Chapter 3
Section 3.1 (pg. 78-84)
Homework: Lewis Symbols
Pg. 82 #2 – 4 Pg. 84 # 2, 4, 5, 7-10
Objectives:
1) Define valence electron, electronegativity, and ionic bond
2) Use the Periodic Table and Lewis structures to support and explain ionic bonding
3) Explain how an ionic bond results from the simultaneous attraction of oppositely charged ions.
Bonding Theory: Valence Electrons & Orbitals
 To describe where electrons exist in the atom, chemists
created the concept of an orbital.
 Orbital – region of space around an atom’s nucleus where an electron
may exist
 An “orbital” is not a definite race track, it is a 3-D space that defines
where an electron may be (like a rain drop in a cloud)
 For bonding study we are only concerned with an atom’s valence
orbitals (the volume of space that can be occupied by electrons in an
atom’s highest energy level)
 WHY? Bonding only involves valence e-’s because lower energy levels are held so
strongly by their positively charged nucleus
FYI Read pg. 78-79 for the history on Bonding Theory
Bonding Theory: Valence Electrons & Orbitals
 According to bonding theory, valence electrons are
classified in terms of orbital occupancy.
(0 = empty, 1 = half filled , 2 = full)
 An atom with a valence orbital that has a single electron
can theoretically share that electron with another atom

Such an electron is called a BONDING ELECTRON
 An atom with a full valence orbital (2 e-’s), repels nearby
orbitals and wants to be alone

Such a pairing is called a LONE PAIR
The Four Rules of Bonding Theory
1.
The first energy level has room for only one orbit
- can only hold 2 e-’s max
2.
Energy levels above the first have room for
four orbitals = 8 electrons max
2e2 p+
He
- Noble gases have this structure; their lack or reactivity indicates
that eight electrons filling a valence orbital is very stable
(Remember the OCTET RULE)
FYI: Only C, N, O,
and F atoms
always obey the
octet rule when
bonding
8e8e2e18 p+
Ar
2e-
2e-
2e-
2eEXCEPTIONS:
B = stable with 6 valence e- (3 orbitals)
P = stable with 10 valence e- (5 orbitals)
S = stable with 12 valence e- (6 orbitals)
2e-
The Four Rules of Bonding Theory
3. An orbital can be unoccupied, or it may contain one or two
electrons – but never more than two (Pauli Exclusion Principle)
4. Electrons “spread out” to occupy any empty valence orbitals
before forming electron pairs
“Aluminum has three half-filled
orbitals and one vacant orbital.”
How would you describe Sulfur?
Never more than
2e- in an orbital
Atomic Models: LEWIS SYMBOLS
(aka Lewis Dot Diagrams, Electron Dot Diagrams, LDD, Lewis Models
• Named after Gilbert Lewis who is responsible for the Octet Rule. He
reasoned that all atoms strive to be like the nearest noble gas.
• Used dots or ‘x’ to represent the valence electrons
• The inner electrons and the nucleus are represented by the element symbol
How to draw Lewis Symbols:
1.
2.
3.
4.
Write the element symbol
Add a dot to represent each valence eStart by placing valence e-’s singly into each of the four valence orbitals (4 sides)
If additional e-’s need to be placed, start filling each of the orbitals with a second e- up to 8
Q: Which element has 4 bonding e-’s?
Which has 3 lone pairs and 1 bonding e-?
Practice
 Draw the Lewis Symbols for the elements in Period 3
 For each one indicate how many lone pairs or bonding
electrons are present
 It is important to remember that the Lewis symbols do not mean that
electrons are dots or that they are stationary.
 The four sides represent the four orbitals that may be occupied by electrons;
it is a simplistic 2-D diagram of a complex 3-D structure
Electronegativity
-A measure of the force that an atom exerts on electrons of
other atoms; (the “pull” on bonding electrons)
-Each atom is assigned a value between 0.0 – 4.0; the larger the number
the greater the “pulling” force
-Example: Fluorine has an EN = 4.0 and francium has an EN = 0.7
-This means fluorine wants to pull on other electrons very strongly
-This means francium doesn’t want to pull on other electrons
Q: Does lithium (EN = 1.0) want to lose or gain an electron to be stable?
Q: Does fluoride (EN = 4.0) want to lose or gain an electron to be stable?
Do you see any relation to their electronegativity numbers?
So how do we assign each atom an electronegativity number?
a) The farther away from the nucleus that electrons are, the weaker
their attraction to the nucleus
.
EN = 0.8
Cesium's valence electrons are not
held as tightly by its nucleus
because the atom is larger
EN = 2.6
b) Inner electrons shield valence electrons from the attraction of
the positive nucleus
1 e8e8e2e19p+
K
EN = 0.8
Potassium’s valence electrons are not
attracted to its nucleus as much as
Nitrogen’s valence electrons because their
are more inner electrons present in K
EN = 3.0
5e2e7p+
N
c) The greater the number of protons in the nucleus, the greater the
attraction for more electrons
14p+
Si
EN =1.9
Bromine has more protons (+ charge)
which attracts the negative charge of
electrons more so than silicon’s 14 protons
EN = 3.0
35p+
Br
Electronegativity
In this 3-D image, the
electronegativity scale is
vertical.
Q: What is the EN trend
within a period and a
group?
Q: Which element has the highest
EN? Give three reasons why?
Why do we care about electronegativity??
 Imagine that two atoms, each with an orbital containing one bonding
electron, collide in such a way that these half-filled orbitals overlap.
 As the two atoms collide, the nucleus of each atom attracts and
attempts to “capture” the bonding electrons of the other atoms
 A “Tug of War” over the bonding electrons occurs
 Which atom wins?
 By comparing the electronegativities of the two atoms we can
predict the result of the contest = 3 different types of bonds result
Covalent Bonding
 Both atoms have a high EN so neither atom “wins”
 The simultaneous attraction of two nuclei for a shared pair of
bonding electrons = covalent bond
Cl2 = diatomic
 EN difference can be zero = Cl – Cl
EN = 3.2 EN = 3.2
 EN difference can be small = H - Cl
EN = 2.2 EN = 3.2

