1. Chapter 8 Chemical Bonding

2.  Na Cl + 1s22s22p63s1 1s22s22p63s23p5 Na+ Cl– + 1s22s22p6

3. Ionic Bond
Cl–
electrostatic attraction between oppositely charged ions
non-directional
Na+

4. Ionic Bond
Cl–
electrostatic attraction between oppositely charged ions
non-directional
Na+
Na+
Cl–

5. Ionic Bond
Cl–
electrostatic attraction between oppositely charged ions
non-directional
Na+
Cl–
Na+
Na+
Cl– =

6. Ionic Bond
Cl–
electrostatic attraction between oppositely charged ions
non-directional
Na+
Cl–
Cl–
Na+
Na+ =
Cl– =
Na+

7. Ionic Bond
Cl–
electrostatic attraction between oppositely charged ions
non-directional
strength directly proportional to charges of ions and inversely proportional to distance between ion centers
Na+

8. E  Q c a d
E = ionic bond strength
Qc = charge on cation
Qa = charge on anion (absolute value)
d = distance between centers of ions

9. E  Q c a d
NaCl Na Cl–

10. E  Q c a d
NaCl Na Cl–
CaCl Ca Cl–

11. E  Q c a d
NaCl Na Cl–
CaCl Ca Cl–
CaS Ca Cl–

12. E  Q c a d
NaCl Na Cl–
CaCl Ca Cl–
CaS Ca S–2
Al2S Al S2–

13. E  Q c a d d
Na+ F–

14. E  Q c a d d d
Cl–
Na+ F–
Na+

15. E  Q c a d d d
Cl–
Na+ F–
Na+ d
Br–
Na+

16. E  Q c a d d d
Cl–
Na+ F–
Na+ d d I–
Br–
Na+
Na+

17. Lattice Energy
energy required to separate 1 mol of an ionic compound into its gaseous ions
eg MgF2(s)  Mg2+(g) + 2F–(g)
Hlattice = 2910 kJ/mol
for NaF, Hlattice = kJ/mol
for KF, Hlattice = kJ/mol
WHY???

18. Ionic Bond Formation Born Haber Cycle
Breaks formation of an ionic compound from its elements into a series of theoretical steps and considers the energetics of each
Example: consider formation of NaCl
Na(s) + 1/2Cl2(g)  NaCl(s) H° = –410.9 kJ

19. 725.4 E
(kJ)
376.4
229.4
107.7
Na(s) + 1/2Cl2(g)
–410.9

20. E (kJ) Na(g) + 1/2Cl2(g) Hsub, Na Na(s) + 1/2Cl2(g) 725.4 376.4 229.4
107.7
Hsub, Na
Na(s) + 1/2Cl2(g)
–410.9

21. E (kJ) Na(g) + Cl(g) Na(g) + 1/2Cl2(g) Hatom,Cl Hsub, Na
725.4 E
(kJ)
376.4
Na(g) + Cl(g)
229.4
Na(g) + 1/2Cl2(g)
Hatom,Cl
107.7
Hsub, Na
Na(s) + 1/2Cl2(g)
–410.9

22. Na+(g) + Cl(g) + e– E (kJ) IENa Na(g) + Cl(g) Na(g) + 1/2Cl2(g)
725.4 E
(kJ)
IENa
376.4
Na(g) + Cl(g)
229.4
Na(g) + 1/2Cl2(g)
Hatom,Cl
107.7
Hsub, Na
Na(s) + 1/2Cl2(g)
–410.9

23. Na+(g) + Cl(g) + e– EACl E (kJ) Na+(g) + Cl–(g) IENa Na(g) + Cl(g)
725.4
EACl E
(kJ)
Na+(g) + Cl–(g)
IENa
376.4
Na(g) + Cl(g)
229.4
Na(g) + 1/2Cl2(g)
Hatom,Cl
107.7
Hsub, Na
Na(s) + 1/2Cl2(g)
–410.9

