What I Absolutely Have to Know
about IMFs
to Survive the AP* Chemistry Exam
Objective
To review the student on the concepts, processes and problem solving
strategies necessary to successfully answer questions on solids, liquid,
phases changes and intermolecular forces.
Standards
Intermolecular forces are addressed in the topic outline of the College
Board AP* Chemistry Course Description Guide as described below.
I. Structure of Matter
B. Chemical Bonding
1. Binding forces a. Types: ionic, covalent, metallic, hydrogen bonding, van
der Waals (including London dispersion forces)
II. States of Matter
B. Liquids and Solids
1. Liquids and solids from the kinetic-molecular
viewpoint
2. Phase diagrams of one-component systems
3. Changes of state, including critical points and
triple points
4. Structure of solids; lattice energies
What I Absolutely Have to Know about IMFs &
Bonding to Survive the AP* Exam
 The following might indicate that the question deals
with intermolecular forces:
Boiling points; vapor pressure; melting points; network
solid; crystalline solids; metallic solids; sea of electrons;
delocalized electrons; triple point, critical
point/temp/press; sublimation; deposition;
condensation; boiling; melting; Freezing;
intermolecular forces; etc…
Key Formulas and Relationships
 When answering questions about melting points
of ionic compounds, justify your response using
Coulomb’s law:
Lattice Energy k = (Q 1 Q 2 ) / d
 If the charges are greater and distances similar,
the greater charged compound will have more ionion attraction; thus it will require more energy to
dissociate. This is useful in justifying melting points,
solubility, and lattice energy differences between
two ionic compounds.
Basic Types of Intermolecular Forces
 An intermolecular force (IMF) is an attraction that
occurs BETWEEN atoms or molecules.
 IMF’s are not bonds; bonds are known as
INTRAmolecular forces.
INTERmolecular forces are merely attractive forces
between molecules; think magnetic attraction!!
 Without these forces there would be no liquids or
solids − everything would be in the gas phase!
 On the AP exam, if the question is about phase
changes you must explain the phenomenon in terms of
IMF’s
London Dispersion forces
 Exist between atoms and/or non-polar compounds
(particles).
 In the example on the next slide six particles are shown in a
form depicting the symmetric distribution of electron density.
 In the second group of six, some of the particles have formed
instantaneous dipoles.
 The instantaneous dipoles result from an unequal distribution
of electrons.
 One particle with an instantaneous dipole will affect other
particles adjacent to it producing a short range attractive
interaction. The larger the particle, the more electrons, the more
polarizable its electron cloud, the stronger the
force of attraction, the stronger the London Dispersion forces.
Dipole-dipole forces
Exist between polar molecules
 The molecules align themselves such that the
opposite poles align.
 These dipoles result from the unequal
distribution of electron density in the molecule
(as determined from the electronegativity of
each atom in the molecule and its molecular
shape).
 The larger the dipoles, the stronger the force of
attraction between the two molecules, the
stronger the Dipole-Dipole force
Hydrogen-bonding forces
 Occur in polar molecules in which a hydrogen
atom is covalently bonded to a very electronegative
element, specifically N, O, or F.
 Just as in dipole-dipole interactions, the two
molecules are attracted and align themselves
such that the opposite charge resulting from the
unequal sharing of electrons form an attractive
interaction
When comparing similar sized
particles,
the Hydrogen bonding forces >
dipole-dipole forces > London
dispersion forces.
WATCH OUT: when non-polar substances with only
London dispersion forces have a considerably larger (thus
very polarizable) electron cloud than the polar molecules,
the London dispersion forces can be quite substantial and
can be stronger than Hydrogen bonding forces or dipoledipole forces…
Substances with Hydrogen bonding forces and/or dipoledipole forces also have London dispersion forces; however,
the London dispersion forces are not very significant
compared to the strength of the Hydrogen bonding forces
and/or the dipole-dipole forces.
Properties of Liquids
Surface Tension
Molecules in the interior of a liquid are attracted by the molecules
surrounding it; whereas a molecule at the surface of a liquid is
attracted only by the molecules below it and on each side. This
leads to an increase in its surface area (polar molecules). High
surface tension indicates strong IMF’s.
Capillary Action
Described by spontaneous rising of a liquid in a narrow tube.
Adhesive forces between the molecules and the glass overcome the
cohesive forces (IMF’s) between molecules themselves. Water has a
higher attraction for glass than itself so its meniscus is inverted or
concave, while Hg has a higher attraction for other Hg molecules,
thus its meniscus is convex.
Viscosity Molecules with larger IMFs and more complexity have
more resistance to flow, i.e. they are more viscous.
