You might find the following useful. CHEMISTRY 1A Fall 2008 EXAM 3 Key CHAPTERS 7, 8, 9 & part 10 1
For each of the following, write the word, words, or number in each blank that best completes each sentence. (1½ points each) 1. The condition of an atom that has at least one of its electrons in orbitals that do not represent the lowest possible potential energy is called a(n) excited state. 2. Paramagnetic means having a net magnetic field due to having at least one unpaired electron. 3. Third electron affinity is the energy associated with adding one mole of electrons to one mole of isolated and gaseous 2 ions to form one mole of isolated and gaseous 3 ions. 4. A(n) antibonding molecular orbital is formed from out-of-phase interaction of two atomic orbitals. This leads to a decrease in negative charge between two nuclei where the atomic orbitals overlap and leads to less +/ attraction between the negative charge generated by the electrons and the nuclei. 5. Isomers are compounds that have the same molecular formula but different molecular structures. 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. A(n) unsaturated fat is a triglyceride that has one or more carbon carbon double bonds. 8. 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. 9. A polar molecule or ion (or a portion of a molecule or polyatomic ion) that is attracted to water is called hydrophilic. 10. Of the two elements carbon, C, and fluorine, F, fluorine has the more favorable first electron affinity. 11. Of the two ionic bonds, Al-S (in Al 2 S 3 ) or Cs-S (in Cs 2 S), the Al-S bond has the most covalent character. 12. Of the two ions, S 2 and Ca 2+, the S 2 is larger. 2
13. 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 trioxide, SO 3. (8 points) a. Draw a reasonable Lewis structure for SO 3 that minimizes formal charges. This structure does not have resonance. Draw the geometric sketch, including representative bond angles, predicted from this structure. Predict whether, based on this structure, the molecule is polar or nonpolar. nonpolar (symmetrical distribution of polar bonds) b. Draw a reasonable Lewis structure for SO 3 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. Draw the geometric sketch, including representative bond angles, predicted from this structure. Predict whether, based on this structure, the molecule is polar or nonpolar. nonpolar (symmetrical distribution of polar bonds) 14. Draw geometric sketches, including representative bond angles for the three isomers of difluoroethene, C 2 H 2 F 2. Label the cis and trans isomers. (6 points) 3
15. Write abbreviated electron configurations for each of the following, (4 points each) a. Copper atom, Cu [Ar] 3d 10 4s 1 b. Iridium(II) ion, Ir 2+ [Xe] 4f 14 5d 7 16. Compare the valence bond model for covalent bonding with the molecular orbital Theory for covalent bonding by answering the following questions. (8 points) a. Draw the best Lewis structure you can for carbon monoxide, CO. b. Evaluate the stability of this Lewis structure in terms of our criteria for stable structures according to the valence bond model. Unstable rare bonding paterns a plus formal charge on the second most electronegativity. c. Draw a molecular orbital diagram for CO. d. Evaluate the stability of CO in terms of our molecular orbital theory criteria for stable molecules. (What is the bond order for CO?) Bond order = ½(10 4) = 3 Stable e. Carbon monoxide has very strong bonds and forms anytime air is heated to high temperature, such as in your car s engine. Which model, valence bond or molecular orbital, best explains this stability? 4
17. Consider the following Lewis Structure for the resonance hybrid for nitric acid, HNO 3. (10 points) a. What is the hybridization for the left oxygen atom? sp 3 b. What is the hybridization for the right oxygen atom? sp 2 c. What is the hybridization for the top oxygen atom? sp 2 d. What is the hybridization for the nitrogen atom? sp 2 e. 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-N bond due to sp 3 -sp 2 overlap 2 sigma N-O bonds due to sp 2 -sp 2 overlap 1 delocalized pi system due to 3 p orbitals overlapping, one on the nitrogen atom, one on the top oxygen atom, and one on right oxygen atom f. What is the name of the electron group geometry around the left O? tetrahedral g. What is the name of the electron group geometry around the nitrogen? trigonal planar h. Draw a sketch with bond angles. 5
