Chapter 7
COVALENT BONDING
COVALENT BONDING
7.1 Lewis Structures; The Octet Rule
7.2 Molecular Geometry
Valence-Shell Electron-Pair Repulsion (VSEPR)
7.3 Polarity of Molecules
7.4 Atomic Orbitals; Hybridization
The steps in converting a molecular formula into a Lewis structure.
Molecular
formula
Step 1
Atom
placement
Place atom
with lowest
EN in center
Step 2
Sum of
valence e-
Add A-group
numbers
Step 3
Remaining
valence e-
Draw single bonds.
Subtract 2e- for each bond.
Step 4
Lewis
structure
Give each
atom 8e(2e- for H)
Sum of
valence e-
:
: F:
: F:
:
Atom
placement
For NF3
:
Molecular
formula
N
Remaining
valence eLewis
structure
:
: F:
N
5e-
F 7e- X 3 = 21eTotal
26e-
SAMPLE PROBLEM
Write a Lewis structure for CCl2F2, one of the compounds
responsible for the depletion of stratospheric ozone.
.
SOLUTION:
Make bonds and fill in remaining valence
electrons placing 8e- around each atom.
F
:
F
:
: Cl :
:
Steps 2-4:
C has 4 valence e-, Cl and F each have 7. The
sum is 4 + 4(7) = 32 valence e-.
Cl C
:Cl C
:
Step 1: Carbon has the lowest EN and is the central atom.
The other atoms are placed around it.
Cl
: F:
F:
:
PLAN: Follow the steps outlined in Figure 10.1
:
PROBLEM:
Writing Lewis Structures for Molecules with
One Central Atom
SAMPLE PROBLEM
Hydrogen can have only one bond so C and O must be next
to each other with H filling in the bonds.
There are 4(1) + 4 + 6 = 14 valence e-.
C has 4 bonds and O has 2. O has 2 pair of nonbonding e-.
H
:
SOLUTION:
Write the Lewis structure for methanol (molecular formula
CH4O), an important industrial alcohol that is being used as a
gasoline alternative in car engines.
H
C
O
:
PROBLEM:
Writing Lewis Structure for Molecules with
More than One Central Atom
H
H
SAMPLE PROBLEM
Writing Lewis Structures for Molecules with
Multiple Bonds.
Write Lewis structures for the following:
(a) Ethylene (C2H4), the most important reactant in the
manufacture of polymers
(b) Nitrogen (N2), the most abundant atmospheric gas
PROBLEM:
PLAN:
For molecules with multiple bonds, there is a Step 5 which follows the
other steps in Lewis structure construction. If a central atom does not
have 8e-, an octet, then two e- (either single or nonbonded pair)can
be moved in to form a multiple bond.
SOLUTION: (a) There are 2(4) + 4(1) = 12 valence e-. H can have only
one bond per atom.
H
H
:
H
C
C
H
H
H
H
C
C
H
(b) N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make
the octet around each N.
N
.
:
N
.
:
:
:.
N
.
N
N
:
.:
N
.
Resonance: Delocalized Electron-Pair Bonding
O3 can be drawn in 2 ways -
O
O
O
O
O
O
Neither structure is actually correct but can be drawn to represent a structure
which is a hybrid of the two - a resonance structure.
B
B
O
O
O
A
O
C
O
O
O
O
A
O
C
Resonance structures have the same relative atom placement but a
difference in the locations of bonding and nonbonding electron pairs.
is used to indicate that resonance occurs.
SAMPLE PROBLEM
PROBLEM:
PLAN:
Write resonance structures for the nitrate ion, NO3-.
After Steps 1-4, go to 5 and then see if other structures can be
drawn in which the electrons can be delocalized over more than
two atoms.
SOLUTION:
O
Writing Resonance Structures
Nitrate has 1(5) + 3(6) + 1 = 24 valence e-
O
O
O
N
N
N
O
O
O
O
O
N does not have an
octet; a pair of ewill move in to form
a double bond.
O
O
O
O
N
N
N
O
O
O
O
O
Formal Charge: Selecting the Best Resonance Structure
An atom “owns” all of its nonbonding electrons and half of its bonding electrons.
Formal charge is the charge an atom would have if the bonding electrons were
shared equally.