24. Na+(g) + Cl(g) + e– EACl E (kJ) Na+(g) + Cl–(g) IENa Na(g) + Cl(g)
725.4
EACl E
(kJ)
Na+(g) + Cl–(g)
IENa
376.4
Na(g) + Cl(g)
–Hlattice
229.4
Na(g) + 1/2Cl2(g)
Hatom,Cl
107.7
Hsub, Na
Na(s) + 1/2Cl2(g)
NaCl(s)
–410.9

25. H°f = 107.7kJ + 121.7kJ + 496.0kJ + –349.0kJ – 787.3 kJ
H°f = Hsublimation, Na + Hatom, Cl + IE1, Na + EACl – Hlattice, NaCl
H°f = 107.7kJ kJ kJ + –349.0kJ – kJ
H°f = –410.9 kJ

26. Lewis Symbols atomic symbol indicates element
symbol considered to have 4 sides
valence electrons indicated by dots
maximum 2 dots per side

27. Examples Na 

28. Examples Na   Na

29. Examples  Na   Na Na

30. Examples  Na   Na Na Na 

31. Examples  Na   Na Na Na     Ca Ca    Ca   Ca Ca 

32. Examples  Na   Na Na Na     Ca Ca    Ca   Ca Ca     Al      Ca 

33.   Na  + Cl   

34. –  +      Na  + Na + Cl Cl       

35. –  +      Na  + Na + Cl Cl          Cl    Ca +     Cl   

36. –  +      Na  + Na + Cl Cl        –      Cl   2+  Ca + Ca +   –       Cl  Cl      

37. 1s22s22p5
1s22s22p5       + F F      

38. 1s22s22p5
1s22s22p5       + F F      

39. 1s22s22p5
1s22s22p5
1s22s22p4
1s22s22p6 –        +       + + F F F F             

40. 1s22s22p5
1s22s22p5
1s22s22p4
1s22s22p6 –        +       + + F F F  F             –        +      + +  F F F F             

41. 1s22s22p5
1s22s22p5
1s22s22p4
1s22s22p6 –        +       + + F F F F              –        +      + +  F F F F                     F F       
shared e– pair

42. y y x x
2px
2px y y x

43.        F F       
shared e– pair       F F      

44. Covalent Bond
attractive force between atoms resulting from sharing electron pair(s)
very strong
highly directional

45. single bond: sharing 1 e– pair
double bond: sharing 2 e– pairs
triple bond: sharing 3 e– pairs

46.          O   O      O  O       

47.          O    O     O    O            O O  

48.          O   O      O O               O O        N   N       N N       

49.          O   O      O O               O O        N   N       N N          N N  

50. +  H  H  H H 

51. +  H  H  H H  y y x        F F       

52. Pure Covalent Bond +  H  H  H H  y y x        F F       

53. y y x        Br F   

54. y y x
+
–        Br F   

55. y y x
+
–        Br F   
Polar Covalent Bond
Unequal Sharing of Electron Pair(s)

56. Bonding Continuum
pure
ionic
polar
covalent
pure
covalent
BrF
CsF H2

57. Electronegativity
The ability of an atom to attract electron density towards itself
increases from L to R across a period and up a group

58. Bonding Continuum
pure
covalent
pure
ionic
polar
covalent
nonpolar
covalent
ionic
2.0
0.5
CsF (3.9)
BrF (1.3)
H2 (0)
 E’neg

59. KCl
 E,neg. = 2.9 – 0.9 = 2.0
ionic
HCl
 E,neg. = 2.9 – 2.1 = 0.8
polar covalent
BrCl
 E,neg. = 2.9 – 2.8 = 0.1
nonpolar covalent
Br2
 E,neg. = 2.9 – 2.9 = 0
pure covalent

60. Dipole Moment results from separation of charge polar bonds have
increases as  E’neg. increases

61. Dipole Moment results from separation of charge polar bonds have
increases as  E’neg. increases
indicated by crossed arrow pointing from positive end to negative end of dipole

