View
5.566
Download
0
Category
Tags:
Preview:
Citation preview
Chem 5Chapter 11
Chemical Bonding I:Basic Concepts
Part 2
November 18, 2002
We covered the following concepts last time:
Lewis structure - the octet ruleIonic bond and covalent bondPolar bond, electronegativity and dipole moment Molecular shapes - VSEPRBond length and bond energy
(g) 6 N N (g) + 12 O=C=O (g)
+ 10 H-O-H (g) + O=O (g)
≡4
∆H = ∆H (bond breakage) + ∆H (bond formation)= ΣBE (reactants) - ΣBE (products)
= 4x(5xBEC-H +2BEC-C +3BEC-O+6BEN-O +3BEN=O)
- (6xBEN=N+24xBEC=O+20xBEO-H+BEO=O)
To predict molecular shape with VSEPR
An example of a polyatomic molecule with no single central atom:
ETHANOL
Apply VSEPR to each atom with more than one electron pair.
OCH H
H
H
C
H
H
CH3CH2OH
HH
H
C
H
HC
H
O
The atoms around the carbons form a
tetrahedral arrangement.
The atoms around the oxygen form a
V-shaped structure.
ETHANOL C2H5OH
The structure of ethanol dehydrogenase
Ethanol
The shape matters!
Ethanol Dehydrogenase
OAn enzyme that catalyzes the hydrolysis of ethanol.
CH3CH2OH CH3C-H
A close look at the active site
New concepts today
Formal chargeResonanceExceptions to the octet rule
Ozone destruction and polymers
Two possible Lewis structures for HCN
H N CH C N Why not
Two possible Lewis structures for formaldehyde
OH
H
C
Why not
CH
H
O
gives an indication of the extent to which atomsFORMAL CHARGE
have gained or lost electrons in the process of covalent bond formation.Each atom is assigned all of its lone electrons and
Formal charge = #valence
electrons_ #unshared
electrons_ 1/2#shared
electrons{ } { }half of the bonding electrons.
Structures with the lowest formal charges are the most stable.
All possible Lewis structures with stable electronic configurations for HCN and HNC.
H C N
Calculate formal charge for
_ 1/2#shared electrons{ }Formal
charge = #valence electrons
_ #unshared electrons{ }
FC on C = 4 - 0 - 8/2 = 0
H C N
..
FC on N = 5 - 2 - 6/2 = 0
FC on H = 1 - 0 - 2/2 = 0
All possible Lewis structures with stable electronic configurations for HNC.
H N C
Calculate formal charge for
FC on C = 4 - 2 - 6/2 = -1
_ 1/2#shared electrons{ }Formal
charge = #valence electrons
_ #unshared electrons{ }
H N C
..
FC on N = 5 - 0 - 8/2 = +1
FC on H = 1 - 0 - 2/2 = 0
Four Rules for Formal Charge
• The sum of F.C. must be the net charge of the molecule or ion.
• F.C. should be as small as possible.
• Negative F.C. usually occurs on the most electronegative atoms.
• F.C. of the same sign on adjacent atoms is unlikely.
All possible Lewis structures with stable electronic configurations for HCN and HNC.
0 0 0H C N
0 +1 -1H N C
FORMAL CHARGES
We choose the structure based onF.C. of each atom being zero
The most electronegative atom (N) should not have positive F.C.
It should beH C N
Another Example:
H2CO vs. H2OC
H
HOC
H
HCO
FC on C =4 – 0 – 2x2/2 – 4/2 = 0
FC on O = 6 – 4 – 4/2 = 0
FC on O = 6 – 0 – 2x2/2 –4/2 = 2
FC on C = 4 – 4 – 4/2 = - 2
Not stable!
Let’s look at the carbonate anion CO32-
C has four 1s22s22p2
O has six 1s22s22p4
Count up valence electrons
Plus two extra for negative charge
Valence electrons = 4 + 3 x 6 + 2 = 24
Put a pair between each atomCarbonate anion CO3
2-
C
O
O O
24 valence electrons
18 left
Add remaining electrons to terminal atoms to complete octets...
Carbonate anion CO3 2-
24 valence electrons
2-
C
O
O O
DO NOT FORGET CHARGE!!!!
The oxygen atoms have their octet but...The carbon atom does not!
So we form a double bond by sharing a pair from one of the oxygen atoms...
Carbonate ion CO32-
2-
C
O
O O
24 valence electrons
Form a double bond by sharing a pair from one of the oxygen atoms...
2-
C
O
O OBut…
C
O
O O
2-
FORM A DOUBLE BOND BETWEEN O AND C
C
O
O O
2-
Here is another!
