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#Chem 161-2007 homework 11th week

#Chem 161-2007 homework 11th week - Chem 161-2007 11th week...

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Chem 161-2007 11 th week Hill & Petrucci Homework Problems Chapter 9 problems: 5,10,20,23,25,27,29,33,35,37,39,41,45,47,49,51,53,55,57,63,67,74,95,98,99,100 Self-Assessment Questions 5. Use Lewis symbols to represent formation of the ionic compound between (a) calcium and bromine atoms (b) barium and oxygen atoms (c) aluminum and sulfur atoms (a) . . . . . .-1 . .-1 :Br: :Br: :Br: :Br: . . . . . . . . Ca Ca 2+ . . . .-2 (b) :O: :O: . . . . Ba Ba 2+ (c) . . . . . . . .2- . .2- . .2- :S: :S: :S: :S: :S: :S: . . . . . . . . . . Al . . Al Al 3+ Al 3+ 10. Using only the periodic table (inside front cover), indicate which element in each set is more electronegative. (a) Br or F F (b) Br or Se Br (c) Cl or As Cl (d) N or H N 20. Given the bond energies, N-to-O bond in NO, 628 kJ/mol; H-H, 435 kJ/mol; N-H, 389 kJ/mol; O-H, 464 kJ/mol, calculate ∆H for the reaction 2 NO(g) + 5H 2 (g) → 2NH 3 (g) + 2H 2 O(g) Note: I do this problem in the same way that we have done ∆Hreaction = ΣH formationproducts - ΣH formationreactants but here we use “bond formation” rather than “formation”. ΔH reaction = ΣH bondformationproducts - ΣH bondformationreactants The enthalpy of bond formation is the reverse of enthalpy of bond energy. One is the energy to form a bond; the other is the energy to break a bond. 1
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Reactants Products ∆H bf ∆H bf 2 NO = 2 x -628 6 NH = 6 x -389 5 H 2 = 5 x -435 4 OH = 4 x -464 ((6 x -389) + (4 x -464)) – ((2 x -628) + (5 x -435)) = -759 kJ Hill & Petrucci: ΔH reaction = ΣH bondsbroken - ΣH bondsformed Reactants Products ∆H be ∆H be 2 NO = 2 x +628 6 NH = 6 x +389 5 H 2 = 5 x +435 4 OH = 4 x +464 ΔH reaction = ((2 x 628) + (5 x 435)) – ((6 x 389) + (4 x 464)) = -759 kJ Note: This answer is different than in the Solutions Manual. The Solutions Manual is using the formula backwards, and getting the reverse answer (+759 kJ). Ions and Ionic Bonding 23. Which of the following ions have noble-gas electron configurations? What are the electron configurations of the ions that do not? (a) Mo 6+ (b) Te 2- (c) Ti 4+ (d) Zn 2+ (e) La 3+ (f) Hg 2+ (a) 42 Mo 6+ = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 4 → 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 = electron configuration of Kr. (b) 52 Te 2- = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 4 → 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 = electron configuration of Xe. (c) 22 Ti 4+ = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 → 1s 2 2s 2 2p 6 3s 2 3p 6 = electron configuration of Ar. (d) 30 Zn 2+ = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 → 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 = not a noble-gas electron configuration. (e) 57 La 3+ = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 4f 3 → 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 3 4f 3 = not a noble-gas electron configuration. Note: The answer in the Solutions Manual is incorrect. To form a cation, first fill up all of the orbitals of the atom. Then remove the electrons from the valence shell, even if the valence shell was not the last one filled. The Solutions Manual removed the electrons from the last orbital filled (4f orbital), not the valence shell (5p orbital).
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