Vapor Pressure
Vapor pressure is the pressure resulting from the particles of a substance
that exist in the vapor phase above the liquid in a closed container. The
weaker the IMF, the higher the vapor pressure of the liquid will be. Why?
Because the substance will more easily overcome those IMF’s and break
away into the vapor phase, increasing the number of molecules that are in
the vapor phase (at that temperature), thus increasing the pressure above
the liquid.
Boiling point is the temperature at which the vapor pressure of a liquid
equals the atmospheric pressure; at this point all the molecules have enough
energy to overcome the IMF’s and thus enter the vapor phase.
The normal boiling point is the temperature at which the vapor pressure
of the liquid equals 1 atmosphere.
 Please note: As you increase the temperature of the liquid, the vapor
pressure of the liquid increases at an increasing rate, again due to more
molecules having enough energy in being able to overcome the IMF’s and
move into the vapor phase.
Five Types of Solids
1. Molecular Solids
Consist of molecules with London dispersion, dipole-dipole or Hydrogen
bonding intermolecular forces. The solid contains arrangements of atoms
or molecules that are organized in an orderly three-dimensional pattern
and are typically soft and have relatively low melting points. Example:
Solid H2O
2. Ionic Solids
Consist of cations and anions distributed throughout in an orderly three
dimensional pattern − a crystal lattice. Ionic solids are more complicated
in their structure but can be thought of as an orderly pattern of one ion,
generally the anion, with cations positioned in 'holes' between the anions.
The occupation of these 'holes' depends on the formula of the ionic
compound. Ionic solids are characterized as hard, brittle substances and
have high melting points. The high melting points are the result of the
electrostatic attractions of the ionic bonds, which are stronger than the
intermolecular forces for molecular solids.
3. Covalent Network Solids
Consist of atoms that are held together in large networks
containing extended covalent bonds. Example: Quartz (SiO2);
consists of SiO2 groups that are covalently bonded to other
SiO2 groups in all three dimensions. Other common
examples are carbon in the form of graphite and diamond.
4. Atomic Solids
Consist of individual atoms held together by weak London
forces; tend to have low melting points. Example: Noble
gases
5. Metallic Solids
Consist of a group of metallic atom’s nuclei that are
surrounded by the freely moving electrons of those
atoms. This is explained in terms of band theory, which
states that the atomic orbitals of these atoms mix to
form a range of molecular orbitals that encompass all
energy levels. Since these empty orbitals are
degenerate (similar energy level), the electrons from
the atoms can move freely from orbital to orbital,
throughout the metal. This movement of electrons
explains why metals are good conductors of heat and
electricity and why they are shiny. Most metals have
high melting points.
IMF’s and Solids Example
Indicate the attractive forces found between the molecules in each
of the following substances. If the substance is a solid, identify the
type of solid it represents.
1)CH3OH(l) 2)Xe(l) 3)H2S(l) 4)C(diamond) 5)Ca(NO3)2(s)
1) CH3OH(l) − Hydrogen bonding forces; as the molecule is
polar covalent and has H bonded to oxygen.
2) Xe(l) − London dispersion forces
3) H2S(l) − Dipole-dipole forces; as the molecule is covalent
and polar
4) C(diamond) − Network covalent bonds; as this is a
Network covalent solid
5) Ca(NO3)2(s) − Ionic bonds; as this is an ionic solid
Heating and Cooling Curves
Graphically represents the relationship of a pure substance in terms of how the
temperature changes over time…
Different substances have different phase change points (melting/freezing and
vaporization/condensation) but the shapes of their heating and/or cooling
curves are very similar.
A to B represents the substance as a solid (q = mCΔT )
B to C represents the process of melting/freezing (ΔHfusion): at this point the energy is overcoming the
IMF’s of the solid so the substance can change from a solid to a liquid (q = ΔHfusion)
C to D represents the substances as a liquid (q = mCΔT )
D to E represents the process of vaporization/condensation (ΔHvap): at this point the energy is
overcoming the IMF’s of the molecules so the substance can move to the vapor phase (q = ΔHvap)
E to F and beyond represents the substance in the vapor/gas phase (q = mCΔT )
Remember !!
…on the slopes (lines going up or down) the
substance is all in the same phase… q = mCΔT
During the phase changes (flat lines) the
energy is being used to overcome IMF’s
(increasing potential energy), not to
increase KE… q = ΔHfusion
Solution Formation and IMF’s
In order to dissolve a substance…
1. Overcome (requires energy)
 Solute-solute IMF’s
Solvent-solvent IMF’s
2. Form solute-solvent attractive forces upon mixing
(releases energy)
Never use “like dissolves like” on the AP exam.
