1 Tim Berners-Lee Université McGill VISUEL VISUEL CLIPS SONORES CLIPS SONORES FORMAT DES FORMAT DES PRÉSENTATIONS PRÉSENTATIONS DÉMONSTRATIONS DÉMONSTRATIONS EXEMPLES PERTINENTS EXEMPLES PERTINENTS SUPPORTS PÉDAGOGIQUES SUPPORTS PÉDAGOGIQUES VISUEL VISUEL CLIPS SONORES CLIPS SONORES FORMAT DES FORMAT DES PRÉSENTATIONS PRÉSENTATIONS DÉMONSTRATIONS DÉMONSTRATIONS EXEMPLES PERTINENTS EXEMPLES PERTINENTS SUPPORTS PÉDAGOGIQUES SUPPORTS PÉDAGOGIQUES 200 DIAPOSITIVES 200 DIAPOSITIVES COURS DE 75 MINUTES COURS DE 75 MINUTES VISUEL VISUEL CLIPS SONORES CLIPS SONORES FORMAT DES FORMAT DES PRÉSENTATIONS PRÉSENTATIONS DÉMONSTRATIONS DÉMONSTRATIONS EXEMPLES PERTINENTS EXEMPLES PERTINENTS SUPPORTS PÉDAGOGIQUES SUPPORTS PÉDAGOGIQUES
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1
Tim Berners-Lee
Université McGill
VISUELVISUEL
CLIPS SONORESCLIPS SONORES
FORMAT DES FORMAT DES PRÉSENTATIONS PRÉSENTATIONS
DÉMONSTRATIONSDÉMONSTRATIONS
EXEMPLES PERTINENTSEXEMPLES PERTINENTS
SUPPORTS PÉDAGOGIQUESSUPPORTS PÉDAGOGIQUES
VISUELVISUEL
CLIPS SONORESCLIPS SONORES
FORMAT DES FORMAT DES PRÉSENTATIONS PRÉSENTATIONS
DÉMONSTRATIONSDÉMONSTRATIONS
EXEMPLES PERTINENTSEXEMPLES PERTINENTS
SUPPORTS PÉDAGOGIQUESSUPPORTS PÉDAGOGIQUES
200 DIAPOSITIVES200 DIAPOSITIVESCOURS DE 75 MINUTESCOURS DE 75 MINUTES
VISUELVISUEL
CLIPS SONORESCLIPS SONORES
FORMAT DES FORMAT DES PRÉSENTATIONS PRÉSENTATIONS
DÉMONSTRATIONSDÉMONSTRATIONS
EXEMPLES PERTINENTSEXEMPLES PERTINENTS
SUPPORTS PÉDAGOGIQUESSUPPORTS PÉDAGOGIQUES
2
ΔE = Δm x c2
1905
VISUELVISUEL
CLIPS SONORESCLIPS SONORES
FORMAT DES FORMAT DES PRÉSENTATIONS PRÉSENTATIONS
DÉMONSTRATIONSDÉMONSTRATIONS
EXEMPLES PERTINENTSEXEMPLES PERTINENTS
SUPPORTS PÉDAGOGIQUESSUPPORTS PÉDAGOGIQUES
TRANSITIONSTRANSITIONSÉLECTRONIQUESÉLECTRONIQUES
VISUELVISUEL
CLIPS SONORESCLIPS SONORES
FORMAT DES FORMAT DES PRÉSENTATIONS PRÉSENTATIONS
DÉMONSTRATIONSDÉMONSTRATIONS
EXEMPLES PERTINENTSEXEMPLES PERTINENTS
SUPPORTS PÉDAGOGIQUESSUPPORTS PÉDAGOGIQUES
LE PRINCIPELE PRINCIPEDE LEDE LEDE LEDE LE
CHATELIERCHATELIER
3
Ca5(PO4)3OH
ApatiteApatite
5 Ca+2 + 3 PO4-3 + OH-
Ca5(PO4)3OH
ApatiteApatite
5 Ca+2 + 3 PO4-3 + OH+ OH-- + H+ H++
Ca5(PO4)3OH
ApatiteApatite
5 Ca+2 + 3 PO4-3 + OHOH-- + H+ H++
HH22OO
Ca5(PO4)3OH
ApatiteApatite
HH22OO
5 Ca+2 + 3 PO4-3 + OHOH-- + H+ H++
CHFBrCl
REPRÉSENTATIONS DE CRAM
CHFBrClCHFBrCl
4
CHFBrCl
REPRÉSENTATIONS DE CRAM
STÉRÉOISOMÈRESMolécules où les atomes ont la même connectivité mais différentes organisations spatiales
ISOMÈRES DE CONSTITUTION (STRUCTURE)
Molécules ou les atomes la même formule brute mais différentes connectivités
Molécules ou les atomes la même formule brute mais différentes connectivités
ISOMÈRES DE CONSTITUTION (STRUCTURE)
5
H
CH3
CH3
H
HO
H
H
AndrostenolAndrostenol
VISUELVISUEL
CLIPS SONORESCLIPS SONORES
FORMAT DES FORMAT DES PRÉSENTATIONS PRÉSENTATIONS
DÉMONSTRATIONSDÉMONSTRATIONS
EXEMPLES PERTINENTSEXEMPLES PERTINENTS
SUPPORTS PÉDAGOGIQUESSUPPORTS PÉDAGOGIQUES