18. Consider the following Lewis Structure for C 2 H 4. (11 points) a. With reference to the assumptions of the valence bond model of covalent bonding, and using orbital diagrams for the valence electrons of carbon, explain how carbon atoms are able to form the bonds in C 2 H 4. Describe each bond as a sigma or pi bonds and tell how each is formed from the overlap of atomic orbitals. Only the highest energy electrons participate in bonding. Covalent bond form to pair unpaired electrons. Two sp 2 atomic orbitals for each carbon overlap 1s orbitals of the hydrogen atoms to form 2 sigma bonds. One sp 2 atomic orbital for each carbon overlaps an sp 2 orbital of the other carbon to form a sigma bond. The unhybridized 2p atomic orbitals of the carbons overlap 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. 6
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 higher melting point and boiling point. (3 points each) a. Iodine trifluoride, IF 3 type of attraction dipole-dipole and London forces or ethyl amine, C 2 H 5 NH 2 type of attraction hydrogen bonds and London forces higher melting point and boiling point? C 2 H 5 NH 2 b. 2-aminopropane, CH 3 CH(NH 2 )CH 3 type of attraction hydrogen bond and London forces or 2-aminodecane, CH 3 CH(NH 2 )(CH 2 ) 7 CH 3 type of attraction hydrogen bond and London forces higher melting point and boiling point? CH 3 CH(NH 2 )(CH 2 ) 7 CH 3 c. Ammonium hydroxide, NH 4 OH type of attraction ionic bonds or octane, C 8 H 18 type of attraction London forces stronger attractions? NH 4 OH 20. Low-pressure sodium lamps have a sodium vapor pressure of 0.7 Pa at 260 C. How many micrograms of Na must evaporate at 260 C to yield a pressure of 0.7 Pa in a 27-mL bulb? (6 points) g PV = RT M 22.9898 g 0.7 Pa ( 27 ml ) 6 PVM 1 mol 1 kpa 10 μg 1 L g = = = 3 3 RT 8.3145 L kpa 0.1 μg 10 Pa 1 g 10 ml 533 K K mol 7
Answer the following in short answer form. (6 points each) 21. Using quantum numbers, explain why the 3d sublevel can have a maximum of 10 electrons. The Pauli exclusion principle states that no two electrons in the same atom can have the same unique set of four quantum numbers. The quantum numbers for the 3d sublevel are 3,2. When l=2, m l can be 2, 1, 0, 1, or 1, and m s can be ½ or ½. Therefore, there are 10 unique sets of four quantum numbers that include 3,2. 3,2,2,½ 3,2,2, ½ 3,2,1,½ 3,2,1, ½ 3,2,0,½ 3,2,0, ½ 3,2, 1,½ 3,2, 1, ½ 3,2, 2,½ 3,2, 2, ½ 22. Explain why decreased volume for the gasoline air mixture in the cylinders of a gasoline engine, increased number of moles of gas, and increased temperature lead to an increase in pressure in the cylinders. (Answers that are something like, because volume is inversely proportional to pressure or because of Boyle s Law will not receive any credit.) Decreased volume leads to increased concentration of gas. This leads to more particles near the walls and therefore more collisions per second with the walls. This creates a greater force per unit area (pressure) against the walls. A spark ignites the mixture of compressed gases in a cylinder, and the hydrocarbon compounds in the gasoline react with the oxygen in the air to form carbon dioxide gas and gaseous water. A typical reaction is 2C 8 H 18 (g) + 25O 2 (g) 16CO 2 (g) + 18H 2 O(g) + energy In the representative reaction shown above, a total of 27 moles of gas are converted into 34 moles of gas. This increase in moles of gas leads to an increase in number of collisions per second with the walls of the cylinder, which creates greater force acting against the walls and a greater gas pressure in the cylinder. The pressure is increased even more by the increase in the temperature of the gas due to the energy released in the reaction. The increased temperature increases the average velocity of the gas particles, which leads to more frequent collisions with the walls and a greater average force per collision. 8