Intermolecular Forces and Liquids and Solids

1 Intermolecular Forces and Liquids and Solids

y Kinetic Molecular Theory of Liquids and Solids (12.1) y Intermolecular Forces (12.2) y Properties of Liquids (12.3) y Crystal Structures (12.4) y Bonding in Solids (12.5) y Phase Changes (12.6) y Phase Diagrams (12.7) 12.1 The Kinetic Molecular Theory of Liquids and Solids

y How are solids and liquids modeled on the particle-level? y How does this help explain macroscopic properties of solids and liquids? Intermolecular Forces and Liquids and Solids H2O(g)

H2O(l)

H2O(s)

Figure 1.2, p. 4 12.1 The Kinetic Molecular Theory of Liquids and Solids

Table 12.1, p. 403 12.1 The Kinetic Molecular Theory of Liquids and Solids

y Does the molecular structure of the substance on the particle level affect function or properties on the macroscopic level?

We finally connect between the abstract (the bonding and structure of substances) to the observed (the function and properties of substances) 12.2 Intermolecular Forces y What are intramolecular forces: ◦ Forces that hold atoms together in a molecule y What are intermolecular forces: ◦ attractive forces between molecules y How do we model these? Intermolecular force (between molecules) ◦ Let’s consider water:

Intramolecular force (covalent bonds in molecules) 12.2 Intermolecular Forces Intermolecular vs Intramolecular •41 kJ to vaporize 1 mole of water (inter) • 930 kJ to break all O‐H bonds in 1 mole of water (intra) Intermolecular force Generally, intermolecular (between molecules) forces are much weaker than intramolecular forces. Intramolecular force (covalent bonds in molecules) 12.2 Intermolecular Forces

“Measure” of intermolecular force: boiling point melting point Δ Δ Δ Hvap Hfus Hsub Intermolecular force (between molecules)

Intramolecular force (covalent bonds in molecules) 12.2 Intermolecular Forces

Focus on pure substances first: y How do we model pure substances? y How are intermolecular forces separated into categories?

We will use structure and size to assist in approximating intermolecular forces and connect this to physical properties 12.2 Intermolecular Forces

y Can all substances be liquefied? y Why? y Dispersion Forces: ◦ Attractive forces that arise as a result of a temporary dipole induced in atoms or molecules IMF – Substances Pure

Figure 12.5, p. 406 12.2 Intermolecular Forces

y Polarizability: ◦ The ease with which the electron distribution in the atom or molecule can be distorted y Polarizability increases with greater number of electrons or more diffuse electron cloud

Dispersion forces usually increase with molar mass. As dispersion forces increase, intermolecular forces increase. IMF – Substances Pure Therefore, it takes more energy to overcome these and the melting points and boiling points increase 12.2 Intermolecular Forces

y What are the states of the halogens under standard conditions? IMF – Substances Pure

Table 12.2, p. 406 12.2 Intermolecular Forces

y Permanent vs. induced dipole moments y Dipole Forces: ◦ Attractive forces between polar molecules IMF – Substances Pure

Figure 12.1, p. 404 12.2 Intermolecular Forces

y What is the trend of boiling points of HI, HBr, HCl and HF?

350 300 250 200 150

IMF – Substances Pure 100

Boiling point (K) 50 0 HF HCl HBr HI 12.2 Intermolecular Forces

y Specific type of dipole force y Hydrogen bonding (IMF – not covalent bond): ◦ Special type of dipole-dipole interaction between the hydrogen atom in a polar bond, such as N-H, O-H or F-H and an electronegative O, N, or F atom. IMF – Substances Pure

Figure, p. 409 12.2 Intermolecular Forces IMF – Substances Pure

Figure 12.7, p. 405 12.2 Intermolecular Forces Hydrogen bonding Greater IMF Higher boiling pt

Decreasing molar mass Decreasing boiling point No hydrogen bonding IMF – Substances Pure

Figure 12.6, p. 408 12.2 Intermolecular Forces

Dispersion forces

Dipole forces

Hydrogen bonding IMF – Substances Pure 12.2 Intermolecular Forces y Using IMF with pure substances: Approximating relative melting or boiling points 1. Determine the intermolecular forces of the substances (usually comparing between two different pure substances) 2. Consider the number of intermolecular forces: If they are different, then the substance with the higher number of intermolecular forces has the higher boiling point If they are the same, then the substance with the IMF – Substances Pure higher number of electrons has the higher boiling point 12.2 Intermolecular Forces

y Using IMF with pure substances: Which has the higher boiling point? 1. Determine the IMF 2. Consider the number of intermolecular forces: If they are different, then the substance with the higher number of intermolecular forces has the higher boiling point If they are the same, then the substance with the higher number of electrons has the higher boiling point

Boiling Substance Point, oC IMF – Substances Pure C2H4 -104

NF3 71 NCl3

C2H4 States at room temperature? 12.2 Intermolecular Forces

Boiling H H H H H H Substance Point, oC

IMF – Substances Pure H C C H C H C C O H O butanol 157

H H H H methanol 65 butanol methanol States at room temperature? 12.2 Intermolecular Forces IMF – Mixtures

Figure 12.7, p. 408 12.2 Intermolecular Forces

Now focus on mixtures: y How do we model mixtures? y How do the intermolecular forces and sizes of molecules affect solubility?

