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Chapter 8 Concepts of Chemical Bonding

Chapter 8 Concepts of Chemical Bonding

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Chapter 8 Concepts of Chemical Bonding

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  1. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 8Concepts of Chemical Bonding John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

  2. General outline • Tuesday: quiz, time to look over chapter 8 • Wednesday morning: revisit topics struggling with • Thursday: Quiz 8.1-8.4; after quiz – students sheet with stuff to fill in and work on (grade quizzes), revisit points of confusion, Lewis structures practice, formal charge practice, resonance practice

  3. Bell work (A) Metallic bonding (B) Network covalent bonding (C) Hydrogen bonding (D) Ionic bonding (E) London dispersion forces 1) Solids exhibiting this kind of bonding are excellent conductors of heat 2) This kind of bonding is the reason that water is more dense than ice. 3) This kind of bonding exists between atoms with very different electronegativities. 4) The stability exhibited by diamonds is due to this kind of bonding

  4. Agenda • Bell work • 8.1-8.4 quiz • Overview of chapter 8 • White board practice: Lewis structures, formal charge, resonance

  5. After you finish your quiz • Begin on the 8.5-8.8 overview

  6. Chemical Bonds • Three basic types of bonds: • Ionic • Electrostatic attraction between ions • Covalent • Sharing of electrons • Metallic • Metal atoms bonded to several other atoms

  7. Ionic Bonding

  8. Energetics of Ionic Bonding As we saw in the last chapter, it takes 495 kJ/mol to remove electrons from sodium.

  9. Energetics of Ionic Bonding We get 349 kJ/mol back by giving electrons to chlorine.

  10. Energetics of Ionic Bonding • But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic!

  11. Energetics of Ionic Bonding • There must be a third piece to the puzzle. • What is as yet unaccounted for is the electrostatic attraction between the newly formed sodium cation and chloride anion.

  12. Q1Q2 d Eel =  Lattice Energy • This third piece of the puzzle is the lattice energy: The energy required to completely separate a mole of a solid ionic compound into its gaseous ions. Always positive value of ΔH • The energy associated with electrostatic interactions is governed by Coulomb’s law:

  13. Lattice Energy • Lattice energy, then, increases with the charge on the ions. • It also increases with decreasing size of ions.

  14. Energetics of Ionic Bonding By accounting for all three energies (ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.

  15. Energetics of Ionic Bonding • These phenomena also helps explain the “octet rule.” • Metals, for instance, tend to stop losing electrons once they attain a noble gas configuration because energy would be expended that cannot be overcome by lattice energies.

  16. Covalent Bonding • In these bonds atoms share electrons. • There are several electrostatic interactions in these bonds: • Attractions between electrons and nuclei • Repulsions between electrons • Repulsions between nuclei

  17. Polar Covalent Bonds • Although atoms often form compounds by sharing electrons, the electrons are not always shared equally. • Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does. • Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.

  18. Electronegativity: • The ability of atoms in a molecule to attract electrons to itself. • On the periodic chart, electronegativity increases as you go… • …from left to right across a row. • …from the bottom to the top of a column.

  19. Polar Covalent Bonds • When two atoms share electrons unequally, a bond dipole results. • The dipole moment, , produced by two equal but opposite charges separated by a distance, r, is calculated:  = Qr • It is measured in debyes (D).

  20. Polar Covalent Bonds The greater the difference in electronegativity, the more polar is the bond.

  21. Bonding • You will NOT be given electronegativity values on the AP exam, so you MUST know the trends! • Some values that would be good to know offhand: B (2.0), C (2.5), N (3.0), O (3.5), F (4.0), and H (2.1)

  22. Bonding • Covalent • ONLY NM to NM • Polar covalent • it is safe to assume that any different NM bonded together form a polar covalent bond • Non-polar covalent • NO difference in electronegativities (usually same kind of atom, ie: O2, N2), equally sharing of electrons • Ionic bonds • ALWAYS metal with NM, large diff in EN (greater than 1.7)

  23. - • - • - • -

  24. - • - • - • -

  25. Each magnesium atom loses one electron and each fluorine atom gains two electrons. • Each magnesium atom loses two electrons and each fluorine atom gains one electron. • Each magnesium atom gains one electron and each fluorine atom loses two electrons. • Each magnesium atom gains two electrons and each fluorine atom loses one electron.