General ChemistryPrinciples and Modern Applications
Atoms surround themselves with eight electrons to achieve noble
t bilitgas stability
Al2O3 Al2O3
WRITING LEWIS STRUCTURES
1. Count total # of valence e-
For +ve charge remove e-For +ve charge, remove e
For –ve charge, add e-
10
2. Draw a plausible skeleton
Symmetrical structures arepreferred
Least electronegative element usually in centre
3. Achieve an octet around each atom
Convert lone pairs into double or triple bonds, if necessary
NO3-
N 5 valence e- x 1 = 5O 6 valence e- x 3 = 18
1 (-)+
24 e-
NO3-
NO3- NO3
-
11
NO3-
CH3NCO
H H?..
CN
CH
H
H
O
CN
CH
H
H
OC
NC
H
H
H
O?....
..
......
..
..
Tells us where electrons are located.
Formal Charges
Helps us predict the most plausible structure.
Tells us where electrons are located
Formal Charges
F.C. = # of valence e- – # of l.p. e-
– ½ # of bonding e-
N O
+
....
NO+
F.C. (N) = 5 – 2 – ½ 6= 0
F.C. (O) = 6 – 2 – ½ 6= +1
N O
+
....
NO+
Note: F.C. (N) + F.C. (O) = +1
The sum of the formal charges must be equal to overall charge of the species
12
RESONANCE FORMS
Several Lewis structures can be written but the “true” one is awritten, but the “true” one is a hybrid
RESONANCE FORMS
Resonance forms have the same connectivity but different electronconnectivity but different electron distribution
O3
Determining the "best"Resonance form
1. When writing resonance structures the connectivity cannot be altered (only lone pair electrons and electrons in double and triple multiple bond can be moved).
f
Writing resonance structures
2. A structure with the lowest magnitudes of formal charges is preferred (greater contribution to the hybrid).
3. A structure with a negative charge on the most electronegative atom is preferred.
N2O
13
Consider of the four forms that can be Consider of the four forms that can be written for Nwritten for N2200
What is the relationship between structure What is the relationship between structure 1 and 4 ?1 and 4 ?
Res
onance
form
...
Iden
tical
Structu
ral is
o...
0% 0%0%
:05
1.1. Resonance Resonance formsforms
2.2. IdenticalIdentical3.3. Structural Structural
isomersisomers
STRUCTURAL ISOMERS
Structural isomers have the same molecular formula but differentmolecular formula but different connectivities
What is the relationship between structure What is the relationship between structure 1,2 and 3 ?1,2 and 3 ?
Res
onance
form
...
Iden
tical
Structu
ral is
o...
0% 0%0%
:05
1.1. Resonance Resonance formsforms
2.2. IdenticalIdentical3.3. Structural Structural
isomersisomers
RESONANCE FORMS
Resonance forms have the same connectivity but different electronconnectivity but different electron distribution
N2O
-1-1+1 +10 0
-1 -1+1+1 +2-2
14
Which of the four forms that can be written Which of the four forms that can be written for Nfor N220 is (are) the most plausible0 is (are) the most plausible
1.1. (1)(1)
2.2. (2)(2)
-1 -2 -3 -4
0% 0%0%0%
:05
3.3. (3)(3)
4.4. (4)(4)
N2O
-1-1+1 +10 0
BEST
-1 -1+1+1 +2-2
H H..
