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Bonds

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									 Chapter Eight

BONDING: GENERAL
   CONCEPTS



    講義
                                 Assignment


                      21, 29, 43, 57, 73, 81, 91, 94, 99




Chapter 8 | Slide 2
Chemical bonds
˙Hold groups of atoms together
˙Occur when a group of atoms can lower its total energy by aggregating
˙Types of chemical bonds
 ˙Ionic: electrons are transferred to form ions
 ˙Covalent: equal sharing of electrons
 ˙Polar covalent: unequal electron sharing
˙Percent ionic character of a bond X─Y:
                Measured dipole moment of X─Y
                                                     100%
                Calculated dipole moment for X + Y 


˙Electronegativity: the relative ability of an atom to attract shared electrons
 ˙The polarity of a bond depends on the electronegativity difference of the bonded atoms
˙The spacial arrangement of polar bonds in a molecule determines whether the molecule has a dipole moment
Ionic bonding
˙An ion has a different size than its parent atom
 ˙An anion is larger than its parent ion
 ˙A cation is smaller htan its parent atom
˙Lattice energy: the change in energy when ions are packed together to form an ionic solid
Bond energy
˙ The energy necessary to break a covalent bond
˙Increases as the number of shared pairs increases
˙Can be used to estimate the enthalpy change for a chemical reaction
Lewis structures
˙Show how the valence electron pairs are arranged among the atoms in a molecule or polyatomic ion
˙Stable molecules usually contain atoms that have their valence orbitals filled
 ˙Leads to a duet rule for hydrogen
 ˙Leads to an octet rule for second-row elements
 ˙The atoms of elements in the third row and beyond can exceed the octet rule
˙Several equivalent Lewis structures can be drawn for some molecules, a concept called resonance
˙When several nonequivalent Lewis structures can be drawn for a molecule, formal charge is often used to choose the most appropriate structure(s)
VSEPR model
˙Based on the idea that electron pairs will be arranged around a central atom in a way that minimizes the electron repulsions
˙Can be used to predict the geometric structure of most molecules
   Chapter 8 | Slide 3
Chapter 8 | Slide 4
               How atoms are linked together?




  A+BAB
  EA+EB>EAB

  Generally,
  An+Bm+Ck+…AnBmCk…

  Ionic bonds
  Covalence bonds

Chapter 8 | Slide 5
                      Questions to Consider

•          What is meant by the term “chemical
           bond”?
•          Why do atoms bond with each other to
           form compounds?
•          How do atoms bond with each other to
           form compounds?




Chapter 8 | Slide 6
                      A Chemical Bond

•          No simple way to define this.
•          Forces that hold groups of atoms
           together and make them function as a
           unit.
•          A bond will form if the energy of the
           aggregate is lower than that of the
           separated atoms.


                                                   8.1
Chapter 8 | Slide 7
The Interaction
of Two
Hydrogen
Atoms



                      8.1
Chapter 8 | Slide 8
 The Interaction of Two Hydrogen Atoms




                                    8.1
Chapter 8 | Slide 9
                       Key Ideas in Bonding

•         Ionic Bonding – electrons are
          transferred
•         Covalent Bonding – electrons are
          shared equally
•         What about intermediate cases?




                                              8.1
Chapter 8 | Slide 10
                       Polar Covalent Bond

•         Unequal sharing of electrons between
          atoms in a molecule.
•         Results in a charge separation in the
          bond (partial positive and partial
          negative charge).




                                                  8.1
Chapter 8 | Slide 11
              The Effect of an Electric Field on
               Hydrogen Fluoride Molecules




Chapter 8 | Slide 12
                       Polar Molecules




                                         8.1
Chapter 8 | Slide 13
                       Concept Check
• What is meant by the term “chemical bond?”
• Why do atoms bond with each other to form
  molecules?
• How do atoms bond with each other to form
  molecules?




                                               8.1
Chapter 8 | Slide 14
                        Electronegativity
 The ability of an atom in a molecule to attract
 shared electrons to itself.
 On the periodic table, electronegativity generally
 increases across a period and decreases down
 a group.

                        (H-H bond energy)+(X-X bond energy)
(H-X)expected =                          2

Δ=(H-X)actual -(H-X)expected                   (Linus Pauling)
                                                              8.2
 Chapter 8 | Slide 15
      The Pauling Electronegativity Values




                                             8.2
Chapter 8 | Slide 16
                       Concept Check
• If lithium and fluorine react, which has more
  attraction for an electron? Why?
• In a bond between fluorine and iodine, which has
  more attraction for an electron? Why?




                                                8.2
Chapter 8 | Slide 17
                       Concept Check
• What is the general trend for electronegativity
  across rows and down columns on the periodic
  table?
• Explain the trend.




                                                    8.2
Chapter 8 | Slide 18
        Bigger difference in electronegativity
   between bonding partners means larger polarity




                                              8.2
Chapter 8 | Slide 19
              Bigger difference in electronegativity
         between bonding partners means larger polarity


 Bond Polarity:          H-H < S-H < Cl-H < O-H < F-H
 Electronegativity:    (2.1)(2.1) (2.5)(2.1) (3.0)(2.1) (3.5) (2.1) (4.0) (2.1)

 Electronegativity
 Difference:             0           0.5        1.1        1.4         1.9
                               Polarity increases
                       Covalent bond  polar covalent bond




Chapter 8 | Slide 20
                       Exercise
Arrange the following bonds from most to least
polar:
a)     N-F        O-F        C-F
b)     C-F        N-O        Si-F
c)     Cl-Cl      B-Cl       S-Cl




                                                 8.2
Chapter 8 | Slide 21
                           Concept Check
Which of the following bonds would be the least
polar yet still be considered polar covalent?

                       Mg-O C-O O-O Si-O N-O




                                                  8.2
Chapter 8 | Slide 22
                          Concept Check
Which of the following bonds would be the most
polar without being considered ionic?

                       Mg-O C-O O-O Si-O N-O




                                                 8.2
Chapter 8 | Slide 23
                         Dipole Moment
     •          Property of a molecule whose charge
                distribution can be represented by a
                center of positive charge and a center
                of negative charge.
     •          Use an arrow to represent a dipole
                moment.
             – Point to the negative charge center with
               the tail of the arrow indicating the positive
               center of charge.

                                                               8.3
Chapter 8 | Slide 24
       Figure 8.4 An Electrostatic Potential
                   Map of HF




Chapter 8 | Slide 25
                               1 Debye



                            r=100 pm (=1 °A)

                       +e                      -e
                               μ=1 D




Chapter 8 | Slide 26
               Polar Bonds Forming Polar Molecules




Chapter 8 | Slide 27
          Polar Bonds Forming Nonpolar Molecules

                        Total dipole moment = 0


Partial negative charge                            Partial negative charge
                            Partial positive charge




 Chapter 8 | Slide 28
                   Charge distribution of CO2




Chapter 8 | Slide 29
The shape of a molecule governs whether
it is polar or not.




Chapter 8 | Slide 30
Chapter 8 | Slide 31
                       Dipole Moment




                                       8.3
Chapter 8 | Slide 32
         Figure 8.6 a-c The Structure and
        Charge Distribution of the Ammonia
                     Molecule




Chapter 8 | Slide 33
                       No Net Dipole Moment
                         (Dipoles Cancel)




                                              8.3
Chapter 8 | Slide 34
                       e.p. Diagram HCL




Chapter 8 | Slide 35
                       e.p.Diagram SO3




Chapter 8 | Slide 36
                       e.p. Diagram CH4




Chapter 8 | Slide 37
                       e.p. Diagram H2S




Chapter 8 | Slide 38
Table 8.2 Types of Molecules with Polar
Bonds but No Resulting Dipole Moment




Chapter 8 | Slide 39
        Linear Molecules with Two Identical
                      Bonds




Chapter 8 | Slide 40
     Planar Molecules with Three Identical
          Bonds 120 Degrees Apart




Chapter 8 | Slide 41
        Tetrahedral Molecules with Four
      Identical Bonds 109.5 Degrees Apart




Chapter 8 | Slide 42
              Nonpolar   Polar
Chapter 8 | Slide 43
Chapter 8 | Slide 44
                               Chapter 8 | Slide 45
Molecular shape and polarity
Chapter 8 | Slide 46
                       BF3




Chapter 8 | Slide 47
                       SF4




Chapter 8 | Slide 48
                       SF6
                                          ..
                                         :F:
                                         |
                                               ..
                              ..               :F    ..
                             :F -   ..   S     ..
                                                    -F:
                              ..    :F               ..
                                    ..
                                          |
                                          ..
                                         :F
                                          ..




