10-Ch9(ikatan kimia).ppt

CHEMICAL CHEMICAL BONDINGBONDING

CocaineCocaine

Water

22

Chapter 9 OutlineChapter 9 Outline

• Ionic -vs- CovalentIonic -vs- Covalent

• Bonding in moleculesBonding in molecules

• Valance electronsValance electrons

• Lewis StructuresLewis Structures– Oxidation NumbersOxidation Numbers

– Formal ChargesFormal Charges

• VESPRVESPR

Chemical BondingChemical BondingProblems and questions —Problems and questions —

How is a molecule or polyatomic ion How is a molecule or polyatomic ion held together?held together?

Why are atoms distributed at strange Why are atoms distributed at strange angles?angles?

Why are molecules not flat?Why are molecules not flat?

Can we predict the structure?Can we predict the structure?

How is structure related to chemical How is structure related to chemical and physical properties.and physical properties.

Forms of Chemical BondsForms of Chemical Bonds• There are 2 extreme forms of There are 2 extreme forms of

connecting or bonding atoms:connecting or bonding atoms:

• IonicIonic—complete transfer of —complete transfer of electrons from one atom to electrons from one atom to anotheranother

• CovalentCovalent—electrons shared —electrons shared between atomsbetween atoms

• Most bonds are somewhere in Most bonds are somewhere in between.between.

Ionic BondsIonic BondsComplete electron Complete electron

transfer from an transfer from an

element of low element of low IEIE (metal)(metal)

to an element of high to an element of high EAEA (nonmetal)(nonmetal)

2 Na(s) + Cl2 Na(s) + Cl22(g) ---> 2 Na(g) ---> 2 Na++ + 2 Cl + 2 Cl--

results in a results in a metalmetal//nonmetalnonmetal or ionic compound. or ionic compound.

Covalent BondingCovalent Bonding

Covalent bond forms by the sharing ofCovalent bond forms by the sharing of

VALENCE ELECTRONSVALENCE ELECTRONS, between , between

two nonmetalstwo nonmetals..

Valence ElectronsValence Electrons

Electrons are divided between Electrons are divided between corecore and and valence valence electronselectrons..

Na 1sNa 1s22 2s 2s22 2p 2p66 3s 3s11

Core = [Ne] and Core = [Ne] and

valence = 3svalence = 3s11

Br [Ar] 3dBr [Ar] 3d1010 4s 4s22 4p 4p55

Core = [Ar] 3dCore = [Ar] 3d1010 and and

valence = 4svalence = 4s22 4p 4p55

35

Br [Ar] 3d10 4s24p5

: ::

.

88

Valence ElectronsValence ElectronsValence ElectronsValence Electrons

Covalent BondingCovalent BondingThe bond arises from the mutual attraction of The bond arises from the mutual attraction of

2 nuclei for the same electrons.2 nuclei for the same electrons.

AA Bond Bond is a balance of attractive and is a balance of attractive and repulsive forces.repulsive forces.

Chemical Bonding Chemical Bonding ObjectivesObjectives

ObjectivesObjectives are to understand: are to understand:

1. e - distribution in molecules 1. e - distribution in molecules and ions.and ions.

2. molecular structures.2. molecular structures.

3. bond properties and their 3. bond properties and their effect on molecular properties.effect on molecular properties.

Electron Electron Distribution in Distribution in

MoleculesMolecules

Electron Electron Distribution in Distribution in

MoleculesMolecules

• Electron distribution Electron distribution is depicted with is depicted with

Lewis electron Lewis electron dot structuresdot structures

• Electrons are Electrons are distributed as distributed as shared or shared or BOND BOND PAIRSPAIRS and and

unshared unshared or or LONE LONE PAIRS.PAIRS.

G. N. Lewis G. N. Lewis 1875 - 19461875 - 1946

Bond and Lone PairsBond and Lone Pairs

Electrons are distributed as shared or Electrons are distributed as shared or

BOND PAIRSBOND PAIRS and unshared or and unshared or

LONE PAIRS.LONE PAIRS.

H Cl••

••

••

lone pair (LP)shared orbond pair

This is called a LEWIS This is called a LEWIS ELECTRON DOT structure.ELECTRON DOT structure.

1313

Lewis Dot Electron ConfigsLewis Dot Electron Configs

Bond FormationBond Formation

This type of overlap places bonding This type of overlap places bonding electrons in a electrons in a MOLECULAR ORBITALMOLECULAR ORBITAL along the line between the two atoms and along the line between the two atoms and

forms a forms a SIGMA BONDSIGMA BOND ( ().).

A bond can result from a “head-to-head” A bond can result from a “head-to-head” overlapoverlap of atomic orbitals on neighboring atoms. of atomic orbitals on neighboring atoms.

H H Cl••

••

•• Cl

••

••

••

+

Overlap of H (1s) and Cl (3p)

Rules of the GameRules of the Game

Observation that atoms want to Observation that atoms want to

obtain a filled noble gas electron obtain a filled noble gas electron

configuration is called theconfiguration is called the

OCTET RULEOCTET RULE• There are exceptions to these There are exceptions to these

rules!!!rules!!!

4.4. Remaining electrons go to Remaining electrons go to the central atom to form LONE the central atom to form LONE PAIRS to complete octet as PAIRS to complete octet as needed.needed.

H H

H

N

Building a Dot StructureBuilding a Dot Structure

H••

H

H

N

3.3. Form a sigma bond Form a sigma bond between the central atom between the central atom and surrounding atoms.and surrounding atoms.

3 BOND PAIRS and 1 LONE 3 BOND PAIRS and 1 LONE PAIR. Note that N has a share PAIR. Note that N has a share in 4 pairs (8 electrons), while in 4 pairs (8 electrons), while H shares 1 pair.H shares 1 pair.