This is called a polar covalent bond – because
one side pulls on the electrons more but we
will learn more about this in Section 3.3
Ionic Bonding
 The EN of the two atoms are quite different
 The atom with the higher EN will remove the bonding
e- from the other atom
electron transfer occurs
 Positive and negative ions are formed which electrically
attract each other
EN = 0.9
EN = 3.2
Metallic Bonding
 Both atoms have a relatively low EN so atoms share valence
electrons, but no actual chemical reaction takes place
 In metallic bonding:
a) e-’s are not held very strongly by their atoms
b) the atoms have vacant valence orbitals
- This means the electrons are free to move around between
the atoms and the (+) nuclei on either side will attract them
Analogy: The positive nuclei are held
together by a glue of negative e-’s
Metallic bonding visual
This diagram
represents Mg atoms
that have released
their electrons and are
embedded in a sea (or
glue) of electrons.
Note: These metal atoms don`t have to be in a particular arrangement
to attract each other therefore they are flexible, malleable and
ductile = useful alloys (Brass, Stainless Steel, etc.)
Summary of Bonding Theory:
Chemical Bond = competition for bonding electrons
1) Atoms with equal EN = electrons shared equally
If both have high EN = covalent bond (equal = non-polar)
If both have a low EN = metallic bond
2) Atoms with unequal EN = covalent bond (unequal = polar)
3) Atoms with unequal EN = ionic bond
metallic
The nature of chemical bonds changes
in a continuous way, creating a broad
range of characteristics.
PRACTICE
Copy pg. 84 – Bonding Theory Summary into your Notes
Pg. 82 #2 - 4
Pg. 84 # 2, 4, 5, 7-10
Explaining Molecular
Formulas
Pg. 89 #5 (a-f), 6 (a-e)
Pg. 90 #1-4, 6
Section 3.2 (pg. 85-90)
Explaining Molecular
Formulas
Objectives:
1) Draw electron dot diagrams of atoms and molecules, writing structural formulas for
molecular substances using Lewis structures to predict bonding in simple molecules
2) Illustrate, by drawing or building models, the structure of simple molecular substances
3) Explain why the formulas for molecular substances refer to the number of atoms of each
constituent element
Section 3.2 (pg. 85-90)
Molecular Elements
 Many molecular elements are
diatomic and some are polyatomic
 You will need to memorize the
formulas of the 9 molecular elements
as they will not be given to you:
Name
Symbol
hydrogen
H2(g)
nitrogen
N2(g)
oxygen
O2(g)
fluorine
F2(g)
chlorine
Cl2(g)
iodine
I2(g)
bromine
Br2(g)
phosphorous
P4(g)
sulfur
S8(g)
Why are they diatomic?
 Remember fluorine has 7 valence e-’s and needs 1 more
e- to be stable?
 Well 2 fluorine atoms could obtain a stable octet of e-’s
if they shared a pair with each other
F-F
In structural formulas,
lone pairs are not shown
Remember: This is a simplified 2-D version,
not where the electrons actually are
F2
Diatomic Elements
Would sharing only 1
electron each work?
 What about oxygen and nitrogen?
O=O
Each oxygen atom only has 6 valence electrons. So by sharing 2 electrons each,
the two oxygen atoms can create a full octet. This creates a double bond
NΞN
Each nitrogen atom only has 5 valence electrons. So by sharing 3 electrons each,
the two nitrogen atoms can create a full octet. This creates a triple bond
Molecular Compounds
 Background:
 Molecular compounds have covalent bonds (shared electrons)
between non-metals and non-metals
 Can be solid, liquid or gas as SATP
 May or may not be soluble in water (more later)
 Don’t ever conduct electricity - even when (aq)
 Generally have lower m.p. and b.p than ionic compounds
Review from Section 1.5 Notes
Molecular Compounds
 Background:
 Empirical Formulas – show the simplest whole number ratios
of atoms in a compound