Chapter 8 | Slide 49
                       A Bauxite Mine




Chapter 8 | Slide 50
                       Lithium Fluoride




Chapter 8 | Slide 51
                       Stable Compounds

•         Atoms in stable compounds usually
          have a noble gas electron configuration.




                                                     8.4
Chapter 8 | Slide 52
                       Isoelectronic Series

•         A series of ions/atoms containing the
          same number of electrons.

                O2-, F-, Ne, Na+, Mg2+, and Al3+




                                                   8.4
Chapter 8 | Slide 53
                            Concept Check
Choose an alkali metal, an alkaline metal, a noble
gas, and a halogen so that they constitute an
isoelectronic series when the metals and halogen
are written as their most stable ions.

        –    What is the electron configuration for each species?
        –    Determine the number of electrons for each species.
        –    Determine the number of protons for each species.
        –    Rank the species according to increasing radius.
        –    Rank the species according to increasing ionization energy.



                                                                           8.4
Chapter 8 | Slide 54
                       Ionic Radii




                                     8.4
Chapter 8 | Slide 55
    We can “read” from the periodic table:
• Trends for:
        – Atomic size, ion radius, ionization energy,
          electronegativity
• Electron configurations
• Formula prediction for ionic compounds
• Covalent bond polarity ranking



                                                        8.4
Chapter 8 | Slide 56
  Table 8.3 Common Ions with Noble
 Gas Configurations in Ionic Compounds




Chapter 8 | Slide 57
         Figure 8.8 Sizes of Ions Related to
          Positions of the Elements on the
                   Periodic Table




Chapter 8 | Slide 58
                 A Chemical Reaction May Involve Many Steps:
                                     +



                sublimation (atomization)



                ionization                                   +

               dissociation of gas (atomization)



                formation of anion                       -
                (affinity)
                                             +       -
                                         -        +
                formation of solid               - + -
Chapter 8 | Slide 59                         +
 Figure 8.9 The Energy Changes Involved in the Formation
            of Lithium Fluoride from Its Elements

                                                     dissociation of gas (atomization)
                                                                                         -
                                                                           formation of anion
                                     ionization                            (affinity)
                         +

                                        sublimation (atomization)



                                                                          formation of solid
                                                                        +  -
                                                                      - +
                                                                         -  -
                                                                       + +
                                              Li(s) + ½F2 (g)  LiF (s)
                             +
                       The overall reaction

Chapter 8 | Slide 60
Figure 8.10
a & b The
Structure of
Lithium
Fluoride



Chapter 8 | Slide 61
                         Lattice Energy

• The change in energy that takes place
  when separated gaseous ions are packed
  together to form an ionic solid.

                Lattice energy = k(Q1Q2/r)




                                             8.5
Chapter 8 | Slide 62
A two-dimensional slice of an ionic crystal. The greater the distance between two ions, the
weaker their attractions. Therefore, the strongest attractions to an ion are those of the adjacent
ions of opposite charge. The blue circles denote the distances over which the two closest
repulsions occur, and the yellow circles denote the two closest attractions.




                                                   N N             N N            N N

    Vij                               E  1 [| Vij |   | Vij |   | Vij |]
                   qi q j
                                           2
Chapter 8 | Slide 63
                       rij                         j 1 i 1         j 1 i 1        j 1 i 1
                       Born-Haber Cycle for NaCl




                                                   8.5
Chapter 8 | Slide 64
Considerable energy is needed to produce cations and anions from neutral gas-phase
atoms: the ionization energy of the metal atoms must be supplied, and it is only partly
recovered from the electron affinity of the nonmetal atoms. The overall lowering of energy
that drags the ionic solid into existence is due to the strong attraction between cations
and anions in the solid. It takes 145 kJ/mol to produce the ions from the elements, and the
solid compound is 787 kJ/mol lower in energy than the separated ions.




Chapter 8 | Slide 65
                       Formation of an Ionic Solid
 1.Sublimation of the solid metal
         • M(s)  M(g)              [endothermic]
 2.Ionization of the metal atoms
         • M(g)  M+(g) + e             [endothermic]
 3.Dissociation of the nonmetal
         •     1/2X (g)
                   2        X(g)     [endothermic]
4. Formation of X ions in the gas phase:
        •       X(g) + e  X(g)        [exothermic]
5. Formation of the solid MX                             8.5
        • M+(g) + X(g)  MX(s) [quite exothermic]
Chapter 8 | Slide 66
Figure 8.11
Comparison of
the Energy
Changes
                                                    -
Involved in the
Formation of
Solid Sodium
Fluoride and
Solid Magnesium           +      -
Oxide           -             +
                              - +-
                          +
                                                     +




                                        +
                                     The overall reaction
   Chapter 8 | Slide 67
This sequence of images illustrates why ionic solids are brittle. (a) The original solid
consists of an orderly array of cations and anions. (b) A hammer blow can push the ions
into positions where cations are next to cations and anions are next to anions; there are
now strong repulsive forces acting (as depicted by the double-headed arrows). (c) As a
result of these repulsive forces, the solid springs apart in fragments. (d) This chunk of
calcite consists of several large crystals joined together. (e) The blow of a hammer has
shattered the crystal, leaving flat, regular surfaces consisting of planes of ions.




Chapter 8 | Slide 68
     These micrographs show the porous structure of bone. The calcium in bone is extracted
     by the body if the level of calcium in the diet is low. (a) Healthy bone tissue. (b) Bone that
     has suffered calcium loss through osteoporosis. The overlay shows the regular
     arrangement of the calcium and phosphate ions in healthy bone.




     (a)                                               (b)




Chapter 8 | Slide 69
                       Partial Ionic Character of
                            Covalent Bonds
• None of the bonds reaches 100% ionic character even
  with compounds that have a large electronegativity
  difference (when tested in the gas phase).
            Chemical bonds
             ˙Ionic: electrons are transferred to form ions
             ˙Covalent: equal sharing of electrons
             ˙Polar covalent: unequal electron sharing

Percent ionic character of a bond X─Y:

  Measured dipole moment of X─Y
                                + 
                                    100%
  Calculated dipole moment for X Y     8.6
Chapter 8 | Slide 70
Figure 8.12 a-c The
Three Possible Types of
Bonds: (a) covalent
bond, (b) polar covalent
bond with both ionic and
covalent components,
(c) ionic bond, no
electron sharing.


Chapter 8 | Slide 71
            Percent ionic character of a bond X─Y:


Measured dipole moment of X─Y
                              + 
                                  100%
Calculated dipole moment for X Y




                                                     8.6
Chapter 8 | Slide 72
The relationship between the ionic character of a covalent bond
   and the electronegativity difference of the bonded atoms
                        Where are authentic ionic compounds?




                                             Not exactly straight line:
                                             Inexact correlation between
                                             electronegative difference and
                                             ionic character



                       This diagram is worth your meditation!             8.6
Chapter 8 | Slide 73
                          Ionic Compound
                       (GENERAL DEFINITON)
• Any compound that conducts an electric
  current when melted will be classified as
  ionic.
This definition would avoid the ambiguity arising from problems
such as a polyatomic ion that contains covalent bonds.
If you think a little bit hard, then you’ll find the problem
remains! What is called ‘conducts electric current’?
My offer of solution: scrap ‘ionic compound’ altogether and just
use ionic degree, but who would follow? I just use this example
to show that many problems do NOT have standard answer, the
questions are ALWAYS open. With a deepened understanding,
                                                                 8.6
Progress is made…that’s all about scientific endeavor.
Chapter 8 | Slide 74
Molten, NaCl Conducts an
Electric Current, Indicating
the Presence of Mobile Na+
and Cl- Ions




 Chapter 8 | Slide 75
                  The Covalent Chemical Bond



                What is a chemical bond? That’s a big question.




Chapter 8 | Slide 76
                       Models

• Models are attempts to explain how
  nature operates on the microscopic level
  based on experiences in the macroscopic
  world.




