Chemistry: Atoms First Julia Burdge & Jason Overby

1. Chapter 12 Chemistry: Atoms First Julia Burdge & Jason Overby Intermolecular Forces and the Physical Properties of Liquids and Solids Kent L. McCorkle Cosumnes River College Sacramento, CA

2. 12 Intermolecular Forces and the Physical properties of Liquids and Solids 12.1 Intermolecular Forces Dipole-Dipole Interactions Hydrogen Bonding Dispersion Forces Ion-Dipole Interactions 12.2 Properties of Liquids Surface Tension Viscosity Vapor Pressure 12.3 Crystal Structure Unit Cells Packing Spheres Closest Packing 12.4 Types of Crystals Ionic Crystals Covalent Crystals Molecular Crystals Metallic Crystals 12.5 Amorphous Solids 12.6 Phase Changes Liquid-Vapor Phase Transition Solid-Liquid Phase Transition Solid-Vapor Phase Transition 12.7 Phase Diagrams

3. Intermolecular Forces 12.1 Intermolecular forcesare attractive forces that hold particles together in the condensed phases. The magnitude (and type) of intermolecular forces is what determines whether the particles that make up a substance are a gas, liquid, or solid. Gas Liquid Solid

4. Intermolecular Forces Attractive forces that act between atoms or molecules in a pure substance are collectively called van der Waals forces. Dipole-dipole interactions are attractive forces that act between polar molecules. The magnitude of the attractive forces depends on the magnitude of the dipole.

5. Intermolecular Forces

6. Hydrogen Bonding Hydrogen bondingis a special type of dipole-dipole interaction. Hydrogen bonding only occurs in molecules that contain H bonded to a small, highly electronegative atom such as N, O, or F. F H F

7. Hydrogen Bonding

8. Dispersion Forces Dispersion forces or London dispersion forcesresult from the Coulombic attractions between instantaneous dipolesof non-polar molecules.

9. Intermolecular Forces

10. Worked Example 12.1 What kind(s) of intermolecular forces exist in (a) CCl4(l), (b) CH3COOH(l), (c) CH3COCH3(l), and (d) H2S(l). StrategyDraw Lewis dot structures and apply VSEPR theory to determine whether each molecule is polar or nonpolar. Nonpolar molecules exhibit dispersion forces only. Polar molecules exhibit dipole-dipole interactions and dispersion forces. Polar molecules with N–H , F–H, or O–H bonds exhibit dipole-dipole interactions (including hydrogen bonding) and dispersion forces. (a) (b) (c) (d)

11. Worked Example 12.1 (cont.) (a) (b) (c) (d) Solution(a) CCl4 is nonpolar, so the only intermolecular forces are dispersion forces. (b) CH3COOH is polar and contains an O–H bond, so it exhibits dipole-dipole interactions (including hydrogen bonding) and dispersion forces. (c) CH3COCH3 is polar but does not contain N–H , F–H, or O–H bonds, so it exhibits dipole-dipole interactions and dispersion forces. (d) H2S is polar but does not contain N–H , F–H, or O–H bonds, so it exhibits dipole-dipole interactions and dispersion forces. Think About ItBeing able to draw correct Lewis structures is, once again, vitally important. Review, if you need to, the procedure for drawing them.

12. Ion-Dipole Interactions Ion-dipole interactionsare Coulombic attractions between ions (either positive or negative) and polar molecules.

13. Properties of Liquids 12.2 Surface tensionis the amount of energy required to stretch or increase the surface of a liquid by a unit area. The stronger the intermolecular forces, the higher the surface tension.

14. Properties of Liquids Capillary actionis the movement of a liquid up a narrow tube. Two types of forces bring about capillary action: • cohesion is the attraction between like molecules • adhesion is the attraction between unlikemolecules Adhesive forces are greater than cohesive forces Cohesive forces are greater than adhesive forces

15. Properties of Liquids Viscosityis a measure of a fluid’s resistance to flow. The higher the viscosity the more slowly a liquid flows. Liquids that have higher intermolecular forces have higher viscosities.

16. Properties of Liquids Vapor pressure is also dependent on intermolecular forces. If a molecule at the surface of a liquid has enough kinetic energy, it can escape to the gas phase in a process called vaporization. T1 < T2 The number of molecules with enough kinetic energy to escape.

17. Properties of Liquids The vapor pressure increases until the rate of evaporation equals the rate of condensation. H2O(l) ⇌ H2O(g) Evaporation: H2O(l) → H2O(g) Condensation: H2O(l) ← H2O(g) When the forward process and reverse process are occurring at the same rate, the system is in dynamic equilibrium.

18. Properties of Liquids The vapor pressure increases until the rate of evaporation equals the rate of condensation. H2O(l) ⇌ H2O(g)

19. Properties of Liquids The vapor pressure increases with temperature.

20. Properties of Liquids The Clausius-Clapeyron equationrelates the natural log of vapor pressure and the reciprocal of absolute temperature. ln P = natural log of vapor pressure ΔHvap = the molar heat of vaporization R = the gas constant (8.314 J/K•mol) T = the kelvin temperature C is an experimentally determined constant

21. Properties of Liquids The Clausius-Clapeyron equation: Plotting ln P versus 1/T is a line with a slope of −ΔH/R. ΔH is assumed to be independent of temperature.

