# 01_ - P1 PBU/OVY P2 PBU/OVY JWDD052-01 JWDD052-Solomons-v2...

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-01 JWDD052-Solomons-v2 March 9, 2007 22:2 1 THE BASICS: BONDING AND MOLECULAR STRUCTURE SOLUTIONS TO PROBLEMS Another Approach to Writing Lewis Structures When we write Lewis structures using this method, we assemble the molecule or ion from the constituent atoms showing only the valence electrons (i.e., the elec- trons of the outermost shell). By having the atoms share electrons, we try to give each atom the electronic structure of a noble gas. For example, we give hydrogen atoms two electrons because this gives them the structure of helium. We give car- bon, nitrogen, oxygen, and ±uorine atoms eight electrons because this gives them the electronic structure of neon. The number of valence electrons of an atom can be obtained from the periodic table because it is equal to the group number of the atom. Carbon, for example, is in group IV A and has four valence electrons; ±uo- rine, in group VII A, has seven; hydrogen, in group 1A, has one. As an illustration, let us write the Lewis structure for CH 3 F. In the example below, we will at ²rst show a hydrogen’s electron as x, carbon’s electrons as o’s, and ±uorine’s electrons as dots. Example A C F H H H or C F H H H 3 H , C , and F are assembled as If the structure is an ion, we add or subtract electrons to give it the proper charge. As an example, consider the chlorate ion, ClO 3 . Example B Cl , and O and an extra electron × are assembled as Ο Ο Ο Cl Ο Ο Ο Cl or 1

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-01 JWDD052-Solomons-v2 March 9, 2007 22:2 2 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1.1 H (a) F F (b) (c) C H H H H (d) O N O O O O H H S (e) (f ) B H H H H H (g) O OH P H H H C (h) (i) C H N F F O O O O O - 1.2 H (b) N H O (a) N O C H C (f) (e) O O H S O O (d) (c) C N O O O H C - - - - - - 1.3 C H + + H C H H H H O H H (a) H H H H O C H H N H H C H H H (e) O H C H H (c) H O C O - - + + 1.4 C O O H C O O H (a) (b) and (c). Since the two resonance structures are equivalent, each should make an equal contribution to the overall hybrid. The C—O bonds should therefore be of equal length (they should be of bond order 1.5), and each oxygen atom should bear a 0.5 negative charge.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-01 JWDD052-Solomons-v2 March 9, 2007 22:2 THE BASICS: BONDING AND MOLECULAR STRUCTURE 3 1.5 (a) + + CH CH CH 3 CH OH CH CH CH 3 CH OH CH CH OH CH 3 CH + (b) CH CH CH CH 2 CH 2 + CH CH CH CH 2 CH 2 CH CH CH CH 2 CH 2 + + + + + (c) (d) CH CH 2 Br CH CH 2 Br + (e) + + CH 2 + CH 2 + CH 2 CH 2 + CH 2 CH CH CH 3 CH OH δ + δ + δ + CH CH CH 2 CH CH 2 δ + δ + δ + δ + δ + δ + CH Br CH 2 δ δ + CH 2 δ + δ + δ + δ + (f ) CH 3 CH 2 O C CH 3 CH 2 O (g) SC H 2 + CH 3 S H 2 CH 3 + C O CH 3 CH 2 C δ δ CH 2 CH 3 δ + δ +

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-01 JWDD052-Solomons-v2 March 9, 2007 22:2 4 THE BASICS: BONDING AND MOLECULAR STRUCTURE (h) + CH 3 N O O + CH 3 N O O O 2 + CH 3 N O (minor) N CH 3 + O O δ δ 1.6 CH 2 N(CH 3 ) 2 + (a) because all atoms have a complete octet (rule 6b), and there are more covalent bonds (rule 6a).
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## This note was uploaded on 10/20/2011 for the course CHEMISTRY ES110 taught by Professor Te.li during the Spring '11 term at Central Texas College.

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01_ - P1 PBU/OVY P2 PBU/OVY JWDD052-01 JWDD052-Solomons-v2...

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