EXPLAIN in terms of structure, IMF’s, and
energy….
Polar/Polar Solution Formation:
The ΔH required to overcome IMF’s in both the polar/ionic solute
particles and the polar water molecules is quite large; however, the
ΔH released due to the interactions between the polar/ionic solute
particles and the polar water molecules is very large. Thus
polar/ionic particles dissolve readily in polar solvents such as water.
In other words, strong IMF’s between solute-solute and solventsolvent must be overcome, which requires a significant input of
energy. However, the solute and solvent particles, as they mix, form
strong interactions between each other, releasing an equal amount
of energy. Simply speaking, the solute can dissolve because it gets
as much energy “back” from the interactions as was required to
overcome the IMF’s…
Non-polar/Non-polar Solution Formation:
The ΔH required to overcome IMF’s in both the non-polar solute
particles and the non-polar solvent molecules is quite
small; the ΔH released due to the interactions between the non-polar
solute particles and the non-polar solvent molecules is also small.
Thus non-polar solute particles dissolve readily in non-polar solvents
such as benzene.
In other words, only weak IMF’s between solute-solute and solventsolvent particles must be overcome. When the solute and solvent
particles mix, they form weak interactions between each other,
releasing an equal amount of energy.
Simply speaking, the solute can dissolve because it gets enough
energy “back” from the interactions as was required to overcome
the IMF’s…
Non-polar/Polar Solution Formation:
The ΔH required to overcome IMF’s in both the polar/ionic
solute particles is very large but for the non-polar solvent
molecules, the ΔH required is quite small; therefore the ΔH
released due to the interactions between the polar/ionic
solute particles and the non-polar solvent molecules is very
small. The solute CANNOT dissolve because the energy
required to overcome the initial IMF’s is not provided by the
solute-solvent interactions…
Solution Formation Example
What solution is more likely to dissolve the alkane C20H42; liquid
methanol (CH3OH) or liquid hexane (C6H14)? Justify your answer.
Alkane compounds are non-polar and exhibit only London dispersion
forces.
Liquid methanol is polar and exhibits Hydrogen bonding forces.
Liquid hexane is non-polar and exhibits only London dispersion forces.
Thus the alkane C 20H 42 will more likely dissolve in the hexane.
The interactions between the alkane molecules and the methanol
molecules are not energetically strong enough to overcome the
intermolecular attractive forces between the methanol molecules, thus
they will not readily dissolve.
However, the interactions between the alkane molecules and the
hexane molecules are energetic enough to overcome the weaker
intermolecular attractive forces between the hexane molecules allowing
them to dissolve.
IMPORTANT
Melting points, vapor pressure, boiling points (really all
phase change processes) are all about the strength of the
intermolecular attractive forces. The physical change that
accompanies any of these processes will require particles
to overcome the attractive forces holding them together.
REMEMBER: If you are asked to compare 2 substances
(higher boiling point, lower vapor pressure, which melts
first, etc…) it’s all about Coulomb’s Law for ionic
substances and IMF’s for molecular compounds
Connections to Other Chapters
Periodicity − especially electronegativity
Bonding
Lewis Structures and Geometry
AP Chemistry Exam Connections
Topics relating to intermolecular forces, solids, and liquids are tested
every year on the multiple choice and in most years on the free
response portion of the exam. The list below identifies free response
questions that have been previously asked over intermolecular
forces, solids, and liquids. These questions are available from the
College Board and can be downloaded free of charge from AP
Central. http://apcentral.collegeboard.com.
Free Response Questions
2008 Question 6
2006 Question 6
2005 Question 7 (a & b)
2004 Question 7 (a, b, & d)
2003 Question 8 (b)
2002 Question 6 (d)
2001 Question 8
2005 B Question 8 (d & e)
MC Section Strategies
 75 questions. 90 minutes (Periodic Table for reference with no
calculator, no equations)
No heirarchy of difficulty
Write on Test
Answer at rate of 30 sec/ques, 20 ques in 10 min,
and all 75 in 40 min.
If you cannot answer in 30 secs, mark Y for those
you know, but need more time and N for those you
don’t know how to solve.
Solve the Y questions in next 40 min.
Can you eliminate 1+ answers on N
question?...guess.
MC Questions 1-4 refer to the following types of solids.
(A) An ionic solid
(B) A metallic solid
(C) A network solid with covalent bonds
(D) A molecular solid with hydrogen bonding forces
(E) A molecular solid with London dispersion forces
1. Pt wire
2 .CO2 (s) 3. CH3OH(s)
4. SiO2,(s)
1. Pt wire
B Platinum is a pure metal and exhibits metallic bonding.
