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Chemistry-140 Lecture 33

Chemistry-140 Lecture 33. Chapter 13: Intermolecular Forces: Liquids & Solids. Chapter Highlights phases of matter & kinetic molecular theory intermolecular forces metallic & ionic solids molecular & network solids

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Chemistry-140 Lecture 33

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  1. Chemistry-140 Lecture 33 Chapter 13: Intermolecular Forces: Liquids & Solids • Chapter Highlights • phases of matter & kinetic molecular theory • intermolecular forces • metallic & ionic solids • molecular & network solids • Will not include Sections 13.3, 13.6 & 13.7 on physcial properties of liquids and solids

  2. Chemistry-140 Lecture 33 • Intermolecular forces are attractive, electrostatic interactions that occur between molecules, atoms, or ions. • Liquids: • intermolecular forces are strong enough to hold the molecules together yet weak enough to allow the molecules to move. • generally denser than gases, fairly incompressible, have definite volumes & can flow Kinetic Molecular Description of Liquids & Solids

  3. Chemistry-140 Lecture 33 • Solids: • intermolecular forces are strong enough to prevent the molecules from moving. • generally more dense than liquids, incompressible and rigid, do not take the shape of their containers, do not flow. • Crystalline solids: molecules or ions are arranged in repeating patterns possessing high symmetry. • Amorphous solids: molecules are arranged in a random fashion. Kinetic Molecular Description of Liquids & Solids

  4. Chemistry-140 Lecture 33 • Intermolecular forces: those between molecules, atoms, or ions that hold the particles together. • At a given temperature these forces are • solids > liquids > gases • In general, the stronger the intermolecular forces, the higher the melting point and the boiling point Intermolecular Forces

  5. Ion Dipole 40 - 600 kJ/mol Chemistry-140 Lecture 33 • Ion-dipole forces: electrostatic attractions that form between an ion and an oppositely charged pole of a polar molecule. Intermolecular Forces d- d+ Polar water molecule attracted to a cation

  6. Cl H Co O Chemistry-140 Lecture 33 • Hydrated cobalt(II) chloride, [CoCl2 . 6H2O], is a solid salt. The structure is actually [CoCl2(H2O)4]. 2H2O in which 4 of the water molecules are attached to the Co2+ ion via ion dipole forces. Intermolecular Forces

  7. Dipole Dipole 5 - 25 kJ/mol including H-bonding Chemistry-140 Lecture 33 • Dipole-dipole forces: between the d+ pole of one polar molecule and the d- pole of another polar molecule. If, equal in mass and size, the attraction increases with increasing polarity. Intermolecular Forces

  8. Chemistry-140 Lecture 33 • Dipole-dipole attractions between two molecules of ClBr. Cl is more electronegative than Br which creates the permanent dipole (d+ and d-). Intermolecular Forces

  9. Chemistry-140 Lecture 33 • Hydrogen bonding: a special type of dipole-dipole interaction which exists between the hydrogen atom in a polar bond and the unbonded electron pair of an electronegative atom or ion. • The strength (4 - 25 kJ/mol) increases in the order • N-H….Y < O-H….Y < F-H….Y • X-H….F < X-H….O < X-H….N • These trends are explained in terms of increasing electronegativities of N < O < F. Intermolecular Forces

  10. Chemistry-140 Lecture 33 Intermolecular Forces • Hydrogen bonding between HF molecules. A lone-pair on F (d-) interacts with an H atom (d+) to give a linear H-bond. This repeats in the solid to form a zig-zag chain of HF molecules

  11. Chemistry-140 Lecture 33 Water is a very unique substance!! High specific heat 4.18 J/ g K High heat of fusion 333 J/g High heat of vapourization 2250 J/g High dielectric constant 80 @ 20 oC High surface tension 7.2 x 109 N/m density of water = 1.00 g/mL density of ice = 0.917 g/mL. HOW? WHY?.... Intermolecular Forces

  12. O H Hydrogen bond Chemistry-140 Lecture 33 Intermolecular Forces Hydrogen Bonding !!!

  13. Hydrogen bond O H Chemistry-140 Lecture 33 Intermolecular Forces

  14. Induced Dipole Dipole 2 - 10 kJ/mol Chemistry-140 Lecture 33 • Dispersion forces: can exist between formally nonpolar molecules and polar molecules due to the polarizability of the electron clouds on the formally nonpolar molecule. Induced dipoles, d+ and d-, are smaller than permanent dipoles in polar molecules. Intermolecular Forces

  15. Induced Dipole Induced Dipole 0.05 - 40 kJ/mol Chemistry-140 Lecture 33 • Dispersion forces: can also exist between formally nonpolar molecules due to the polarizability of electron clouds. All liquids and solids possess dispersion forces that tend to increase with increasing formula weight. Intermolecular Forces

  16. Chemistry-140 Lecture 33 Intermolecular Forces Question: List the substances BaCl2, H2, CO, HF and Ne in order of increasing boiling point.

