1. Chapter 8
Chemical Bonding

2. Bonds
Forces that hold groups of atoms together and make them function as a unit.
We will consider three major categories of chemical bonds: Ionic Bonds, Covalent Bonds, and Metallic Bonds.

3. Examples of Bonding Types

4. Lewis Structures
Since bonds hold atoms close together, then the valence electrons are responsible for bonding since they are on the outside of an atom.
It has been recognized for a long time that the noble gases have great chemical stability. With few exceptions they are unreactive or inert.
The noble gases have 8 valence electrons with the exception of He which has 2.

5. Lewis Structures He 1s2 Ne 1s22s22p6 Ar 1s22s22p63s23p6
Kr 1s22s22p63s23p64s23d104p6
Xe 1s22s22p63s23p64s23d104p65s24d105p6

6. Lewis Structures
The electronic configuration of the noble gases is described as being energetically stable.
We can draw a Lewis diagram to illustrate the number of valence electrons an atom has.
In a Lewis diagram valence electrons are represented by dots placed above, below and to the left and right of the atoms symbol. E
e.g. element with 4 valence electrons

7. Lewis Structures E E    
There are two simple rules to keep in mind when drawing Lewis diagrams:
Place one dot in each of the four locations before doubling up.
There can be only a maximum of 2 dots in any one location. E   E  

8. Lewis Structures H What is the Lewis diagram for H?
First write the electron configuration:
1s1
Identify the number of valence electrons.
1 valence electron.
For a representative element it is easy to identify the number of valence electrons as this is equal to the group number.

9. Lewis Structures S What is the Lewis diagram for S?
First write the electron configuration:
[Ne]3s23p4
Identify the number of valence electrons.
6 valence electrons
Alternatively you can recognize that S is in group VIA so has six valence electrons S 9

10. Lewis Structures S What is the Lewis diagram for S?
First write the electron configuration:
[Ne]3s23p4
Identify the number of valence electrons.
6 valence electrons
Alternatively you can recognize that S is in group VIA so has six valence electrons S

11. Lewis Structures S What is the Lewis diagram for S?
First write the electron configuration:
[Ne]3s23p4
Identify the number of valence electrons.
6 valence electrons
Alternatively you can recognize that S is in group VIA so has six valence electrons S

12. Lewis Structures S What is the Lewis diagram for S?
First write the electron configuration:
[Ne]3s23p4
Identify the number of valence electrons.
6 valence electrons
Alternatively you can recognize that S is in group VIA so has six valence electrons S

13. Lewis Structures S What is the Lewis diagram for S?
First write the electron configuration:
[Ne]3s23p4
Identify the number of valence electrons.
6 valence electrons
Alternatively you can recognize that S is in group VIA so has six valence electrons S

14. Lewis Structures S What is the Lewis diagram for S?
First write the electron configuration:
[Ne]3s23p4
Identify the number of valence electrons.
6 valence electrons S
Alternatively you can recognize that S is in group VIA so has six valence electrons

15. LEWIS STRUCTURES OF THE ELEMENTSIA
IIA
SKIP
B’S
IIIA
IVA VA
VIA
VIIA H He Li B C N Be O F Ne Na Mg Al Si P S Cl Ar

16. LEWIS STRUCTURES OF IONS (AFTER REMOVAL OR ADDITION OF ELECTRONS)IA
IIA
SKIP
B’S
IIIA
IVA VA
VIA
VIIA 2+ 4- 3- 2- 1- 1+ H He Li B C N F Be O Ne 3+ Na Mg Si P S Cl Ar Al

17. Lewis Structures The octet rule states that:
“Atoms interact in order to obtain a stable octet of eight valence electrons”
The octet rule works extremely well at describing the interactions of the representative elements.

18. Lewis Structures
One way in which atoms can interact to satisfy the octet rule is by transferring electrons between each other.
Transferring of electrons results in the atoms acquiring net positive and negative charges.
When an atom loses or gains electrons a simple ion is formed.
Cations have more protons than electrons and are positive.
Anions have more electrons than protons and are negative.