2. CO2 (s)
E Carbon dioxide is a molecular compound and a non-polar molecule; thus its
intermolecular attractive forces are London dispersion forces.
3. CH3OH(s)
D Solid methanol is a molecular compound that is a polar molecule; the H atoms
bonded to O allow it to form hydrogen bonding forces.
4. SiO2,(s)
C Silicon dioxide or quartz is a network solid with covalent bonds that are really
made of SiO4 tetrahedra with 2 shared oxygen atoms rather than discrete SiO2
molecules.
Questions 5 and 6 refer to the following graph.
5. Which of the following statements best account for the general
trend represented by the graph above ?
I. Boiling point increases as group number of the hydride increases.
II. Boiling point increases as the number of electrons in the molecule
increases.
III. Boiling point increases as the number of hydrogen atoms in the
molecule increases.
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
5. Which of the following statements best account for the general trend
represented by the graph?
I. Boiling point increases as group number of the hydride increases.
II. Boiling point increases as the number of electrons in the molecule
increases.
III. Boiling point increases as the number of hydrogen atoms in the
molecule increases.
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
D Generally, within any group on the periodic table the boiling
point of a nonmetal hydride increases with molar mass since the
electron cloud becomes more polarizable.
6. Which of the following statements best accounts for
the difference in boiling points between water and
methane?
(A) Methane is nonpolar molecule.
(B) Water exhibits hydrogen bonding.
(C) Methane exhibits hydrogen bonding.
(D) Water has a higher molar mass than methane.
(E) At room temperature, methane is a gas and water is a
liquid.
6. Which of the following statements best accounts for
the difference in boiling points between water and
methane?
(A) Methane is nonpolar molecule.
(B) Water exhibits hydrogen bonding.
(C) Methane exhibits hydrogen bonding.
(D) Water has a higher molar mass than methane.
(E) At room temperature, methane is a gas and water is a
liquid.
B Generally, within any group on the periodic table the boiling
point of a nonmetal hydride increases with molar mass since the
electron cloud becomes more polarizable. Water, ammonia and
hydrogen fluoride defy this trend due to the presence of
hydrogen bonding.
7. What type of attractive force is being
overcome when liquid oxygen boils at 90 K?
(A) Ionic bonds
(B) Covalent bonds
(C) Hydrogen bonds
(D) Dipole-dipole forces
(E) London dispersion forces
7. What type of attractive force is being
overcome when liquid oxygen boils at 90 K?
(A) Ionic bonds
(B) Covalent bonds
(C) Hydrogen bonds
(D) Dipole-dipole forces
(E) London dispersion forces
E The oxygen molecule is a non-polar molecule.
The only attractive forces it
exhibits are London dispersion forces.
8. A slush of liquid water and ice is at equilibrium at 0ºC when a
small amount of heat is removed so that more ice is formed. The
system will experience
(A) an increase in temperature since the formation of ice from water
is an exothermic process
(B) a change in vapor pressure above the mixture of ice and water
(C) a small decrease in temperature since the formation of ice
produces a colder system
(D) a large decrease in temperature since more ice will form
(E) no change in temperature
E The temperature will not change
because the system in equilibrium.
9. When a substance is converted from a liquid to a gas at its
normal boiling point, which of the following is/are correct?
I. The potential energy of the system increases.
II. The distance between the molecules increases.
III. The vapor pressure of the liquid is less than 1 atm.
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II and III
9. When a substance is converted from a liquid to a gas at its
normal boiling point, which of the following is/are correct?
I. The potential energy of the system increases.
II. The distance between the molecules increases.
III. The vapor pressure of the liquid is less than 1 atm.
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II and III
C As energy is added to the liquid this energy is being used to
break the intermolecular forces, which represents an increase in
potential energy. The gas molecules are farther apart than those
in the liquid state and at the substance’s normal boiling point the
vapor pressure of the liquid is 1 atm.
10. The temperature and pressure at which all three
states of matter can exist in equilibrium is called the
(A) critical point
(B) boiling point
(C) triple point
(D) equilibrium point
(E) flash point
10. The temperature and pressure at which all three
states of matter can exist in equilibrium is called the
(A) critical point
(B) boiling point
(C) triple point
(D) equilibrium point
(E) flash point
C By definition, there is only one set of pressure and
temperature conditions at which all three states of
matter can exist in equilibrium and that is known as the
triple point.
11. Which of the following sequences correctly lists the
compounds below in decreasing order of their solubility in
water?
