# 01+ - P1 PBU/OVY JWCL234-01 P2 PBU/OVY QC PBU/OVY T1 PBU...

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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 fluorine 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 IVA and has four valence electrons; fluo- rine, in Group VIIA, 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 first show a hydrogen’s electron as x, carbon’s electrons as o’s, and fluorine’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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2 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1.1 14 N, 7 protons and 7 neutrons; 15 N, 7 protons and 8 neutrons 1.2 (a) one (b) seven (c) four (d) three (e) eight (f ) five 1.3 (a) covalent (b) ionic (c) covalent (d) covalent 1.4 O O O • • • • O P 1.5 H (a) • • • • • • • • F • • • • F • • • • F • • • • F • • • • (b) (c) C H H H H (d) O • • O N O (f) B H H H H • • • • H (g) O • • • • O • • • • O • • • • • • O • • • • O O H P H H H C (h) O O O H H S (e) 1.6 O O H C H H (a) H (b) N H C H C (f ) (e) (c) C N O O O H C C O H (d)
THE BASICS: BONDING AND MOLECULAR STRUCTURE 3 1.7 C H + + H C H H H H (b) O H H (a) H (f ) H H H O C H H N H H C H H H (e) + + O H (d) C H H - (c) H O C O - (g) H H C H C N (h) H H C H N N + 1.8 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. 1.9 (a) H H C O + H H C Ο (b) H C H C H O O C C H H H (c) + + H H N H C H C N H H H H (d) C N C H H C C N H H

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4 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1.10 (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 H 2 C O C CH 3 H 2 C O (g) S CH 2 + CH 3 S S CH 2 CH 3 + C O CH 3 H 2 C C δ
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