Building a Dot StructureBuilding a Dot Structure

Ammonia, NHAmmonia, NH33

1. Decide on the central atom; never H.1. Decide on the central atom; never H.

Central atom is atom of lowest affinity for Central atom is atom of lowest affinity for electrons. Therefore, N is the central atom.electrons. Therefore, N is the central atom.

2. Count valence electrons2. Count valence electrons

H = 1 and N = 5H = 1 and N = 5

Total = (3 x 1) + 5 Total = (3 x 1) + 5

= 8 electrons / 4 pairs= 8 electrons / 4 pairs

Sulfite ion, SOSulfite ion, SO332-2-Sulfite ion, SOSulfite ion, SO332-2-

Step 1. Central atom = SStep 1. Central atom = SStep 2. Count valence electrons Step 2. Count valence electrons

S = 6S = 6 3 x O = 3 x 6 = 183 x O = 3 x 6 = 18 Negative charge = 2Negative charge = 2

TOTAL = 26 e- or 13 pairsTOTAL = 26 e- or 13 pairs

Step 3. Form sigma bondsStep 3. Form sigma bonds

Sulfite ion, SOSulfite ion, SO332-2-

Step 1. Central atom = SStep 1. Central atom = S

Step 2. Count valence electrons Step 2. Count valence electrons S = 6S = 6

3 x O = 3 x 6 = 183 x O = 3 x 6 = 18

Negative charge = 2Negative charge = 2

TOTAL = 26 e- or 13 pairsTOTAL = 26 e- or 13 pairs

Step 3. Form sigma bondsStep 3. Form sigma bonds

10 pairs of electrons are 10 pairs of electrons are

now left.now left.O O

O

S

Sulfite ion, SOSulfite ion, SO332-2-

Each atom is surrounded by an octet of Each atom is surrounded by an octet of electrons.electrons.

Remaining pairs become lone pairs, Remaining pairs become lone pairs, first on outside atoms and then first on outside atoms and then remaining pairs on central atom.remaining pairs on central atom.

•• O O

O

S

••

••

•• ••

••

••

•• •

•

••

-2

Carbon Dioxide, COCarbon Dioxide, CO22

This leaves 6 pairs.This leaves 6 pairs.

4. Place lone pairs on outer atoms.4. Place lone pairs on outer atoms.

O OC

••O OC

•• ••

••••••

1. Central atom = _______1. Central atom = _______

2. Valence electrons = __ or __ pairs2. Valence electrons = __ or __ pairs

3. Form sigma bonds.3. Form sigma bonds.16 16 88

CC


The second bonding pair forms a The second bonding pair forms a pi (pi ()) bond.bond.

••O OC

•• ••

••••••

••O OC

•• ••

••••••

••O OC

•• ••

••

5. So that C has an octet, we form 5. So that C has an octet, we form DOUBLE BONDS between C and O.DOUBLE BONDS between C and O.

4. Place lone pairs on outer atoms.4. Place lone pairs on outer atoms.

Double and even Double and even triple bonds are triple bonds are commonly commonly observed for C, observed for C, N, P, O, and SN, P, O, and S HH22COCO

SOSO33CC22FF44

••O OS

••

••

••

••••••

bring inleft pair

OR bring inright pair

Sulfur Dioxide, SOSulfur Dioxide, SO22Sulfur Dioxide, SOSulfur Dioxide, SO22

1. Central atom = S1. Central atom = S

2. Valence electrons = 18 or 9 pairs2. Valence electrons = 18 or 9 pairs

••O OS

••

••

••

••••••

3. Form pi (3. Form pi () bond so that S has an octet ) bond so that S has an octet — but note that there are two ways of — but note that there are two ways of doing this.doing this.

Sulfur Dioxide, SOSulfur Dioxide, SO22

This leads to the following structures.This leads to the following structures.

••O OS

••

••

••

••••

••

bring inleft pair

OR bring inright pair

••O OS

••

••

••

••

••

••O OS

••

••

••

••

••

These equivalent structures are calledThese equivalent structures are called

RESONANCE STRUCTURESRESONANCE STRUCTURES. The true . The true electronic structure is a electronic structure is a HYBRIDHYBRID of the two. of the two.

Urea, (NHUrea, (NH22))22COCO

1. Arrangement, is C or O the central atom?1. Arrangement, is C or O the central atom?

2. Number of valence electrons = 24 e-2. Number of valence electrons = 24 e-

3. Draw sigma bonds.3. Draw sigma bonds.

CN N H

HH

H

O

C O


4. Place remaining electron pairs 4. Place remaining electron pairs in the molecule.in the molecule.

••O

CN N H

HH

H

••

•• ••

••

5. Carbon needs an Octet, take from the Oxygen


Exceptions to the Octet RuleExceptions to the Octet RuleOccurs with B (the only 2Occurs with B (the only 2ndnd period period

exception) and elements of 3exception) and elements of 3rd rd - 7- 7thth periods. periods.

BF3 SF4

Boron TrifluorideBoron Trifluoride• Central atom = Central atom =

• Valence electrons = Valence electrons =

• or electron pairs = or electron pairs =

• Assemble dot structureAssemble dot structure

••F

••

•

F••

••F

B••

••

••

• ••

The B atom has a share in only 6 electrons (or 3 pairs). The B atom in many molecules is electron deficient.

Sulfur Tetrafluoride, SFSulfur Tetrafluoride, SF44

• Central atom = Central atom =

• Valence electrons = ___ or ___ pairs.Valence electrons = ___ or ___ pairs.