                                             8.7
Chapter 8 | Slide 77
          Fundamental Properties of Models

1. A model does not equal reality.
2. Models are oversimplifications, and are
   therefore often wrong.
3. Models become more complicated and
   are modified as they age.
4. We must understand the underlying
   assumptions in a model so that we don’t
   misuse it.
                                             8.7
Chapter 8 | Slide 78
            Table 8.4 Average Bond Energies (kJ/mol)




Chapter 8 | Slide 79
     Table 8.5 Bond Lengths for Selected Bonds




Chapter 8 | Slide 80
           The strengths and lengths of bonds




Chapter 8 | Slide 81
(a) The three normal vibrational modes
   of H2O.
                                         (b) The four normal vibrational modes
                                             of CO2.



   Chapter 8 | Slide 82
         Infrared spectroscopy measures normal
                  vibrational frequencies




     The infrared spectrum of tryptophan (an amino acid).
Chapter 8 | Slide 83
                         Bond enthalpy ΔHB

                        H2(g)2H(g)   ΔHo=+436 kJ/mol



                         ΔHB(H-H)= 436 kJ/mol


                       ΔHB(F-CF3)= 484 kJ/mol


                       ΔHB(H-CH3)= 412 kJ/mol
Chapter 8 | Slide 84
Chapter 8 | Slide 85
     The bond enthalpies, in kilojoules per mole (kJ/mol), of diatomic
     nitrogen, oxygen and fluorine molecules. Note how the bonds
     weaken as they change from a triple bond in N2 to a single bond in F2.




Chapter 8 | Slide 86
The bond enthalpies, in kilojoules per mole (kJ/mol), of hydrogen halide
molecules. Note how the bonds weaken as the halogen atom becomes
larger.




                                                     F


                                                   Cl


                                                Br


                                               I

Chapter 8 | Slide 87
                           Bond strengths
                       in diatomic molecules
• The bond strength between two atoms is
  measured by the bond enthalpy.
• The bond enthalpy typically increases as
  the order of the bond increases, decreases
  as the number of lone pairs on neighboring
  atoms increases, and decreases as the
  atom radius increases.



Chapter 8 | Slide 88
   Bond strengths in polyatomic molecules




     The bond strength between a given pair of atoms varies slightly.
     The average bond enthalpy is a guide to the strength of a bond in
     any molecule.
Chapter 8 | Slide 89
                       Multiple bond strength < bond order * single bond strength

                       Double bonds are shorter than single bonds, leading to
Chapter 8 | Slide 90   bond strengths larger than double of single-bond strengths.
     The pairs of electrons in a multiple bond repel each other and can
     weaken the bond. As a result, a double bond between carbon atoms
     is not twice as strong as two single bonds would be.




         The reason of
         Multiple bond strength < bond order * single bond strength
Chapter 8 | Slide 91
     The bond enthalpies for bonds between hydrogen and the p-block
     elements. The bond strengths decrease from top to bottom of each
     group as the atoms increase in size.




Chapter 8 | Slide 92
           How to use average bond enthalpies
                 Average bond enthalpies can be used to estimate reaction
                 enthalpies and to predict the stability of a molecule.

                                      Step 1: decide which bonds are broken
                                      and which formed. Calculate the
                                      change in enthalpy when the bonds
                                      are broken in the reactants.
                                      Step 2: calculate the bond enthalpies
                                      for the new product bonds.
                                      Step 3: calculate the bond formation
                                      enthalpies for the product bonds from
                                      the bond dissociation enthalpies
                                      obtained in step 2 and reversing the
                                      sign.
                                      Step 4: add the enthalpy change required
Chapter 8 | Slide 93
                                       to break the reactant bonds.
                       Example of using bond enthalpies
                        to estimate a reaction enthalpy

• Decide whether the following reaction is
  exothermic or endothermic:
                 CH2CH3I (g) + H2O (g)CH3CH2OH (g) + HI (g)

Bonds broken: C-I (238 kJ/mol) , O-H (463 kJ/mol)

                       ΔHo=238 + 463 = 701 (kJ/mol)

Bond formed: H-I(299 kJ/mol), C-O (360 kJ/mol)
                       ΔHo=299 + 360 = 659 (kJ/mol)
 Te reaction enthalpy: ΔHr=701 - 659 = +42 (kJ/mol) endothermic
Chapter 8 | Slide 94
                             Classroom exercise

         Which of the following reactions is
         exothermic?                    Tables 9.2,3

                       (a) CCl3CHCl2 (g) + HF(g)CCl3CHClF (g) + HCl (g)
                       (b) CCl3CHCl2 (g) + HF(g)CCl3CCl2F (g) + H2 (g)

  (a) Bonds broken: Cl-C (338 kJ/mol), H-F(565 kJ/mol)
      bonds formed: C-F (484 kJ/mol), H-Cl(431 kJ/mol)


     (b) Bonds broken: H-C (412 kJ/mol), H-F(565 kJ/mol)
         bonds formed: C-F (484 kJ/mol), H-H(436 kJ/mol)

Chapter 8 | Slide 95
                       Bond Energies

• To break bonds, energy must be added to
  the system (endothermic).
• To form bonds, energy is released
  (exothermic).




                                        8.8
Chapter 8 | Slide 96
                         Bond Energies


             ΔH = ΣD(bonds broken) – ΣD(bonds formed)

                D represents the bond energy per
                mole of bonds (always has a positive
                sign).



                                                        8.8
Chapter 8 | Slide 97
                       Localized Electron Model

• A molecule is composed of atoms that
  are bound together by sharing pairs of
  electrons using the atomic orbitals of the
  bound atoms.




                                                  8.9
Chapter 8 | Slide 98
                       Localized Electron Model

• Electron pairs are assumed to be
  localized on a particular atom or in the
  space between two atoms:
         Lone pairs – pairs of electrons localized on
          an atom
         Bonding pairs – pairs of electrons found in
          the space between the atoms



                                                         8.9
Chapter 8 | Slide 99
                        Localized Electron Model

1. Description of valence electron
   arrangement (Lewis structure).
2. Prediction of geometry (VSEPR model).
3. Description of atomic orbital types used
   to share electrons or hold lone pairs.




                                                   8.9
Chapter 8 | Slide 100
                        Lewis Structure

• Shows how valence electrons are
  arranged among atoms in a molecule.
• Reflects central idea that stability of a
  compound relates to noble gas electron
  configuration.




                                              8.10
Chapter 8 | Slide 101
Figure 8.14
G.N. Lewis




Chapter 8 | Slide 102
        A Diamond Anvil Cell Used to Study
         Materials at Very High Pressures




Chapter 8 | Slide 103
                        Covalent Bonds
         • In covalent bond formation, atoms go as far as
           possible toward completing their octets (duplets) by
           sharing electron pairs.




Chapter 8 | Slide 104
     Figure 8.13 The formation of a covalent bond between two hydrogen atoms. (a) Two
     separate hydrogen atoms, each with one electron. (b) The electron cloud that forms when
     the spins pair and the orbitals merge is most dense between the two nuclei. (c) The
     boundary surface that we shall use to depict a covalent bond.




Chapter 8 | Slide 105
                      Covalent Bonds
    • In covalent bond formation, atoms go as far as
      possible toward completing their octets
      (duplets) by sharing electron pairs.

      H+H  H:H or H-H
          ..         ..           ..     ..                 ..      ..
      :
          F  : F. : F : F : or
               .                                        :
                                                            F F:
          ..         ..           ..     ..                 ..      ..

