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New Way Chemistry for Hong Kong A-Level Book 11

Covalent BondingCovalent Bonding

8.18.1 Formation of Covalent BondsFormation of Covalent Bonds8.28.2 Dative Covalent BondsDative Covalent Bonds8.38.3 Bond EnthalpiesBond Enthalpies

8.48.4 Estimation of Average Bond Enthalpies usingEstimation of Average Bond Enthalpies usingData from EnergeticsData from Energetics

8 8


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8.58.5 Use of Average Bond Enthalpies to Estimate theUse of Average Bond Enthalpies to Estimate theEnthalpy Changes of ReactionsEnthalpy Changes of Reactions

8.68.6 Bond Enthalpies, Bond Lengths and CovalentBond Enthalpies, Bond Lengths and Covalent

RadiiRadii8.78.7 Shapes of Covalent Molecules and PolyatomicShapes of Covalent Molecules and PolyatomicIonsIons8.88.8 Multiple BondsMultiple Bonds8.98.9 Covalent CrystalsCovalent Crystals


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8.8.

11 Formation of Formation of Covalent BondsCovalent Bonds


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H H

Sharedelectrons

The shared electronpair spends most of the

time between the twonuclei.

e-e-

Attraction betweenoppositely charged nucleiand shared electrons( _____________ in nature)electrostatic

Overlapping of atomic orbitals covalent bond formation

Overlapping of atomic orbitals covalent bond formation

8.1 Formation of Covalent Bonds (SB p.213)

A. Electron Sharing in Covalent BondsA. Electron Sharing

in Covalent Bonds


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A hydrogen molecule is achieved byA hy

drogen molecule is achieved by

partial overlapping of 1s orbitals

partial overlapping of 1s orbitals



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Thus electronsare shared between the

two atoms.

Compare electron-density-map for ionic compounds:

There issubstantialelectron densityat all points alongthe internuclear

axis.

Electron density map for covalent compoundsElectron density

map for covalent compounds8.1 Formation of Covalent Bonds (SB p.214)


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Dot and cross diagramDot and cross diagram


B. Covalent Bonds in ElementsB. Covalent Bonds in Elements Hydrogen molecule


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Chlorine molecule

Oxygen molecule


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Nitrogen molecule


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C. Covalent Bonds in CompoundsC. Covalent Bonds in Compounds


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8.1 Formation of Covalent Bonds (SB p.216 217)

D. Octet Rule and its LimitationsD. Octet Rule and its Limitations

In forming chemical bonds, atoms tend toachieve the stable noble gas electronicconfiguration with 8 electrons in the valence

shell (except helium which has 2 electrons inthe valence shell) by gaining, losing or sharing of electrons.

In forming chemical bonds, atoms tend toachieve the stable noble gas electronicconfiguration with 8 electrons in the valenceshell (except helium which has 2 electrons inthe valence shell) by gaining, losing or sharing of electrons.


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Why doesnt B form ionic compounds with F?Why doesnt B form ionic compounds with F?

B: small atomicsize high I.E.srequired tobecome a cation.

B: small atomicsize high I.E.srequired tobecome a cation.

electrons from F

not fullfillingoctect (electrondeficient)


1. Boron Trifluoride (BF1. Boron Trifluoride ( BF33))


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Why Phosphorus can expand its octet to form PCl 5?Why Phosphorus can expand its octet to form PCl 5?

There is low-lyingvacant d-orbital inP.

There is low-lying vacant d-orbital in

P.

electrons from Cl


2. Phosphorus Pentachloride (PCl2. Phosphorus Pentachloride (PCl 55))

Check PointCheck Point8-18-1


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8.8.

22Dative Covalent Dative Covalent BondsBonds


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8.2 Dative Covalent Bonds (SB p.218)

Dative Covalent BondsDative Covalent Bonds

A dative covalent bond is formed by theoverlapping of an empty orbital of an atomwith an orbital occupied by a lone pair of

electrons of another atom.

