1. Chapter 8 Concepts of Chemical Bonding
Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten
Chapter 8 Concepts of Chemical Bonding
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2. 8.1 Chemical Bonds Three basic types of bonds Ionic Covalent Metallic
Electrostatic attraction between ions
Covalent
Sharing of electrons
Metallic
Metal atoms bonded to several other atoms
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3. Lewis Symbols
Electrons involved in chemical bonding are the valence electrons.
•G.N. Lewis( ) suggested a simple way of showing the valence electrons in an atom.
•Lewis electron-dot structures: consist of the chemical symbol for the element plus a dot for each valence electron.
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4. Octet Rule
Atoms tend to gain, lose, or share electrons until they are surrounded by eight valence electrons.
•An octet of electrons consists of full s and p-subshells in an atom.
•There are many exceptions to the octet rule, but it provides a useful framework for many important concepts of bonding.
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5. Write the Lewis symbol for atoms of each of the following elements:Al Br Ar Sr
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6. 8.2 Ionic Bonding
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7. Energetics of Ionic Bonding
As we saw in the last chapter, it takes 495 kJ/mol to remove electrons from sodium.
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8. Energetics of Ionic Bonding
We get 349 kJ/mol back by giving electrons to chlorine.
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9. Energetics of Ionic Bonding
But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic!
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10. Energetics of Ionic Bonding
There must be a third piece to the puzzle.
What is as yet unaccounted for is the electrostatic attraction between the newly-formed sodium cation and chloride anion.
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11. Lattice Energy This third piece of the puzzle is the lattice energy:
The energy required to completely separate a mole of a solid ionic compound into its gaseous ions.
NaCl(s)  Na+ (g) + Cl-(g) ∆Hlattice= kJ/mol
The energy associated with electrostatic interactions is governed by Coulomb’s law:
Eel = 
Q1Q2 d
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12. Lattice Energy
Lattice energy, then, increases with the charge on the ions.
It also increases with decreasing size of ions.
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13. Energetics of Ionic Bonding
These phenomena also helps explain the “octet rule.”
Metals, for instance, tend to stop losing electrons once they attain a noble gas configuration because energy would be expended that cannot be overcome by lattice energies.
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14. Example
Arrange the following ionic compounds in order of increasing lattice energy.
NaF, CsI, and CaO
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15. Solution CsI < NaF < CaO
CaO has greatest ionic charge at +2 and -2
Cs+ is larger than Na+
I-is larger than F-
The distance between the Na+ and F- ions in NaF will be less than the distance between Cs+ and I- ions in CsI. As a result, the lattice energy of NaF should be greater than that of CsI.
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16. Example
Which substance would you expect to have the greatest lattice energy?
AgCl
CuO
CrN
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17. Solution
CrN
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18. Calculation of Lattice Energies
The Born-Haber Cycle:The formation of an ionic compound as occurring in a series of well-defined steps. We use Hess’s Law to put these steps together in a way that gives us the lattice energy for the compound
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19. Energetics of Ionic Bonding
By accounting for all three energies (ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.
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20. Born-Haber Cycle
The enthalpy change for the direct route (red arrow) is the heat of formation of NaCl(s)
Na(s) + 1/2 Cl2(g)  NaCl(s)
∆Hf [NaCl(s)]= -411 kJ
Values are found using Appendix C!
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21. Born-Haber Cycle
The indirect route consists of five steps, shown by the green arrows.
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22. Step 1 We generate gaseous atoms of sodium by vaporizing sodium metal:
Na(s) Na(g) ∆Hf [Na(g)] = 108 kJ
Endothermic required energy
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23. Step 2
Then we form gaseous atoms of chlorine by breaking the bonds in the Cl2 molecules.
1/2 Cl2(g) Cl(g) ∆Hf [Cl(g)] = 122 kJ
Endothermic  requires energy
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24. Steps 3 and 4 Removing an electron from Na (g) to form Na+ (g)
and then adding the electron to Cl (g) to form Cl-(g).
Na(g)  Na+ + e- ∆Hf = I1(Na) = 496 kJ
Cl(g) + e- Cl-(g) ∆Hf = E(Cl) = -349 kJ
Ionization energy found on pg 268
Electron affinity found on pg 271
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25. Step 5
Finally, we combine the gaseous sodium and chloride ions to form solid sodium chloride.
Because this process is just the reverse of the lattice energy (breaking a solid into gaseous ions), the enthalpy change is the negative of the lattice energy, the quantity we want to determine:
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26. Step 5 con’t Na+ (g) + Cl- (g) NaCl (s) ∆H = -∆Hlattice= ?
The sum of the five steps in the indirect path gives us the NaCl(s) from Na(s) and 1/2 Cl2(g).
Thus from Hess’s law we know the sum of the enthalpy changes for these five steps equals that for the direct path.
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27. Solution -411 kJ = 108 kJ + 122 kJ + 496 kJ –349 kJ – ∆Hlattice
∆H°f [NaCl(s)] = ∆H°f [Na(g)] + ∆H°f [Cl(g)] + I1(Na) + E(Cl) –∆Hlattice
-411 kJ = 108 kJ kJ kJ –349 kJ – ∆Hlattice
Solving for ∆Hlattice
∆Hlattice= 108 kJ kJ kJ –349 kJ kJ
∆Hlattice= 788 kJ
Thus the lattice energy of NaCl is 788 kJ/mol
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28. Explain the following trends in lattice energy.
CaF2 > BaF2
NaCl > RbBr
BaO > KF
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29. 8.3 Covalent Bonding
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30. 8.3 Covalent Bonding In covalent bonds atoms share electrons.
There are several electrostatic interactions in these bonds:
Attractions between electrons and nuclei
Repulsions between electrons
Repulsions between nuclei
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31. Covalent Bonds
The attractions between nuclei and the electrons cause electron density to concentrate between the nuclei. As a result, the overall electrostatic interactions are attractive.
•A shared pair of electrons in any covalent bond acts as a kind of “glue” to bind atoms together.
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32. Why should two atoms share electrons?
A covalent bond is a chemical bond in which two or more electrons are shared by two atoms.
Why should two atoms share electrons?
7e-
7e-
8e-
8e- F F + F
Lewis structure of F2
lone pairs F
single covalent bond
single covalent bond F

