The Shapes of Molecules
1
Lewis Structures
2
The Lewis Model of Chemical
Bonding
In 1916 G. N. Lewis proposed that atoms
combine in order to achieve a more stable
electron configuration.
►
Maximum stability results when an atom
is isoelectronic with a noble gas.
►
An electron pair that is shared between
two atoms constitutes a covalent bond.
►
3
Covalent Bonding in
H2
Two hydrogen atoms, each with 1 electron,
H.
.H
can share those electrons in a covalent bond.
H: H
► Sharing
the electron pair gives each hydrogen
an electron configuration analogous to helium.
4
Covalent Bonding in
F2
Two fluorine atoms, each with 7 valence electrons,
..
..
. F:
: ..F .
..
can share those electrons in a covalent bond.
.. ..
: ..
F : ..
F:
► Sharing
the electron pair gives each fluorine
an electron configuration analogous to neon.
5
The Octet Rule
In forming compounds, atoms gain, lose, or
share electrons to give a stable electron
configuration characterized by 8 valence
electrons.
.. ..
: ..
F : ..
F:
► The
octet rule is the most useful in cases
involving covalent bonds to C, N, O, and F.
6
Example
Combine carbon (4 valence electrons) and
four fluorines (7 valence electrons each)
.
..
.
. C.
:
F
..
.
to write a Lewis structure for CF4.
..
.. : ..F: ..
:
:
F
: ..F: C
..
..
: ..F:
The octet rule is satisfied for carbon and
each fluorine.
7
Example
It is common practice to represent a covalent
bond by a line. We can rewrite
..
.. : ..F: ..
:
: ..F: C
.. : ..F
: ..F:
..
: F:
as
..
: ..
F
C
: ..F:
8
..
..F:
Lewis Structures with
Double and Triple Bonds
9
Inorganic examples
..
..
: O: : C : : O:
..
:O
C
..
O:
C
N:
Carbon dioxide
H : C : :: N:
H
Hydrogen cyanide
10
Organic examples
H
.. H
..
H: C : : C:H
H
Ethylene
H
C
H
H
H : C : :: C:H
11
Acetylene
H
C
C
C
H
Rules for Lewis Structures
►
12
1. Make certain that the bond is a covalent bond
then set up the skeleton structure as follows:
 The atom with the lowest E.N. will tend to go in
middle
 Place all the other atoms around this central atom
 Attach these atoms to the central atom in
reasonable fashion with single bonds
Rules for Lewis Structures
2. Sum valence electrons (adding for negative charge
and subtracting for positive)
► 3. Complete octets of peripheral atoms
► 4. Place leftover e- on central atom
► 5. If necessary use multiple bonds to fill center
atom's octet unless the central atom is a metal close
to the metalloids.
►
13
Sum of
valence e-
:
: F:
: F:
:
Atom
placement
For NF3
:
Molecular
formula
N
Remaining
valence eLewis
structure
14
:
: F:
N
5e-
F 7e- X 3 = 21eTotal
26e-
SAMPLE PROBLEM:
Writing Lewis Structures for Molecules with
One Central Atom
PROBLEM:
Write a Lewis structure for CCl2F2, one of the compounds
responsible for the depletion of stratospheric ozone.
PLAN:
Follow the steps outlined previously
Cl
Cl C
F
SOLUTION:
F
:
:Cl C
:
: F:
Make bonds and fill in remaining valence
electrons placing 8e- around each atom.
15
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-.
:
Step 1: Carbon has the lowest EN and is the central atom.
The other atoms are placed around it.
SAMPLE PROBLEM:
PROBLEM:
PLAN:
Writing Lewis Structures for Molecules with
Multiple Bonds.
Write Lewis structures for the following:
Nitrogen (N2), the most abundant atmospheric gas
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 e- can be moved in to form a multiple bond,
except for which atoms?
SOLUTION:
N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make the
octet around each N.
N
.
:
N
.
:
:
:.
.:
16
N
.
N
N
:
N
.