A dative covalent bond is formed by theoverlapping of an empty orbital of an atomwith an orbital occupied by a lone pair of electrons of another atom.

Remarks(1) The atom that supplies the shared pair of electrons is

known as the donor while the other atom involved in thedative covalent bond is known as the acceptor .

(2) Once formed, a dative covalent bond cannot be distinguished from a normal covalent bond.


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8.2 Dative Covalent Bonds (SB p.218 219)

A. NHA. NH33BFBF33 moleculemolecule


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AlCl 3

Al: relative smallatomic size; highI.E.s required tobecome a cationof +3 charge.

Al: relative smallatomic size; highI.E.s required tobecome a cationof +3 charge.


D. Aluminium Chloride Dimer (AlD. Aluminium Chloride Dimer (Al22ClCl66))


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Why doesnt Al form ionic compounds with Cl?Why doesnt Al form ionic compounds with Cl?(a dimer of AlCl 3)


D. Aluminium Chloride Dimer (AlD. Aluminium Chloride Dimer (Al22ClCl66))



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8.8.

33Bond EnthalpiesBond Enthalpies


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8.3 Bond Enthalpies (SB p.220)

Bond EnthalpyBond Enthalp y

Bond enthalpy is the energy associated with achemical bond. When a chemical bond isbroken or formed, a certain amount of energy

is absorbed from or released to thesurroundings.

Bond enthalpy is the energy associated with achemical bond. When a chemical bond isbroken or formed, a certain amount of energyis absorbed from or released to thesurroundings.


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Example:

Combustion of methane


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Standard enthalpy changes of combustion of thehomologous series of alkanes and alkanols


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e.g. H-H(g) 2H(g) H = +431 kJ mol -1


Bond Dissoication EnthalpyBond Dissoication Enthalp y

Bond dissociation enthalpy is the enthalpychange when one mole of a particular bond ina particular environment is broken under

standard conditions.

Bond dissociation enthalpy is the enthalpychange when one mole of a particular bond ina particular environment is broken under standard conditions.

d h l ( )


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Why do successive B.D.E. of C-H differ?Why do successive B.D.E. of C-H differ?

(Average) bond enthalpy; E(C-H)

4

335)(425)(480)(422)(

+++++++=

= +415.5 kJ mol -1

CH 4(g ) CH 3(g ) + H( g ) H = +422 kJ mol -1

CH 3(g ) CH 2(g ) + H( g ) H = +480 kJ mol-1

CH 2(g ) CH( g ) + H( g ) H = +425 kJ mol -1

CH( g ) C( g ) + H( g ) H = +335 kJ mol -1


8 3 B d E h l i (SB 222)


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Average Bond EnthalpiesAverag e Bond Enthalpies

Average bond enthalpy is the average of thebond dissociation enthalpies required to breaka particular chemical bond.

Average bond enthalpy is the average of thebond dissociation enthalpies required to breaka particular chemical bond.


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8 4 E i i f A B d E h l i i D f E i


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8.4 Estimation of Average Bond Enthalpies using Data from Enegetics(SB p.223)

A. Derived from the Enthalpy ChangeA. Derived from the Enthalp y Changeof Atomization of a Compoundof Atomization of a Comp ound

Atomization of a compound means the

breaking down of one mole of the gaseouscompound into its constituent atoms in thegaseous state.

Atomization of a compound means the

breaking down of one mole of the gaseouscompound into its constituent atoms in thegaseous state.

8 4 E ti ti f A B d E th l i i D t f E ti


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Example:

Atomization of methane

C(g ) + 4H( g ) H = +1 662 kJ mol -1

E(C-H) = +415.5 kJ mol -1E(C-H) = +415.5 kJ mol -1

The atomization of methane involves the breaking of a four C-H bonds. Assume that all four C-H bonds are equal instrength.

