AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
Chapter 8
Concepts of Chemical Bonding
8.1 Chemical Bonds
 Three basic types of bonds
___________
•
Electrostatic attraction between ions
•
Sharing of electrons
•
Metal atoms bonded to several other atoms
___________
___________
Lewis Symbols
Electrons involved in chemical bonding are the _____________________
electrons.
 G.N. _____________________(1875-1946) 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.
Octet Rule
Atoms tend to ______________________________________________ electrons
until they are surrounded by eight valence electrons.
 An octet of electrons consists of full _______________________________subshells in an atom.
 There are many exceptions to the octet rule, but it provides a useful
framework for many important concepts of bonding.
Write the Lewis symbol for atoms of each of the following elements:
 Al
 Br
 Ar
 Sr
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
8.2 Ionic Bonding
Energetics of Ionic Bonding
As we saw in the last chapter, it takes _______________ to remove electrons
from sodium.
Energetics of Ionic Bonding
We get ______________________ back by giving electrons to chlorine.
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 ______________________!
Energetics of Ionic Bonding
 There must be a third piece to the puzzle.
 What is as yet unaccounted for is the __________
___________________between the newly-formed sodium cation and chloride
anion.
Lattice Energy
•
This third piece of the puzzle is the lattice energy:
The energy required to completely separate a mole of a s olid ionic
compound into its _____________________
NaCl(s)  Na
+
•
(g) + Cl-(g) ∆Hlattice= + 788 kJ/mol
The energy associated with electrostatic interactions is governed by Coulomb’s
law:
Lattice Energy
 Lattice energy, then,
__________________________ with
the charge on the ions.
 It also increases with decreasing size
of ions.
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
Energetics of Ionic Bonding.
 These phenomena also helps explain the “_________________.”
 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.
Example
Arrange the following ionic compounds in order of increasing lattice energy.
NaF, CsI, and CaO
Solution:
Example
Which substance would you expect to have the greatest lattice energy?
 •AgCl
 •CuO
 •CrN
Solution:
3
AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
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.
Energetics of Ionic Bonding

By accounting for all three
energies (ionization energy, electron
affinity, and
______________________________), we
can get a good idea of the energetics
involved in such a process.
Born-Haber Cycle

The enthalpy change for the
direct route (red arrow) is the heat of
formation of NaCl(s)
 _____________________________________
 ∆Hf [NaCl(s)]= -411 kJ
 Values are found using Appendix C!
Born-Haber Cycle
The indirect route consists of five steps, shown by the green arrows.
Step 1
 We generate gaseous atoms of sodium by vaporizing sodium metal:
 Na(s) Na(g)
∆Hf [Na(g)] = 108 kJ
 ______________________ required energy
Step 2
 Then we form gaseous atoms of chlorine by breaking the bonds in the Cl2
molecules.
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
 1/2 Cl2(g) Cl(g) ∆Hf [Cl(g)] = 122 kJ
 Endothermic  requires energy
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) = ___________
 Cl(g) + e- Cl-(g)
∆Hf = E(Cl) = -349 kJ
 Ionization energy found on pg 268
 Electron affinity found on pg 271
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:
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.
Solution
H°f [NaCl(s)] = ∆H°f [Na(g)] + ∆H°f [Cl(g)] + I1(Na) + E(Cl) –∆Hlattice
 -411 kJ = 108 kJ + 122 kJ + 496 kJ –349 kJ – ∆Hlattice
Solving for ∆Hlattice
5
AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
 ∆Hlattice= 108 kJ + 122 kJ + 496 kJ –349 kJ + 411 kJ
 ∆Hlattice= 788 kJ
 Thus the lattice energy of NaCl is _____________________
Explain the following trends in lattice energy.
 CaF2 > BaF2
 NaCl > RbBr
 BaO > KF
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
8.3 Covalent Bonding
8.3 Covalent Bonding
In covalent bonds atoms __________________electrons.
There are several electrostatic interactions in these bonds:
o Attractions between electrons
and nuclei
o Repulsions between electrons
o Repulsions between nuclei
Covalent Bonds
 The attractions between nuclei and the
electrons cause electron density to concentrate
_______________________. 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.
A covalent bond is a chemical bond in which
two or more electrons are shared by two atoms
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
Lewis Structures
 The formation of covalent bonds can be represented using Lewis symbols

Examples: Draw:

