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Homework3_answerkey - S117 Homework 2 Out of 50 Points...

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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: S 2- , Cl - , K + , Ca 2+ , Sc 3+ . 22. a. Mg 2+ : [Ne] Sn 2+ : [Kr]5s 2 K + : [Ar] Al 3+ : [Ne] Tl + : [Xe]6s 2 As 3+ : [Ar]4s 2 b. N 3- : [Ne] O 2- : [Ne] F - : [Ne] Te 2- : [Xe] c. Be 2+ : [He] Rb + : [Kr] Ba 2+ : [Xe] Se 2- : [Kr] I - : [Xe] 31. A thermodynamic cycle can be used. Note that the sizes of the arrows are not to scale. Δ H f is the heat of formation, Δ H I-I is the bond energy of iodine (strictly, half the bond energy), Δ H sublimation 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. Two routes are shown to go from to . The overall change in energy for each must be the same, so , so Comment [DK1]: Under standard conditions, iodine is actually I 2 (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) + I 2 (s) and LiI(s). For the purpose of consistency with the book, we’ll just pretend it s right.
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