Chem 141 2009_Sample Exam 2 Solns

Chem 141 2009_Sample Exam 2 Solns - Chem 141 Sample Exam 2...

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Chem 141 Sample Exam 2 Solutions 2009 Ia. PCl 2 (CH 3 ) 3 > SF 2 Cl 2 > [ICl 4 ] - . The structures based on VSEPR principles are shown below and the corresponding Cl-E-Cl bond angles are indicated. H 3 CP CH 3 3 Cl 180° S F F <120° I 90° Ib. Al > O > He. Electrons with l =1 are in p orbitals. Based on the ground state electronic configurations, the total number of p electrons for each atom is indicated in the table below. Atom Electronic Configuration # of p electrons He (1 s ) 2 0 O (1 s ) 2 (2 s ) 2 (2 p ) 4 4 Al (1 s ) 2 (2 s ) 2 (2 p ) 6 (3 s ) 2 (3 p ) 1 7 Ic. Br > Cl > F. The standard enthalpies of the reactions are given by Δ H o rxn = 2 Δ H f o (HX) - Δ H f o (H 2 ) - Δ H f o (X 2 ) For F 2 (g) and Cl 2 (g), Δ H f o (X 2 ) = 0 since these represent the elemental states of fluorine and chlorine, and Δ H o rxn = 2 Δ H f o (HX) = -542 kJ mol -1 for F and -184 kJ mol -1 for Cl. For Br 2 (g), Δ H o rxn = 2 Δ H f o (HBr) - Δ H f o (Br 2 ,g) = -103 kJ mol -1 , thus establishing the ordering given above. Id. P(CH 3 ) 3 > PBr 3 > [PO 4 ] 3- . The softness of an atom increases as softer atoms are bound to it. Thus, the softer the atoms bound to P, the softer the P atom becomes. Based on these criteria, the P atom in P(CH 3 ) 3 is the softest, since CH 3 - is a soft base.
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Chem 141 2009_Sample Exam 2 Solns - Chem 141 Sample Exam 2...

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