Which of the structures Which of the structures best best represent represent methyl methyl isocyanateisocyanate
CN
CH
H
H
O
CN
CH
H
H
OC
NC
H
H
H
O....
..
..
..
..
..
..0
0
000
0
0
0
-1
-1+1
+1
AC
B
Which of the structures Which of the structures best best represent represent methyl methyl isocyanateisocyanate
1.1. AA
2.2. BB
A B C All
0% 0%0%0%
:05..
2.2. BB
3.3. CC
4.4. AllAll
CH3NCO
H H..
CN
CH
H
H
O
CN
CH
H
H
OC
NC
H
H
H
O....
..
..
..
..
..
..0
0
000
0
0
0
-1
-1+1
+1
BEST
Exceptions to the Octet Rule
1. Incomplete Octet
15
F B F
BF3
.... ....
..
..
F B FF
6 e-
.. ....
F B F
BF3
.... ....0
00
..
..F B F
F
6 e-
.. ....0
BETTER!
Exceptions to the Octet Rule
1. Incomplete Octet
2 E d d O t t2. Expanded Octet
SO42-
SO42-
BETTER!
Exceptions to the Octet Rule
1. Incomplete Octet
2 E d d O t t2. Expanded Octet
3. Paramagnetic Species
16
NO NO
Odd # of electrons
GENERAL CHEMISTRYCHEM-110
V.S.E.P.R.
VSEPR Theory
• Molecules adopt the geometry which maximizes the distance between electron pairs around abetween electron pairs around a central atom, and thus minimizes electrostatic repulsions.
VSEPR Theory
Molecular shape depends on:
• The number of electron pairs
• The type of electron pairs
17
VSEPR
Molecules containing only bonding pairs around the central atom
The shape depends only on the number of electron pairs
2 e- pairs
Linear
Formula: AB2
Bond Angle: 180°
2 e- pairs Linear
Example: BeI2
1800
2 e- pairs
Linear3 e- pairs
Trigonal
Formula: AB3
Bond Angle: 120°
3 e- pairs
Trigonal
Example: BF3
1200
2 e- pairs
Linear4 e- pairs
Tetrahedral3 e- pairs
Trigonal
18
4 e- pairs
Tetrahedral
Formula: AB4
Bond Angle: 109.5°Example: CH4 109.50
2 e- pairs
Linear4 e- pairs
Tetrahedral3 e- pairs
Trigonal
5 e- pairs
Trigonal bipyramidal
5 e- pairs
Trigonal bipyramidal
Formula: AB5
Bond Angle: 90°, 120°Example: PF5
1200
900
2 e- pairs
Linear4 e- pairs
Tetrahedral3 e- pairs
Trigonal
5 e- pairs
Trigonal bipyramidal6 e- pairs
Octahedral
6 e- pairs
O t h d l
Formula: AB6
Bond Angle: 90°Example: SF6
Octahedral900
900
VSEPR Examples
Predict the geometry of the following molecules
19
BeCl2
Lewis Structure::
:
:
:
:
: BeCl
Cl
What is the molecular geometry of BeClWhat is the molecular geometry of BeCl22??
electron pairs geometry: trigonal bipyramidalXeF2 22 e-
electron pairs geometry: trigonal bipyramidal
molecular geometry: linear
XeF2
22 e-
electron pairs geometry: trigonal bipyramidal
molecular geometry: linear
XeF2
Molecules and Ions containing Multiple Bondsp
For geometrical considerations, a multiple bond can be treated as if it pwere a single bond
CH3NCO
Recall
25
CH3NCO
Recall e- groups: 4
e- geometry:e geometry:tetrahedral
molecular geometry:
tetrahedral
Angle: ~109.5°
CH3NCO
Recall
CH3NCO
Recalle- groups: 3
CH3NCO
Recalle- groups: 3
Moleculargeometry: bent
Angle: ~120°
CH3NCO
Recall
CH3NCO
Recalle- groups: 2
26
CH3NCO
Recalle- groups: 2
e- geometry:linear
moleculargeometry: linear
Angle: ~180°
CH3NCO
GENERAL CHEMISTRYCHEM-110
VALENCE BOND THEORY
Valence Bond Theory
The covalent bond results from the l f t i bit l t i ioverlap of atomic orbitals containing
one unpaired electron each
Hydrogen H2
1sH
1sH
27
Covalent Bond
• The electrons are no longer confined to a single s orbital around one atomone atom.