33. Lewis structure of water
single covalent bonds
2e-
8e-
2e- H + O + H O H or
Double bond – two atoms share two pairs of electrons
8e-
8e-
8e-
double bonds O C or O C
double bonds
Triple bond – two atoms share three pairs of electrons
8e-
triple bond N
8e- or N
triple bond
9.4

34. Lewis Structures
The formation of covalent bonds can be represented using Lewis symbols
Examples: Draw:
H2,
Cl2,
NH3,
CH4
H H  H:H
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35. Lewis Structures Multiple Bonds: Doubles and Triples
Examples: Draw CO2 and N2
As a rule, the distance between bonded atoms decreases as the number of shared electron pairs increases.
Multiple bonds are always shorter and stronger than single bonds between the same two elements.
The added strength makes it more difficult to break these molecules apart than those with only single bonds between the same two elements
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36. 8.4 Bond Polarity & Electronegativity
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37. 8.4 Polar Covalent Bonds H F F H d+ d-
electron rich
region
electron poor
region
e- poor
e- rich F H d+ d-
Although atoms often form compounds by sharing electrons, the electrons are not always shared equally.
Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does in the molecule HF.
Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.
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38. Polarity
Nonpolar covalent bond is one in which the electrons are shared equally between two atoms, as in the Cl2 and N2 examples we just drew.
Polar covalent bond, one of the atoms exerts a greater attraction for the bonding electrons than the other. If the difference in relative ability to attract electrons is large enough, an ionic bond is formed.
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39. Electronegativity
The ability of atoms in a molecule to attract electrons to themselves.
On the periodic chart, electronegativity increases as you go…
…from left to right across a row.
…from the bottom to the top of a column.
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40. Vocabulary
Electronegativity – the ability of an atom IN A MOLECULE to attract electrons to itself.
Ionization energy – how strongly an atom holds on to its electrons.
Electron affinity – the measure of how strongly an atom attracts additional electrons.
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41. Electron Affinity - measurable, Cl is highest
Another way to define electronegativity -Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond.
Electron Affinity - measurable, Cl is highest
X (g) + e X-(g)
Electronegativity - relative, F is highest

