chapter 8 jan13

8/12/2019 Chapter 8 Jan13

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CHAPTER 8

CONCEPTS OFCHEMICAL BONDING


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CONTENTS

8.1 Basic Concepts of Chemical Bonding8.2 Ionic Bonding

8.3 Sizes of Ions8.4 Covalent Bonding8.5 Lewis Structures: Comment about

the Octet Rule8.6 Strength of Covalent Bonds


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Learning outcomes

Able to determine size of ions & lattice energy

Able to draw and use Born Haber cyclefor

calculations

Able to draw Lewis structureto fulfill octet

rule

Able to identify resonance structuresand

calculate formal charge Able to calculate enthalpies(energy) from

Lewis structure of compounds


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8.1 Basic Concepts of ChemicalBonding

1. Why do atoms in an element or molecule(except noble gas) prefers to undergo

chemical reaction?

- to attain the octet configuration of the

inert noble gas


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Cont: 8.1 Basic Concepts ofChemical Bonding

2. Why Octet configuration?

- very stable electron arrangement.

- high ionization energy.

- low affinity for additional electrons.

- lack of chemical reactivity


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3. What is this octet configuration orrule?

Octet rule- an atom is stable whensurrounded by 8 electrons at the valence

orbital i.e. an octet of 4 pairs of valenceelectrons arranged around the atoms.


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4. How do they attain this Octetconfiguration?

- by forming chemical bonds.

5. What is a chemical bond?

- chemical bond is viewed as forces thatcauses a group of atoms to behave as aunit.


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6. Why do chemical bond occurs?

- bond results from a tendency of a system to

seek its lowest possible energy. There are 3types of bonding: ionic, covalent and metallic.

7. How are chemical bonds formed?

- by gaining, losingor sharingelectron valenceas the case might be between the atoms.


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8.1.1 Types of Chemical Bonds

i) Ionic Bond

- electronic forces that exist between ions ofoppositecharge.

- ions are formed from atoms by the transferof one or more electronsfrom one atom to

another.- results from the interaction of metalswithnon metals.


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Cont: 8.1.1 Types of ChemicalBonds

ii) Covalent Bond

- sharingof electrons between 2 atomse.g. interactions of non-metallicelements

with one another. (e.g. Cl2)


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Cont: 8.1.1 Types of ChemicalBonds

iii) Metallic Bonds- found in metals

- each metal atom is bonded to severalneighboring atoms.- bonding electrons are delocalised

- high heat & electrical conductivity.- high melting & boiling point- luster


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Cont:


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8.2 Ionic Bonding

Consider the reaction between sodium andchlorine:

Na(s) + 1/2Cl2

(g)NaCl(s)Hof = - 410.9 kJ/mol

The reaction is exothermic. Formation of NaCl:

- Na has lost one electron Na+- Cl has gained the electron to become Cl-.- Electron lost by Na is gained by Cl.


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Cont: 8.2 Ionic Bonding

The octet of Na+is [He] 2s22p6

Na : 1s22s22p63s1 - e [He] 2s22p6

The octet of Cl-[Ar] 3s23p6

Cl: [Ne] 3s23p5+e [Ne] 3s23p6


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Sodium chloride has a crystal structure.

Each Na+ion is surrounded by six Cl- ions.

Each Cl- ion is surrounded by sixNa+ions. This ionic compoundis stable- the

attractionbetween ions of unlike charges.

This attraction draws the ions together.


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Energy Involved in Ionic Bonding

Ionic compounds are stable (why?):- because the attraction between ions of unlike

charges.- The strong interactions cause most ionic

materials to be hard, brittle materials with highmelting point.

How?- this stabilization to form a solid array or lattice

results from the release of energy (lattice

energy).


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8.2.1 Lattice Energy

What is lattice energy?

- the energy requiredto completely

separate a moleof a solidionic compoundinto its gaseousions.

Example:NaCl(s) Na+(g) + Cl- (g) Hlattice= +788kJ/mol


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Cont: 8.2.1 Lattice Energy

The magnitude of the lattice energy depends on:i) charges of the ions ii) sizes of the ions

iii) arrangement of the ions in the solid

The potential energy between 2 interactingcharged particles:

E = k(Q1Q2)d

Q1, Q2- charges on the particles

d - distance between their centres

k - constant, 8.99 x 109J-m/C2


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7.


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Example 1

Arrange the following ionic compounds in

order of increasing lattice energy:

NaF, CsI andCaO.


