What I Absolutely Have to Know about IMFs to Survive the AP* Chemistry Exam

1. What I Absolutely Have to Know about IMFs to Survive the AP* Chemistry Exam

2. 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

3. 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…

4. 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 chargesare 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.

5. 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

6. 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.

7. symmetric distribution of electron density. instantaneous dipoles result from an unequal distribution of electrons

8. 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

9. opposite poles align

10. 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

11. opposite poles align

12. When comparing similar sized particles, the Hydrogen bonding forces > dipole-dipole forces > London dispersion forces.

13. 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.

14. 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.

15. 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.

18. 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.

19. 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

20. 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.

21. 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

22. 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 )

23. 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

24. Solution Formation and IMF’s • In order to dissolve a substance… • 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….

25. 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…

26. 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 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…

27. 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…

28. 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.

29. 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

30. Connections to Other Chapters Periodicity − especially electronegativity Bonding Lewis Structures and Geometry

31. 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 2005 B Question 8 (d & e) 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

32. MC Section Strategies • 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. • 75 questions. 90 minutes (Periodic Table for reference with no calculator, no equations)

33. 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 • 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.

34. Questions 5 and 6 refer to the following graph.

35. 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

36. 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.

37. 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.

38. 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.

39. 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

40. 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.

41. 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.

42. 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

43. 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.

44. 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

45. 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.

46. 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

47. 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.

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