Chemistry 101
Prof. Hammon
CHAPTER 9 - CHEMICAL BONDS
LEWIS STRUCTURE - How atoms are connected together.
LEWIS STRUCTURE FOR ELEMENTS – Valence
electrons are represented as dots surrounding the symbol of an
element.
To write a Lewis Structure for an element:
1. Determine # of valence electrons
Example:
Fluorine
F electron configuration 1s22s22p5 (outer shell has 7 e-)
 Group # is equal to valence electrons (for main group
elements, except helium: 2 electrons)
2. Draw dots around element symbol to represent valence
electrons. Draw a maximum of 2 dots per side.
..
:F:
 The maximum valence electrons is 8 (an octet).
Write the Lewis structure for Magnesium
Chemistry 101
Prof. Hammon
COVALENT BONDS
 Two atoms share a pair of electrons to form a covalent
bond.
H.
i.e. hydrogen
H.
H.
H ..H
H2
same as helium
1s2
1s2
i.e. chlorine, Cl2
..
..
:Cl:
+ :Cl:
[Ne]3s23p5
..
..
:Cl . . Cl:
[Ne] 3s23p6
(octet)
Bonding Electrons – electrons shared between atoms represent a
covalent bond. Bonding electrons are drawn as a line between
atoms.
Lone Pair Electrons - Non-bonding electrons, located on only
one atom.
Bonding pair
i.e.
H-H
..
..
:Cl
– Cl:
Lone pair
Chemistry 101
Prof. Hammon
Double and Triple bonds
Double Bond: 2 electron pairs are shared between atoms
(stronger and shorter than single bonds)
Example
..
.. ..
.O=O.
Triple Bond: 3 electron pairs are shared between atoms
(stronger and shorter than double bonds)
Example
Rules for Writing Lewis Structures for Covalent Compounds:
1. Write the skeletal structure for the molecule.
(hydrogen is always placed around the central atom)
(place highest electron affinity (more electronegative) atoms
in terminal positions)
i.e. H2O
H
O
H
Chemistry 101
Prof. Hammon
2. Count the total number of electrons for the molecule by
adding the valence electrons (VE) for each atom.
H.
H.
:O:
Total e- for molecule:
1+1+6 = 8
3. Insert the electrons in the skeletal structure making sure no
atom gets no more than eight electrons (octet) (hydrogen gets
no more than two electrons: duet). Start with bonding
electrons, then lone pairs on terminal atoms, and then lone
pairs on the central atom.
H :O : H
non-bonding pair of electrons
(lone pair)
H -O - H
bonding pair of electrons
4. Form double or triple bonds if atoms do not have eight
electrons around them.
Chemistry 101
Example:
Write the Lewis structure for CO2
1) Write skeletal structure
2) Calculate total # of electrons for molecule
3) Place electrons in skeletal structure
4) Form double or triple bonds, if necessary.
Prof. Hammon
Chemistry 101
Write the Lewis Structure for carbon monoxide.
Prof. Hammon
Chemistry 101
Prof. Hammon
Lewis Structures for Polyatomic Ions
1) Write skeletal structure (i.e. CN-)
2) Calculate total # of electrons for polyatomic ion. Add 1
electron for each negative charge and subtract one electron
for each positive charge.
3) Place electrons in skeletal structure
4) Form double or triple bonds, if necessary.
5) Put brackets around Lewis Structure and put charge in upper
right corner outside brackets.
Write a Lewis structure for the hydroxide ion, OH-
Chemistry 101
Prof. Hammon
EXCEPTIONS TO THE OCTET RULE
Expanded Octet – more than 8 electrons around an atom.
 Elements with atomic number 15 or greater may have more
than 8 valence electrons.
 Same rules + place additional electrons on central atom
PCl5
..
:Cl:
..
:Cl
:Cl:
P
..
Cl:
:Cl:
Write a Lewis structure for XeF4
Chemistry 101
Prof. Hammon
EXCEPTIONS TO THE OCTET RULE (cont.)
Incomplete Octet – less than 8 electrons around an atom
 Boron or Beryllium involved
 Same rules + deficient electrons on B or Be
Write a Lewis structure for BF3
Write a Lewis structure for BeH2
Chemistry 101
Prof. Hammon
ELECTRONEGATIVITY & BOND POLARITY:
Electronegativity: a measure of the ability of an atom to attract
electrons to itself within a bond.
Example: H2O
Oxygen is more electronegative than hydrogen, so it can attract
the shared electrons to itself within the bond.
Numerical values of electronegativity are assigned to the
elements. (Refer to Figure 9.8)
Electronegativity increases to the right across the periodic table
and decreases down a column.
Classifying Compounds based on Electronegativity
Differences
Nonpolar Covalent Compounds – Electronegativity difference
between atoms is 0-0.40. Electrons are evenly shared between
atoms resulting in a pure covalent bond.
i.e. Cl2
electronegativity difference is 3.0 – 3.0 = 0
Chemistry 101
Prof. Hammon
Polar Covalent Compounds - Electronegativity difference
between atoms is 0.41-1.99. Electrons are not evenly shared
between atoms.
i.e. HF
electronegativity difference is 4.0 – 2.1 = ____
Ionic Compounds - Electronegativity difference between atoms
is 2.00+. Electrons are completely transferred between atoms.
i.e. NaCl
electronegativity difference is ____ – ____ = _____
Electronegativity Difference
(EN)
0-0.40
0.41-1.99
≥ 2.00
Bond Type
Covalent
Polar Covalent
Ionic
Classify the bonds between the following atoms as nonpolar
covalent, polar covalent or ionic
Br - Br
Cs - Br
P-O
Chemistry 101
Prof. Hammon
RESONANCE & FORMAL CHARGE
Resonance Structures: equivalent Lewis structures for the
same molecule
 Drawn with a double headed arrow between equivalent
structures.