• Instead, they can move over both orbitals for the entire molecule.
Covalent Bond
• The electrons are no longer confined to a single s orbital around one atomone atom.
• Instead, they can move over both orbitals for the entire molecule.
GREATER STABILITY!
Valence Bond Theory Examplesp
1sH
Hydrogen H2
1sHs-s overlap
1sH
Hydrogen H2
1sHs-s overlap
1sH
Hydrogen H2
1sHs-s overlap
σ bond
28
F
Fluorine F2
p1s 2s 2p
1s 2s 2pF
p-p overlap
p
p-p head to head overlap
x
σ bond
p-p head to head overlap
Does diatomic helium HeDoes diatomic helium He22 exist?exist?
1.1. YesYes2.2. NoNo33 No IdeaNo Idea
Yes No
No Id
ea
0% 0%0%
:05
3.3. No IdeaNo Idea
Helium
He 1s
He 1sHe 1s
Helium
He 1s
He 1sHe 1s
Filled orbital ∴ no possible overlap
He2
X
29
Multiple Bonds
In OIn O2 2 the atoms are linked athe atoms are linked a ??
1.1. Single bond Single bond 2.2. Double bondDouble bond33 TrripleTrriple bondbond
Single
bond
Double
bond
Trriple
bond
No id
ea
0% 0%0%0%
:05
3.3. TrripleTrriple bond bond 4.4. No ideaNo idea
1 double covalent bond
Oxygen
.O. . . O .. ..
Oxygen
O1s 2s 2p
1s 2s 2pO
x
px px
p-p head to head overlap
x
σ bond
p-p head to head overlap
O O
30
Oxygen
O
1 σ bond
1s 2s 2p
1s 2s 2pO
y y
py py
p-p side to side overlap
y
p-p side to side overlap p-p side to side overlap
x
Overlap above and belowthe bond axis
p-p side to side overlap
π bond
Oxygen
O
1 σ bond
1s 2s 2p
1s 2s 2pO
1 π bond
31
Double Bond
1 σ bond
Maximum electron density along the the bond axis
In NIn N2 2 the atoms are linked athe atoms are linked a ??
1.1. Single bond Single bond 2.2. Double bondDouble bond33 Triple bondTriple bond
Single
bond
Double
bond
Triple
bond
No id
ea
0% 0%0%0%
:05
3.3. Triple bond Triple bond 4.4. No ideaNo idea
1 triple covalent bond
Nitrogen
..N N..
Nitrogen
N1s 2s 2p
1s 2s 2pN
x
σ bond
p-p head to head overlap
N N
Nitrogen
N
1 σ bond
1s 2s 2p
1s 2s 2pN
32
Nitrogen
N
1 σ bond
1s 2s 2p
1s 2s 2pN
2 π bonds p-p side to side overlap
zVertical plane
x
Overlap above and belowthe bond axis
p-p side to side overlap
π bond
x
Overlap above and belowthe bond axis
yHorizontal plane
π bond
Nitrogen
N
1 σ bond
1s 2s 2p
1s 2s 2pN
2 π bonds
Summary
Single bonds: Always σ
s-s overlap
p-p overlap head to head
s-p overlap head to head
33
Summary
Double bonds: 1 σ + 1 π
Single bonds: Always σ
Double bonds: 1 σ 1 π
Triple bonds: 1 σ + 2 π
Summary
π bonds
Only after σ bondsOnly after σ bonds
Summary
π bonds
Only after σ bondsOnly after σ bonds
Only from p-p side to side overlap
Which one of the following best describes Which one of the following best describes the bonding in hydrogen cyanide, HCNthe bonding in hydrogen cyanide, HCN
1. 1σ and 1 1 π bond
2 2 d 11 b d
0% 0% 0%0%0%
:05
2. 2σ and 1 1 π bond
3.3. 2 2 σ, 1 1 π bond and 1 lone pair
4.4. 2 2 σ, 2 2 π bond and 1 lone pair
5.5. 2 2 σ, 1 π bond and 2 lone i
Describe the bonding in HCN
C NH .C N.H
2 σ bonds2 π bonds
..
1 lone pair198
STEREOCHEMISTRY
34
Same molecular formulas but different connectivity.
STRUCTURAL ISOMERSSame molecular formulas but different connectivity.