1. CHEMICAL BONDING
Water
Cocaine
CHAPTER 9

2. Chapter 9 Outline Ionic -vs- Covalent Bonding in molecules
Valance electrons
Lewis Structures
Oxidation Numbers
Formal Charges
VESPR

3. Chemical Bonding Problems and questions —
How is a molecule or polyatomic ion held together?
Why are atoms distributed at strange angles?
Why are molecules not flat?
Can we predict the structure?
How is structure related to chemical and physical properties.

4. Forms of Chemical Bonds
There are 2 extreme forms of connecting or bonding atoms:
Ionic—complete transfer of electrons from one atom to another
Covalent—electrons shared between atoms
Most bonds are somewhere in between.

5. Ionic Bonds Complete electron transfer from an
element of low IE (metal)
to an element of high EA (nonmetal)
2 Na(s) + Cl2(g) ---> 2 Na Cl-
results in a metal/nonmetal or ionic compound.
Ionic Bonds

6. Covalent Bonding
Covalent bond forms by the sharing of VALENCE ELECTRONS, between two nonmetals.

7. Valence Electrons
Electrons are divided between core and valence electrons.
Na 1s2 2s2 2p6 3s1
Core = [Ne] and
valence = 3s1
Br [Ar] 3d10 4s2 4p5
Core = [Ar] 3d10 and
valence = 4s2 4p5 35 Br
[Ar] 3d10 4s24p5 : .

8. Valence Electrons

9. Covalent Bonding
The bond arises from the mutual attraction of 2 nuclei for the same electrons.
A Bond is a balance of attractive and repulsive forces.

10. Chemical Bonding Objectives
Objectives are to understand:
1. e - distribution in molecules and ions.
2. molecular structures.
3. bond properties and their effect on molecular properties.

11. Electron Distribution in Molecules
Electron distribution is depicted with Lewis electron dot structures
Electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS.
G. N. Lewis

12. Bond and Lone Pairs This is called a LEWIS ELECTRON DOT structure.
Electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS.
This is called a LEWIS ELECTRON DOT structure.

13. Lewis Dot Electron Configs

14. Bond Formation H Cl H Cl + Overlap of H (1s) and Cl (3p)
A bond can result from a “head-to-head” overlap of atomic orbitals on neighboring atoms. •• •• H Cl • H Cl • + • • •• ••
Overlap of H (1s) and Cl (3p)
This type of overlap places bonding electrons in a MOLECULAR ORBITAL along the line between the two atoms and forms a SIGMA BOND ().

15. Rules of the Game
Observation that atoms want to obtain a filled noble gas electron configuration is called the OCTET RULE
There are exceptions to these rules!!!

16. Building a Dot Structure
Ammonia, NH3
1. Decide on the central atom; never H.
Central atom is atom of lowest affinity for electrons. Therefore, N is the central atom.
2. Count valence electrons
H = 1 and N = 5
Total = (3 x 1) + 5
= 8 electrons / 4 pairs

17. Building a Dot Structure
3. Form a sigma bond between the central atom and surrounding atoms.
4. Remaining electrons go to the central atom to form LONE PAIRS to complete octet as needed. H •• N
3 BOND PAIRS and 1 LONE PAIR. Note that N has a share in 4 pairs (8 electrons), while H shares 1 pair.

18. Sulfite ion, SO32- Step 1. Central atom = S
Step 2. Count valence electrons S = 6
3 x O = 3 x 6 = 18
Negative charge = 2
TOTAL = 26 e- or 13 pairs
Step 3. Form sigma bonds

19. Sulfite ion, SO32- O S Step 1. Central atom = S
Step 2. Count valence electrons S = 6
3 x O = 3 x 6 = 18
Negative charge = 2
TOTAL = 26 e- or 13 pairs
Step 3. Form sigma bonds
10 pairs of electrons are
now left. O S

20. Sulfite ion, SO32-
Remaining pairs become lone pairs, first on outside atoms and then remaining pairs on central atom. • O S •• -2
Each atom is surrounded by an octet of electrons.

21. Carbon Dioxide, CO2 C 1. Central atom = _______
2. Valence electrons = __ or __ pairs
3. Form sigma bonds.
This leaves 6 pairs.
4. Place lone pairs on outer atoms.

22. Carbon Dioxide, CO2 4. Place lone pairs on outer atoms.
5. So that C has an octet, we form DOUBLE BONDS between C and O.
The second bonding pair forms a pi (p) bond.