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Example 1 (Answer)

NaFconsist of Na+and F-ions

CsIconsist of Cs+and I-ions

CaOconsist of Ca2+and O2-ions

Q1and Q2for CaOhigher CaObe the greatest of the

three.

Q1,Q2for CsIand NaFare the same.

Radius of Naatom is less than that of a Csatom.

Na+is smaller than Cs+,F-is a smaller ion than I-.

CsI< NaF< CaO

{CsI= 600 kJ/mol, NaF= 910 kJ/mol, CaO= 3414 kJ/mol}


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Example 2

Explain why MgO > CaO > SrOAnswer1. Q1and Q2are the same.

2. Radius different(Highest radius, lowestlattice energy)

All 3 from Group 2A, radius decreases from Sr >Ca > Mg

MgO - 3795 kJ/molCaO - 3414 kJ/molSrO - 3217 kJ/mol


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8.2.2 The Born-Haber Cycle:Calculation of Lattice Energies

Direct experimental determination of latticeenergy of an ionic solid is difficult.

Lattice energy can be indirectly determinedfrom experiment by means of a thermochemicalcycle called Born-Haber cycle.

Born-Haber cycle use Hesss Law to put thesteps together.


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Cont: 8.2.2 The Born-HaberCycle

Hesss Law states that if a reaction is

carried out in a series of steps, H forthereaction will be equal to the sum of theenthalpy changes for the individual step.


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Consider the formation of NaCl(s) fromNa(s) and Cl2(g) by 2 different routes:

i) direct

ii) indirect


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Direct route

Na(s) + 1/2Cl2(g)NaCl(s)

Hof [NaCl(s)]= -410.9kJ/mol


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Indirect route

Step 1- Change Na metal to vapor.

Na(s) Na(g) Ho

f [Na(g)] = 107.7kJ

Step 2- Decomposition of Cl2 moleculeto Clatom

1/2Cl2(g) Cl(g) Hof [Cl(g)] = 121.7kJ

Both are endothermic


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Step 3- Ionization of Na(g) to form Na+(g)

Na(g) Na+(g) + e- H= IE1(Na) = 496 kJ

Step 4- Formation of Cl(g) to Cl-(g) ion

Cl(g) + e-

Cl-

(g) H

= EA1(Cl) = -349 kJ


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Step 5- Formation of NaCl (s) from ions ofNa+(g) and Cl-(g)

Na+

(g) + Cl-

(g) NaCl(s) H

= - H

lattice

this process is the reverse of the latticeenergy (breaking of solids into ions), thereforeH= -Hlattice


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Na(s)

Na(g)

Cl2(g)

Cl(g) Cl-(g)

Na+(g)

NaCl(s)+

+

Step 1

H f= +107.7 kJ/mol

Step 2H

f

= +121.7 kJ/mol

Step 3H=I

1

= +496 kJ/mol

Step 4H=EA

1

= -349 kJ/mol

Direct route

Step 5= - H

lattice

-410.9kJ/mol


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Summation of Hesss Law on the 5 steps

Ho, kJ/mol

Step 1Na(s) Na(g) 107.7Step 2 1/2Cl2(g) Cl(g) 121.7Step 3 Na(g) Na+(g) + e- 496

Step 4 Cl(g) + e-

Cl-

(g) -349Step 5 Na+(g) + Cl-(g) NaCl(s) -Hlattice

Na(s) + 1/2Cl2(g)NaCl(s) 376.4 -Hlattice


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Direct route

Na(s) + 1/2Cl2(g)NaCl(s)Hof [NaCl(s)]= -410.9 kJ/mol

Indirect route-410.9 kJ/mol = 376.4 -Hlattice

Hlattice= 787.3 kJ/mol


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8.2.3 Electron Configuration ofIons

Metals from:

Group 1A cation +1 charge (Na+) Group 2A cation +2 charge (Ca2+)

Group 3A cation +3 charge (Al3+)


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Cont: 8.2.3 ElectronConfiguration of Ions

Nonmetals from:

Group 5A anion 3 charge (N3-) Group 6A anion -2 charge (S2-)

Group 7A anion -1 charge (Cl-)


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Group 4A

C (nonmetal), Si (metalloid), Ge (metalloid)

CO2, CO, SiO2, SiCl4

Sn (metal), Pb (metal)

Divalent cations (Sn

2+

, Pb2+

)


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Transition metal ions:

Loss up to 3 electrons from atoms.