I. HO−CH2−CH2−CH2−CH2−CH2−OH
II. CH3−CH2−CH2−CH2−CH2−CH2−OH
III. CH3−CH2−CH2−CH2−CH2−CH2−CH3
(A) I > II > III
(B) II > I > III
(C) II > III > I
(D) III > I > II
(E) III > II > I
11. Which of the following sequences correctly lists the
compounds below in decreasing order of their solubility in
water?
I. HO−CH2−CH2−CH2−CH2−CH2−OH
II. CH3−CH2−CH2−CH2−CH2−CH2−OH
III. CH3−CH2−CH2−CH2−CH2−CH2−CH3
(A) I > II > III
(B) II > I > III
(C) II > III > I
(D) III > I > II
(E) III > II > I
A I. 2 opportunities for hydrogen bonding
II. 1 opportunity for hydrogen bonding
III. London dispersion forces only
Therefore I is the most soluble and III the least.
6 questions. 95 minutes
(1st 3 w/ calculator, PT, equations sheet
& SRP chart)
AP PREP Sessions
Mr. J. Hnatow (pronounced “NATO”)
….A High School AP CHEMISTRY TEACHER
who helped to write the test and who
grades the tests!!
ANYTHING YOU LEARN TODAY
OR ON YOUR OWN WILL HELP
YOU TREMENDOUSLY!!!
STRIVE TO GET BETTER!
MULTIPLE-CHOICE
AND
FREE-RESPONSE QUESTIONS IN
PREPARATION FOR THE AP CHEMISTRY
EXAMINATION.
USE THIS BOOK: Go through
all of it B4 the May Exam!!!
…ON YOUR OWN (IF YOUR TEACHER DOESN’T!)
It has most excellent information about the
types of exam questions, and has practice
exams!!!!!
What I Absolutely Have to Know
about Bonding
to Survive the AP* Exam
Objective
To review the student on the concepts, processes and problem solving strategies
necessary to successfully answer questions over the principles of bonding.
Standards
Bonding is addressed in the topic outline of the College Board AP Chemistry Course
Description Guide as described below.
I. Structure of Matter
B. Chemical Bonding
1. Binding forces
a. Types: ionic, covalent, metallic
b. Relationships to states, structure, and properties of matter
c. Polarity of bonds, electronegativities
2. Molecular Models
a. Lewis structures
b. Valence bond: hybridization of orbitals,resonance, sigma
and pi bonds
c. VSEPR
3. Geometry of molecules and ions, structural isomerism of
simple organic molecules and coordination complexes;
dipole moments of molecules; relation of properties to
structure
AP Chemistry Exam Connections
The principles of bonding are tested every year on the multiple choice and
in most years on the free response portion of the exam. The list below
identifies free response questions that have been previously asked over
bonding. These questions are available from the College Board and an be
downloaded free of charge from AP Central
http://apcentral.collegeboard.com.
Free Response Questions
2008 Question 5 (d−f)
2007 Question 6 (a−d)
2006 Question 7
2005 Question 6
2004 Question 8 (a & b)
2003 Question 8
2002 Question 6 (c)
2000 Question 7 (d)
1999 Question 8
2006 B Question 7 (b)
2006 B Question 6
2005 B Question 8 (a−c)
2002 B Question 6
What I Absolutely Have to Know to Survive the AP
Exam
The following might indicate that the question deals
with bonding and/or molecular geometry:
Electronic or molecular geometry; type of bond;
VSEPR; Lewis diagram; hybridization; polar or nonpolar; dipole moment; shape of the molecule; bond
angle; resonance; bond length/strength; sigma/pi
bonds
BOND, CHEMICAL BOND…
The attractive forces that hold groups of atoms
together are called chemical bonds.
The system is achieving the lowest possible energy
state by bonding.
IMPORTANT
Energy is RELEASED when a bond is formed Energy is
REQUIRED to break a bond
TYPES OF CHEMICAL BONDS
Ionic
Electrostatic attraction between ions
Typically high melting & boiling points
Usually in the solid state since the electrostatic
attraction is SO SO VERY strong.
Conductors of electricity
in (aq) or (l) states
Covalent
Electrons are shared by nuclei;
careful, sharing is hardly ever equal!
Made of molecules (can be solids, liquids, or gases)
have low melting & boiling points
Poor conductors of electricity
Ionic Bonding
 The final result of ionic bonding is a solid, regular array of cations
and anions called a crystal lattice.
 Coulombs’ law indicates the more highly charged the ions the
stronger the attraction; and smaller ions will form stronger
attractions than larger ions.