19. Ionic Bonds
Formed from electrostatic attractions of closely packed, oppositely charged ions.
Formed when an atom that easily loses electrons reacts with one that has a high electron affinity.

20. Ionization A Review Na Na+ + 1e- Cl + 1e- Cl-
Consider a Na atom what happens if it loses one electron?
I.E. Na
Na e-
[Ne]3s1
[Ne]
11 P and 11 e-
11 P and 10 e-
Consider a Cl atom would you expect it to lose or gain electrons?
[Ne]3s23p5 Cl
E.A.
Cl-
+ 1e-
[Ne]3s23p6
17 P and 17 e-
17 P and 18 e-

21. +1 The ion formed would be Li+
Metals tend to lose electrons forming positively charged ions called cations.
A representative metal will lose its group number of electrons to obtain a stable octet.
Na → Na+ + 1e- ( Isoelectronic with Ne)
Mg → Mg2+ + 2e- (isoelectronic with Ne)
What would the charge be of the ion formed by a Li atom?
And which Noble gas is it isoelectronic with?
+1 The ion formed would be Li+
Isoelectronic with He

22. -1 The ion formed would be I-
Noble Stability
Non-metals tend to gain electrons forming negatively charged ions called anions.
A representative non-metal will gain (8 - group number) electrons to obtain a stable octet.
O + 2e- → O2- (isoelectronic with Ne)
S + 2e- → S2- (isoelectronic with Ar)
What would the charge be of the ion formed by a I atom?
Which Noble gas is it isoelectronic with?
-1 The ion formed would be I-
Isoelectronic with Xe

23. Lewis Structure of NaCl
Na+Cl-Na+Cl-Na+Cl-
Cl-Na+Cl-Na+Cl-Na+
Forces between oppositely charged ions are called
Ionic bonds. Each ion is surrounded by an octet of
Electrons, thus making the ions stable.

24. Crystal Lattice of NaCl
Ionic compounds do not exist as discrete molecules. Instead they exist as crystals where ions of opposite charges occupy positions known as lattice sites.
Ions combine in the ratio that results in zero charge to form ionic compounds.
Which ions are the smaller ones?
Crystal Lattice of NaCl

25. Crystal Lattice of NaCl
Ionic compounds do not exist as discrete molecules. Instead they exist as crystals where ions of opposite charges occupy positions known as lattice sites.
Ions combine in the ratio that results in zero charge to form ionic compounds.
Which ions are the smaller ones? Sodium
Crystal Lattice of NaCl

26. Sodium Chloride Lattice

27. Molecular Compounds
In our early lectures we defined a molecule as “as a compound
Made of nonmetals.”
Molecules exist as particles containing the number of atoms specified by their formula.
e.g. a water molecule is a particle containing 2 hydrogen atoms and one oxygen atom and has the formula H2O.

28. Molecular Compounds
Non-metals may also complete their octets by sharing electrons.
This may occur between non-metal atoms of the same type:
e.g. H2, O2, N2, Cl2, F2, I2, etc
Or between different types of non-metal atoms:
e.g. CO2, H2O, CH4, etc

29. Covalent Bond Formation+ - + -
Consider two hydrogen atoms separated by a large distance. Each has 1 electron in a 1s atomic orbital.
Why does the electron stay around the nucleus?
Now lets bring the two atoms together so there orbitals overlap.

30. + -
The atomic orbitals overlap to form a new molecular orbital. This is a stable configuration as each H atom can have a full 1s susbshell (like He) where the electrons spend most of their time shared between the atoms. In this arrangement each nucleus feels an inwards attraction to the two electrons. This is called covalent bonding.

31. + -
This new arrangement of protons and electrons is more stable than separate hydrogen atoms since the attraction of a proton to two electrons is a stronger attraction compared to one proton to one electron of a hydrogen atom.