• Form sigma bonds and distribute Form sigma bonds and distribute electron pairs.electron pairs.

F

••

••

••

F

F

S••

••

••

••

•• F

••

••

••

••

••

5 pairs around the S atom. A common occurrence outside the 2nd period.

5 pairs around the S atom. A common occurrence outside the 2nd period.

Draw Lewis Structures for:Draw Lewis Structures for:

NN22

CHCH44

II22COCO

HCOOHHCOOH

NONO22--

XeFXeF44

Oxidation NumberOxidation Number• Oxidation Number is assigned based on Oxidation Number is assigned based on

a set of rules.a set of rules.

• These rules are based on the Lewis These rules are based on the Lewis Structures of the compounds or ions.Structures of the compounds or ions.

• Oxidation Number = Group no. Oxidation Number = Group no. - - ((no. assigned electrons)no. assigned electrons)- (no. of LP electrons)- (no. of LP electrons)

Sample Problem

Formal Atom ChargesFormal Atom Charges• Atoms in molecules often bear a Atoms in molecules often bear a

charge (+ or -). charge (+ or -).

• The predominant resonance structure The predominant resonance structure of a molecule is the one with charges of a molecule is the one with charges as close to 0 as possible.as close to 0 as possible.

• Formal charge = Group no. Formal charge = Group no. - - 1/21/2 (no. bond electrons) (no. bond electrons)- (no. of LP electrons)- (no. of LP electrons)

••O OC

••••

••

4 - (1/ 2)(8) - 0 = 0

6 - (1/ 2)(4) - 4 = 0

Carbon Dioxide, COCarbon Dioxide, CO22Formal charge calculationsFormal charge calculations


C atom C atom charge is 0.charge is 0.

Which is the predominant resonance structure?Which is the predominant resonance structure?

••O OC

••

••

••

6 - (1/2)(6) - 2 = +1

6 - (1/2)(2) - 6 = -1


Actual partial charges.Actual partial charges.

+1.46-0.73 -0.73

Boron Trifluoride, BFBoron Trifluoride, BF33

F••

••

••

F

F

B••

••

••

••

••

••

What if we form a B—F double What if we form a B—F double bond to satisfy the B atom octet?bond to satisfy the B atom octet?

Boron Trifluoride, BFBoron Trifluoride, BF33

• To have +1 charge on F, with To have +1 charge on F, with its very high affinity for its very high affinity for electrons, is not good.electrons, is not good.

• Negative charges are best Negative charges are best placed on atoms with high placed on atoms with high affinity for electrons.affinity for electrons.

F••

••

F

F

B••

••

••

••

••

••

fc = 7 - 2 - 4 = +1

fc = 3 - 4 - 0 = -1

Thiocyanate ion, SCNThiocyanate ion, SCN--

Which of three possible resonance Which of three possible resonance structures is the most important?structures is the most important?

Calculate the formal charge for each element.

-0.52-0.32

-0.16

Calculated partial chargesCalculated partial charges

Thiocyanate ion, SCNThiocyanate ion, SCN--

PRACTICE PROBLEMPRACTICE PROBLEMOx # Ox # == group # group # -- lone pair electrons lone pair electrons -- assigned assigned

electrons electrons

F.C. F.C. == group # group # -- lone pair electrons lone pair electrons -- 1/2 bonding 1/2 bonding electronselectrons

Determine oxidation numbers and formal Determine oxidation numbers and formal charges for the atoms in SOcharges for the atoms in SO33

-2-2..

Practice Problem

MOLECULAR GEOMETRYMOLECULAR GEOMETRY

MOLECULAR GEOMETRYMOLECULAR GEOMETRY

VSEPR VSEPR • VValence alence SShell hell EElectron lectron

PPair air RRepulsion theory.epulsion theory.

• Most important factor in Most important factor in determining geometry is determining geometry is relative repulsion between relative repulsion between electron pairs.electron pairs.

Molecule adopts the shape that minimizes the electron pair repulsions.

Figure 9.11

Figure 9.12

H

HH

H

tetrahedral

109

C4

120

planar trigonal

FFB

F3

180

linear2

GeometryExample

No. of e- PairsAround CentralAtom

F—Be—F

Notice where the central atom is

4848

4 electron pair groups4 electron pair groupsElectron pair geometryElectron pair geometry Molecular Shape nameMolecular Shape name

Figure 9.14

Electron pair geometryElectron pair geometry Molecular nameMolecular name

5050

Brief Summary of VESPRBrief Summary of VESPRElectron Pair Electron Pair GeometryGeometry

Molecular GeometryMolecular Geometry

2 electron pairs2 electron pairs LinearLinear LinearLinear

3 electron pairs3 electron pairs Trigonal PlanarTrigonal Planar

Trigonal PlanarTrigonal Planar

Trigonal PlanarTrigonal Planar

““V” BentV” Bent

4 electron pairs4 electron pairs TetrahedralTetrahedral

TetrahedralTetrahedral



Trigonal PyramidalTrigonal Pyramidal

““V” BentV” Bent

5 electron pairs5 electron pairs Trigonal Trigonal Bipyramidal Bipyramidal

““look all 4 of them up”look all 4 of them up”

6 electron pairs6 electron pairs OctahedralOctahedral ““look all 4 of them up”look all 4 of them up”

14 possible combinations….learn them all!

Structure Determination Structure Determination by VSEPRby VSEPR

2. Count BP’s and LP’s around central 2. Count BP’s and LP’s around central N atom = 4 N atom = 4

(Called the number of structural pairs.)(Called the number of structural pairs.)