23. Double and even triple bonds are commonly observed for C, N, P, O, and S
H2CO
SO3
C2F4

24. Sulfur Dioxide, SO2 O S 1. Central atom = S
2. Valence electrons = 18 or 9 pairs
3. Form pi (p) bond so that S has an octet — but note that there are two ways of doing this. • O S ••
bring in
left pair
OR bring in
right pair

25. Sulfur Dioxide, SO2 O S O S This leads to the following structures.• O S ••
bring in
left pair
OR bring in
right pair
This leads to the following structures. • O S ••
These equivalent structures are called
RESONANCE STRUCTURES. The true electronic structure is a HYBRID of the two.

26. Urea, (NH2)2CO 1. Arrangement, is C or O the central atom?
2. Number of valence electrons = 24 e-
3. Draw sigma bonds. C O

27. Urea, (NH2)2CO
4. Place remaining electron pairs in the molecule. •• O • • C N H •• ••
5. Carbon needs an Octet, take from the Oxygen

28. Urea, (NH2)2CO

29. Exceptions to the Octet Rule
Occurs with B (the only 2nd period exception) and elements of 3rd - 7th periods.
BF3
SF4

30. Boron Trifluoride B F Central atom = Valence electrons =
or electron pairs =
Assemble dot structure •• F • B
The B atom has a share in only 6 electrons (or 3 pairs). The B atom in many molecules is electron deficient.

31. Sulfur Tetrafluoride, SF4
Central atom =
Valence electrons = ___ or ___ pairs.
Form sigma bonds and distribute electron pairs.
5 pairs around the S atom. A common occurrence outside the 2nd period.

32. Draw Lewis Structures for:N2
CH4
I2CO
HCOOH
NO2-
XeF4

33. Oxidation Number
Oxidation Number is assigned based on a set of rules.
These rules are based on the Lewis Structures of the compounds or ions.
Oxidation Number = Group no. - (no. assigned electrons) - (no. of LP electrons)
Sample Problem

34. Formal Atom Charges
Atoms in molecules often bear a charge (+ or -).
The predominant resonance structure of a molecule is the one with charges as close to 0 as possible.
Formal charge = Group no /2 (no. bond electrons) - (no. of LP electrons)

35. Carbon Dioxide, CO2 Formal charge calculations

36. Carbon Dioxide, CO2 O C O C atom charge is 0.6 - ( 1 / 2 ) ( 2 ) - 6 = - 1
C atom charge is 0. • • • O C O • • • • • 6 - ( 1 / 2 ) ( 6 ) - 2 = + 1
Which is the predominant resonance structure?

37. Carbon Dioxide, CO2
+1.46
-0.73
Actual partial charges.

38. Boron Trifluoride, BF3
What if we form a B—F double bond to satisfy the B atom octet?

39. Boron Trifluoride, BF3
To have +1 charge on F, with its very high affinity for electrons, is not good.
Negative charges are best placed on atoms with high affinity for electrons.

40. Thiocyanate ion, SCN-
Which of three possible resonance structures is the most important?
Calculate the formal charge for each element.

41. Thiocyanate ion, SCN-
-0.16
-0.32
-0.52
Calculated partial charges

42. PRACTICE PROBLEM
Ox # = group # - lone pair electrons - assigned electrons
F.C. = group # - lone pair electrons - 1/2 bonding electrons
Practice Problem
Determine oxidation numbers and formal charges for the atoms in SO3-2.

43. MOLECULAR GEOMETRY

44. VSEPR MOLECULAR GEOMETRY Valence Shell Electron Pair Repulsion theory.
Molecule adopts the shape that minimizes the electron pair repulsions.
VSEPR
Valence Shell Electron Pair Repulsion theory.
Most important factor in determining geometry is relative repulsion between electron pairs.

45. Figure 9.11

46. Figure 9.12

47. Notice where the central atom is
No. of e- Pairs
Around Central
Atom
Example
Geometry 2
F—Be—F
linear
180° F
planar
trigonal 3 B F F
120°
109° H 4
tetrahedral C H H H
Notice where the central atom is

48. 4 electron pair groups
Electron pair geometry
Molecular Shape name

49. Electron pair geometry
Molecular name
Figure 9.14

50. Brief Summary of VESPR 14 possible combinations….learn them all!
Electron Pair Geometry
Molecular Geometry
2 electron pairs
Linear
3 electron pairs
Trigonal Planar
“V” Bent
4 electron pairs
Tetrahedral
Trigonal Pyramidal
5 electron pairs
Trigonal Bipyramidal
“look all 4 of them up”
6 electron pairs
Octahedral
14 possible combinations….learn them all!