32. Lewis Structures
A single bond results when two atoms share one pair of electrons.
A lone pair, or unshared pair, of electrons is a pair of electrons that is not shared.
A bonding pair of electrons is a pair of electrons shared between two atoms.

33. Multiple Bonds
A double bond results when two atoms share two pairs of electrons.
A triple bond results when two atoms share three pairs of electrons
Bond length is the distance between the nuclear centers of the two atoms jointed together in a bond.

34. Covalent Bond Formation
We can draw Lewis diagrams showing the arrangement of valence electrons in covalent compounds. In these diagrams we represent each pair of electrons between atoms as a line.
So for the H2 molecule discussed previously the Lewis diagram would be:
H – H
All other electrons are represented by dots as described previously.

35. Molecular Compounds H H H O H O H N H H N H H H
Draw Lewis Structures of the following molecular compounds H H
Nonbonding
electons
a. H2O H O H O
Note each element has a Noble gas structure by electron sharing
b. NH3 H N H H N H H H
Covalent bonding e’s

36. Simplified Lewis Structures
Straight lines are used to indicate a shared pair, or a covalent bond. H O H
Nonbonding electrons

37. Lewis Structure Construction
Step 1 Connect each element with a single line
Step 2 Use the “P” formula to determine the number of extra bonds
Step 3 Insert the extra bonds, to make double or triple bonds.
Step 4 Give each atom an octet of electrons, except hydrogen
Step 5 Determine the formal charge of each element
N = number of atoms in molecule
Q = number of hydrogen atoms
V = total number of valence electrons
P = 8(n-q) +2q - 2(n-1) - v
Examples: Give Lewis Structures for the following
CO2
H2CO3
SO3
NO2+

38. Lewis Structure of Carbon Dioxide
First, connect atoms with lines O C O
Second, use “p” formula to determine the number of extra bonds.
P = 8(n-q) + 2q – 2(n-1) - v
P = 8(3-0) + 2(0) – 2(3-1) - 16
P = –
4 extra bonding electrons
2 extra bonds
2 extra lines
P = 4

39. Lewis Structure of Carbon Dioxide
Third, add extra lines O C O O C O O C O
Fourth, give each atom an octet of electrons

40. Lewis Structure of Carbon Dioxide
Third, add extra lines O C O O C O O C O
Fourth, give each atom an octet of electrons O C O O C O O C O

41. Lewis Structure of Carbon Dioxide
Third, add extra lines O C O O C O O C O
Fourth, give each atom an octet of electrons O C O O C O O C O
Fifth, give each an atom a formal charge
If the element owns less than its valence, then it is positive
If the element has more than its valence, then it is negative O C O O C O O C O

42. Lewis Structure of Carbon Dioxide
Third, add extra lines O C O O C O O C O
Fourth, give each atom an octet of electrons O C O O C O O C O
Fifth, give each an atom a formal charge
If the element owns less than its valence, then it is positive
If the element has more than its valence, then it is negative - O C O O C O O C O

43. Lewis Structure of Carbon Dioxide
Third, add extra lines O C O O C O O C O
Fourth, give each atom an octet of electrons O C O O C O O C O
Fifth, give each an atom a formal charge
If the element owns less than its valence, then it is positive
If the element has more than its valence, then it is negative - + O C O O C O O C O

44. Lewis Structure of Carbon Dioxide
Third, add extra lines O C O O C O O C O
Fourth, give each atom an octet of electrons O C O O C O O C O
Fifth, give each an atom a formal charge
If the element owns less than its valence, then it is positive
If the element has more than its valence, then it is negative - + - O C O O C O O C O

45. Lewis Structure of Carbon Dioxide
Third, add extra lines O C O O C O O C O
Fourth, give each atom an octet of electrons O C O O C O O C O
Fifth, give each an atom a formal charge
If the element owns less than its valence, then it is positive
If the element has more than its valence, then it is negative - + - + O C O O C O O C O

46. Practice
Draw the most stable Lewis structure for CO2.

47. Practice
Draw the most stable Lewis structure for CO2. O C O

48. Sulfur Trioxide Lewis Structure
There are actually three possible Lewis structures for SO3. - 2+ - 2+ - - 2+ S O S O S O - -
Each of these three structures is equivalent. We say they are in “resonance” or that they are “resonance structures”.