 Represent an average between the Lewis Structures.
Example
SO2
.. .. ..
.. .. ..
.O=S-O.
.O-S=O.
Write a Lewis Structure for ozone, O3 including resonance
structures.
Chemistry 101
Prof. Hammon
BOND ORDER
Bond Order: # of bonds for single, double & triple bonds
Type of Bond
Single
Double
Triple
Bond Order
1
2
3
For Resonance Structures:
Bond Order =
Example
# of Bonds
# of Bonding Situations
SO2
.. .. ..
.. .. ..
.O=S-O.
.O-S=O.
Bond Order for SO2 = # of bonds / # of bonding sit. = 3 / 2 = ___
What is bond order for CO32-?
Chemistry 101
Prof. Hammon
FORMAL CHARGE
Formal charge of an atom is the charge it would have if all
bonding electrons were shared equally between the bonded
atoms.
To determine Formal Charge of an atom:
Formal Charge = #VE - #of non-bonding e- -1/2(# of bonding e-)
SO42-
..
:O:
..
:O
S
2..
O:
:O:
To calculate Formal charge on S
# valence electron (VE) for S = ____
# of non-bonding electrons on S = ___
# of bonding electrons for S = _____
Formal charge on “S” = ___ – ___ – ½(___) = _____
Calculate the Formal Charge on each “O” in SO42-
Chemistry 101
Prof. Hammon
FORMAL CHARGE (cont.)
Formal charge is used to determine the “best” (closest to reality)
resonance structure.
Rules to pick the “best” structure:
1. Want structure with the smallest total of the absolute values
of the formal charge.
2. If total formal charges on structures are equal, then the
negative formal charge should be on the more
electronegative atom.
Which one is the “best” resonance structure for SO42-?
..
:O:
..
:O
S
2..
O:
..
:O:
↔
2-
O = S =O
:O:
:O:
To determine:
1) Calculate the formal charge on each atom:
2) Add together the absolute values of all formal charges:
/-1/+/-1/+/-1/+/-1/+/+2/ = ___
/-1/+0 +0+/-1/+0 = ____
Chemistry 101
Prof. Hammon
3)The one with the smallest total value is the “best” resonance
structure.
..
:O:
ANSWER:
2-
O = S =O
:O:
Which one is the “best” resonance structure?
..
:O
C = N: ↔
..
.. O=C=N
↔
.. :O = C N:
Chemistry 101
Prof. Hammon
BOND ENERGY
Bond Energy is the energy required to break 1 mole of the bond
in the gas phase.
Breaking bonds requires energy
Breaking bonds is an _________________process (H is +).
Making bonds is an _________________ process (H is -).
Bond Energies vary for polyatomic molecules:
C-H Bond breaking
H3C-H(g)  H3C(g) + H(g)
F3C-H(g)  F3C(g) + H(g)
Br3C-H(g)  Br3C(g) + H(g)
Bond Energy (kJ/mol)
438
446
402
Ave. Bond Energy: Ave of the bond energies for that bond in
numerous compounds (refer to Table 9.3).
The larger the bond energy the stronger the bond.
Chemistry 101
Prof. Hammon
Estimating Enthalpy of Reaction (Hrxn) from Bond
Energies
Hrxn =  (H bonds broken ) +  (H bonds formed)
Example
H2(g) +
Cl2(g)

2 HCl(g)
Bonds Broken Bond En. (H)
H-H
Cl-Cl
436 kJ/mol
243 kJ/mol
Bonds Formed Bond En.(H)
H-Cl
-431 kJ/mol
Hrxn = [1 mol(436kJ/mol) + 1 mol(243 kJ/mol)] + [2 mol(-431 kJ/mol)]
Hrxn =
Is above reaction exothermic or endothermic?
Complete the Diagram
Chemistry 101
Prof. Hammon
Estimate the Hrxn for the following reaction using bond
energies:
CH4(g)
+
2 H2O(g) 
4 H2(g)
+
CO2(g)
Hint: To solve draw Lewis Structures & then determine the
bonds that are broken and formed.
Chemistry 101
Prof. Hammon
Bond Enthalpies – The Strength of Covalent Bonds
Sharing electrons between atoms is a favorable process, which is why
chemical bonds form! Therefore, breaking these bonds will require energy,
as shown in the following reaction:
Therefore, breaking bonds is an _________________process.
Making bonds must is an _________________ process.
The value of H for the reaction shown above is called the ____________,
which is related to the _______________of the covalent bond.
There is a table of average bond energies on page 392 of your text book.
Sometimes, Hf° is not known for a compound, and we may not have time
to measure this value using calorimetry in a lab. We can use the bond
enthalpies to ESTIMATE Hrxn in this case.

Hrxn = (Hbonds broken ) – (Hbonds formed)
Example: Use Table 9.3 on page 392 to estimate Hrxn for the reaction
CH4 (g) + Cl2 (g) CH3Cl (g) + HCl (g) Hrxn = _____________ kJ
(book example)
What does this equation, Hrxn =  (Hbonds broken ) –  (Hbonds formed),
physically mean? Fill in the Diagram with the enthalpy changes.