H••

H

H

N


1. Draw electron dot structure1. Draw electron dot structure




2. Count BP’s and LP’s = 42. Count BP’s and LP’s = 4

3. The 4 electron pairs are at the 3. The 4 electron pairs are at the corners of a tetrahedron.corners of a tetrahedron.

H••

H

H

N

H

H

H

lone pair of electronsin tetrahedral position

N



There are 4 electron pairs at the corners of There are 4 electron pairs at the corners of a tetrahedron.a tetrahedron.

The The ELECTRON PAIR GEOMETRYELECTRON PAIR GEOMETRY is is tetrahedral.tetrahedral.

H

H

H


NH••

H

H

N



The electron pair geometry is tetrahedral.The electron pair geometry is tetrahedral.

The The MOLECULAR GEOMETRY or SHAPEMOLECULAR GEOMETRY or SHAPE —— the positions of the atoms — the positions of the atoms —

is is TrigonalTrigonal PYRAMIDALPYRAMIDAL..

H

H

H


N

5555

VESPR Step-by-stepVESPR Step-by-stepDeterminationDetermination

12



H O H••

••

Water, HWater, H22OO



Water, HWater, H22OO1. Draw electron dot structure1. Draw electron dot structure


3. The 4 electron pairs are at the 3. The 4 electron pairs are at the corners of a tetrahedron.corners of a tetrahedron.

H

HO

The electron pair The electron pair geometry is geometry is TETRAHEDRAL.TETRAHEDRAL.

H O H••

••


Water, HWater, H22OO

H

HO

The electron pair The electron pair geometry is geometry is TETRAHEDRAL.TETRAHEDRAL.

The molecular The molecular geometry is geometry is bent 109.5bent 109.5

H O H••

••


Formaldehyde, CHFormaldehyde, CH22OO


••

C HH

O••





(the double bond is treated as a (the double bond is treated as a

“lump” of electrons or one “lump” of electrons or one

pair)pair)

••

C HH

O••


Formaldehyde, CHFormaldehyde, CH22OO1. Draw electron dot structure1. Draw electron dot structure


3. There are 3 electron pairs are at the 3. There are 3 electron pairs are at the corners of a planar triangle.corners of a planar triangle.

The electron pair The electron pair geometry is TRIGONAL geometry is TRIGONAL PLANAR with 120PLANAR with 120oo bond bond angles.angles.

CHH

••

••O

••

C HH

O••



The electron pair The electron pair geometry is geometry is TRIGONAL PLANAR TRIGONAL PLANAR

The molecular The molecular geometry is also geometry is also TRIGONAL PLANAR. TRIGONAL PLANAR.

CHH

••

••O


2. Define bond angles 1 and 22. Define bond angles 1 and 2

H

H1 2

••

••H—C—O—H

Methanol, CHMethanol, CH33OHOH1. Draw electron dot structure1. Draw electron dot structure


Methanol, CHMethanol, CH33OH OH

Define bond angles 1 and 2Define bond angles 1 and 2

In both cases the atom In both cases the atom

is surrounded by is surrounded by

4 electron pairs.4 electron pairs.

C

H

H

H

O H

1

2


Acetonitrile, CHAcetonitrile, CH33CNCN

Define bond angles 1 and 2Define bond angles 1 and 2

Angle 1 = 109Angle 1 = 109oo

Angle 2 = 180Angle 2 = 180oo

One C is surrounded by 4 electron One C is surrounded by 4 electron “lumps” and the other by 2 “lumps”“lumps” and the other by 2 “lumps”

Phenylalanine, Phenylalanine, an amino acidan amino acid

Determine the bond angles. 120o

109o

120o

109o

109o:

Compounds with 5 or 6 Pairs Compounds with 5 or 6 Pairs Around the Central AtomAround the Central Atom

F

Trigonal bipyramidF

F

FF 120

90

P 5 electron pairs5 electron pairs

Compounds with 5 or 6 Pairs Compounds with 5 or 6 Pairs Around the Central AtomAround the Central Atom

F

OctahedralF

F F

FF90

90

S6 electron pairs6 electron pairs

6969

STRUCTURES STRUCTURES WITH CENTRAL WITH CENTRAL

ATOMS THAT DO ATOMS THAT DO NOT OBEY THE NOT OBEY THE

OCTET RULEOCTET RULE

STRUCTURES STRUCTURES WITH CENTRAL WITH CENTRAL

ATOMS THAT DO ATOMS THAT DO NOT OBEY THE NOT OBEY THE

OCTET RULEOCTET RULE

7070

Deviations from the Octet RuleDeviations from the Octet Rule

Usually occurs with Group 3A elements Usually occurs with Group 3A elements and with those of 3rd period and higher.and with those of 3rd period and higher.

Consider boron trifluoride, BFConsider boron trifluoride, BF33

F••

••

••

F

F

B••

••

••

••

••

••

7171

Deviations from the Octet RuleDeviations from the Octet Rule

Consider boron trifluoride, BFConsider boron trifluoride, BF33

The B atom is surrounded by The B atom is surrounded by

only 3 electron pairs.only 3 electron pairs.

Bond angles are 120Bond angles are 120oo

F••

••

••

F

F

B••

••

••

••

••

••

Geometry described as planar trigonal

Sulfur Tetrafluoride, SFSulfur Tetrafluoride, SF44Sulfur Tetrafluoride, SFSulfur Tetrafluoride, SF44

• Number of valence electrons = Number of valence electrons =

• Central atom = Central atom =

• Dot structureDot structure

Sulfur Tetrafluoride, SFSulfur Tetrafluoride, SF44Sulfur Tetrafluoride, SFSulfur Tetrafluoride, SF44

• Number of valence electrons = 34Number of valence electrons = 34

• Central atom = SCentral atom = S

• Dot structure Dot structure

F

F

F

F

••• •••

••••••

•••

••

••••

••

••••••

••

S

F

F

F

F

••• •••

••••••

•••

••

••••

••

••••••

••

S


Number of valence electrons = 34Number of valence electrons = 34

Central atom = SCentral atom = S

Dot structureDot structure

Electron pair geometry = ?Electron pair geometry = ?