The average bond enthalpy of C-H bonds

= x (+1 662) kJ mol -1 = +415.5 kJ mol -1

8 4 E ti ti f A g B d E th l i i g D t f E g ti


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Two ways to determine the enthalpy change

of atomization of methane:1. From successive bond dissociationenthalpies

2. From enthalpy cycle and Hesss law

8 4 Estimation of Average Bond Enthalpies using Data from Enegetics


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The enthalpy change of atomization of butane

(C 4H10 ) and pentane (C 5H12 ) are +5165 kJ mol-1

and+6337 kJ mol -1 respectively. Find a values for thebond enthalpies of C-H and C-C based on theabove data.

B. Derived from the Enthalpy ChangesB. Derived from the Enthalp y Changesof Atomization of Two Compoundsof Atomization of Two Comp ounds

8.4 Estimation of Average Bond Enthalpies using Data from Enegetics(SB p.224 225)

8 4 Estimation of Average Bond Enthalpies using Data from Enegetics


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For butane,

3 E(C-C) + 10 E(C-H) = +5 165 kJ mol -1 .(1)

For pentane,

4 E(C-C) + 12 E(C-H) = +6 337 kJ mol -1 .(2)

Solving simultaneous equations (1) and (2), weobtain the following bond enthalpy values.

E (C-H) = +412.25 kJ mol -1

E (C-C) = +347.5 kJ mol -1

B. Derived from the Enthalpy ChangesB. Derived from the Enthalp y Changesof Atomization of Two Compoundsof Atomization of Two Comp ounds

8.4 Estimation of Average Bond Enthalpies using Data from Enegetics(SB p.224 225)


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8.8.55Use of AverageUse of Average

Bond EnthalpiesBond Enthalpiesto Estimate theto Estimate the

Enthalpy ChangesEnthalpy Changes

of Reaction sof Reaction s

8 5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes


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Sum of bondenthalpies of products

Enthalpy change

of reaction=

Sum of bondenthalpies of reactants

-

8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changesof Reactions (SB p.225)

Reaction between ethene and hydrogenReaction between ethene and hy drogen



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Enthalpy level diagram for the reactionEnthalp y level diagram for the reactionbetween ethene and hydrogenbetween ethene and hy drogen



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Reaction between methane and oxygenReaction between methane and oxy gen



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8.8.

66Bond Enthalpies,Bond Enthalpies,Bond Lengths and Bond Lengths and

Covalent Radii Covalent Radii

8 6 Bond Enthalpies Bond Lengths and Covalent Radii (SB p 227)


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8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.227)

A. Bond Enthalpies as an Indication ofA. Bond Enthalp ies as an Indication of

the Strength of Covalent Bondsthe Streng th of Covalent Bonds Gives a direct measure of the strength of a

covalent bond

It is the energy required to break the bond Not in proportion to the bond order

(The number of bonding electrons divided by

two)



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B. Bond LengthsB. Bond Lengths The distance between the two bonded nuclei

Inversely related to bond strength

Not constant Depends on the local environment of that

particular bond

Determined experimentally by electrondiffraction, X-ray diffraction or spectroscopictechniques




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Any conclusion for the relationshipbetween bond

length & bondenthalpy?

Any conclusion for the relationshipbetween bond

length & bondenthalpy?

Usually a longer bond lengthcorresponds toa lower value of bond enthalpy(weaker bond).

Usually a longer bond lengthcorresponds toa lower value of bond enthalpy(weaker bond).

Bond Bond length(nm)

Bond enthalpy(kJ mol -1 )

H-HCl-ClBr-Br

I-IH-FH-ClH-Br H-I

0.0740.1990.2280.2660.0920.1270.1410.161

436242193151565431364299

C. Relationship between Bond LengthsC. Relationship between Bond Lengths

and Bond Enthalpiesand Bond Enthalp ies




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Bond Bond Length /nm Bond Enthalpy / kJ mol -1

F-F 0.142 158

Cl-Cl 0.199 242

Br-Br 0.228 193

I-I 0.266 151

Special Situation for FSpecial Situation for F 22


Explain why the bond enthalpy of F-F is smaller than thatof Cl-Cl even though the bond length of F-F is theshortest among the halogens.