Very useful for IONIC compounds

Formula Unit – the ratio of ions that repeats in a pattern within the
crystal; the chemical formula of ionic compounds represents the
formula unit
Na242Cl242  Na16Cl16  NaCl

Not useful for MOLECULAR compounds
CH  C2H2  acetylene
 C6H6  benzene
 C8H8  octene
All are extremely
different compounds but
the empirical formula
would be the same
Molecular Compounds
 Background:
 Molecular Formulas – a molecular formula shows the actual number
of atoms that are covalently bonded to make-up each molecule
 We use this because chemical formulas for molecular compounds result from
sharing electrons, therefore a variety of compounds are possible (which we
determine empirically through experiments)

Often the symbols are written in a sequence that helps you determine how
the atoms are bonded
C2H4O2  CH3COOH
Empirical formula:
CH2O
incorrect for molecular compounds
Review of Molecular Compound Formulas
See
Pg. 88
Q: How do we know how molecular
compounds bond?
(Aka: How do we draw Lewis Formulas?)
Where do these come from???
Determining Lewis
Formulas

Bonding Capacity: the maximum number of single
covalent bonds that an atom can form

REMINDER: How many e-’s does an atom want in its valence
energy level to be satisfied?
H = 2 e-

C, N, O, F, P, S, Cl, etc. = 8 e-
REMINDER: What types of covalent bonds are possible?
F – F single = sharing one e- pair
O = O double = sharing two e- pairs
N Ξ N triple = sharing three e- pairs
Determining Lewis
Formulas
 So why do we care about bonding capacity?
 If we know how many bonding e-’s an atom has, we can
predict what structure a molecular compound will have
Atom
H
Number of
Number of
Bonding capacity
valence electrons bonding electrons
carbon
4
4
4
nitrogen
5
3
3
oxygen
6
2
2
halogens
7
1
1
hydrogen
1
1
1
I.e. Carbon can form 4 single bonds, 2 double bonds, 1 triple and 1 single, or 1 double and 2 singles
Lewis Formulas- Guided
Ex. #1
 Determine the Lewis formula and structural formula for sulfur
trioxide, SO3(g)
1. Count the number of valence electrons there are in total? (If
polyatomic ions are included, subtract or add electrons to
account for the net charge)
Oxygen = 3 atoms x 6 valence e-’s each = 18 valence e-’s
Sulfur = 1 atom x 6 valence e-’s each = 6 valence e-’s
24 valence e-’s
Lewis Formulas – Guided
Ex. #1
2. Choose your central atom
 In our course, we will limit our formulas to ones with one
central atom (unless extra info is provided)
 So how do you know which is the central atom?