Cations with +1, +2, +3 charges.

In forming ions, transition metals lose thevalence-shell selectrons first, then as many delectrons as required.


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1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10

Ionization (positive ion is formed) process:1st electron is lost from the subshell withlargest value of n.


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Example 3

Write the electron configuration for Fe2+andFe3+ions

Fe (Z = 26) : 1s22s22p63s23p64s23d6

or [Ar] 4s23d6

Fe2+: [Ar] 3d6

Fe3+: [Ar] 3d5


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8.3 Sizes of Ions

Ionic size:- determine the structure

- stability of ionic solid- properties of ions in solution

Size of ion depends on:- nuclear charge- number of electrons- the outermost/valence shell


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Cont:


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Cont: 8.3 Sizes of Ions

Positive ions (cations):

formed by removing one or more electronsfrom the outermost region.

decreases the total electron-electronrepulsion.

Result:cations aresmaller than their parentatoms.


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Cont: 8.3 Sizes of Ions

Negative ions (anions):

electrons are added to form ion. increase electron-electron repulsion.

Result:electrons are more spread out inspaceanions are larger than their parentatoms.

9_6


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Cont:


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Cont:


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Example 4

Arrange the ions S2-, Cl-, K+and Ca2+in

order of decreasing size.


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Example 4 (Answer)Isoelectronic ions: (ions possess same number

of electrons).

All ions have 18 electrons.

Sizedecreasesas nuclear charge (atomicnumber) of ion increases.

Atomic number S(16), Cl(17), K(19) and Ca(20).

Ions decrease in size: S2- > Cl-> K+> Ca2+.


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8.4 Covalent Bonding Chemical bond formed by sharinga pair of

electrons with other atoms (overlapping ofelectron cloud)

Commonly formed between non-metaland non-metal

Forms an octet in its valence shell

H Cl ClH+


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8.4.1 Drawing Lewis Structureof Compounds

Lewis structure

Valence electrons are arranged around theatom.

Maximum 8 electrons (except Hydrogen)

X

Cl


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Cont: 8.4.1 Drawing LewisStructure of Compounds

Rules for Drawing Lewis StructuresStep 1- Sum total valence electrons of all theatoms (for cation, subtract an electron for each

positive charge and for anion, add an electron foreach negative charge).Step 2- Write the symbols for the atoms toshow which atoms are attached to which, and

connect them with a single bond.Step 3Complete the octets of all the atomsexcept the central atom (hydrogen can have onlytwo electrons).


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Cont: 8.4.1 Drawing LewisStructure of Compounds

Step 4- Calculate number of remainingelectrons (if any) and place them at the central

atom.Step 5- Use remaining electrons to achieveoctet configuration. If not enough electrons togive the central atom an octet, try multiplebonds.Step 6- Check number of electrons aroundeach atom.


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8.4.2 Bond Polarity andElectronegativity

The bonding electrons are not shared equally. Bond polarity- describing the sharing of

electron between atoms.

A nonpolarcovalent bond - the electrons areshared equallybetween 2 atoms.

A polarcovalent bond one atom exerts agreater attractionfor the bonding electronsthan the other.

An ionic bond - the difference in relativeability to attract electron is larger.


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Cont: 8.4.2 Bond Polarity andElectronegativity

Electronegativity Electronegativity is the abilityof an atom ina

moleculeto attract bondingelectrons to itself.

Electronegativity is used to estimate a givenbond - non polar covalent, polar covalentorionic.

The greateran atoms electronegativitythegreaterits ability to attract electronstoitself.


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Cont:


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Cont: 8.4.2 Bond Polarity andElectronegativity

Difference in electronegativity between 2 atomscan be used to gauge the polarity of the bonding.

General rule:

Electonegativity Difference, EN Type of Bond

EN< 0.5 non polar

0.5 EN


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8.4.3 Dipole Moment

Molecules like HF have centres of positiveand negative charge that do not coincide.

These are polar molecules. Polarity of molecules can be indicated in 2

ways: Positive end (+) and negative end (-)

An arrow over the bond line. The arrow pointstowards the more electronegative element.


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Cont: 8.4.3 Dipole Moment

We can quantify the polarity of themolecule:

When charges are separated by a distance, adipoleis produced.

The dipole moment is the quantitative measureof the magnitude of the dipole ().

= Qr The magnitude of the dipole moment is given in

Debyes.


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8.4.4 Molecular Polarity

We can extend this to polyatomic molecules.

