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Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction.

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Presentation on theme: "Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction."— Presentation transcript:

1 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

2 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

3 Phases and Phase Diagrams condensation vaporization deposition sublimation fusion freezing incompressible d ≈ 1 – 10 g mL −1 very compressible d ≈ 1 – 10 g L −1 at SATP

4 Phases and Phase Diagrams A few definitions: STP vs SATP STP = Standard Temperature and Pressure (0 o C, 100 kPa) SATP = Standard Ambient Temperature and Pressure (25 o C, 100 kPa) 100 kPa = 1 bar

5 Phases and Phase Diagrams Differences between the different states of Matter: Gas: Molecules move randomly and the intermolecular separations are large (i.e. most of a gas is empty space). Liquids and Solids: the molecular motions are quite restricted and the intermolecular separations are small. Solids: The molecules are often, but not always, arranged in regular, repeating patterns. Substances exist in different phases and phase changes occur because molecules exert forces on each other. (Without intermolecular forces, all substances would behave as ideal gases!!) It takes energy to overcome the attractive intermolecular forces that cause molecules to aggregate. Therefore, sublimation, fusion and evaporation are all endothermic processes A given substance will exist as a solid, liquid or gas depending on the temperature and pressure of the sample. A phase diagram shows the stable phases at each temperature and pressure.

6 Phases and Phase Diagrams Phase diagram of I 2

7 Phases and Phase Diagrams Take note of the following points: 1. Solid is the most stable phase at low T and high P. Gas is the stable phase at high T and low P.

8 Phases and Phase Diagrams Take note of the following points: 2. The S-L line shows the T’s and P’s at which both solid and liquid are stable and can coexist. It also shows us how the melting temperature changes with pressure.

9 Phases and Phase Diagrams Take note of the following points: 3. The L-G line curve shows the T’s and P’s at which both liquid and gas are stable and can coexist. It also shows us how the boiling temperature changes with pressure.

10 Phases and Phase Diagrams Take note of the following points: 4. For most substances, the S-L line has a positive slope, but for a few substances (most notably, water but also bismuth and antimony), it has a negative slope! For most substances

11 Phases and Phase Diagrams Take note of the following points: 4. For most substances, the S-L line has a positive slope, but for a few substances (most notably, water but also bismuth and antimony), it has a negative slope! For water

12 Phases and Phase Diagrams Take note of the following points: 5. At the triple point, all three phases are stable and coexist.

13 Phases and Phase Diagrams Take note of the following points: 6. The G-L line ends abruptly at the critical point (T c, P c )

14 Phases and Phase Diagrams Phase diagram of I 2 What is the phase of I 2 at 25 o C and 1 atm? We are dealing with a solid. 25 o C 25 o C and 1 atm

15 Phases and Phase Diagrams Phase diagram of CO 2 What is the phase of CO 2 at 25 o C and 1 atm? We are dealing with a gas. 25 o C 25 o C and 1 atm 1 atm

16 Phases and Phase Diagrams Phase diagram of H 2 O What is the phase of H 2 O at 25 o C and 1 atm? We are dealing with a liquid. 25 o C 25 o C and 1 atm 1 atm

17 Phases and Phase Diagrams Take note of the following points: 4. For most substances, the S-L line has a positive slope, but for a few substances (most notably, water but also bismuth and antimony), it has a negative slope! The slope of the S-L line is negative!

18 Phases and Phase Diagrams Take note of the following points: 1 m 10 m A column of water 1 m ×1 m × 10 m occupies a volume of 10 m 3 or 10,000 L. 1 L of water weighs 1 kg. 10,000 L of water weigh 10,000 kg. The pressure exerted by 10,000 kg of water equals: 9.8 m 2 /s×10,000 kg/(1 m×1 m)  10 5 Pa  1 atm 10 m of water generates a 1 atm additional pressure.

19 Phases and Phase Diagrams Take note of the following points: We find liquid water at the bottom of the ocean. The slope of the S-L line is negative!

20 Phases and Phase Diagrams Take note of the following points: Polymorphism: The existence of a solid substance in more than one form. Other forms of ice obtained at several thousands of atmospheres

21 Phases and Phase Diagrams To summarize, a typical phase diagram looks like this: T supercritical fluid P Liquid Gas Solid TcTc PcPc T vap T fus 1 atm A B solid-liquid coexistence line liquid-vapour coexistence line triple point “normal” boiling point “normal” melting point critical point

22 Phases and Phase Diagrams Supercritical fluid: T supercritical fluid P (L) (G) (S) TcTc PcPc A B solid-liquid coexistence line liquid-vapour coexistence line critical point As one moves from A to B, the pressure increases and the density of the gas increases until it equals the density of the liquid. At this point, gas and liquid are indistinguishable, the interface between liquid and gas vanishes and we have a supercritical fluid. If a gas is at T > T C (Point A in diagram), increasing the pressure of the gas does not yield a liquid but rather a supercritical fluid (Point B). To take a gas at T > T C and transform it into a liquid, the temperature must first be reduced below T C. Then the pressure is increased to pass the liquid-vapor coexistence curve.

23 Phases and Phase Diagrams Supercritical fluid: T supercritical fluid P (L) (G) (S) TcTc PcPc A B solid-liquid coexistence line liquid-vapour coexistence line critical point The phase boundary between liquid benzene and its vapour disappears at T c. Below T c, the phase boundary is clearly visible. Just below T c, the phase boundary is barely visible. At T c, the phase boundary disappears. T increases from below T C to above T C

24 Phases and Phase Diagrams Supercritical fluid: Did you know? Supercritical CO 2 is used to extract caffeine from coffee beans. The extracted caffeine can be sold to pharmaceutical or beverage companies. The critical point for CO 2 is fairly low (T c = 31 o C) and so, supercritical CO 2 can be used at ambient temperatures without causing decomposition or “denaturing” of other compounds. Because it has low toxicity, a low critical temperature and is nonflammable, supercritical CO 2 is becoming an increasingly important industrial and commercial solvent. T supercritical fluid P (L) (G) (S) TcTc PcPc solid-liquid coexistence line liquid-vapour coexistence line critical point

25 Phases and Phase Diagrams Examples: For a particular substance, the S-L coexistence curve has a negative slope. a)What phase changes are possible if the pressure is increased at constant temperature T? Assume that T is less than T tp, where T tp is the triple point temperature. Gas  deposition  solid  melting  liquid T supercritical fluid P (L) (G) (S) TcTc PcPc solid-liquid coexistence line liquid-vapour coexistence line critical point T PTPT

26 Phases and Phase Diagrams Examples: For a particular substance, the S-L coexistence curve has a negative slope. b)What phase changes are possible if the pressure is increased at constant temperature T, assuming T tp < T < T c, where T tp and T c are the triple point and critical point temperatures, respectively. Gas  condensation  liquid T supercritical fluid P (L) (G) (S) TcTc PcPc solid-liquid coexistence line liquid-vapour coexistence line critical point T PTPT

27 Phases and Phase Diagrams Examples: For a particular substance, the S-L coexistence curve has a negative slope. c)True or False?The melting temperature increases as the pressure increases. False T supercritical fluid P (L) (G) (S) TcTc PcPc solid-liquid coexistence line liquid-vapour coexistence line critical point T PTPT TmTm

28 Phases and Phase Diagrams Examples: For a particular substance, the S-L coexistence curve has a negative slope. d) True or False?The solid is more dense than the liquid. At a given temperature, when we increase the pressure, the density increases and the solid becomes a liquid. False T supercritical fluid P (L) (G) (S) TcTc PcPc solid-liquid coexistence line liquid-vapour coexistence line critical point T PTPT

29 Phases and Phase Diagrams a) What is the phase of this substance at 25 o C and 73 atm? We are dealing with a liquid. 25 o C 25 o C and 73 atm 73 atm Examples: For a particular substance, the triple point is at  57 ºC and 5.1 atm, and the critical point is at 31 o C and 73 atm.