22. Properties of Liquids The Clausius-Clapeyron equation can be rearranged into a two point form:

23. P1 P2 ΔHvap R 1 T2 1 T1 ln = − Worked Example 12.2 Diethyl ether is a volatile, highly flammable organic liquid that today is used mainly as a solvent. (It was used as an anesthetic during the nineteenth century and as a recreational intoxicant early in the twentieth century during prohibition, when ethanol was difficult to obtain.) The vapor pressure of diethyl ether is 401 mmHg at 18°C, and its molar heat of vaporization is 26 kJ/mol. Calculate its vapor pressure at 32°C. StrategyGiven the vapor pressure at one temperature, P1, use the equation below to calculate the vapor pressure at a second temperature, P2. Temperature must be expressed in kelvins, so T1 = 291.15 K and T2 = 305.15 K. Because the molar heat of vaporization is given in kJ/mol, we will have to convert it to J/mol for the units of R to cancel properly: ΔHvap = 2.6×104 J/mol. The inverse function of ln x is ex.

24. P1 P2 2.6×104 J/mol 8.314 J/K∙mol 1 305.15 K 1 291.15 K ln = − Worked Example 12.2 (cont.) Solution = −0.4928 P1 P2 = e−0.4928 = 0.6109 P1 0.6109 = P2 401 mmHg 0.6109 P2 = = 6.6×102 mmHg Think About ItIt is easy to switch P1 and P2 or T1 and T2 accidentally and get the wrong answer to a problem such as this. One way to help safeguard against this common error is to verify that the vapor pressure is higher at the higher temperature.

25. 12.3 Crystal Structure A crystalline solidpossess rigid and long-range order; its atoms, molecules, or ions occupy specific positions. A unit cellis the basic repeating structural unit of a crystalline solid.

26. Crystal Structure There are seven types of unit cells.

27. Crystal Structure The coordination numberis defined as the number of atoms surrounding an atom in a crystal lattice. The value of the coordination number indicates how tightly the atoms are packed together. The basic repeating unit in the array of atoms is called a simple cubic cell.

28. Crystal Structure There are three types of cubic cells.

29. Crystal Structure In a body-centered cubic cell (bcc)the spheres in each layer rest in the depressions between spheres in the previous layer. The coordination number is 8.

30. Crystal Structure In a face-centered cubic cell (fcc)the coordination number is 12.

31. Crystal Structure A corner atom is shared by eight unit cells. An edge atom is shared by four unit cells. A face-centered atom is shared by two unit cells. Most of a cell’s atoms are shared by neighboring cells.

32. Crystal Structure A simple cubic cell has the equivalent of only one complete atom contained within the cell.

33. Crystal Structure A body-centered cubic cell has two equivalent atoms: A face-centered cubic cell contains four complete atoms:

34. Crystal Structure Hexagonal close-packed (hcp) structure: Site directly over an atom in layer A Close packing starts with a layer of atoms (A) Atoms in the second layer (B) fit into the depressions of the first layer Hexagonal close-packed structure.

35. Crystal Structure Cubic close-packed (ccp) structure: Site directly over an atom in layer A (hcp) Site NOT directly over an atom in layer A (ccp) Cubic close-packed structure

36. Crystal Structure Closest packing: Hexagonal close-packing (hcp) Cubic close-packing (ccp) corresponds to a face-centered cubit cell.

37. Crystal Structure Edge length (a) and radius (r) are related: Simple cubic Body-centered cubic Face-centered cubic

38. Worked Example 12.3 Gold crystallizes in a cubic close-packed structure (face-centered cubic unit cell) and has a density of 19.3 g/cm3. Calculate the atomic radius of an Au atom in angstroms (Å). StrategyUsing the given density and the mass of gold continued within a face-centered cubic unit cell, determine the volumes of the unit cell. Then, use the volume to determine the value of a, and use the equation a = √8r to find r. Be sure to use consistent units for mass, length, and volume. The face-centered cubic unit cell contains a total of four atoms of gold [six faces, each shared by two unit cells, and eight corners, each shared by eight unit cells]. d = m/V and V = a3. SolutionFirst, we determine the mass of gold (in grams) contained within a unit cell: m = × × = 1.31×10-21 g/unit cell 4 atoms unit cell 1 mol 6.022×1023 atoms 197.0 g Au 1 mol Au

39. Think About ItAtomic radii tend to be on the order of 1 Å, so this answer is reasonable. Worked Example 12.3 (cont.) Solution Then we calculate the volume of the unit cell in cm3: V = = = 6.78×10-23 cm3 Using the calculated volume and the relationship V = a3 (rearranged to solve for a), we determine the length of a side of a unit cell: a = = √6.78×10-23 cm3 = 4.08×10-8 cm Using the relationship provided a = √8r (rearranged to solve for r), we determine the radius of a gold atom in centimeters. r = = = 1.44×10-8 cm Finally, we convert centimeters to angstroms: 1.44×10-8 cm × × = 1.44 Å m d 1.31×10-21 g 19.3 g/cm3 3 a √8 4.08×10-8 cm √8 1×10-2 m 1 cm 1 Å 1×10-10 m

40. Types of Crystals 12.4 Ionic crystals are composed of charged ions that are held together by Coulombic attraction. The unit cell of an ionic compound can be defined be either the positions of the anions or the positions of the cations.

