Drawing Lewis Structures

1. Drawing Lewis Structures • Find total # of valence e-. • Arrange atoms - singular atom is usually in the middle. • Form bonds between atoms (2 e-). • Distribute remaining e- to give each atom an octet (recall exceptions). • If there aren’t enough e- to go around, form double or triple bonds.

2. Polyatomic Ions • To find total # of valence e-: • Add 1e- for each negative charge • Subtract 1e- for each positive charge • Place brackets around the ion and label the charge

3. VSPER Model • 1. Draw the Lewis structure • 2. Count total number of electron pairs around the CENTRAL ATOM. Arrange them to minimize the electron shell repulsion • Note: a double or triple bond is counted as one bonding pair when predicting geometry • 3. Describe the molecular geometry in terms of the angular arrangement of the bonding pairs

4. Polarity of Molecules The "charge distribution" of a molecule is chosen by: • The shape of the molecule • The polarity of its bonds • A Polar Molecule: • The center of the overall negative charge on the molecule does not coincide with the center of overall positive charge on the molecule • The molecule can be oriented such that one end has a net negative charge and the other a net positive charge, i.e. the molecule is a dipole (di-pole means two poles i.e. positive and negative)

5. Polarity of Molecules • A Nonpolar molecule • Has no charges on the opposite ends of the molecule • Or, has charges of the same sign on the opposite ends of the molecule

6. Determining Polarity • Each polar bond in a polyatomic molecule will have a dipole • The overall dipole of the molecule will be the sum of the individual dipoles Vectors • In carbon dioxide the oxygens have a partial negative charge and the carbon a partial positive charge, the molecule has no dipole, due to the center of positive and negative charges being the same, canceling out • Water has a dipole and will orient in an electrical field

7. Ch. 12 - Liquids & Solids I. Intermolecular Forces(pages 411- 414)

8. Intramolecular and Intermolecular Forces • Covalent & Ionic bonds - the force that holds atoms together making molecules. • These are intramolecular forces. • There are also forces that cause molecules to attract each other. These are called intermolecular forces.

9. Definition of IMF • Attractive forces between molecules. • Much weaker than chemical bonds within molecules. • a.k.a. van der Waals forces

10. Types of IMF

11. London Dispersion Forces • The distribution of electrons around an atom, at a given instant in time, may not be symmetrical • Both electrons may be on 1 side of the nucleus • Atom would have an apparent dipole moment at that instant in time (i.e. aninstantaneousdipole moment) View animation online.

12. London Dispersion Forces • Due to electron repulsion, a temporary dipole on one atom can induce a similar dipole on a neighboring atom • This will cause the neighboring atoms to be attracted to one another • This is called the London dispersion force (or just dispersion force) • It is significant only when the atoms are close together

13. + - Dipole-Dipole Forces • Polar molecules attract one another when the partial positive charge on one molecule is near the partial negative charge on the other molecule View animation online.

14. Hydrogen Bonding • A hydrogen atom in a polar bond (e.g. H-N, H-O or H-F) can experience an attractive force with a neighboring highly electronegative molecule or ion which has an unshared pair of electrons • Hydrogen bonds are considered to be dipole-dipole type interactions • A bond between hydrogen and an electronegative atom such as N, O or F is quite polar:

15. Determining IMF • NCl3 • polar = london dispersion, dipole-dipole • CH4 • nonpolar = london dispersion • HF • H-F bond = london dispersion, dipole-dipole, hydrogen bonding

16. Ionic Bonds: Ion-Dipole • Involves an interaction between a charged ion and a polar molecule (i.e. a molecule with a dipole) • Cations are attracted to the negative end of a dipole • Anions are attracted to the positive end of a dipole • Ion-dipole forces are important in solutions of ionic substances in polar solvents (e.g. a salt in aqueous solvent)

17. Ch. 12 - Liquids & Solids II. Physical Properties(p. 415 - 424)

18. Physical Properites •  The physical properties of a substance depends upon its physical state. • Water vapor, water and ice have the same chemical properties, but the physical properties are different. • Covalent bondsdetermine: • 1.molecular shape • 2. bond energies • 3. chemical properties • Intermolecular forces(IM)(non-covalent bonds) influence physical properties of liquids and solids

19. LIQUIDS Stronger than in gases Y high N slower than in gases SOLIDS Very strong N high N extremely slow Liquids vs. Solids IMF Strength Fluid Density Compressible Diffusion

20. Kinetic Molecular Theory • The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the intermolecular forces • Kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance • Intermolecular forces try to draw the particles together

21. Kinetic Molecular Theory • Gases have weaker IM forces than liquids & solids due to the kinetic energy of the molecules being greater than any attractive forces between the molecules • Solids and liquids have particles that are fairly close to one another, and are thus called "condensed phases" to distinguish them from gases

22. Changing State Temperature • Heating and cooling can change the kinetic energy of the particles in a substance, so we can change the physical state of a substance by heating or cooling it. • Cooling a gas may change the state to a liquid • Cooling a liquid may change the state to a solid

23. Changing State Pressure • Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces • Increasing the pressure on a gas may change the state to a liquid • Increasing the pressure on a liquid may change the state to a solid

24. Liquid Properties • Surface Tension • attractive force between particles in a liquid that minimizes surface area • energy required to increase the surface area of a liquid by a unit amount

