General Chemistry
M. R. Naimi-Jamal
Faculty of Chemistry
Iran University of Science & Technology
‫فصل هشتم‪:‬‬
‫پیوند کوواالنسی‬
Chemical Bonding
Problems and questions:
How is a molecule or polyatomic ion held together?
Why are atoms distributed at strange angles?
Why are molecules not flat?
Can we predict the structure?
How is structure related to chemical and physical
properties?
Forms of Chemical Bonds
• There are 2 extreme forms of connecting or
bonding atoms:
• Ionic—complete transfer of electrons from one
atom to another
• Covalent—electrons shared between atoms
• Most bonds are somewhere in between.
Covalent Bonding
Covalent bond forms by the sharing of VALENCE
ELECTRONS, the electrons at the outer edge of the
atom.
The bond arises from the mutual attraction of 2
nuclei for the same electrons.
Valence Electrons
Electrons are divided between core and valence
electrons.
Na 1s2 2s2 2p6 3s1
Core = [Ne] and valence = 3s1
Br [Ar] 3d10 4s2 4p5
Core = [Ar] 3d10 and valence = 4s2 4p5
Chemical Bonding Objectives
Objectives are to understand:
1. e- distribution in molecules and ions.
2. molecular structures
3. bond properties and their effect on molecular
properties.
Electron Distribution in Molecules
• Electron distribution is
depicted with Lewis electron
dot structures
• Electrons are distributed as
shared or BOND PAIRS and
unshared or LONE PAIRS.
G. N. Lewis
1875 - 1946
Lewis Symbols
• A chemical symbol represents the nucleus
and the core e-.
• Dots around the symbol represent valence e-.
•
• Si •
•
• Al •
•
• As •
•
P•
•
••
• Se
•
•
.
••
••
• Bi •
•
• Sb •
•
••
I
••
•
••
Ar
••
•
••
••
•N•
•
••
••
••
••
Lewis Structures for Ionic Compounds
• O•
••
2+ ••
Ba
O
••
2-
••
Ba•
••
••
BaO
•
••
••
Mg
••
• Cl
••
••
2 Cl ••
••
2+
••
Mg •
••
MgCl2
•
• Cl
••
Bond and Lone Pairs
• Electrons are distributed as shared or BOND PAIRS
and unshared or LONE PAIRS.
••
H
Cl
•
•
••
shared or
bond pair
lone pair (LP)
This is called a LEWIS ELECTRON DOT structure.
Bond Formation
A bond can result from a “head-to-head” overlap of
atomic orbitals on neighboring atoms.
••
H
+
Cl
••
••
•
•
H
Cl
•
•
••
Overlap of H (1s) and Cl (2p)
This type of overlap places bonding electrons in a
MOLECULAR ORBITAL along the line between the
two atoms and forms a SIGMA BOND ().
Covalent Bonding
Coordinate Covalent Bonds
+
H
H
Cl
H N H
H
••
Cl
••
-
••
••
H N
H
••
H
Multiple Covalent Bonds
•O
••
••
•
•
•
••
•
••
••
O C O
•
•
•
••
•
••
O C O
••
• C•
•
••
••
O•
•
••
••
••
••
O C O
••
•
••
•
Multiple Covalent Bonds
•
•
••
••
N N
•
•
•
N N
••
••
•
N N
••
•
••
•N
•
N•
••
•
••
•
Polar Covalent Bonds
δ+
H
Cl
δ-
Electronegativity
xA-xB = √ D/23.06
Pauling electronegativity
D = 2 E(A-B) – E(A-A) – E(B-B)
=
I+A
2
I = Ionization Energy,
A = Electron Affinity
Mulliken electronegativity
Electronegativity
Dipole Moments
Dipole Moments
‫گشتاور دو قطبی‬
‫• گشتاور دو قطبی = فاصله‬
‫‪x‬‬
‫بار‬
‫• واحد گشتاور دو قطبی = دبی = ‪3.34 x 10 -30 C.m‬‬
Percent Ionic Character
WritingLewis
LewisStructures
Structures
Writing
• No. of valence electrons of an atom = group number
• For groups 1A-4A, no. of bond pairs = group number
• For groups 5A-7A, BP’s = 8 - gr. no.
• Except for H (and atoms of 3rd and higher periods),
BP’s + LP’s = 4
This observation is called the
OCTET RULE
Writing Lewis Structures
• All the valence e- of atoms must appear.
• Usually, the e- are paired.
• Usually, each atom requires an octet.
– H only requires 2 e-.
• Multiple bonds may be needed.
– Readily formed by C, N, O, S, and P.
Skeletal Structure
• Identify central and terminal atoms.
• C2H5OH
H
H C
H
H
C O H
H
Skeletal Structure
• Hydrogen atoms are always terminal atoms.
• Central atoms are generally those with the lowest
electronegativity.
• Carbon atoms are always central atoms.
• Generally structures are compact and symmetrical.
Building a Dot Structure
Ammonia, NH3
1. Decide on the central atom;
Central atom is generally atom of lowest affinity for
electrons, but never H, here N is central.
2. Count valence electrons
H = 1 and N = 5
Total = (3 x 1) + 5
= 8 electrons / 4 pairs
Building a Dot Structure
3.Form a sigma bond between the
central atom and surrounding atoms.
H
N
H
H
4.Remaining electrons form LONE
PAIRS to complete octet as needed.