62. H I
2.1
2.2

63. +
– H I
2.1
2.2

64. +
– H I
 E’neg. = 0.1

65. +
– H I
 E’neg. = 0.1 H Cl
2.1
2.9

66. +
– H I
 E’neg. = 0.1
+
– H Cl
2.1
2.9

67. +
– H I
 E’neg. = 0.1
+
– H Cl
 E’neg. = 0.8

68. +
– H I
 E’neg. = 0.1
+
– H Cl
 E’neg. = 0.8 H F
2.1
4.1

69. +
– H I
 E’neg. = 0.1
+
– H Cl
 E’neg. = 0.8
+
– H F
2.1
4.1

70. +
– H I
 E’neg. = 0.1
+
– H Cl
 E’neg. = 0.8
+
– H F
 E’neg. = 2.0

71. Lewis Structures
shorthand notation for placement of atoms and valence electrons in molecules

72. Drawing Lewis Structures
arrange atoms in formula such that element with only 1 atom in formula is in center surrounded by other atoms in formula
connect central atom to outer atoms by single covalent bonds
total valence electrons in molecule
add valence e– of all atoms
add 1 e– for each – charge
subtract 1 e– for each + charge
subtract 2 e– for each bond

73. use remaining e– to complete octets of outer atoms
place remaining e– on central atom, in pairs when possible
check
form multiple bonds if necessary

74. Octet Rule all atoms require 8 valence electrons EXCEPTIONS H, He 2 e–
B, Al 6 e–
Z  or more e–

75. Examples
draw Lewis structures for each of the following

76. Methane, CH4

77. Methane, CH4 H H C H H

78. Methane, CH4 H H C H H

79. Methane, CH4 C
4 e– H H
4 H
4 e– C
8 e– H H

80. Methane, CH4 C
4 e– H H
4 H
4 e– C
8 e– H H
4 bonds
– 8 e–
0 e–

81. SnCl3– Cl Sn Cl Cl

82. SnCl3– Cl Sn Cl Cl

83. SnCl3– Cl Sn
4 e–
3 Cl
21 e– Sn
25 e– Cl Cl

84. SnCl3– Cl Sn
4 e–
3 Cl
21 e– Sn
25 e– Cl Cl
1 –
1 e–
26 e–

85. SnCl3– Cl Sn
4 e–
3 Cl
21 e– Sn
25 e– Cl Cl
1 –
1 e–
26 e–
3 bonds
– 6 e–
20 e–

86. SnCl3–     Cl   Sn
4 e–
3 Cl
21 e– Sn
25 e–     Cl Cl        
1 –
1 e–
26 e–
3 bonds
– 6 e–
20 e–
– 18 e–
2 e–

87. SnCl3–     Cl   Sn
4 e– 
3 Cl
21 e– Sn 
25 e–     Cl Cl        
1 –
1 e–
26 e–
3 bonds
– 6 e–
20 e–
– 18 e–
2 e–
– 2 e–

88. SnCl3–     Cl   Sn
4 e– 
3 Cl
21 e– Sn 
25 e–     Cl Cl        
1 –
1 e–
26 e–
3 bonds
– 6 e–
20 e–
– 18 e–
2 e–
– 2 e–