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As the size of fluorine atom is very small, the repulsion

between the non-bonding pairs of electrons on the fluorineatoms weaken the F-F bond.

FF

Non-bonding e -

/ lone pair of e -




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D. Covalent RadiiD. Covalent Radii

Half the internuclear distance between twoatoms in a covalently bonded molecule

Generally taken as half of the bond length of homoatomic covalent molecules (whereidentical atoms are bonded together)



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p , g ( p )

The covalent radius of an atom is taken as half ofthe bond length of a homoatomic molecule



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p , g ( p )

The covalent radii (in nm) of some elements



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Predicting bond length of A-B if r A & r B are knownPredicting bond length of A-B if r A & r B are known

Bond lengthof a covalentbond A-B

=Covalentradius of atom A

+Covalentradius of atom B

p , g ( p )

8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.229


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Bond Calculated bond length(nm)

Experimentally determinedbond length (nm)

C-OC-FC-ClC-Br

C-CH-ClC-HN-Cl

0.1500.1490.1760.191

0.1540.1360.1140.173

0.1430.1380.1770.193

0.1540.1280.1090.174

By what technique can the bond lengthsbe determined experimentally?

By what technique can the bond lengthsbe determined experimentally?

Similar electronegativity

p , g ( p230)Calculated and experimentally determinedCalculated and exper imentally determinedbond lengthbond leng th

8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.229


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Bond Calculated bond length(nm)

Experimentally determinedbond length (nm)

C-OC-FC-ClC-Br

C-CH-ClC-HN-Cl

0.1500.1490.1760.191

0.1540.1360.1140.173

0.1430.1380.1770.193

0.1540.1280.1090.174

p g ( p230)Calculated and experimentally determinedCalculated and exper imentally determinedbond lengthbond leng th

Quite differentelectronegativity



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8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.231)


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Shapes of Covalent Molecules andShap es of Covalent Molecules and

Polyatomic IonsPolyatomic Ions Geometric arrangement of atoms within the

molecules or ions

The non-bonding electrons (i.e. the lone pair electrons) are not taken into account


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A. Molecules and Polyatomic Ions withoutA. Molecules and Polyat omic Ions withoutLone Pair Electrons on the Central AtomLone Pair Electrons on the Central Atom Examples:

1. Beryllium Chloride (BeCl 2) Molecule

2. Boron Trifluoride (BF 3) Molecule

3. Methane (CH 4) Molecule

4. Ammonium Ion (NH 4+)

5. Phosphorus Pentachloride (PCl 5) Molecule

6. Sulphur Hexafluoride (SF 6) Molecule



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1. Beryllium Chloride Molecule (BeCl1. Beryllium Chloride Molecule (BeCl22))

BeCl Cl

Electronic Diagram

Shape in word Linear

Bond angle= angle between

2 bonds

Bond angle

= angle between2 bonds

Shape in Diagram



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2. Boron Trifluoride Molecule (BF2. Boron Trifluoride Molecule (B F33))

Shape in word

Trigonal planar

B

F F

F

Electronic Diagram Shape in Diagram



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3. Methane (CH3. Methane (C H44) Molecule) Molecule

Tetrahedral

CH H

H

HShape in word



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Trigonal bipyramidal


5. Phosphorus Pentachloride (PCl5. Phosphorus Pentachloride (PCl 55) Molecule) Molecule

P

Shape in wordCl

Cl

Cl Cl

Cl



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Octahedral


6. Sulphur Hexafluoride (SF6. Sulp hur Hexafluoride (SF 66) Molecule) Molecule

Shape in word

S

F

F

F

F

F

F



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B. Molecules and Polyatomic Ions withB. Molecules and Polyat omic Ions withLone Pair Electrons on the Central AtomLone Pair Electrons on the Central Atom