3 BOND PAIRS and 1 LONE PAIR.
Note that N has a share in 4 pairs (8
electrons), while H shares 1 pair.
••
H
N
H
H
Sulfite ion, SO32Step 1. Central atom = S
Step 2. Count valence electrons
S= 6
3 x O = 3 x 6 = 18
Negative charge = 2
TOTAL = 26 e- or 13 pairs
Step 3. Form sigma bonds
Sulfite ion, SO32-
O
O
S
O
10 pairs of electrons are now left.
Sulfite ion, SO32Remaining pairs become lone pairs, first on outside
atoms and then on central atom.
••
•
•
O
••
•
•
O
••
•
•
••
S
••
O
•
•
••
Each atom is surrounded by an octet of electrons.
Carbon Dioxide, CO2
1. Central atom = C
2. Valence electrons = 16 or 8 pairs
3. Form sigma bonds.
O
C
O
This leaves 6 pairs.
Carbon Dioxide, CO2
4. Place lone pairs on outer atoms.
••
••
•
•
O
••
C
O
•
•
••
BUT C doesn’t obey the octet rule!
5. So that C has an octet, we shall form
DOUBLE BONDS between C and O.
Carbon Dioxide, CO2
The second bonding pair forms a pi (p) bond.
••
••
•
•
O
••
C
O
••
•
•
•
•
O
••
C
O
••
•
•
H2CO
Double and even triple
bonds are commonly
observed for C, N, P,
O, and S
SO3
C2F4
Sulfur Dioxide, SO2
1. Central atom = S
2. Valence electrons = 18 or 9 pairs
3. Form sigma bonds.
O
S
O
This leaves 7 pairs.
4. Place 7 lone pairs on outer atoms.
••
•
•
O
••
••
S
••
O
••
•
•
Sulfur Dioxide, SO2
5. Form pi (p) bond so that S has an octet —
but note that there are two ways of doing
this.
OR bring in
right pair
bring in
left pair
••
•
•
O
••
••
S
••
O
••
•
•
Sulfur Dioxide, SO2
This leads to the following structures:
••
•
•
O
••
S
••
•
•
O
••
••
•
•
••
O
••
S
•
•
O
••
These equivalent structures are called:
RESONANCE STRUCTURES
The true electronic structure is a HYBRID of the two.
Urea, (NH2)2CO
Urea,
Urea,(NH
(NH2)22)CO
2CO
1. Central atom = C
2. Number of valence electrons = 24 e3. Draw sigma bonds
O
H N
H
C
N H
H
Urea, (NH2)2CO
4. Place remaining electron pairs in the
molecule.
••
•
•
O
••
H N
H
•
•
••
C
N
H
H
Urea,
Urea,(NH
(NH2)22)CO
2CO
5. Complete C atom octet with double
bond.
••
O
••
H N
H
•
•
••
C
N H
H
Violations of the Octet Rule
Usually occurs with B and elements of
higher periods.
..
BF3
SF4
Boron Trifluoride
• Central atom = B
• Valence electrons = 24 or electron
pairs = 12
• Assemble dot structure
••
•
•
•
•
F
••
•
•
F
••
B
•
•
•
•
F
••
The B atom has a share in only 6
pairs of electrons (or 3 pairs).
B atom in many molecules is
electron deficient.
Sulfur Tetrafluoride, SF4
• Central atom = S
• Valence electrons = 34 or 17 pairs.
• Form sigma bonds and distribute
electron pairs.
••
•
•
••
F
••
••
•
•
F
••
S
••
F
••
••
F
••
•
•
•
•
5 pairs around the S atom.
A common occurrence
outside the 2nd period.
Exceptions to the Octet Rule
• Odd e- species:
••
••
•
N=O
••
H
••
•
O—H
••
•
H—C—H
Exceptions to the Octet Rule
••
• Incomplete octets:
••
••
F
B
F
F
Exceptions to the Octet Rule
• Expanded octets:
••
F
Cl
Cl
S
P
PCl5
F
F
••
Cl
F
••
••
••
Cl
Cl
F
••
••
••
••
SF6
F
Formal Charge
FC = #valence e- - #lone pair e- -
1
2
#bond pair e-
Carbon Dioxide, CO2
6 - (1/ 2)(4) - 4
•
•
O
••
C
4 - (1/ 2)(8) - 0
O
••
•
•
=
0
=
0
Carbon Dioxide, alternative lewis structure
6 - (1/ 2)(2) - 6
=
-1
••
•
•
O
C
6 - (1/ 2)(6) - 2
O
•
•
••
= +1
Which is the predominant resonance structure?
Boron Trifluoride, BF3
••
•
•
•
•
F
••
•
•
F
••
B
•
•
•
•
F
••
What if we form a B—F double
bond to satisfy the B atom octet?
Boron
Trifluoride,
BF
3
Boron Trifluoride, BF
3
••
•
•
F
FC = 7 - 2 - 4 = +1
B
FC = 3 - 4 - 0 = -1
••
•
•
F
••
•
•
•
•
F
••
• To have +1 charge on F, with its very
high affinity for electrons, is not good.
• Negative charges are best placed on
atoms with high affinity for electrons.
Exceptions to the Octet Rule
• Incomplete octets.
••
••
••
••
••
B
B
F
+ F
••
F
F
F
F
••
••
••
+
-
F
F
-
B
F
Formal Charges
&
Lewis Structure
Chapter 7 Questions
6, 8, 18, 21, 31
32, 38, 44, 48
52