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Chapter 9 Chemical Bonding

Chapter 9 Chemical Bonding. 1 type of Potential Energy : Gravitational P.E. Ball On Top of a Hill. m. P.E. = mgh. h. Section 9.1: Why does bonding occur in the first place?. Bonding lowers the potential energy between positive and negative particles (p341). What is potential energy?.

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Chapter 9 Chemical Bonding

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  1. Chapter 9Chemical Bonding

  2. 1 type of Potential Energy: Gravitational P.E. Ball On Top of a Hill m P.E. = mgh h Section 9.1: Why does bonding occur in the first place? Bonding lowers the potential energy between positive and negative particles (p341). What is potential energy? Energy changes forms: P.E.  Kinetic Energy (K.E.)

  3. Energy changes forms Mechanical Energy Friction Motor Generator Engines Electrical Energy Heat (Thermal) Energy Solar Heater Battery FIre Battery Charger Light (Radiant) Energy Chemiluminescence Chemical Energy Photosynthesis Section 9.1: Why does bonding occur in the first place? Bonding lowers the potential energy between positive and negative particles (p341).

  4. Section 9.1: Why does bonding occur in the first place? Bonding lowers the potential energy between positive and negative particles (p341). When chemical bonds form: Chemical P.E. changes to Heat Energy & Light Energy Mechanical Energy Electrical Energy Heat (Thermal) Energy Light (Radiant) Energy Chemical Energy

  5. Section 9.1: Why does bonding occur in the first place? Bonding lowers the potential energy between positive and negative particles (p341). Energy changes forms: Chemical P.E.  Heat & Light Energy http://chemsite.lsrhs.net/chemKinetics/PotentialEnergy.html

  6. Section 9.1: Three Type of Bonds Ionic bonding: Metal + Nonmetal (Valence e- transferred) Covalent bonding: Nonmetal + Nonmetal (Valence e- shared) Metallic bonding: Metal + Metal (“Sea” of e-) http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch10/non.php

  7. Concept Check Review: Valence Electrons – e- involved in forming compounds (Ch 8, p315) Boron (B) Magnesium (Mg) Hydrogen (H) How many valence e-? How many needed for full outer shell? Total valance e-:

  8. Representative Elements For REPRESENTATIVE elements: • period (row) = shell # (n = 1, 2, 3, 4….n) • group (column) = # of e- in outer shell Group ## of valence e- Transition Elements IA 1 IIA 2 IIIA 3 IVA 4 VA 5 VIA 6 Shells of an atom VIIA 7 VIIA 8 Section 9.1: Two Bond Types With Localized Electrons Ionic & Covalent Bonding

  9. Section 9.1: Two Bond Types With Localized Electrons Ionic & Covalent Bonding: Why do ionic bonds form instead of covalent bonds, and vice versa? Covalent Bonds Ionic Bonds “Bonding Continuum” nonmetals + nonmetal metal + nonmetal Nonpolar Covalent Bond Polar Covalent Bond Ionic Bond Electrons are shared unequally. Electrons are transferred.

  10. Extent of electron sharing in Covalent Bonds e-’s shared between atoms of the same element: Equal Sharing e-’s shared between atoms of different elements: Unequal Sharing Unequal sharing – occurs because one of the atoms in a bond has a stronger attraction for the pair of e-’s than does the other atom Why does one atom have a stronger attraction for e-?

  11. Electronegativity Definition: electronegativity (E.N) – the ability of an atom to attract the shared electrons Increasing E.N. Decreasing E.N. Rule for Bond Formation The atom with the greater E.N. pulls the shared electrons closer to its nucleus resulting in (1) – charge on high E.N. atom (2) + charge on low E.N. atom More later: Section 9.5

  12. Why do ionic bonds form instead of covalent bonds, and vice versa? e- sharing 2 nonmetals e- transfer metal + nonmetal “Bonding Continuum” Covalent Bonds Ionic Bonds Polar Colvalent Nonpolar Colvalent 0.4 < E.N. 0.4 < E.N. < 1.7 E.N. difference > 1.7 1.7 Answer: Electronegativity Differences Example: Oxygen (O) bonds with Magnesium (Mg): MgO E.N. of O = 3.5 E.N. of Mg = 1.2 E.N. difference = 2.3

  13. Section 9.1: “The Other” Bond Type With Delocalized Electrons Metallic Bonding Metallic Bonding - Delocalized Covalent Bonding, Ionic Bonding - Delocalized A messy “sea” of electrons Electrons fit neatly into shells.

