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CH. 9 MOLECULAR SHAPE Bonding patterns Molecular shape VSEPR, shape, angle Polarity - Dipole Bond...

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  • CH. 9 MOLECULAR SHAPEBonding patterns

    Molecular shape

    VSEPR, shape, anglePolarity - Dipole

    Bond order Effects of repulsion on bond : lone pair - l.p. > l.p. - bonding pair > b.p - b.p.

    Look at: Lewis structure, resonance, formal charge, radicals

  • Equationsformal charge (f.c.) = # val. e- - (# unbond e- + 0.5 # bond e-)

  • FORMAL CHARGEf.c. = # val. e- - (# unshare e- + 0.5 # share e-)What does the atom own???* all unbonded e- pairs* 0.5 of bonding e-Have 2+ possible arrangements, which structure imprt??* smaller f.c. to large* f.c. not side-by-side* more -f.c. on more -EN atom

  • Lets look at O3Oa: 6 val e- 4 unbonded 4 bonded b

    a g 6 - [4 + .5(4)] 6 - (4 + 2) 6 - 6 = 0Ob: 6 val e- 2 unbonded 6 bonded 6 - [2 + .5(6)] 6 - (2 + 3) 6 - 5 = +1Og: 6 val e- 6 unbonded 2 bonded 6 - [6 + .5(2)] 6 - (6 + 1) 6 - 7 = -1 +1 -1 0

  • LEWIS DIAGRAMS ^ not show shapeLook at: 1 central atom 2+ central atoms multiple bonds resonanceHelps in understanding bonding in cmpds/molecules

  • BONDING PATTERNS/REQUIREMENTSHydrogen (1) Nitrogen (3) Oxygen (2) Halogens (1) carbon (4) X = F, Cl, Br, INitrogen Ion, N+1, (4)P : 3S : 3

  • LEWIS STRUCTURESQuick Overview1. Sum the total number of valence electrons from all atoms in subst.

    2. Identify central atom. Show which atoms are bonded to each other

    3. Place 2 e-s to show a bond between each atom

    4. Complete the octet rule for each atom

    5. If not enough e-s to give central atom octet, try multiple bonds

  • LEWIS STRUCTURES1 central atom, single bondsDetermine central atom * forms 2+ bonds, octet rule * lower group #, lower EN N - O N - C P - Cl N C P * from same group, the higher period # N - P S - O Br - F P S Br# Val. e- * add total # e- of all atoms C O S 4 6 6* add 1 e- for each - charge C-4 O-2 Cl-1 8 8 8

    * substr 1 e- for each + charge C+4 N+1 2 4 Halogens F, Cl, Br, IF never central atom

  • Bonding central atom * place single bond to each atom bonded to central atom * substr 2 e- for each single bond from total val e- count Remaining e- * place pairs of e- to complete octet rule to each attached atom * any remaining e- place pairs around central atomCBrFI2bond formed: 4 1 1 1 # val e-: 4 7 7 (2*7) = 32 e-central atom: Cattached: 1 Br 1 F 2 I

  • Place 2 e-s to show bonds32 - 8 = 24 e-s left to account for- 8 = 4Final step, replace bonding pairs with lineto represent the bond(s) betweenNo e-s left, no multiple bonds24 - 12 = 12- 4 = 0PCl3 cen.atom P attach 3 Ps 26 val e-26 - 6 = 20..- 18 = 2- 2 = 0

  • NHI2bonds 3 1 1OF2 H2Sval e- 5 1 2*7 = 2020 - 6 = 14- 12 = 2..- 2 = 0 ..H--S--H .. .. .. ..:F--O--F : .. .. ..

