CH%20301%20Chapter%203%20notes%20part2%20example%20worksheets_answers

CH%20301%20Chapter%203%20notes%20part2%20example%20worksheets_answers

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 Cl F Other orbitals not shown!! σ -bond Oxygen, O 2 Valence e = 2x6 = 12 O O No central atom, so linear NON POLAR covalent bond; NON POLAR molecule 3p O [Ne] 3s Each O has two unpaired e O O σ -bond Other orbitals not shown!! π -bond Beryllium Dichloride BeCl 2 Valence e = 2 + (2x7) = 16 Be Cl Cl RHED = 2 Electronic geometry = LINEAR. No lone pairs on Be Molecular geometry = LINEAR δ - δ + δ - POLAR covalent bonds; NON POLAR molecule [He] 2s Be Possible ? 2p 3p Cl [Ne] 3s Each Cl has one unpaired e [He] 2s Be Need 2 unpaired e for Be. . 2p [He] sp Be NO. . Use this. . 2p Using a 2s and a 2p orbital for Be and the 3p orbital for each Cl, we would get: 3p Cl 2p Not very symmetrical! HYBRIDIZE the 2s and one of the 2p orbitals to get TWO sp hybrid orbitals 3p Cl 3p Cl The two sp orbitals are arranged in space 180º apart, forming a symmetrical BeCl 2 molecule: sp sp Be 1s σ -bond σ -bond BeCl 2 : example of hybridization: Be 1s 2s
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

CH%20301%20Chapter%203%20notes%20part2%20example%20worksheets_answers

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online