This preview shows page 1. Sign up to view the full content.
Unformatted text preview: S117 Homework 2 Out of 50 Points Chapter 13 (not to turn in) 20. All of these ions are isoelectronic with argon, so only the charge of the nucleus determines the size. A larger atomic number attracts the electrons closer to the nucleus. From largest to smallest: S2-, Cl-, K+, Ca2+, Sc3+. 22. a. Mg2+: [Ne] Tl+: [Xe]6s2 b. N3-: [Ne] c. Be2+: [He] Sn2+: [Kr]5s2 As3+: [Ar]4s2 O2-: [Ne] Rb+: [Kr] K+: [Ar] F-: [Ne] Ba2+: [Xe] Al3+: [Ne] Te2-: [Xe] Se2-: [Kr] I-: [Xe] 31. A thermodynamic cycle can be used. Note that the sizes of the arrows are not to scale. Hf is the heat of formation, HI-I is the bond energy of iodine (strictly, half the bond energy), Hsublimation is the heat of sublimation, IE is ionization energy, EA is electron affinity, and LE is lattice energy. the degree sign indicates that these are the values under standard conditions. Comment [DK1]: Under standard conditions, iodine is actually I2(s), but the book seems to think otherwise. Therefore, the heat of formation of LiI(s) is actually the difference in energy between Li(s) + I2(s) and LiI(s). For the purpose of consistency with the book, we'll just pretend it's right. Two routes are shown to go from each must be the same, so to . The overall change in energy for , so . 33. a. The lattice energy of Mg2+O2- is enormous because the lattice energy depends on the product of the charges on the components, so despite the large energies involved in forming Mg2+ and O2-, it is still energetically favorable for the solid to have those charge states. In other words, Mg2+O2- is of a lower energy than Mg+O-. b. Both Mg+ and O- are paramagnetic, whereas Mg2+ and O2- are diamagnetic. Unpaired electrons show attraction to magnetic fields, so if a paramagnetic compound is weighed in the presence of a magnetic field, its mass will be different than when it is weighed in the absence of a magnetic field. (Diamagnetic materials are slightly repelled by magnetic fields, but the force is weaker than it is for paramagnetic materials.) Therefore, by taking the mass of MgO in the presence and absence of a magnetic field, one can conclude that the compound exists as Mg2+O2-. 39. In the reaction , one C--N bond and one N C bond is broken, and one C--C bond and one C N bond is made. The difference in enthalpy, therefore, is . 43. The change in enthalpy is given by the difference in enthalpy between the bonds broken and the bonds formed; thus, . Note that there are two ways of looking at this kind of problem. Either look only at the regions which change and calculate those energies (as was done in question 39.), or break every bond of the reactants and make every bond of the products. Chapter 13 (to turn in) 22. (5 points) a. Mg2+: [Ne] Tl+: [Xe]6s2 b. N3-: [Ne] c. Be2+: [He] Sn2+: [Kr]5s2 As3+: [Ar]4s2 O2-: [Ne] Rb+: [Kr] K+: [Ar] F-: [Ne] Ba2+: [Xe] Al3+: [Ne] Te2-: [Xe] Se2-: [Kr] I-: [Xe] 34. (10 points) Comment [DK2]: Under standard conditions, sulfur is actually S8(s), but the book seems to think otherwise. Therefore, the heat of formation of M2S(s) is actually the difference in energy between 2M(s) + S8(s) and M2(s). For the purpose of consistency with the book, we'll just pretend it's right. The preceding thermodynamic cycle can be used to calculate the second electron affinity of sulfur, where the M stands for any of the metals shown in the question. As with problem 31, two routes are shown to go from to . The change in enthalpy must be the same in both. For an example of how to deal with this type of problem, see the solution to 31. Here, results will just be shown. Metal Used Na K Rb Cs Second Electron Affinity of Sulfur (kJ/mol) 553 576 529 493 Average 538 One way of estimating error is to calculate the standard deviation of a data set. Excel can do this easily with the STDEV() formula. The standard deviation of this data set is that the second electron affinity of sulfur is . . We can say 40. (10 points) The change in enthalpy is given by the difference in enthalpy between the bonds broken and the bonds formed; thus, . 44. (5 points) a. The intermediates in the reaction are irrelevant in calculating the overall change in energy. As was explained in other types of problems, the difference in energy between two systems (reactants and products here) is always the same regardless of the route taken to get from one to the other. All we need to do, then, is to calculate the energy required to break the bonds in the reactants and the energy released to form the bonds in the products. . b. . c. . 47. (5 points) The notation for this problem will be as follows: is the change in enthalpy from breaking the A--B bond, is the ionization energy of A, and is the electron affinity of A. In order to get from reactants to products, we must break the bond(s) in the reactants; we must ionize hydrogen, and we must add an electron to the other component of the reaction. a. b. c. d. 51. (10 points) a. . . . . b. c. d. e. f. g. h. i. Notice a few things about these structures. First, any location where bonds come together but where no atom is explicitly shown is the location of a carbon atom. This is a convention in organic chemistry since carbon is such a common atom. Also, no lone pairs are shown. Make sure you know how many lone pairs there are on every single atom by counting electrons and ensuring that all atoms but H obey the octet rule.
-2 52. (5 points) a. -3 - b. -2 -3 - - - c. As with question 51, lone pairs are not shown. Make sure you know where they go. ...
View Full Document
This note was uploaded on 01/17/2012 for the course S 117 taught by Professor Stephenjacobson during the Fall '11 term at Indiana.
- Fall '11