  • LEWIS STRUCTURES2 central atoms, single bondsDetermine central atomsonly C & O can have multiple bondsH only 1 bondCH3OHO..H..# Val. e- 4 + 3 + 6 + 1 = 14 e-O....H

  • LEWIS STRUCTURES2 central atoms, single bondsDetermine central atomsonly N & O can have multiple bondsH only 1 bondNH3OO..H..# Val. e- 5 + 3 + 6 = 14 e-

  • LEWIS STRUCTURES2+ central atoms, OH bondDetermine central atomsonly C & O can have multiple bondsH only 1 bondC2H6O# Val. e- 8 + 6 + 6 = 20 e-20 - 16 = 4bonding e-4 - 4 = 0nonbonding e-w/o OH bond

  • HCNbonds 1 4 3val e- 1 4 5 = 10....LEWIS STRUCTURESmultiple bonds

    H--C--N :

    ----:EXCEPTIONcentral atom not octet, move l.p. to b.p. to central atom

  • bonds 4 2CO2 val e- 4 2*6 = 1616 - 4 = 12- 12 = 0

  • RESONANCEdbl bonds next to single bondse- pair vibrate back-forth bet atoms,fills octet ruleOZONE, O3 O=O=OToo many O bondsSoooooo

  • results from e--pair delocalizationCan show as bond order = 3/2 = 1.5

  • BENZENE -- C6H6or Bond order = 9/6 = 1.5

  • POLYATOMIC IONSShow as [ ]chargeNO3-1 5 3*6= 5 + 18= 23 + 1 = 24

    4/3 = 1.333Bond order = ???

  • EXCEPTIONSdeficient atoms Be B 2, 4 3* e--deficient atoms* odd # e- atoms* expanded val shellsOdd # free radicals contain unpair e- paramagneticNO2More imprt due to way reacts, as freeradicals react w/ each other to pair e-

  • EXPANDED SF6 PCl5Bond order: 6/4 = 1.5Exothermicuse empty d orbitalsperiod 3+ (nonmetals) P S I 5 6 7H2SO4-2H ----> SO4-2More imprt, observedbond lengths

  • P: 5 - [2 + .5(6)] 5 - (2 + 3) = 0

    O: 6 - [4 + .5(4)] 6 - (4 + 2) = 0

    Cl: 7 - [6 + .5(2)] 7 - (6 + 1) = 0POCl3 ClO2Draw most likely structue for:*32 e- e- P: 5 - [0 + .5(10)] 5 - (0 + 5) = 0

    O: 7 - [6 + .5(2)] 7 - (6 + 1) = 0

    Cl: 6 - [4 + .5(4)] 6 - (4 + 2) = 0

  • Cl: 7 - [3 + .5(8)] 7 - (3 + 4) = 0

    O: 6 - [4 + .5(4)] 6 - (4 + 2) = 0ClO2 . . . . .: O--Cl-- O : . . . . . .

  • VSEPR pg334 - 43Valence Shell Electron Pair RepulsionLewis: 2-D, shows relative placement of atoms, the building plans, not shapeVSEPR: molecular shape; minimizes e- repulsion, val e- around central atom will locate as far away as possible from other e- to minimize repulsionsassigned designation AXmEncentral atomsurrounding atomm indicates # ofnonbonding e-n indicates # of

  • 2 0 Linear AX2 1800

    3 0 Trigonal Planar AX3 12002 1 Angular (Bent or V) AX2E < 1200

    4 0 Tetrahedral AX4 109.503 1 Pyramidal AX3E 10702 2 Bent or V AX2E2 1050

    5 0 Trigonal Bipyramidal AX5 900, 12004 1 Seesaw AX4E 1730, 10203 2 T-shaped AX3E2 87.502 3 Linear AX2E3 1800

    6 0 Octahedral AX6 9005 1 Square Pyramidal AX5E 8504 2 Square Planar AX4E2 900 GEOMETRY AROUND CENTRAL ATOM pg 337 Linear

    Trigonal Planar (Planar Triangular)

    Tetrahedral

    Trigonal Bipyramidal

    Octahedral

  • Draw Lewis structure for phosphorus trichloride, PCl3 Total valence e-s P: 5 Cl: 3*7 =21 total=26Central atom: P Attached: 3 ClsGEOMETRYAX3E Pyramidal 4 possible bonding sites, tetrahedralin this case; 3 bonds, 1 lone pair3-D DRAWING

  • Use VSEPR to Predict LewisStructureMolecular Shape assigne- groupBondAnglePolarity: take shape into acct; polar bonds present but counterbalanced results in NP (no dipole moment, m)

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