Chapter 9: The Basics of Chemical Bonding Chemistry: The Molecular Nature of Matter, 6E Brady/Jespersen/Hyslop

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 2 Chemical Bonds  Attractive forces that hold atoms together in complex substances  Molecules and ionic compounds Why study?  Changes in these bonding forces are the underlying basis of chemical reactivity  During reaction:  Break old bonds  Form new bonds

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 3 Two Classes of Bonds  Ionic Bonding  Occurs in ionic solid  e  ’s transferred from 1 atom to another  Covalent bonding  Occurs in molecules  Sharing of e  ’s

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 4 Ionic Bonds  Result from attractive forces between oppositely charged particles  Metal - nonmetal bonds are ionic because:  Metals have  Low ionization energies  Easily lose e – to be stable  Non-metals have very exothermic electron affinities Na + Cl –

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 5 Ionic Compounds  Formed from metal and nonmetal Ionic crystals:  3-dimensional array of cations and anions = lattice structure

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 6 Covalent Compounds  Form individual separate molecules  Atoms bound by sharing e – ’s  Do not conduct electricity  Often low melting point Covalent Bonds  Shared pairs of e – ’s between 2 atoms  2 H atoms come together, Why?

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 7 Covalent Bond  Attraction of valence e – of 1 atom by nucleus of other atom  Shifting of e – density  As distance between nuclei ,  probability of finding either e – near either nucleus  Pulls nuclei closer together

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Which species is most likely covalently bonded? A. CsCl B. NaF C. CaF 2 D. CO E. MgBr 2 8

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 9 Electronegativity and Bond Polarity  2 atoms of same element form bond  Equal sharing of e  ’s  2 atoms of different elements form bond  Unequal sharing of e  ’s

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 10 Electronegativity and Bond Polarity  Leads to concept of Partial charges  +   H——Cl  + on H =   on Cl =  0.17

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 11 Electronegativity (  )  Relative attraction of atom for e  s in bond  Ability of bonded atom to attract e  s to itself  Quantitative basis  Difference in electronegativity estimate of bond polarity   = |  1   2 |  Ex. N—H Si—F    +  +  

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 12 Electronegativity Table

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 13 Electronegativities  Must know for 2 nd row and H.   (H) ~ 2.0   (F) ~ 4.0    by 0.5 for each element as go to left H  2 LiBeBCNOF  PSCl

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Which of the following species has the least polar bond? A. HCl B. HF C. HI D. HBr 14

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 15 Octet Rule  Works well with  Group IA and IIA metals  Al  Non-metals  H and He can't obey  Limited to 2 e  's in n = 1 shell  Doesn't work with  Transition metals  Post transition metals

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 16 Lewis Symbols  Electron bookkeeping method  Way to keep track of e – ’s  Write chemical symbol surrounded by dots for each e – Group #IAIIAIIIAIVA Valence e – 's1234 e – conf'nns 1 ns 2 ns 2 np 1 ns 2 np 2

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 17 Lewis Symbols Group #VAVIAVIIAVIIIA Valence e - 's5678 e - conf'nns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 For the representative elements Group # = # valence e – ’s

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 18 Lewis Symbols  Can use to diagram e – transfer in ionic bonding

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 19 Lewis Structures Diatomic Gases:  H and Halogens H 2  H· + ·H  H:H or H  H  Each H has 2 e – ’s through sharing  Can write shared pair of e – ’s as line (  )  : =   covalent bond

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 20 Lewis Structures Diatomic Molecules: F2F2  Each F has complete octet  Only need to form one bond to complete octet  Pairs of e – ’s not included in covalent bond are called Lone Pairs  Same for Cl 2, Br 2, I 2, HCl, HBr, HI.

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 21 Lewis Structures  Many nonmetals form more than 1 covalent bond Needs 4 e - ’s Forms 4 bonds Needs 3 e - ’s Forms 3 bonds Needs 2 e - ’s Forms 2 bonds methaneammoniawater

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 22 Double Bonds  2 pairs of e – ’s shared between 2 atoms Ex. CO 2 Trile bond  3 pairs of e – ’s shared between 2 atoms Ex. N 2

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Which species is most likely to have multiple bonds ? A. CO B. H 2 O C. PH 3 D. BF 3 E. CH 4 23

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 24 Method for Drawing Lewis Structure 1.Decide how atoms are bonded  Skeletal structure = arrangement of atoms.  Central atom  Usually given first  Usually least electronegative 2.Count all valence e  s. (All atoms) 3.Place 2 e  s between each pair of atoms  Draw in single bonds

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 25 Method for Drawing Lewis Structure 4.Complete octets of terminal atoms (atoms attached to central atom) by adding e  s in pairs 5.Place any remaining e  ’s on central atom in pairs 6.If central atom does not have octet  Form double bonds  If necessary, form triple bonds

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 26 Ex. N 2 F 2 2 N = 2  5e  = 10 e  2F = 2  7e  = 14 e  Total = 24 e  single bonds  6 e  18 e  F lone pairs  12 e  6 e  N electrons  6 e  0 e  Skeletal Structure Complete terminal atom octets Put remaining e  s on central atom

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 27 Ex. N 2 F 2  Not enough electrons to complete N octets  Must form double bond between N’s to satisfy both octets.