Resonance and Formal
Charge
17
SAMPLE PROBLEM:
PROBLEM:
PLAN:
Nitrate has 1(5) + 3(6) + 1 = 24 valence e-
O
O
O
N
N
N
O
O
O
18
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
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
Resonance and Formal Charge
Formal charge of atom =
# valence e- = (# unshared electrons + 1/2 # shared
electrons)
Three criteria for choosing the more important
resonance structure:
1. Smaller formal charges (either positive or negative) are
preferable to larger charges;
2. Avoid like charges (+ + or - - ) on adjacent atoms;
3. A more negative formal charge should exist on an atom with a
larger EN value.
19
Nitric
acid
H
..
O
..
..
O:
N
:O
.. :
► We
will calculate the formal charge for
each atom in this Lewis structure.
20
Nitric
acid
Formal charge of H
H
..
O
..
..
O:
N
:O
.. :
► Hydrogen
shares 2 electrons with oxygen.
► Assign 1 electron to H and 1 to O.
► A neutral hydrogen atom has 1 electron.
► Therefore, the formal charge of H in nitric
acid is 0.
21
Nitric
acid
Formal charge of O
..
H O
..
..
O:
N
:O
.. :
► Oxygen
has 4 electrons in covalent bonds.
► Assign 2 of these 4 electrons to O.
► Oxygen has 2 unshared pairs. Assign all 4 of
these electrons to O.
► Therefore, the total number of electrons
assigned to O is 2 + 4 = 6.
22
Nitric
acid
Formal charge of O ..
H O
..
..
O:
N
:O
.. :
► Electron
count of O is 6.
► A neutral oxygen has 6 electrons.
► Therefore, the formal charge of oxygen is 0.
23
Nitric
acid
H
..
O
..
..
O:
Formal charge of O
N
:O
.. :
► Electron
count of O is 6 (4 electrons from
unshared pairs + half of 4 bonded electrons).
► A neutral oxygen has 6 electrons.
► Therefore, the formal charge of oxygen is 0.
24
Nitric
acid
H
..
O
..
..
O:
N
Formal charge of O
:
:O
..
► Electron
count of O is 7 (6 electrons from
unshared pairs + half of 2 bonded electrons).
► A neutral oxygen has 6 electrons.
► Therefore, the formal charge of oxygen is -1.
25
Nitric
acid
H
► Electron
..
O
..
..
O:
N
Formal charge of N
–
:
:O
..
count of N is 4 (half of 8 electrons in
covalent bonds).
► A neutral nitrogen has 5 electrons.
► Therefore, the formal charge of N is +1.
26
Nitric
acid
Formal charges
H
►A
..
O
..
..
O:
N+
–
:
:O
..
Lewis structure is not complete unless
formal charges (if any) are shown.
27
Formal Charge
An arithmetic formula for calculating formal charge.
Formal charge =
Number of
valence
electrons
28
number of
number of
–
–
bonds
unshared electrons
"Electron Counts" and Formal
Charges in NH4+ and BF4-
1
H
+
H
4
29
N
H
H
..
: F:
..
– ..
: ..
F B ..F:
: ..F:
7
4
Formal Charge: Selecting the Best Resonance Structure
An atom “owns” all of its nonbonding electrons and half of its bonding electrons.
Formal charge of atom =
# valence e- - (# unshared electrons + 1/2 # shared electrons)
B
For OA
# valence
e-
O
# nonbonding
# bonding
e-
# valence e- = 6
O
=6
e-
For OC
=4
= 4 X 1/2 = 2
Formal charge = 0
A
For OB
O
# nonbonding e- = 6
C
# valence
# bonding e- = 2 X 1/2 = 1
Formal charge = -1
e-
=6
# nonbonding e- = 2
# bonding e- = 6 X 1/2 = 3
Formal charge = +1
30
Resonance and Formal Charge
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 important 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.
31
Exceptions to the Octet
Rule
32
Exceptions to the Octet Rule
► Electron
deficient – have fewer than eight
 Ex: BeCl2, BF3
 may attain an octet by coordinate covalent bond
► Odd
number of electrons – aka free radicals
 Ex: NO2
 May attain an octet by pairing with another free
radical
► Expanded
33
Octets – only on period 3 and
higher
 Expanded octets form when an atom can decrease
(or maintain at 0) it’s formal charge
 Ex: SF6, PCl5, SO2, SO3, SO4
SAMPLE PROBLEM:
Writing Lewis Structures for
Exceptions to the Octet Rule.