The valence shell electron pair repulsiontheory states

Electrostatic repulsion decreases in thefollowing order:

Lone pair lone pair > Lone pair bond pair > Bond pair bond pair



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B. Molecules and Polyatomic Ions withB. Molecules and Polyat omic Ions withLone Pair Electrons on the Central AtomLone Pair Electrons on the Central Atom Examples:

1. Ammonia (NH 3) Molecule

2. Water (H 2O) Molecule

3. Amide Ion (NH 2 )



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Trigonal pyramidal

H H

H

N

lp-lp repulsion > lp-bp repulsion

> bp-bp repulsion


> bp-bp repulsion


1. Ammonia (NH1. Ammonia (NH33) Molecule) Molecule

Shape in word



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V-shaped


> bp-bp repulsionlp-lp repulsion > lp-bp repulsion

> bp-bp repulsion


2. Water (H2. Water ( H22O) MoleculeO) Molecule

Shape in word

H HO



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V-shaped


> bp-bp repulsionlp-lp repulsion > lp-bp repulsion

> bp-bp repulsion


3. Amide Ion (NH3. Amide Ion ( NH22))

Shape in word

H HN

8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237 238)


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238)

ExampleExample8-7A8-7A

ExampleExample8-7B8-7B



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8.8.

88Multiple BondsMultiple Bonds

8.8 Multiple Bonds (SB p.238)


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Single BondsSingle Bonds

A covalent bond with two shared electrons

Multiple BondsMultiple Bonds

Some atoms share more than two electronsin a bond

e.g. double bond, triple bond



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Bond Bond order Bond length(nm)

Bond enthalpy(kJ mol -1 )

C C

C = CC C

1

23

0.154

0.1340.120

+348

+612+837

N NN = N

N

N

12

3

0.1460.120

0.110

+163+409

+944C OC = O

12

0.1430.122

+360+743

Comparison of bond lengths and bondComparison of bond lengths and bondenthalpies between single and multiple bondsenthalpi es between single and multiple bonds



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Effect of Multiple Bonding on Shapes ofEffect of Multip le Bonding on Shapes ofMoleculesMolecules Predict the shapes of molecules or polyatomic

ions with multiple bonds

Examples:1. Ethene (CH 2 = CH 2) Molecule

2. Ethyne (CH CH) Molecule

3. Carbon Dioxide (CO 2) Molecule

4. Sulphur Dioxide (SO 2) Molecule



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Trigonal planar

C C

H

HH

H


1. Ethene (CH1. Ethene ( CH22 = CH= CH22) Molecule) Molecule

Shape in word



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Linear


2. Ethyne (CH2. Ethy ne (CH CH) MoleculeCH) Molecule

C C HH

Shape in word



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Linear


3. Carbon dioxide (CO3. Carbon dioxide ( CO22) Molecule) Molecule

Shape in word

CO O



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V-shaped


4. Sulphur dioxide (SO4. Sulp hur dioxide (SO 22) Molecule) Molecule

Shape in word

SO O



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8.9 Covalent Crystals (SB p.240)


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Covalent CrystalsCovalent Crystals

May have simple molecular structures or giant covalent structures



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Substances with Simple MolecularSubstances with Simp le Molecular

StructuresStructures Consist of discrete molecules held together

by weak intermolecular forces

Atoms in a molecule are held together bystrong covalent bonds

Examples: H 2 , O 2 , H 2O, CO 2, I2



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Substances with Giant CovalentSubstances with Giant Covalent

StructuresStructures Consist of millions of atoms bonded

covalently together in a structural network

No simple molecules present

Examples: diamond, graphite and quartz(silicon(IV) oxide)



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1. Diamond1. Diamond

Each C atom is covalently bonded to 4other C atoms to form a three-dimensionalnetwork

The C C bonding pattern accounts for thehigh m.p. , stability and extreme hardness

Applications: scratch proof cookware,watch crystals, ball bearings and razor blade


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The structure of diamond



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2. Graphite2. Grap hite

Each C atom is covalently bonded to 3other C atoms in the same layer. A networkof coplanar hexagons is formed

Weak van der Waals forces hold the layerstogetherDelocalized e- free to move withinlayers

Properties: soft and slippery (used as pencillead), conductor



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Graphite



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The structure of graphite



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Why graphite has a high m.p. than that of diamond?Why graphite has a high m.p. than that of diamond?