  14. Outer e- Inner e- Section 9.1: “The Other” Bond Type With Delocalized Electrons Metallic Bonding Metallic Bonding - Delocalized A messy “sea” of electrons

  15. Li Different elements can have the same number of dots Be Mg Same Group (Column) Lewis Electron-Dot Symbols Two parts: (1) Element symbol – nucleus + inner electrons Ex: The element lithium has an element symbol Li (2) Surrounding dots – valence electrons (outer most shell)

  16. Mg2+ Cl- Review: Ions Ion – charged particles that form when an atom gains or loses one or more electrons (Ch2, p60) Element Ion Ion Type Mg Cation Cl Anion

  17. Mg Review: Electron Configuration and Orbital Diagrams (Ch8, p304-317) Example:

  18. Concept Check • End of Chapter Problems in-class (for now): 9.7, 9.9, 9.13, 9.15 Write the ion for the following elements: K, Br, Sr, Ar, O For example, the ion for Mg is Mg2+. • Suggested Optional Practice Problems (for outside of class): 9.6, 9.8, 9.10, 9.12, 9.14 (Answers in back of book or online)

  19. O Examples: Water (H2O) Carbon Dioxide (CO2) H H O O C Section 9.2: Ionic Bonding Central idea: Electrons are transferred from metal atoms to nonmetal atoms to form ions that come together in a solid ionic compound. Solid Ionic compound Na – metal Cl - nonmetal Sodium chloride (NaCl) Contrast with molecules formed during covalent bonding (more later).

  20. Cl- Section 9.2: Ionic Bonding Rule: The total number of e- lost by the metal atom equals the total number gained by the nonmetal atom. Na+ lost gained

  21. Behavior of Ionic Compounds Why is the melting point of MgO higher than the melting point of KCl?

  22. Lattice Energy (∆Hºlattice)

  23. Section 9.2: Lattice Energy Definition – The enthalphy change that occurs when 1 mol of ionic solid separates into gaseous ions. For Review of Enthalpy: Ch6, p243 Lattice Energy denoted as: ∆Hºlattice ∆Hºlattice cannot be measured directly, BUT it can be calculate using the: Born-Haber cycle

  24. Section 9.2: Born-Haber Cycle Uses Hess’s Law: Total enthalpy of an overall reaction is the sum of the enthalpy changes of individual reactions. (∆Htotal = ∆Hrxn1 + ∆Hrxn2 +……….) *Not actual steps.

  25. Section 9.2: Trends in Lattice Energy Coulomb’s Law (Ch2)

  26. Section 9.2: Trends in Lattice Energy

  27. Behavior of Ionic Compounds So, why is the melting point of MgO higher than the melting point of KCl?

  28. Concept Check • End of Chapter Problems in-class (for now): 9.27, 9.30 • Suggested Optional Practice Problems (for outside of class): 9.26, 9.28 (Answers in back of book)

  29. Problem 9.30

  30. Contrast with ionic solids formed during ionic bonding (discussed previously). Na – metal Cl - nonmetal Sodium chloride (NaCl) Section 9.3: Covalent Bonding e- sharing – primary way that atoms interact Nonmetal + Nonmetal Examples: Water (H2O) Carbon Dioxide (CO2) Organic Compounds O O O H H H C C C C H H H H H H H

  31. Section 9.3: Covalent Bonding Why do covalent bonds form? Lower P.E. = More stable

  32. Section 9.3: Covalent Bonding How are the electrons distributed? Electron Density In order for each atom to have a full outer shell (2 e- for H, He; 8 e- for others), the electrons arrange themselves in certain configurations: • Bonding Pairs & Lone Pairs • Bond Type – double, single, triple

  33. Bond Energy – energy needed to overcome attraction and break the bond Section 9.3: Covalent Bonding Bond Energy (B.E.) – aka Bond Enthalpy or Bond Strength Covalent Bond Strength – depends on strength of attraction between nuclei and shared electrons

  34. Section 9.3: Covalent Bonding Bond Energy (B.E.) Bond formation is exothermic: ∆Hº always + Bond breakage is endothermic: ∆Hº always - Absolute value of B.E. – Each bond has its own unique B.E. due to variations in: (1) e- density (2) charge (3) atomic radii

  35. Section 9.3: Covalent Bonding Strength of Bond different than E required to pull atoms apart (B.E.) Less E needed to break. Lower B.E. Weaker Bonds = Higher Energy “Shallow Energy Well” Stronger Bonds = Lower Energy “Deeper Energy Well” More E needed to break. Higher B.E.

  36. Section 9.3: Covalent Bonding Bond Energy (B.E.) and Bond Length Bond Length – sum of the radii of the bonded atoms (analogous to distance in Coloumb’s Law) At minimum E point.