We will use structure and size to assist in IMF – Mixtures approximating intermolecular forces and connect this to physical properties Think about this like a “substitution” 12.2 Intermolecular Forces IMF – Mixtures

Margin Figure, p. 409 12.2 Intermolecular Forces

y Using IMF with mixtures: Approximating solubility 1. Determine the intermolecular forces of the substances (usually considering the solubility of two pure substances) 2. Determine the relative size of the two molecules (same or different) If both (IMF and size) are the same, then we would IMF – Mixtures approximate that the substances are soluble If both (IMF and size) are different, then we could approximate that the substances are insoluble If one (IMF or size) are the same, then we would approximate that the substances are partially soluble 12.2 Intermolecular Forces

y Using IMF with mixtures: What is the solubility of the two substances? 1. Determine the IMF 2. Determine the relative size of the two molecules (same or different) If both (IMF and size) are the same, then the substances are soluble If both (IMF and size) are different, then the substances are insoluble If one (IMF or size) are the same, then the substances are partially soluble IMF – Mixtures

H H

C H Solubility (g in 100g) H O Infinitely (miscible) methanol water What does this look like? 12.2 Intermolecular Forces

H H H H

H C C H Solubility (g in 100g) C C O 8.88 H H H H butanol water What does this look like? 12.2 Intermolecular Forces

Alcohol Solubility (g in 100g H2O)

Infinitely (miscible)

8.88

2.73

IMF – Mixtures 0.602

0.174

What do these look like? 12.2 Intermolecular Forces

y Using IMF with mixtures: Determining IMF that exist between the substances: ◦ Dispersion forces (or induced dipole-induced dipole) between all substances in a solution. ◦ A nonpolar and a polar substance would have induced dipole-dipole forces ◦ A polar and a polar substance would have dipole-dipole forces IMF – Mixtures ◦ A ionic and a nonpolar substance would have ion-induced dipole forces ◦ A ionic and a polar substance would have ion-dipole forces 12.2 Intermolecular Forces

y Using IMF with mixtures: What are the IMF between the two substances? Determine the IMF x Dispersion forces (or induced dipole-induced dipole) between all substances in a solution. x A nonpolar and a polar substance would have induced dipole-dipole forces x A polar and a polar substance would have dipole-dipole forces x A ionic and a nonpolar substance would have ion-induced dipole forces x A ionic and a polar substance would have ion-dipole forces

IMF – Mixtures H H

C H H O

methanol water

dipole-dipole forces dispersion forces 12.2 Intermolecular Forces

Salt, NaCl water

ion-dipole forces dispersion forces Which molecule does not have the intermolecular force of hydrogen bonding as a pure I II III substance? (A) Only I(B)Iand II (C) I and III (D) II and III Chapter 12 –Chapter Practice 12.3 Properties of Liquids y How is structure related to other properties of liquids (surface tension, viscosity)? y What is surface tension? ◦ The amount of energy required to stretch or increase the surface of a liquid by a unit area ◦ Would you expect higher intermolecular forces, higher surface tension

Figure 12.8, p. 410 12.3 Properties of Liquids y What is cohesion? Adhesion and ◦ The IMF attraction between like cohesion molecules capillary action y What is adhesion? ◦ An attraction between unlike molecules ◦ Would expect similar intermolecular forces, better adhesion y How can these work together? Capillary action

Figure 12.10, p. 411 12.3 Properties of Liquids y What is cohesion? Only Cohesion ◦ The IMF attraction between like no capillary molecules action y What is adhesion? ◦ An attraction between unlike molecules ◦ Would expect similar intermolecular forces, better adhesion y What if these don’t work together?

Figure 12.10, p. 411 12.3 Properties of Liquids y What is viscosity? ◦ A measure of a fluid’s resistance to flow ◦ Would expect higher intermolecular forces, higher viscosity

Table 12.3, p. 411 12.3 Properties of Liquids y How is water unique?