F

F

F

F

••• •••

••••••

•••

••

••••

••

••••••

••

S

F

F

F

F

••• •••

••••••

•••

••

••••

••

••••••

••

S


• Number of valence Number of valence electrons = 34electrons = 34

• Central atom = SCentral atom = S

• Dot structureDot structure

F

F

F

F

••• •••

••••••

•••

••

••••

••

••••••

••

S

Electron pair geometryElectron pair geometry

= = trigonal bipyramidtrigonal bipyramid

(because there are 5 (because there are 5 pairs around the S) pairs around the S)

What is the Molecular Geometry?What is the Molecular Geometry?

90 F

F

F

F•• 120S


Minimize lone pair bond pair repulsions at Minimize lone pair bond pair repulsions at the 90the 90oo angle. angle.

Not all repulsions are equivalent.Not all repulsions are equivalent.L.P. - L.P. > L.P. - B.P. > B.P. - B.P.L.P. - L.P. > L.P. - B.P. > B.P. - B.P.

Structural DeterminationStructural Determination

Determine the geometries for PFDetermine the geometries for PF33, ,

NONO22--, NO, NO22

++, ICl, ICl22--, ClF, ClF33,, ClFClF55, and IBr, and IBr44

--, , giving VSEPR class, shape name, 3-D giving VSEPR class, shape name, 3-D diagram, and bond angles. diagram, and bond angles.

SolutionsSolutions

Bond PropertiesBond Properties• What is the effect of bonding and What is the effect of bonding and

structure on molecular properties?structure on molecular properties?

Buckyball in HIV-proteaseBuckyball in HIV-protease

Bond OrderBond Order• The number of bonds between a The number of bonds between a

pair of atoms.pair of atoms.

singleBO = 11

H

H

H

C C NC

double, BO = 21 and 1

1 and 2 triple, BO = 3

Bond OrderBond Order

The N—O bond order = 1.5The N—O bond order = 1.5

O O O O

N••

••••

••••

••••••••••

••••

••N

Bond order = 3 e- pairs in N — O bonds

2 N — O bonds

Bond order = Total # of e- pairs used for a type of bond

Total # of bonds of that type

Fractional bond orders Fractional bond orders occur in molecules with occur in molecules with resonance structures.resonance structures.

Consider NOConsider NO22--

Bond OrderBond Order Bond order is proportional to two Bond order is proportional to two

important bond properties: important bond properties:

(a) (a) bond strengthbond strength

(b)(b) bond lengthbond length

745 kJ745 kJ

414 kJ414 kJ 123 pm123 pm

110 pm110 pm

Bond LengthBond Length Bond length is the distance between Bond length is the distance between

the nuclei of two bonded atoms.the nuclei of two bonded atoms.

Bond LengthBond Length Bond length Bond length

depends on depends on size size of bonded atomsof bonded atoms..

H—FH—F

H—ClH—Cl

H—IH—I

Bond distances measured in Angstroms or pm where 1 A = 100 pm.

Bond distances measured in Angstroms or pm where 1 A = 100 pm.

Bond LengthBond Length Bond length Bond length

depends on depends on bond orderbond order..

Different C-O Bond Lengths

C-C bond lengthsC-N bond lengths

Bond StrengthBond Strength• Bond strength is measured by the energy Bond strength is measured by the energy

required to break a bond. See Table 9.9required to break a bond. See Table 9.9

C-C bond strengths

Bond StrengthBond Strength BOND BOND STRENGTH (kJ/mol)STRENGTH (kJ/mol)

H—HH—H 436436

C—CC—C 346346

C=CC=C 602602

CCC C 835835

NNNN 945945

The GREATER the number of bonds The GREATER the number of bonds (bond order) the HIGHER the bond (bond order) the HIGHER the bond strength and the SHORTER the bond.strength and the SHORTER the bond.

Bond Strengths for O-OBond Strengths for O-OBondBond OrderOrder LengthLength StrengthStrength

HO—OHHO—OH 1 1 147 pm147 pm 210 kJ/mol 210 kJ/mol

1.51.5 128128

O=OO=O 2 2 121 121 498 498

O O•••••••

••••

••O

??394

Using Bond EnergiesUsing Bond Energies

HHoorxnrxn = = H (bonds broken) H (bonds broken) --

H (bonds formed) or H (bonds formed) or

HHoorxnrxn = = energy input ( energy input (++) ) ++

energy output ( energy output (--).). Net energy = Net energy = HHrxn, bondrxn, bond = =

energy required to break bond energy required to break bond -- energy energy evolved when bonds are formed.evolved when bonds are formed.

Using Bond EnergiesUsing Bond EnergiesEstimate the energy of the reactionEstimate the energy of the reaction H—H + Cl—Cl ----> 2 H—Cl H—H + Cl—Cl ----> 2 H—Cl

H—H = 436 kJ/molH—H = 436 kJ/molCl—Cl = 242 kJ/molCl—Cl = 242 kJ/molH—Cl = 432 kJ/molH—Cl = 432 kJ/mol

Sum of H-H + Cl-Cl bond energies = Sum of H-H + Cl-Cl bond energies = 436 kJ + 242 kJ = +678 kJ436 kJ + 242 kJ = +678 kJ2 mol H-Cl bond energies = 864 kJ2 mol H-Cl bond energies = 864 kJNet = Net = H = +678 kJ - 864 kJ = -186 kJH = +678 kJ - 864 kJ = -186 kJ


Estimate the energy of the reactionEstimate the energy of the reaction

2 H—O—O—H ----> O=O + 2 H—O—H2 H—O—O—H ----> O=O + 2 H—O—H

Is the reaction exo- or endothermic?Is the reaction exo- or endothermic?