  That’s why F2 is so reactive.               Lone pairs
  The Lewis structure gives not only bond positions (shared electron pairs)
  But also reactivity-how reactive is the molecules and which part(s) is(are)
  reactive (lone pairs).
Chapter 8 | Slide 106
         The Structures of Ployatomic Species
   H                 H
                   |
H:C:H             H- C -H                                
                   |
                                              :F O F:
   H                 H                                   


                           
                H         :O:          H
                |         |        |
            [H- N-H]+[:O S  O:]2-[H- N-H]+
                |         |        |
                H         :O:          H
                           

                     • Single, double, triple bonds may be
Chapter 8 | Slide 107  formed.
                     How to write Lewis structure of a
                           polyatomic species
• Count the total number of valence electrons on each atom
  and divide by 2 to obtain the number of electron pairs. If the
  species is polyatomic ion, add one more electron for each
  negative charge or subtract one electron for each positive
  charge.
• Write the chemical symbols of the atoms to show their
  layout in the molecule. One can predict the most likely
  arrangements of atoms by using common patterns and the
  rules of thumb (largely correct but exceptions possible)
  given earlier.
• Place one electron pair between each pair of bonded
  atoms.
• Complete the octet (duplet for H) of each atom by placing
  any remaining electron pairs as lone pairs around the
  atoms. If there are not enough electrons to form octets,
  form multiple bonds.
• Care for the possibility of cations (anions) and resonance.
 Chapter 8 | Slide 108
    How to write Lewis structure of a polyatomic species:
            another summary (from Wikipedia)
•   Step one: Nitrogen is the least electronegative atom, so it is the central atom by
    multiple criteria.
•   Step two: Count valence electrons. Nitrogen has 5 valence electrons; each oxygen has 6,
    for a total of (6 × 2) + 5 = 17. The ion has a charge of −1, which indicates an extra electron,
    so the total number of electrons is 18.
•   Step three: Place ion pairs. Each oxygen must be bonded to the nitrogen, which uses four
    electrons — two in each bond. The 14 remaining electrons should initially be placed as 7 lone
    pairs. Each oxygen may take a maximum of 3 lone pairs, giving each oxygen 8 electrons
    including the bonding pair. The seventh lone pair must be placed on the nitrogen atom.
•   Step four: Satisfy the octet rule. Both oxygen atoms currently have 8 electrons assigned to
    them. The nitrogen atom has only 6 electrons assigned to it. One of the lone pairs on an
    oxygen atom must form a double bond, but either atom will work equally well. We therefore
    must have a resonance structure.
•   Step five: Tie up loose ends. Two Lewis structures must be drawn: one with each oxygen
    atom double-bonded to the nitrogen atom. The second oxygen atom in each structure will be
    single-bonded to the nitrogen atom. Place brackets around each structure, and add the
    charge (−) to the upper right outside the brackets. Draw a double-headed arrow between the
    two resonance forms.




    Chapter 8 | Slide 109
                                   More Examples

                         H     NH3:
               H- N :
                         |
                               3+5=8 valence electrons = 4 pairs.
                         |     3 pairs to form three bonds,
                         H     leaving one lone pair
                                                               :O:
                                                                 ||
                                                           H- C
                                                              |
             H  O Br :                                     : O -H
                                                           

Hypobromous acid HBrO:                             Formic acid HCOOH
1+6+7=14 valence electrons = 7 pairs               2+12+4=18 valence electrons
Two pairs to form two bonds,                       = 9 pairs
Leaving 5 lone pairs.                              5 pairs to form five bonds,
                                                   leaving 4 lone pairs.
 Chapter 8 | Slide 110
                         Classroom Exercise

    • Write the Lewis structure of CH3COOH



                           H  :O:
                           |     ||
                        H- C -  C
                           |     |
                          H : O -H
                              
Chapter 8 | Slide 111
                  Resonance: Blending of Structures



                                                  
     :O                            :O:               :O:
    ||                         |             | 
[:O N  O:]-                 [:O  N  O:]-    [:O N  O:]-
                                    


All valid. We cannot find two bond lengths (hypothetical N-O vs N=O)


 Chapter 8 | Slide 112
                   Resonance: Blending of Structures




Chapter 8 | Slide 113
                   Resonance: Blending of Structures




Chapter 8 | Slide 114
                                         Quiz
       Write the Lewis structure of the following compounds:

       (1) Formic acid (2) Nitrate (3) Carbon Monoxide

 Answer:
                                                             
    :O:                       :O               :O:              :O:
      ||                     ||            |            | 
H- C
       |
                         [:O N  O:]-    [:O  N  O:]-   [:O N  O:]-
                                                      
  : O -H
    



 Chapter 8 | Slide 115
  Electrons in an Atom and in a Molecule:
  from a freeman to a slave or otherwise?
 • From spherically symmetrical to less or none of symmetry,
   from pure to hybrid, from localized to delocalized, from
   individual to collective, from neutral to charged, from simple
   happy primitive tribe to adversary yet opportune land…
 • From freedom to restriction or just opposite? from
   independent to correlated or other way around?

  Electrons have rich feeling; they try their best to keep the changes
  minimized when transferred from an atom to a molecule so that an
  atom in a molecule also looks the same as a free atom as possible.
Atoms seem to have some good happy memory about their free, neutral
state—after put in molecules, they tend to stay in their original state of
zero net charge by keeping their valence electrons unchanged (or if
forced 8to change, minimizing it).
   Chapter | Slide 116
                             Formal Charge
• The real number of valence electrons each atom in a molecule
  “owns” may be different from the number of normal valence
  electrons of that atom, rendering it positively or negatively charged.




    Formal charge (FC) = number of valence electrons on the free atom (V)
                       - number of electrons present as lone pairs (L)
                       - ½ number of electrons shared in bonds (S/2)


                          FC  V  L  S             1
                                                     2
Formal charge can be understood as surplus or deficit of valence electrons after
deducting lone pairs and shared pairs.

Typically, the most stable Lewis structures are those in which the formal charge
of the individual nonmetal atoms are close to zero

  Chapter 8 | Slide 117
                  Plausibility of a structure
                                                         
       OCO                               NO N
                                                         
       0 0 0                                -1 +2 -1
                        
                                                             
       OOC
                                        N NO
                                                             
                0 +2 -2


       0 +2 -2                              -1 +1 0
A molecule is more favored by God than others if more atoms in it
have formal charges closest to zero. An atom feels guilty if it takes
electrons from others and it feels angry if it has to give up electrons to
            So
others.| Slide 118atoms are like you, aren’t they?
  Chapter 8
                              Classroom Exercise
• Suggest a plausible structure for the poisonous
  gas phosgene, COCl2.
                  Resonance
                  states


                                                      0    0
   :Cl:           0
                            ..O     :Cl:                 ..
0
   |                    ||    |                 :Cl-O - C 0
O C              0     :Cl C :Cl C                  ..  |
   |                   ..   |  .. ||                       :Cl: 0
   :Cl:           0                 :O                        
                            :Cl:     
                                       Formal charges are all zero, too.
                                           But why is this one not favored?
                                           Experiment shows the oxygen has
                                           double bond rather than single bond.
Chapter 8 | Slide 119
                        Exceptions to the Octet Rule
•     B, C, N, O, F observe the octet rule.
•     P, S, Cl can have more than 8 valence electrons.
•     Radicals: odd-electron species, highly reactive.
•     Biradicals: with two unpaired electrons.




             
CH 3 -CH 3  H 3C  +  CH 3
                             stress
                                                             
                                              O  O O: :O O  O
          
O2 (g)  2O(g)          sunlight                               



             
O+O 2  O3  O+O 2
Chapter 8 | Slide 120
                                      UV
Chapter 8 | Slide 121
                              O2 is a biradical!


                                
                    OO  Wrong!

                                


                                                   ..
                                          ..        ..
                   O O
                                 
                                          .
                                              ..   O
                                                   ..   2
                                                             .


Chapter 8 | Slide 122
                        Classroom Exercise
                                                           .
Write a Lewis structure for the hydrogenperoxyl radical HOO ,
which plays an important role in atmospheric chemistry and
which in the body has been implicated in the degeneration of
neurons.




Chapter 8 | Slide 123
 Case Study 8 (a) The equipment in the illustration monitors air quality from a rooftop in
 Los Angeles. The concentration of NO2 in the air due to automobile traffic increases
 during the day, contributing to the typical brown color of the afternoon sky.




 Exhaust: N 2 (g)+O 2 (g)  2NO(g)
 +Atmosphere: 2NO(g)+O2 (g)  2NO2 (g)
                                    
  +Sunlight:  NO 2  NO 2 +  O  (smog)
Chapter 8 | Slide 124
                      UV (< 400 nm)
      Case Study 8 (b) The gas NO (left) is colorless. When it is exposed to air (right), it is
      rapidly oxidized to brown NO2.




  Oxygen molecules are
  actually biradicals.




2NO(g)+O2 (g)  2NO 2 (g)
 Chapter 8 | Slide 125
                        Expanded valence shells
• Nonmetal atoms in Period 3 or higher can
  accommodate 10, 12 or more electrons (expanded
  valence shell). e.g., P, S, Cl.