30 Phases and Phase Diagrams b)What phase changes occur if the pressure is decreased from 75 atm to atm at −60 o C? Assume that the solid-liquid line has a positive slope. Solid  sublimation  gas -60 o C -60 o C and 75 atm Examples: For a particular substance, the triple point is at  57 ºC and 5.1 atm, and the critical point is at 31 o C and 73 atm. -60 o C and atm

31 Phases and Phase Diagrams condensation vaporization deposition sublimation fusion freezing incompressible d ≈ 1 – 10 g mL −1 very compressible d ≈ 1 – 10 g L −1 at SATP REVIEW

32 Phases and Phase Diagrams To summarize, a typical phase diagram looks like this: T supercritical fluid P Liquid Gas Solid REVIEW

33 Phases and Phase Diagrams To summarize, a typical phase diagram looks like this: T supercritical fluid P Liquid Gas Solid TcTc PcPc T vap T fus 1 atm A B solid-liquid coexistence line liquid-vapour coexistence line triple point “normal” boiling point “normal” melting point critical point REVIEW

34 Phases and Phase Diagrams Examples 12-45: Which substances listed in the table can exist as liquids at room temperature (~ 20 o C)? SubstanceT c, KP c, atm H2H N2N O2O CH CO HCl NH SO H2OH2O T supercritical fluid P (L) (G) (S) TcTc PcPc solid-liquid coexistence line liquid-vapour coexistence line critical point T = 20 o C

35 Phases and Phase Diagrams Examples 12-45: Which substances listed in the table can exist as liquids at room temperature (~ 20 o C)? SubstanceT c, KP c, atm H2H N2N O2O CH CO HCl NH SO H2OH2O supercritical fluid P (L) (G) (S) T c =  240 o C PcPc solid-liquid coexistence line liquid-vapour coexistence line critical point T = 20 o C

36 Phases and Phase Diagrams Examples 12-45: Which substances listed in the table can exist as liquids at room temperature (~ 20 o C)? SubstanceT c, KP c, atm H2H N2N O2O CH CO HCl NH SO H2OH2O T supercritical fluid P (L) (G) (S) TcTc PcPc solid-liquid coexistence line liquid-vapour coexistence line critical point T = 20 o C 20 o C K Gases that can be liquified at room temperature are said to be “non-permanent gases”. Gases that cannot be liquified at room temperature are said to be “permanent gases”.

37 Phases and Phase Diagrams Examples 12-51: Phase diagram of phosphorous a) Indicate the phases present in the regions labeled with a question mark? T P (L) (G) (S) 590 o C 43 atm ? ?

38 Phases and Phase Diagrams Examples 12-51: Phase diagram of phosphorous b) A sample of solid red phosphorous cannot be melted by heating in a container open to the atmosphere. Explain why this is so? T P (L) (G) (S) 590 o C 43 atm 1 atm Solid phosphorous can only be sublimed (S  G) if it is heated at P = 1 atm.

39 Phases and Phase Diagrams Examples 12-51: Phase diagram of phosphorous c) Trace the phase changes that occur when the pressure on a sample is reduced from Point A to B, at constant temperature. T P (L) (G) (S) 590 o C 43 atm Solid  condensation  Liquid  vaporization  Gas A B

40 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

41 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

42 Liquids and Liquid Properties T supercritical fluid P (L) (G) (S) solid-liquid coexistence line liquid-vapour coexistence line critical point We know that if the temperature of a gas is lowered sufficiently, the gas will condense to a liquid. Why is this? As T is lowered, the average kinetic energy of the molecules decreases. At some point, the molecules will no longer have enough kinetic energy to overcome the attractive forces that draw the molecules together. Consequently, the molecules cluster together to form a liquid. Condensation

43 Liquids and Liquid Properties T supercritical fluid P (L) (G) (S) solid-liquid coexistence line liquid-vapour coexistence line critical point The freezing of a liquid can be explained in the same way: If the temperature of a liquid is lowered sufficiently, the molecules will not have enough kinetic energy to overcome attractive forces that draw the molecules closer together  the liquid freezes. Freezing

44 Liquids and Liquid Properties The physical properties of a liquid depend on the strength and nature of the intermolecular forces. We shall examine why the following physical properties are different from substance to substance. vapour pressure=equilibrium pressure of vapour that forms above a liquid in a closed container normal boiling point (T vap )=temperature at which the vapour pressure of the liquid equals 1 atm surface tension (  )=energy required to increase the surface area of a liquid viscosity (η)  provides a measure of a fluid’s resistance to flow; the speed of flow through a tube is inversely proportional to the viscosity In general, the stronger the intermolecular attractions, the higher the boiling point, the greater the surface tension, the higher the viscosity and the lower the vapour pressure.

45 Liquids and Liquid Properties vapour pressure=equilibrium pressure of vapour that forms above a liquid in a closed container to vacuum liquid air

46 Liquids and Liquid Properties vapour pressure=equilibrium pressure of vapour that forms above a liquid in a closed container to vacuum liquid Liquid N 2 air

47 Liquids and Liquid Properties vapour pressure=equilibrium pressure of vapour that forms above a liquid in a closed container to vacuum solid Liquid N 2 air

48 Liquids and Liquid Properties vapour pressure=equilibrium pressure of vapour that forms above a liquid in a closed container to vacuum solid Liquid N 2 vacuum

49 Liquids and Liquid Properties vapour pressure=equilibrium pressure of vapour that forms above a liquid in a closed container to vacuum solid Liquid N 2 vacuum

50 Liquids and Liquid Properties vapour pressure=equilibrium pressure of vapour that forms above a liquid in a closed container to vacuum solid vacuum

51 Liquids and Liquid Properties vapour pressure=equilibrium pressure of vapour that forms above a liquid in a closed container to vacuum liquid vacuum

52 Liquids and Liquid Properties vapour pressure=equilibrium pressure of vapour that forms above a liquid in a closed container to vacuum liquid

53 Liquids and Liquid Properties The physical properties of a liquid depend on the strength and nature of the intermolecular forces. We shall examine why the following physical properties are different from substance to substance. vapour pressure=equilibrium pressure of vapour that forms above a liquid in a closed container normal boiling point (T vap )=temperature at which the vapour pressure of the liquid equals 1 atm surface tension (  )=energy required to increase the surface area of a liquid viscosity (η)  provides a measure of a fluid’s resistance to flow; the speed of flow through a tube is inversely proportional to the viscosity In general, the stronger the intermolecular attractions, the higher the boiling point, the greater the surface tension, the higher the viscosity and the lower the vapour pressure.

54 Liquids and Liquid Properties T supercritical fluid P (L) (G) (S) solid-liquid coexistence line liquid-vapour coexistence line critical point The L-G line shows us how 1. the vapour pressure of a liquid changes with temperature 2. the boiling temperature of a liquid changes with pressure Clausius-Clapeyron equation

55 Liquids and Liquid Properties T supercritical fluid P (L) (G) (S) solid-liquid coexistence line liquid-vapour coexistence line critical point Along the L-G line, both liquid and gas co-exist. At equilibrium, the rate of evaporation equals the rate of condensation Clausius-Clapeyron equation Liquid X(l) Vapour X(g)

56 =standard enthalpy of vaporization R= J K −1 mol −1 Liquids and Liquid Properties Clausius-Clapeyron equation The variation of vapour pressure with temperature is modeled reasonably well by the Clausius-Clapeyron equation: The quantities appearing in this equation are described below. P 2 =vapour pressure at temperature T 2 P 1 =vapour pressure at temperature T 1 Extremely important: Pay attention to the units!

57 Liquids and Liquid Properties T supercritical fluid P (L) (G) (S) solid-liquid coexistence line liquid-vapour coexistence line critical point Clausius-Clapeyron equation (P 1, T 1 ) (P 2, T 2 )

58 Liquids and Liquid Properties T supercritical fluid P (L) (G) (S) solid-liquid coexistence line liquid-vapour coexistence line critical point Clausius-Clapeyron equation (P 1, T 1 ) (P 2, T 2 ) Example: a)If the vapour pressure of P 4 (l) is 10 Torr at 128 o C and 400 Torr at 251 o C, then what is  vap H o ?

59 Liquids and Liquid Properties Clausius-Clapeyron equation Example: b) What is the normal boiling point of P 4 (l)? P 1 = 10 Torr T 1 = 128 o C P 2 = 400 Torr T 2 = 251 o C  vap H o = 52.4 kJ/mol

60 Liquids and Liquid Properties Clausius-Clapeyron equation Example: c) What is the vapour pressure at 200 o C? P 1 = 10 Torr T 1 = 128 o C P 2 = 400 Torr T 2 = 251 o C  vap H o = 52.4 kJ/mol

61 Liquids and Liquid Properties The physical properties of a liquid depend on the strength and nature of the intermolecular forces. We shall examine why the following physical properties are different from substance to substance. vapour pressure=equilibrium pressure of vapour that forms above a liquid in a closed container normal boiling point (T vap )=temperature at which the vapour pressure of the liquid equals 1 atm surface tension (  )=energy required to increase the surface area of a liquid viscosity (η)  provides a measure of a fluid’s resistance to flow; the speed of flow through a tube is inversely proportional to the viscosity In general, the stronger the intermolecular attractions, the higher the boiling point, the greater the surface tension, the higher the viscosity and the lower the vapour pressure. REVIEW

62 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

63 Intermolecular Forces Dipole moments and polarizabilities of a few molecules. (Based on data from Physical Chemistry, 6 th Edition, by P. Atkins, published by Freeman, 1998). μ (in debye) α/α He H2H HF HCl HBr CO CO H2OH2O NH He01 Ar08.4 CH CCl The purpose of this section is to understand how molecules interact with each other and how these interactions help us to understand trends in the physical properties of compounds (e.g. boiling points, vapour pressures, viscosity and surface tensions of liquids; densities and melting points of solids; deviations from ideal gas behaviour, etc.) Generally speaking, differences in the physical properties of substances can often be rationalized by considering how molecules interact with each other at the molecular level. We’ll look at some of the types of intermolecular forces that act between pairs of molecules.