Which is larger: energy req’d to break Which is larger: energy req’d to break

bonds or energy evolved on making bonds or energy evolved on making

bonds?bonds?



Energy required to break bonds:Energy required to break bonds:

break 4 mol of O—H bonds = 4 (463 kJ)break 4 mol of O—H bonds = 4 (463 kJ)

break 2 mol O—O bonds = 2 (146 kJ)break 2 mol O—O bonds = 2 (146 kJ)

TOTAL ENERGY to break bonds = 2144 kJTOTAL ENERGY to break bonds = 2144 kJ



Energy evolved on making bonds:Energy evolved on making bonds:

make 1 mol of O=O bonds = 498 kJmake 1 mol of O=O bonds = 498 kJ

make 4 mol O—H bonds = 4 (463 kJ)make 4 mol O—H bonds = 4 (463 kJ)

TOTAL ENERGY evolved on making bonds = TOTAL ENERGY evolved on making bonds =

2350 kJ2350 kJ



Net energy = +2144 kJ - 2350 kJ = - 206 kJNet energy = +2144 kJ - 2350 kJ = - 206 kJ

The reaction is exothermic!The reaction is exothermic!

More energy is evolved on making bonds than is expended in breaking bonds.

More energy is evolved on making bonds than is expended in breaking bonds.

Calculate Calculate HHoorxnrxn for the reaction for the reaction

below using bond energies.below using bond energies.

CHCH4 4 (g) + 2 O(g) + 2 O2 2 (g) (g) ____________> 2 H> 2 H22O (g) + COO (g) + CO2 2 (g)(g)

H = [4(C-H) + 2(O=O)] - [4(H-O) + 2(C=O)] H = [4(C-H) + 2(O=O)] - [4(H-O) + 2(C=O)]

H = [4(413kJ) + 2(498kJ)] - [4(463kJ) + 2(732kJ)]H = [4(413kJ) + 2(498kJ)] - [4(463kJ) + 2(732kJ)]

H = -1166 kJH = -1166 kJ

More Sample Problems!More Sample Problems!

Molecular PolarityMolecular Polarity

Why are water molecules attracted Why are water molecules attracted to a balloon that has a static to a balloon that has a static electric charge?electric charge?

Bond PolarityBond Polarity

HCl is HCl is POLARPOLAR because it because it has a positive end and has a positive end and a negative end.a negative end.

Polarity arises because Polarity arises because Cl has a greater share Cl has a greater share in bonding electrons in bonding electrons than does H.than does H.

Cl

-+

•••H••

••

Bond PolarityBond Polarity This model, calc’d using CAChe software, This model, calc’d using CAChe software,

shows that H is + (red) and Cl is - (yellow). shows that H is + (red) and Cl is - (yellow). Calc’d charge is + or - 0.20. Calc’d charge is + or - 0.20.

This model shows that the electron density This model shows that the electron density is greater around Cl than around H.is greater around Cl than around H.


BONDBOND ENERGY ENERGY

““pure” bondpure” bond339 kJ/mol calc’d339 kJ/mol calc’d

real bondreal bond 432 kJ/mol measured432 kJ/mol measured

Difference = 92 kJ. This difference is Difference = 92 kJ. This difference is proportional to the difference in proportional to the difference in

ELECTRONEGATIVITY, ELECTRONEGATIVITY, ..

Cl

-+

•••H••

•• Due to the bond polarity, the H—Cl Due to the bond polarity, the H—Cl

bond energy is GREATER than bond energy is GREATER than expected for a “pure” covalent bond.expected for a “pure” covalent bond.

Don’t confuse with mole fraction from Ch 12


Partial charge on atom APartial charge on atom A ==

Group Number of A Group Number of A -- # #

lone pair electrons lone pair electrons -- AA/(/(AA+ +

BB)( )( # bonding e# bonding e-- shared by shared by A ) A )

Cl

-+

•••H••

••

Electronegativity, Electronegativity, is a measure of the ability of is a measure of the ability of

an atom in a molecule to an atom in a molecule to attract electrons to itself. attract electrons to itself.

Concept proposed byConcept proposed by Linus Linus

PaulingPauling 1901-1901-

19941994

Linus Pauling, 1901-1994Linus Pauling, 1901-1994

The only person to receive two unshared The only person to receive two unshared Nobel prizes (for Peace and Chemistry). Nobel prizes (for Peace and Chemistry).

Chemistry areas: bonding, electronegativity, Chemistry areas: bonding, electronegativity, protein structureprotein structure

• F has maximum F has maximum ..

• Atom with lowest Atom with lowest is the center atom is the center atom in most molecules.in most molecules.

• Relative values of Relative values of determine determine BOND BOND POLARITYPOLARITY (and point of attack on a (and point of attack on a molecule).molecule).

Electronegativity, Electronegativity,

See See Figure Figure 9.109.10

See See Figure Figure 9.109.10

Figure 9.9

Electronegativity,Electronegativity,

Basic Trend =


OH is more polar than OFOH is more polar than OF

and polarity is “reversed.”and polarity is “reversed.”

+ -+-O—FO—H

Which bond is more polar (or DIPOLAR)?Which bond is more polar (or DIPOLAR)?