Here is one
Here is another!
C
O
O O
2-
C
O
O O
2-
All three are perfect! But, none alone is correct!
Because experiment shows all three bonds are the same.
All three bond lengths the same!
Longer than double bonds,
Shorter than single bondsO O
O
C
And bond angles 120°
2
RESONANCEWe use a double headed arrow between the
structures..
The electrons involved are said to be DELOCALIZED over the structure.
The blended structure is a RESONANCE HYBRID
O O
O
CCO O
O2 2
CO O
O2
We interpret the experimental structure with a partial double bond. The bond order is (2+2x1)/3 =1.33.
Another example of resonance structure:
OO O
LEWIS STRUCTUREO3Make double bond...
OO O
OO O
Experiment shows that both O-O bonds are equivalent.
The O3 molecule is a hybrid of the two resonance forms.
Ozone is generated by photochemistry in the stratosphere (15-60 Km),forming a protective layerfor all life on earth.
Ozone absorption spectrum
O2 (g) → 2 O (g)hν
O2 (g) + O (g) → O3 (g)
• O3 absorbs UV light 220-300 nm• N2 and O2 do not absorb at those wavelengths• O3 layer protects DNA from photodamage
Ozone Reduction
The ozone destruction is related to human activity!
Cl + O3 = ClO + O2
ClO + O = Cl + O2
O3 + O = 2 O2
chloroflorocarbon (CFC) → Cl hν
The Heroes
Sherwood Rowland Mario Molina
The billion-dollar CFC industry and some government agencies attackedthem viciously, even tried to get them fired from their professor posts. However, the scientific community provided moral support and overwhelming experimental evidence.The truth prevailed! Rowland and Molina were awarded the Nobel Prize in 1995.
International Ban on CFC
CFC is replaced by fluorocarbon.
The ozone level is expected to recover in 50 years.
Professor Jim Andersonto teach Chem7
The flying Chem Lab
EXCEPTIONS TO THE OCTET RULE…
• Molecules with more than an octet around the central atom
• Molecules with less than an octet around the central atom
• Molecules with unpaired electrons
A CENTRAL ATOM WITH MORE THAN AN OCTET
Elements in the third or higher periods can exceed the octet due to d or f orbitals
EXAMPLE : PF5F
PFF
F F
P
PFF
F
F
FBond angle900
1200
The shape of PF5 is trigonal bipyramidal.
PF5
AXIAL
EQUATORIAL
Five electron pairs around the phosphorus atom.
Other examples based on 5 pairs of electrons…All with empty d orbitals
ClF3
2 lone pairs
T-shaped
ClF
F
F
SF4
1 lone pair
See-saw shaped
SF
F
F
F
XeF2
3 lone pairs
Linear
F
F
F
Xe
Lone pairs occupy the trigonal plane (the “equator”) first to minimize the number of 90° repulsions
Another Example I3-
I I INumber of valence electrons: 7+7+7+1=22
Add electrons... 4 used in two bonds
F
I
I
I
-I II
Linear shapetrigonal bipyramidalwith three lone pairs in the equator
Draw the Lewis structure for SF6
6 + 6 x 7 = 48Total number of valence electrons =Why? F has seven 1s22s22p5
1s22s22p63s22p4S has six
Place S in middle of 6 fluorine atoms
F
S
F
F
F
F
F
There are six electron pairs around the sulfur atom.
Shape of SF6
All bond angles 900
S
F
F
F
F
F
F
6 electron pairs octahedral
A CENTRAL ATOM WITH LESS THAN AN OCTET..
BF
F
..
....F..
. ...
....
BF
F
..
. ...
..+1
-1
..
F.C.
..F....
BF
F
..
..
..
+
...._
. ...
with tworesonancestructures
with tworesonancestructures..
F....F.C.Not stable Ionic
FREE RADICALSHave unpaired electrons.
NO2 Is a free radical
Total number of valence electrons=5+6+6=17
O N O
RESONANCE
Form double bond to get N close to octet
O N O O N O
SummaryFormal Charge:apparent charges on certain atoms in a Lewis structure that arise when atoms have not contributed equal numbers of electrons to the covalent bonds joining them
- Four rules to select Lewis structures based on F.C.
Resonance Structure:More than one plausible Lewis structure can be written for a species. The true structure is a resonance hybrid of two or morecontributing structures.
Exceptions to the Octet RuleAt times, a molecule may have unpaired electrons or the valence shell of the central atom must be expanded to 10 (trigonal bipyramidal) or 12 electrons (octahedral).
Recommended