51. Structure Determination by VSEPR
Ammonia, NH3
1. Draw electron dot structure H •• N
2. Count BP’s and LP’s around central N atom = 4
(Called the number of structural pairs.)

52. Structure Determination by VSEPRH •• N
Ammonia, NH3
1. Draw electron dot structure
2. Count BP’s and LP’s = 4
3. The 4 electron pairs are at the corners of a tetrahedron.

53. Structure Determination by VSEPR
Ammonia, NH3
There are 4 electron pairs at the corners of a tetrahedron.
The ELECTRON PAIR GEOMETRY is tetrahedral. H •• N

54. Structure Determination by VSEPR
Ammonia, NH3
The electron pair geometry is tetrahedral.
The MOLECULAR GEOMETRY or SHAPE — the positions of the atoms — is Trigonal PYRAMIDAL.

55. VESPR Step-by-step Determination1 2

56. Structure Determination by VSEPR
Water, H2O
1. Draw electron dot structure H O ••
2. Count BP’s and LP’s = 4

57. Structure Determination by VSEPR
Water, H2O
1. Draw electron dot structure
2. Count BP’s and LP’s = 4
3. The 4 electron pairs are at the corners of a tetrahedron. H O ••
The electron pair geometry is TETRAHEDRAL.

58. Structure Determination by VSEPR
Water, H2O
The electron pair geometry is TETRAHEDRAL. H O ••
The molecular geometry is bent 109.5

59. Structure Determination by VSEPR
Formaldehyde, CH2O
1. Draw electron dot structure • C H O

60. Structure Determination by VSEPR
Formaldehyde, CH2O
1. Draw electron dot structure
2. Count BP’s and LP’s = 3
(the double bond is treated as a “lump” of electrons or one pair) • C H O

61. Structure Determination by VSEPR• C H O
Formaldehyde, CH2O
1. Draw electron dot structure
2. Count BP’s and LP’s = 3
3. There are 3 electron pairs are at the corners of a planar triangle. C H • O
The electron pair geometry is TRIGONAL PLANAR with 120o bond angles.

62. Structure Determination by VSEPR
Formaldehyde, CH2O C H • O
The electron pair geometry is TRIGONAL PLANAR
The molecular geometry is also TRIGONAL PLANAR.

63. Structure Determination by VSEPR
Methanol, CH3OH
1. Draw electron dot structure H 1 2 ••
H—C—O—H
2. Define bond angles 1 and 2

64. Structure Determination by VSEPR
Methanol, CH3OH
Define bond angles 1 and 2
In both cases the atom
is surrounded by
4 electron pairs.

65. Structure Determination by VSEPR
Acetonitrile, CH3CN
Define bond angles 1 and 2
Angle 1 = 109o
Angle 2 = 180o
One C is surrounded by 4 electron “lumps” and the other by 2 “lumps”

66. Phenylalanine, an amino acid:
109o
Determine the bond angles.
120o
109o
109o
120o

67. Compounds with 5 or 6 Pairs Around the Central Atom

68. Compounds with 5 or 6 Pairs Around the Central AtomF
Trigonal bipyramid
120°
90° P
5 electron pairs F
Octahedral
90° S
6 electron pairs

69. STRUCTURES WITH CENTRAL ATOMS THAT DO NOT OBEY THE OCTET RULE

70. Deviations from the Octet Rule
Usually occurs with Group 3A elements and with those of 3rd period and higher.
Consider boron trifluoride, BF3 F •• • B

71. Deviations from the Octet Rule
Consider boron trifluoride, BF3
The B atom is surrounded by only 3 electron pairs.
Bond angles are 120o
Geometry described as planar trigonal

72. Sulfur Tetrafluoride, SF4
Number of valence electrons =
Central atom =
Dot structure

73. Sulfur Tetrafluoride, SF4
Number of valence electrons = 34
Central atom = S
Dot structure

74. Sulfur Tetrafluoride, SF4
Number of valence electrons = 34
Central atom = S
Dot structure
Electron pair geometry = ?

75. Sulfur Tetrafluoride, SF4
Number of valence electrons = 34
Central atom = S
Dot structure F • •• S
Electron pair geometry
= trigonal bipyramid
(because there are 5 pairs around the S)
What is the Molecular Geometry?