O—HO—H O—FO—F

3.5 - 2.13.5 - 2.1 3.5 - 4.03.5 - 4.0

1.41.4 0.50.5

Bond PolarityBond Polarity Relative polarity in a bond is determined Relative polarity in a bond is determined

by by , where the approximate % ionic , where the approximate % ionic character is give by:character is give by:

= 1.0, 20%; = 1.0, 20%; = 1.5, 40%; = 1.5, 40%;

= 2.0, 60%; = 2.0, 60%; = 2.5, 80%. = 2.5, 80%.

Molecular PolarityMolecular PolarityMolecules—such as HCl and HMolecules—such as HCl and H22O— can be O— can be POLARPOLAR

(or dipolar). (or dipolar).

They have a They have a DIPOLE MOMENTDIPOLE MOMENT. The polar HCl . The polar HCl

molecule will turn to align with an electric field.molecule will turn to align with an electric field.

Molecular PolarityMolecular PolarityThe magnitude of the The magnitude of the

dipole is given in dipole is given in Debye units. Debye units.

Named for Peter Named for Peter Debye (1884 - 1966). Debye (1884 - 1966).

Received the 1936 Received the 1936 Nobel prize for work Nobel prize for work on x-ray diffraction on x-ray diffraction and dipole and dipole moments. moments.

Molecular PolarityMolecular PolarityMolecules will be polar ifMolecules will be polar if

1. bonds are polar 1. bonds are polar ANDAND

2. the molecule is NOT “symmetric”2. the molecule is NOT “symmetric”

Symmetric moleculesSymmetric molecules

Molecular Polarity Molecular Polarity

Unsymmetrical

Carbon DioxideCarbon Dioxide• COCO22 is NOT polar even though the CO bonds are is NOT polar even though the CO bonds are

polar.polar.

• COCO22 is symmetrical. is symmetrical.

Positive C Positive C atom is reason atom is reason COCO22 + H + H22O O gives Hgives H22COCO33

Consequences Consequences of Hof H22O PolarityO Polarity

Microwave Microwave ovenoven

B—F bonds in BFB—F bonds in BF33 are polar. are polar. But molecule is symmetrical But molecule is symmetrical and NOT polarand NOT polar

Molecular Polarity Molecular Polarity More ExamplesMore Examples

F

F FB

B atom is positive and F atoms are negative.

B—F and B—H bonds in HBFB—F and B—H bonds in HBF22 are polar. But molecule is are polar. But molecule is NOT symmetrical and is polarNOT symmetrical and is polar

Molecular PolarityMolecular PolarityB atom is positive but H & F atoms are negative.

H

F FB

Polarity of Methane, CHPolarity of Methane, CH44

Methane is symmetrical and is NOT polar.Methane is symmetrical and is NOT polar.

H

HH

HC

Polarity of CHPolarity of CH33FF

C—F bond is very polar. C—F bond is very polar. Molecule is not symmetrical Molecule is not symmetrical and so is polar.and so is polar.

F

HH

HC

Substituted EthyleneSubstituted Ethylene

• C—F bonds are MUCH more polar than C—F bonds are MUCH more polar than C—H bonds. C—H bonds.

• Because both C—F bonds are on same Because both C—F bonds are on same side of molecule, molecule is side of molecule, molecule is POLARPOLAR..

Substituted EthyleneSubstituted Ethylene

• C—F bonds are MUCH more polar than C—F bonds are MUCH more polar than C—H bonds. C—H bonds.

• Because both C—F bonds are on opposing Because both C—F bonds are on opposing ends of molecule, molecule is ends of molecule, molecule is NOT NOT POLARPOLAR..

Practice ProblemsPractice Problems

1. Give the VSEPR class, bond angle, shape, 1. Give the VSEPR class, bond angle, shape, and polarity of KrFand polarity of KrF22, CO, and NO, CO, and NO22

--..

2. Calculate the heat of reaction for2. Calculate the heat of reaction for

CC22HH44 (g) + 3 O (g) + 3 O22 (g) --> 2 CO (g) --> 2 CO22 (g) + 2 H (g) + 2 H22O (g)O (g)

3. Calculate the carbon-carbon triple bond 3. Calculate the carbon-carbon triple bond energy in Cenergy in C22HH22(g). (g).

Heat of formation of CHeat of formation of C22HH22 = 226 kJ/mole) = 226 kJ/mole)

4. Determine the number of 4. Determine the number of and and bonds in bonds in acetic acid.acetic acid.

5. Arrange in order of increasing polarity:5. Arrange in order of increasing polarity:

Sr-Br, P-F, S-F, Al-OSr-Br, P-F, S-F, Al-O

Practice Problems AnswersPractice Problems Answers

1. AX1. AX22EE33, 180, 18000, linear, nonpolar, linear, nonpolar

AXE, 180AXE, 18000, linear, polar, linear, polar

AXAX22E, 120E, 12000, bent, polar, bent, polar

2. -1032 kJ2. -1032 kJ

3. 818 kJ3. 818 kJ

4. 7 4. 7 and 1 and 1

5. S-F, P-F, Al-O, Sr-Br,5. S-F, P-F, Al-O, Sr-Br,

Ionic BondingIonic Bonding

Ionic bonds form by the transferring ofIonic bonds form by the transferring of

VALENCE ELECTRONSVALENCE ELECTRONS, the , the

electrons at the outer edge of the atom.electrons at the outer edge of the atom.