Usually the one in lesser quantity (SO3(g))
OR

The one with the higher bonding capacity

Carbon usually – because 4 is the highest bonding capacity
So which is the central atom?
Lewis Formulas – Guided
Ex. #1
3. Arrange peripheral atoms around central
atom and place one pair of valence e-’s
between them
4. Place lone pairs on all peripheral atoms
to complete their octet
5. Place any remaining valence e-’s on the
central atom as lone pairs.
For this example, all 24 have been assigned
Lewis Formulas – Guided
Ex. #1
6. If the central atom’s octet is not
complete, move a lone pair from a
peripheral atom to a new position
between the peripheral and
central atom.
7. Show the structural formula but
omit lone pairs and replace every
bond with a line
(Count the electrons around each atom
to confirm the octet rule. Each atom
should have 8 e-’s around it; exception:
H)
Lewis Formulas – Guided
Ex. #2
Determine the Lewis formula & structural formula for
the nitrate ion, NO31. Count the valence electrons (*look for a net charge if an ion).
nitrogen = 1 x 5 valence e-’s = 5
oxygen = 3 x 6 valence e-’s = 18
23 + 1 (b/c net charge is -1) = 24
2. Which is the central atom? Nitrogen (in lesser quantity)
3. Arrange peripheral atoms around central atom and
place 1 pair of valence e-’s between them
N
Lewis Formulas – Guided
Ex. #2
4. Place lone pairs on all peripheral atoms
to complete their octet
N
5. Place any remaining valence e-’s on the
central atom as lone pairs.
6. If the central atom’s octet is not
complete, move a lone pair from a
peripheral atom to a new position
between the peripheral and central
atom.
7. If the entity is a polyatomic ion, place
square brackets around the entire Lewis
formula and then write the net charge
outside the bracket on the upper left.
N
N
Practice
 Pg. 89 #5 (a-f), 6 (a-e)
 Watch 5 (f) there is an exception noted. The central atom
does not follow the octet rule.
 We will go through these answers as a class.
 Pg. 90 #1-4, 6
 The theory presented today is not absolute – there are
exceptions. But rather than presenting a more detailed theory,
your textbook will always note such exceptions.
Tomorrow...
 Molecular Model Investigation
 Thought Lab Investigation
 Morse Code Assignment
Section 3.3 – Part A
Pg. 91-96
Objective:
1) Apply VSEPR theory to predict molecular shapes
Molecular Shapes
 Stereochemistry – is the study of the 3-D spatial
configuration of molecules and how this affects their
reactions.
Solid = in plane of page Dashed = behind (away) Wedge = ahead (toward)
 The shape of molecules is determined by the repulsion that happens
between electron pairs
 The theory behind molecular shapes is called VSEPR Theory (Valence
Shell Electron Pair Repulsion)
VSEPR
 General Rule:
 Pairs of electrons in the valence shell of an atom stay as
far apart as possible because of the repulsion of their
negative charges
 The type, number and direction of bonds to the central
atom of a molecule determine the shape of the resulting
molecule.
 So how do we predict these molecular shapes?
Using VSEPR to Predict Molecular Shapes
 We will be using the following compounds to analyze the 6
shapes possible
 BeH2(s), BH3(g), CH4(g), NH3(g), H2O(l), HF(g)
 To start, draw a Lewis formula for each of the molecules
and then consider the arrangement of all pairs of valence
electrons.
 (Remember – all pairs of valence e-’s repel each other
and want to get as far apart as possible)
Shape #1 = Linear
Lewis
Formula
Be
Bond
Pairs
2
Lone Total
Pairs Pairs
0
2
General
Formula
AX2
Electron Pair
Arrangement
Stereochemical
Formula
linear
X–A–X
linear
* A is the central atom; X is another atom
• This Lewis formula indicates that BeH2(s) has two bonds and
no lone pairs on the central atom.
• VSPER theory suggests that the two bond pairs will be farthest
apart by moving to opposite sides to a bond angle of 180°
• This gives the molecule a linear orientation
*Exception* Beryllium does not follow OCTET RULE
Shape #2 = Trigonal Planar
Lewis
Formula
B
Bond
Pairs
3
Lone Total
Pairs Pairs
0
3
General
Formula
Electron Pair
Arrangement
AX3
trigonal
planar
Stereochemical
Formula
* A is the central atom; X is another atom
