Chapter 9: The Basics
of Chemical Bonding
Chemistry: The Molecular Nature
of Matter, 6E
Brady/Jespersen/Hyslop
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
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Chemistry: The Molecular Nature of Matter, 6E
2
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
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Chemistry: The Molecular Nature of Matter, 6E
3
Ionic Bonds
 Result from attractive forces between
oppositely charged particles
Na+
Cl –
 Metal - nonmetal bonds are ionic because:
 Metals have
 Low ionization energies
 Easily lose e– to be stable
 Non-metals have very exothermic electron
affinities
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Chemistry: The Molecular Nature of Matter, 6E
4
Ionic Compounds
 Formed from metal and nonmetal
Na + Cl
Na+ + Cl
NaCl(s)
e
Ionic crystals:
 3-dimensional array of
cations and anions =
lattice structure
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Chemistry: The Molecular Nature of Matter, 6E
5
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?
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Chemistry: The Molecular Nature of Matter, 6E
6
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
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Chemistry: The Molecular Nature of Matter, 6E
7
Your Turn!
Which species is most likely covalently bonded?
A. CsCl
B. NaF
C. CaF2
D. CO
E. MgBr2
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Chemistry: The Molecular Nature of Matter, 6E
8
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
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Chemistry: The Molecular Nature of Matter, 6E
9
Electronegativity and Bond Polarity
 Leads to concept of
Partial charges
+

H——Cl
+ on H = +0.17
 on Cl = 0.17
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Chemistry: The Molecular Nature of Matter, 6E
10
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





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+
+

Chemistry: The Molecular Nature of Matter, 6E
11
Electronegativity Table
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Chemistry: The Molecular Nature of Matter, 6E
12




Electronegativities
Must know for 2nd row and H.
(H) ~ 2.0
(F) ~ 4.0
 by 0.5 for each element as go to left
H
2

Li Be B
C
N
O
F
1
1.5
2
2.5
3
3.5
4

P
S
Cl
2
2.5 3.0
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Chemistry: The Molecular Nature of Matter, 6E
13
Your Turn!
Which of the following species has the least
polar bond?
A. HCl
B. HF
C. HI
D. HBr
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Chemistry: The Molecular Nature of Matter, 6E
14
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
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15
Lewis Symbols
 Electron bookkeeping method
 Way to keep track of e–’s
 Write chemical symbol surrounded by dots for each
e–
Group #
Valence e–'s
e– conf'n
IA
1
ns1
IIA
2
ns2
H
Li
He
Na
Brady/Jespersen/Hyslop
Be
Mg
Chemistry: The Molecular Nature of Matter, 6E
IIIA
3
ns2np1
IVA
4
ns2np2
B
C
Al
Si
16
Lewis Symbols
Group #
Valence e-'s
e- conf'n
VA
VIA
VIIA
VIIIA
5
6
7
8
ns2np3
ns2np4
ns2np5
ns2np6
He
N
O
F
Ne
P
S
Cl
Ar
For the representative elements
Group # = # valence e–’s
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Chemistry: The Molecular Nature of Matter, 6E
17
Lewis Symbols
 Can use to diagram e– transfer in ionic
bonding
Na
Mg
Brady/Jespersen/Hyslop
+ Cl
+
O
Na+
+ Cl
2+
Mg
Chemistry: The Molecular Nature of Matter, 6E
+ O

2
18
Lewis Structures
Diatomic Gases:
 H and Halogens
H2





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
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Chemistry: The Molecular Nature of Matter, 6E
19
Lewis Structures
Diatomic Molecules:
F2
F
+
F
FF
F
F
 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 Cl2, Br2, I2, HCl, HBr, HI.
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Chemistry: The Molecular Nature of Matter, 6E
20
Lewis Structures
 Many nonmetals form more than 1 covalent bond
C
O
N
Needs 4 e-’s
Forms 4 bonds
H
Needs 3 e-’s
Forms 3 bonds
H N H
H C H
H
H
Needs 2 e-’s
Forms 2 bonds
O H
H
H
H C H
H N H
H
H
methane
ammonia
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Chemistry: The Molecular Nature of Matter, 6E
O H
H
water
21
Double Bonds
 2 pairs of e–’s shared between 2 atoms
Ex. CO2
O
C
O
O C O
O C O
Trile bond
 3 pairs of e–’s shared between 2 atoms
Ex. N2
N
N
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N
N
Chemistry: The Molecular Nature of Matter, 6E
N
N
22
Your Turn!
Which species is most likely to have multiple
bonds ?
A. CO
B. H2O
C. PH3
D. BF3
E. CH4
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Chemistry: The Molecular Nature of Matter, 6E
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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
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Chemistry: The Molecular Nature of Matter, 6E
24
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
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Chemistry: The Molecular Nature of Matter, 6E
25
Ex. N2F2
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
F
N
F
Complete terminal
atom octets
F
N
N
F
Put remaining es on
central atom
F
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N
Chemistry: The Molecular Nature of Matter, 6E
N
N
F
26
Ex. N2F2
 Not enough electrons to complete N octets
 Must form double bond between N’s to
satisfy both octets.
F
N
Brady/Jespersen/Hyslop
N
F
Chemistry: The Molecular Nature of Matter, 6E
F
N
N
F
27
Ex. SiF4
1 Si = 1 
= 4
4F = 4  7e = 28 e
Total = 32 e
single bonds  8 e
24 e
F lone pairs  24 e
0 e
4e
Skeletal Structure
F
e
F
Si
F
F
Complete terminal
atom octets
F
F
Si
F
F
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Ex. H2CO3
 CO32 oxoanion, so C central, and O’s
around, H+ attached to two O’s
O
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
Brady/Jespersen/Hyslop
H
O
C
O
H
O
H
O
H
O
C
 But C only has 6 e
Chemistry: The Molecular Nature of Matter, 6E
29
Ex. H2CO3 (cont.)
 Too few e
 Must convert 1 of
lone pairs on O to
2nd bond to C
 Form double bond
between C and O
O
H
C
O
H
O
H
O
H
Brady/Jespersen/Hyslop
O
Chemistry: The Molecular Nature of Matter, 6E
O
C
30
Ex. CO2
O C O
1 C = 1  4e = 4 e
2 O = 2  6e = 12 e
O C O

