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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/molMg + Br2 ? MgBr2 DH = -561 kJ/molWhy 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