76. Sulfur Tetrafluoride, SF4
Not all repulsions are equivalent.
L.P. - L.P. > L.P. - B.P. > B.P. - B.P.
Minimize lone pair bond pair repulsions at the 90o angle.
90° F •
120° S

77. Structural Determination
Determine the geometries for PF3,
NO2-, NO2+, ICl2-, ClF3, ClF5, and IBr4-, giving VSEPR class, shape name, 3-D diagram, and bond angles.
Solutions

78. Bond Properties
What is the effect of bonding and structure on molecular properties?
Buckyball in HIV-protease

79. Bond Order The number of bonds between a pair of atoms. 1 s and 2 p
triple, BO = 3 H C N
single
BO = 1 1 s
double, BO = 2 1 s
and 1 p

80. Bond Order •• N • O Bond order = 3 e - pairs in N — O bonds 2 N
Fractional bond orders occur in molecules with resonance structures.
Consider NO2- O N •• •
The N—O bond order = 1.5
Bond
order =
Total #
of e -
pairs used for a type of bond
of bonds of that type
Bond
order =
3 e -
pairs in N —
O bonds
2 N

81. Bond Order (a) bond strength (b) bond length
Bond order is proportional to two important bond properties:
(a) bond strength
(b) bond length
745 kJ
414 kJ
123 pm
110 pm

82. Bond Length
Bond length is the distance between the nuclei of two bonded atoms.

83. Bond Length H—F H—Cl H—I Bond length depends on size of bonded atoms.
Bond distances measured in Angstroms or pm where 1 A = 100 pm.

84. Bond Length Bond length depends on bond order.
Different C-O Bond Lengths

85. C-C bond lengths
C-N bond lengths

86. Bond Strength
Bond strength is measured by the energy required to break a bond. See Table 9.9

87. C-C bond strengths

88. Bond Strength
BOND STRENGTH (kJ/mol)
H—H 436
C—C 346
C=C 602
CC
NN 945
The GREATER the number of bonds (bond order) the HIGHER the bond strength and the SHORTER the bond.

89. Bond Strengths for O-O Bond Order Length Strength
HO—OH pm kJ/mol
O=O
394 ?

90. DHorxn = S H (bonds broken) - S H (bonds formed) or
Using Bond Energies
DHorxn = S H (bonds broken) S H (bonds formed) or
DHorxn = energy input (+) energy output (-).
Net energy = DHrxn, bond = energy required to break bond energy evolved when bonds are formed.

91. Using Bond Energies
Estimate the energy of the reaction H—H + Cl—Cl ----> 2 H—Cl
H—H = 436 kJ/mol
Cl—Cl = 242 kJ/mol
H—Cl = 432 kJ/mol
Sum of H-H + Cl-Cl bond energies = 436 kJ kJ = +678 kJ
2 mol H-Cl bond energies = 864 kJ
Net = DH = kJ kJ = -186 kJ

92. Estimate the energy of the reaction
Using Bond Energies
Estimate the energy of the reaction
2 H—O—O—H > O=O H—O—H
Is the reaction exo- or endothermic?
Which is larger: energy req’d to break bonds or energy evolved on making bonds?

93. Using Bond Energies 2 H—O—O—H ----> O=O + 2 H—O—H
Energy required to break bonds:
break 4 mol of O—H bonds = 4 (463 kJ)
break 2 mol O—O bonds = 2 (146 kJ)
TOTAL ENERGY to break bonds = 2144 kJ

94. Using Bond Energies 2 H—O—O—H ----> O=O + 2 H—O—H
Energy evolved on making bonds:
make 1 mol of O=O bonds = 498 kJ
make 4 mol O—H bonds = 4 (463 kJ)
TOTAL ENERGY evolved on making bonds = kJ

95. Using Bond Energies 2 H—O—O—H ----> O=O + 2 H—O—H
Net energy = kJ kJ = kJ
The reaction is exothermic!
More energy is evolved on making bonds than is expended in breaking bonds.