H2, H . + . H  H:H

Cl2,

NH3,

CH4
Lewis Structures
 Multiple Bonds: Doubles and Triples
Examples: Draw CO2 and N2
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
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
8.4 Bond Polarity & Electronegativity
8.4 Polar Covalent Bonds
 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
 Although atoms often form compounds by
sharing electrons, the electrons are not
always shared equally.
Polarity
•____________________________ covalent bond is one in which
the electrons are shared equally between two atoms, as in the Cl2
and N2 examples we just drew.
•___________________ 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
_____________________is formed.
Electronegativity
•
•
The ability of atoms ________________________
to attract electrons to themselves.
On the periodic chart, electronegativity _________
as you go…
 …from left to right across a row.
 …from the bottom to the top of a column.
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
Vocabulary
 ___________________________– the ability of an atom IN A MOLECULE to
attract electrons to itself.
 _________________________________ how strongly an atom holds on to
its electrons.
 __________________________________– the measure of how strongly an
atom attracts additional electrons.
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
Question
•
What is the trend in electronegativity going from left to right in a row on the
periodic table?
11
AP Chemistry (BLM 11th)
•
•
Chapter 8 - Chemical Bonding
How do electronegativity values change as you move down a group of
elements on the periodic table?
How do periodic trends in electronegativity relate to those for ionization
energy & electron affinity?
Polar Covalent Bonds
The ___________________________ the
difference in electronegativity, the more
polar is the bond.
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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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
Example
B-Cl or C-Cl
 •Which bond is more polar?
 •Indicate in each case which atom has the partial negative charge.
Solution
 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 ____________________ carries the
partial negative charge because it has a higher electronegativity.
Polar Covalent Bonds
 When two atoms share
electrons unequally, a bond
dipole results.
 The ______________________,
, produced by two equal but
opposite charges separated by a
distance, r, is calculated:

 It is measured in debyes (D).
 A unit that equals 3.34 x 10-34 Coulomb-meters
What this means
 For molecules, we usually measure charge in units of electronic charge, e,1.6
x 10-19 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
 •m= Q•r
 •μ=(1.60 x 10-19 C)•[(1A)(10-10 m/1Å)( 1D/3.34x10-30 C-m)]
 μ= ____________________
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
8.5 Drawing Lewis Structures
8.5 Lewis Structures
 __________________________ are representations of molecules showing
all electrons, bonding and nonbonding.
Writing Lewis Structures
1. Find the __________of valence electrons of
all atoms in the polyatomic ion or molecule.

If it is an ___________ add one
electron for each negative charge.

If it is a __________, subtract one
electron for each positive charge.
Writing Lewis Structures
2. The central atom is the ______
__electronegative element that isn’t
hydrogen. Connect the outer atoms to it by
single bonds.
3. Fill the octets of the outer
atoms.
4. Fill the octet of the central atom.
Keep track of the electrons:
26 - 6 = 20
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
Writing Lewis Structures
5. If you run out of electrons
before the central atom has an octet…
…form _____________ until it does.
Writing Lewis Structures
Then assign __________________________.
 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.
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 – 2e4 + (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
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
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
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
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
SO3
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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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
8.6 Resonance –
 We sometimes encounter molecules
and ions in which the experimentally
determined arrangement of atoms in
_______adequately described by a
_________________________.
This is the Lewis structure we would
draw for ozone, O3.
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.
Resonance
 One Lewis structure cannot accurately
depict a molecule like ozone.
 We use multiple structures, _________
____________ to describe the molecule.
Resonance
Just as green is a synthesis of blue and
yellow…
…ozone is a
synthesis of these
two resonance
structures.
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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
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 _____________________________
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 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–.
19
AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
Resonance
 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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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
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.
1. EX: Fe 2+  [Ar] 3d6
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:
1. Ions or molecules with an ________________________ of electrons
2. Ions or molecules with ______________________________
3. Ions or molecules with more than eight valence electrons (an expanded
octet)
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 ____________________ of electrons such as ClO2,
NO, NO2, O2- . Complete pairing of electrons is impossible and an octet
around each atom cannot be achieved.
 For example, NO contains 5 + 6 =11 Electrons. The two most important
Lewis structures for this molecule are:
Fewer Than Eight Electrons
•
Consider BF3:
 Giving boron a filled octet places a ______________charge on the boron
and a __________________ charge on fluorine.
 This would not be an accurate picture of the distribution of electrons in
BF3.
21
AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
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
_________ electrons
.
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.
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.
•
The only way PCl5 can exist is if phosphorus
has ________________________ 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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AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
More Than Eight Electrons
1. Other examples of molecules & ions with “__________________”
valence shells include:
2. SF4, AsF6-, & ICl43. Element from the 3rd period and beyond have ns, np, and unfilled nd
orbitals that can be used in bonding.
4. 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.
5. Expanded valence shells occur most often when the central atom is
bonded to the smallest and most electronegative atoms, such as
_____________________.
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
___________ ________between the phosphorus and one of the oxygens.
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.
23
AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
8.8 Strengths of Covalent Bonds
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
_________________________ to break the bond.
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 _______________________________
 The bond enthalpy for a Cl-Cl bond, D(Cl-Cl), is measured to be 242 kJ/mol.
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 _______________________.
Average Bond Enthalpies
 NOTE: These are ____________
_______________________, not absolute bond enthalpies; the C-H bonds in
methane, CH4, will be a bit different than the C-H bond in chloroform, CHCl3.
24
AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
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)
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.
Enthalpies of Reaction
So,
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
 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
25
AP Chemistry (BLM 11th)
Chapter 8 - Chemical Bonding
Using Table 8.4, estimate ΔH for the reaction:
H = [D(N-H) + D(N-N)] - [D(N=N) + D(H-H)]
∆H =
∆H =
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.
26