64 The molecules are not organized into perfect straight line because the molecules are in motion (i.e. each one possesses kinetic energy) and they “jiggle” out of perfect alignment. Intermolecular Forces μAμA μBμB d A. Dipole-dipole forces Digging Deeper For a pair of interacting polar molecules with dipole moments of μ A and μ B : Some molecules possess a permanent dipole moment, μ, because the bond dipoles do not cancel out. Such molecules are said to be “polar” because one end of the molecule is slightly positive and the other end is slightly negative. The charge distribution of a polar molecule can be represented by an arrow that points from the positive end to the negative end. The dipole moments of a few molecules were given in the previous table. Polar molecules (i.e. dipoles) tend to orient themselves in a “head-to-tail” manner, as shown below:

65 Intermolecular Forces A. Dipole-dipole forces Example:Iodine chloride, ICl, and bromine, Br 2, have exactly the same number of electrons, and it is reasonable to assume that these molecules are essentially the same size. Yet the boiling points of ICl(l) and Br 2 (l) are quite different, 97 o C and 59 o C, respectively. Use your knowledge of dipole-dipole interactions to explain why ICl(l) has a higher boiling point than Br 2 (l).

66 Intermolecular Forces A. Dipole-dipole forces Example:Iodine chloride, ICl, and bromine, Br 2, have exactly the same number of electrons, and it is reasonable to assume that these molecules are essentially the same size. Yet the boiling points of ICl(l) and Br 2 (l) are quite different, 97 o C and 59 o C, respectively. Use your knowledge of dipole-dipole interactions to explain why ICl(l) has a higher boiling point than Br 2 (l). Chlorine is more electronegative than bromine. Consequently, I-Cl shows a dipole moment: (  ) I  Cl (  ) Br 2 on the other hand cannot generate a dipole moment because the molecule is made of two identical bromine atoms.

67 Intermolecular Forces B. London dispersion forces The electrons in a molecule are in constant motion and at any particular instant, there may be an asymmetric distribution of electrons in the molecule (i.e. with a greater number of electrons at one end than at the other end). The asymmetric distribution of electrons gives rise to an instantaneous and temporary dipole moment (  inst ). The formation of  inst in a molecule causes (or induces) the formation of a dipole in neighbouring molecules. The induced dipole moment is  ind. (Notice the head-to-tail arrangement of the instantaneous and induced dipole moments.) Molecule A Molecule B There is an attraction between  inst and  ind. The strength of the interaction increases as the “polarizabilities” of the molecules increase.  inst  ind

68 Intermolecular Forces B. London dispersion forces The charge cloud of a large molecule is diffuse and easily polarized. The charge cloud of a small, compact molecule is not easily polarized. The polarizabilities of a few molecules were given earlier. Note that the larger the molecule, the larger the polarizability. Polarizability (α) provides a measure of the extent to which the charge cloud of a molecule can be distorted (i.e. polarized) by another molecule. Digging Deeper For a pair of molecules with polarizabilities of α A and α B : αAαA αBαB d

69 Intermolecular Forces B. London dispersion forces Remarks: London dispersion forces are most attractive when the molecules are large because large molecules have larger, more diffuse (i.e. more polarizable) charge clouds. London dispersion forces always contribute to the molecular interactions because all molecules have charge clouds and are therefore polarizable to some extent Molecule A Molecule B  inst  ind

70 Intermolecular Forces B. London dispersion forces The C-Cl bonds are very polar, but the bond dipoles cancel. The CCl 4 molecule is nonpolar. Example: Methane (CH 4 ) and carbon tetrachloride (CCl 4 ) are both nonpolar molecules. Use your knowledge of London dispersion forces to explain why the boiling point of CCl 4 (l) is much higher than that of CH 4 (l). μ (in debye) α/α He H2H HF HCl HBr CO CO H2OH2O NH He01 Ar08.4 CH CCl Carbon tetrachloride (CCl 4 ) is a molecule that is much larger than methane (CH 4 ). Consequently, CCl 4 contains many more electrons which can generate an instantaneous dipole moment more readily than CH 4. These dipole moments generate strong intermolecular forces between CCl 4 molecules which are more difficult to break than in CH 4. Consequently, T b of CCl 4 (78 o C) is larger than T b of CH 4 (  162  o C).

71 Intermolecular Forces C. Hydrogen bonding forces A special type of bond forms between molecules when the molecules contain a hydrogen atom bonded to N, O, or F. When H is bonded to N, O or F, the H atom carries a significant positive charge and it is strongly attracted to a lone pair on another molecule! When a hydrogen atom which is covalently bonded to one atom is simultaneously attracted to the lone pair on another atom, it is “bridging” two molecules, as shown below. Such a bond is called a hydrogen bond. An intermolecular hydrogen bond “bridges” two molecules. An intramolecular hydrogen bond bridges two parts of the same molecule. Did you know? “inter” means “between” and “intra” means “within”. X H Y intermolecular hydrogen bond

72 Intermolecular Forces C. Hydrogen bonding forces Note carefully: H is covalently bonded to X but is simultaneously attracted to a lone pair of electrons on Y. Both X and Y must be N, O or F Hydrogen bonds are the strongest type of intermolecular force (but they are still weak in comparison to covalent and ionic bonding forces) X H Y intermolecular hydrogen bond dipole-dipole & LDFs H bonds covalent & ionic bonds kJ mol −1 10−40 kJ mol −1 100’s or 1000’s kJ mol −1 intermolecular forces chemical bonding forces

73 Intermolecular Forces C. Hydrogen bonding forces Hydrogen bonds are important! H bonds between H 2 O’s in ice give the solid an open structure H bonds between H 2 O’s in water give the liquid a high BP, high surface tension and a large heat capacity. Hydrogen bonding in water. This is Figure 12-7 of Petrucci 10e. Used with permission.

74 Intermolecular Forces C. Hydrogen bonding forces Hydrogen bonds are important! H bonds are especially important in biology (e.g. H bonds keep the two helices of DNA together; the structures and functions of proteins and enzymes are determined by H bonds) The helical structures of proteins (above) and DNA (on the right) are stabilized by hydrogen bonds. These are Figures and of Petrucci 10e. Used with permission.

75 Intermolecular Forces C. Hydrogen bonding forces Hydrogen bonds are important! H bonds are especially important in biology (e.g. H bonds keep the two helices of DNA together; the structures and functions of proteins and enzymes are determined by H bonds)

76 Intermolecular Forces A.Dipole-dipole forces Polar molecules with a dipole moment . B. London dispersion forces The formation of  inst in a molecule causes (or induces) the formation of a dipole in neighbouring molecules. C. Hydrogen bonding forces A special type of bond forms between molecules when the molecules contain a hydrogen atom bonded to N, O, or F. X H Y intermolecular hydrogen bond Molecule A Molecule B  inst  ind REVIEW

77 Intermolecular Forces C. Hydrogen bonding forces A special type of bond forms between molecules when the molecules contain a hydrogen atom bonded to N, O, or F. When H is bonded to N, O or F, the H atom carries a significant positive charge and it is strongly attracted to a lone pair on another molecule! X H Y intermolecular hydrogen bond Electronegativity Scale

78 Intermolecular Forces C. Hydrogen bonding forces (Z)-1,2-dichloroethene, also called cis-1,2-dichloroethene (E)-1,2-dichloroethene, also called trans-1,2-dichloroethene Example: Dichloroethene, C 2 H 2 Cl 2, has several isomeric forms. Use your knowledge of intermolecular forces to predict whether (Z)-1,2-dichloroethene or (E)-1,2-dichloroethene has the higher boiling point. Lewis structures are given below. The chlorine atom being more electronegative than carbon induces a dipole moment in the C-Cl bond. However, those dipole moments are opposite in (E)-1,2-dichloroethane and they cancel each other. The dipole moments of the C-Cl bond do not cancel each other in (Z)-1,2-dichloroethane. Thus the Z-isomer has a permanent dipole moment which induces strong intermolecular forces. The Z-isomer has a higher boiling point. T b = 60 o C T b = 48 o C A word of warning! Don’t over generalize the results of this example. You might end up making the wrong prediction!