49. Resonance Form Rules

50. Resonance Form Rules

51. Resonance Form Rules

52. Resonance Structures

53. Practice Determine the most stable structure for the
phosphite ion by calculating formal charge for each ion. - O
O P O 3- O
O P O 3- - - - - -

54. Practice Problem
Give the most stable resonance form of H2SO4

55. Practice Problem O H O S O H O
Give the most stable resonance form of H2SO4
Step #1 Connect atoms with single bonds in the most symmetrical arrangement possible O H O S O H O

56. Practice Problem O H O S O H O
Give the most stable resonance form of H2SO4
Step #2 Find the number of extra bonding electrons with the “P” formula
P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines O H O S O H O

57. Practice Problem O H O S O H O
Give the most stable resonance form of H2SO4
Step #3 Give each atom an octet of electrons
P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines O H O S O H O

58. Practice Problem O H O S O H O
Give the most stable resonance form of H2SO4
Step #4 Give each atom a formal charge
P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines O - H O + + S O H O -

59. Practice Problem O H O S O H O
Give the most stable resonance form of H2SO4
Step #4 Give each atom a formal charge
P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines O - H O + + S O H O -
According to resonance rules this should not be very stable

60. Practice Problem O H O S O H O
Give the most stable resonance form of H2SO4
Step #4 Give each atom a formal charge
P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines O - H O + + S O H O -
Moving electrons around might improve stability

61. Practice Problem O H O S O H O
Give the most stable resonance form of H2SO4
Step #4 Give each atom a formal charge
P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines O - H O + + S O H O -
Moving electrons around might improve stability

62. Practice Problem O H O S O H O
Give the most stable resonance form of H2SO4
Step #4 Give each atom a formal charge
P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines O - H O + S O H O
This arrangement is better, but still not the best.

63. Practice Problem O H O S O H O
Give the most stable resonance form of H2SO4
Step #4 Give each atom a formal charge
P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines O - H O + S O H O
This arrangement is better, but still not the best.

64. Practice Problem O H O S O H O
Give the most stable resonance form of H2SO4
Step #4 Give each atom a formal charge
P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines O H O S O H O
Since sulfur is a period three element, then it can accommodate more than eight electrons. This is the most stable arrangement.

65. Exceptions to the Octet Rule
Some molecules have less than eight electrons in a Lewis structure.
BF3 is an example
These are known as electron-deficient compounds
Other molecules have more than an octet.
SF6 is an example
Some molecules have an odd number of electrons when summing up the total number of electrons. These don’t obey the octet rule.
NO is an example
Free radicals are molecules having an odd number of valence electrons.

66. Example
What is the Lewis dot structure of SO32- ion? Does this ion have resonance structures?

67. Metallic Bonds
A metallic bond consists of the nuclei of metal atoms surrounded by a “sea” of evenly spaced shared electrons. The nuclei is then attracted to each electron by the same amount, but in different directions, thus making the nuclei stay in a fixed position.

68. Electron Sea Model

69. Electronegativity
Electronegativity is a measure of an element’s ability to attract bonding electrons. It is used for predicting the degree to which bonding pairs of electrons are shared unequally.

70. Electronegativity Trends

71. Atomic Size Relationship
Changes in electronegativity are related to increasing atomic size.
The size of the valence orbitals increases as the value of the principle quantum number (n) increases.
Therefore, the atomic size increases and electronegativity decreases as you go down a group.

