Resonance and Formal
Charge
1
Resonance and Formal Charge:
At the conclusion of our time together,
you should be able to:
1. Define resonance
2. Determine resonance structures for a
molecule
3. Calculate the formal charge for an atom
4. Determine the resonance structure that
contributes the most to a compound by
using formal charge
2
Redneck Jet Skiing
3
SAMPLE PROBLEM:
Writing Resonance Structures
PROBLEM:
Write resonance structures for the nitrate
ion, NO3-.
PLAN:
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: Nitrate has 1(5) + 3(6) + 1 = 24
valence e-
O
4
O
O
O
N
N
N
O
O
O
O
O
N does not
have an octet;
a pair of e- will
move in to
form a double
bond.
O
5
O
O
O
N
N
N
O
O
O
O
O
Four criteria for choosing the more
important resonance structure:
1. Smaller formal charges (either positive or
negative) are preferable to larger charges;
2. A more negative formal charge should exist
on an atom with a larger EN value.
3. Get unlike charges as close together as
possible
6
4. Avoid like charges (+ + or - - ) on adjacent
atoms
Resonance and Formal
Charge
Formal charge of atom =
# valence e- =
(# unshared electrons +
1/2 the # shared electrons)
7
Nitric acid
H
..
O
..
..
O:
N
:O
.. :
► We
will calculate the formal charge for each
atom in this Lewis structure.
8
9
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.
10
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.
11
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.
12
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.
13
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.
14
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.
15
Nitric acid
Formal charges
H
►A
..
O
..
..
O:
N+
–
:
:O
..
Lewis structure is not complete unless
formal charges (if any) are shown.
16
Sure!! And I suppose the
Pope is Jewish???
17
Formal Charge
An arithmetic formula for calculating formal charge.
Formal charge =
Number of
valence
electrons
18
number of
number of
–
–
bonds
unshared electrons
"Electron Counts" and Formal
Charges in NH4+ and BF4-
1
H
+
H
4
19
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
20
For OB
# valence e- = 6
# nonbonding e- = 2
# bonding e- = 6 X 1/2 = 3
Formal charge = +1
B
O
O
For OA
A
# valence e = 6
# nonbonding e- = 4
# bonding e- = 4 X 1/2 = 2
Formal charge = 0
21
O
C
For OC
# valence e- = 6
# nonbonding e- = 6
# bonding e- = 2 X 1/2 = 1
Formal charge = -1
Resonance and Formal Charge
EXAMPLE: NCO- has 3 possible resonance
forms -
N C
A
22
O
N C
B
O
N C
C
O
Now Determine Formal Charges
-2
0
N C
+1
O
-1
0
N C
0
0
0
O
N C
-1
O
Forms B and C have smaller formal charges;
this makes them more important than form A.
(rule 1)
Form C has a negative charge on O which is the
more electronegative element, therefore C
contributes the most to the resonance hybrid.
(rule 2)
23
Redneck NASCAR Racing
24
Exceptions to the Octet
Rule
25
Exceptions to the Octet Rule
► 4b.
Expanded 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
► 5a.
Electron deficient – have fewer than 8
 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
26
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
27
Cl
Resonance and Formal Charge:
Let’s see if you can:
1. Define resonance
2. Determine resonance structures for a
molecule
3. Calculate the formal charge for an atom
4. Determine the resonance structure that
contributes the most to a compound by
using formal charge
28
Your Turn
Now determine the formal charges
and best structure for the 2
examples at the bottom of page 11.
29
Your Turn
Now determine the formal charges
and best structure for the middle
example on page 12.
30
-1
0
-1
O
O
O
N
O
0
1
N
1
N
O
O
O
-1
-1
-1
Is there a better structure??
No!!
31
1
O
O
-1
0
The base is under
assault!!!
32