79 Intermolecular Forces C. Hydrogen bonding forces Example: Consider the trans and cis isomers of C 4 H 4 O 4. Which one has the higher melting point? Fumaric acid generates intermolecular H-bonds which leads to the formation of a network where all the molecules are associated with another. Maleic acid forms intramolecular H-bonds and the intermolecular forces between molecules are weaker. Consequently, maleic acid has a lower melting point than fumaric acid. T m = 300 o CT m = 140 o C Fumaric acid (trans)Maleic acid (cis)

80 Intermolecular Forces C. Hydrogen bonding forces Example: Consider the trans and cis isomers of C 4 H 4 O 4. Which one has the higher melting point? Fumaric acid generates intermolecular H-bonds which leads to the formation of a network where all the molecules are associated with another. Maleic acid forms intramolecular H-bonds and the intermolecular forces between molecules are weaker. Consequently, maleic acid has a lower melting point than fumaric acid. T m = 300 o CT m = 140 o C Fumaric acid (trans)Maleic acid (cis)

81 Intermolecular Forces C. Hydrogen bonding forces Example:Use your knowledge of intermolecular forces to predict whether CH 3 COCH 3 (l) or CH 3 CH 2 CH 2 OH(l) has the higher boiling point. Acetone and 1-propanol have a similar number of C- and O-atoms. They both induce a dipole moment since O is more electronegative than C. However, 1-propanol can form H-bonds and acetone cannot. Consequently, acetone boils at a lower temperature than 1-propanol. acetone 1-propanol T b = 56 o CT b = 97 o C

82 Intermolecular Forces C. Hydrogen bonding forces Example:Which one of the liquids, HO-CH 2 CH 2 -OH or CH 3 CH 2 OH, has the highest vapour pressure at room temperature? A single ethylene glycol (EG) can form more H-bonds with other EG molecules than ethanol can form with other ethanol molecules. Consequently, EG boils at a higher temperature. EG has a lower vapor pressure than ethanol. Ethylene glycol Ethanol T b = 196 o CT b = 78 o C

83 Intermolecular Forces C. Hydrogen bonding forces Example:Use your knowledge of intermolecular forces to predict the order of boiling points for H 2 O, H 2 S (hydrogen sulfide), H 2 Se (hydrogene selenide), and H 2 Te (hydrogen telluride). All these molecules are bent, and as such, generate a dipole moment. 

84 Intermolecular Forces C. Hydrogen bonding forces Example:Use your knowledge of intermolecular forces to predict the order of boiling points for H 2 O, H 2 S, H 2 Se and H 2 Te. All these molecules are bent, and as such, generate a dipole moment. H 2 S T b =  60 o C H 2 Se T b =  41 o C H 2 Te T b =  2 o C H 2 O T b =  60 o C T b o C o C -2 o C 

85 Intermolecular Forces C. Hydrogen bonding forces Example:Use your knowledge of intermolecular forces to predict the order of boiling points for H 2 O, H 2 S, H 2 Se and H 2 Te. All these molecules are bent, and as such, generate a dipole moment. H 2 S T b =  60 o C H 2 Se T b =  41 o C H 2 Te T b =  2 o C H 2 O T b =  o C  T b +100 o C o C o C -2 o C

86 Intermolecular Forces C. Hydrogen bonding forces H 2 S T b =  60 o C H 2 Se T b =  41 o C H 2 Te T b =  2 o C H 2 O T b =  o C Example:Use your knowledge of intermolecular forces to predict the order of boiling points for H 2 O, H 2 S, H 2 Se and H 2 Te. By increasing the mass and the size of the molecule H 2 X, one increases the polarizability, and thus the strength of the interactions between the H 2 X molecules. The exception is H 2 O because water forms H-bonds:H 2 S < H 2 Se < H 2 Te < H 2 O T b +100 o C o C o C -2 o C

87 Intermolecular Forces C. Hydrogen bonding forces Example:Use your knowledge of intermolecular forces to predict the order of boiling points for H 2 O, H 2 S, H 2 Se and H 2 Te. It should be noted that similar arguments can be applied to the hydrides of the group 15 and group 17 elements. Thus, in order of increasing BP, we have: PH 3 < AsH 3 < SbH 3 < NH 3 and HCl < HBr < HI < HF

88 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

89 Heating Curves A heating curve shows us how the temperature varies with the amount of heat added. Consider heating a sample of ice from t i = –10 o C to t = 150 o C at constant pressure. Q T titi t fus t vap melting solid boiling liquid  H fus  H vap warming liquid (slope 2 ) warming solid (slope 1 ) warming vapour (slope 3 ) q liq q sol tftf q gas Review this section on your own.

90 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

91 Introduction to Solids Solids can be classified as crystalline or as amorphous, depending on whether we have regular or irregular packing of the atoms, molecules or ions that make up the solid. Crystalline and amorphous solids have rather different physical characteristics. Crystalline solids regular repeating patterns “sharp” melting points Amorphous solids irregular packing melt over a temperature range Examples of amorphous solids include rubber, polystyrene, window glass, candle wax, and cotton candy. We can get irregular packing (amorphous solids) if there are impurities in the sample when the liquid freezes, or if the molecules are large or have flexible structures. For molecules that are large or very flexible (i.e. polymers!), it is statistically most probable that the molecules will be irregularly packed when the liquid freezes. We shall focus exclusively on crystalline solids.

92 Introduction to Solids For crystalline solids, we use the following concepts to characterize the pattern of the packing arrangement. three dimensional array of points that shows how the structural units are arranged in space. Crystal lattice Each structural unit could be a single atom, a molecule or an ion. In C(s), the structural unit is a carbon atom. In I 2 (s), the structural unit is an I 2 molecule. In NH 4 NO 3 (s), the structural units are NH 4 + and NO 3 − ions. the smallest building block that possesses the symmetry of the crystal lattice; a solid sample of any size can be “built” by stacking together unit cells. Unit cell

93 Introduction to Solids Crystalline solids are classified according to the nature of the bonding, or according to the geometry and symmetry of the packing arrangement. Let’s focus first on the nature of the bonding. Crystalline solids ionicNetwork covalent molecularmetallic In an ionic solid, positive and negative ions are held in their lattice positions by (strong) ionic bonding forces. NaCl T m = 800 o C LiF T m = 848 o C ZnO T m = 1974 o C In a network covalent solid, atoms are held in their lattice positions by (strong) covalent bonds. SiO 2 T m = 1600 o C SiCT m =2830 o C C diamond T m = 4440 o C (at P = 12.4 GPa) In a molecular solid, molecules are held in their lattice positions by (weak) intermolecular forces. H 2 OT m = 0 o C S 8 T m = 113 o C I 2 T m = 114 o C In a metallic solid, metal cations are held in their lattice positions by (strong) metallic bonding forces. CuT m = 1083 o C AgT m = 962 o C WT m = 3422 o C

94 Introduction to Solids Crystalline solids are classified according to the nature of the bonding, or according to the geometry and symmetry of the packing arrangement. Let’s focus first on the nature of the bonding. Crystalline solids ionicNetwork covalent molecularmetallic In an ionic solid, positive and negative ions are held in their lattice positions by (strong) ionic bonding forces. NaCl T m = 800 o C LiF T m = 848 o C ZnO T m = 1974 o C In a network covalent solid, atoms are held in their lattice positions by (strong) covalent bonds. SiO 2 T m = 1600 o C SiCT m =2830 o C C diamond T m = 4440 o C (at P = 12.4 GPa) In a molecular solid, molecules are held in their lattice positions by (weak) intermolecular forces. H 2 OT m = 0 o C S 8 T m = 113 o C I 2 T m = 114 o C In a metallic solid, metal cations are held in their lattice positions by (strong) metallic bonding forces. CuT m = 1083 o C AgT m = 962 o C WT m = 3422 o C

95 Introduction to Solids The Periodic Table can help understand the differences in T m between NaCl, LiF, and ZnO. NaCl T m = 800 o C LiF T m = 848 o C ZnO T m = 1974 o C