72. Bond Types

73. Bond Polarity
Bond polarity is a measure of the extent to which bonding electrons are shared between two atoms in a covalent bond.

74. Polar Covalent Bonds
Compounds that contain two or more different elements may contain polar covalent bonds.
:C O:
A polar molecule contains bonds that have an uneven distribution of charge because electrons in the bonds are not shared equally by the two atoms.

75. Ionization vs. Electronegativities

76. Practice Which of the following bonds in each pair are more polar?
C-S or C-O
Cl-Cl or O=O
N-H or C-H

77. Different Elemental Forms
Oxygen is found in two forms: oxygen, O2, and ozone, O3.
Carbon is found in many different forms: soot, diamond, graphite, nanotubes, etc.
These different forms of an element are known as allotropes.

78. Evidence for Resonance Structures
Ozone bond lengths were both found to be 128pm
A single bond would have a longer bond length than a double bond, but in ozone we find the bond lengths are equal meaning they are a mixture of single and double bonds.

79. Bond Lengths
Bond length depends on the identity of the atoms as well as on the number of bonds between them.
The bond order is the number of bonds between two atoms.
1 for a single bond
2 for a double bond
3 for a triple bond

80. Covalent Bond Lengths

81. Examples of Bond Length

82. Bond Strengths

83. Bond Energies
The energy needed to break 1 mole of covalent bonds in the gas phase is the bond energy of that bond.
Breaking bonds consumes energy whereas forming bonds releases energy.
Bond energies can be used to estimate Hrxn.

84. Energy of Reaction Calculation
Calculation ΔH for the following reaction.
CH4 + O CO2 + HOH

85. Energy of Reaction Calculation
Calculation ΔH for the following reaction.
CH O CO HOH
First energy must be added to break the reactant bonds. Bonds to break are 4 C-H bonds and 2 O=O double bond. Next energy is released when 2 C=O bonds are formed and 4 H-O bonds are formed.

86. Energy of Reaction Calculation
Calculation ΔH for the following reaction.
CH O CO HOH
First energy must be added to break the reactant bonds. Bonds to break are 4 C-H bonds and 2 O=O double bond. Next energy is released when 2 C=O bonds are formed and 4 H-O bonds are formed.
Bonds broken (endothermic)
Bonds formed (exothermic)
2 C=O 2(799) = 1598 kj
4 H-O 4(464) = 1856 kj
4 C-H bonds 4(414) = kj
2 O=O bonds 2(498) = 996 kj
3452 kj
2652 kj
ΔH = = 800 kj

87. Example

88. The End

89. Review Questions

90. Click to launch animation
ChemTour: Bonding
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This ChemTour shows how ionic and covalent bonds form.

91. ChemTour: Lewis Dot Structures
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Students learn to draw and use Lewis dot structures to visualize and represent molecular structures and the locations of valence electrons. Includes Practice Exercises.

92. ChemTour: The Periodic Table
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This ChemTour offers a guided tour of the trends summarized by the periodic table (metallic properties, subshells, electronegativity, and atomic radius), and explains how to use this tool to predict an element’s characteristics, including its bonding capacity.

93. Click to launch animation
ChemTour: Resonance
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Molecules that cannot be represented by a single Lewis dot structure are said to exhibit resonance. This ChemTour describes resonance and explains how to represent it visually.

94. ChemTour: Expanded Valence Shells
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In this ChemTour, students explore the exceptions to the octet rule, and learn to identify the conditions under which an element will expand its outer electron shell to hold more than 8 electrons. Includes Practice Exercises.

95. ChemTour: Estimating Enthalpy Changes
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In this ChemTour, students learn how to use average bond energies to estimate the energy released during a combustion reaction. Includes Practice Exercises.