89. XeF4 F F Xe F F

90. XeF4 Xe
8 e– F F
4 F
28 e– Xe
36 e– F F
4 bonds
– 8 e–
28 e–

91. XeF4        Xe
8 e–  F    F 
4 F
28 e– Xe
36 e–     F F        
4 bonds
– 8 e–
28 e–
– 24 e–
4 e–

92. XeF4        Xe
8 e–  F   F    
4 F
28 e– Xe  
36 e–     F F        
4 bonds
– 8 e–
28 e–
– 24 e–
4 e–
– 4 e–
0 e–

93. SF6

94. SF6         S
6 e– F   F       
6 F
42 e–  F S F      
48 e–     F F        
6 bonds
–12 e–
36 e–
– 36 e–
0 e–

95. O2 O O
2 O
12 e–
1 bond
– 2 e–
10 e–

96. O2       O O    
2 O
12 e–
1 bond
– 2 e–
10 e–
– 10 e–
0 e–

97.       O O           O O           O O    

98. Form Multiple Bond       O O    

99. Form Multiple Bond       O O    

100. Form Multiple Bond       O O           O O  

101. N2 N N
2 N
10 e–
1 bond
– 2 e–
8 e–

102. N2       N N  
2 N
10 e–
1 bond
– 2 e–
8 e–
– 8 e–
0 e–

103. Form Multiple Bonds       N N  

104. Form Multiple Bonds       N N       N N    

105. Form Multiple Bonds       N N       N N       N N  

106. CO2       O C O   C
4 e–    
2 O
12 e–
16 e–
2 bonds
– 4 e–
12 e–
– 12 e–
0 e–

107.       O C O       O C O         O C O    

108. Resonance more than one valid Lewis structure
each structure called resonance structure

109.               O C O O C O O C O          

110. Resonance Hybrid average of resonance structures
written in square brackets

111. Bond Order
1, single bond
2, double bond
3, triple bond

112.               O C O O C O O C O          
B.O.left bond = = 2 3

113.               O C O O C O O C O          
B.O.left bond = = 2 3
B.O.right bond = = 2 3

114.               O C O O C O O C O          
B.O.left bond = = 2 3
B.O.right bond = = 2 3       O C O  

115.               O C O O C O O C O          
B.O.left bond = = 2 3
B.O.right bond = = 2 3       O C O  

116. CO32–

117. CO32–           O O O       C C C          

118. CO32–           O O O       C C C          
B.O.each bond = =
1.3 3

119. CO32–           O O O       C C C          
B.O.each bond = =
1.3 3 2– O C O O

120. Bond Strength and Bond Length
triple < double < single
bond strength
single < double < triple

121. Length (Å) C C
1.54 C C
1.34 C C
1.20

122. Length (Å)
Strength (kJ/mol) C C
1.54
348 C C
1.34
614 C C
1.20
839

123. Formal Charge an accounting tool for electron ownership
(# valance e– in free atom) – (# e– in lone pairs on atom) – 1/2(# bonded e– on atom)

124. Best structure has zero FC on all atoms lowest FC possible
–FC on most electronegative atoms and +FC on least electronegative atoms

125. For example, consider thiocyanate ion

126. For example, consider thiocyanate ion
SCN–

127. For example, consider thiocyanate ion
SCN–             N C S C S N S N C            

128. For example, consider thiocyanate ion
SCN–             N C S C S S N C N            
FCN = 5 – 4 – 2 = –1

129. For example, consider thiocyanate ion
SCN–             N C S C S C N S N            
FCN = 5 – 4 – 2 = –1
FCC = 4 – 0 – 4 = 0

130. For example, consider thiocyanate ion
SCN–             N C C S C S S N N            
FCN = 5 – 4 – 2 = –1
FCC = 4 – 0 – 4 = 0
FCS = 6 – 4 – 2 = 0

131. For example, consider thiocyanate ion
SCN– –1             N C C S C S S N N            
FCN = 5 – 4 – 2 = –1
FCC = 4 – 4 – 2 = –2
FCS = 6 – 0 – 4 = +2

132. For example, consider thiocyanate ion
SCN– –1 –2 +2 –1             N C S C S N S N C            
FCN = 5 – 0 – 4 = +1
FCC = 4 – 4 – 2 = –2
FCS = 6 – 4 – 2 = 0

133. For example, consider thiocyanate ion
SCN– –1 –2 +2 –1 +1 –2             N C S C S N S N C            

134. For example, consider thiocyanate ion
SCN– –1 –2 +2 –1 +1 –2             N C S C S N S N C            
best structure because lowest FC and
–FC on most electronegative atom