What I Absolutely Have to Know about IMFs

to Survive the AP* Chemistry Exam

ObjectiveTo review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions on solids, liquid, phases changes and intermolecular forces.

StandardsIntermolecular 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 Bonding1. Binding forces a. Types: ionic, covalent, metallic, hydrogen bonding, van der Waals (including London dispersion forces)

II. States of MatterB. Liquids and Solids

1. Liquids and solids from the kinetic-molecular viewpoint

2. Phase diagrams of one-component systems3. Changes of state, including critical points and

triple points4. 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 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.

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 asINTRAmolecular 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 theforce of attraction, the stronger the London Dispersion forces.

Dipole-dipole forcesExist between polar molecules The molecules align themselves such that the opposite poles align. These dipoles result from the unequaldistribution of electron density in the molecule(as determined from the electronegativity ofeach atom in the molecule and its molecularshape). The larger the dipoles, the stronger the force of attraction between the two molecules, thestronger the Dipole-Dipole force

Hydrogen-bonding forces Occur in polar molecules in which a hydrogenatom is covalently bonded to a very electronegative element, specifically N, O, or F.

Just as in dipole-dipole interactions, the twomolecules are attracted and align themselvessuch that the opposite charge resulting from theunequal sharing of electrons form an attractiveinteraction

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 dipole-dipole forces…

Substances with Hydrogen bonding forces and/or dipole-dipole 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 LiquidsSurface TensionMolecules 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 ActionDescribed 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 PressureVapor 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 Solids1. Molecular SolidsConsist 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 SolidsConsist 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 SolidsConsist 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 SolidsConsist of individual atoms held together by weak London forces; tend to have low melting points. Example: Noble gases

5. Metallic SolidsConsist 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 havehigh melting points.

IMF’s and Solids ExampleIndicate 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 forces3) H2S(l) − Dipole-dipole forces; as the molecule is covalent and polar4) C(diamond) − Network covalent bonds; as this is a Network covalent solid5) Ca(NO3)2(s) − Ionic bonds; as this is an ionic solid

Heating and Cooling CurvesGraphically 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 toincrease KE… q = ΔHfusion

Solution Formation and IMF’sIn 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 solvent-solvent 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 quitesmall; 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 solvent-solvent 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 ExampleWhat 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.

IMPORTANTMelting 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 ConnectionsTopics 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 Questions2008 Question 6 2005 B Question 8 (d & e)2006 Question 62005 Question 7 (a & b)2004 Question 7 (a, b, & d)2003 Question 8 (b)2002 Question 6 (d)2001 Question 8

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 wireB 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 moleculeincreases.III. Boiling point increases as the number of hydrogen atoms in themolecule 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 moleculeincreases.III. Boiling point increases as the number of hydrogen atoms in themolecule 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 itexhibits 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−OHII. CH3−CH2−CH2−CH2−CH2−CH2−OHIII. 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−OHII. CH3−CH2−CH2−CH2−CH2−CH2−OHIII. 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 bondingII. 1 opportunity for hydrogen bondingIII. London dispersion forces onlyTherefore 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

ObjectiveTo review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions over the principles of bonding.StandardsBonding is addressed in the topic outline of the College Board AP Chemistry Course Description Guide as described below. I. Structure of MatterB. Chemical Bonding

1. Binding forcesa. Types: ionic, covalent, metallicb. Relationships to states, structure, and properties of matterc. Polarity of bonds, electronegativities

2. Molecular Modelsa. Lewis structuresb. Valence bond: hybridization of orbitals,resonance, sigma and pi bondsc. 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 ConnectionsThe 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 Questions2008 Question 5 (d−f) 2006 B Question 7 (b)2007 Question 6 (a−d) 2006 B Question 62006 Question 7 2005 B Question 8 (a−c)2005 Question 6 2002 B Question 62004 Question 8 (a & b)2003 Question 82002 Question 6 (c)2000 Question 7 (d)1999 Question 8

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 non-polar; 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.

IMPORTANTEnergy 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 electrostaticattraction is SO SO VERY strong.

Conductors of electricityin (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 Repulsive ForcesProton−electron attraction Electron−electron repulsion

Proton−proton repulsionWhen 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 periodWhy? 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 bondsHUGE, 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 Bond LengthSigma (σ) bonds are stronger Single Bonds are the longest than pi (π) bonds;

Pi bonds never exist alone Double bonds are shorter than single bonds

Combinations of σ and π are stronger Triple bonds are the shortest of all than σ alone

Resonance StructuresWhen 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!

PolarityLike 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 linearOctahedral 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 ofbonding 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 anglesyou 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 theirshape, 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 ChaptersPeriodicity − 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 Choice1. Which compound below contains exactly 2 pi bonds?(A) PCl3

(B) OF2

(C) HCN(D) NO3

-

(E) O3

Multiple Choice1. 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 momentII. 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 thus there will be a dipole moment.

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 bypyramidalII. T-shapedIII. 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 geometryE 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 hybridizedB 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 CO3

2-.

10. The molecule that contains only one lone pair in the entire moleculeD The nitrogen atom has a lone pair on it.

6 questions. 95 minutes (1st 3 w/ calculator, PT, equations sheet & SRP chart)