• This Lewis formula indicates that BH3(g) has three bonds and
no lone pairs on the central atom.
• VSPER theory suggests that the three bond pairs will be farthest
apart by moving to a bond angle of 120° to each other.
• This gives the molecule a trigonal planar orientation.
*Exception* - Boron Does not follow OCTET RULE
Practice
 Draw the Lewis Formula for BF3
F
Trigonal Planar
F
F
Does not obey the octet rule
Shape #3 =Tetrahedral
Lewis
Formula
Bond
Pairs
4
Lone Total General
Pairs Pairs Formula
0
4
AX4
Electron Pair
Arrangement
tetrahedral
* A is the central atom; X is another atom
• This Lewis formula indicates that CH4(g) has four bonds
and no lone pairs on the central atom.
• VSPER theory suggests that the four bond pairs will be
farthest apart by arranging in three dimensions so that
every bond makes an angle of 109.5° with each other.
• This gives the molecule a tetrahedral orientation.
Stereochemical
Formula
Practice
 Draw the Lewis Formula for SiH4
H
Tetrahedral
H
H
H
Shape #4 =Trigonal Pyramidal
Lewis
Formula
Bond
Pairs
3
Lone Total General
Pairs Pairs Formula
1
4
AX3E
Electron Pair
Arrangement
Stereochemical
Formula
tetrahedral
Trigonal pyramidal
* A is the central atom; X is another atom, E is a lone pair of electrons
• This Lewis formula indicates that NH3(g) has three bonds and
one lone pair on the central atom.
• VSPER theory suggests that the four groups of e-’s should repel
each other to form a tetrahedral shape (bond angle = 109.5°)
• But the lone pair is very repulsive, thus pushes the atoms more
to a 107.3° bond angle
• This gives the molecule a trigonal pyramidal orientation.
Practice
 Draw the Lewis Formula for PCl3
Cl
Cl
Cl
Trigonal pyramidal
Shape #5 =Angular (Bent)
Lewis
Formula
Bond
Pairs
2
Lone Total General
Pairs Pairs Formula
2
4
AX2E2
Electron Pair
Arrangement
Stereochemical
Formula
tetrahedral
* A is the central atom; X is another atom, E is a lone pair of electrons
• This Lewis formula indicates that H2O(l) has two bonds and
two lone pairs on the central atom.
• VSPER theory suggests that the four groups of e-’s should repel
each other to form a tetrahedral shape (bond angle = 109.5°)
• But the TWO lone pairs are very repulsive, thus pushes the atoms
more to a 105° bond angle
• This gives the molecule an angular (bent) orientation.
Angular
(Bent)
Practice
 Draw the Lewis Formula for OCl2
Angular (bent)
Shape #6 =Linear (Tetrahedral)
Lewis
Formula
H
F
Bond
Pairs
1
Lone Total General
Pairs Pairs Formula
3
4
AXE3
Electron Pair
Arrangement
Stereochemical
Formula
Linear
(Tetrahedral)
* A is the central atom; X is another atom, E is a lone pair of electrons
• This Lewis formula indicates that H2O(l) has two bonds and
two lone pairs on the central atom.
• VSPER theory suggests that the four groups of e-’s should repel
each other to form a tetrahedral shape (bond angle = 109.5°)
• But since there are only two atoms with one covalent bond holding
them together, by definition, the shape is linear, as is the shape of every
other diatomic molecule.
Practice
 Draw the Lewis Formula for HCl
Summary
 VSEPR theory describes, explains, and predicts the geometry
of molecules by counting pairs of electrons that repel each
other to minimize repulsion. The process for predicting the
shape of a molecule is summarized below:
 Step 1: Draw the Lewis formula for the molecule, including the
electron pairs around the central atom.
 Step 2: Count the total number of bonding pairs (bonded
atoms) and lone pairs of electrons around the central atom.
 Step 3: Refer to Table 7, and use the number of pairs of
electrons to predict the shape of the molecule.
Pg. 95
Practice
 Draw the Lewis and stereochemical formulas for a sulfate
ion, SO42- and predict the shape
 See pg. 95
 Draw the Lewis and stereochemical formulas for a
chlorate ion, ClO3- and predict the shape
 See pg. 96
 On your own: Pg. 96 #3
Multiple Bonds in VSEPR Models
 It is important to remember that a double or triple bond is one bond, and to
treat it as such, when predicting the VSEPR shapes of molecules.
 Example: Predict the shape of C2H4(g)
H H
 Draw the Lewis formula for the molecule
 Count the # of pairs of


