Coverage of Chapter 991 All92 All93 All94 All95 Omit Hybridization Involving d Orbitals96 All97 and 98 Omit ALLnnThe shape of a molecule plays an important role in its reactivityBy knowing ID: 897107
Download Pdf The PPT/PDF document "Chapter 9Molecular Geometriesand Bonding..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
1 Chapter 9Molecular Geometriesand Bonding
Chapter 9Molecular Geometriesand Bonding Theories Coverage of Chapter 99.1 All9.2 All9.3 All9.4 All9.5 Omit H
2 ybridization Involving d Orbitals9.6 A
ybridization Involving d Orbitals9.6 All9.7 and 9.8 Omit ALL \n \nThe shape of a molecule plays an im
3 portant role in its reactivity.By knowi
portant role in its reactivity.By knowing the number of bonding and nonbonding electron pairs we can predict the shap
4 e of the molecule. Two (2) Theories for
e of the molecule. Two (2) Theories for \r1.Valence hell lectron air epulsion\n \n \n
5 \n THEORY2.Thealence ond(VB)THEOR
\n THEORY2.Thealence ond(VB)THEORY Lewis StructuresFormal ChargeFormal charge is a charge assigned to each atom
6 in a Lewis structure that helps to dist
in a Lewis structure that helps to distinguish among competing structures. What is the correct formula forHypo Chloro
7 usAcid HClO(aq)H ClOorH O Cl Hypo c
usAcid HClO(aq)H ClOorH O Cl Hypo chlorite ionClONumber of Valence eNumber of Nonbonding e½Number of Bonding eFor
8 mal Charge| O Cl 6 7-6 -6 -
mal Charge| O Cl 6 7-6 -6 -1 -1 -1 0 Where does the H go on HClO?|O Cl H or H O Cl V
9 alence eNonbonding ½Bonding eO Cl O Cl
alence eNonbonding ½Bonding eO Cl O Cl 6 7 6 7-6 -4 -4-6 -1 -2 -2
10 -1 -1 -10 0 Electron p
-1 -1 -10 0 Electron pairs are referedto as electrondomainsSingle, double or triple bonds all count
11 as one electron domain. The atom A in t
as one electron domain. The atom A in this molecule, has four electron domains. The First MOLECULAR GEOMETRYtheoryVS
12 EPRalence hell lectron air epulsion theo
EPRalence hell lectron air epulsion theory VSEPR VSEPR Theory Theory1.To predict molecular shape, assume the valence e
13 lectrons repel each other2.The electron
lectrons repel each other2.The electrons adopt an arrangement in space to minimize erepulsion3.The molecule adopts
14 whichever 3D geometry minimized this re
whichever 3D geometry minimized this repulsion. What Determines the Shape of a Molecule?Four electron domains on N3 b
15 onding and 1 nonbonding What Determines
onding and 1 nonbonding What Determines the Shape of a Molecule?Electrons, whether they be bonding or non-bonding, rep
16 el each other. So electrons are placed a
el each other. So electrons are placed as far as possible from each other Two (2) Different Typesof Molecules1.Molec
17 ules with NO nonBondingelectrons on the
ules with NO nonBondingelectrons on the central atom2.Molecules with nonBondingelectrons on the central atom Electron
18 Domains & NonBondingElectronsExample
Domains & NonBondingElectronsExample 1 CO | O = C = O | How many electron domains on CHow many NonBondingele
19 ctrons on C Electron Domains & NonBo
ctrons on C Electron Domains & NonBondingElectronsExample 2 H..H O HHow many electron domains on OHow many N
20 onBondingelectrons on O There are five f
onBondingelectrons on O There are five fundamental geometries :1.Linear2.TrigonalPlanar3.Tetrahedral4.Trigonalbepyrami
21 dal5.OctahedralMolecular Geometries for
dal5.OctahedralMolecular Geometries for molecules with no nonbonding electrons on central atom Only consider Three in
22 detail1.Linear2.TrigonalPlanar3.Tetrahed
detail1.Linear2.TrigonalPlanar3.Tetrahedral In order to determine geometryFirst Draw Lewis Dot Formula MOLECULES IN
23 WHICH THE CENTRAL ATOM HAS NO LONE PAIR
WHICH THE CENTRAL ATOM HAS NO LONE PAIRS ZINC CHLORIDEZn ClZn (30) [Ar] 3d10 4sClZn -ClB A B AB2 = LINEAR
24 ABMolecules Such as COare Linear (Mole
ABMolecules Such as COare Linear (Molecules With NO UnPairedElectrons On the Central Atom). . . .: O = C
25 = O : B A B Molecular Shape and Molec
= O : B A B Molecular Shape and Molecular Shape and Molecular Polarity Molecular Polarity ABMolecules Such as BFar
26 e Planar (Molecules With NO UnPairedE
e Planar (Molecules With NO UnPairedElectrons On the Central Atom)Formula B FNumber of Va
27 lence e3 21 = 24 totalLewis Structur
lence e3 21 = 24 totalLewis Structure F B F F AB(Molecules With NO UnPairedElectrons On the Central
28 Atom) Such as BFare Planar ABMolecules
Atom) Such as BFare Planar ABMolecules Such as CHare Tetrahedral(Molecules With NO UnPairedElectrons On the Central A
29 tom)Formula C
tom)Formula C HNumber of Valence e4 4 = 8 totalLewis Structure H C H H AB
30 Such as CHare Tetrahedral(Molecules With
Such as CHare Tetrahedral(Molecules With NO UnPairedElectrons On the Central Atom) ABMolecules Such as CClare Tetrahed
31 ral Carbon Carbon TetraChloride TetraC
ral Carbon Carbon TetraChloride TetraChloride ABSuch as PClare Triangular bipyramidalName ?Number of Bonds ?Lewis
32 dot structure ?ClClP ClClCl ABSuch as
dot structure ?ClClP ClClCl ABSuch as SFare OctahedralName ?Number of Bonds ?Lewis dot structure ? Molecules Wi
33 th NO UNPairede Molecules with NO unpai
th NO UNPairede Molecules with NO unpaired eon Central Atom1.2 Bonds ABor AX2 e.g. CO2.3 Bonds ... ABor A
