ground-state electron configuration of a Po atom in the gas phase, and state the number of unpaired electrons in the atom. You may use a shorthand “noble gas core” notation if you wish. Electron configuration of Po: [Xe]6s 2 4f 14 5d 10 6p 4 # of unpaired electrons = 2 (c) (6 pts.) Polonium has both a lower ionization energy (812 vs. 869 kJ/mol) and a lower electron affinity (180 vs. 190 kJ/mol) than tellurium (Te). Explain why. As one goes down the periodic table, the valence orbitals have a higher value of the principal quantum number n and are hence higher in energy and farther from the nucleus. This means that electrons are less stable in the valence orbitals of the heavier Po, making them easier to remove and harder to add. (d) (12 pts.) Gaseous polonium consists of diatomic molecules Po 2 . In the space below, sketch the molecular orbital diagram for Po 2 . Remember to place the relevant atomic orbitals on the outside of the diagram, the molecular orbitals on the inside, and clearly label all orbitals. Finally, fill the MO diagram with electrons as appropriate for Po 2 in its lowest- energy configuration. (You may assume that the gap between the s and p orbitals in Po is the same as in O.) 6s 6p z 6p y 6p x π * 6px π * 6py π 6px π 6py σ 6s σ∗ 6s σ 6p σ∗ 6p Po Po Po 2 6s 6p x 6p y 6p z
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Chem 10181, Fall 2007 PRACTICE FINAL EXAM KEY (e) (10 pts.) Polonium dioxide is known to form PoO 2 molecules in the gas phase. Draw a Lewis structure for PoO 2 , including all significant resonance structures (if any). Remember to include formal charges, if needed, and indicate the relative importance of any resonance structures you draw. Also indicate the hybridization of the Po atom and describe the geometry of the molecule. Po O O Po O O Both structures are equally major. The Po is sp 2 hybridized and the molecule is bent. (f) (14 pts.) Compare the following two aspects of the absorption of infrared light by the symmetric stretching mode of gaseous PoO 2 vs. gaseous CO 2 . [Recall that the symmetric stretching mode is one where each of the X-O bonds expands while the other expands, and contracts while the other contracts (X = Po or C).] (i) The intensity of absorption of IR light by the symmetric stretch of PoO 2 vs. CO 2 . CO 2 is linear while PoO 2 is bent. Because it is linear and symmetrical, CO 2 has no dipole moment and the dipole moment remains zero as the C-O bonds stretch symmetrically.
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Practice Final Exam Answer Key - Chem 10181 Fall 2007(1...

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