96. Calculate DHorxn for the reaction below using bond energies.
CH4 (g) + 2 O2 (g) ______> 2 H2O (g) + CO2 (g)
DH = [4(C-H) + 2(O=O)] - [4(H-O) + 2(C=O)]
DH = [4(413kJ) + 2(498kJ)] - [4(463kJ) + 2(732kJ)]
DH = kJ
More Sample Problems!

97. Molecular Polarity
Why are water molecules attracted to a balloon that has a static electric charge?

98. Bond Polarity
HCl is POLAR because it has a positive end and a negative end.
Polarity arises because Cl has a greater share in bonding electrons than does H.

99. Bond Polarity
This model, calc’d using CAChe software, shows that H is + (red) and Cl is - (yellow). Calc’d charge is + or
This model shows that the electron density is greater around Cl than around H.

100. Bond Polarity
Due to the bond polarity, the H—Cl bond energy is GREATER than expected for a “pure” covalent bond.
BOND ENERGY
“pure” bond 339 kJ/mol calc’d
real bond 432 kJ/mol measured
Difference = 92 kJ. This difference is proportional to the difference in ELECTRONEGATIVITY, .

101. Bond Polarity
Partial charge on atom A = Group Number of A # lone pair electrons cA/(cA+ cB)( # bonding e- shared by A )

102. Electronegativity, 
 is a measure of the ability of an atom in a molecule to attract electrons to itself.
Concept proposed by Linus Pauling

103. Linus Pauling,
The only person to receive two unshared Nobel prizes (for Peace and Chemistry).
Chemistry areas: bonding, electronegativity, protein structure

104. Electronegativity,  See Figure 9.10 F has maximum .
Atom with lowest  is the center atom in most molecules.
Relative values of  determine BOND POLARITY (and point of attack on a molecule).

105. Electronegativity, 
Figure 9.9
Basic Trend =

106. Bond Polarity Which bond is more polar (or DIPOLAR)? O—H O—F 
OH is more polar than OF
and polarity is “reversed.”

107. Bond Polarity
Relative polarity in a bond is determined by D c, where the approximate % ionic character is give by:
Dc = 1.0, 20%; Dc = 1.5, 40%; Dc = 2.0, 60%; Dc = 2.5, 80%.

108. Molecular Polarity
Molecules—such as HCl and H2O— can be POLAR (or dipolar).
They have a DIPOLE MOMENT. The polar HCl molecule will turn to align with an electric field.

109. Molecular Polarity
The magnitude of the dipole is given in Debye units.
Named for Peter Debye ( ).
Received the 1936 Nobel prize for work on x-ray diffraction and dipole moments.

110. Molecular Polarity Molecules will be polar if 1. bonds are polar AND
2. the molecule is NOT “symmetric”
Symmetric molecules

111. Molecular Polarity
Unsymmetrical

112. Carbon Dioxide Positive C atom is reason CO2 + H2O gives H2CO3
CO2 is NOT polar even though the CO bonds are polar.
CO2 is symmetrical.
Positive C atom is reason CO2 + H2O gives H2CO3

113. Consequences of H2O Polarity
Microwave oven
Consequences of H2O Polarity

114. Molecular Polarity More Examples
B atom is positive and F atoms are negative.
B—F bonds in BF3 are polar. But molecule is symmetrical and NOT polar

115. Molecular Polarity B atom is positive but H & F atoms are negative.
B—F and B—H bonds in HBF2 are polar. But molecule is NOT symmetrical and is polar

116. Polarity of Methane, CH4 H C
Methane is symmetrical and is NOT polar.

117. Polarity of CH3F F H C
C—F bond is very polar. Molecule is not symmetrical and so is polar.

118. Substituted Ethylene C—F bonds are MUCH more polar than C—H bonds.
Because both C—F bonds are on same side of molecule, molecule is POLAR.

119. Substituted Ethylene C—F bonds are MUCH more polar than C—H bonds.
Because both C—F bonds are on opposing ends of molecule, molecule is NOT POLAR.

120. Practice Problems
1. Give the VSEPR class, bond angle, shape, and polarity of KrF2, CO, and NO2-.
2. Calculate the heat of reaction for
C2H4 (g) O2 (g) --> 2 CO2 (g) H2O (g)
3. Calculate the carbon-carbon triple bond energy in C2H2(g).
Heat of formation of C2H2 = 226 kJ/mole)
4. Determine the number of p and s bonds in acetic acid.
5. Arrange in order of increasing polarity:
Sr-Br, P-F, S-F, Al-O

121. Practice Problems Answers
1. AX2E3, 1800, linear, nonpolar
AXE, 1800, linear, polar
AX2E, 1200, bent, polar kJ kJ
4. 7 s and 1 p
5. S-F, P-F, Al-O, Sr-Br,

122. Ionic Bonding
Ionic bonds form by the transferring of VALENCE ELECTRONS, the electrons at the outer edge of the atom.