Total = 16 e
single bonds  4 e
O C O
O C O
12 e
O lone pairs  12 e
O C O
O O O

0e
 C only has 4 e
O C O
O C O
 Must form 2 double
bonds to O to complete  Which of these is correct?
C’s octet
 3 ways you can do this
Brady/Jespersen/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
31
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
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Chemistry: The Molecular Nature of Matter, 6E
32
Formal Charge (FC)
 Apparent charge on atom
 Does not represent real charges
FC = # valence e  [# bonds to an atom +
# unshared e ]
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Chemistry: The Molecular Nature of Matter, 6E
33
CO2 Which Structure is Best
 Use FC to determine
which structure is best
1
+1
FCC = 4  (4 + 0) = 0
FCO(s) = 6  (1 + 6) = 1
FCO(d) = 6  (2 + 4) = 0
FCO(t) = 6 – (3 + 2) = +1
O
C
O
O
C
O
-1
 Central structure Best
 All FC’s = 0
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Chemistry: The Molecular Nature of Matter, 6E
+1
O
C
O
34
FC = #valence e  [#bonds to atom
+ # unshared e ]
O
H
O
-1
+2
S
O
O
H
-1
O
H
O
S
O
Brady/Jespersen/Hyslop
O
H
Structure 1
FCS = 6  (4 + 0) = 2
FCH = 1  (1 + 0) = 0
FCO(s) = 6  (1 + 6) =  1
FCO(d) = 6  (2 + 4) = 0
Structure 2
FCS = 6  (6 + 0) = 0
FCH = 1  (1 + 0) = 0
FCO(s) = 6  (2 + 4) = 0
FCO(d) = 6  (2 + 4) = 0
Chemistry: The Molecular Nature of Matter, 6E
35
Your Turn!
What is the formal charge on Xe for the
following ?
A. +2, +4
B. +2, +3
C. +4, 0
D. +4, +2
Brady/Jespersen/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
36
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
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Chemistry: The Molecular Nature of Matter, 6E
37
 Lewis structure predicts 1 bond shorter than
other 2

O
o Experimental observation:
 All 3 N—O bond lengths are same
 All shorter than N—O single bonds
N
O
O
 Have to modify Lewis Structure
 e can't distinguish O atoms

O
1
O
N
+1
1
Brady/Jespersen/Hyslop
O
1
O
N +1
O
O
1
Chemistry: The Molecular Nature of Matter, 6E

1
O
N
O
1

+1
O
38
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
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Chemistry: The Molecular Nature of Matter, 6E

O
N
O
O
39
C = 2.5
Ex. CO32
Brady/Jespersen/Hyslop
O
O = 3.5
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 only has 6 e so form
double bond
Chemistry: The Molecular Nature of Matter, 6E

C
O
O

O
C
O
O

O
C
O
O
40
Three Equivalent Resonance
Structures

O
O
C
O
1
1

C
O
1
O
O
1

C
O
1
O 1
O
 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
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Chemistry: The Molecular Nature of Matter, 6E
41
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.
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Ex. NCO
N
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
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Chemistry: The Molecular Nature of Matter, 6E
C
O


N
N
N
N
C
C
C
C
O
O
O
O



43
Ex. NCO
FCN = 5 – 2 – 3 = 0
FCC = 4 – 0 – 4 = 0
FCO = 6 – 6 – 1 = –1
FCN = 5 – 4 – 2 = –1
FCC = 4 – 0 – 4 = 0
FCO = 6 – 4 – 2 = 0
FCN = 5 – 6 – 1 = –2
FCC = 4 – 0 – 4 = 0
FCO = 6 – 2 – 3 = +1
Brady/Jespersen/Hyslop
N
1
C
O

Best
1
N
C
2
N
Chemistry: The Molecular Nature of Matter, 6E
O
+1
C
O

OK
 Not
Acceptable 44