You might find the following useful. CHEMISTRY 1A Spring 2010 EXAM 3 Key CHAPTERS 7-10 For each of the following, write the word, words, or number in each blank that best completes each sentence. (1½ points each) 1. Ground state is the condition of an atom whose electrons are in the orbitals that give it the lowest possible potential energy. 2. The bonding molecular orbital is formed from in-phase interaction of two atomic orbitals. This leads to an increase in negative charge between two nuclei where the atomic orbitals overlap and leads to more +/ attraction between the negative charge generated by the electrons and the nuclei. 3. Valence electronsare the highest energy s and p electrons for an atom. They are the electrons that are most important in the formation of chemical bonds. 1
4. Isomers are compounds that have the same molecular formula but different molecular structures. 5. A(n) delocalized pi system is a system of overlapping p orbitals used to describe the bonding in resonance hybrids. 6. A(n) trans isomer is a structure that has like groups on different carbons (which are linked by a double bond) and on different sides of the double bond. 7. Hydrogenation is a process by which hydrogen is added to an unsaturated triglyceride to convert double bonds to single bonds. This can be done by combining the unsaturated triglyceride with hydrogen gas and a platinum catalyst. 8. A nonpolar molecule (or a portion of a molecule or polyatomic ion) that is not expected to mix with water is called hydrophobic. 9. A(n) catalyst is a substance that speeds a chemical reaction without being permanently altered itself. 10. The stratosphere is the second layer of the earth s atmosphere, which extends from about 10 km to about 50 km. 11. UV-C is the ultraviolet radiation in the range of about 40 to 290 nm wavelengths. Almost all of it is filtered out by our atmosphere. Because DNA and proteins absorb radiation in this range, it could cause crop damage and general ecological disaster if it were to reach the earth s surface in significant quantities. 12. Of the two elements magnesium or sulfur, magnesium has the larger atoms. 13. Of the two ions, Al 3+ and Mg 2+, Mg 2+ is larger. 2
14. Consider the element thallium, Tl. (8 points) a. Write the abbreviated electron configurations for the Tl + and Tl 3+ ions. Tl + [Xe]4f 14 5d 10 6s 2 Tl 3+ [Xe]4f 14 5d 10 b. Explain why atoms of the element thallium form +1 and +3 ions but not +2 ions. It takes energy to remove an electron from an uncharged atom, it takes more energy to remove an electron from a +1 ion, and it takes even more energy to remove another electron from a +2 ion to form a +3 ion. This factor favors lower charges. Energy is released when ions come together to form ionic bonds, and the more highly charged the ions release more energy. This factor favors higher charges. Ions form to yield the best balance of these two factors, and the best balance often happens when the ions form stable electron configurations for the ions. This minimizes the energy necessary to form the ions. Tl + forms a stable d 10 s 2 configuration, and Tl 3+ forms a stable d 10 configuration. Tl 2+ has the [Xe]4f 14 5d 10 6s 1 configuration, which is not particularly stable, so it would yield a less favorable balance. 15. (1) Draw a reasonable Lewis structure for each of the following. If the molecule has resonance, draw all the resonance structures and the resonance hybrid. (2) For each, sketch the molecular geometry, including bond angles. (4 points each) a. CHOCH 2 NH 2 b. Iodine pentafluoride, IF 5. 3
16. There are disagreements among chemists as to how Lewis structures should be drawn. Some feel that we should try to minimized formal charges, others feel we should emphasize the octet rule, and others feel that for most of the uses for Lewis structures, it doesn t make any difference. As an example, consider the prediction of the molecular geometry and molecular polarity of sulfur dioxide, SO 2. (8 points) a. (1) Draw a reasonable Lewis structure for SO 2 that minimizes formal charges. This structure does not have resonance. (2) Draw the geometric sketch, including representative bond angles, predicted from this structure. (3) Predict whether, based on this structure, the molecule is polar or nonpolar. Polar (asymmetrical distribution of polar bonds) b. (1) Draw a reasonable Lewis structure for SO 2 that has eight total electrons around each of the atoms. Identify all formal charges. This structure has resonance, so draw all the resonance structures and the resonance hybrid. (2) Draw the geometric sketch, including representative bond angles, predicted from this structure. (3) Predict whether, based on this structure, the molecule is polar or nonpolar. Polar (asymmetrical distribution of polar bonds) 4