For each bond in polyatomic molecule, we can

consider the bond dipole. The dipole moment due only to the two atoms in

the bond is the bond dipole.

Because bond dipoles and dipole moments arevector quantities, the orientation of theseindividual dipole moments determines whetherthe molecule has an overall dipole moment.


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Cont: 8.4.4 Molecular Polarity

Examples:

In CO2, each+CO-dipole is canceled

because the molecule is linear. In H2O,

+HO-the dipoles do not cancelbecause the molecule is bent.

It is possible for a molecule with polarbonds to be either polar or nonpolar.

CANCELATION OF

Type of Molecules with Polar Bonds but No Resulting Dipole Moment


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A

B

BB

B

A BB

B

A

B B120

o

Linear moleculeswith two identical

bonds

Planar moleculeswith three identical

bonds 120

o

apart

Tetrahedral molecules

with four identicalbonds 109.5

oapart

CO2

SO3

CCl4

TYPECANCELATION OFPOLAR BONDS EXAMPLE


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8.4.5 Formal Charge

Formal Charge

Lewis structure describes distribution ofelectron in a molecule that obeys the octetrule

sometimes more then one structure can exist

which one is most reasonable ?


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Cont: 8.4.5 Formal Charge

How to calculate formal charge?

Step 1- all unshared (non-bonded) electronsare assigned to each isolated atom

Step 2- half bonding electrons are assigned toeach atom in the bond

Step 3- subtract number of assignedelectrons from the valence electrons in theisolated atom (Valence e-- Assigned e-)


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General Rule - the most stable Lewisstructure will be:

1. The atoms bear the smallest formal charges;

2. Any negative charges reside on the moreelectronegative atoms.


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E.g1. Carbon dioxide (CO2)

Set 1 Set 2

valence e- 6 4 6 6 4 6

-(e- assigned) 6 4 6 7 4 5

Form. Charge 0 0 0 -1 0 +1

Note : Set I is the preferred one

C OO C OO


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thiocyanide ion (NCS -).

There are 3 possible Lewis structures

5 4 6 5 4 6 5 4 6

7 4 5 6 4 6 5 4 7

-2 0 +1 -1 0 0 0 0 -1Set 1 Set 2 Set 3

Note : Set 2is the preferred one because it has thelowest formal charge and N is more electronegativethen C or S

N C S N C S N C S


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8.4.6 Resonance Structures

Sometimes the arrangement of atoms in moleculesand ions is not adequately described by a singleLewis structure.

Example: Ozone O3

None of the Lewis structures above can representO3. Why?

O

O

O O

O

O


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Cont: 8.4.6 ResonanceStructures

O=O double bond shorter than O-O.

In drawing the Lewis structure we can put O=Obond on the left or on the right.

We have two alternative Lewis structure forozone. This equivalent Lewis structures are calledresonance structure

O

O

O O

O

O


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Write both Lewis structures - the real molecule isdescribed by an average of the two resonancestructures:

The double-headed arrow () indicates theresonance structures.

O

O

O O

O

O

O

O

O


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The molecule (blend of resonance) has its

own identity separate from the individual

resonance structure.

Example: Nitrate ion, NO3-

O

N

O O O

N

O

O O

N

O

O O

N

O

O

=


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Benzene (C6H6):

aromatic molecule (organic molecule).

2 equivalent Lewis structures for benzene.

C

C

C C

C

C

H

HH

H

H H

C

C

C C

C

C

H

HH

H

H H


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The delocalised electron structure of benzene

C

C

C C

C

C

H

HH

H

H H

C

C

C C

C

C

H

HH

H

H H

=

i


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8.5 Lewis Structures :Comments about the Octet Rule

The second row elements: C, N, O and F shouldalways be assumed to obeythe Octet Rule intheir Lewis structures.

The second row elements: B and Be have fewerthan 8 electronsaround them in their Lewisstructures. These electron deficientcompounds are very reactive.

The second row elements never exceed theoctet rule, since their valence orbital (2sand2p) can accommodate only 8 electrons.

C 8 i S


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Cont: 8.5 Lewis Structures :Comments about the Octet Rule

Third row and heavier elements often satisfythe octet rule but can exceed the octet rule byusing their empty valence dorbitals.

When writing the Lewis structure for amolecule, satisfy the octet rule for the atomfirst. If there are electrons remaining after

the octet rule has been satisfied, place themon the elements having available d orbitals(elements in period 3 or beyond).

