chem notes 11-30

chem notes 11-30 - -hybridized sp^2 sp^2 sp^2-shape...

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November 30, 2007 Dr. Stevens Gen. Chem. 1 Mon: homework 10.1 Wed: job apps Fri: unit 4 exam; 2:30 x-mas party Dipole Moments & Geometry -3 polar B-H bonds in BH3 -molecule is non-polar overall -3 polar N-H bonds in H3 -bonds do not cancel out => polar molecule -dipole: -+---------> pos. neg. Valence Bond Theory (VBT) -theory used to explain observed bonding -attempts to explain geometry around central atom -B => [He] 2s2 2p1 -shape of orbitals does NOT explain observations -Note: s (& p) orbitals are not equal/120 degrees apart -duration bond formation -orbitals of B must hybridize so they become equal -hybridized pulls s and p together and are uniformly distributed (one per box) -notation: 1s + 2p = 1 part s, 2 parts p = 1/3s + 2/3p => s^1 p^2 = sp^2 -reflects input orbital composition -3 orbitals => each exactly the same -unhybridized: s + p + p
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Unformatted text preview: -hybridized: sp^2 + sp^2 + sp^2-shape affected by partial input-character circular ‘s’ orbital + 2 infinite ‘p’ orbitals-big lobe and small lobe make morphed ‘8’-1 e- per each sp^2 orbital to share with each H e--each B-H bond is overlap of B hybridization sp^2 orbital with H s orbital-all initial bonds formed from unhybridized + unhybridized or hybridized + unhybridized are called sigma-subsequent bonds are all called pi-B-H are sigma bonds-H-C= - C – C –H(3)-H-C bonds are all sigma, C-C is sigma, middle C= -C bond is sigma, outer 2 = pi-6 sigma bonds, 2 pi bonds-n=2: sp-n=3: sp^2-n=4: sp^3-n=5: sp^3d-n=6: sp^3d^2-C2H4:-n=3-Chyb: sp^2-Carbons have sp^2 + sp^2 + sp^2 + p-p orbitals make pi bond-pi bonds result from side-to-side interaction of UNhybridized p orbitals adjacent to one another...
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This note was uploaded on 04/23/2008 for the course CH 151 taught by Professor Stevens during the Fall '08 term at Whitworth University.

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chem notes 11-30 - -hybridized sp^2 sp^2 sp^2-shape...

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