                        
P 4 (g)+6Cl2 (g)  PCl3 (l)
                              limited supply of chlorine




                        
 P 4 (g)+10Cl2 (g)  4PCl5 (s)
                               excessive chlorine


Chapter 8 | Slide 126
     Figure 8.16 (a) A model using small spheres to represent atoms and
     (b) a space-filling model of PCl5, showing how closely the chlorine
     atoms must pack around the central phosphorus atom. (c) A nitrogen
     atom is significantly smaller than a phosphorus atom, and five
     chlorine atoms cannot pack around it.




Chapter 8 | Slide 127
     Figure 8.17 Phosphorus trichloride is a colorless liquid. When it
     reacts with chlorine (the pale yellow-green gas in the flask), it forms
     the very pale yellow solid phosphorus pentachloride (at the bottom of
     the flask).




Chapter 8 | Slide 128
Chapter 8 | Slide 129
                        +



Chapter 8 | Slide 130
Figure 8.18 Two circumstances in which a central atom assumes an expanded
octet. First, there are too many electrons to be accommodated in octets
because either (a) there are too many atoms attached, or (b) the central atom
must accommodate additional lone pairs. (c) Second, a resonance structure
with multiple bonds has a favorable energy.




 Chapter 8 | Slide 131
                  S accommodates 10 electrons




Chapter 8 | Slide 132
                        How many electrons does Xe
                             accommodate?




Chapter 8 | Slide 133
                        Which is most plausible?




Chapter 8 | Slide 134
                        12 valence electrons!


Chapter 8 | Slide 135
                         Classroom exercise:
                        which is more plausible?




Chapter 8 | Slide 136
                        Answer




Chapter 8 | Slide 137
                        One more…




Chapter 8 | Slide 138
            The unusual structures of group 13
                         halides

• Compounds of boron and aluminum may
  have unusual Lewis structures in which
  boron and aluminum have incomplete
  octets or in which halogen atoms as
  bridges.




Chapter 8 | Slide 139
                        Incomplete octet




Chapter 8 | Slide 140
Chapter 8 | Slide 141
                        +   F-




Chapter 8 | Slide 142
                           Coordinate covalent bond




                                 This covalent bond is
                                 formed by two electrons
                                 from nitrogen.


     BF3 (g)+NH3 (g)  NH3BF3 (s)



Both electrons shared in the bond come from the same atom.
The coordinate bond is more extended (nonlocal) than an ordinary covalent bond.
   Chapter 8 | Slide 143
                        Coordinate covalent bond




Chapter 8 | Slide 144
Which atom contributes the electrons that form the coordinate covalent bond?




                          
           AlCl3 (s)  Al2Cl6 (g)
 Chapter 8 | Slide 145
                            sublimes at 180 o C
                                 >200 o C
Which atom contributes the electrons that form the coordinate covalent bond?




                          
           AlCl3 (s)  Al2Cl6 (g)
 Chapter 8 | Slide 146
                            sublimes at 180 o C
                                 >200 o C
                        Writing Lewis Structures

1. Sum the valence electrons.
2. Place bonding electrons between pairs
   of atoms.
3. Atoms usually have noble gas
   configurations. Arrange remaining
   electrons to satisfy the octet rule (or
   duet rule for hydrogen).


                                                   8.10
Chapter 8 | Slide 147
                        Concept Check
Draw a Lewis structure for each of the following
molecules:

             H2
             F2
             HF




                                                   8.10
Chapter 8 | Slide 148
                        Concept Check
Draw a Lewis structure for each of the following
molecules:

                H2O
                NH3




                                                   8.10
Chapter 8 | Slide 149
                        Exceptions

• When it is necessary to exceed the octet
  rule for one of several third-row (or higher)
  elements, place the extra electrons on
  the central atom.




                                             8.11
Chapter 8 | Slide 150
                        Resonance

• More than one valid Lewis structure can
  be written for a particular molecule.
• Actual structure is an average of the
  resonance structures.




                                            8.12
Chapter 8 | Slide 151
                                Concept Check
     Draw a Lewis structure for each of the
     following molecules:

                        CO           CO2
                        CH3OH        BF3
                        PCl5         NO3-
                        SF6



                                                8.10-8.12
Chapter 8 | Slide 152
                        Formal Charge

• Nonequivalent Lewis structures.
• Atoms in molecules try to achieve formal
  charges as close to zero as possible.
• Any negative formal charges are
  expected to reside on the most
  electronegative atoms.



                                             8.12
Chapter 8 | Slide 153
                        Formal Charge

• Formal charge = # valence e– on free
  atom – # valence e– assigned to the atom
  in the molecule.
• Assume:
        – Lone pair electrons belong entirely to the
          atom in question
        – Shared electrons are divided equally
          between the two sharing atoms

                                                       8.12
Chapter 8 | Slide 154
                        VSEPR Model

• VSEPR: Valence Shell Electron-Pair
  Repulsion.
• The structure around a given atom is
  determined principally by minimizing
  electron pair repulsions.




                                         8.13
Chapter 8 | Slide 155
  Figure 8.15 The Molecular Structure of
                Methane




Chapter 8 | Slide 156
Balloons Tied
Together
Naturally Form
Tetrahedral
Shape



Chapter 8 | Slide 157
Figure 8.16 a-c The Molecular Structure
   of Ammonia is a Trigonal Pyramid




Chapter 8 | Slide 158
          Figure 8.17 a-c The Tetrahedral
         Arrangement of Oxygen In a Water
                      Molecule




Chapter 8 | Slide 159
       Figure 8.18 The Bond Angles In the
          CH4, NH3, and H2O Molecules




Chapter 8 | Slide 160
   Some of the basic geometrical shapes that are used to
        describe the shapes of simple molecules




Chapter 8 | Slide 161
                   Tetrahedral   Octahedral
Chapter 8 | Slide 162
                        Shape names and bond angles
                           (lone pairs not included)




                                             T-shaped




Chapter 8 | Slide 163
                         VSEPR Model
            (Valence-Shell Electron-Pair Repulsion model)


      Bonding electrons and lone pairs take
      up positions as far from one another as
      possible, for then they repel each other
      the least.




Chapter 8 | Slide 164
                        VSEPR Model (1):
                        locate the high concentrations
  The positions taken up by regions of high electron concentration
  (green) around a central atom.




    Electron pairs or bonds are as far away from one
    another as possible and so experience the minimal
    repulsion from other electrons
Chapter 8 | Slide 165
VSEPR Model (2): determine
the shapes


                                Two ‘highs’        Three ‘highs’       Four ‘highs’




                                Five ‘highs’        Six ‘highs’    Seven ‘highs’
   A summary of the positions taken up by regions of high electron
   concentration (other atoms and lone pairs) around a central atom.
   The locations of these regions are given by straight lines sticking out
   of the central atom.
  Use this chart to identify the arrangement of a given number of
  atoms and lone pairs and then use Fig. 9.2 to identify the shape
  of the molecule from the location of its atoms.
  Chapter 8 | Slide 166
    VSEPR Model: Example

     Electron pairs or bonds are as far away from one another as
     possible and so experience the minimum repulsion from other
     electrons

        ..               ..
:Cl-Be-Cl:
        ..               ..




       Shape: linear (two ‘highs’)             Shape: trigonal planar
Chapter 8 | Slide 167
                                               (three ‘highs’)
    VSEPR Model: Example

     Electron pairs or bonds are as far away from one another as
     possible and so experience the minimum repulsion from other
     electrons




                        Shape: trigonal bipyramidal (five ‘highs’)
Chapter 8 | Slide 168
                        VSEPR Model


• In order to reduce repulsions, bonding
  pairs and lone pairs take up positions
  around an atom that maximize their
  separations. The shape of the molecule
  is determined by the location of the
  atoms attached to the central atom.



Chapter 8 | Slide 169
                        Classroom Exercise



         Predict the structure of SiCl4 and AsF5




Chapter 8 | Slide 170
                           Classroom Exercise
       Predict the structure of SiCl4 and AsF5

           Cl
                        Si: 3s23p2                            AsF5
        |                                         F
    Cl- Si -Cl                           F
        |
           Cl
                                             As       F
                                     F
                                                  F

                                             As: 3d104s24p3
                           SiCl4




Chapter 8 | Slide 171
             Molecules with Multiple Bonds
    The VSEPR model does not distinguish single and multiple
    bonds. A multiple bond is treated as just another region of
    high electron concentration.
      ..                     ..
 O =C= O
      ..                     ..




