CH 301 Chapter 3 notes part 2

CH 301 Chapter 3 notes part 2 - Beryllium Dichloride BeCl2...

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Beryllium Dichloride BeCl 2 Valence e = 2 + (2x7) = 16 RHED = 2 (2 individual Be-Cl bonds) Electronic geometry is LINEAR. No lone pairs on Be so molecular geometry is LINEAR Be Cl Cl 180º δ + δ - δ - POLAR covalent bonds; NON POLAR molecule 3p [He] 2s Be Cl [Ne] 3s Need 2 unpaired e for Be. . How about this? NO- Use this. . See over. . Cl already has one unpaired e in a p orbital, ready to bond 2p 2p 2p [He] 2s Be [He] sp Be
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Using a 2s and a 2p orbital for Be and the 3p orbital for each Cl, we would get: 3p 3p Cl Cl Be 1s 2s 2p Not very efficient, and very uneven. Instead, we HYBRIDIZE the single 2s orbital and one of the 2p orbitals, (this is basically mixing two equations to get two new ones). The resulting sp hybridized orbitals are degenerate (of the same energy) and look like this: The two sp orbitals are arranged in space 180º apart, and so we can now build our symmetrical BeCl 2 molecule: 3p Cl 3p Cl sp sp Be 1s
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Boron Trichloride, BCl 3 Valence e = 3 + (3x7) = 24 RHED = 3 (3 individual B-Cl bonds) Electronic geometry is TRIGONAL PLANAR. No lone pairs on Be so molecular geometry is TRIGONAL PLANAR B Cl Cl Cl POLAR covalent bonds; NON POLAR molecule [He] 2s 3p Need 3 unpaired e for B. . B Cl [Ne] 3s Cl already has one unpaired e in a p orbital, ready to bond 2p review by yourself
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3p Each individual sp 2 orbital looks like this - they are aligned 120º to each other We create THREE sp 2 hybrid orbitals, using the single 2s and two of the 2p orbitals [He] sp 2 B Cl [Ne] 3s 2p Cl already has one unpaired e in a p orbital, ready to bond We use all three sp 2 orbitals for B, they are half full and ready to overlap with the available 3p orbital for each Cl: 3p Cl 3p 3p Cl Cl B sp 2 sp 2 sp 2 review by yourself
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Methane , CH 4 Valence e = 4 + (4x1) = 8 RHED = 4 (4 individual C-H bonds) Electronic geometry is TETRAHEDRAL. No lone pairs on Be so molecular geometry is TETRAHEDRAL C H H H H CH H H Need 4 unpaired e for C. . H already has one unpaired e in a s orbital, ready to bond 2p H Very slightly POLAR covalent bonds; NON POLAR molecule [He] 2s 1s C H
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We create FOUR sp 3 hybrid orbitals, using the single 2s and three of the 2p orbitals Each individual sp 3 orbital looks like this, they are aligned 109.5º to each other H already has one unpaired e in a p orbital, ready to bond [He] sp 3 H H 1s H H C sp 3 sp 3 sp 3 1s 1s 1s 1s C H sp 3 We use all four sp 3 orbitals for C, all are half full and ready to overlap with the available 1s orbital for each H:
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Ammonia , NH 3 Valence e = 5 + (3x1) = 8 NH RHED = 4 (3 individual N-H bonds, and ONE lone pair) Electronic geometry is TETRAHEDRAL. ONE lone pairs on N so molecular geometry is TRIGONAL PYRAMIDAL H H Need 3 unpaired e for N. . But, we have 4 RHED so we need 4 equivalent orbitals and tetrahedral electronic geometry. . POLAR covalent bonds; due to lone pair, POLAR molecule H already has one unpaired e in a s orbital, ready to bond 2p [He] 2s N H 1s N H H H
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We create FOUR sp 3 hybrid orbitals, using the single 2s and three of the 2p orbitals Each individual sp 3 orbital looks like this and they are aligned 109.5º to each other H already has one unpaired e
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CH 301 Chapter 3 notes part 2 - Beryllium Dichloride BeCl2...

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