Chapter 1
Electronic Structure
and
Bonding
Acids and Bases
~ 0.1 nm
Anders Jöns Ångström
(1814-1874)
1 Å = 10 picometers = 0.1 nanometers =
10-4 microns = 10-8 centimeters
• 1 nm = 10 Å
• An atom vs. a nucleus
~10,000 x larger
Nucleus =
1/10,000
of the atom
Question 1
• What is the electronic configuration of
carbon?
•
A) 1s2 2s2 2px2
•
B) 1s2 2s2 2px1 2py12pz0
•
C) 1s2 2s2 2px12py12pz1
•
D) 1s2 1px1 1py12s2
Electron Configurations
Noble Gases and The Rule of Eight
•
When two nonmetals react to form a
covalent bond: They share electrons to
achieve a Noble gas electron
configuration.
•
When a nonmetal and a metal react to
form an ionic compound: Valence
electrons of the metal are lost and the
nonmetal gains these electrons.
G.N. Lewis
Photo Bancroft Library, University of California/LBNL Image Library
Footnote:
G.N. Lewis, despite his insight and contributions
to chemistry, was never awarded the Nobel prize.
Notes from Lewis’s notebook and his “Lewis” structure.
http://chemconnections.org/organic/Movies%20Org%20Flash/LewisDotStructures.swf
Ionic Compounds
• Ionic compounds are formed when electron(s) are
transferred.
• Electrons go from less electronegative element to the
more electronegative forming ionic bonds.
Covalent Compounds
•Share electrons.
•1 pair = 1 bond.
•Octet rule (“duet” for hydrogen)
•Lewis structures:
Notice the charges:
In one case they balance, can you name the compound?
In the other they do not, can you name the polyatomic ion?
More about “formal” charge to come.
Question 2
• Select the correct Lewis structure for
methyl fluoride (CH3F).
•
A)
B)
•
•
C)
D)
Important Bond Numbers
(Neutral Atoms / Normal electron distribution)
one bond
two bonds
three bonds
four bonds
H
F
Cl
Br
O
N
C
I
Question 3
• What is the correct Lewis structure of
formaldehyde (H2CO)?
•
A)
B)
•
C)
D)
Question 4
• Which of the following contains a triple
bond?
•
A) SO2
•
B) HCN
•
C) C2H4
•
D) NH3
Formal Charge
Formal charge is the charge of an atom in
a Lewis structure which has a different
than normal distribution of electrons.
Important Bond Numbers
(Neutral Atoms / Normal electron distribution)
one bond
two bonds
three bonds
four bonds
H
F
Cl
Br
O
N
C
I
Important Bond Numbers
(Neutral Atoms / Normal electron distribution)
Organic Chemistry
C
H
O
N
# of Valence e s
4
1
6
5
Total # of Bonds
(neutral atom)
4
1
2
3
-
Combinations of bonds
(neutral atom):
# of single bonds
4
2
1
1
2
0
3
1
0
# of double bonds
0
1
0
0
0
1
0
1
0
# of triple bonds
0
0
1
0
0
0
0
0
1
Total Bonds
4
4
4
1
2
2
3
3
3
# of Free Pairs of
electrons
0
0
0
0
2
2
1
1
1
Formal Charge
Formal charge = number of valence electrons –
(number of lone pair electrons +1/2 number of bonding electrons)
•
•
Equals the number of valence electrons
(Group Number of the free atom) minus [the
number of unshared valence electrons in
the molecule + 1/2 the number of shared
valence electrons in the molecule].
Moving/Adding/Subtracting atoms and
electrons.
HNO3 Nitric Acid
Complete the following table. It summarizes the formal charge on a
(“central”) atom for the most important species in organic chemistry.
Question 5
• What is the formal charge of the carbon
atom in the Lewis structure?
•
A) -1
•
B) 0
•
C) +1
C
•
D) +2
Question 6
• What is the formal charge of the oxygen
atom in the Lewis structure?
•
A) -1
•
B) 0
•
C) +1
•
D) +2
Resonance
Resonance
Resonance is a very important intellectual concept that
was introduced by Linus Pauling in 1928 to explain
experimental observations.
Eg. SO2
Bond order  1.5
Bond length > double bond; < single bond
TUTORIAL
Resonance
•Two or more Lewis structures may be
legitimately written for certain compounds
(or ions) that have double bonds and/or
free pairs of non-bonded electrons
•It is a mental exercise in “pushing” or
moving electrons.
•Refer to Table 1.6
Rules of Resonance
Step 1:
The atoms must stay in the same position. Atom
connectivity is the same in all resonance
structures. Only electrons move.
NON-Example: The Lewis formulas below are
not resonance forms. A hydrogen atom has
changed position.
H
N
H
OH
O
N
C
H
H
C
H
Rules of Resonance
•
Step 2:
Each contributing structure must have
the same total number of electrons and
the same net charge.
•
Example:
All structures have 18 electrons and a
net charge of 0.