          (Two ‘highs’)
Chapter 8 | Slide 172                           (Three ‘highs’)
 The VSEPR model does not distinguish single and multiple bonds. A
 multiple bond is treated as just another region of
 high electron concentration.




                          Two central atoms




Chapter 8 | Slide 173
                               Example

                HC  CH
           Two central atoms, no lone pairs.

            The most possible VSEPR arrangement:




Chapter 8 | Slide 174
                        Formaldehyde


     CH2O                                H
                                          |  ..
                                       H- C =O
                                             ..


                        CH2
                         O



Chapter 8 | Slide 175
                        Classroom Exercise
Predict the structure of HCN



                                 H-C  N:


Chapter 8 | Slide 176
Different resonance structures correspond to a
single shape




  Chapter 8 | Slide 177
    How about if the central atoms have lone
                     pairs?


Lone pairs on attached atoms have little effect on molecular shape,
but the lone pairs on central atoms may have significant effect.


    • A: central atom
    • X: atom bonded to the central atom
    • E: lone pair on the central atom

The single nonbonding electron on radicals is treated as a ‘lone pair’.
  Chapter 8 | Slide 178
                    Lone Pair on the Central Atom


                               AX3E




Chapter 8 | Slide 179
                          Repulsion Order:




Lone pair/lone pair > lone pair/bonding pair > bonding pair/bonding pair
  Chapter 8 | Slide 180
                           Example
       H2O

            ..
 H-O-H
            ..


              AX2E2




                        Shape: Angular (not tetrahedron!)
Chapter 8 | Slide 181
              Three views of water molecular shape




Chapter 8 | Slide 182
                        NH3

                H
     |
  H- N-H
                   ..



               AX3E


                              Triangular pyramid
Chapter 8 | Slide 183         (NOT tetrahedron!)
                         Classroom Exercise
Predict the shape of
  NO2-


    ..                  .. -                     AEX2

  O = N  O:
    ..             ..   ..

                                              NO2-
                                        Smaller
                                        angle


Chapter 8 | Slide 184
                                Angular (NOT planar triangle!)
                                      AX4E




                    Axial lone pair          Equatorial lone pair
Chapter 8 | Slide 185
                              AX4E2




                        Square planar
                        (NOT octahedron!)
Chapter 8 | Slide 186
 How to Use VSEPR Model


1. Write Lewis structure
and determine the number of
electron pairs

2. Maximize the separations.

3. Decide the positions
of lone pairs (on the central atom).

4a. Name the shape (without
considering the lone pair).
 4b. Consider distortion using
 repulsion order.

Lone pair/lone pair > lone pair/bonding pair > bonding pair/bonding pair
  Chapter 8 | Slide 187
                        Example: SF4


                               AX4E



                                        Equatorial lone pair
                                        Bent seesaw



                                T-shaped
Chapter 8 | Slide 188           (NOT triangular bipyramid!)
                                       ClF3


                         ..

               ..
                        :F:
                         |        ..
             :F - Cl - F:
               ..            ..   ..



                                              T-shaped
                                              (NOT triangular bipyramid!)
                        AX3E
Chapter 8 | Slide 189
                        Classroom Exercise: XeF4
                   ..
                 :F:
                  |
..
                        ..
                             ..      AX4E2
:F - .. Xe - F:
..                           ..
                  |
                 :F:
                   ..




         Square planar
         (NOT octahedron!)
Chapter 8 | Slide 190
                          Quiz




         • Write the VSEPR formula of water and
           sulfur tetrafluoride, draw and name their
           structures.


Chapter 8 | Slide 191
                                Answer
    • Write the VSEPR formula of water and
      sulfur pentafluoride. Draw and name
      their structures.
          ..
H-O-H
          ..

AX2E2

                                           AX4E
                                         T-shaped
            Angular (not tetrahedron!)
Chapter 8 | Slide 192                    (not triangular bipyramid!)
Chapter 8 | Slide 193
                        VSEPR: Two Electron Pairs




                                                    8.13
Chapter 8 | Slide 194
                   VSEPR: Three Electron Pairs




                                                 8.13
Chapter 8 | Slide 195
                        VSEPR: Four Electron Pairs




                                                     8.13
Chapter 8 | Slide 196
                    VSEPR: Iodine Pentafluoride




                                                  8.13
Chapter 8 | Slide 197
                        VSEPR




                                8.13
Chapter 8 | Slide 198
                        NH3




Chapter 8 | Slide 199
                        PH3




Chapter 8 | Slide 200
                  Predicting a VSEPR Structure

1. Draw Lewis structure.
2. Put electron pairs as far apart as
   possible.
3. Determine positions of atoms from the
   way electron pairs are shared.
4. Determine the name of molecular
   structure from positions of the atoms.

                                                 8.13
Chapter 8 | Slide 201
                        Queen Bee




Chapter 8 | Slide 202
                        Concept Check
Determine the shape for each of the following
molecules, and include bond angles:

                HCN
                PH3
                SF4
                O3
                KrF4

                                                8.13
Chapter 8 | Slide 203
                        Concept Check
True or false:
A molecule that has polar bonds will always be
polar.
       -If true, explain why.
       -If false, provide a counter-example.




                                                 8.13
Chapter 8 | Slide 204
                        Concept Check
True or false:
Lone pairs make a molecule polar.
     -If true, explain why.
     -If false, provide a counter-example.




                                             8.13
Chapter 8 | Slide 205
Chemical bonds
˙Hold groups of atoms together
˙Occur when a group of atoms can lower its total energy by aggregating
˙Types of chemical bonds
 ˙Ionic: electrons are transferred to form ions
                                                                                              Review
 ˙Covalent: equal sharing of electrons
 ˙Polar covalent: unequal electron sharing
˙Percent ionic character of a bond X─Y:
                Measured dipole moment of X─Y
                                                     100%
                Calculated dipole moment for X + Y 


˙Electronegativity: the relative ability of an atom to attract shared electrons
 ˙The polarity of a bond depends on the electronegativity difference of the bonded atoms
˙The spacial arrangement of polar bonds in a molecule determines whether the molecule has a dipole moment
Ionic bonding
˙An ion has a different size than its parent atom
 ˙An anion is larger than its parent ion
 ˙A cation is smaller htan its parent atom
˙Lattice energy: the change in energy when ions are packed together to form an ionic solid
Bond energy
˙ The energy necessary to break a covalent bond
˙Increases as the number of shared pairs increases
˙Can be used to estimate the enthalpy change for a chemical reaction
Lewis structures
˙Show how the valence electron pairs are arranged among the atoms in a molecule or polyatomic ion
˙Stable molecules usually contain atoms that have their valence orbitals filled
 ˙Leads to a duet rule for hydrogen
 ˙Leads to an octet rule for second-row elements
 ˙The atoms of elements in the third row and beyond can exceed the octet rule
˙Several equivalent Lewis structures can be drawn for some molecules, a concept called resonance
˙When several nonequivalent Lewis structures can be drawn for a molecule, formal charge is often used to choose the most appropriate structure(s)
VSEPR model
˙Based on the idea that electron pairs will be arranged around a central atom in a way that minimizes the electron repulsions
˙Can be used to predict the geometric structure of most molecules
   Chapter 8 | Slide 206
Chapter Eight

Bonding: General
   Concepts




案例/討論
   Figure 8.1 a & b (a) The Interaction of Two
     Hydrogen Atoms (b) Energy Profile as a
  Function of the Distance Between the Nuclei of
                the Hydrogen Atoms




Chapter 8 | Slide 208
     Figure 8.2 The Effect of an Electric
    Field on Hydrogen Fluoride Molecules




Chapter 8 | Slide 209
                        Figure 8.3 The Pauling
                        Electronegativity Vaules




Chapter 8 | Slide 210
       Figure 8.4 An Electrostatic Potential
                   Map of HF




Chapter 8 | Slide 211
  Figure 8.5 a-c The Charge Distribution
           in the Water Molecule




Chapter 8 | Slide 212
         Figure 8.6 a-c The Structure and
        Charge Distribution of the Ammonia
                     Molecule




Chapter 8 | Slide 213
         Figure 8.7 a-c The Carbon Dioxide
                      Molecule