Attractive Forces
Proton−electron attraction
Repulsive Forces
Electron−electron repulsion
Proton−proton repulsion
When the attractive forces offset the repulsive forces, the energy
of the two atoms decreases and a bond is formed. This happens
when attraction outweighs repulsion!
 Remember, nature is striving for a LOWER ENERGY STATE
THINK GOLDILOCKS!
If the atoms are too close together the repulsive forces outweigh the
attractive forces and the atoms do not reach a lower energy state;
therefore they DO NOT form a chemical bond!
If the atoms are too far apart the two atoms do not effectively interact;
i.e. the attractive forces are not sufficient enough to reach a lower energy
state; therefore they DO NOT form a chemical bond!
 If the atoms are just right the attractive forces offset the repulsive
forces and the atoms reach a lower energy state; therefore they form a
chemical bond! The distance between the 2 nuclei where the energy is at a
minimum between the two nuclei represents the bond length.
Bond Polarity and Electronegativity
 Remember, the electrons are seldom shared
50−50
 The attraction or “pull” on the bonded electron
pair is what determines polarity(electronegativity)
 When 2 nuclei are the same (or have the same
electronegativity), the sharing is equal − a
NONPOLAR COVALENT BOND. Why? The
two
atoms have the same attraction for the
bonding
electrons.
 When the 2 nuclei are different the electrons are
typically not shared equally, One atom’s
nucleus
has a greater attraction on the
bonding
electrons, which distorts the electron
cloud, creating slight +/− poles; this is a POLAR
COVALENT BOND.
 When the electrons are shared unequally to a
much greater extent, it is an IONIC BOND.
Covalent Bonding and Lewis Structures
 Most compounds are covalently bonded, especially carbon compounds.
 Electron pairs are assumed to be localized on a particular atom [lone
pairs] or in the space between two atoms [bonding pairs].
 Typically atoms bond covalently in a manner that allows for an complete
octet (8) of electrons in the atom’s valence shell
There are exceptions:
< 8 − H can have a maximum of 2 electrons (can form only one bond); BeH2,
has only 4 valence electrons around Be (only two bonds); other
Boron compounds of the type BX3, have only 6 valence electrons
(three bonds)!
> 8 − can only happen if the central atom is from the 3rd or higher period
Why? d orbitals are needed for the expansion − the combination of 1
s orbital and 3 p orbitals provides the four bonding sites that make
up the octet rule; the 2 additional “d”orbitals allow for expansion to
either 5 or 6 bonding sites.
Covalent Bonding and Lewis Structures (cont.)
 Odd-electron compounds − A few stable compounds have
an odd number of valence electrons; thus cannot obey
the octet rule. NO, NO2, and ClO2 are common
examples.
 Lewis Structures describe the valence electron arrangement
and allow the geometry of the molecule to be predicted,
as well as the polarity of the molecule as a whole, and
provide a description of the type of atomic orbitals
“blended” by the atoms in order to share electrons or to
hold lone pairs [hybridization].
 VSEPR − Lewis structures should be drawn geometrically so
that the molecular shape minimizes electron pair
repulsions
Single and Multiple Bonds
 Single bond − one pair of electrons shared; called a sigma
(σ) bond
 MULTIPLE BONDS ARE MOST OFTEN FORMED by
C,N,O,P and S ATOMS — say “C-NOPS”
 Double bond − two pairs of electrons shared: one σ bond
and one π bond
 Triple bond − three pairs of electrons shared: one σ bond
and two π bonds
HUGE, CRITICAL, IMPORTANT!
 Multiple bonds increase the electron density between two
nuclei
 This decreases the repulsions between the 2 nuclei and the
added electrons enhance the attractions between both
nuclei and the increased electron density—
 The nuclei can move closer together; thus the bond length
is shorter for a double than a single, and triple is
shortest of all!
Bond Strength
Sigma (σ) bonds are stronger
than pi (π) bonds;
Bond Length
Single Bonds are the longest
Pi bonds never exist alone
Double bonds are shorter than
single bonds
Combinations of σ and π are stronger
than σ alone
Triple bonds are the shortest of all
Resonance Structures
When a molecule has equally different positions where a double or triple bond can
be placed, you must draw resonance structures. In terms of “bond properties” it
is as if the multiple bond “resonates” between all the possible positions, giving
the bond length and bond strengths a value somewhere between that of a pure
single or double bond.
The bonds are more equivalent to a “bond and ½” in terms of length and strength.
We use the double edged arrows to indicate resonance. We also bracket the
structures just as we do for polyatomic ions.