  37. Section 9.3: Covalent Bonding Bond Energy (B.E.) and Bond Length This relationship holds, in general, ONLY for single bonds.

  38. Section 9.3: Covalent Bonding Bond Type (Single, Double, Triple) also matters Same two elements, different B.E. Nuclei more attracted to 2 shared pairs of e- than one shared pair of e-. Higher bond order = Shorter bond length = Higher Bond Energy

  39. Section 9.3: Covalent Bonding Periodic Table Trends Without Detailed Bond Lengths The closer the atoms, the stronger the bond. Bond Energy: C—F > C—Cl > C—Br

  40. O solid  liquid  gas H H Strong covalent bonding forces Hold atoms together Weak intermolecular forces Hold molecules together (More in Chapter 12) - + + Chemical Reaction Phase Change Section 9.3: Covalent Bonding Covalent Bonds are stronger than Ionic Bonds So why, then, do covalent compounds have lower melting points than ionic compounds? Example: CCl4 m.p. = -23 ºC NaCl m.p. = 800 ºC

  41. Section 9.4: Bond Energy and Chemical Change Where does the heat that is released come from? http://chemsite.lsrhs.net/chemKinetics/PotentialEnergy.html

  42. solid  liquid  gas Kinetic Energy (K.E.) Three types: (1) Vibrational (2) Rotational (3) Translational • Does not change during chemical reaction (depends on T). Changes during a Phase Change (Chapter 12). Section 9.4: Bond Energy and Chemical Change Total energy of a chemical system = K.E. + P.E. Example of a chemical system A container filled with molecules. http://www.landfood.ubc.ca/courses/fnh/301/water/motion.gif

  43. = Bond Energy Where does the heat that is released come from? The energy released or absorbed during a chemical change is due to the differences between the reactant bond energies and the product bond energies. B.E.reactants - B.E.products = Heat Section 9.4: Bond Energy and Chemical Change This leaves us with changes in P.E. during chemical reactions. P.E. contributions can from electrostatic forces between: Separate Vibrating Atoms Nucleus & Electrons in Atoms Protons & Neutrons in Nucleus Nuclei and Shared Electron Pair in Each Bond

  44. Analogous to ionic compound formation: Lattice Energy, ∆Hºlattice Born-Haber cycle (∆Hºtotal = ∆Hºrxn1 + ∆Hºrxn2 +……+ ∆Hºlattice) Section 9.4: Bond Energy and Chemical Change Heat of reaction, ∆Hºrxn Exothermic reaction: - ∆Hºrxn Endothermic reaction: + ∆Hºrxn ∆Hºrxn = ∆Hºreactantbonds broken + ∆Hºproductbonds formed ∆Hºrxn = ∆BEreactantbonds broken – ∆BEproductbonds formed

  45. Section 9.4: Bond Energy and Chemical Change Example: H2 + F2 2 HF Weaker Bonds Less Stable, More Reactive H2 and F2 Stronger Bond More Stable, Less Reactive HF

  46. Section 9.4: Bond Energy and Chemical Change Another way to looks at this reaction: H2 + F2 2 HF Heat of reaction, ∆Hºrxn 2 H + 2 F H2 + F2 HF ∆Hºrxn = ∆Hºreactantbonds broken + ∆Hºproductbonds formed

  47. Section 9.4: Bond Energy and Chemical Change Use bond energies to calculate ∆Hºrxn (Table 9.2) H2 + F2 2 HF 9.39, 9.47, 9.49 Optional Homework Problems: 9.38, 9.46, 9.48, 9.50

  48. ∆Hºrxn = ∆BEreactantbonds broken – ∆BEproductbonds formed Energy Released = B.E.(fuel + O2) – B.E.(CO2 + H2O) Fuels with more weak bonds yield more energy than fuels with fewer weak bonds. Fats: More C-H C-C Carbs: More O-H C-O Food fuels the body: Section 9.4: Bond Energy and Chemical Change Application: Energy Released From Combustion of Fuel

  49. e- sharing 2 nonmetals e- transfer metal + nonmetal “Bonding Continuum” Covalent Bonds Ionic Bonds Polar Colvalent Nonpolar Colvalent 0.4 < E.N. 0.4 < E.N. < 1.7 E.N. difference > 1.7 1.7 Section 9.5: Between the Extremes Scientific models are idealized descriptions of reality. Electronegativity – the relative ability of a bonded atom to attract the shared e-

  50. atomic size E.N. Section 9.5: Between the Extremes Electronegativity – inversely related to atomic size (radius) WHY?

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