Property Diamond Graphite

Density (g cm -3 )

HardnessMelting point ( C)Colour Electrical conductivity

3.51

10 (hardest)3 827ColourlessNone

2.27

< 1 (very soft)3 652 (sublime)Shiny blackHigh

Comparison of the properties ofComparison of the properties of

diamond and graphitediamond and g raphite


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Quartz



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The structure of quartz


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The END



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(a) How many lone pair and bond pairelectrons are present in NH 3 and H 2 Omolecules respectively?(a) Ammonia has one lone pair and three bond pairs of electrons.Water has two lone pairs and two bond pairs of electrons.

Answer



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(b) The electronic configuration of nitrogen is 1 s2

2s2

2 p x 1

2 py 1

2 pz 1

. Itsoutermost shell electrons are filled in the second quantum shell.There are no lowlying d orbitals available for nitrogen to expandoctet. It has a maximum of three half-filled p orbitals to form threebonds,i.e. NCl

3.

(b) Nitrogen can only form one chloride,NCl 3 , while phosphorus can form twochlorides, PCl 3 and PCl 5 . Explain briefly.Answer


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(b) Phosphorus has low-lying d orbitals which allow it to expand octet(contain more than eight outermost shell electrons) whereas nitrogenhas not.

Back



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(a) Draw a dot and cross diagram for theproduct formed in the reaction between anammonia molecule and a hydrogen chloridemolecule. Answer

(a)



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(b) There is a dative covalent bond presentin a HNO 3 molecule. Draw a dot and crossdiagram of the molecule.

Answer(b)



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(c) A dative covalent bond is covalent bond in which the shared pair of electrons is supplied by only one of the bonded atoms, whereaselectrons in an ordinary covalent bond come from both bondedatoms.

(c) State the major difference between anordinary and a dative covalent bond.

Answer

Back


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(a) Referring to Table 8-2 on page 222,calculate the enthalpy change for thefollowing reactions and state whether thereactions are endothermic or exothermic.

(i) Reaction between nitrogen and hydrogen.N 2 (g) + 3H 2 (g) 2NH 3 ( g) Answer(a) (i)

Sum of average bond enthalpies of reactants

= E(N N) + 3 E(H H)

= [+944 + 3 (+436)] kJ mol 1

= +2 252 kJ mol 1



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(a) (i) Sum of average bond enthalpies of products

= 6 E(N H)

= 6 (+388) kJ mol 1

= +2 238 kJ mol 1

H = [+2 252 (+2 328)] kJ mol 1

= 76 kJ mol 1

The reaction is exothermic.



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(a) (ii) H H + Cl Cl 2H Cl

Sum of average bond enthalpies of reactants= E(H H) + E(Cl Cl)

= (+436 + 242) kJ mol 1

= +678 kJ mol 1

(a) (ii) Reaction between hydrogen andchlorine.

H 2 (g) + Cl 2 (g) 2HCl(g) Answer

Back



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(a) (ii) Sum of average bond enthalpies of products

= 2 E(N Cl)

= 2 (+431) kJ mol 1

= +862 kJ mol 1

H = [+678 (+862)] kJ mol 1

= 184 kJ mol 1




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(a) (iii) Complete combustion of hydrogen.Answer

(a) (iii)


= 2 E(H H) + E(O = O)

= [2 (+436) + 496] kJ mol 1

= +1 368 kJ mol 1



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(a) (iii) Sum of average bond enthalpies of products

= 4 E(O H)