96 Introduction to Solids Crystalline solids are classified according to the nature of the bonding, or according to the geometry and symmetry of the packing arrangement. Let’s focus first on the nature of the bonding. Crystalline solids ionicNetwork covalent molecularmetallic In an ionic solid, positive and negative ions are held in their lattice positions by (strong) ionic bonding forces. NaCl T m = 800 o C LiF T m = 848 o C ZnO T m = 1974 o C In a network covalent solid, atoms are held in their lattice positions by (strong) covalent bonds. SiO 2 T m = 1600 o C SiCT m =2830 o C C diamond T m = 4440 o C (at P = 12.4 GPa) In a molecular solid, molecules are held in their lattice positions by (weak) intermolecular forces. H 2 OT m = 0 o C S 8 T m = 113 o C I 2 T m = 114 o C In a metallic solid, metal cations are held in their lattice positions by (strong) metallic bonding forces. CuT m = 1083 o C AgT m = 962 o C WT m = 3422 o C

97 Introduction to Solids Crystalline solids are classified according to the nature of the bonding, or according to the geometry and symmetry of the packing arrangement. Let’s focus first on the nature of the bonding. Crystalline solids ionicNetwork covalent molecularmetallic In an ionic solid, positive and negative ions are held in their lattice positions by (strong) ionic bonding forces. NaCl T m = 800 o C LiF T m = 848 o C ZnO T m = 1974 o C In a network covalent solid, atoms are held in their lattice positions by (strong) covalent bonds. SiO 2 T m = 1600 o C SiCT m =2830 o C C diamond T m = 4440 o C (at P = 12.4 GPa) In a molecular solid, molecules are held in their lattice positions by (weak) intermolecular forces. H 2 OT m = 0 o C S 8 T m = 113 o C I 2 T m = 114 o C In a metallic solid, metal cations are held in their lattice positions by (strong) metallic bonding forces. CuT m = 1083 o C AgT m = 962 o C WT m = 3422 o C

98 Introduction to Solids Crystalline solids are classified according to the nature of the bonding, or according to the geometry and symmetry of the packing arrangement. Let’s focus first on the nature of the bonding. Crystalline solids ionicNetwork covalent molecularmetallic In an ionic solid, positive and negative ions are held in their lattice positions by (strong) ionic bonding forces. NaCl T m = 800 o C LiF T m = 848 o C ZnO T m = 1974 o C In a network covalent solid, atoms are held in their lattice positions by (strong) covalent bonds. SiO 2 T m = 1600 o C SiCT m =2830 o C C diamond T m = 4440 o C (at P = 12.4 GPa) In a molecular solid, molecules are held in their lattice positions by (weak) intermolecular forces. H 2 OT m = 0 o C S 8 T m = 113 o C I 2 T m = 114 o C In a metallic solid, metal cations are held in their lattice positions by (strong) metallic bonding forces. CuT m = 1083 o C AgT m = 962 o C WT m = 3422 o C

99 Introduction to Solids Metallic solids can be viewed as an array of metal cations bathing in a sea of valence electrons A “sea” of delocalized valence electrons Electron sea model of metallic bonding

100 Introduction to Solids When we focus on the geometry of the packing arrangements, we find that there are seven basic shapes for unit cells. The shape of each unit cell is described in terms of three lengths (a, b and c) and three angles ( , , and  ). We will focus primarily on cubic unit cells. The others are mentioned only to emphasize that there are other types/shapes of unit cells besides cubic unit cells. cubic a = b = c  =  =  = 90 o trigonal a = b = c  =  =  ≠ 90 o tetragonal a = b ≠ c  =  =  = 90 o hexagonal a = b ≠ c  =  = 90 o ; and  = 120 o monoclinic a ≠ b ≠ c  =  = 90 o and  ≠ 90 o triclinic a ≠ b ≠ c  ≠  ≠  ≠ 90 o orthorhombic a ≠ b ≠ c  =  =  = 90 o Crystalline solids a b c   

101 Introduction to Solids When we focus on the geometry of the packing arrangements, we find that there are seven basic shapes for unit cells. The shape of each unit cell is described in terms of three lengths (a, b and c) and three angles ( , , and  ). We will focus primarily on cubic unit cells. The others are mentioned only to emphasize that there are other types/shapes of unit cells besides cubic unit cells. cubic a = b = c  =  =  = 90 o trigonal a = b = c  =  =  ≠ 90 o tetragonal a = b ≠ c  =  =  = 90 o hexagonal a = b ≠ c  =  = 90 o ; and  = 120 o monoclinic a ≠ b ≠ c  =  = 90 o and  ≠ 90 o triclinic a ≠ b ≠ c  ≠  ≠  ≠ 90 o orthorhombic a ≠ b ≠ c  =  =  = 90 o Crystalline solids a a a

102 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

103 Introduction to Solids For crystalline solids, we use the following concepts to characterize the pattern of the packing arrangement. three dimensional array of points that shows how the structural units are arranged in space. Crystal lattice Each structural unit could be a single atom, a molecule or an ion. In C(s), the structural unit is a carbon atom. In I 2 (s), the structural unit is an I 2 molecule. In NH 4 NO 3 (s), the structural units are NH 4 + and NO 3 − ions. the smallest building block that possesses the symmetry of the crystal lattice; a solid sample of any size can be “built” by stacking together unit cells. Unit cell REVIEW

104 Cubic Packing Arrangements To investigate the cubic packing arrangements, we are going to imagine packing together identical, hard spheres each having a radius R. A. Simple cubic packing Let’s consider ”building” a simple cubic packing arrangement on top of a table. We generate the first layer by arranging the spheres as shown (on the left) below. Subsequent layers are added by lining up spheres with those in the previous layer as shown (on the right). Layer 1 Layer 2 Layer 3

105 Cubic Packing Arrangements A. Simple cubic packing (i) What is the relationship between the edge length of the unit cell (a) and the atom’s radius (R)? a a a R a = 2 × R

106 Cubic Packing Arrangements A. Simple cubic packing (ii) What is the number of spheres contained in one unit cell? a a a R Each cell contains eight one eighth of a sphere. Thus there are:

107 Cubic Packing Arrangements A. Simple cubic packing (iii) What is the coordination number of a sphere? a a a R Each sphere touches four spheres in the plane, one on top, and one underneath. Thus each sphere touches 6 other spheres. The coordination number equals 6.

108 Cubic Packing Arrangements A. Simple cubic packing (iv) What is the fraction of empty space in a simple cubic crystal? a a a R The unit cell occupies a volume V UC = (a) 3 = (2×R) 3 = 8×R 3. The unit cell contains one sphere of volume: The fraction of empty volume = (V UC – V sphere )/V UC = 1 –  /6 = 0.476

109 Cubic Packing Arrangements B.Body-centred cubic packing (i) What is the relationship between the edge length of the unit cell (a) and the atom’s radius (R)? Since the atoms are touching along the cube’s diagonal (body diagonal), its length equals: R + 2×R + R = 4×R R a

110 Cubic Packing Arrangements B.Body-centred cubic packing (i)What is the relationship between the edge length of the unit cell (a) and the atom’s radius (R)? a Body Diagonal = 4×R 4×R a a What is the square diagonal?

111 Cubic Packing Arrangements B.Body-centred cubic packing (i)What is the relationship between the edge length of the unit cell (a) and the atom’s radius (R)? a Body Diagonal = 4×R (Square Diagonal) 2 = a 2 + a 2 = 2 × a 2 a a a Square Diagonal 4×R

112 Cubic Packing Arrangements B.Body-centred cubic packing (i)What is the relationship between the edge length of the unit cell (a) and the atom’s radius (R)? a Body Diagonal = 4×R 4×R (Square Diagonal) 2 = a 2 + a 2 = 2 × a 2 Square Diagonal (Body Diagonal) 2 = a 2 + (  2 × a) 2 = 3 ×a 2 Body Diagonal =  3 ×a = 4×R or a

113 Cubic Packing Arrangements B.Body-centred cubic packing (ii) What is the number of spheres contained in one unit cell? Each corner is occupied by one eighth of an atom and there is a full atom at the center of the unit cell. Thus the cell contains: R a

114 Cubic Packing Arrangements B.Body-centred cubic packing (iii) What is the coordination number of a sphere? The atom at the center of the unit cell touches 8 other atoms. The coordination number equals 8. R a

115 Cubic Packing Arrangements B.Body-centred cubic packing (iv) What is the fraction of empty space in a body centered cubic crystal? R a The unit cell occupies a volume V UC = (a) 3 = The unit cell contains two spheres, each sphere of volume: The fraction of empty volume = (V UC – 2 × V sphere )/V UC =

116 Cubic Packing Arrangements C. Face-centred cubic packing (i) What is the relationship between the edge length of the unit cell (a) and the atom’s radius (R)? R a

117 Cubic Packing Arrangements C. Face-centred cubic packing (i) What is the relationship between the edge length of the unit cell (a) and the atom’s radius (R)? a a a (Square Diagonal) 2 = a 2 + a 2 = 2 ×a 2 Square Diagonal =  2 ×a = 4×R or

118 Cubic Packing Arrangements C. Face-centred cubic packing (ii) What is the number of spheres contained in one unit cell? R a Each one of the eight corners is occupied by one eighth of an atom and each one of the 6 faces is occupied by one half of an atom.