Chapter 8 | Slide 214
                        e.p. Diagram HCL




Chapter 8 | Slide 215
                        e.p.Diagram SO3




Chapter 8 | Slide 216
                        e.p. Diagram CH4




Chapter 8 | Slide 217
                        e.p. Diagram H2S




Chapter 8 | Slide 218
         Figure 8.8 Sizes of Ions Related to
          Positions of the Elements on the
                   Periodic Table




Chapter 8 | Slide 219
          Figure 8.9 The Energy Changes
        Involved in the Formation of Lithium
             Fluoride from Its Elements




Chapter 8 | Slide 220
Figure 8.10
a & b The
Structure of
Lithium
Fluoride



Chapter 8 | Slide 221
Figure 8.11
Comparison of
the Energy
Changes
Involved in the
Formation of
Solid Sodium
Fluoride and
Solid
Magnesium
Oxide

Chapter 8 | Slide 222
Figure 8.12
a-c The
Three
Possible
Types of
Bonds



Chapter 8 | Slide 223
   Figure 8.13 The Relationship Between the
   Ionic Character of a Covalent Bond and the
   Electronegativity Difference of the Bounded
                      Atoms




Chapter 8 | Slide 224
  Figure 8.15 The Molecular Structure of
                Methane




Chapter 8 | Slide 225
Figure 8.16 a-c The Molecular Structure
   of Ammonia is a Trigonal Pyramid




Chapter 8 | Slide 226
          Figure 8.17 a-c The Tetrahedral
         Arrangement of Oxygen In a Water
                      Molecule




Chapter 8 | Slide 227
       Figure 8.18 The Bond Angles In the
          CH4, NH3, and H20 Molecules




Chapter 8 | Slide 228
Figure 8.19 a &
b In a Bonding
Pair of
Electrons the
Electrons are
Shared by Two
Nuclei (b) In a
Lone Pair, Both
Electrons Must
Be Close to a
Single Nucleus
Chapter 8 | Slide 229
        Figure 8.20 a & b Possible Electron
            Pair Arrangements for XeF4




Chapter 8 | Slide 230
        Figure 8.21 a-c Three Possible
     Arrangements of the Electron Pairs in
                   the I3- Ion




Chapter 8 | Slide 231
Figure 8.22
a-c The
Molecular
Structure of
Methanol



Chapter 8 | Slide 232
                        Quartz




Chapter 8 | Slide 233
        Linear Molecules with Two Identical
                      Bonds




Chapter 8 | Slide 234
     Planar Molecules with Three Identical
          Bonds 120 Degrees Apart




Chapter 8 | Slide 235
        Tetrahedral Molecules with Four
      Identical Bonds 109.5 Degrees Apart




Chapter 8 | Slide 236
                        A Bauxite Mine




Chapter 8 | Slide 237
                        Lithium Fluoride




Chapter 8 | Slide 238
Molten, NaCl
Conducts an
Electric
Current,
Indicating the
Presence of
Mobile Na+
and Cl- Ions

Chapter 8 | Slide 239
Figure 8.14
G.N. Lewis




Chapter 8 | Slide 240
        A Diamond Anvil Cell Used to Study
         Materials at Very High Pressures




Chapter 8 | Slide 241
Balloons Tied
Together
Naturally
Form
Tetrahedral
Shape



Chapter 8 | Slide 242
                        Queen Bee




Chapter 8 | Slide 243
                        NH3




Chapter 8 | Slide 244
                        PH3




Chapter 8 | Slide 245
      Table 8.1 The Relationship Between
        Electronegativity and Bond Type




Chapter 8 | Slide 246
Table 8.2 Types of Molecules with Polar
Bonds but No Resulting Dipole Moment




Chapter 8 | Slide 247
  Table 8.3 Common Ions with Noble
 Gas Configurations in Ionic Compounds




Chapter 8 | Slide 248
          Table 8.4 Average Bond Energies
                      (kj/mol)




Chapter 8 | Slide 249
     Table 8.5 Bond Lengths for Selected
                   Bonds




Chapter 8 | Slide 250
Table 8.6
Arrangements
of Electron
Pairs Around
an Atom
Yielding
Minimum
Repulsion

Chapter 8 | Slide 251
Table 8.7
Structures of
Molecules
that Have
Four Electron
Pairs Around
the Central
Atom

Chapter 8 | Slide 252
 Table 8.8
 Structures of
 Molecules
 with Five
 Electron Pairs
 Around the
 Central Atom


Chapter 8 | Slide 253
Chapter Eight


Bonding: General
   Concepts



問題
                        Question

• Which of the following elements forms the
  most ionic bond with chlorine?
        –K
        – Al
        –P
        – Kr
        – Br



Chapter 8 | Slide 255
                        Answer
•a) K

•Section 8.2, Electronegativity

•The most ionic bond results from the largest dipole
moment and the greatest difference in
electronegativity. With chlorine, potassium has the
greatest difference in electronegativity.



Chapter 8 | Slide 256
                                   Question

•As a general pattern, electronegativity is
inversely related to
             –          ionization energy.
             –          atomic size.
             –          the polarity of the atom.
             –          the number of neutrons in the nucleus.




Chapter 8 | Slide 257
                        Answer
• b) atomic size.

• Section 8.2, Electronegativity

• Electronegativity increases from left to right and
  from bottom to top in the periodic table. Atomic
  size increases from top to bottom in the periodic
  table. Thus electronegativity and atomic size are
  inversely related.


Chapter 8 | Slide 258
                        Question

• Which of the following is the most polar
  bond without being considered ionic?
        – C—O
        – Mg—O
        – N—O
        – O—O
        – O—F



Chapter 8 | Slide 259
                        Answer
•a) C—O

•Section 8.3, Bond Polarity and Dipole Moments

•Given that O is common to all of the bonds, the relative
difference in electronegativity for C, Mg, F, and N will
determine the polarity. Decreasing polarity results from
lowering the difference in electronegativity. While the
greatest difference in electronegativity is between Mg and O,
this bond is considered to be ionic, not polar.



Chapter 8 | Slide 260
                        Question

• Which of the following is the least polar
  bond, yet still considered polar covalent?
        – C—O
        – Mg—O
        – N—O
        – O—O
        – O—F



Chapter 8 | Slide 261
                         Answer
•c) N—O

•Section 8.3, Bond Polarity and Dipole Moments

•Given that O is common to all of the bonds, the relative
difference in electronegativity for C, Mg, F, and N will
determine the polarity. Decreasing polarity results from
lowering the difference in electronegativity. While the
smallest difference in electronegativity is between O and O
(both have the same electronegativity value, so the
difference is zero), this bond is not considered to be polar,
yet is perfectly covalent.



Chapter 8 | Slide 262
                        Question

• What is the correct order of the following
  bonds in terms of decreasing polarity?
        – N—Cl, P—Cl, As—Cl
        – P—Cl, N—Cl, As—Cl
        – As—Cl, N—Cl, P—Cl
        – P—Cl, As—Cl, N—Cl
        – As—Cl, P—Cl, N—Cl



Chapter 8 | Slide 263
                        Answer
•e) As—Cl, P—Cl, N—Cl

•Section 8.3, Bond Polarity and Dipole Moments

•Given that Cl is common to all of the bonds, the
relative difference in electronegativity for As, P, and
N will determine the polarity. Decreasing polarity
results from lowering the difference in
electronegativity.




Chapter 8 | Slide 264
                        Question

 • In which case is the bond polarity incorrect?
         a) +H—F-
         b) +K—O-
         c) +Mg—H-
         d) +Cl—I-
         e) +Si—S-




Chapter 8 | Slide 265
                         Answer
•d)      +Cl—I-



•Section 8.3, Bond Polarity and Dipole Moments

•Bond polarity results from the more electronegative element
pulling the electron density toward it, resulting in a partial
negative charge. Because F, O, H, Cl, and S are more
electronegative than H, K, Mg, I, and Si, respectively, they
will carry the – charge.




Chapter 8 | Slide 266
                        Question

• Which of the following has the largest
  radius?
        – S2-
        – Cl-
        – Ar
        – K+
        – Ca2+



Chapter 8 | Slide 267
                        Answer
•a)              S2-

•Section 8.4, Ions: Electron Configurations and
Sizes

•All of these ions have similar sizes as atoms, so
the charge determines the largest radius. Adding
electrons increases the size.