Molecular and Structural Pair (Electronic) Geometry
 Molecular geometry − the arrangement in space of the atoms bonded to a central
atom; not necessarily the same as the structural pair geometry
 Since lone pair electrons experience an attraction or “pull” from only one nucleus
(as opposed to two nuclei for bonding electron pairs), the lone pairs have a
greater repulsive effect and take up more space around an atom.
 Each lone pair or bonding pair repels all other lone pairs and bond pairs − i.e. they
try “to avoid” each other, making as wide an angle as possible.
HUGE CONCEPT:
 When lone pairs are NOT present, the molecular and structural pair (or electronic)
geometry is the same
 When lone pairs are present, the structural pair (or electronic) geometry is different!
 Lone pair electrons in molecules with EXPANDED VALENCE
 Two possible arrangements exist; the lone pairs should be found as far apart as
possible.
Axial — lone pairs are located “top and bottom”
Equatorial — shared pairs surround the central atom.
Lone pairs prefer maximum separation—use this in your determinations!
Polarity
Like with bonds, molecules can be polar or non-polar; i.e. they exhibit an
electron cloud that is not distributed symmetrically about the molecule. This
creates the presence of a dipole moment − one side of the molecule has more
electrons than the other, thus that side is more negative than the other.
Determining molecular polarity
 If the terminal atoms are different, the molecule IS polar, regardless of its
molecular shape.
 If lone pairs are present on the central atom the molecule is typically polar.
 There are a couple of exceptions to this.
Trigonal bypyramidal structures that have a molecular shape that is
linear
Octahedral structures that have a molecular shape that is square
planar
 Why do lone pair electrons on the central atom make the molecule polar?
Their presence creates increased electron repulsion and thus, an
unequal distribution of electron density.
Hybridization
 Description of the atomic orbitals “blended” by the atoms
to share electrons or hold lone pairs.
 The central atom forms these hybrid orbitals with the
terminal atoms depending on the number of
bonding pairs and lone pairs about the central atom.
 The valence orbitals used for bonds are the 1−s orbital,
3−p orbitals, and if the molecule is expanded, 2 of the
d orbitals.
Molecular Shapes
 You MUST KNOW the molecular shapes!
 It all revolves around what is on the central atom!
 The charts on the following pages give examples of the
shapes, names, hybridizations, and bond angles
you must know.
Key Formulas and Relationships
When answering questions about Ionic bond strength,
justify your response using Coulomb’s law:
Lattice Energy k = (Q 1 Q 2 ) / d
If the charges are greater and distances similar, the greater
charged compound will have more ion-ion attraction; thus it
will require more energy to dissociate. This is useful in
justifying melting points, solubility, and lattice energy
differences between two ionic compounds.
Key Concepts and Phrases
Be able to determine what type of bonding is present by
looking at the chemical formula; Never ever forget that ionic
bonds are merely electrostatic attractions (forces)
Be able to sketch Lewis structures and determine their
shape, bond angle, polarity, and hybridization
You MUST memorize the structural pair (electronic) and
molecular geometries
Breaking bonds takes in energy (endothermic; +ΔH) Forming
bonds RELEASES energy (exothermic; -ΔH)
Connections to Other Chapters
Periodicity − especially electronegativity
Atomic Structure − especially understanding orbitals and
valence electrons
What to write, or not to write: That is the question…
 NEVER use the term “happy” when referring to atoms or
molecules. Everything is about energy, not emotion!!
 When justifying polarity, indicate there is either “an
asymmetrical distribution of electron density”, “unequal
distribution of charge on the molecule”, or “presence of a
dipole moment”… DO NOT refer to the molecule as being
asymmetrical or unbalanced.
 When lone pairs are present on the central atom, they will
distort the “expected” bond angle. Explain your response
by indicating… lone pairs have more repulsive forces
compared to bonding pairs since they are only attracted
to one nuclei.
What to write, or not to write: That is the question…(cont.)
 When discussing “expanded valence” recall only the
elements in Period 3 and below can expand their valence
shell. Be sure to explain that elements that do not have
“d” sublevels available (elements in Periods 1 and 2)
cannot have an expanded octet. They need d orbitals to
have sp3d (trigonal bipyramidal) and sp3d2
(octahedral)arrangements.
 In the trigonal bipyramidal structure, when lone pairs are
present on the central atom, they will locate themselves
on the equatorial plane (around the triangle) because
they best minimize repulsion at 120°.
 In the octahedral structure, when lone pairs are present on
the central atom, they will locate themselves on the axial
position (on “top” and “bottom”).
And finally……..
