Chemistry Name:_____________ 112 Final
You must answer all questions 1-9. You may chose any 3 questions from 10-15 to answer. 1. (12 points) Give the electron configuration for the following atoms or ion Ge [Ar] 4s2 3d10 4p2 Cu [Ar] 4s1 3d10 (this is one of the oddballs you have to memorize)
S2[Ne] 3s2 3p6 -or- [Ar]
Mn2+ [Ar] 4s2 3d3
2 A (5 points) Arrange the atoms N, F, Ga, As from smallest to largest ionization energy. Smallest: Ga, As, N, F : Largest B (5 points) Arrange the atoms P, Cl, Rb, Sb in size from smallest to largest. Smallest: Cl, P, Sb, Rb : Largest 3A (4 points) When we talked about the reactivity of the groups IA metals, I said for gas or solid reactions, the reactivity of K>Na>Li. Why? Reactivity of group 1A metals in gas or solid form correlates with 1st ionization energy; that is, the lower the ionization energy, the greater the reactivity. Since the lower ionization energy is lower on the period table, the reactivity goes in the order shown. B (4 points) On the other hand, in solution chemistry when you rate these three metals for their reducing power Li>K>Na. Why? When these metals react in solution, you have an additional interaction between the metal atoms and the water. This interaction is strongest for the smallest atom, Li and smallest for the larger atom, Na. This interaction dominates the reduction potential, so you get the reactivity given in this problem. C (4 points) On yet the other hand, when you react the pure metal with water K >Na >> Li. Why? Seeing as how this is a reduction of H2O in solution you would expect it to correlate with the reduction potential order given in 3B. However there is yet one more additional factor. This reaction gets HOT. Enough heat is generated to melt both the K and the Na. Once these metals melt, they form a liquid, and the liquid has a larger surface to react with the water, hence they become much more reactive. The reaction is not hot enough to melt the Li, so it stays in a solid chink, and react more slowly.
2 4. (10 points) Rank the ions Na+, Al3+, S2-, Cl- by size from smallest to largest. of these ions Cl- and S2- have the larger [Ar] electron configuration while Na+ and Al3+ have the smaller [Ne] configuration. For ions with the same electron the configuration, smaller ions are the ones with the larger atomic number because they have the most positive charge in the nucleus and have the largest `pull' on the electrons. Smallest: Al3+, Na1+, Cl-, S2- : largest 5. (10 points) Rank the compounds KBr, CaSe, and RbI from smallest to largest lattice energy. smallest: RbI, KBr, CaSe: largest lattice energy has the general form
QQ Lattice Energy = k 1 2 r
Lets look at r, the distance between the atoms K, Br, Ca, and Se are in the 4th period and are smaller than Rb and I that are in the 5th period, this gives them a stronger lattice energy. For KBr vs CaSe, the charge for K and Br is +/- 1, while the charge for Ca and Se is +/- 2 so the CaSe has the larger lattice energy 6. (15 points) Using the following bond energy information: H-H 432 C-H 413 H-F 565 C-C 347 H-Cl 427 C-F 485
C-Cl 339 Cl-Cl 239 F-F 154
Calculate the energy for the reaction: CH4(g) + 2Cl2(g) + 2F2(g) 6 CF2Cl2 (g)+ 2HF(g) + 2 HCl(g) (CF2Cl2 is Freon-12 a common refrigerant) )H = energy required to break bonds in reactants - energy released when bonds form in products = 4(C-H) + 2(Cl-Cl) + 2(F-F) - [2(C-F) + 2(C-Cl) + 2(H-F) + 2(H-Cl)] = 4(413) + 2(239) + 2(154) - [2(485) + 2(339) + 2(565) + 2(427)] = 1652 + 478 + 308 - [970 + 678 + 1130 + 854] = 2,428-3,632 = -1,194 kJ/mol
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7. (16 points) Draw Lewis Struct...