ch11-2web-Revised

Chem 5 Chapter 11 Chemical Bonding I: Basic Concepts Part 2 November 18, 2002

We covered the following concepts last time:

Lewis structure - the octet rule Ionic bond and covalent bond Polar bond, electronegativity and dipole moment Molecular shapes - VSEPR Bond length and bond energy

4

(g)

6 N≡ N (g) + 12 O=C=O (g) + 10 H-O-H (g) + O=O (g)

∆H = ∆H (bond breakage) + ∆H (bond formation) = ΣBE (reactants) - ΣBE (products) = 4x(5xBEC-H +2BEC-C +3BEC-O+6BEN-O +3BEN=O) - (6xBEN=N+24xBEC=O+20xBEO-H+BEO=O)

To predict molecular shape with VSEPR An example of a polyatomic molecule with no single central atom: CH3CH2OH

ETHANOL

Apply VSEPR to each atom with more than one electron pair.

H

H

H

C

C

H

H

O

H

ETHANOL

C2H5OH

H O

H

The atoms around the carbons form a

C

tetrahedral arrangement.

C H

H

The atoms around the oxygen form a

H H

V-shaped structure.

The shape matters!

Ethanol Dehydrogenase An enzyme that catalyzes the hydrolysis of ethanol. O

Ethanol

CH3CH2OH Æ CH3C-H A close look at the active site

The structure of ethanol dehydrogenase

New concepts today

Formal charge Resonance Exceptions to the octet rule Ozone destruction and polymers

Two possible Lewis structures for HCN H

C

N

Why not

H

N

C

Two possible Lewis structures for formaldehyde Why not

H C H

H

O

O H

C

FORMAL CHARGE gives an indication of the extent to which atoms have gained or lost electrons in the process of covalent bond formation. Each atom is assigned all of its lone electrons and half of the bonding electrons. Formal #valence _ charge = electrons

{

#unshared electrons

_

} {

1/2#shared electrons

}

Structures with the lowest formal charges are the most stable.

All possible Lewis structures with stable electronic configurations for HCN and HNC. H Formal #valence _ charge = electrons

C

{

#unshared electrons

_

} { H

FC on C

=4 -0

- 8/2

=0

FC on N

=5 -2

- 6/2

=0

FC on H

=1 -0

- 2/2

=0

1/2#shared electrons

C

} N

..

Calculate formal charge for

N

All possible Lewis structures with stable electronic configurations for HNC. H Formal #valence _ charge = electrons

N

{

C

#unshared electrons

} {

1/2#shared electrons

H

FC on C

=4 -2

- 6/2

= -1

FC on N

=5 -0

- 8/2

= +1

FC on H

=1 -0

- 2/2

=0

N

} C

..

Calculate formal charge for

_

Four Rules for Formal Charge • The sum of F.C. must be the net charge of the

molecule or ion. • F.C. should be as small as possible. • Negative F.C. usually occurs on the most electronegative atoms. • F.C. of the same sign on adjacent atoms is unlikely.

All possible Lewis structures with stable electronic configurations for HCN and HNC. H

0

0

C

0

N

H

N

0

+1

C

-1

FORMAL CHARGES We choose the structure based on F.C. of each atom being zero The most electronegative atom (N) should not have positive F.C. It should be H C

N

Another Example: H2CO H C

vs.

H2OC H

O

H

O

C

H

FC on C =4 – 0 – 2x2/2 – 4/2 = 0 FC on O = 6 – 0 – 2x2/2 –4/2 = 2 FC on O = 6 – 4 – 4/2 = 0

FC on C = 4 – 4 – 4/2 = - 2

Not stable!

Let’s look at the carbonate anion CO32Count up valence electrons C has four

1s22s22p2

O has six

1s22s22p4

Plus two extra for negative charge Valence electrons =

4 + 3 x 6 + 2 = 24

Carbonate anion CO32Put a pair between each atom 24 O valence electrons O O C 18 left Add remaining electrons to terminal atoms to complete octets...

Carbonate anion CO3 2-

-

2

24 valence electrons

O O

C

DO NOT FORGET CHARGE!!!!

O

The oxygen atoms have their octet but... The carbon atom does not! So we form a double bond by sharing a pair from one of the oxygen atoms...

Carbonate ion CO3

22-

24 valence electrons

O O

C

O

Form a double bond by sharing a pair from one of the oxygen atoms...

-

-

2

2

O

O O

C

O

O

C

O But…

FORM A DOUBLE BOND BETWEEN O AND C 2-

Here is one O O

O

C

Here is another!

