Chem 5Chapter 11

Chemical Bonding I:Basic Concepts

Part 2

November 18, 2002

We covered the following concepts last time:

Lewis structure - the octet ruleIonic bond and covalent bondPolar bond, electronegativity and dipole moment Molecular shapes - VSEPRBond length and bond energy

(g) 6 N N (g) + 12 O=C=O (g)

+ 10 H-O-H (g) + O=O (g)

≡4

∆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:

ETHANOL

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

OCH H

H

H

C

H

H

CH3CH2OH

HH

H

C

H

HC

H

O

The atoms around the carbons form a

tetrahedral arrangement.

The atoms around the oxygen form a

V-shaped structure.

ETHANOL C2H5OH

The structure of ethanol dehydrogenase

Ethanol

The shape matters!

Ethanol Dehydrogenase

OAn enzyme that catalyzes the hydrolysis of ethanol.

CH3CH2OH CH3C-H

A close look at the active site

New concepts today

Formal chargeResonanceExceptions to the octet rule

Ozone destruction and polymers

Two possible Lewis structures for HCN

H N CH C N Why not

Two possible Lewis structures for formaldehyde

OH

H

C

Why not

CH

H

O

gives an indication of the extent to which atomsFORMAL CHARGE

have gained or lost electrons in the process of covalent bond formation.Each atom is assigned all of its lone electrons and

Formal charge = #valence

electrons_ #unshared

electrons_ 1/2#shared

electrons{ } { }half of the bonding electrons.

Structures with the lowest formal charges are the most stable.

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

H C N

Calculate formal charge for

_ 1/2#shared electrons{ }Formal

charge = #valence electrons

_ #unshared electrons{ }

FC on C = 4 - 0 - 8/2 = 0

H C N

..

FC on N = 5 - 2 - 6/2 = 0

FC on H = 1 - 0 - 2/2 = 0

All possible Lewis structures with stable electronic configurations for HNC.

H N C

Calculate formal charge for

FC on C = 4 - 2 - 6/2 = -1

_ 1/2#shared electrons{ }Formal

charge = #valence electrons

_ #unshared electrons{ }

H N C

..

FC on N = 5 - 0 - 8/2 = +1

FC on H = 1 - 0 - 2/2 = 0

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.

0 0 0H C N

0 +1 -1H N C

FORMAL CHARGES

We choose the structure based onF.C. of each atom being zero

The most electronegative atom (N) should not have positive F.C.

It should beH C N

Another Example:

H2CO vs. H2OC

H

HOC

H

HCO

FC on C =4 – 0 – 2x2/2 – 4/2 = 0

FC on O = 6 – 4 – 4/2 = 0

FC on O = 6 – 0 – 2x2/2 –4/2 = 2

FC on C = 4 – 4 – 4/2 = - 2

Not stable!

Let’s look at the carbonate anion CO32-

C has four 1s22s22p2

O has six 1s22s22p4

Count up valence electrons

Plus two extra for negative charge

Valence electrons = 4 + 3 x 6 + 2 = 24

Put a pair between each atomCarbonate anion CO3

2-

C

O

O O

24 valence electrons

18 left

Add remaining electrons to terminal atoms to complete octets...

Carbonate anion CO3 2-

24 valence electrons

2-

C

O

O O

DO NOT FORGET CHARGE!!!!

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 CO32-

2-

C

O

O O

24 valence electrons

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

2-

C

O

O OBut…

C

O

O O

2-

FORM A DOUBLE BOND BETWEEN O AND C

C

O

O O

2-

Here is another!

Here is one

Here is another!

C

O

O O

2-

C

O

O O

2-

All three are perfect! But, none alone is correct!

Because experiment shows all three bonds are the same.

All three bond lengths the same!

Longer than double bonds,

Shorter than single bondsO O

O

C

And bond angles 120°

2

RESONANCEWe use a double headed arrow between the

structures..

The electrons involved are said to be DELOCALIZED over the structure.

The blended structure is a RESONANCE HYBRID

O O

O

CCO O

O2 2

CO O

O2

We interpret the experimental structure with a partial double bond. The bond order is (2+2x1)/3 =1.33.

Another example of resonance structure:

OO O

LEWIS STRUCTUREO3Make double bond...

OO O

OO 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 layerfor all life on earth.

Ozone absorption spectrum

O2 (g) → 2 O (g)hν

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!

Cl + O3 = ClO + O2

ClO + O = Cl + O2

O3 + O = 2 O2

chloroflorocarbon (CFC) → Cl hν

The Heroes

Sherwood Rowland Mario Molina

The billion-dollar CFC industry and some government agencies attackedthem 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 Andersonto 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 : PF5F

PFF

F F

P

PFF

F

F

FBond angle900

1200

The shape of PF5 is trigonal bipyramidal.

PF5

AXIAL

EQUATORIAL

Five electron pairs around the phosphorus atom.

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

ClF3

2 lone pairs

T-shaped

ClF

F

F

SF4

1 lone pair

See-saw shaped

SF

F

F

F

XeF2

3 lone pairs

Linear

F

F

F

Xe

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

Another Example I3-

I I INumber of valence electrons: 7+7+7+1=22

Add electrons... 4 used in two bonds

F

I

I

I

-I II

Linear shapetrigonal bipyramidalwith three lone pairs in the equator

Draw the Lewis structure for SF6

6 + 6 x 7 = 48Total number of valence electrons =Why? F has seven 1s22s22p5

1s22s22p63s22p4S has six

Place S in middle of 6 fluorine atoms

F

S

F

F

F

F

F

There are six electron pairs around the sulfur atom.

Shape of SF6

All bond angles 900

S

F

F

F

F

F

F

6 electron pairs octahedral

A CENTRAL ATOM WITH LESS THAN AN OCTET..

BF

F

..

....F..

. ...

....

BF

F

..

. ...

..+1

-1

..

F.C.

..F....

BF

F

..

..

..

+

...._

. ...

with tworesonancestructures

with tworesonancestructures..

F....F.C.Not stable Ionic

FREE RADICALSHave unpaired electrons.

NO2 Is a free radical

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

O N O

RESONANCE

Form double bond to get N close to octet

O N O O N O

SummaryFormal 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 morecontributing structures.

Exceptions to the Octet RuleAt 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).