  26. Each magnesium atom loses one electron and each fluorine atom gains two electrons. • Each magnesium atom loses two electrons and each fluorine atom gains one electron. • Each magnesium atom gains one electron and each fluorine atom loses two electrons. • Each magnesium atom gains two electrons and each fluorine atom loses one electron.

  27. Rh • Tc • Ru • Pd

  28. Rh • Tc • Ru • Pd

  29. There are no attractive forces in He2; that’s why it doesn’t exist. • Attractive forces are between between the two entire atoms; repulsive forces are only between the electron clouds. • Attractive forces are between each electron and either nucleus; repulsive forces are those between the two nuclei and those between the two electrons. • Attractive forces are between each atom’s two electrons and between the two nuclei; repulsive forces are those between the nuclei and the two electrons.

  30. There are no attractive forces in He2; that’s why it doesn’t exist. • Attractive forces are between between the two entire atoms; repulsive forces are only between the electron clouds. • Attractive forces are between each electron and either nucleus; repulsive forces are those between the two nuclei and those between the two electrons. • Attractive forces are between each atom’s two electrons and between the two nuclei; repulsive forces are those between the nuclei and the two electrons.

  31. The repulsive forces are greater than the attractive forces. • The attractive and repulsive forces are equal and thus balance each other out. • The attractive forces are greater than the repulsive forces.

  32. The repulsive forces are greater than the attractive forces. • The attractive and repulsive forces are equal and thus balance each other out. • The attractive forces are greater than the repulsive forces.

  33. C–O in carbon monoxide is a single bond. • C–O in carbon monoxide is a double bond. • C–O in carbon monoxide is a triple bond.

  34. C–O in carbon monoxide is a single bond. • C–O in carbon monoxide is a double bond. • C–O in carbon monoxide is a triple bond.

  35. EN and EA are the same; they both measure the same characteristic. • EA measures the energy released when an isolated atom gains an electron to form a 1– ion; EN measures the ability of an atom to hold onto its own electrons and attract electrons from other atoms. • EN values of neutral atoms are just the negative of EA values of neutral atoms. • EN measures the energy released when an isolated atom gains an electron to form a 1– ion; EA measures the ability of an atom to hold onto its own electrons and attract electrons from other atoms.

  36. EN and EA are the same; they both measure the same characteristic. • EA measures the energy released when an isolated atom gains an electron to form a 1– ion; EN measures the ability of an atom to hold onto its own electrons and attract electrons from other atoms. • EN values of neutral atoms are just the negative of EA values of neutral atoms. • EN measures the energy released when an isolated atom gains an electron to form a 1– ion; EA measures the ability of an atom to hold onto its own electrons and attract electrons from other atoms.

  37. nonpolar • polar covalent • ionic

  38. nonpolar • polar covalent • ionic

  39. ClF • IF • They have the same dipole moment. • Neither has a dipole moment; they are both nonpolar.

  40. ClF • IF • They have the same dipole moment. • Neither has a dipole moment; they are both nonpolar.

  41. The large size of the I and the Br make the charge density less noticeable. • The dipole moments in both HI and HBr are in the opposite direction of HF and HCl. • Because HI and HBr are both strong acids, less negative charge forms on the halogens. • In HI and HBr the electronegativity differences are too small to lead to large charge separations in the molecules.

  42. The large size of the I and the Br make the charge density less noticeable. • The dipole moments in both HI and HBr are in the opposite direction of HF and HCl. • Because HI and HBr are both strong acids, less negative charge forms on the halogens. • In HI and HBr the electronegativity differences are too small to lead to large charge separations in the molecules.

  43. OsO4 • MoO3 • Both compounds have the same melting point.

  44. OsO4 • MoO3 • Both compounds have the same melting point.

  45. Lewis Structures Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.

  46. Find the sum of valence electrons of all atoms in the polyatomic ion or molecule. If it is an anion, add one electron for each negative charge. If it is a cation, subtract one electron for each positive charge. PCl3 Writing Lewis Structures 5 + 3(7) = 26

  47. Writing Lewis Structures • The central atom is the least electronegative element that isn’t hydrogen. Connect the outer atoms to it by single bonds. Keep track of the electrons: 26  6 = 20

  48. Writing Lewis Structures • Fill the octets of the outer atoms. Keep track of the electrons: 26  6 = 20  18 = 2

  49. Writing Lewis Structures • Fill the octet of the central atom. Keep track of the electrons: 26  6 = 20  18 = 2  2 = 0