41. Types of Crystals Crystal structures of three ionic compounds: CsCl Simple cubic lattice ZnS Zincblende structure (based on FCC) CaF2 fluorite structure (based on FCC)

42. Think About ItMake sure that the ratio of cations to anions that you determine for a unit cell matches the ratio expressed in the compound’s empirical formula. 1 8 1 2 Worked Example 12.4 How many of each ion are contained within a unit cell of ZnS? StrategyDetermine the contribution of each ion in the unit cell based on its position. Referring to the figure, the unit cell has four Zn2+ ions completely contained within the unit cell, and S2- ions at each of the eight corners and at each of the six faces. Interior ions (those completely contained within the unit cell) contribute one, those at the corners each contribute one-eighth, and those on the faces contribute one-half. SolutionThe ZnS unit cell contains four Zn2+ ions (interior) and four S2- ions [8 × (corners) and 6 × (faces)]

43. 1 g 6.022×1023 amu Worked Example 12.5 The edge length of the NaCl unit cell is 564 pm. Determine the density of NaCl in g/cm3. StrategyUse the number of Na+ and Cl- ions in a unit cell (four of each) to determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is mass divided by volume (d = m/V). Be careful to use units consistently. The masses of Na+ and Cl- ions are 22.99 amu and 35.45 amu, respectively. The conversion factor from amu to grams is so the masses of the Na+ and Cl- ions are 3.818×10-23 g and 5.887×10-23 g, respectively. The unit cell length is 564 pm × × = 5.64×10-8 cm 1×10-12 m 1 pm 1 cm 1×10-2 m

44. 3.882×10-22 g 1.794×10-22 cm3 Worked Example 12.5 (cont.) SolutionThe mass of the unit cell is 3.882×10-22 g (4 × 3.818×10-23 g + 4 × 5.887×10-23 g). The volume of a unit cell is 1.794×10-22 cm3 [(5.64×10-8 cm)3]. Therefore, the density is given by d = = 2.16 g/cm3 Think About ItIf you were to hold a cubic centimeter (1 cm3) of salt in your hand, how heavy would you expect it to be? Common errors in this type of problem include errors of unit conversion–especially with regard to length and volume. Such errors can lead to results that are off by many orders of magnitude. Often you can use common sense to gauge whether or not a calculated answer is reasonable. For instance, simply getting the centimeter-meter conversion upside down would result in a calculated density of 2.16×1012 g/cm3! You know that a cubic centimeter of salt doesn’t have a mass that large. (That’s billions of kilograms!) If the magnitude of a result is not reasonable, go back and check your work.

45. Types of Crystals In covalent crystals, atoms are held together in an extensive three-dimensional network entirely by covalent bonds.

46. Types of Crystals In molecular crystals, the lattice points are occupied by molecules; the attractive forces between them are van der Waals forces and/or hydrogen bonding.

47. 1 2 1 8 1 g 6.022×1023 amu Worked Example 12.6 The metal iridium (Ir) crystallizes with a face-centered cubic unit cell. Given that the length of the edge of a unit cell is 383 pm, determine the density of iridium in g/cm3. StrategyA face-centered metallic crystal contains four atoms per unit cell [8 × (corners) and 6 × (faces)]. Use the number of atoms per cell and the atomic mass to determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is then mass divided by volume (d = m/V). Be sure to make all necessary unit conversions. The mass of an Ir atom is 192.2 amu. The conversion factor from amu to grams is so the mass of an Ir atom is 3.192×10-22 g. The unit cell length is 383 pm × × = 3.83×10-8 cm 1×10-12 m 1 pm 1 cm 1×10-2 m

48. 1.277×10-21 g 5.62×10-23 cm3 Worked Example 12.6 (cont.) SolutionThe mass of the unit cell is 1.277×10-21 g (4 × 3.192×10-22 g). The volume of a unit cell is 5.618×10-23 cm3 [(3.83×10-8 cm)3]. Therefore, the density is given by d = = 22.7 g/cm3 Think About ItMetals typically have high densities, so common sense can help you decide whether or not your calculated answer is reasonable.

49. Types of Crystals In metallic crystals, every lattice point is occupied by an atom of the same metal. Electrons are delocalized over the entire crystal. Delocalized electrons make metals good conductors. Large cohesive force resulting from delocalization makes metals strong.

50. Types of Crystals