PROBLEM:
Write the Lewis structure for BFCl2.
PLAN:
Draw the Lewis structures for the molecule and
determine if there is an element which can be an
exception to the octet rule.
SOLUTION:
BFCl2 will have only 1
Lewis structure.
F
B
Cl
34
Cl
VSEPR Theory
35
Molecular Shapes
Lewis structures show which atoms are
connected where, and by how many
bonds, but they don't properly show 3D shapes of molecules.
To find the actual shape of a molecule,
first draw the Lewis structure, and then
use VSEPR Theory.
36
Valence Shell Electron-Pair Repulsion
Theory or VSEPR
►
►
►
37
Molecular Shape is determined by the repulsions of
electron pairs
 Electron pairs around the central atom stay as far
apart as possible.
Electron Pair Geometry - based on number of regions
of electron density
 Consider non-bonding (lone pairs) as well as
bonding electrons.
 Electron pairs in single, double and triple bonds
are treated as single electron clouds.
Molecular Geometry - based on the electron pair
geometry, this is the shape of the molecule
Electron-group Repulsions And The Five Basic
Molecular Shapes.
38
The Single Molecular Shape Of The Linear Electrongroup Arrangement.
Examples:
CS2, HCN, BeF2
39
The Two Molecular Shapes Of The Trigonal Planar
Electron-group Arrangement.
Class
Examples:
SO2, O3, PbCl2, SnBr2
Shape
Examples:
SO3, BF3, NO3-, CO32-
40
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
real
Sn
Cl
Cl
950
41
C
H
greater
electron
density
Effect of Nonbonding Pairs
Lone pairs repel bonding
pairs more strongly than
bonding pairs repel each
other
H
larger EN
C
H
1220
O
The Three Molecular Shapes Of The Tetrahedral
Electron-group Arrangement.
Examples:
CH4, SiCl4,
SO42-, ClO4-
NH3
H 2O
PF3
OF2
ClO3
SCl2
H 3 O+
42
The Four Molecular Shapes Of The Trigonal Bipyramidal
Electron-group Arrangement.
PF5
SF4
AsF5
XeO2F2
SOF4
IF4+
IO2F2-
ClF3
XeF2
BrF3
I3 -
IF2-
43
The Three Molecular Shapes Of The Octahedral
Electron-group Arrangement.
SF6
IOF5
BrF5
TeF5
-
XeOF4
44
XeF4
ICl4-
Sample Problems
45
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)
46
Review of Lewis Structures
47
►
Step 1:
Count the number of valence electrons.
For a neutral molecule this is equal to
the number of valence electrons of the
constituent atoms.
►
Example (CH3NO2):
Each hydrogen contributes 1 valence
electron. Each carbon contributes 4,
nitrogen 5, and each oxygen 6 for a total
of 24.
Review of Lewis Structures
►
Step 2:
Connect the atoms by a covalent bond
represented by a dash.
►
Example:
Methyl nitrite has the partial structure:
H
H
C
H
48
O
N
O
Review of Lewis Structures
►
Step 3:
Subtract the number of electrons in
bonds from the total number of valence
electrons.
►
49
Example:
24 valence electrons – 12 electrons in
bonds. Therefore, 12 more electrons to
assign.
Review of Lewis Structures
►
Step 4: Add electrons in pairs so that as
many atoms as possible have 8
electrons. Start with the most
electronegative atom.
►
Example: The remaining 12 electrons in
methyl nitrite are added as 6 pairs.
H
H
C
H
50
..
O
..
N
..
..
:
O
..
Review of Lewis Structures
►
Step 5: If an atom lacks an octet, use
electron pairs on an adjacent atom to
form a double or triple bond.
►
Example: There are 2 ways this can be
done.
H
H
C
H
51
H
O
..
N
..
..
:
O
..
H
C
H
..
O
..
N
..
..
O:
Review of Lewis Structures
►
Step 6: Calculate formal charges.
►
Example: The left structure has formal
charges that are greater than 0.
Therefore it is a less stable Lewis
structure.
H
H
C
H
52
+
O
..
N
..
.. –
O
.. :
H
H
C
H
..
O
..
N
..
..
O:
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
53
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
(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
54
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
55
Br
F
F