42. The Electronegativities of Common Elements
9.5

43. Variation of Electronegativity with Atomic Number –
within a period there is a steady increase in electronegativity from left to right.
9.5

44. Question
What is the trend in electronegativity going from left to right in a row on the periodic table?
It increases
How do electronegativity values change as you move down a group of elements on the periodic table?
It decreases
How do periodic trends in electronegativity relate to those for ionization energy & electron affinity?
Generally, electronegativity increases as ionization energy increases and electron affinity becomes more negative
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45. Polar Covalent Bonds
The greater the difference in electronegativity, the more polar is the bond.
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46. Electronegativity & Bond Polarity
F2 4.0 –4.0 =0 Nonpolar
HF 4.0 –2.1 = 1.9 Polar
Covalent
LiF 4.0 –1.0 = 3.0 Ionic
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47. Classification of bonds by difference in electronegativity
Bond Type
Nonpolar Covalent
 2
Ionic
0 < and <2
Polar Covalent
Increasing difference in electronegativity
Covalent
share e-
Polar Covalent
partial transfer of e-
Ionic
transfer e-
9.5

48. Classify the following bonds as ionic, polar covalent,
or covalent: The bond in CsCl; the bond in H2S; and
the NN bond in H2NNH2.
Cs – 0.7
Cl – 3.0
3.0 – 0.7 = 2.3
Ionic
H – 2.1
S – 2.5
2.5 – 2.1 = 0.4
Polar Covalent
N – 3.0
N – 3.0
3.0 – 3.0 = 0
Covalent
9.5

49. Example B-Cl or C-Cl Which bond is more polar?
Indicate in each case which atom has the partial negative charge.
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50. Solution
Use Figure 8.6
The difference in the electronegativities of chlorine and boron is 3.0 –2.0 = 1.0
The difference between chlorine and carbon is 3.0 –2.5 = 0.5
Therefore B-Cl is more polar. The chlorine atom carries the partial negative charge because it has a higher electronegativity.
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51. Polar Covalent Bonds
When two atoms share electrons unequally, a bond dipole results.
The dipole moment, , produced by two equal but opposite charges separated by a distance, r, is calculated:
 = Qr
It is measured in debyes (D).
A unit that equals 3.34 x Coulomb-meters
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52. What this means
= Qr
For molecules, we usually measure charge in units of electronic charge, e,1.6 x C, and distance in units of angstroms.
Suppose we have a charge of 1+ and 1- and are separated by a distance of 1 Angstrom. A dipole moment produced is
= Q•r
μ=(1.60 x C)•[(1A)(10-10 m/1Å)( 1D/3.34x10-30 C-m)]
μ= 4.79 D
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53. 8.5 Drawing Lewis Structures
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54. 8.5 Lewis Structures
Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.
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55. Writing Lewis Structures
Find the sum of valence electrons of all atoms in the polyatomic ion or molecule.
If it is an anion, add one electron for each negative charge.
If it is a cation, subtract one electron for each positive charge.
PCl3
(7) = 26
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56. Writing Lewis Structures
The central atom is the least electronegative element that isn’t hydrogen. Connect the outer atoms to it by single bonds.
Keep track of the electrons:
= 20
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57. Writing Lewis Structures
Fill the octets of the outer atoms.
Keep track of the electrons:
= 20; = 2
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58. Writing Lewis Structures
Fill the octet of the central atom.
Keep track of the electrons:
= 20; = 2; = 0
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59. Writing Lewis Structures
If you run out of electrons before the central atom has an octet…
…form multiple bonds until it does.
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60. Writing Lewis Structures
Then assign formal charges.
For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms.
Subtract that from the number of valence electrons for that atom: the difference is its formal charge.
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61. Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4)
-2 charge – 2e-
4 + (3 x 6) + 2 = 24 valence electrons
Step 3 – Draw single bonds between C and O atoms and complete
octet on C and O atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
Step 5 - Too many electrons, form double bond and re-check # of e-
2 single bonds (2x2) = 4
1 double bond = 4
8 lone pairs (8x2) = 16
Total = 24 O C
9.6

62. Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F, put N in center
Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)
5 + (3 x 7) = 26 valence electrons
Step 3 – Draw single bonds between N and F atoms and complete
octets on N and F atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons F N

63. FOR EACH OF THE FOLLOWING MOLECULES OR IONS OF SULFER AND OXYGEN, WRITE A SINGLE LEWIS STRUCTURE THAT OBEYS THE OCTET RULE, AND CALCULATE THE OXIDATION NUMBERS AND FORMAL CHARGE FOR ALL OF THE ATOMS:
SO2
SO3

64. Writing Lewis Structures
The best Lewis structure…
…is the one with the fewest charges.
…puts a negative charge on the most electronegative atom.
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65. 8.6 Resonance
We sometimes encounter molecules and ions in which the experimentally determined arrangement of atoms in not adequately described by a single Lewis structure.
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66. 8.6 Resonance This is the Lewis structure we would draw for ozone, O3.+ -
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67. Resonance
But this is at odds with the true, observed structure of ozone, in which…
…both O-O bonds are the same length.
…both outer oxygens have a charge of -1/2.
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68. Resonance
One Lewis structure cannot accurately depict a molecule like ozone.
We use multiple structures, resonance structures, to describe the molecule.
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69. Resonance Just as green is a synthesis of blue and yellow…
…ozone is a synthesis of these two resonance structures.
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70. Resonance
In truth, the electrons that form the second C-O bond in the double bonds below do not always sit between that C and that O, but rather can move among the two oxygens and the carbon.
They are not localized; they are delocalized.
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71. Which is predicted to have the shorter sulfur–oxygen bonds, SO3or SO32–?
The sulfur atom has six valence electrons, as does oxygen. Thus, SO3 contains 24 valence electrons. In writing the Lewis structure, we see that three equivalent resonance structures can be drawn:
The actual structure of SO3 is an equal blend of all three. Thus, each S—O bond distance should be about one-third of the way between that of a single and that of a double bond. That is, they should be shorter than single bonds but not as short as double bonds.
The SO32– ion has 26 electrons, which leads to a Lewis structure in which all the S—O bonds are single bonds:
There are no other reasonable Lewis structures for this ion. It can be described quite well by a single Lewis structure rather than by multiple resonance structures. Our analysis of the Lewis structures leads us to conclude that SO3 should have the shorter S—O bonds and SO32– the longer ones. This conclusion is correct: The experimentally measured S—O bond lengths are 1.42 Å in SO3 and 1.51 Å in SO32–.
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72. Resonance in Benzene
The organic compound benzene, C6H6, has two resonance structures.
It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring.
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73. 8.7 Exceptions to the Octet Rule
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74. 8.7 Exceptions to the Octet Rule
The Octet rule is simple & useful for most basic bonding concepts. However, in some situations, it does have some limitations:
In the bonding of ionic compounds of involving some transition metals.
EX: Fe 2+  [Ar] 3d Fe 3+  [Ar] 3d5.
Some covalent bonding doesn’t follow the octet rule either. These exceptions to the octet rule are of three main types.
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75. 8.7 Exceptions to the Octet Rule
Ions or molecules with an odd number of electrons
Ions or molecules with less than an octet
Ions or molecules with more than eight valence electrons (an expanded octet)
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76. Odd Number of Electrons
In the vast majority of molecules and polyatomic ions, the total number of valence electrons is even.
Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons such as ClO2, NO, NO2, O2- .
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77. Odd Number of Electrons
For example, NO contains =11 Electrons. The two most important Lewis structures for this molecule are:
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78. Fewer Than Eight Electrons
Consider BF3:
Giving boron a filled octet places a negative charge on the boron and a positive charge on fluorine.
This would not be an accurate picture of the distribution of electrons in BF3.
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79. Fewer Than Eight Electrons