H
e-’s
H
around the central carbon atoms.
The carbon atoms have 3 bonds (2 single, 1 double) and no lone pairs.
This is the same as a trigonal planar configuration.
 Practice: Predict the shape for C2H2(g).
Answer: See pg. 97
Homework
1) Finish pg. 96 #1-3
2) Pg. 98 #6-7 (Multiple Bond Practice)
 For 7 c, d, e - If there is more than one central atom involved, tell me the
shape around each of the central atoms
 Example:
trigonal planar—first two carbons
tetrahedral—third carbon
3) Pg. 104 #1, 2, 3

#2: If there is more than one central atom involved, tell me the
shape around each of the central atoms
Practice
 Draw the Lewis Formula for PCl3
Dipole Theory
Section 3.3 – Part B
Pg. 98 - 104
1)
Determine the polarity of a molecule based on simple structural shapes
and unequal charge distribution
2) Describe bonding as a continuum ranging from complete electron transfer
to equal sharing of electrons.
Polarity
 Chemists believe that molecules are made up of charged particles
(electrons and nuclei).
 A polar molecule is one in which the negative
(electron) charge is not distributed symmetrically
among the atoms making up the molecule.
 Thus, it will have partial positive and negative charges on opposite
sides of the molecule.
 A molecule with symmetrical electron distribution is a nonpolar
molecule.
 The existence of polar molecules can be demonstrated by
running a stream of water past a charged object.
 Demo: See Figure 9
TESTING A LIQUID WITH A CHARGED OBJECT:
 In a liquid,
molecules are able
to rotate freely.
 Polar molecules in a
liquid will rotate so
that their positive
sides are closer to a
negatively charged
material.
 Near a positively
charged material
they become
oriented in the
opposite direction.
EMPIRICAL RULES FOR POLAR
AND NONPOLAR MOLECULES
Type
Polar
Description of
molecule
diatomic with different
atoms
containing nitrogen and
other atoms
Examples
OxAy
containing oxygen and
other atoms
H2O(l),
OCl2(g)
Cx A y B z
containing carbon and
two other kinds of
atoms
all elements
CHCl3(l),
C2H5OH(l)
containing carbon and
only one other kind of
atom (except CO(g))
CO2(g),
CH4(g)
AB
Nx A y
Nonpolar
Ax
Cx A y
HCl(g), CO(g)
NH3(g),
NF3(g)
Cl2(g), N2(g)
Pg. 99
When the water test was repeated with a large number of pure
liquids, this provided the set of empirical rules above.
PREDICTING AND EXPLAINING POLARITY
 Linus Pauling explained polarity by
creating the concept of
electronegativity.
 Introduced in Section 3.1
 Electronegativity increases as you go up
or to the right on the periodic table.
PREDICTING AND EXPLAINING POLARITY
 Pauling explained the polarity of a covalent bond
as the difference in electronegativity of the bonded
atoms.
Cl2(g)
 If the bonded atoms have the same electronegativity,
they will attract any shared electrons equally and form
a nonpolar covalent bond.
 If the atoms have different electronegativities, they will
form a polar covalent bond.
 The greater the electronegativity difference, the more
polar the bond will be.
 For a very large electronegativity difference, the
difference in attraction may transfer one or more
electrons resulting in ionic bonding.
We use the
Greek symbol
delta to show
partial charges
PREDICTING AND EXPLAINING POLARITY
 Pauling liked to think of chemical bonds as being different in
degree rather than different in kind.
 According to him, all chemical bonds involve a sharing of electrons,
with ionic bonds and nonpolar covalent bonds being just the two
extreme cases
~EN difference:
nonpolar (< 0.4)
polar (0.4+)
ionic (m + nm)
 The bonding in substances therefore ranges anywhere along a
continuum from nonpolar covalent to polar covalent to ionic.
 For polar covalent bonds, the greater the electronegativity
difference of the atoms, the more polar the bond.
PRACTICE
 See pg. 100 Sample Problem 3.4
 Try on your own
 pg. 100 #9 (a-c), 10, 11(a only)