ALWAYS ALWAYS ALWAYS DRAW THE
LEWIS STRUCTURE
Even if it’s not “required” it helps answer
many questions; such as shape, bond angle,
polarity, type of IMF, etc…
MC Section Strategies
 75 questions. 90 minutes (Periodic Table for reference with no calculator, no
equations)
No heirarchy of difficulty
Write on Test
Answer at rate of 30 sec/ques, 20 ques in 10 min,
and all 75 in 40 min.
If you cannot answer in 30 secs, mark Y for those
you know, but need more time and N for those you
don’t know how to solve.
Solve the Y questions in next 40 min.
Can you eliminate 2+ answers on N
question?...guess.
Multiple Choice
1. Which compound below contains exactly 2 pi
bonds?
(A) PCl3
(B) OF2
(C) HCN
(D) NO3(E) O3
Multiple Choice
1. Which compound below contains exactly 2 pi
bonds?
(A) PCl3
(B) OF2
(C) HCN
(D) NO3(E) O3
C HCN This molecule contains a single and a triple
bond; 2 sigma and 2 pi bonds.
2. Identify which substance(s) below is/are nonpolar?
I. PH3 II. SF6 III. XeF4 IV. HCN
(A) I only
(B) II only
(C) II and III only
(D) II and IV only
(E) I, II, III, and IV
2. Identify which substance(s) below is/are nonpolar?
I. PH3 II. SF6 III. XeF4 IV. HCN
(A) I only
(B) II only
(C) II and III only
(D) II and IV only
(E) I, II, III, and IV
C I. PH3; this structure is pyramidal, the lone pair of electrons
determines there will be a dipole moment
II. SF6; this structure is octahedral; since all 6 bonds cancel each
other there is no net dipole moment (i.e. the electron
density is symmetrically distributed).
III. XeF4; this structure is square planar, even though it has 2 lone
electron pairs, they and the 4 bonds cancel each other so
there is no net dipole moment (i.e. the electron density is
symmetrically distributed).
IV. HCN; this structure is linear, but has different terminal atoms
there will be a dipole moment.
thus
3. I3- is a linear compound. The hybridization of the
central iodine atom is
(A) sp
(B) sp2
(C) sp3
(D) sp3d
(E) sp3d2
D Since the central iodine atom has 5 regions of electron density
(2 atoms and 3 lone electron pairs), the hybridization is sp3d
4. How many pi bonds are found in the structure above?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
4. How many pi bonds are found in the structure above?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
D The structure has 6 sigma and 3 pi bonds.
5. Determine the geometry of the molecule at carbon #2?
(A) linear
(B) trigonal planar
(C) trigonal pyramidal
(D) bent
(E) tetrahedral
B The #2 carbon has three areas of electron density, thus it is trigonal
planar about that carbon atom.
6. All the following substances are linear EXCEPT:
(A) BeF2
(B) O2
(C) SO2
(D) Cl3(E) CS2
6. All the following substances are linear EXCEPT:
(A) BeF2
(B) O2
(C) SO2
(D) Cl3(E) CS2
C The SO2 molecule is electronically trigonal planar
and its molecular shape is bent.
7. The central atom of a certain compound is sp3d hybridized. Identify all
possible geometric shapes for this structure.
I. Trigonal bypyramidal
II. T-shaped
III. See-saw
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III
E If the central atom is hybridized sp3d then its structural pair
(electronic) geometry is trigonal bypyramidal (5 areas of electron
density). Therefore, its molecular geometries could be trigonal
bypyramidal, see-saw, T-shaped, and linear.
Questions 8−10 refer to the species below.
(A) N2H2F2
(B) CO32(C) CH3OH
(D) CH3NH2
(E) ClF5
8. The molecule that has a square pyramidal geometry
9. The molecule that contains one atom that is sp2 hybridized
10. The molecule that contains only one lone pair in the entire
molecule
Questions 8−10 refer to the species below.
(A) N2H2F2
(B) CO32(C) CH3OH
(D) CH3NH2
(E) ClF5
8. The molecule that has a square pyramidal geometry
E To be square pyramidal the molecule must have 5 atoms and one lone pair about
the central atom. Only E is possible.
9. The molecule that contains one atom that is sp2 hybridized
B If the central atom is hybridized sp2, then its structural pair (electronic) geometry
is trigonal planar (3 areas of electron density). This means the atom either has 3
bonded atoms or 2 bonded atoms and a lone pair of electrons. The only choice is
CO32-.
10. The molecule that contains only one lone pair in the entire molecule
D The nitrogen atom has a lone pair on it.
6 questions. 95 minutes
(1st 3 w/ calculator, PT, equations sheet
& SRP chart)