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Chemical Bonding I: Lewis Theory

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Chemical Bonding I: Lewis Theory

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    1. Chemical Bonding I: Lewis Theory Chapter 9

    2. Chemical Bonding Atoms gain, lose, or share electrons in order to achieve a full outer shell electron configuration.

    3. Bonds Ionic Bonds Composed of ions that have gained or lost electrons to achieve a full outer shell Electrostatic attractive forces Crystalline solids – no discrete molecules - formula units Identified by empirical formulas Metal + non-metal Covalent Bonds Composed of atoms that are sharing electrons to achieve a full outer shell Shared electron bonds Discrete molecules, forms gases, liquids, and solids Identified by molecular formulas Non-metal + non-metal

    4. Molecular compounds boil at low temperatures because only weak intermolecular forces must be disrupted. Figure: 09-07 Title: Intermolecular and Intramolecular Forces Caption: The covalent bonds between atoms of a molecule are much stronger than the interactions between molecules. To boil a molecular substance, you simply have to overcome the relatively weak intermolecular forces, so molecular compounds generally have low boiling points.Figure: 09-07 Title: Intermolecular and Intramolecular Forces Caption: The covalent bonds between atoms of a molecule are much stronger than the interactions between molecules. To boil a molecular substance, you simply have to overcome the relatively weak intermolecular forces, so molecular compounds generally have low boiling points.

    6. Ionic Bonds Ions are held together by ionic attraction where the force of attraction is governed by Coulomb’s Law.     Makes sense Large Z ? strong attraction ? larger E Large d ? charge felt less ? smaller E

    7. Figure: 09-05 Title: Lattice Energy Caption: The lattice energy of an ionic compound is the energy associated with forming a crystalline lattice of the compound from the gaseous ions.Figure: 09-05 Title: Lattice Energy Caption: The lattice energy of an ionic compound is the energy associated with forming a crystalline lattice of the compound from the gaseous ions.

    9. Lattice energy dependence on atomic sizes Figure: 09-06-02UN Title: Group I metal chlorides with radii Caption: The relative sizes of the ions will determine the lattice energy.Figure: 09-06-02UN Title: Group I metal chlorides with radii Caption: The relative sizes of the ions will determine the lattice energy.

    10. Lattice energy dependence on ionic charges Figure: 09-06-02UN Title: Group I metal chlorides with radii Caption: The relative sizes of the ions will determine the lattice energy.Figure: 09-06-02UN Title: Group I metal chlorides with radii Caption: The relative sizes of the ions will determine the lattice energy.

    11. Formation of ionic compounds Get elements as atoms (generally requires energy) Form ions (anions are energetically favorable, cations are unfavorable) Bring ions together (favorable) Condense to solid phase (favorable)

    12. Energetics of NaCl Formation Na(s) ??Na(g) +107.3 kJ/mol Na(g) ?? Na+ + 1e? +495.8 kJ/mol 1/2 Cl2(g) ?? Cl (g) +122 kJ/mol Cl(g) ?? Cl?(g) ?348.6 kJ/mol Na+(g) + Cl?(g) ? NaCl(s) ?787 kJ/mol ==================================== Na(s) + 1/2 Cl2(g) ?? NaCl(s) ?411 kJ/mol

    14. Figure: 09-06 Title: Born-Haber Cycle for Sodium Chloride Caption: The sum of the steps is the formation of NaCl from elemental Na and Cl. The enthalpy change of the last step is the lattice energy.Figure: 09-06 Title: Born-Haber Cycle for Sodium Chloride Caption: The sum of the steps is the formation of NaCl from elemental Na and Cl. The enthalpy change of the last step is the lattice energy.

    15. Determine the energy of formation of MgBr2 from the elements.

    19. magnesium bromide Mg(s) ?? Mg(g) +147.7 kJ/mol Mg(g) ?? Mg+ + 1e? +737.7 kJ/mol Mg+(g) ?? Mg2+(g) + 1e? +1450.7 kJ/mol Br2(g) ?? 2 Br (g) +193 kJ/mol 2Br(g) ?? 2Br?(g) 2(?325 kJ/mol) = ?650 kJ/mol Mg2+(g) + 2Br?(g) ? MgBr2(s) ?2440 kJ/mol ==================================== Mg(s) + Br2(g) ?? MgBr2(s) ? 561 kJ/mol Notice that Hfor MgCl2 = ?642 kJ/mol as compared to ?561 kJ/mol for MgBr2. Why the difference? Show how lattice energies are related to d & Z (Table 6.3)   Br is larger than Cl, therefore it packs into a crystal less efficiently, Also charges cannot get as close together.   Br gets less back when it attracts an electron, energy level is farther out and ? not as strong attraction (more shielded) Notice that Hfor MgCl2 = ?642 kJ/mol as compared to ?561 kJ/mol for MgBr2. Why the difference? Show how lattice energies are related to d & Z (Table 6.3)   Br is larger than Cl, therefore it packs into a crystal less efficiently, Also charges cannot get as close together.   Br gets less back when it attracts an electron, energy level is farther out and ? not as strong attraction (more shielded)