119 Cubic Packing Arrangements C. Face-centred cubic packing (iii) What is the coordination number of a sphere? The atom at the center of the face touches 4 atoms in the plane shown on the figure.

120 Cubic Packing Arrangements C. Face-centred cubic packing (iii) What is the coordination number of a sphere? The atom at the center of the face touches 4 more atoms in the plane shown on the figure.

121 Cubic Packing Arrangements C. Face-centred cubic packing (iii) What is the coordination number of a sphere? The atom at the center of the face touches 4 more atoms in the plane shown on the figure.

122 Cubic Packing Arrangements C. Face-centred cubic packing (iii) What is the coordination number of a sphere? In total, the atom at the center of the face touches 12 atoms. The coordination number equals 12.

123 Cubic Packing Arrangements C. Face-centred cubic packing (iv) What is the fraction of empty space in a face-centered cubic crystal? R a The unit cell occupies a volume V UC = (a) 3 = The unit cell contains four spheres, each sphere of volume: The fraction of empty volume = (V UC – 4 × V sphere )/V UC =

124 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

125 Closest Packed Structures Same layer as the layer with the grey balls. Different layer from the layer with the grey balls.

126 Closest Packed Structures “x” type dimple (vertically aligned with a sphere in layer 1) “y” type dimple (not vertically aligned with a sphere in layer 1 or layer 2) There are two types of “dimples” to choose from when forming layer 3!!

127 Closest Packed Structures choose “x” only choose “y” only Spheres in layer 3 are vertically aligned with those in layer 1. That is, layer 3 is a repeat of layer 1. ABAB... closest-packing Spheres in layer 3 are not aligned vertically with those in layer 1 or layer 2. Layer 3 is a distinct layer. ABCABC... closest-packing “x” type dimple (vertically aligned with a sphere in layer 1) “y” type dimple (not vertically aligned with a sphere in layer 1 or layer 2)

128 Spheres in layer 3 are vertically aligned with those in layer 1. That is, layer 3 is a repeat of layer 1. ABAB... closest-packing Spheres in layer 3 are not aligned vertically with those in layer 1 or layer 2. Layer 3 is a distinct layer. ABCABC... closest-packing Closest Packed Structures The unit cell for ABAB... closest-packing is “hexagonal”. Thus, this type of closest-packing is also called “hexagonal closest-packing” or hcp for short. The unit cell for ABCABC... closest- packing is “face-centred cubic”. Thus, this type of closest-packing is also called “cubic closest-packing” or ccp for short.

129 Closest Packed Structures The unit cell for ABAB... closest-packing is “hexagonal”. Thus, this type of closest-packing is also called “hexagonal closest-packing” or hcp for short. The unit cell for ABCABC... closest- packing is “face-centred cubic”. Thus, this type of closest-packing is also called “cubic closest-packing” or ccp for short. You are expected to remember that cubic closest-packing (ccp) and face-centered cubic (fcc) are the same!!

130 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

131 Density of a Crystalline Solid Keep in mind: The formula d = n cell M atom /(N A a 3 ) should be used only for identical hard spheres in a cubic lattice. If you have more than one type of sphere (e.g. of different sizes or masses) or unit cell that is not cubic, then you should start from d = m cell / V cell. The density of a solid depends on the microscopic details of the packing arrangement. mass of one atom M atom is molar mass, in g mol −1 N A = 6.022×10 23 mol −1

132 Density of a Crystalline Solid Example: Tungsten, W, crystallizes in one of the three cubic arrangements. If the edge length of the unit cell is 321 pm and the density is 18.5 g cm -3, then what type of crystal lattice does tungsten have? What is the radius of a tungsten atom? M atom = 184 g.mol 

133 Density of a Crystalline Solid Example: Tungsten, W, crystallizes in one of the three cubic arrangements. If the edge length of the unit cell is 321 pm and the density is 18.5 g cm -3, then what type of crystal lattice does tungsten have? What is the radius of a tungsten atom? M atom = 184 g.mol 

134 Density of a Crystalline Solid Example: Tungsten, W, crystallizes in one of the three cubic arrangements. If the edge length of the unit cell is 321 pm and the density is 18.5 g cm -3, then what type of crystal lattice does tungsten have? What is the radius of a tungsten atom? Simple cubic Body-centered cubic Face-centered cubic # of atoms/unit cell Relationship between a and R R = a/2

135 Density of a Crystalline Solid The edge length can be determined experimentally using x-ray diffraction. When x-rays are passed through a crystalline solid, the x-rays are deflected from their paths by the atoms of the solid and interfere with each other to produce an interference pattern – a “diffraction pattern” – that can be analyzed to determine the geometry of the crystal lattice. Simple cubic Body-centered cubic Face-centered cubic # of atoms/unit cell Relationship between a and R R = a/2

136 Introduction to Solids For crystalline solids, we use the following concepts to characterize the pattern of the packing arrangement. three dimensional array of points that shows how the structural units are arranged in space. Crystal lattice Each structural unit could be a single atom, a molecule or an ion. In C(s), the structural unit is a carbon atom. In I 2 (s), the structural unit is an I 2 molecule. In NH 4 NO 3 (s), the structural units are NH 4 + and NO 3 − ions. the smallest building block that possesses the symmetry of the crystal lattice; a solid sample of any size can be “built” by stacking together unit cells. Unit cell REVIEW

137 Introduction to Solids When we focus on the geometry of the packing arrangements, we find that there are seven basic shapes for unit cells. The shape of each unit cell is described in terms of three lengths (a, b and c) and three angles ( , , and  ). We will focus primarily on cubic unit cells. The others are mentioned only to emphasize that there are other types/shapes of unit cells besides cubic unit cells. cubic a = b = c  =  =  = 90 o trigonal a = b = c  =  =  ≠ 90 o tetragonal a = b ≠ c  =  =  = 90 o hexagonal a = b ≠ c  =  = 90 o ; and  = 120 o monoclinic a ≠ b ≠ c  =  = 90 o and  ≠ 90 o triclinic a ≠ b ≠ c  ≠  ≠  ≠ 90 o orthorhombic a ≠ b ≠ c  =  =  = 90 o Crystalline solids a b c    REVIEW

138 Introduction to Solids When we focus on the geometry of the packing arrangements, we find that there are seven basic shapes for unit cells. The shape of each unit cell is described in terms of three lengths (a, b and c) and three angles ( , , and  ). We will focus primarily on cubic unit cells. The others are mentioned only to emphasize that there are other types/shapes of unit cells besides cubic unit cells. cubic a = b = c  =  =  = 90 o trigonal a = b = c  =  =  ≠ 90 o tetragonal a = b ≠ c  =  =  = 90 o hexagonal a = b ≠ c  =  = 90 o ; and  = 120 o monoclinic a ≠ b ≠ c  =  = 90 o and  ≠ 90 o triclinic a ≠ b ≠ c  ≠  ≠  ≠ 90 o orthorhombic a ≠ b ≠ c  =  =  = 90 o Crystalline solids a a a REVIEW

139 Density of a Crystalline Solid The edge length can be determined experimentally using x-ray diffraction. When x-rays are passed through a crystalline solid, the x-rays are deflected from their paths by the atoms of the solid and interfere with each other to produce an interference pattern – a “diffraction pattern” – that can be analyzed to determine the geometry of the crystal lattice. Simple cubic Body-centered cubic Face-centered cubic # of atoms/unit cell Relationship between a and R R = a/2 REVIEW R = 0.43 × a R = 0.35 × a R = 0.50 × a

140 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

141 Ionic Solids and Interstitial Sites Up to this point, we’ve focused on packing together identical hard spheres. Now, we’ll extend that model so that we can describe the structures of binary ionic solids. For many ionic solids, it is often the case that one of the ions forms a cubic lattice and the other ion occupies “holes” in that lattice. In order to understand the structures of ionic crystals, we must first examine the types of holes we can have. “Holes” (or interstitial sites) are named according to the “coordination number” of a small sphere that just fits into that hole. The types of holes that interest us the most are cubic, tetrahedral and octahedral holes.