Formal charge of atom =
# valence e- - (# unshared electrons + 1/2 # shared electrons)
B
# valence
e-
e-
O
=6
# nonbonding
# bonding
For OC
O
For OA
e-
=4
= 4 X 1/2 = 2
A
For OB
O
C
# valence e- = 6
Formal charge = 0
# valence e- = 6
# nonbonding e- = 6
# bonding e- = 2 X 1/2 = 1
Formal charge = -1
# nonbonding e- = 2
# bonding e- = 6 X 1/2 = 3
Formal charge = +1
Resonance (continued)
Three criteria for choosing the more important resonance structure:
Smaller formal charges (either positive or negative) are preferable
to larger charges.
Avoid like charges (+ + or - - ) on adjacent atoms.
A more negative formal charge should exist on an atom with a
larger EN value.
Resonance (continued)
EXAMPLE: NCO- has 3 possible resonance forms -
N C
O
N C
A
N C
O
B
O
C
formal charges
-2
0
N C
+1
O
-1
0
N C
0
O
0
0
N C
-1
O
Forms B and C have negative formal charges on N and O; this makes them
more preferred than form A.
Form C has a negative charge on O which is the more electronegative
element, therefore C contributes the most to the resonance hybrid.
SAMPLE PROBLEM
PROBLEM:
PLAN:
Writing Lewis Structures for Octet Rule Exceptions
Write Lewis structures for (a) H3PO4 (pick the most likely
structure); (b) BFCl2.
Draw the Lewis structures for the molecule and determine if there is
an element which can be an exception to the octet rule. Note that
(a) contains P which is a Period-3 element and can have an
expanded valence shell.
SOLUTION:
(a) H3PO4 has two resonance forms and formal charges
indicate the more important form.
-1
0
O
0 H O
P
O
0
H
0
+1
O H 0
0
0
0 H O
0
O
0
P
O H 0
0
0 O
H more stable
0
lower formal charges
(b) BFCl2 will have only 1
Lewis structure.
F
B
Cl
Cl
VSEPR
VSEPR - Valence Shell Electron Pair Repulsion Theory
Each group of valence electrons around a central atom is located as far
away as possible from the others in order to minimize repulsions.
These repulsions maximize the space that each object attached to the
central atom occupies.
The result is five electron-pair (group) arrangements of minimum energy seen
in a large majority of molecules and polyatomic ions.
The electron-pairs, Np, are defining the object arrangement or electron pair
geometry, but the molecular shape is defined by the relative positions of the
atomic nuclei.
Because valence electrons can be bonding or nonbonding, the same
electron-pair arrangement can give rise to different molecular shapes.
A - central atom
X -surrounding atom
AXmEn
integers
E -nonbonding valence electron-group
Electron-group repulsions and the five basic molecular shapes.
linear
trigonal bipyramidal
tetrahedral
trigonal planar
octahedral
The single molecular shape of the linear electron-group
arrangement. Number of Electron pairs, Np=2.
Examples:
CS2, HCN, BeF2
The two molecular shapes of the trigonal planar electron
group arrangement. Np=3.
Class
Examples:
SO2, O3, PbCl2, SnBr2
Shape
Examples:
SO3, BF3, NO3-, CO32-
Factors Affecting Actual Bond Angles
Bond angles are consistent with theoretical angles when the atoms
attached to the central atom are the same and when all electrons are
bonding electrons of the same order.
H
Effect of Double Bonds
1200
ideal
1200
O
1160
C
H
greater
electron
density
real
Effect of Nonbonding(Lone) Pairs
Lone pairs repel bonding pairs
more strongly than bonding pairs
repel each other.
H
larger EN
C
H
1220
Sn
Cl
Cl
950
O
The three molecular shapes of the tetrahedral electrongroup arrangement. Np=4.
Examples:
CH4, SiCl4,
SO42-, ClO4-
NH3
H 2O
PF3
OF2
ClO3
SCl2
H 3 O+
Lewis structures and molecular shapes.
The four molecular shapes of the trigonal bipyramidal
electron-group arrangement. Np=5.