= 4 (+463) kJ mol 1

= +1 852 kJ mol 1

H = [+1 368 (+1 852)] kJ mol 1

= 484 kJ mol 1




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(a) (iv) Complete combustion of ethanol.Answer(a) (iv)


= E(C C) + E(C O) + E(O H) + 5 E(C H) + 3 E(O = O)

= [+348 + 360 + 463 + 5 (+412) + 3 (+496)] kJ mol 1

= +4 719 kJ mol 1



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(a) (iv) Sum of average bond enthalpies of products

= 4 E(C = O) + 6 E(O H)

= [4 (+743) + 6 (+463)] kJ mol 1

= +5 750 kJ mol 1

H = [+4 719 (+5 750)] kJ mol 1

= 1031 kJ mol 1




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(a) (v) Complete combustion of octane.Answer

(a) (v)


= 14 E(C C) + 36 E(C H) + 25 E(O = O)

= [14 (+348) + 36 (+412) + 25 (+496)] kJ mol 1

= +32 104 kJ mol 1



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(a) (iv) Sum of average bond enthalpies of products

= 32 E(C = O) + 36 E(O H)

= [32 (+743) + 36 (+463)] kJ mol 1

= +40 444 kJ mol 1

H = [+32 104 (+40 444)] kJ mol 1

= 8 340 kJ mol 1




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(b) Calculate the enthalpy change for thereaction

CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g)

using the following bond enthalpies.E(C H in CH 4 ) = +435 kJ mol 1

E(C O in CO) = +1 078 kJ mol 1

E(H H in H 2 ) = +436 kJ mol 1

E(H O in H 2 O) = +464 kJ mol 1Answer



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(b) CH 4(g) + H 2O CO(g) + 3H 2(g)


= 4 E(C H) + 2 E(O H)

= [4 (+435) + 2 (+464)] kJ mol 1

= +2 668 kJ mol 1

Sum of average bond enthalpies of products

= E(C O) + 3 E(H H)

= [+1 078 + 3 (+436)] kJ mol 1

= +2 386 kJ mol 1

H = [+2 668 (+2 386)] kJ mol 1 = +282 kJ mol 1

Back



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Why does the covalent radius of a given elementchange from one compound to another compound?

Back

The size of an atom (its covalent radius) is not fixed. It is because thesize of an atom is determined by its electron cloud which has a diffuseshape. In different compounds, the electron cloud of a given atom mayvary slightly due to the different internal environment (i.e. the atom thatis bonded to).

Answer



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(a) Predict the approximate bond lengths of

Si H, P H, S H and H Cl from thefollowing data:

(Hint: Assume that covalent radii are

additive.)

Bond Bond length (nm)

H HSi Si

P P (P 4)

S S (S 4)

Cl Cl

0.0740.2350.2210.2070.199

Answer



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(a) Bond length of Si H =

= 0.154 5 nm

Bond length of P H == 0.147 5 nm

Bond length of S H =

= 0.140 5 nm

Bond length of H Cl =

= 0.136 5 nm

nm2

0.074nm

20.235

+

nm2

0.074nm

2

0.221+

nm2

0.074nm

20.207

+

nm2

0.199nm

20.074

+



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(b) The bond enthalpies of Si H, P H, S H and H Cl are given in the following table:

Assume the actual bond lengths are very closeto that calculated in (a), describe therelationship between bond length and bondenthalpy.

Bond Bond enthalpies (kJ mol 1 )

Si HP HS HCl H

+318+322+338+431

Answer



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(b) The bond enthalpy of a covalent bond is related to the length. Thelarger the bond length, the weaker the attractive force between thetwo bonded atoms and the smaller is the bond enthalpy.

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(a) Explain why a molecule of CCl 4 is tetrahedral,but a molecule of NCl 3 is trigonal pyramidal inshape.(a) In a CCl 4 molecule, there are four bond pairs of electrons on the central

carbon atom. The bond pairs have to stay as far away as possible. Theytake up the shape of a tetrahedron and thus the molecule is tetrahedralin shape. The four electron pairs in a NCl 3 molecule take up the shapeof a tetrahedron as well. However, one of the electron pairs is a lonepair and the other three are bond pairs. The shape of a NCl 3 molecule isthus trigonal pyramidal.