DOES BOND POLARITY = MOLECULAR POLARITY??
 NO! Chemists have found that the existence of polar bonds in a
molecule does not necessarily mean that you have a polar molecule.
 Example:
 Carbon dioxide is found to be a nonpolar molecule, although each of the CO
bonds is a polar bond. WHY??
 According to VSEPR (two bonds, no lone pairs) = linear arrangement
 We will start showing bond polarity as arrows pointing in the negative
direction (where e-’s want to go) = bond dipole

Points from lower to higher electronegativityδ–
δ+
3.4
2.6
δ–
O = C = O
3.4
 The arrows are vectors and when added together, the equal but opposite
bond dipoles equal zero.
 Non-polar molecules are ones where the bond dipoles balance each
producing a molecular dipole (vector sum) of zero
other;
Prediction Molecular
Polarity
 Step 1: Draw a Lewis formula for the molecule.
 Step 2: Use the number of electron pairs and VSEPR rules to determine
the shape around each central atom.
 Step 3: Use electronegativities to determine the polarity of each bond.
 Step 4: Add the bond dipole vectors to determine whether the final
result is zero (nonpolar molecule) or nonzero (polar molecule).
Guided Practice #1
Go to Learning
Tip pg. 102
 Predict the polarity of the water molecule.
1)
Draw the Lewis formula
2)
VSEPR: Draw the stereochemical formula
O
H
H
Angular (bent)
3)
Assign the EN of the atoms, assign δ– and δ+ to the bonds
4)
Draw in the bond dipoles
• The bond dipoles (vectors) do not balance.
• Instead, they add together to produce a nonzero
molecular dipole (shown in red).
• This results in a polar molecule (explains bending water)
Guided Practice #2
 Predict the polarity of the methane molecule.
1)
Draw the Lewis formula
Tetrahedral
2)
VSEPR: Draw the stereochemical formula
3)
Assign the EN of the atoms, assign δ– and δ+ to the bonds
4)
Draw in the bond dipoles
• Notice how all the bond dipoles point into the central atom.
• There are no positive or negative areas on the outer part of
the molecule.
• A tetrahedral molecule is symmetrical in 3-D and four
equal tetrahedral bond diploes always sum to zero
Practice
 Predict the bond polarity of the ammonia, NH
molecule. Include your reasoning.
Answer: See pg. 102
3(g)
FYI: DO YOU REMEMBER “LIKE DISSOLVES LIKE”?
 Means:
 “Polar substances are soluble in polar substances;
Non-polar substances are soluble in non-polar
substances”
 Mixing non-polar and polar substances results in
them forming layers, with the least dense one on
top.
 This occurs because polar molecules attract each
Two clear liquids formed layers in this tube: nonpolar
other more strongly; thus they stay close together
hexane C6H14(l)on top, and polar water, H2O(l) below.
excluding nonpolar molecules
Nonpolar dark orange liquid bromine, Br2(l) was then
added. The bromine dissolves much more readily in
the nonpolar hexane.
WHY DO WE CARE ABOUT POLAR MOLECULES?
 Cleaning!! Water (polar) is useless at removing
oil (nonpolar) so detergents are artificially
created molecules that overcome this problem
 Detergents have long, nonpolar sections which are attracted to
(dissolve in) a tiny oil droplet.
 The polar end of each of these detergent
molecules helps form a polar “layer”
the droplet, which attracts polar
molecules.
 This allows them to pull the oil droplet
from a stained area of fabric and
suspended in the wash water.
around
water
away
hold it
Homework:
 Pg. 102-103 #13-16
 Pg. 104 # 4, 5, 10
Section 3.4
Pg. 105-117
Explain intermolecular forces, London
(dispersion) forces, dipole-dipole attractions and
hydrogen bonding
1)
2)
Relate properties of substances to the predicted
intermolecular bonding in the substance.
BACKGROUND
• All chemical changes (reactions) are accompanied by energy changes
▫ Energy is mostly heat, light, or electrical energy
▫ Energy can be released slowly (battery) or quickly (fireworks)
▫ Two types of energy changes are possible:

EXOTHERMIC – energy is released into the surroundings
- the product’s bonds have less energy than the reactant’s bonds

ENDOTHERMIC – energy is absorbed from the surroundings
- the product’s bonds have more energy than the reactant’s bonds
▫ Bond Energy – the energy required to break a chemical bond or the
energy released when a bond is formed
BACKGROUND
• There are three types of forces in matter:
1)
Intranuclear force (bond) – bonds within the nucleus between protons and
neutrons (very strong)
2) Intramolecular force (bond) – bonds between atoms within the molecule or
between ions within the crystal lattice (quite strong)
3) Intermolecular force (bond) – bonds between molecules (quite weak); are
electrostatic (involve positive and negative charges)
There are 3 types of intermolecular bonds:
Weakest
a) Dipole-Dipole Forces (a.k.a. Polar Forces)
Medium
b) London Force (a.k.a. London Dispersion Force, Dispersion Force)
Strongest c) Hydrogen
Bonding
Note: “Van der Walls force” – includes London and dipole-dipole forces
1) Dipole-Dipole Force
• The simultaneous attraction between oppositely charged
ends of polar molecules.
▫ Simply put, the attraction between diploes
Dipole: a partial separation of positive and negative charges
within a molecule, due to electronegativity differences
▫ Dipole-dipole forces are among the weakest
intermolecular forces, but still control important
properties (i.e. Solubility because water is polar))
1) Dipole-Dipole Force
In a liquid, polar molecules
can move and rotate to
maximize attractions and
minimize repulsions. The
net effect is greater overall
attraction.
The strength of the dipoledipole force is dependent
on the overall polarity of
the molecule
Note: If a molecule is
polar it will be soluble
in water? Why?
1) Dipole-Dipole Forces
In a liquid:
In a solid:
2) London Force
 Simultaneous attraction between a momentary dipole in a
molecule and the momentary dipoles in surrounding molecules
momentary dipole: an uneven distribution of electrons around a
molecule, resulting in a temporary charge difference between its ends
They last for just
the instant that
the electrons are
not distributed
perfectly even.
2) London Force
• Fritz London also showed that momentary dipoles occurring in
adjacent molecules would result in an overall attraction
• The strength of the London force is directly related to the number
of electrons in the molecule, and inversely related to the distance
between the molecules.
▫ Increase electrons = Increase force (directly related)..
▫ Increase distance = Decrease force (inversely related)
2) London Force
• The key point is that:
▫ the more electrons a molecule has, the more easily
momentary dipoles will form, and the greater the effect
of the London force will be.
• London forces are present between all molecules, whether
any other type of attraction is present.
Why do we care about intermolecular forces?
 We can use Dipole-Dipole and London Forces to predict Boiling Points
Compound (at SATP)
Electrons
Boiling Point (°C)
CH4(g)
10
-164
SiH4(g)
18
-112
GeH4(g)
36
-89
SnH4(g)
54
-52
A higher boiling point temperature means more energy has to be
added, thus we assume the intermolecular forces are stronger.
(see Learning Tip pg. 109)
Remember (if all other factors are equal):
1) The more polar the molecule = The stronger the dipole-dipole force
2) Increase the number of electrons = Increase the strength of London Force
Example #1
• Use Intermolecular force theory to predict which of the
following hydrocarbons has the highest boiling point:
▫ methane (CH4), ethane (C2H6), propane (C3H8), butane (C4H10)
1)
Are the molecules polar or non-polar?
non-polar (no dipole-dipole force)
2) Which has more electrons?
butane: greatest # of e-’s = greatest London force
Check:
Alkane
Boiling Point (°C)
methane
-162
ethane
-89
propane
-42
butane
-0.5
Example #2
• Use Intermolecular force theory to predict which of the following
has the highest boiling point:
▫ bromine (Br2 )
1)
or
iodine monochloride (ICl)
Which has more electrons?
They are isoelectronic: have the same number of electrons (70 e-’s)
-Therefore the London force is the same (or nearly the same)
1)
Are the molecules polar or non-polar?
-Bromine is non-polar (has no dipole-dipole force; only London forces)
- Iodine monochloride is polar (has dipole-dipole forces and London forces)
- This extra attraction between ICl molecules produces a higher boiling
point
Check:
Substance
Electrons
Boiling Point (°C)
bromine
70
59
iodine monochloride
70
97
You cannot predict boiling points when:
 One molecule has a stronger dipole-dipole force and the other
has a stronger London force
 The two molecules differ significantly in shape
 The central atom of either molecule has an incomplete octet
Practice
 Pg. 109 #1-4
 4) a) Boron is stable with 6 valence e-
b) chloromethane (CH3Cl)
3) Hydrogen Bonding
• Occurs when a hydrogen atom bonded to a strongly
electronegative atom, (N, O and F) is attracted to a
lone pair of electrons in an adjacent molecule.
▫ Hydrogen nucleus (proton) is simultaneously attracted to two
pairs of electrons; one closer (in the same molecule) and one
further away (on the next molecule)
Why do you need a strongly
electronegative atom?
It pulls the hydrogen’s
electron away making it
“unshielded”, so the lone
pair on the other side can
come much closer
3) Hydrogen Bonding
• Hydrogen bonds are momentary attractive forces
between passing mobile molecules but are the
strongest of the intermolecular forces.
• Hydrogen bonds only act as continuous bonds
between molecules in solids, where the molecules are
moving slowly enough to be locked into position.
• Hydrogen force would have been a better name.
3) Hydrogen Bonding
 In ice, hydrogen bonds
between the molecules
result in a regular
hexagonal crystal
structure.
 The ···H– represents a
hydrogen nucleus
(proton) being shared
unequally between two
pairs of electrons
3) Hydrogen Bonding
• Do lakes freeze from the
bottom-up or the top-down?
• Top–down, because water is
unique in that its solid form
(ice) is less dense than its
liquid form. Why??
• The hydrogen bonds hold
water molecules in a
hexagonal lattice with open
space in the center, which
explains the low density
(mass/volume) of ice.
Hydrogen Bonding in DNA
• FYI: The double helix of the DNA
molecule owes its unique structure
largely to hydrogen bonding.
• The red bonds are hydrogen bonds.
• If the helix were held together by
covalent bonds, the DNA molecule
would not be able to unravel and
replicate and life could not
continue!!
Why do we care about intermolecular forces?
• Explains surface tension, shape of a
meniscus, volatility and capillary action
1)
Surface Tension
▫ Molecules within a liquid are attracted
by other molecules in all directions
equally, but right at the surface,
molecules are only attracted
downwards and sideways. This
means the net pull is downward so the
surface tends to stay intact
▫ The stronger the intermolecular force
the stronger the surface tension.
This shows water
adhering to the faucet
gaining mass until it is
stretched to a point
where the surface
tension can no longer
bind it to the faucet.
It then separates and
surface tension forms
the drop into a sphere.
Why do we care about intermolecular forces?
2) Capillary Action – due to adhesion (attraction between unlike
molecules) and cohesion (attraction of like molecules)
▫ The adhesion between water and glass is greater than the cohesion
between water molecules.
▫ The cohesion between mercury molecules is greater than the
adhesion between mercury and glass
Hg clip
Meniscus
In a sense, water is
pulled up the tube
by the
intermolecular
forces between
water and glass
Practice
 Pg. 117 # 1, 4, 5
 #1 – use pg. 99 table to determine polarity
 #1 – look for NH2, NH, OH2, OH, to determine if hydrogen bonding is
possible
 Ex. CH3CHOHCH3 will it have hydrogen bonding?