    20. Mg + Cl2 ? MgCl2 DH = -642 kJ/mol Mg + Br2 ? MgBr2 DH = -561 kJ/mol Why are they different?

    21. Calculate the energy released in kJ/mol in the reaction   Na(s) + 1/2 I2(s) ?? NaI(s)   The energy of vaporization of Na(s) is 107 kJ/mol. The sum of the enthalpies of dissociation and vaporization of I2(s) is 214 kJ/mol, and the lattice energy of NaI is 704 kJ/mol. From book IE Na = 496 kJ/mol EA I = ?300 kJ/mol   1st vaporize Na +107 kJ/mol 2nd ionize Na +496 kJ/mol 3rd vaporize and dissociate I2 +107 kJ/mol divide given value by 2 because it gives 2 I atoms, we only want 1 4th add electron to I atom ?295 kJ/mol 5th let everything come together ?704 kJ/mol   Total ?289 kJ/mol   From book IE Na = 496 kJ/mol EA I = ?300 kJ/mol   1st vaporize Na +107 kJ/mol 2nd ionize Na +496 kJ/mol 3rd vaporize and dissociate I2 +107 kJ/mol divide given value by 2 because it gives 2 I atoms, we only want 1 4th add electron to I atom ?295 kJ/mol 5th let everything come together ?704 kJ/mol   Total ?289 kJ/mol  

    22. Calculate the energy released in kJ/mol when LiH is formed in the reaction Li(s) + ˝ H2(g) ? LiH(s)   Heat of vaporization, Li 161 kJ/mol Dissociation energy, H2 436 kJ/mol Lattice energy, LiH -917 kJ/mol Ionization energy, Li 520 kJ/mol Electron affinity, H -73 kJ/mol   Answer: -91 kJ/mole net change

    23. Covalent Bonding Shared electron bonds Due to overlap of atomic orbitals (Valence Bond Theory) Allows each atom to fill valence shell with electrons

    24. Representing Atoms, Ions, and Molecules as Lewis Electron Dot Structures Use dots to represent valence electrons

    25. Figure: 09-06-09UN Title: Double bond for the octet Caption: In covalent bonding, double bonds are generally formed when two electrons from each atom participating in the bond are shared by both atoms.Figure: 09-06-09UN Title: Double bond for the octet Caption: In covalent bonding, double bonds are generally formed when two electrons from each atom participating in the bond are shared by both atoms.

    26. Polar Covalent Bonds and Electronegativity

    27. Figure: 09-07-01UN Title: Bond polarity in HF Caption: Hydrogen and fluorine do not share their electrons equally; the electrons are around F more than H.Figure: 09-07-01UN Title: Bond polarity in HF Caption: Hydrogen and fluorine do not share their electrons equally; the electrons are around F more than H.

    28. Bond Polarity

    29. Electronegativity The ability of an atom in a bond to attract electrons toward itself. Electron greed Note that electronegativity increases up and to the right as do the ionization energy and the electron affinity

    31. Lewis Electron Dot Structures Bonding electrons pairs – electron pairs involved in bonds Lone electron pairs – electron pairs that do not participate in bonding Bond order = number of bonds

    32. Writing Lewis Dot Structures Decide which atoms are bonded together - draw a skeleton structure Count the total number of valence electrons available. Find the number of electrons needed to give an octet around all atoms -- (remember H needs 2, all else need 8).

    33. Writing Lewis Dot Structures Determine number of electrons short. Number of bonds needed = number of electrons short/2. Distribute bonds -- (1st hook atoms together and then add double bonds where appropriate). Calculate number of electrons used in bonds.

    34. Writing Lewis Dot Structures Calculate electrons remaining. Distribute remaining electrons to give all atoms an octet. Done!!

    35. Formal Charge The result of a method of electron bookkeeping that tells whether an atom in a molecule has gained or lost electrons compared to an isolated atom. Formal charge = # valence electrons – (# bonds + # electrons as lone pairs)

    36. Expanded Octets Elements beyond neon have available d orbitals that may be used to accept additional electrons if necessary. If you the number of bonds necessary to hook all atoms together is greater than the number needed to give all an octet then put in necessary bonds and distribute extra electrons on atoms that have available d orbitals in which to expand.

    37. Lewis Structures of ions for anions add the extra electrons to the number available for cations subtract the lost electrons from the number available

    38. Resonance In some Lewis structures, the multiple bonds can be written in several equivalent locations. All structures have the exact same energy. Which is the correct Lewis structure?? Answer : None alone are correct – the true molecule is a hybrid of the possible structures. The electrons are delocalized.

    39. Bond length - the optimum distance between nuclei in a covalent bond.

    40. Bond dissociation energy – (Bond Strength) the amount of energy necessary to break a chemical bond in an isolated molecule in the gaseous state the amount of energy released when a bond forms Average bond dissociation energies are tabulated in the book. Table 9.3

    41. It is experimentally found that there is a direct correlation between the bond length and the bond strength. As the bond length decreases, the bond strength increases.

    42. Definitions Sigma bond the first bond to form between any two atoms forms between atoms Pi bond second or third bond to form between two atoms forms above and below plane of the molecule

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