CHEM131_Lecture_4-03-14

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    Lecture April 3

    1

    HW Chap. 10 - Due April 8Quiz Chap. 10 - Due April 10

    More Valence Bond TheoryLook and the figures in the text!

    Summary Table 10.3

    Next Exam April 10:acid-base reactions, oxidation-reduction (Chap. 4)

    Chap. 9 and 10

    MO Theory Sec. 10.8

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    sp2Hybrids

    2

    CH2CO formaldehyde

    O

    CH

    H

    Determining thegeometry - countthe #!bonds and

    the #lone pairs

    about the atom

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    Review Multiple Bonds

    3

    CO2

    linear

    sp hybrid

    CO32-

    carbonate ion

    trigonal planarsp2hybrid

    2 sigma bonds

    3 sigma bonds1 "bond

    2 "bonds

    leaves 2 porbitals free

    central C is sp hybridized

    central C is sp2hybridized

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    More Multiple Bonds

    4

    CH2Cl-CH2Cl CHCl=CHCl

    OR

    C-C rotates freely rigid C=Ctwo distinct structures

    cis trans

    sp3 sp2

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    C6H12 vs C6H6

    5

    cyclohexane

    sp3

    flexible - boat andchair geometries

    sp2

    rigid planar geometry

    benzene

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    Resonance - Benzene

    6

    The !bonds aredelocalized

    O3ozoneO=O-O

    O-O=O

    2 !and 1 "See p. 445

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    Molecular Orbitals

    7

    We will start with the individual orbital shapesLinear Combination of Atomic Orbitals (LCAO)

    bonding - energy islower than the

    individual orbitals

    but have to end upwith the same energy

    you started with sosomething has to also

    be higher

    sum oforbitals

    di#erence

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    Complete Energy Diagram

    8

    !1s

    !*1s

    Figure p.435

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    9

    Creating MO Bonds

    1) Create the MO energy level diagram (given)2) Fill in the number of valence electrons -

    starting with the lowest energy and still

    filling UNPAIREDfirst3) Find the BOND ORDER=

    1/2 (# bonding electrons-#antibonding electrons)

    From H2find BO = 1/2(2) = 1

    A BO = 0 means that no energetic advantageto being bound

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    Another Example

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    BO = 1/2 (2-1) = 1/2So conclude He2+is bound and exists

    But He2does NOT BO=0

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    The p Orbitals

    11

    px orbital in the!

    orientation

    pzorbital in the"orientation

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    MO Diagrams with p orbitals

    1210

    "2p

    !2p

    "*2p

    "2p

    !2p

    orderinverted

    !*2p !*2p

    "*2p

    order NOTinverted

    B2, C2, N2O2, F2, Ne2

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    1311

    Paramagnetic

    unsharedelectrons

    O2

    $% %$

    $%

    $%

    $% $ $ $%$ $

    %

    %%

    % %

    $

    $

    $

    BO=1/2(6-2) = 2

    Example with ps

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    Summary of 2s-2p MOs

    14

    Bondenergy

    increaseswith BO

    Bondlength

    decreases

    with BO

    1 2 3 2 1 0

    Figure 10.15

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    Quiz of the Day

    15

    Give the number of !and "bonds in thefollowing molecule. Give your answer with ,

    between and NO spaces.

    O

    C

    H

    O H

    H

    H

    CCH3-COOH

    acetic acid

    Answer: 7,1