142 Ionic Solids and Interstitial Sites The small sphere has a coordination number of three. trigonal hole

143 Ionic Solids and Interstitial Sites tetrahedral hole coordination # = 4

144 Ionic Solids and Interstitial Sites octahedral hole coordination # = 6 side view top view

145 Ionic Solids and Interstitial Sites cubic hole coordination # = 8

146 Ionic Solids and Interstitial Sites Example: Within the face-centered cubic (fcc) unit cell, there are both octahedral and tetrahedral holes. Examine the fcc unit cell to identify the locations of the octahedral and tetrahedral holes. How many holes of each type are there per unit cell? ¼ of an octahedral hole One octahedral hole

147 Ionic Solids and Interstitial Sites ¼ of an octahedral hole One octahedral hole There is ¼ of an octahedral hole (per edge) and an octahedral hole at the centre of the cell. For the fcc cell, there are 4 spheres per cell and 4 octahedral holes holes. The ratio of spheres-to-holes is: Example: Within the face-centered cubic (fcc) unit cell, there are both octahedral and tetrahedral holes. Examine the fcc unit cell to identify the locations of the octahedral and tetrahedral holes. How many holes of each type are there per unit cell? # octahedral holes = ¼ hole/edge × 12 edges + 1 = 4 (per fcc unit cell) # spheres : # octahedral holes = 4 : 4 = 1 : 1

148 Ionic Solids and Interstitial Sites a b c d tetrahedral hole Example: Within the face-centered cubic (fcc) unit cell, there are both octahedral and tetrahedral holes. Examine the fcc unit cell to identify the locations of the octahedral and tetrahedral holes. How many holes of each type are there per unit cell? There is one tetrahedral hole associated with each corner of the fcc cell (that is, the sphere at each corner can be considered to be the “cap” of a tetrahedron). Thus, For the fcc cell, there are 4 spheres per celland 8 tetrahedral holes. The ratios of spheres-to-holes are: # spheres : # tetrahedral holes = 4 : 8 = 1 : 2 # tetrahedral holes = 8 (per fcc cell)

149 Ionic Solids and Interstitial Sites a b c d tetrahedral hole Example: Within the face-centered cubic (fcc) unit cell, there are both octahedral and tetrahedral holes. Examine the fcc unit cell to identify the locations of the octahedral and tetrahedral holes. How many holes of each type are there per unit cell? In general: In a closest-packed structure containing N spheres, where N is a very large number, there are N octahedral holes and 2N tetrahedral holes.

150 Ionic Solids and Interstitial Sites How big are tetrahedral, octahedral and cubic holes? Octahedral holes. Let r o be the radius of a sphere that just fits into the octahedral hole. 2R2R 2R2R 2ro2ro Let r o be the radius of a sphere that just fits into the octahedral hole.

151 Ionic Solids and Interstitial Sites How big are tetrahedral, octahedral and cubic holes? Cubic holes: Let r o be the radius of a sphere that just fits into the cubic hole. Let r c be the radius of a sphere that just fits into the cubic hole. This sphere has its centre at the midpoint of the body- diagonal. Therefore, we need to find the length of the body-diagonal. 2R2R

152 Ionic Solids and Interstitial Sites How big are tetrahedral, octahedral and cubic holes? Cubic holes: Let r o be the radius of a sphere that just fits into the cubic hole. a a Diagonal 2 =Diagonal =

153 Ionic Solids and Interstitial Sites How big are tetrahedral, octahedral and cubic holes? Cubic holes: Let r o be the radius of a sphere that just fits into the cubic hole. a

154 Ionic Solids and Interstitial Sites How big are tetrahedral, octahedral and cubic holes? Tetrahedral holes: Let r o be the radius of a sphere that just fits into the tetrahedral hole. b cd a b c d a 2×ro2×ro

155 Ionic Solids and Interstitial Sites How big are tetrahedral, octahedral and cubic holes? Tetrahedral holes: Let r o be the radius of a sphere that just fits into the tetrahedral hole. b cd a b c d a 2×ro2×ro 2×R2×R x ?

156 Ionic Solids and Interstitial Sites How big are tetrahedral, octahedral and cubic holes? Tetrahedral holes: Let r o be the radius of a sphere that just fits into the tetrahedral hole. b cd a b c d a x x 2 ×R

157 Ionic Solids and Interstitial Sites How big are tetrahedral, octahedral and cubic holes? Tetrahedral holes: Let r o be the radius of a sphere that just fits into the tetrahedral hole. b cd a b c d a 2×ro2×ro 2×R2×R x =

158 Ionic Solids and Interstitial Sites How big are tetrahedral, octahedral and cubic holes? Tetrahedral holes: Let r o be the radius of a sphere that just fits into the tetrahedral hole. b cd a b c d a 2×ro2×ro 2×R2×R x =

159 HolesRelationship TrigonalToo small to worry about Tetrahedral Octahedral cubic Ionic Solids and Interstitial Sites Summary of hole types and the radius ratio rules Trigonal Tetrahedral Octahedral cubic

160 Ionic Solids and Interstitial Sites Radius Ratio Rules for Ionic Solids Let R + be the radius of the positive ion in a binary ionic solid and let R − be the radius of the negative ion. Positive ions occupy tetrahedral sites in the lattice of negative ions. The coordination number of each positive ion is 4. Positive ions occupy octahedral sites in the lattice of negative ions. The coordination number of each positive ion is 6. Positive ions occupy cubic sites in the lattice of negative ions. The coordination number of each positive ion is 8. Note: The rules above are just guidelines. There are many exceptions. Note: The positive ions are too big for the tetrahedral sites and thus the negative ions are forced apart a little  this increases the attraction between the “+” and “−” ions and decreases the repulsion between the “−” ions.

161 Ionic Solids and Interstitial Sites Sodium Chloride Structure (also called “rock salt structure”) Cl − ions form a face-centered cubic (fcc) lattice and Na + ions occupy 100% of the octahedral holes in the chloride lattice. Cl  Na 

162 Ionic Solids and Interstitial Sites Cesium Chloride Structure Cl  Cs  Cl − ions form a simple cubic lattice and Cs + ions occupy cubic holes in the chloride lattice.

163 Ionic Solids and Interstitial Sites Zinc Blende Structure S 2− ions form a face-centered cubic (fcc) lattice and Zn 2+ ions occupy half (50%) of the tetrahedral hole. Why are only half of the tetrahedral holes occupied? How many spheres:4 S  How many tetrahedral holes:8 for only 4 Zn 2+

164 Ionic Solids and Interstitial Sites Zinc Blende Structure S 2− ions form a face-centered cubic (fcc) lattice and Zn 2+ ions occupy half (50%) of the tetrahedral hole. Why are only half of the tetrahedral holes occupied? There are 4 S 2− ions per cell. There must be four Zn 2+ ions per cell because ZnS is a 1:1 salt. Therefore, only 4 of 8 tetrahedral holes are occupied by Zn 2+ ions.

165 Introduction to Solids For crystalline solids, we use the following concepts to characterize the pattern of the packing arrangement. three dimensional array of points that shows how the structural units are arranged in space. Crystal lattice Each structural unit could be a single atom, a molecule or an ion. In C(s), the structural unit is a carbon atom. In I 2 (s), the structural unit is an I 2 molecule. In NH 4 NO 3 (s), the structural units are NH 4 + and NO 3 − ions. the smallest building block that possesses the symmetry of the crystal lattice; a solid sample of any size can be “built” by stacking together unit cells. Unit cell REVIEW

166 Introduction to Solids When we focus on the geometry of the packing arrangements, we find that there are seven basic shapes for unit cells. The shape of each unit cell is described in terms of three lengths (a, b and c) and three angles ( , , and  ). We will focus primarily on cubic unit cells. The others are mentioned only to emphasize that there are other types/shapes of unit cells besides cubic unit cells. cubic a = b = c  =  =  = 90 o trigonal a = b = c  =  =  ≠ 90 o tetragonal a = b ≠ c  =  =  = 90 o hexagonal a = b ≠ c  =  = 90 o ; and  = 120 o monoclinic a ≠ b ≠ c  =  = 90 o and  ≠ 90 o triclinic a ≠ b ≠ c  ≠  ≠  ≠ 90 o orthorhombic a ≠ b ≠ c  =  =  = 90 o Crystalline solids a b c    REVIEW