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 28 Ex. SiF 4 1 Si = 1  4e  = 4 e  4F = 4  7e  = 28 e  Total = 32 e  single bonds  8 e  24 e  F lone pairs  24 e  0 e  Skeletal Structure Complete terminal atom octets

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 29 Ex. H 2 CO 3  CO 3 2  oxoanion, so C central, and O’s around, H + attached to two O’s 1 C = 1  4e  = 4 e  3 O = 3  6e  = 18 e  2 H = 2  1e  = 2 e  Total = 24 e  single bonds  10 e  14 e  O lone pairs  14 e  0 e   But C only has 6 e 

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 30 Ex. H 2 CO 3 (cont.)  Too few e   Must convert 1 of lone pairs on O to 2 nd bond to C  Form double bond between C and O

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 31 Ex. CO 2 1 C = 1  4e  = 4 e  2 O = 2  6e  = 12 e  Total = 16 e  single bonds  4 e  12 e  O lone pairs  12 e  0 e   C only has 4 e   Must form 2 double bonds to O to complete C’s octet  3 ways you can do this  Which of these is correct?

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! What is wrong with the following structure? A. Too few total electrons B. Too many total electrons C. Lack of octet around nitrogen D. Too many electrons around the N atom E. Structure is correct as written 32

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 33 Formal Charge (FC)  Apparent charge on atom  Does not represent real charges FC = # valence e   [# bonds to an atom + # unshared e  ]

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 34 CO 2 Which Structure is Best  Use FC to determine which structure is best FC C = 4  (4 + 0) = 0 FC O(s) = 6  (1 + 6) =  1 FC O(d) = 6  (2 + 4) = 0 FC O(t) = 6 – (3 + 2) = +1  Central structure Best  All FC’s = 0 11 +1

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 35 FC = #valence e   [#bonds to atom + # unshared e  ] Structure 1 FC S = 6  (4 + 0) = 2 FC H = 1  (1 + 0) = 0 FC O(s) = 6  (1 + 6) =  1 FC O(d) = 6  (2 + 4) = 0 Structure 2 FC S = 6  (6 + 0) = 0 FC H = 1  (1 + 0) = 0 FC O(s) = 6  (2 + 4) = 0 FC O(d) = 6  (2 + 4) = 0 +2

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! What is the formal charge on Xe for the following ? A. +2, +4 B. +2, +3 C. +4, 0 D. +4, +2 36

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 37 What are Resonance Structures?  Multiple Lewis Structures for single molecule  No single Lewis structure is correct  Structure not accurately represented by any 1 Lewis Structure  Actual structure = "average" of all possible structures  Double headed arrow between resonance structures used to denote resonance

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 38  Lewis structure predicts 1 bond shorter than other 2 o Experimental observation:  All 3 N—O bond lengths are same  All shorter than N—O single bonds  Have to modify Lewis Structure  e  can't distinguish O atoms 11 +1 11 11 11 11 11

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 39 Resonance Structures  Lewis structures assume e  is localized between 2 atoms  In cases where need resonance structures, e  are delocalized  Smeared out over all atoms  Can move around entire molecule to give equivalent bond distances Resonance Hybrid  Way to depict resonance delocalization

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 40 Ex. CO 3 2  1 N = 1  5e  = 5 e  3 O = 3  6e  = 18 e   1 charge = 1 e  Total = 24 e  single bonds  6 e  18 e  O lone pairs  18 e  0 e   C = 2.5  O = 3.5 C only has 6 e  so form double bond

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 41 Three Equivalent Resonance Structures  All have same net formal charges on C and O’s  FC =  1 on singly bonded O’s  FC = O on doubly bond O and C 11 11 11 11 11 11

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 42 How Do We Determine Which are Good Contributors? 1. All Octets are satisfied 2. All atoms have as many bonds as possible 3a. FC   1  3b. Any negative charges are on electronegative atoms. 4. As little charge separation as possible.

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 43 1 C = 1  4e  = 4 e  1 N = 1  5e  = 5 e  1 O = 1  6e  = 6 e   1 charge = 1 e  Total = 16 e  single bonds  4 e  12 e  O lone pairs  12 e  0 e  Ex. NCO 

Brady/Jespersen/Hyslop Chemistry: The Molecular Nature of Matter, 6E 44 Ex. NCO  FC N = 5 – 2 – 3 = 0 FC C = 4 – 0 – 4 = 0 FC O = 6 – 6 – 1 = –1 FC N = 5 – 4 – 2 = –1 FC C = 4 – 0 – 4 = 0 FC O = 6 – 4 – 2 = 0 FC N = 5 – 6 – 1 = –2 FC C = 4 – 0 – 4 = 0 FC O = 6 – 2 – 3 = +1 OK Best Not Accept- able 2 11 11