Na and FNa and F

Na 1sNa 1s222s2s222p2p663s3s11 F 1sF 1s222s2s222p2p55

NaNa1+1+ 1s 1s222s2s222p2p66 FF1-1- 1s 1s222s2s222p2p66

NaFNaF


Mg and OMg and O

Mg 1sMg 1s222s2s222p2p663s3s22 O 1sO 1s222s2s222p2p44

MgMg2+2+ 1s 1s222s2s222p2p66 OO2-2- 1s 1s222s2s222p2p66

MgOMgO


Mg and FMg and F

Mg 1sMg 1s222s2s222p2p663s3s22 F 1sF 1s222s2s222p2p55

MgMg2+2+ 1s 1s222s2s222p2p66 FF1-1- 1s 1s222s2s222p2p66

MgFMgF22 FF1-1- 1s 1s222s2s222p2p66

F 1sF 1s222s2s222p2p55


Na and ONa and O

Na 1sNa 1s222s2s222p2p663s3s11 O 1sO 1s222s2s222p2p44

NaNa22OO

NaNa1+1+ 1s 1s222s2s222p2p663s3s11 OO2-2- 1s 1s222s2s222p2p66

NaNa1+1+ 1s 1s222s2s222p2p663s3s11

Na 1sNa 1s222s2s222p2p663s3s11


Cl and FCl and F

Cl . . F

. FCl .


S and FS and F

. F

. F

S ..

S ... F


O and OO and O

O O

. O.O ..

O - H- H

Sample ProblemsSample ProblemsSample ProblemsSample ProblemsDraw dot structures for the following:Draw dot structures for the following:

HH22O O CO CO NO NO221-1-

O C

- O N O

- O N O

-1

-1


NN22

CHCH44

II22COCO

HCOOHHCOOH

NONO22--

XeFXeF44

NN


NN22

CHCH44

II22COCO

HCOOHHCOOH

NONO22--

XeFXeF44

HC

H

H

H


NN22

CHCH44

II22COCO

HCOOHHCOOH

NONO22--

XeFXeF44

IC

O

I


NN22

CHCH44

II22COCO

HCOOHHCOOH

NONO22--

XeFXeF44

OC

O

H H

O


NN22

CHCH44

II22COCO

HCOOHHCOOH

NONO22--

XeFXeF44

ON

O O N

-

-

O


NN22

CHCH44

II22COCO

HCOOHHCOOH

NONO22--

XeFXeF44

FXe

F

F

F....

Sample ProblemsSample ProblemsSample ProblemsSample ProblemsDetermine the oxidation number for each Determine the oxidation number for each

atom in the following:atom in the following:

COCO22 SO SO332-2-

O = -2

C = 4

O = -2

S = 4

Sample ProblemsSample ProblemsSample ProblemsSample ProblemsDetermine the formal charge for each atom Determine the formal charge for each atom

in the following:in the following:

COCO NONO221-1-

single bond O = -1

double bond O = 0

N = 0

O = 1

C = -1

Structural DeterminationStructural DeterminationStructural DeterminationStructural Determination

PFPF33 AXAX33EE trigonal pyramidaltrigonal pyramidal

..

PFPF33


NONO22++ AXAX22 linearlinear

NONO22++


IClICl22-- AXAX22EE33 linearlinear

180180oo

..

....120o

IClICl22--


ClFClF33 AXAX33EE22 T - shapedT - shaped

..

..120o

ClFClF33


ClFClF55 AX AX55EE square pyramidalsquare pyramidal

ClFClF55

..


IBrIBr44-- AXAX44EE22 square planarsquare planar

..

..

IBrIBr44--



C (g) + 2 ClC (g) + 2 Cl2 2 (g) (g) ____________> CCl> CCl4 4 (g)(g)

H = [2(Cl-Cl)] - [4(C-Cl)] H = [2(Cl-Cl)] - [4(C-Cl)]

H = [2(242 kJ)] - [4(339 kJ)] H = [2(242 kJ)] - [4(339 kJ)]

H = - 872 kJH = - 872 kJ



C (s) + 2 ClC (s) + 2 Cl2 2 (g) (g) ____________> CCl> CCl4 4 (g)(g)

H = [ H = [ HHsubsubC + 2(Cl-Cl)] - [4(C-Cl)] C + 2(Cl-Cl)] - [4(C-Cl)]

H = [ 717 kJ + 2(242 kJ)] - [4(339 kJ)] H = [ 717 kJ + 2(242 kJ)] - [4(339 kJ)]

H = -155 kJH = -155 kJ

Calculate Calculate HHooff for CHfor CH44 using using

bond energies.bond energies.

C (s) + 2 HC (s) + 2 H22 (g) --> CH (g) --> CH44 (g) (g)

HHff = [ = [HHsubsub C + 2(H-H)] - [4(C-H)] C + 2(H-H)] - [4(C-H)]

HHff = [ 717 kJ + 2(436 kJ)] - [4(413kJ)] = [ 717 kJ + 2(436 kJ)] - [4(413kJ)]

HHff = - 63 kJ = - 63 kJ

Given Given HHooff for Cfor C22HH44 (g) is 52 kJ/mole. (g) is 52 kJ/mole.

Using bond energies, calculate the Using bond energies, calculate the carbon carbon bond energy in Ccarbon carbon bond energy in C22HH44 (g). (g).

2 C (s) + 2 H2 (g) -- > C2H4 (g)

Hf = [ 2(HsubC) + 2(H-H) ] - [ 4(C-H) + 1(C=C) ]

52kJ = [ 2(717kJ) + 2(436kJ)] - [4(413kJ) + 1(C=C)]

(C=C) = 602 kJ

Bond PolarityBond PolarityBond PolarityBond Polarity

IonicIonic Polar Polar Nonpolar Nonpolar

Covalent Covalent CovalentCovalent

<--------------------------------------------------------------><-------------------------------------------------------------->

Li Li :: F F C C :: F F F F :: F F

Documents

10-Ch9(ikatan kimia).ppt