Workshop 6 Key-1

Workshop 6 Key-1 - () () ) = 89 kJ/mol = E 1 IE for K ( ()...

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Workshop #6 Chem 1A Fall 2010 1. Draw Lewis structures for the following molecules. Include all resonance structures, formal charges, and indicate the preferred structure for each molecule. a. boron tribromide, BBr 3 Br B Br Br FC Br = 0 FC B = 0 Br B Br Br Br B Br Br Br B Br Br b. xenon tetrafluoride, XeF 4 Xe F F F F FC Xe = 0 FC F = 0 c. acetate, CH 3 CO 2 1- O O H H H O O H H H structures are equivalent FC C = 0 FC O = 1- FC H = 0 d. sulfur trioxide, SO 3 O S O O 2+ O S O O 2+ O S O O structures are equivalent, central S has octet O S O O FC S = 0 FC O = 0 O S O O 1+ O S O O 1+ O S O O structures are equivalent, central S has octet FC S = 1+ FC O = 1- FC S = 2+ FC O = 2- e. phosgene, Cl 2 CO Cl O Cl FC C = 0 FC O = 0 FC Cl = 0
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f. thiocyanate, SCN 1- S C N FC C = 0 FC N = 0 FC S = 1- S C N FC C = 0 FC N = 1- FC S = 0 g. phosphate, PO 4 3- O P O O O FC P = 0 FC O = 0, 1- O P O O O O P O O O O P O O O FC P = 0 FC O = 0, 1- FC P = 0 FC O = 0, 1- FC P = 0 FC O = 0, 1- structures are equivalent 2. Determine the lattice energy for KF (s) given the following data: Enthalpy of sublimation of K (
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Unformatted text preview: () () ) = 89 kJ/mol = E 1 IE for K ( () () ) = 419 kJ/mol = E 2 Dissociation of F 2 molecules (F 2 (g) 2F (g) ) = 79 kJ/mol = E 3 EA for F ( () () ) = -238 kJ/mol = E 4 Enthalpy of formation of KF (s) ( H KF ) = -567.23 kJ/mol = E 5 Energy K(s) + 1 / 2 F 2 KF(s) + H KF K (s) K (g) = E 1 K (g) K 1+ (g) + e-= E 2 F 2 (g) 2F(g) = E 3 F(g) + e- F 1-(g) = E 4 E lattice H KF = E 5 E 1 + E 2 + E 3 + E 4 + E lattice = E 5 E 5- E 1 - E 2- E 3- E 4 = E lattice -567.23 kJ/mol(89 kJ/mol)- (419 kJ/mol) (79 kJ/mol) (-238 kJ/mol) = E lattice-877 kJ/mol = E lattice...
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Workshop 6 Key-1 - () () ) = 89 kJ/mol = E 1 IE for K ( ()...

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