CH14 Notes Part 2

CH14 Notes Part 2 - Chlorine Fluoride ClF Valence e = 2x7 =...

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Chlorine Fluoride ClF Valence e = 2x7 = 14 F Cl No central atom, so linear δ + δ - POLAR covalent bond; POLAR molecule 3p Cl [Ne] 3s Each Cl has one unpaired e [He] 2s F 2p Each F has one unpaired e Each Atom has 3 RHED. Each Atom has one unpaired e BUT: These p orbitals do not give the correct geometry so we HYBRIDIZE: [Ne] sp 3 Cl lone pairs!! [He] sp 3 F lone pairs!! Cl sp 3 sp 3 sp 3 sp 3 F sp 3 Chlorine Fluoride ClF Boron Trichloride, BCl 3 Valence e = 3 + (3x7) = 24 B Cl Cl Cl RHED = 3 Electronic geometry = TRIGONAL PLANAR. No lone pairs on B Molecular geometry = TRIGONAL PLANAR POLAR covalent bonds; NON POLAR molecule 3p Cl [Ne] 3s Cl has 4 RHED, the p orbital has the wrong geometry [He] 2s B B has 3RHED, one unpaired e and wrong geometry 2p B sp 2 sp 2 sp 2 [He] sp 2 B 2p Use all three half full B sp 2 orbitals to overlap with half full Cl sp 3 orbitals: [Ne] sp 3 Cl lone pairs!! Cl sp 3 sp 3 sp 3 sp 3 Cl sp 3 sp 3 sp 3 sp 3 Cl sp 3 sp 3 sp 3 sp 3
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Methane , CH 4 Valence e = 4 + (4x1) = 8 C H H H H RHED = 4 Electronic geometry = TETRAHEDRAL. No lone pairs on C Molecular geometry = TETRAHEDRAL Very slightly POLAR covalent bonds; NON POLAR molecule [He] 2s C H 1s Need 4 unpaired e for C. . Each H has one unpaired e 2p C H H H H [He] sp 3 C H 1s Create FOUR sp 3 hybrid orbitals for C Each H has one unpaired e Use all four half full C sp 3 orbitals to overlap with half full H 1s orbitals: H H 1s H H C sp 3 sp 3 sp 3 1s 1s 1s sp 3 Ammonia , NH 3 Valence e = 5 + (3x1) = 8 N H H H RHED = 4 Electronic geometry = TETRAHEDRAL. ONE lone pair on N Molecular geometry = TRIGONAL PYRAMIDAL POLAR covalent bonds; due to lone pair, POLAR molecule H 1s Each H has one unpaired e [He] 2s N Have 3 unpaired e for N. .
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This note was uploaded on 11/09/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.

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CH14 Notes Part 2 - Chlorine Fluoride ClF Valence e = 2x7 =...

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