Chapter 10 The Shapes of Molecules

Chapter 10 The Shapes of Molecules - Chapter 10.1 10 . . ....

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Chapter 10 The Shapes of Molecules 10.1 b) He and d) H cannot serve as a central atom in a Lewis structure. Both have an outermost shell with maximum of two electrons. F cannot serve as a central atom because its valence electrons are in the n = 2 level. F cannot have an expanded outlet because the n = 2 level only contains s and p orpitals, so F can form only 1 bond. 10.2 When more than one canonical structure can be drawn for the same molecule. For NO 2 , four structures can be drawn: In "averaging" these four structures, the N-O bond order is 1.5, so we would expect it to be intermediate between N-O and N=O. 10.3 a, b, d, e, f and h obey the octet rule; c and g do not. 10.4 Atoms must have available d orbitals in their valence shell, i.e., must have n = 3 or larger. Se, Cl, S and Al can have expanded shells; F and H cannot. 10.5 a) b) c) 10.6 a) b) c) 10.7 a) b) c) 10.8 a) b) c) 10.9 a) See 10.3 above. b) 177 F C F .. .. .. .. O H P H H H + Cl C Cl .. .. Cl .. .. .. H F Si F F F .. .. .. .. .. .. Cl Se Cl .. .. .. .. .. .. H Sb H .. H F P F F .. .. .. .. .. .. O C O .. .. O .. .. H H S C S .. .. Cl S S Cl .. .. .. .. .. .. .. .. O N O .. .. O N O .. F F . . . . .. .. .. .. .. H C H H S H .. .. .. . O N O .. .. O N O . .. .. .. O N O . .. .. .. .. .. O N O . .. ..
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10.10 a) b) (3 equivalent forms) 10.11 a) b) 10.12 a) b) (3 equivalent forms) 10.13 a) F +7 - 6 - 2/2 = 0 I +7 - 2 - 10/2 = 0 b) H +1 - 0 - 2/2 = 0 Al +3 - 8/2 = - 1 10.14 a) S +6 - 4 - 4/2 = 0 C +4 - 0 - 8/2 = 0 O +6 - 4 - 4/2 = 0 b) N +5 - 3 - 4/2 = 0 O +6 - 4 - 4/2 = 0 10.15 a) C +4 - 2 - 6/2 = - 1 N +5 - 2 - 6/2 = 0 b) Cl +7 - 6 - 2/2 = 0 O +6 - 6 - 2/2 = - 1 10.16 a) F +7 - 6 - 2/2 = 0 B +3 - 0 - 8/2 = - 1 178 H Al H H H - O N O .. O N O O O H H . . . . . . . . . . . . . . . . F I F F F F .. .. .. .. .. .. .. . . As O O O O H As O O O H 2 - 2 - .. .. . . .. . . .. .. .. .. .. O .. .. S C O .. .. O N .. O N .. .. . .. .. C N C N - - .. - B F F F F .. .. .. .. .. .. N N N N N N N N N .. .. .. .. .. .. .. .. - .. O N O .. .. O N O .. - .. .. .. .. .. .. H C O H C O O O - .. .. .. .. .. Br O O O O H Br O O O H .. .. .. .. .. .. .. .. .. .. O .. .. - O Cl . . . . . . . .
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b) Cl +7 - 6 - 2/2 = 0 N +5 - 2 - 6/2 = 0 O +6 - 4 - 4/2 = 0 10.17 a) O +6 6 2/2 = –1 O +6 - 4 - 4/2 = 0 Br +7 - 2 - 10/2 = 0 Net formal charge = - 1 O.N.: O - 2 each; Br +5 b) S +6 - 2 - 8/2 = 0 O +6 - 6 - 2/2 = - 1 ea O +6 4 4/2 = 0 Net formal charge = - 2 O.N.: S +4; O - 2 10.18 a) As +5 - 0 - 8/2 = +1 O +6 - 6 - 2/2 = - 1 ea = - 4 Net formal charge (+1 - 4) = - 3 O.N.: O - 2 x 4 = - 8 total; As +5 b) Cl +7 - 4 - 6/2 = 0 O +6 - 4 - 4/2 = 0 O* +6 - 6 - 2/2 = - 1 O.N.: O - 2 x 2 = - 4 total; Cl +3 10.19 a) only 6 e - in outer shell b) expanded outer shell to 10 e - c) expanded outer shell to 10 e - 10.20 a) expanded outer shell to 12 e - b) radical – chlorine lacks an e - to complete octet or one of the oxygens lacks the octet c) expanded outer shell to 10 e - 179 H B H H Cl N O .. .. .. .. Se
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This note was uploaded on 01/21/2010 for the course CHEM 1A taught by Professor Kobiashi during the Spring '07 term at Ventura College.

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Chapter 10 The Shapes of Molecules - Chapter 10.1 10 . . ....

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