  17. Chemistry-140 Lecture 33 Answer: Boiling point depends, in part, on attractive forces. BaCl2: ionic bonding H2: dispersion forces (M = 2) CO: dipole-dipole forces (M = 28) HF: dipole-dipole forces (hydrogen bonding) (M = 20) Ne: dispersion forces (M = 20) Intermolecular Forces H2(20 K) <Ne(27 K) < CO(83 K) < HF(293 K) < BaCl2(1813 K)

  18. Chemistry-140 Lecture 33 Final Exam Monday December 10th, 2001 12:00 Noon St. Denis Centre Field House

  19. Chemistry-140 Lecture 34 Chapter 13: Intermolecular Forces: Liquids & Solids • Chapter Highlights • phases of matter & kinetic molecular theory • intermolecular forces • metallic & ionic solids • molecular & network solids • Will not include Sections 13.3 & 13.6 on physcial properties of liquids and solids

  20. Chemistry-140 Lecture 34 • Crystalline solid: one in which the molecules, atoms, or ions are ordered in well-defined, repeating, patterns. • ionic: NaCl • molecular: H2O • metallic: Fe • network: C(diamond) • Amorphous solid: one in which no order exists; may have small regions of order, but no long-range order. • glass: borosilicate glass • polymer: nylon Structures of Solids

  21. Chemistry-140 Lecture 34 Crystalline solids can be described by unit cells. Crystalline Solids Unit cell: the smallest "piece" of the crystal requiredto show the repeating pattern.

  22. a b c b a g Chemistry-140 Lecture 34 • Unit cells are parallelepipeds (6-sided figures with faces that are parallelograms) described in terms of the lengths of the edges (a, b and c) and the angles between these edges (a, b, and g) Crystalline Solids

  23. a b c a b g Chemistry-140 Lecture 34 • The simplest unit cell is cubic (a = b = g = 90o) and • (a = b = c). The primitive cubic unit cell consists only of atoms at the corners of the box. Crystalline Solids

  24. Face Centred Cubic Body Centred Cubic Primitive Cubic Chemistry-140 Lecture 34 Cubic Unit Cells

  25. Chemistry-140 Lecture 34 • In any cubic lattice, a corner atom is shared equally between eight unit cells. Atom Sharing Between Unit Cells • It therefore only contributes 1/8 of an atom • to that particular unit cell

  26. Chemistry-140 Lecture 34 • In any cubic lattice, a face-centred atom is shared equally between two unit cells. Atom Sharing Between Unit Cells • It therefore only contributes 1/2 of an atom • to that particular unit cell

  27. Chemistry-140 Lecture 34 • The body-centred cubic unit cell has an atom at the centre of the box in addition to the eight atoms in the corners. A body-centred atom is not shared and therefore contributes one fullatom to the unit cell. • Atoms may also be found on the edges of the box. An edge-centred atom is shared equally between four unit cells. It therefore only contributes 1/4 of an atom to that particular unit cell. Atom Sharing Between Unit Cells

  28. Chemistry-140 Lecture 34 Example 13.6: Aluminium has a density of 2.699 g/cm3 and the atoms are packed into a face-centred cubic unit cell. Determine the radius of an Al atom. Crystal Lattice Measurements

  29. Chemistry-140 Lecture 34 Known: the density of Al the structural make-up of the basic unit cell Crystal Lattice Measurements We know that: Mass of the unit cell = (# of Al atoms)(atomic weight of Al) Volume of a cube = (length of the side)3

  30. Cell diagonal Cell edge Chemistry-140 Lecture 34 If we know the length of the side then we can determine the radius of a single atom!! Crystal Lattice Measurements

  31. Chemistry-140 Lecture 34 Step 1: We know M, N and d but need to calculate Z. Each atom at the centre of a face contributes 1/2 of an Al atom to the cell. 6 x (1/2) = 3 Al atoms Each atom at the corner contributes 1/8 of an Al atom to the cell. 8 x (1/8) = 1 Al atom Z = 3 + 1 = Crystal Lattice Measurements 4 Al atoms

  32. Chemistry-140 Lecture 34 Step 2: Use the equation: Crystal Lattice Measurements Solve for V using: = 6.640 x 10-23 cm3

  33. Chemistry-140 Lecture 34 Step 3: Use the equation: Crystal Lattice Measurements 4.049 x 10-8 cm =

  34. Chemistry-140 Lecture 34 Step 4: Crystal Lattice Measurements and the equation: 1.432 Å = 1.432 x 10-8 cm =

  35. Cl- Cs+ Chemistry-140 Lecture 34 Ionic Solids CsCl Lattice

  36. Chemistry-140 Lecture 34 Ionic Solids Expanded NaCl Lattice

  37. Cl- Na+ Chemistry-140 Lecture 34 The Cl- anions form a face-centred unit cell and the Na+ cations fill the spaces, such as the centre of the unit cell. There must of course be equal numbers of cations & anions!! (Z = 4) Ionic Solids

  38. Chemistry-140 Lecture 34 • Molecular solids: those in which the particles that comprise the crystal are molecules. Molecular & Network Solids

  39. Chemistry-140 Lecture 34 • Covalent network solids: those in which the crystal itself is a large, covalently bonded, molecule. Molecular & Network Solids

  40. Chemistry-140 Lecture 34 Textbook Questions From Chapter # 13 Intermolecular Forces: 20, 22, 26 Metallic & Ionic Solids: 43, 45, 48 Molecular & Network Solids: 52 General Questions: 60, 69 Conceptual Questions: 88

  41. Chemistry-140 Lecture 33 Final Exam Friday December 10th, 2001 12:00 Noon St. Denis Centre Field House

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