Answer



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(b) Deduce the shape of a molecule of BCl 3 .

(b) A BCl 3 molecule has six outermost shell electrons around the centralboron atom, forming three bond pairs. The shape of the BCl 3 moleculeis thus trigonal planar.

Answer


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The following data refer to the molecules NH 3 ,H 2 O and HF.

Molecule Bond length (nm) Bond angle

NH3 0.101 107

H2O 0.096 104.5

HF 0.092



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(a) Briefly explain the variation in bond length.

Answer(a) The bond lengths of the three molecules decrease as follows:H N H O H F

0.101 nm 0.096 nm 0.092 nm

The atomic radius of H is the same in the three molecules, so the bondlengths of the molecules depend on the size of the N, O and F atoms.N, O and F are in the same period in the Periodic Table. Since atomicsizes decrease across a period owing to the increase in effectivenuclear charge, the bond lengths of the three molecules decreaseaccordingly.



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(b) Explain why the bond angle of H 2 O is lessthan that of NH 3 .

(b) This can be explained by the valence shell electron pair repulsiontheory.

The central oxygen atom in H 2O has two lone pairs and two bond pairsof electrons while the central nitrogen atom in NH 3 has one lone pair and three bond pairs of electrons.

The electrostatic repulsion between electron pairs decreases in thisorder:

lone pair and lone pair > lone pair and bond pair > bond pair and bondpair

Thus, the bond angle of H 2O is less than that of NH 3.

Answer



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(c) Match the following bond enthalpies to thebonds in the above three molecules:+562 kJ mol l , +388 kJ mol l , +463 kJ mol lAnswer

(c) The bond enthalpies are:

H N H O H F

+388 kJ mol l +463 kJ mol l +562 kJ mol l

The bond enthalpies increase as shown owing to the decrease in bondlength and increase in polarity of bonds.

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What are the shapes of a H 2 S molecule and aH 3 O + ion?

Explain their shapes in terms of the valence shellelectron pair repulsion theory. AnswerH2S molecule is V-shaped. In H 2S molecule, there are two bond pairs andtwo lone pairs of electrons in the outermost shell of the central sulphur atom. All three types of electrostatic repulsion (lone pair lone pair, lonepair bond pair, bond pair bond pair) are present. The two lone pairs willstay the furthest apart and the separation between the lone pair and a bond

will be greater that that between the two bond pairs. Therefore, the H S H bond angle in the H 2S molecule is about 104.5 instead of 109.5 intetrahedron.



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H3O + ion has a trigonal pyramidal shape. In H 3O + ion, the central oxygenatom forms two covalent bonds with two hydrogen atoms respectively.Also, one dative covalent bond is formed between the oxygen atom and theremaining hydrogen ion. We can regard the central oxygen atom has threebond pairs and one lone pair of electrons. According to the valence shell

electron pair repulsion theory, the lone pair will stay further away from thethree bond pairs. The three bond pairs are in turn compressed closer together. Thus, the H O H bond angles in the H 3O + ion are about 107 instead of 109.5 in tetrahedron.

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(a) Does sulphur obey the octet rule in forming aSO 2 molecule? Explain your answer. Answer

(a) In the formation of SO 2 molecule, sulphur does not obey the octet rulebecause sulphur has 10 electrons in its outermost shell.


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(b) Draw a dot and cross diagram of thehydrogen cyanide molecule (HCN). Describeand explain the shape of the molecule. Answer

(b)

HCN molecules has a linear shape as the central carbon atom does

not have any lone pair electrons. In order to minimize electrostaticrepulsion, the two electron clouds of the central carbon atom areseparated at a maximum with bond angles of 180 .

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