Figure 12.13, p. 413 12.3 Properties of Liquids

Maximum Density 40C

Which beaker contains water?

Figures 12.131 and 12.13 p. 412-413 12.4 Crystal Structure

y How do we model solids? y What are two general ways solids form? y Can we model both forms?

Figure 12.14, p. 414 12.4 Crystal Structure

Figure 12.15, p. 414 12.4 Crystal Structure

Figure 12.16, p. 415 12.4 Crystal Structure y How do we determine the number of atoms per unit cell?

Number of Number of atoms per unit Fraction of atom atom(s) in that cell in that Position in that position position position On a corner 1/8 8 1 On a face In the center

Figure 12.18, p. 416 12.4 Crystal Structure y How do we determine the number of atoms per unit cell?

Number of Number of atoms per unit Fraction of atom atom(s) in that cell in that Position in that position position position On a corner 1/8 8 1 On a face 1/2 6 3 In the center

Figure 12.18, p. 416 12.4 Crystal Structure y How do we determine the number of atoms per unit cell?

Number of Number of atoms per unit Fraction of atom atom(s) in that cell in that Position in that position position position On a corner 1/8 8 1 On a face 1/2 6 3 In the center 11 1

Figure 12.18, p. 416 12.4 Crystal Structure

y For each simple cubic, elemental structure, we will answer: 1. Number of atoms per unit cell 2. What is the length of one side of this crystal (in terms of the radius of the atom)? 3. What is the volume of this crystal (assuming the atoms are spheres)? 4. What is the packing efficiency of this crystal? 12.4 Crystal Structure

Figure 12.19, p. 416 12.4 Crystal Structure

Figure 12.19, p. 416 Eu crystallized in a body-centered cubic cell. The density of Eu is 5.26 g·cm–3, what is the unit cell edge length in pm? Practice exercise –Practice exercise crystal structures For this we will need to calculate: - The number of atoms per unit cell. - The mass (in g) of these atoms. - The volume of the unit cell in cm3. - The length of one side of the unit cell in cm. Because Eu is body-centered cubic, we can calculate the number of atoms per unit cell as 2.

Using the density, the molar mass, NA, and the number of atoms per unit cell, we can calculate the volume of the unit cell.

3 ⎛⎞cm⎛⎞⎛ 152.0 gmol2 ⎞⎛⎞ atoms −23− 3 ⎜⎟⎜⎟⎜23 ⎟⎜⎟=×9.60 10 cm ⎝⎠5.26 g⎝⎠⎝ mol6.022× 10 atoms ⎠⎝⎠ unit cell

Because this is a cubic unit cell, the length of one side is:

length of one side (BCC) = 3 Volume=×32338 9.60 10−− cm =× 4.58 10 cm Practice exercise –Practice exercise crystal structures Metallic copper crystallizes in a cubic structure. The length of the edge of the unit cell is 361 pm. If copper has a density of 8.96 g·cm–3, which type of unit cell is present in this sample? Practice exercise –Practice exercise crystal structures For this we will need to calculate: - The volume of the unit cell in cm3. - The mass of the unit cell (using density).

- The number of atoms per unit cell (using NA).

Because copper has a unit cell with an edge length of 361 pm, we will first convert into cm (to use the density) and then cube for the volume of the unit cell. The side length is 361 pm = 361x10-12 m = 361x10-10 cm. The volume of the unit cell is (361x10-10 cm)3 = 4.70x10-23 cm3

The number of atoms per unit cell is:

−23 3 23 ⎛⎞4.70×× 10 cm⎛⎞ 8.96 g⎛⎞ mol ⎛ 6.022 10 atoms ⎞ −1 ⎜⎟⎜⎟3 ⎜⎟ ⎜ ⎟=⋅4 atoms unit cell ⎝⎠unit cell⎝⎠cm ⎝⎠ 63.55 gmol ⎝ ⎠

Practice exercise –Practice exercise crystal structures 4 atoms per unit cell indicates a face-centered cubic packing. When aluminum crystallizes, it forms face-centered cubic cells. The atomic radius of an Al atom is 143 pm. What is the density of metallic aluminum? Practice exercise –Practice exercise crystal structures For this we will need to calculate: - The length of one side of the unit cell in cm. - The volume of the unit cell in cm3. - The number of atoms per unit cell. - The mass (in g) of these atoms. Because aluminum is face-centered cubic, using the radius, we can calculate the length of one side of the cube. The radius is 143 pm = 143x10-12 m = 143x10-10 cm. The face-centered cubic unit cell has the length of one side equal to 4.04x10-8 cm. length of one side (FCC) = 8r

The volume of the unit cell is 6.59x10-23 cm3 There are 4 atoms per unit cell. The density is:

