Lecture 1 (1-05-17) Chapters 1 and 2
Goals:
Review Periodic Table, types of bonding for carbon (hybridization), and Lewis structure
Learn how to draw bond-line structures
Problems: (for all formal charge problems, use the table where possible rather than calculating)
Chapter 1: 6, 7, 10, 13, 14, 20, 22, 23, 26, 29, 32, 33, 39, 51-53, 55, 56, 61
Chapter 2: 5-8, 10-14, 17, 18 AND hybridization practice problems on website
Problem set due on Tuesday at beginning of class.
Next time: Continue Chapter 2, focus on the concept of resonance.
Information from the Periodic Table
1A
atomic number =
number of protons
(and electrons in neutral atom)
groups or columns:
similar reactivity
8A
group number = number of valence e-
1
2
H
1s
Hydrogen
1.00797
He
2A
periods or rows:
similar sizes
3
1s, 2s, 2p
1s, 2s, 2p,
3s, 3p
3A
4A
5A
6A
7A
5
6
7
8
9
Li
B
C
N
O
F
Lithium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
6.941
10.811
12.0111
14.0067
15
15.9994
18.9984
11
Na
P
Cl
Sodium
Phosphorus
Chlorine
22.9898
30.9738
35.4527
35
Sc
Br
Scandium
Bromine
44.9559
79.904
s valence orbitals filled
4.0026
17
21
1s, 2s, 2p,
3s, 3p,
4s, 4p, 3d
Helium
d orbitals filled
s and p valence orbitals filled

Lewis structure for atoms shows valence electrons. Number of unpaired electron predicts
number of bonds for uncharged atom (remember Hund’s rule):
Carbon
Nitrogen
Oxygen
Hydrogen
Chlorine
C
N
O
H
Cl
4 bonds
(tetravalent)
3 bonds
(trivalent)
2 bonds
(divalent)
1 bond
(monovalent)
1 bond
(monovalent)
Recall the shapes of the orbitals
(f orbitals not shown)
y
y
y
z
z
z
x
x
x
s
px
dyz, dxz,
dxy, dx2-y2
py
pz
p orbitals
dz2
d orbitals
Covalent Bonding for Carbon

What is the electron configuration for carbon? 1s22s22p2

Based on this, we would expect 2 different types of bonds (from s and p orbitals)

But how many bonds does carbon typically form?
o Predicted # of bonds = 8 – (# of valence electrons) so 8 - 4 = 4 bonds
1.
C—C single bond:



Ethane (below) has only one type of carbon bond: single.
All C—H bonds are the same: 98 kcals/mol bond energy, 1.1 angstrom bond lengths.
This is explained by orbital hybridization.
wedge = bond coming out
of plane of the page
H
C
H
H
 bond: sp3-sp3
dash = bond going into
plane of the page
 bond: sp3-s
H
H
C
H
H
H
H
H
H
H
(tetrahedral)
2p
sp3
2s
1s


 bonds
hybridization
1s
σ (sigma) bond—arises from head-on orbital overlap
sp3 hybridized atoms are tetrahedral and have ca. 109.5 o bond angles. (C—C is 85 kcals)
2. C—C double bond
H
H
C
H
C
H
(trigonal planar)
 bond
2p
2s
1s
2p
sp2  bonds
hybridization
1s

π bond arises from side-on overlap of p orbitals

sp2 hybridization is trigonal planar with 120o bond angles

double bond is shorter and stronger than single bond (C=C 146 kcals)

Note that hybridized orbitals always form  bonds, and p orbitals typically form  bonds
o For above compound—3 hybrid orbitals = 3  bonds per carbon (2 C—H and 1 C—C)
o 1 p orbital = 1  bond per carbon (1 C—C)
3. C—C triple bond
H
C C
(linear)
H
 bonds
2p
2s
1s




2p
sp  bonds
hybridization
1s
sp hybridization is linear with 180o bond angles
triple bond is shorter and stronger than double bond (C≡C 200 kcals)
For above compound—2 hybrid orbitals = 2  bonds per carbon (1 C—H and 1 C—C)
2 p orbital = 2  bond per carbon 2 C—C
Molecular Orbital Theory

Consider H—H

Rules for drawing Lewis dot structures
1. Draw the molecular skeleton. Monovalent atoms (hydrogen and halogens) are always on the
outside. Otherwise, the least electronegative element is usually at the center. e.g. Methane, CH4
H
H
C
H
H
2. Count the number of available valence electrons. Add up all the valence electrons of each atom.
Pay special attention to charged structures. For (+), subtract one electron from the total. For (-),
add one electron. For CH4:
4H
1C
Total
4 X 1 valence electron: 4 e4 valence electrons: 4 e8 e-
3. (The octet rule) Show all covalent bonds as two shared e-, giving as many 2nd row atoms as
possible a surrounding electron octet. Hydrogen, however, should only have only2 e-.
H
H
C
H
H
Unlike this example, double and triple bonds many be necessary to satisfy the octet rule.
4. Assign formal charges to atoms in the molecule:
Charge = (# of valence e- in the free neutral atom) – (# of unshared e- on the atom in the molecule) –
(# of bonds)
H
H
C
H
H
For each H, Charge = 1 – 0 – 1 = 0
For C, Charge = 4 – 0 – 4 = 0
Each atom has (0) formal charge, so the molecule is neutral.
This method does not necessarily give you a single structure. Consider the following formula:
C3H6O
H
C
H
H
O
C
H
H
C
H
H
H
C
H
O
C
H
C H
C
H
HH
H
C
H H
C
O H
C
H H
H
C
H
H
C
H
H
C
H C
H
O
C
O H
H
H
These molecules are called constitutional isomers: same molecular formula, but different
connectivity of atoms
Exceptions to the Lewis structure rules
There are two important violations to the octet rule
1. Some elements, most commonly Boron (B) and Aluminum (Al) don’t have enough valence
electrons to form octets as the neutral compound. They are very reactive and readily form charged
complexes (to give octets).
H
B
H
H
Cl
Cl
Al
Cl
2. The Lewis structure model is not applicable beyond the second row, because elements have d
orbitals, and can often violate the octet rule by filling these orbitals. Ex. H2SO4, C3H9PO
sulfuric acid
H
trimethylphosphine oxide
O
O S O
H
O
-
12 e around sulfur
(possible because sulfur is in period 3,
and so has empty d orbitals)
H
H O
C
P
C
H
H
H
H C
H
H
H
-
10 e around
phosphorus
Bonding patterns and formal charges for common atoms in organic chemistry
(Know these by Tuesday!)
charge
+1
0
-1
+1
0
-1
C
C
C
C
C
+1
N
0
-1
+1
H
H
H
C
N
O
C
# of
bonds
# of nonbonding
electron
pairs
C
N
O
O
N
+1
0
X
X
O
N
C
-1
-1
O
N
N
structure
0
X
X
O
N
X = F, Cl,
Br, I
0
1
0
3
4
3
4
3
2
3
2
1
2
1
0
0
0
1
0
0
1
0
1
2
1
2
3
2
3
4