Chem 161-2009 Homework 11th week

Chem 161-2009 Homework 11th week - Chem 161-2009 11th Week...

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Chem 161-2009 11 th Chapter 8 problems: 69,71,78,82,87 Chapter 9 problems: 5,10,23,25,27,29,33,35,37,39,41,45,47,49,51,95,99 69. Following are the periodic table locations of certain elements. Arrange them in the expected order of increasing ionization energy, I 1 . (a) group 4A, period 4; (b) group 6A, period 3; (c) group 3A, period 6; (d) group 8A, period 2; (e) group 6A, period 4. (a) 32 Ge = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 2 (b) 16 S = 1s 2 2s 2 2p 6 3s 2 3p 4 (c) 81 Tl = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 1 (d) 10 Ne = 1s 2 2s 2 2p 6 (e) 34 Se = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4 32 Ge 16 S 81 Tl 10 Ne Chem 161-2009 homework 11th week 1 32+ 4+ 16+ 6+ 2- 2- 8- 18- 2- 8- 6- X X X X X X X 81+ 3+ 10+ 8+ 2- 3- 8- 8- 18- 18- 32- 2- X X X X X X X X
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34 Se The initial ranking of ionization energy is based on the number of occupied principal shells. Ge has 4; S has 3; Tl has 6, Ne has 2, and Se has 4. Since it’s relatively easy to remove an electron when it’s far from the nucleus: Tl < Ge = Se < S < Ne. Ge and Se can be separated based on ENC, with the highest ENC holding the electron the strongest. Ge < Se. Therefore, Tl < Ge < Se < S < Ne. 71. Complete and balance the following equations. If no reaction occurs, so indicate. (a) Cl 2 (g) + Br - (aq) → (b) I 2 (s) + F - (aq) → (c) Br 2 (l) + I - (aq) → Determine which is a better oxidizing agent, using F 2 and Cl 2 as an example. 9 F = 1s 2 2s 2 2p 5 17 Cl = 1s 2 2s 2 2p 6 3s 2 3p 5 Chem 161-2009 homework 11th week 2 34+ 6+ 2- 8- 18- 6- X X X X
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17 Cl 9 F Based on the number of occupied principal shells (i.e., closeness of valence electrons from the nucleus), fluorine has a higher electron affinity than chlorine. Hence, F will capture electrons from Cl, not the other way around. Hence, F is the best oxidizing agent, followed by Cl, then Br, then I. (a) Cl 2 (g) + Br - (aq) → Since Cl is a better oxidizing agent than Br, it will take electrons from Br. Therefore, Cl 2 (g) + 2Br - (aq) → 2Cl - + Br 2 Note: the oxidation state of Cl 2 is reduced from 0 to -1, so it is reduced, and is the oxidizing agent. (b) I 2 (s) + F - (aq) → No reaction. There would have been a reaction if the equation was: F 2 (s) + I - (aq) → (c) Br 2 (l) + I - (aq) → Since Br is a better oxidizing agent than I, it will take electrons from I. Therefore, Br 2 (g) + 2I - (aq) → 2Br - + I 2 Note: the oxidation state of Br 2 is reduced from 0 to -1, so it is reduced, and is the oxidizing agent. 78. Arrange the following ionization energies in the most probable order of increasing value, and explain your reasoning: I 1 for B, I 1 for Cs, I 2 for In, I 2 for Sr, I 2 for Xe, and I 3 for Ca. 5
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This document was uploaded on 11/02/2011 for the course GEN CHEM 162 at Rutgers.

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Chem 161-2009 Homework 11th week - Chem 161-2009 11th Week...

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