Therefore, structures that put a double bond between boron and fluorine are much less important than the one that leaves boron with only 6 valence electrons.
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80. Fewer Than Eight Electrons
The lesson is: if filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative outer atom, don’t fill the octet of the central atom.
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81. More Than Eight Electrons
The 3rd & largest class of exceptions consists of molecules or polyatomic ions in which there are more that 8 electrons in the valence shell for the atom.
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82. More Than Eight Electrons
The only way PCl5 can exist is if phosphorus has 10 electrons around it.
It is allowed to expand the octet of atoms on the 3rd row or below.
Presumably d orbitals in these atoms participate in bonding.
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83. More Than Eight Electrons
Other examples of molecules & ions with “expanded” valence shells include:
SF4, AsF6-, & ICl4-
Element from the 3rd period and beyond have ns, np, and unfilled nd orbitals that can be used in bonding.
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84. More Than Eight Electrons
Although 3rd period often satisfy the octet rule, as in PCl3, they can exceed an octet by seeming to use their empty d orbital to accommodate additional electrons.
Expanded valence shells occur most often when the central atom is bonded to the smallest and most electronegative atoms, such as F, Cl, & O.
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85. More Than Eight Electrons
Even though we can draw a Lewis structure for the phosphate ion that has only 8 electrons around the central phosphorus, the better structure puts a double bond between the phosphorus and one of the oxygens.
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86. More Than Eight Electrons
This eliminates the charge on the phosphorus and the charge on one of the oxygens.
The lesson is: when the central atom in on the 3rd row or below and expanding its octet eliminates some formal charges, do so.
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87. 8.8 Strengths of Covalent Bonds
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88. 8.8 Covalent Bond Strength
The stability of a molecule is related to the strengths of the covalent bonds it contains.
The strength of a covalent bond between two atoms is determined by the energy required to break the bond.
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89. 8.8 Covalent Bond Strength
Most simply, the strength of a bond is measured by determining how much energy is required to break the bond.
This is the bond enthalpy.
The bond enthalpy for a Cl-Cl bond, D(Cl-Cl), is measured to be 242 kJ/mol.
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90. Average Bond Enthalpies
This table lists the average bond enthalpies for many different types of bonds.
Average bond enthalpies are positive, because bond breaking is an endothermic process.
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91. Average Bond Enthalpies
NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C-H bonds in methane, CH4, will be a bit different than the C-H bond in chloroform, CHCl3.
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92. Enthalpies of Reaction
Yet another way to estimate H for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed.
In other words,
Hrxn = (bond enthalpies of bonds broken) -
(bond enthalpies of bonds formed)
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93. Enthalpies of Reaction
CH4 (g) + Cl2 (g) 
CH3Cl (g) + HCl (g)
In this example, one C-H bond and one Cl-Cl bond are broken; one C-Cl and one H-Cl bond are formed.
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94. Enthalpies of Reaction
H = [D(C-H) + D(Cl-Cl)] - [D(C-Cl) + D(H-Cl)]
= [(413 kJ) + (242 kJ)] - [(328 kJ) + (431 kJ)]
= (655 kJ) - (759 kJ)
= -104 kJ
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95. Using Table 8.4, estimate ΔH for the following reaction (where we explicitly show the bonds involved in the reactants and products
Plan: Among the reactants, we must break six C—H bonds and a C—C bond in C2H6; we also break
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96. Using Table 8.4, estimate ΔH for the reaction
Sample Exercise 8.12 Using Average Bond Enthalpies
Practice Exercise
Using Table 8.4, estimate ΔH for the reaction
H = [D(N-H) + D(N-N)] - [D(N=N) + D(H-H)]
H = [4(391) + 1(163)] - [1(941) + 2(436)
H = [ ] - [ ]
Answer: –86 kJ

97. Bond Enthalpy and Bond Length
We can also measure an average bond length for different bond types.
As the number of bonds between two atoms increases, the bond length decreases.
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