167 Introduction to Solids When we focus on the geometry of the packing arrangements, we find that there are seven basic shapes for unit cells. The shape of each unit cell is described in terms of three lengths (a, b and c) and three angles ( , , and  ). We will focus primarily on cubic unit cells. The others are mentioned only to emphasize that there are other types/shapes of unit cells besides cubic unit cells. cubic a = b = c  =  =  = 90 o trigonal a = b = c  =  =  ≠ 90 o tetragonal a = b ≠ c  =  =  = 90 o hexagonal a = b ≠ c  =  = 90 o ; and  = 120 o monoclinic a ≠ b ≠ c  =  = 90 o and  ≠ 90 o triclinic a ≠ b ≠ c  ≠  ≠  ≠ 90 o orthorhombic a ≠ b ≠ c  =  =  = 90 o Crystalline solids a a a REVIEW

168 Density of a Crystalline Solid The edge length can be determined experimentally using x-ray diffraction. When x-rays are passed through a crystalline solid, the x-rays are deflected from their paths by the atoms of the solid and interfere with each other to produce an interference pattern – a “diffraction pattern” – that can be analyzed to determine the geometry of the crystal lattice. Simple cubic Body-centered cubic Face-centered cubic # of atoms/unit cell Relationship between a and R R = a/2 REVIEW R = 0.43 × a R = 0.35 × a R = 0.50 × a

169 HolesRelationship TrigonalToo small to worry about Tetrahedral Octahedral cubic Ionic Solids and Interstitial Sites Summary of hole types and the radius ratio rules Trigonal Tetrahedral Octahedral cubic REVIEW

170 Ionic Solids and Interstitial Sites Fluorite Structure Ca 2+ ions form a fcc lattice and F − ions occupy all of the tetrahedral holes. Thus, there are 4 Ca 2+ ions per unit cell and 8 F − ions. How many spheres:4 Ca 2+ How many tetrahedral holes:8 F 

171 Ionic Solids and Interstitial Sites Antifluorite Structure Sodium oxide, Na 2 O, adopts the so-called antifluorite structure. The antifluorite structure is the “reverse” of the fluorite structure in the sense that the negative ions form a fcc lattice and the positive ions occupy all of the tetrahedral holes. In the case of Na 2 O, the O 2− ions form a fcc lattice and the Na + ions occupy 100% of the tetrahedral holes. O  Na 

172 Ionic Solids and Interstitial Sites Example: In beryllium oxide, BeO, the oxide ions form a face-centred cubic lattice and the beryllium ions occupy tetrahedral sites in the lattice of O 2- ions. What fraction of the holes is occupied by the Be 2+ ions? (Ans: 50%) How many spheres:4 O 2  How many tetrahedral holes:8 tetrahedral holes for Be 2  With 8 negative charges from O  anions and 16 positive charges from Be 2+ cations, the charge balance would not work. Only 4 Be 2+ generating 8 positive charges can be present. 50% of the tetrahedral holes can be occupied by Be 2+ cations.

173 Ionic Solids and Interstitial Sites Example: To “build” the sodium chloride structure, we would first arrange Cl  ions in a face-centred cubic structure and then insert Na + ions into the octahedral sites of the Cl  lattice. Assume that Na + and Cl  are hard spheres with radii of 97 pm and 181 pm, respectively. Calculate the density of sodium chloride, in g cm  3. (Ans: 2.26 g cm −3 ) Cl  Na  a 4 spheres (Cl  ) 4 octahedral holes (Na + ) a = 2×(97+181) = 556 pm

174 Cl  Cs  Ionic Solids and Interstitial Sites Example: In cesium chloride, the Cl  ions form a simple cubic lattice and the Cs + ions occupy cubic holes in the chloride lattice. However, the Cs + ions force the Cl  ions apart so that none of the chloride ions are in direct contact with each other. If the density of cesium chloride, CsCl, is g cm  3, then what is the distance (R + + R  ) ? (Ans: 357 pm) a 1 spheres (Cl  ) 1 cubic hole (Cs + ) Body Diagonal =  3a where

175 Cl  Cs  Ionic Solids and Interstitial Sites Example: In cesium chloride, the Cl  ions form a simple cubic lattice and the Cs + ions occupy cubic holes in the chloride lattice. However, the Cs + ions force the Cl  ions apart so that none of the chloride ions are in direct contact with each other. If the density of cesium chloride, CsCl, is g cm  3, then what is the distance (R + + R  ) ? (Ans: 357 pm) a Body Diagonal =  3a

176 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle

177 The Born Haber Cycle The stability of an ionic solid is quantified in terms of its lattice energy. Lattice Energy = energy change when gas phase ions combine to form an ionic solid Na + (g) + Cl − (g) → NaCl(s) Lattice energy (  H lattice ) for NaCl(s)  H lattice The lattice energy cannot be measured directly. However, we can obtain it indirectly from other thermochemical data using the Born-Haber cycle. The Born-Haber cycle involves the following steps: Write down the “formation reaction” for the solid from the elements under standard conditions (T = 25 o C and P = 10 5 bar). Convert the elements into gas-phase atoms. Convert the atoms into gas-phase ions. Combine the ions to form the solid.

178 The Born Haber Cycle Born-Haber cycle for NaCl(s) Write down the “formation reaction” for the solid. Convert the elements into gas-phase atoms. Convert the atoms into gas-phase ions. Combine the ions to form the solid. Na(s) + ½ Cl 2 (g)NaCl(s) HfoHfo Na(g) + Cl(g)  H sub o ½ D Cl-Cl Na + (g) + Cl  (g)  H lattice o (dissociation) (sublimation) (first ionization energy) IE(1) (first electron affinity) EA(1)

179 The Born Haber Cycle Born-Haber cycle for NaCl(s) State function does not depend on the path: University of d(home – UW) = constant path(UW  home) varies

180 The Born Haber Cycle Born-Haber cycle for NaCl(s) Write down the “formation reaction” for the solid. Convert the elements into gas-phase atoms. Convert the atoms into gas-phase ions. Combine the ions to form the solid. Na(s) + ½ Cl 2 (g)NaCl(s) HfoHfo Na(g) + Cl(g)  H sub o ½ D Cl-Cl Na + (g) + Cl  (g)  H lattice o (dissociation) (sublimation) (first ionization energy) IE(1) (first electron affinity) EA(1) Apply Hess’ Law:  H f o =  H sub o + IE(1) Na(g) + ½ D Cl-Cl + EA(1) Cl(g) +  H lattice Thus:  H lattice =  (  H sub o + IE(1) Na(g) + ½ D Cl-Cl + EA(1) Cl(g) )   H f o

181 The Born Haber Cycle Born-Haber cycle for MgF 2 (s) Write down the “formation reaction” for the solid. Convert the elements into gas-phase atoms. Convert the atoms into gas-phase ions. Combine the ions to form the solid. Mg(s) + F 2 (g)MgF 2 (s) HfoHfo Mg(g) + 2 F(g)  H sub o D F-F Mg 2+ (g) + 2 F  (g)  H lattice o (dissociation) (sublimation) (first and second ionization energy) IE(1) + IE(2) (first electron affinity) EA(1) Apply Hess’ Law:  H f o =  H sub o + IE(1) Mg(g) +IE(2) Mg(g) + D F-F + 2 × EA(1) F(g) +  H lattice Thus:  H lattice =  (  H sub o + IE(1) Mg(g) +IE(2) Mg(g) + D F-F + 2 × EA(1) F(g) ) +  H f o

182 The Born Haber Cycle Born-Haber cycle for MgF 2 (s) Hess’ Law:  H f o =  H sub o + IE(1) Mg(g) +IE(2) Mg(g) + D F-F + 2 × EA(1) F(g) +  H lattice Born-Haber cycle for NaCl(s) Hess’ Law:  H f o =  H sub o + IE(1) Na(g) + ½ D Cl-Cl + EA(1) Cl(g) +  H lattice By comparing the expressions we wrote for NaCl(s) and MgCl 2 (s), we can see that the expression we obtain for ΔH f o will vary from salt to salt. Therefore, to obtain the correct expression, you must first construct the Born-Haber cycle.

183 Intermolecular Forces: Liquids and Solids ● Phases and Phase Diagrams ● Liquids and Liquid Properties ● Intermolecular Forces ● Heating Curves ● Introduction to Solids ● Cubic Packing Arrangements ● Closest-Packed Structures ● Density of a Crystalline Solid ● Ionic Solids and Interstitial Sites ● The Born-Haber Cycle All Done!


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