17. Consider the following Lewis Structure for formaldehyde, H 2 CO. (10 points) a. With reference to the valence bond model of covalent bonding, explain how the carbon atom is able to form three sigma bonds and one pi bond. Only the highest energy electrons participate in bonding. To yield the four unpaired electrons necessary to explain four bonds, it s as if an electron is promoted from the 2s orbital to an empty 2p orbital. The three electron groups around the carbon leads to a prediction of sp 2 hybridization. Covalent bond form to pair unpaired electrons. The three sp 2 hybrid orbitals overlap other atomic orbitals to form the three sigma bonds, and the unhybridized p orbital will overlap a p orbital from another atom to form a pi bond. b. Explain why one of the bonds in the double bond is weaker than the other. The p orbitals, which form the pi bond by parallel overlap, overlap less than the sp 2 atomic orbitals, which form the sigma bond by end-on overlap. Less overlap leads to less of an increase in negative charge between the nuclei and therefore, less of a stabilization of the molecule. 18. Consider the following Lewis Structure for the resonance hybrid for hydrogen carbonate, HCO 3. (7 points) a. What is the hybridization for the carbon atom? sp 2 b. What is the hybridization for the top and right oxygen atoms? sp 2 c. What is the hybridization for the left oxygen atom? sp 3 d. Write a description of the bonding, stating whether each bond is sigma, pi, or part of a delocalized pi system and by stating which atomic orbitals overlap to form the bonds. 1 sigma O-H bond due to sp 3-1s overlap 1 sigma O-C bond due to sp 3 -sp 2 overlap 2 sigma C-O bonds due to sp 2 -sp 2 overlap 1 delocalized pi system due to 3 p orbitals overlapping, one on the carbon atom, and one on each oxygen atom 5
19. For each of the following, write the name of the type of attraction holding these particles in the solid and liquid form. Indicate the formula in each pair that represents the substance that you would expect to have the stronger attractions among particles. (3 points each) a. Methanol, CH 3 OH type of attraction hydrogen bond and London forces or silicon dioxide, SiO 2 type of attraction covalent bond stronger attractions? SiO 2 b. Propanone (acetone), CH 3 COCH 3 type of attraction dipole-dipole and London forces or benzoic acid, C 6 H 5 CO 2 H type of attraction hydrogen bond and London forces stronger attractions? C 6 H 5 CO 2 H c. Propane, CH 3 CH 2 CH 3 0Btype of attraction London forces or nonane, C 9 H 20 type of attraction London forces stronger attractions? C 9 H 20 20. Methanol, CH 3 OH, a common solvent, is also used to make formaldehyde, acetic acid, and detergents. It is made from the methane in natural gas in two steps, the second of which is run at 200-300 C and 50-100 atm using metal oxide catalysts. In the second reaction, what minimum volume of H 2 gas at 4.0 10 3 kpa and 500 K is necessary to convert 187 m 3 of CO gas at 2.0 10 3 kpa and 500 K? (8 points) 3CH 4 (g) + 2H 2 O(g) + CO 2 (g) 4CO(g) + 8H 2 (g) CO(g) + 2H 2 (g) CH 3 OH(l) 3 3 3 3 10 L K i mol 2.0 x 10 kpa 2 mol H2? m H 2 = 187 m CO 3 1 m 8.3145 L g kpa 500 K 1 mol CO 3 8.3145 L g kpa 500 K 1 m 2 3 3 3 = 1.9 10 m H 2 K i mol 4.0 x 10 kpa 10 L 6
21. Write an explanation for why bonding molecular orbital formed from the overlap of two 1s atomic orbitals is more stable and why antibonding molecular orbitals formed from the overlap of 1s atomic orbitals is less stable than the separate 1s atomic orbitals that form them. (6 points) 7
22. Avogadro s Law states that moles of gas and volume are directly proportional if temperature and pressure are constant. (8 points) a. Describe a simple system that could be used to demonstrate this relationship. Your description should include mention of how the temperature and pressure are held constant. b. Write an explanation for why this relationship is true. 8
23. Explain why the highest concentrations of ozone in the air we breathe are found in large industrial cities with lots of cars and lots of sun. (6 points) The cylinders of cars provide the necessary conditions for the formation of nitrogen oxides. Any time air (which contains nitrogen and oxygen) is heated to high temperature (as occurs in the cylinders of our cars and in many industrial processes), nitrogen oxides are formed (NO and NO2). N 2 (g) + O 2 (g) 2NO(g) 2NO(g) + O 2 (g) 2NO 2 (g) Nitrogen dioxide, NO 2, is a red-brown gas that contributes to the brown haze associated with smog. The radiant energy that passes through the air on sunny days can supply the energy necessary to break covalent bonds between nitrogen atoms and oxygen atoms in NO 2 molecules, converting NO 2 molecules into NO molecules and oxygen atoms. Remember that the shorter the wavelength of light is, the higher the energy. Radiant energy of wavelengths less than 400 nm has enough energy to break N O bonds in NO 2 molecules, but radiant energy with wavelengths longer than 400 nm does not supply enough energy to separate the atoms. The oxygen atoms react with oxygen molecules to form ozone molecules. O(g) + O 2 (g) O 3 (g) 9