⎛⎞⎛26.98 g mol ⎞⎛⎞⎛⎞ 4 atoms unit cell −3 ⎜⎟⎜23 ⎟⎜⎟⎜⎟− 23 3 =⋅2.72 g cm

Practice exercise –Practice exercise crystal structures ⎝⎠⎝mol6.022×× 10 atoms ⎠⎝⎠⎝⎠ unit cell 6.59 10 cm 12.5 Bonding in Solids y What are the main y How do the general classes of crystalline properties help us solids? differentiate between ◦ Ionic these classes? ◦ Covalent ◦ Type of bonding in the ◦ Molecular solid ◦ Metallic ◦ Melting points ◦ Conductivity (heat or electricity) ◦ Solubility (in polar or nonpolar solvents) 12.5 Bonding in Solids

Properties Ionic Solids Type of bonding in the solid electrostatic attraction Melting points 600-2000 oC Conductivity (heat or electricity) poor conductor Solubility (in polar or nonpolar solvents) soluble in polar solvents

CsCl ZnS CaF2 Figure 12.22, p. 420 12.5 Bonding in Solids

Properties Covalent Solids Type of bonding in the solid covalent bonds Melting points >1000 oC Conductivity (heat or electricity) poor conductor Solubility (in polar or nonpolar solvents) insoluble in either type

carbon atoms

Figure 12.24, p. 422 12.5 Bonding in Solids

Properties Molecular Solids Type of bonding in the solid intermolecular forces Melting points Low melting points Conductivity (heat or electricity) poor conductor Solubility (in polar or nonpolar solvents) soluble in nonpolar solvents

Margin Figure p. 421 12.5 Bonding in Solids

Properties Metallic Solids Type of bonding in the solid metallic bonding Melting points Low to high Conductivity (heat or electricity) good conductor Solubility (in polar or nonpolar solvents) insoluble Cross Section of a Metallic Crystal nucleus & inner shell e-

Mobile “sea” of e-

Figure 12.25, p. 422 12.5 Bonding in Solids

Table 12.4, p. 420 12.6 Phase Changes y What are the three phases of matter? ◦ Phases are a homogeneous part of the system in contact with other parts of the system by separated from them by a well-defined boundary. y Consider the liquid- vapor equilibrium on a particle level. How is this an equilibrium?

Figure 12.27, p. 423 12.6 Phase Changes

y What is vapor pressure? ◦ The pressure exerted by gaseous molecules from when a liquid evaporates y Consider an ideal gas, how are pressure and temperature related? y How is vapor pressure and temperature related? 12.6 Phase Changes

y How are vapor pressure and enthalpy of vaporization related?

Table 12.5, p. 425 12.6 Phase Changes y What is the boiling point? ◦ The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure. ◦ The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm. 12.7 Phase Diagrams y What is a phase diagram? ◦ summarizes the conditions under which a substance exists as a solid, liquid or gas. y Consider modeling this for water: ◦ What happens at 100oC and 1 atm? x What happens when temperature changes x What happens when pressure changes ◦ What happens at 0oC and 1 atm? x What happens when temperature changes x What happens when pressure changes 12.7 Phase Diagrams y What is the triple point? ◦ the only temperature and pressure at which all three phase can be in equilibrium with one another. y What are critical temperature (Tc) and critical pressure (Pc)? ◦ above which its gas form cannot be made to liquefy; no matter how great the applied pressure. This is also the highest temperature at which a substance can exist as a liquid. The minimum pressure that must be applied to bring about liquefaction at this critical temperature is the critical pressure 12.7 Phase Diagrams

y How do we model phases for a pure substance over a range of temperatures and pressures?

Figure 12.32, p. 430 12.7 Phase Diagrams

y How do we model phases for a pure substance over a range of temperatures and pressures?

Figure 12.33, p. 430 Bigger Picture y We have finally connected structure (on the particle level) based on: ◦ Atomic structure ◦ Electronic structure ◦ Bonding ◦ Shape ◦ Polarity and function – properties on the macroscopic level We have reason and explanation for observations As the temperature of a liquid increases, the number of particles in the vapor phase ______and the vapor pressure ______. (A) decreases; decreases (B) decreases; increases (C) increases; decreases (D) increases; increases

A metal crystallizes in a body‐centered 4r cubic (bcc) crystal structure. If the atomic a = radius of the metal is 124.1 pm and the 3 density is 7.87 g∙cm–3, what is the metal? (A) Cd (B) F (C) Fe (D) K

Using the phase diagram to the right,

Chapter 12 –Chapter Practice which letter indicates (only) the liquid region? (A) A (B) B (C) C (D) D