PF5
SF4
AsF5
XeO2F2
SOF4
IF4+
IO2F2-
ClF3
XeF2
BrF3
I3 -
IF2-
The three molecular shapes of the octahedral electrongroup arrangement. Np=6.
SF6
IOF5
BrF5
TeF5
-
XeOF4
XeF4
ICl4-
A summary of common molecular shapes with two to six
electron groups.
The steps in determining a molecular shape.
Molecular
formula
Step 1
Lewis
structure
Step 2
Electron-group
arrangement
Count all e- groups around central
atom (A)
Step 3
Bond
angles
Note lone pairs and double
bonds
Count bonding and
Step 4
nonbonding egroups separately.
Molecular
shape
(AXmEn)
SAMPLE PROBLEM
PROBLEM:
Predicting Molecular Shapes with Two, Three,
or Four Electron Groups
Draw the molecular shape and predict the bond angles (relative
to the ideal bond angles) of (a) PF3 and (b) COCl2.
SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair
The shape is based upon the tetrahedral arrangement.
F
P
F
F
P
F
F
F
<109.50
The type of shape is
AX3E
The F-P-F bond angles should be <109.50 due
to the repulsion of the nonbonding electron
pair.
The final shape is trigonal pyramidal.
SAMPLE PROBLEM
Predicting Molecular Shapes with Two, Three,
or Four Electron Groups
continued
(b) For COCl2, C has the lowest EN and will be the center atom.
There are 24 valence e-, 3 atoms attached to the center atom.
Cl
C
O
Cl
The shape for an atom with three atom
attachments and no nonbonding pairs on the
central atom is trigonal planar.
O
O
C
Cl
C does not have an octet; a pair of nonbonding
electrons will move in from the O to make a
double bond.
Cl
The Cl-C-Cl bond angle will
be less than 1200 due to
the electron density of the
C=O.
124.50
C
Cl
1110
Cl
Type AX3
SAMPLE PROBLEM
PROBLEM:
SOLUTION:
Predicting Molecular Shapes with Five or Six
Electron Groups
Determine the molecular shape and predict the bond angles
(relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.
(a) SbF5 - 40 valence e-; all electrons around central
atom will be in bonding pairs; shape is AX5 - trigonal
bipyramidal.
F
F
F
F
Sb
F
F
F
Sb
F
F
F
(b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central
atom. Shape is AX5E, square pyramidal.
F
F
F
Br
F
F
SAMPLE PROBLEM
PROBLEM:
PLAN:
Predicting Molecular Shapes with More Than
One Central Atom
Determine the shape around each of the central atoms in
acetone, (CH3)2C=O.
Find the shape of one atom at a time after writing the Lewis
structure.
SOLUTION:
tetrahedral
H
H C
H
O
C
H
C H
H
tetrahedral
trigonal planar
O
H
C
H C
C
H
HH
>1200
H
<1200
The tetrahedral centers of ethane and ethanol.
ethane
ethanol
CH3CH3
CH3CH2OH
The orientation of polar molecules in an electric field.
Electric field OFF
Electric field ON
SAMPLE PROBLEM
PROBLEM:
Predicting the Polarity of Molecules
From electronegativity (EN) values (button) and their periodic
trends, predict whether each of the following molecules is polar
and show the direction of bond dipoles and the overall
molecular dipole when applicable:
(a) Ammonia, NH3
(b) Boron trifluoride, BF3
(c) Carbonyl sulfide, COS (atom sequence SCO)
PLAN: Draw the shape, find the EN values and combine the concepts to
determine the polarity.
SOLUTION:
(a) NH3
The dipoles reinforce each
other, so the overall
molecule is definitely polar.
ENN = 3.0
H
ENH = 2.1
N
H
H
H
N
H
H
bond dipoles
H
N
H
H
molecular
dipole
SAMPLE PROBLEM
Predicting the Polarity of Molecules
continued
(b) BF3 has 24 valence e- and all electrons around the B will be involved in
bonds. The shape is AX3, trigonal planar.
F
B
F
F
1200
F (EN 4.0) is more electronegative
than B (EN 2.0) and all of the dipoles
will be directed from B to F. Because
all are at the same angle and of the
same magnitude, the molecule is
nonpolar.
(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is
quite polar(DEN) so the molecule is polar overall.
S
C
O