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2
O
SAMPLE PROBLEM: Writing Resonance Structures
PROBLEM: Write resonance structures for the nitrate ion, NO
3
.
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
N
O O
O
N
O O
O
N
O
N does not have an octet; a pair of e will move in to form a double bond.

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O
O
N
O
O
O
N
O
O
O
N
O

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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
4. Avoid like charges (+ + or - - ) on adjacent atoms

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Formal charge of atom =
# valence e =
(# unshared electrons +
1/2 the # shared electrons)

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Nitric acid
H
..
..
N
..
O :
:
►
We will calculate the formal charge for each atom in this Lewis structure.

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8
Nitric acid
H
..
..
N
..
O :
:
Formal charge of H
►
►
►
►
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.

Nitric acid
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Formal charge of O
H
..
..
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.

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Nitric acid
Formal charge of O
H
..
..
N
..
O :
:
►
Electron count of O is 6.
►
A neutral oxygen has 6 electrons.
►
Therefore, the formal charge of oxygen is 0.

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Nitric acid
H
..
..
N
..
O :
:
Formal charge of 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.

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Nitric acid
H
..
..
N
..
O :
:
Formal charge of 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.

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Nitric acid
H
..
..
N
..
O :
:
Formal charge of N
–
►
Electron 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.

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Nitric acid
H
..
..
..
O :
N +
:
–
Formal charges
►
A Lewis structure is not complete unless formal charges (if any) are shown.

An arithmetic formula for calculating formal charge.
Formal charge =
Number of valence electrons
– number of bonds
– number of unshared electrons
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4
+
4
-
1
4
H
H
N
+
H
H
:
..
..
F
B
– ..
F :
..
4
7

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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

For O
B
# valence e = 6
# nonbonding e = 2
# bonding e = 6 X 1/2 = 3
Formal charge = +1
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B
O
O
For O
A
# valence e = 6
A
# nonbonding e = 4
# bonding e = 4 X 1/2 = 2
O
C
Formal charge = 0
For O
C
# valence e = 6
# nonbonding e = 6
# bonding e = 2 X 1/2 = 1
Formal charge = -1

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EXAMPLE: NCO has 3 possible resonance forms -
N C O
A
N C O
B
N C O
C

Now Determine Formal Charges
-2 0 +1
N C O
-1 0 0
N C O
0 0 -1
N C O
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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)

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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: SF
6
, PCl
5
, SO
2
, SO
3
, SO
4
5a. Electron deficient – have fewer than 8
 Ex: BeCl
2
, BF
3
 may attain an octet by coordinate covalent bond
Odd number of electrons – aka free radicals
 Ex: NO
2
 May attain an octet by pairing with another free radical

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SAMPLE PROBLEM: Writing Lewis Structures for
Exceptions to the Octet Rule.
PROBLEM:
PLAN:
SOLUTION:
Write the Lewis structure for BFCl
2
.
Draw the Lewis structures for the molecule and determine if there is an element which can be an exception to the octet rule.
BFCl
2 will have only
1 Lewis structure.
Cl
F
B
Cl

Let’s see if you can:
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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

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0
O
-1
O
1
N
O
-1
O
0
O
N
1
O
-1 -1
Is there a better structure??
No!!
-1
O
-1
O
N
1
O
0

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