8 5 1 E i h O


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8.5.1 Exceptions to the OctetRule

The octet rule is useful but fails in manysituations.

There are threeexceptions to the octet ruleare three types:

i) Molecules with an odd numbers of electrons.

ii) Molecules in which an atom has less than an

octet.iii) Molecules in which an atom has more than an

octet.


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8.5.2 Odd Number of Electrons

A few molecules such as NO and NO2.

NO: 5+6 = 11

NO2: 5+12 =17

Note:

Complete pairing of electrons is impossible


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8.5.3 Less than an Octet

There are fewer than eight electrons around anatom in a molecule or ion.

Most often encountered in compounds of boronand beryllium. E.g boron trifluoride, BF3.

If we follow the rule:

F

B

F F


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Cont: 8.5.3 Less than an Octet

Bond formation with other molecules withunpaired electrons is formed at the boronterminal.

Note:

In this stable compound B has an octet ofelectrons

B

F

F

F

N

H

H

H


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8.5.4 More than an Octet

Consists of molecules or ions in which there aremore than 8electrons in the valence shell of an

atom. Consider PCl5.

We expand the valence shell and place 10electrons around the central phosphorus atom.


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Cont: 8.5.4 More than an Octet

PCl5 P = 5 valence e-

5Cl = 5 (7 valence e-)

= 40 valence e-

Other example with expanded valence shellsare SF4, AsF6

-and ICl4-.

P

Cl

Cl

Cl

Cl

Cl


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Why expanded valence shells are observed onlyfor elements in period 3and beyond in the

periodic table? Element of the 2nd period have only the 2s and

2p valence orbitals.

For the period 3 and beyond, have unfilled ndorbitals that can be used in bonding.


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For example, the orbital diagram for thevalence shell of a phosphorus atom:

[Ne]

3s 3p 3d


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For PCl5: use the empty d orbitals toaccommodate additional electrons.

Expanded valence shells occur most often whenthe central atom is bonded to the smallest andmost electronegative atoms such as F, Cl and O.

8 6 Bond Enthalpies and the


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8.6 Bond Enthalpies and theEnthalpies of Reactions

Bond enthalpy for the bond between chlorineatoms in the Cl2molecule is:

The enthalpy change when a mole of Cl2

isdissociated into chlorine atoms:

Cl-Cl(g) 2Cl(g) H = D (Cl-Cl) = 242 kJ/mol

D(bond type) to represent bond enthalpies.

D-the energy required to break the diatomicmolecule into its component atoms.

Cont: 8 6 Bond Enthalpies and


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Cont: 8.6 Bond Enthalpies andthe Enthalpies of Reactions

Example: Atomization of CH4.

There are 4 equivalent C-H bonds inmethane.

CH4(g) C(g) + 4H(g) H = 1660 kJ/mol

The bond enthalpy is the fraction of H

for the atomization reaction:D(C-H) = 1660/4 kJ/mol= 415 kJ/mol.


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Cont:



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Cont: 8.6 Bond Enthalpies andthe Enthalpies of Reactions

The bond enthalpy is always positive:

Energy is always requiredto breakchemical

bonds. Energy is releasedwhen a bond formsbetween

2 gaseous atoms or molecules fragments.

The greaterthe bond enthalpy, the strongerthe bond (less tendency for reactions).



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Con t: 8.6 Bond Enthalpies andthe Enthalpies of Reactions

We can estimate the enthalpies of reactions inwhich bonds are broken and new bonds areformed.

We can estimate whether the reaction will beendothermic (H>0) or exothermic (H


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Example 5

CH4(g) + Cl-Cl(g) CH3Cl(g) + HCl(g)

Bond broken : 1 mol C-H; 1mol Cl-ClBond formed : 1mol C-Cl; 1mol H-Cl

Hrxn

= [D(C-H) + D(Cl-Cl)] - [D(C-Cl) + D(H-Cl)]

= [(413 kJ + 242 kJ) - (328 kJ + 431 kJ)]

= -104 kJ.


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Example 6

Estimate Hfor the following reaction.

Break: (H-N) 4 = 4 x 391 Formed: NN = 941(N-N) = 163 (H-H)2 = 436 x 2

= 1727 kJ = 1813 kJHrxn = 1727 kJ - 1813 kJ

= -86 kJ.

2N N

H

H

H

H(g)

N N (g) + H H (g)


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END of CHAPTER 8

chapter 8 jan13

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