96. Molecular Potential Energy of N2 and O2
The plot to the left shows the molecular potential energy curve for N2. Which of the plots below shows the correct molecular potential energy curve for O2 (blue solid line) compared to N2 (red dashed line)?
© 2008 W. W. Norton & Company Inc. All rights reserved. A) B) C)
Molecular Potential Energy of N2 and O2

97. Molecular Potential Energy of N2 and O2
Consider the following arguments for each answer and vote again:
The O2 double bond is less stable than the N2 triple bond, so the dissociation energy of O2 should be less than that of N2.
An oxygen atom is smaller than a nitrogen atom, so O2 should have a smaller bond length than N2.
The fact that O2 has two electrons in antibonding Π* orbitals is reflected in the potential energy curve always being above zero.
Answer: A
Molecular Potential Energy of N2 and O2

98. The oxygen molecule, O2, has a bond order of 2 in its ground state.
How many electrons can be removed from O2 without altering the bond order? Assume the electrons are always removed from the highest-occupied molecular orbital.
A) B) C) 6
© 2008 W. W. Norton & Company Inc. All rights reserved.
Bond Order of Oxygen

99. Consider the following arguments for each answer and vote again:
Removing 2 electrons from antibonding orbitals will not affect the bond order.
To retain a bond order of 2, a total of 2 electrons must be removed from a bonding orbital and 2 must be removed from an antibonding orbital.
Removing 6 electrons will leave 4 electrons in 2 bonding orbitals.
Answer: A
Bond Order of Oxygen

100. Stability of He2, CN, and NO
Which of the following three molecules would be made less stable by removing a single electron from the highest-occupied molecular orbital?
A) He B) NO C) CN
© 2008 W. W. Norton & Company Inc. All rights reserved.
Stability of He2, CN, and NO

101. Stability of He2, CN, and NO
Consider the following arguments for each answer and vote again:
The 4 electrons in He2 are incapable of holding 2 helium atoms together, so removing 1 electron will only make matters worse.
Removing an electron from NO would decrease its bond order from 2.5 to 2, thus decreasing its stability.
The highest-occupied molecular orbital for CN is a bonding orbital, so removing an electron will decrease CN's stability.
Answer: C
Stability of He2, CN, and NO

102. Lewis Electron Dot Structure of Formaldehyde
Lewis electron dot structures depict the arrangement of valence electrons around the atoms of a molecule, as shown below for water, H2O.
Which of the following is the correct Lewis electron dot structure for formaldehyde, CH2O?
© 2008 W. W. Norton & Company Inc. All rights reserved. A) B) C)
Lewis Electron Dot Structure of Formaldehyde

103. Lewis Electron Dot Structure of Formaldehyde
Please consider the following arguments for each answer and vote again:
The octet rule is satisfied, and the formal charges of all of the atoms are zero.
The carbon-oxygen triple bond lends extra stability to this structure.
The octet rule is satisfied without the need to form double bonds.
Answer: A
Lewis Electron Dot Structure of Formaldehyde

104. Bonding Structure of NO2-
How many bonds would it take to complete the structure of the nitrite ion, pictured to the left?
A) B) C) 2
© 2008 W. W. Norton & Company Inc. All rights reserved.
Bonding Structure of NO2-

105. Bonding Structure of NO2-
Please consider the following arguments for each answer and vote again:
All of the atoms are attached, so more bonds are not needed.
The correct overall charge and the octet rule are satisfied by 1 single bond and 1 double bond.
Both oxygens are double bonded to the nitrogen in order for their formal charges to be zero.
Answer: B
Bonding Structure of NO2-

106. Molecular Structures of Pentane
Which of the following depicts a molecule that is different from the one shown to the left? A) B) C)
© 2008 W. W. Norton & Company Inc. All rights reserved.
Molecular Structures of Pentane

107. Molecular Structures of Pentane
Please consider the following arguments for each answer and vote again:
This molecule, known as isopentane, has a unique T-shaped arrangement.
In this molecule, only two carbon atoms are bonded to the second carbon.
This molecule is unbranched, whereas the one pictured in the question is not.
Answer: C
Molecular Structures of Pentane