25. water mercury Liquid Properties • Capillary Action • attractive force between the surface of a liquid and the surface of a solid due to • Adhesive forces bind a substance to a surface, i.e. IM force binding different substances together

26. Liquid Properties • Viscosity:The resistance of a liquid to flow is called its viscosity • A measure of the ease with which molecules move past one another • It depends on the attractive force between the molecules • Decreases with increasing temperature - the increasing kinetic energy overcomes the attractive forces and molecules can more easily move past each other

27. decreasing m.p. Types of Solids • Crystalline - repeating geometric pattern • covalent network • metallic • ionic • covalent molecular • Amorphous - no geometric pattern

28. Ch. 12 - Liquids & Solids III. Changes of State(p. 425 – 430)

29. A. Phase Changes

30. Phase Changes • Evaporation • molecules at the surface gain enough energy to overcome IMF • Volatility • measure of evaporation rate • depends on temp & IMF

31. A. Phase Changes • Equilibrium • trapped molecules reach a balance between evaporation & condensation

32. temp v.p. IMF v.p. A. Phase Changes p.478 • Vapor Pressure • pressure of vapor above a liquid at equilibrium v.p. • depends on temp & IMF • directly related to volatility temp

33. Patm b.p. IMF b.p. A. Phase Changes • Boiling Point • temp at which v.p. of liquid equals external pressure • depends on Patm & IMF • Normal B.P. - b.p. at 1 atm

34. IMF m.p. A. Phase Changes • Melting Point • equal to freezing point • Which has a higher m.p.? • polar or nonpolar? • covalent or ionic? polar ionic

35. A. Phase Changes • Sublimation • solid  gas • v.p. of solid equals external pressure • EX: dry ice, mothballs, solid air fresheners

36. Phase Diagrams

37. Phase Diagrams • Equilibrium can exist not only between the liquid and vapor phase of a substance but also between the solid and liquid phases, and the solid and gas phases of a substance. • The "triple point" is the particular condition of temperature and pressure where all three physical states are in equilibrium • The Critical Point: The temperature above which the gas cannot be liquefied no matter how much pressure is applied (the kinetic energy simply is too great for attractive forces to overcome, regardless of the applied pressure)

38. Gas - KE  Boiling - PE  Liquid - KE  Melting - PE  Solid - KE  B. Heating Curves

39. B. Heating Curves • Temperature Change • change in KE (molecular motion) • depends on heat capacity • Heat Capacity • energy required to raise the temp of 1 gram of a substance by 1°C • Specific heat - different values for each substance

40. B. Heating Curves • Phase Change • change in PE (molecular arrangement) • temp remains constant • Heat of Fusion (Hfus) • energy required to melt 1 gram of a substance at its m.p. • Hfus of ice = 6.009 kJ/mol

41. B. Heating Curves • Heat of Vaporization (Hvap) • energy required to boil 1 gram of a substance at its b.p. • Hvap for water = 40.79 kJ/mol • usually larger than Hfus…why? • EX: sweating, steam burns

42. C. Properties of Water • Hfus of ice = 6.009 kJ/mol • Hvap for water = 40.79 kJ/mol • To calculate the amount of heat energy required to melt or freeze a substance, convert grams to moles and multiply by molar heat of fusion or vaporization

43. C. Properties of Water • How much heat energy is required to melt 25 grams of ice at 0oC to liquid water at a temperature of 0oC? ice 6.009 kJ 1 mol H2O 25 g H2O 1 mol H2O 18.02 g H2O = 8.3 kJ

44. C. Properties of Water • How much heat energy is required to change 500.0 grams of liquid water at 100oC to steam at 100oC? steam 40.79 kJ 1 mol H2O 500.0 g H2O 1 mol H2O 18.02 g H2O = 1132 kJ

45. How to Calculate Heat in Changes of State ΔHvap = 40.79 kJ/mol E D q = mCΔT CH2O(g) = 1.87 J/goC C q = mCΔT CH2O(l) = 4.184 J/goC ΔHfus = 6.009 kJ/mol B A q = mCΔT CH2O(s) = 2.108 J/goC

46. Calculating Heat Changes • Example problem: • How many joules are required to heat 10.0grams of ice from -20.0oC to steam at 110.oC? • A + B + C + D + E (make sure they are all same units!)

47. A: Energy Change as ice is heated GIVEN: M = 10.0 g CH2O(s) = 2.108 J/goC TF=0°C Ti = -20 °C Q = ? WORK: q = mCΔT (10.0 g)(2.108 J/goC)(0-(-20 °C)) q=(10.0 g)(2.108 J/goC)(20 °C) q = 421.6 Joules

48. B: Phase Change ΔHfusion : amount of heat energy required to melt a substance 6.009 kJ 1 mol H2O 1000 J 1 kJ 10 g H2O 1 mol H2O 18.02 g H2O ΔHfusion = 3334.5 Joules

49. C: Energy Change as liquid is heated GIVEN: M = 10.0 g CH2O(l) = 4.184 J/goC TF=100°C Ti = 0 °C Q = ? WORK: q = mCΔT (10.0 g)(4.184 J/goC)(100 – 0 °C) q=(10.0 g)(4.184 J/goC)(100 °C) q = 4184 Joules

50. D: Phase Change • ΔHvaporization : energy required to boil 1 gram of a substance at its b.p. 40.79 kJ 1 mol H2O 1000 J 1 kJ 10 g H2O 1 mol H2O 18.02 g H2O ΔHvapor = 22636 Joules