Chapter 8 | Slide 268
                        Question
• Which of the following ionic compounds has the
  largest lattice energy (i.e., the lattice energy most
  favorable to a stable lattice)?
        –    CsI
        –    NaCl
        –    LiF
        –    CsF
        –    MgO




Chapter 8 | Slide 269
                        Answer

•e) MgO

•Section 8.5, Formation of Binary Ionic
Compounds

•Lattice energy is directly proportional to
charge, and the species with the highest
charge is MgO.

Chapter 8 | Slide 270
                                 Question
    •                   Bond    Average Bond Energy (kJ/mol)
    •                   H—H                432
    •                   F—F                154
    •                   H—F                565

    •Given the average bond energies above, estimate ∆H for
    the following reaction:
    •                    H2 + F2  2HF
    •
                    – -21 kJ
                    – 21 kJ
                    – 544 kJ
                    – -544 kJ
                    – 1151 kJ

Chapter 8 | Slide 271
                          Answer

•d)             –544 kJ


•Section 8.8, Covalent Bond Energies and
Chemical Reactions

•∆H = [432 kJ + 154 kJ] – 2(565 kJ) = –544 kJ




Chapter 8 | Slide 272
                        Question

• Which of the following does not contain at
  least one double bond in the Lewis
  structure?
        – H2CO
        – C2H4
        – CO2
        – C3H8




Chapter 8 | Slide 273
                           Answer

•d)               C 3H 8

•Section 8.10, Lewis Structures

•Choice (a) contains a C=O double bond,
choice (b) contains a C=C double bond, and
choice (c) contains two C=O double bonds.


Chapter 8 | Slide 274
                                Question
• Which of the following statements concerning resonance
  structures is correct?
        – The concept of resonance is used because the Lewis structure
          model is incomplete when describing bonding in a molecule.
        – For a species having three resonance structures, it is best to think
          of the species as existing as each of these structures one-third of
          the time.
        – All charged molecules have resonance structures.
        – The octet rule must not be violated in writing resonance structures.




Chapter 8 | Slide 275
                         Answer
•a) The concept of resonance is used because the Lewis
structure model is incomplete when    describing bonding
in a molecule.

•Section 8.12, Resonance; Section 8.11, Exceptions to the
Octet Rule

•Lewis structures show static, localized electrons. These
structures are incomplete because electrons are better
thought of as delocalized in the molecule.



Chapter 8 | Slide 276
                        Question

• Which of the following has a Lewis
  structure most like that of CO32-?
        – CO2
        – SO32-
        – NO3-
        – O3
        – NO2



Chapter 8 | Slide 277
                        Answer
•c) NO3-

•Section 8.10, Lewis Structures; Section 8.11,
Exceptions to the Octet Rule

•The number of atoms and the electron count are
the same for NO3- and CO32-; thus these two ions
have similar Lewis structures.



Chapter 8 | Slide 278
                            Question

  • Which molecule or ion violates the octet
    rule?
          – H2O
          – NO3-
          – PF3
          – I3-
          – None of these



Chapter 8 | Slide 279
                        Answer

•d)               I3-

•Section 8.10, Lewis Structures; Section 8.11,
Exceptions to the Octet Rule

•Violations of the octet rule involve having
more than 8 electrons around an atom, which
is the case with I3-.

Chapter 8 | Slide 280
                        Question

• How many resonance structures does the
  molecule SO2 have?
        –0
        –1
        –2
        –3
        –4



Chapter 8 | Slide 281
                        Answer

•c)               2

•Section 8.10, Lewis Structures; Section 8.12,
Resonance

•The two resonance structures (without the
lone pairs) are O–S=O and O=S–O.


Chapter 8 | Slide 282
                          Question

•         Which of the following molecules is polar?
           –       XeF4
           –       XeF2
           –       BF3
           –       NF3




Chapter 8 | Slide 283
                        Answer
•d) NF3

•Section 8.3, Bond Polarity and Dipole Moments; Section
8.13, Molecular Structure: The VSEPR Model

•A polar molecule results from a species with a nonzero
dipole moment. The square planar (XeF4), linear (XeF2), and
trigonal planar (BF3) geometries have zero dipole moments.
The trigonal pyramidal geometry (NF3) has a nonzero dipole
moment.



Chapter 8 | Slide 284
                                   Question

•Which of the following molecules does not
have a dipole moment?
             –          H2S
             –          H2O
             –          H2Xe
             –          All of the above have a dipole moment.
             –          None of the above has a dipole moment.



Chapter 8 | Slide 285
                        Answer
•c) H2Xe

•Section 8.3, Bond Polarity and Dipole Moments;
Section 8.13, Molecular Structure: The VSEPR
Model

•Dipole moments result from differences in
electronegativity and geometry. Both H2S and H2O
are bent molecules, whereas H2Xe is linear.




Chapter 8 | Slide 286
                        Question

• Which of the following molecules has a
  dipole moment?
        – CF4
        – SF4
        – XeF4
        – All of the above have a dipole moment.
        – None of the above has a dipole moment.



Chapter 8 | Slide 287
                        Answer
•b) SF4

•Section 8.3, Bond Polarity and Dipole Moments; Section
8.13, Molecular Structure: The VSEPR Model

•Dipole moments result from differences in electronegativity
and geometry. Both the tetrahedral (CF4) and square planar
(XeF4) geometries result in dipole moments of zero. The
distorted tetrahedral geometry of SF4 produces a dipole
moment.



Chapter 8 | Slide 288
                                Question

•Which of the following molecules has a
dipole moment?
             –          BCl3
             –          SiCl4
             –          PCl3
             –          Cl2




Chapter 8 | Slide 289
                          Answer
•c) PCl3

•Section 8.3, Bond Polarity and Dipole Moments; Section
8.13, Molecular Structure: The VSEPR Model

•The trigonal planar (BCl3), tetrahedral (SiCl4), and linear
(Cl2) geometries result in dipole moments of zero. The
trigonal pyramidal geometry (PCl3) results in a nonzero
dipole moment.




Chapter 8 | Slide 290
                                    Question

•Which of the following statements about the
species N2, CO, CN–, and NO+ is false?
             –          All are isoelectronic.
             –          Each contains a triple bond.
             –          All are linear.
             –          The bond in each species is polar.




Chapter 8 | Slide 291
                        Answer
•d) The bond in each species is polar.

•Section 8.3, Bond Polarity and Dipole Moments; Section
8.13, Molecular Structure: The VSEPR Model

•The species N2, CO, CN–, and NO+ all have 10 valence
electrons, which requires using a triple bond. Because they
all involve only two atoms, these species are linear.
Because the atoms are the same in N2, it does not have
polar bond and is not a polar molecule.



Chapter 8 | Slide 292
                        Question

• What is the approximate measure of the
  bond angles about the carbon atom in the
  formaldehyde molecule, H2CO?
        – 120°
        – 60°
        – 109°
        – 180°
        – 90°


Chapter 8 | Slide 293
                        Answer

•a)             120°

•Section 8.13, Molecular Structure: The
VSEPR Model

•According to the VSEPR model, C in
formaldehyde has three electron pairs, which
are oriented 120° from one another.

Chapter 8 | Slide 294
                            Question

• What type of structure does the XeOF2
  molecule have?
        – Pyramidal
        – Tetrahedral
        – T-shaped
        – Trigonal planar
        – Octahedral



Chapter 8 | Slide 295
                        Answer
•c) T-shaped

•Section 8.13, Molecular Structure: The VSEPR
Model

•According to the VSEPR model, Xe has five
electron pairs with three bonding pairs, which
means its geometry is T-shaped.



Chapter 8 | Slide 296
                        Question
    • Which of the following statements best describes BF3
      and NF3? (Note: Geometry refers to the electron pair
      arrangement, and shape refers to the atom
      arrangement.)
       – They have variable geometries and shapes due to
         their potential resonance structures.
       – They have the same geometry and different shapes.
       – They have the same geometry and the same shape.
       – They have different geometries and the same
         shape.
       – They have different geometries and different
         shapes.



Chapter 8 | Slide 297
                        Answer
•e) They have different geometries and different
shapes.

•Section 8.13, Molecular Structure: The VSEPR
Model

•BF3 has a trigonal planar geometry and a trigonal planar
shape. NF3, owing to the lone pair on the nitrogen, has a
tetrahedral geometry and a trigonal pyramidal shape.



Chapter 8 | Slide 298

								
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