Resonance and Formal Charge

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Resonance and Formal Charge

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Resonance and Formal Resonance and Formal Charge:Charge:

At the conclusion of our time At the conclusion of our time together, you should be able to:together, you should be able to:

1. Define resonance2. Determine resonance structures for a

molecule3. Calculate the formal charge for an

atom4. Determine the resonance structure that

contributes the most to a compound by using formal charge

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SAMPLE PROBLEM: Writing Resonance Structures

PLAN:

SOLUTION:

PROBLEM: 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.

Nitrate has 1(5) + 3(6) + 1 = 24 valence e-

N

O

O O

N

O

O O

N

O

O O

N does not have an octet; a pair of e- will move in to form a double bond.

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N

O

O O

N

O

O O

N

O

O 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

Four criteria for choosing the more important resonance structure:

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Formal charge of atom =

# valence e-

– # unshared electrons

– 1/2 the # shared electrons)

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Nitric acidNitric acidNitric acidNitric acid

.... ::

....HH OO

OO

OO

NN

::

::....

....

► We will calculate the formal charge for We will calculate the formal charge for each atom in this Lewis structure.each atom in this Lewis structure.

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10


.... ::

....HH OO

OO

OO

NN

::

::....

....

► Hydrogen shares 2 electrons with oxygen.Hydrogen shares 2 electrons with oxygen.► Assign 1 electron to H and 1 to O.Assign 1 electron to H and 1 to O.► A neutral hydrogen atom has 1 valence A neutral hydrogen atom has 1 valence

electron.electron.► Therefore, the formal charge of H in nitric Therefore, the formal charge of H in nitric

acid is 1 – 1 = 0.acid is 1 – 1 = 0.

Formal charge of HFormal charge of H

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.... ::

....HH OO

OO

OO

NN

::

::....

....

► Oxygen has 4 electrons in covalent bonds.Oxygen has 4 electrons in covalent bonds.► Assign 2 of these 4 electrons to O.Assign 2 of these 4 electrons to O.► Oxygen has 2 unshared pairs. Assign all 4 Oxygen has 2 unshared pairs. Assign all 4

of these electrons to O.of these electrons to O.► Therefore, the total number of electrons Therefore, the total number of electrons

assigned to O in the structure is 2 + 4 = 6.assigned to O in the structure is 2 + 4 = 6.

Formal charge of OFormal charge of O

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.... ::

....HH OO

OO

OO

NN

::

::....

....

► Electron count of O is 6.Electron count of O is 6.► A neutral oxygen has 6 valence A neutral oxygen has 6 valence

electrons.electrons.► Therefore, the formal charge of oxygen Therefore, the formal charge of oxygen

is is 6 – 6 = 0.6 – 6 = 0.


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.... ::

....HH OO

OO

OO

NN

::

::....

....

► Electron count of O is 6 (4 electrons from Electron count of O is 6 (4 electrons from unshared pairs + half of 4 bonded electrons).unshared pairs + half of 4 bonded electrons).

► A neutral oxygen has 6 valence electrons.A neutral oxygen has 6 valence electrons.► Therefore, the formal charge of oxygen is Therefore, the formal charge of oxygen is

6 – 6 = 0.6 – 6 = 0.


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.... ::

....HH OO

OO

OO

NN

::

::....

....

► Electron count of O is 7 (6 electrons from Electron count of O is 7 (6 electrons from unshared pairs + half of 2 bonded electrons).unshared pairs + half of 2 bonded electrons).

► A neutral oxygen has 6 valence electrons.A neutral oxygen has 6 valence electrons.► Therefore, the formal charge of oxygen is Therefore, the formal charge of oxygen is

6 – 7 = -1.6 – 7 = -1.


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.... ::

....HH OO

OO

OO

NN

::

::....

....

► Electron count of N is 4 (half of 8 electrons Electron count of N is 4 (half of 8 electrons in covalent bonds).in covalent bonds).

► A neutral nitrogen has 5 valence electrons.A neutral nitrogen has 5 valence electrons.► Therefore, the formal charge of N is Therefore, the formal charge of N is

5 – 4 = +1.5 – 4 = +1.

Formal charge of NFormal charge of N

––

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.... ::

....HH OO

OO

OO

NN

::

::....

....

► A Lewis structure is not complete unless A Lewis structure is not complete unless formal charges (if any) are shown.formal charges (if any) are shown.

Formal chargesFormal charges

––

++

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....

BBFF

FF

FF

FF

....

............:: ::

:: ::

:: ::

....

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44

NN

HH

HH HH

HH

––

"Electron Counts""Electron Counts" and Formal and Formal Charges in NHCharges in NH44

+ + and BFand BF44--

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44

++

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Formal charge of atom =

# valence e-

– # unshared electrons

– 1/2 the # shared electrons)

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Formal ChargeFormal ChargeFormal ChargeFormal Charge

Formal charge = Formal charge =

Number of Number of

valencevalence

electronselectrons

number ofnumber ofbondsbonds

number ofnumber ofunshared electronsunshared electrons

–– ––

An arithmetic formula for calculating formal charge.An arithmetic formula for calculating formal charge.

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OO O

A

B

CFor OA # valence e- = 6

# nonbonding e- = 4

# bonding e- = 4 X 1/2 = 2

Formal charge = 0

For OB

# valence e- = 6

# nonbonding e- = 2

# bonding e- = 6 X 1/2 = 3

Formal charge = +1For OC

# 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

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N C O N C O N C O

Now Determine Formal Charges

-2 0 +1 -1 0 0 0 0 -1

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

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Exceptions to the Octet RuleExceptions to the Octet Rule

► 4b. Expanded Octets – only on period 3 and 4b. Expanded Octets – only on period 3 and higher higher

Expanded octets form when an atom can Expanded octets form when an atom can decrease (or maintain at 0) it’s formal chargedecrease (or maintain at 0) it’s formal charge

Ex: SFEx: SF66, PCl, PCl55, SO, SO22, SO, SO33, SO, SO44

► 5a. Electron deficient – have fewer than 85a. Electron deficient – have fewer than 8 Ex: BeClEx: BeCl22, BF, BF33 may attain an octet by coordinate covalent bondmay attain an octet by coordinate covalent bond

► Odd number of electrons – aka free radicalsOdd number of electrons – aka free radicals Ex: NOEx: NO22 May attain an octet by pairing with another free May attain an octet by pairing with another free

radicalradical

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SAMPLE PROBLEM: Writing Lewis Structures for Exceptions to the Octet Rule.

PLAN:

SOLUTION:

PROBLEM: Write the Lewis structure for BFCl2.

Draw the Lewis structures for the molecule and determine if there is an element which can be an exception to the octet rule.

BFCl2 will have only 1 Lewis structure.

F

BCl Cl

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Resonance and Formal Resonance and Formal Charge:Charge:

Let’s see if you can:Let’s see if you can:

1. Define resonance2. Determine resonance structures for a

molecule3. Calculate the formal charge for an

atom4. Determine the resonance structure that

contributes the most to a compound by using formal charge

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Now determine the formal charges and best structure for the 2 examples at the bottom of page 9.

Your Turn

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Now determine the formal charges and best structure for the middle example on page 13.

Your Turn

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N

O

O O

N

O

O O

N

O

O O

1

-1

0

-1

1 1

-1

-1

00

-1

-1

Is there a better structure??

No!!

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The base is under assault!!!

Date post:	23-Jan-2016
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Resonance and Formal Charge

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