34 X3 e.g. BF3.4 Bonds ABor AX4 e.
X3 e.g. BF3.4 Bonds ABor AX4 e.g. CH4.5 Bonds ... ABor AX5 e.g. PCl5.6 Bonds ... ABor AX6 e.g. SF
35 Polarity Part 2. of VSEPR Theory Part 2
Polarity Part 2. of VSEPR Theory Part 2. of VSEPR TheoryCENTRAL ATOM HAS LONE PAIRS Molecules With UnPai
36 redElectrons On the Central AtomClass Ex
redElectrons On the Central AtomClass Example ABE SO& OAB2 2 ABE NHGeometry BentBent
37 Trigonalpyramidal 1. AB 2 E
Trigonalpyramidal 1. AB 2 E 2. AB 2 E 2 OOO OOO AB(Molecules With UnPairedEl
38 ectrons On the Central Atom) Such as HO
ectrons On the Central Atom) Such as HO are Bent 3. ABE ABE (Molecules With UnPairedElectrons On the Cen
39 tral Atom) Such as NHare NOT Planar Pred
tral Atom) Such as NHare NOT Planar Predict Molecular Shapes1.SiCl2.CHCl3.GeCl4.OF5.NH6.PH____________________________
40 _____________________________ Give the e
_____________________________ Give the electron domain and molecular geometries for(a) N(b) SO(c) PCl(d) NHCl electron
41 domain molecular geometry ____________
domain molecular geometry ____________ ____________________________ ___________________________
42 ___________________________ ___
___________________________ ________________ Examples of ABmoleculesLinear ABHow many bondsCOBent ABE H
43 ow many bondsSOand NOBent AB2 How ma
ow many bondsSOand NOBent AB2 How many bonds Examples of ABmoleculesPlanar ABHow many bondsBFPyramidal ABE
44 How many bondsNHT shape AB2 How man
How many bondsNHT shape AB2 How many bondsClF Two (2) Theories for \r1.Valence hell lectron a
45 ir epulsion\n \n \n \n
ir epulsion\n \n \n \n THEORYNow consider 2.Thealence ond(VB)THEORY Methoduses molecular orbita
46 lsnot Atomic Orbitals \n
lsnot Atomic Orbitals \n \rOrbitalsused in bonding of Molecules CHas an EXAMPLEGround Sta
47 te Electron ConfigurationC (6 e) 1s2s2
te Electron ConfigurationC (6 e) 1s2s2p2 = ( ¯) (¯) () () ( ) Only place for two bonds to form Therefo
48 re would predict CHformation and not
re would predict CHformation and not CHBut CHdoes not exist while CHdoes C (6 e) 1s2s2p2 = ( ¯) (¯) () () ( )
49 Only place for two bonds to form
Only place for two bonds to form Excited State Electron ConfigurationC (6 e) 1s2s2p3 = (¯) () () () (Now
50 a place for four bondsOne electron from
a place for four bondsOne electron from H goes into an s orbitaland Three from H go into the p orbitals The \n in
51 are the\nOne electron in an s or
are the\nOne electron in an s orbital and Three in p orbitalswould create different bonds. Since All the Bonds
52 are Equal, this cannot be correct
are Equal, this cannot be correct HybridizationIn order to made All Bonding
53 sites equal, we must create NEW Or
sites equal, we must create NEW Orbitals.s, p, d, f are ATOMIC ORBITALSMOLECULAR ORBITALS are formed from A
54 tomic orbitals VALENCE SHELL ORBITAL
tomic orbitals VALENCE SHELL ORBITALS HYBRIDIZE THE ORIENTATION OF ALL HYBRID VALENCE SHELL ORBITALS DE
55 TERMINES THE GEOMETRY OF THE MOLECULE ar
TERMINES THE GEOMETRY OF THE MOLECULE are formed from \nAtomic Orbitals one S + one Pone S + two Pone S
56 + three PMolecular Orbitals Two (2) SPTh
+ three PMolecular Orbitals Two (2) SPThree (3) SPFour (4) SP They are calledSP SPSPSPd and SP spHYBRIDIZATION TETRA
57 HEDRAL Bond Angles 109½Methane CH4
HEDRAL Bond Angles 109½Methane CH4 Four Bonds on C sp2 HYBRIDIZATION spHYBRIDIZATIONone S orbital + one P orbita
58 l Carbon is NOT The Only Element That Un
l Carbon is NOT The Only Element That Undergoes spHYBRIDIZATION In CHCOOH, there are three (3) hybridized atoms. Ge