H
O
N
H
H
C
O
N
H
H
O
H
N
C
H
H
C
H
Rules of Resonance
•
Step 3:
Calculate formal charges for each atom
in each structure.
•
Example:
None of the atoms possess a formal
charge in this Lewis structure.
H
O
N
H
C
H
Rules of Resonance
•
Step 4:
Calculate formal charges for the second
and third structures.
•
Example:
These structures have formal charges.
NOTE: They are less favorable Lewis structures.
H
O
N
H
O
H
N
C
H
H
C
H
“Pushing” Electrons
•same atomic positions
H
H
C
•differ in electron positions
H
..
..
+
O
O:
H C O
.. N
..
..
H
more stable
Lewis
structure
N
..
H
less stable
Lewis
structure
.. –
O
.. :
“Pushing” Electrons
•same atomic positions
H
H
C
•differ in electron positions only
H
..
..
+
O
O:
H C O
.. N
..
..
H
more stable
Lewis
structure
N
..
H
less stable
Lewis
structure
.. –
O
.. :
Why use Resonance Structures?
•Delocalization of electrons and charges between
two or more atoms helps explain energetic
stability and chemical reactivity.
•Electrons in a single Lewis structure are
insufficient to show electron delocalization.
•A composite of all resonance forms more
accurately depicts electron distribution. (HYBRID)
NOTE: Resonance forms are not always evenly
weighted. Some forms are better than others.
Resonance Example
•Ozone (O3)
–Lewis structure of
ozone shows one
double bond and
one single bond
••
•O
•
Expect: one short bond and one
long bond
Reality: bonds are of equal length
(128 pm)
+
O
••
•• –
O ••
••
Resonance Example
•Ozone (O3)
–Lewis structure of
ozone shows one
double bond and
one single bond
••
•O
•
+
O
•• –
O ••
••
••
Resonance:
••
•O
•
+
O
••
•• –
O ••
••
– ••
•O
• ••
+
O
••
O ••
••
Resonance Example
•Ozone (O3)
–Electrostatic potential
map shows both end
carbons are equivalent
with respect to negative
charge. Middle carbon
is positive.
••
•O
•
+
O
••
•• –
O ••
••
– ••
•O
• ••
+
O
••
O ••
••
Detailed Resonance
Examples
Question 7
• Which resonance structure contributes
more to the hybrid?
•
A)
B)
VSEPR Model
Valence Shell Electron Pair Repulsion
VSEPR Model
The molecular structure of a given atom is determined
principally by minimizing electron pair (bonded &free)
repulsions through maximizing separations.
Some examples of minimizing interactions.
Predicting a VSEPR Structure
• 1. Draw Lewis structure.
• 2. Put pairs as far apart as possible.
• 3. Determine positions of atoms from the
way electron pairs are shared.
• 4. Determine the name of molecular
structure from positions of the atoms.
Orbital
Geometry
Chem 226 Linear
Molecular
Geometry
Bond Angle
# of lone pairs
Linear
0
Trigonal Planar
Trigonal Planar
0
Trigonal Planar
Bent
1
Tetrahedral
Tetrahedral
0
Tetrahedral
Trigonal Pyramidal
1
Tetrahedral
Bent
2
Trigonal Bipyramidal
Trigonal Bipyramidal
0
Trigonal Bipyramidal
Seesaw
1
Trigonal Bipyramidal
T-shape
2
Trigonal Bipyramidal
Linear
3
Octahedral
Octahedral
0
Octahedral
Square Pyramidal
1
Octahedral
Square Planar
2
Lewis Structures / VSEPR /
Molecular Models
• Computer Generated Models
Ball and stick models of ammonia, water and
methane.
Worksheet 1: Bonds, Formulas, Structures & Shapes
http://chemconnections.org/organic/chem226/226assign-09.html#Worksheets
http://chemconnections.org/organic/chem226/Labs/VSEPR/
Covalent Compounds
•Equal sharing of electrons: nonpolar covalent
bond, same electronegativity (e.g., H2)
• Unequal sharing of electrons between atoms of
different electronegativities: polar covalent bond
(e.g., HF)
Question 8
• Which of the following bonds is the most
polar?
•
•
A)
B)
•
•
C)
D)
Bond Dipole & Dipole Moment
• Dipole moments are experimentally measured.
• Polar bonds have dipole moments.
dipole moment (D) = m = e x d
(e) : magnitude of the charge on the atom
(d) : distance between the two charges
Question 9
• Which of the following bonds have the
greatest dipole moment (m)?
•
A)
B)
•
C)
D)
Bond Polarity
A molecule, such as HF, that has a center
of positive charge and a center of negative
charge is polar, and has a dipole moment.
The partial charge is represented by  and
the polarity with a vector arrow.
H F
+

Question 10
• In which of the compounds below is the +
for H the greatest?
•
A) CH4
•
B) NH3
•
C) SiH4
•
D) H2O
Question 11
• In which of the following is oxygen the
positive end of the bond dipole?
•
A) O-F
•
B) O-N
•
C) O-S
•
D) O-H