123. Ionic Bonding NaF Na and F Na 1s22s22p63s1 F 1s22s22p5
Na1+ 1s22s22p6 F1- 1s22s22p6
NaF

124. Ionic Bonding MgO Mg and O Mg 1s22s22p63s2 O 1s22s22p4
Mg2+ 1s22s22p6 O2- 1s22s22p6
MgO

125. Ionic Bonding MgF2 Mg and F Mg 1s22s22p63s2 F 1s22s22p5 F 1s22s22p5
Mg2+ 1s22s22p F1- 1s22s22p6
MgF2
F1- 1s22s22p6

126. Ionic Bonding Na2O Na and O Na 1s22s22p63s1 O 1s22s22p4

127. Covalent Bonding
Cl and F . F Cl . . F Cl .

128. Covalent Bonding
S and F . F S . . F . F S .

129. Covalent Bonding
O and O O O . O O .

130. - Sample Problems - O N O H O C Draw dot structures for the following:
H2O CO NO21- - O N -1 O - H O C

131. Draw Lewis Structures for:N2
CH4
I2CO
HCOOH
NO2-
XeF4 N N

132. Draw Lewis Structures for:H N2
CH4
I2CO
HCOOH
NO2-
XeF4 H C H H

133. Draw Lewis Structures for:N2
CH4
I2CO
HCOOH
NO2-
XeF4 O I C I

134. Draw Lewis Structures for:N2
CH4
I2CO
HCOOH
NO2-
XeF4 O H C O H

135. Draw Lewis Structures for:N2
CH4
I2CO
HCOOH
NO2-
XeF4 - O N O - O O N O

136. Draw Lewis Structures for:N2
CH4
I2CO
HCOOH
NO2-
XeF4 F .. F Xe .. F F

137. Sample Problems
Determine the oxidation number for each atom in the following:
CO SO32-
O = -2
C = 4
O = -2
S = 4

138. Sample Problems
Determine the formal charge for each atom in the following:
CO NO21-
O = 1
C = -1
single bond O = -1
double bond O = 0
N = 0

139. Structural Determination
PF3
PF AX3E trigonal pyramidal ..

140. Structural Determination
NO2+
NO AX2 linear

141. Structural Determination
ICl2-
ICl AX2E3 linear
180o .. .
120o

142. Structural Determination
ClF3
ClF AX3E2 T - shaped ..
120o

143. Structural Determination
ClF5
ClF AX5E square pyramidal ..

144. Structural Determination
IBr4-
IBr AX4E2 square planar ..

145. Calculate DHorxn for the reaction below using bond energies.
C (g) Cl2 (g) ______> CCl4 (g)
DH = [2(Cl-Cl)] - [4(C-Cl)]
DH = [2(242 kJ)] - [4(339 kJ)]
DH = kJ

146. Calculate DHorxn for the reaction below using bond energies.
C (s) Cl2 (g) ______> CCl4 (g)
DH = [ DHsubC + 2(Cl-Cl)] - [4(C-Cl)]
DH = [ 717 kJ + 2(242 kJ)] - [4(339 kJ)]
DH = kJ

147. Calculate DHof for CH4 using bond energies.
C (s) H2 (g) --> CH4 (g)
DHf = [DHsub C + 2(H-H)] - [4(C-H)]
DHf = [ 717 kJ + 2(436 kJ)] - [4(413kJ)]
DHf = kJ

148. Given DHof for C2H4 (g) is 52 kJ/mole
Given DHof for C2H4 (g) is 52 kJ/mole. Using bond energies, calculate the carbon carbon bond energy in C2H4 (g).
2 C (s) + 2 H2 (g) -- > C2H4 (g)
DHf = [ 2(DHsubC) + 2(H-H) ] - [ 4(C-H) + 1(C=C) ]
52kJ = [ 2(717kJ) + 2(436kJ)] - [4(413kJ) + 1(C=C)]
(C=C) = kJ

149. Bond Polarity Ionic Polar Nonpolar Covalent Covalent
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Li : F C : F F : F