Here is another!

2-

2-

O O

C

O O

All three are perfect!

O

C

O

But, none alone is correct!

Because experiment shows all three bonds are the same. All three bond lengths the same! Longer than double bonds,

O

Shorter than single bonds

C O

And bond angles 120°

2

O

RESONANCE We use a double headed arrow between the structures.. 2

O

O

C O

2

O

C O

O

2

C O

O

O

The electrons involved are said to be DELOCALIZED over the structure. The blended structure is a RESONANCE HYBRID We interpret the experimental structure with a partial double bond. The bond order is (2+2x1)/3 =1.33.

Another example of resonance structure: O3

O

LEWIS STRUCTURE Make double bond...

O

O O

O

O O

O

O

Experiment shows that both O-O bonds are equivalent.

The O3 molecule is a hybrid of the two resonance forms.

Ozone is generated by photochemistry in the stratosphere (15-60 Km), forming a protective layer for all life on earth.

Ozone absorption spectrum

hν O2 (g) → 2 O (g) O2 (g) + O (g) → O3 (g)

• O3 absorbs UV light 220-300 nm • N2 and O2 do not absorb at those wavelengths • O3 layer protects DNA from photodamage

Ozone Reduction

The ozone destruction is related to human activity! chloroflorocarbon (CFC)

hν → Cl

Cl + O3 = ClO + O2 ClO + O = Cl + O2 The Heroes

Sherwood Rowland

O3 + O = 2 O2

Mario Molina

The billion-dollar CFC industry and some government agencies attacked them viciously, even tried to get them fired from their professor posts. However, the scientific community provided moral support and overwhelming experimental evidence. The truth prevailed! Rowland and Molina were awarded the Nobel Prize in 1995.

International Ban on CFC

CFC is replaced by fluorocarbon. The ozone level is expected to recover in 50 years.

Professor Jim Anderson to teach Chem7

The flying Chem Lab

EXCEPTIONS TO THE OCTET RULE…

• Molecules with more than an octet around the central atom • Molecules with less than an octet around the central atom • Molecules with unpaired electrons

A CENTRAL ATOM WITH MORE THAN AN OCTET Elements in the third or higher periods can exceed the octet due to d or f orbitals

EXAMPLE : PF5

F

F

F

P F

F

Five electron pairs around the phosphorus atom.

PF5 P

Bond angle F

AXIAL

900 F

EQUATORIAL

P

F F

1200

F

The shape of PF5 is trigonal bipyramidal.

Other examples based on 5 pairs of electrons… All with empty d orbitals

SF4

ClF3

XeF2

1 lone pair

2 lone pairs

3 lone pairs

F S

F F

F See-saw shaped

F

F

Cl

Xe F

F T-shaped

F F Linear

Lone pairs occupy the trigonal plane (the “equator”) first to minimize the number of 90° repulsions

Another Example I3

I

I

Number of valence electrons: Add electrons...

I

7+7+7+1=22

4 used in two bonds

I

I

I

I I F I

Linear shape

trigonal bipyramidal with three lone pairs in the equator

Draw the Lewis structure for SF6 Total number of valence electrons = 6 + 6 x 7 = 48 Why? F has seven 1s22s22p5 S has six

1s22s22p63s22p4

Place S in middle of 6 fluorine atoms

F

F

F

S F

F

F

There are six electron pairs around the sulfur atom.

Shape of SF6 All bond angles 900 F F F

S

F F

F 6 electron pairs

octahedral

A CENTRAL ATOM WITH LESS THAN AN OCTET

..

.. F ..

B

..

..

. F. ..

.. F .. ..

..

.. F ..

.. F .. ..

B

-1

.. F .. Not stable

F.C.

..

F ..

with two resonance structures

+1 F.C.

..

B

+

.. F .. Ionic

_

..

. F .. . ..

with two resonance structures

FREE RADICALS Have unpaired electrons. NO2

Is a free radical

Total number of valence electrons=5+6+6=17 O

N

O

Form double bond to get N close to octet O

N

O

O RESONANCE

N

O

Summary Formal Charge: apparent charges on certain atoms in a Lewis structure that arise when atoms have not contributed equal numbers of electrons to the covalent bonds joining them - Four rules to select Lewis structures based on F.C. Resonance Structure: More than one plausible Lewis structure can be written for a species. The true structure is a resonance hybrid of two or more contributing structures. Exceptions to the Octet Rule At times, a molecule may have unpaired electrons or the valence shell of the central atom must be expanded to 10 (trigonal bipyramidal) or 12 electrons (octahedral).