New Way Chemistry for Hong Kong A-Level Book 11 Covalent Bonding 8.1Formation of Covalent Bonds 8.2Dative Covalent Bonds 8.3Bond Enthalpies 8.4Estimation

New Way Chemistry for Hong Kong A-Level Book 11

Covalent BondingCovalent Bonding

8.18.1 Formation of Covalent BondsFormation of Covalent Bonds

8.28.2 Dative Covalent BondsDative Covalent Bonds

8.38.3 Bond EnthalpiesBond Enthalpies

8.48.4 Estimation of Average Bond Enthalpies using Estimation of Average Bond Enthalpies using DD

ata from Energeticsata from Energetics

88


8.58.5 Use of Average Bond Enthalpies to Estimate theUse of Average Bond Enthalpies to Estimate the

Enthalpy Changes of ReactionsEnthalpy Changes of Reactions

8.68.6 Bond Enthalpies, Bond Lengths and Covalent Bond Enthalpies, Bond Lengths and Covalent RR

adiiadii

8.78.7 Shapes of Covalent Molecules and Polyatomic Shapes of Covalent Molecules and Polyatomic IoIo

nsns

8.88.8 Multiple BondsMultiple Bonds

8.98.9 Covalent CrystalsCovalent Crystals


8.8.11 Formation of Formation of Covalent BondsCovalent Bonds


H H

Shared electrons

The shared electron pair spends most of the time between the two nuclei.

e-

e-

Attraction between oppositely charged nuclei and shared electrons ( _____________ in nature)electrostatic

Overlapping of atomic orbitals covalent bond formationOverlapping of atomic orbitals covalent bond formation

8.1 Formation of Covalent Bonds (SB p.213)

A. Electron Sharing in Covalent A. Electron Sharing in Covalent BondsBonds


A hydrogen molecule is achieved by A hydrogen molecule is achieved by partial overlapping of 1s orbitalspartial overlapping of 1s orbitals



Thus electrons are shared between the two atoms.

Compare electron-density-map for ionic compounds:

There is substantial electron density at all points along the internuclear axis.

Electron density map for covalent Electron density map for covalent compoundscompounds



Dot and cross diagramDot and cross diagram


B. Covalent Bonds in ElementsB. Covalent Bonds in Elements

• Hydrogen molecule



• Chlorine molecule

• Oxygen molecule



• Nitrogen molecule



C. Covalent Bonds in CompoundsC. Covalent Bonds in Compounds




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 to achieve the stable noble gas electronic configuration with 8 electrons in the valence shell (except helium which has 2 electrons in the valence shell) by gaining, losing or sharing of electrons.

In forming chemical bonds, atoms tend to achieve the stable noble gas electronic configuration with 8 electrons in the valence shell (except helium which has 2 electrons in the valence shell) by gaining, losing or sharing of electrons.


Why doesn’t B form ionic compounds with F?Why doesn’t B form ionic compounds with F?

B: small atomic size high I.E.’s required to become a cation.

B: small atomic size high I.E.’s required to become a cation.

electrons from F

not fullfilling octect (electron deficient)


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


Why Phosphorus can expand its octet to form PCl5?Why Phosphorus can expand its octet to form PCl5?

There is low-lying vacant d-orbital in P.

There is low-lying vacant d-orbital in P.

electrons from Cl


2. Phosphorus Pentachloride (PCl2. Phosphorus Pentachloride (PCl55))

Check Point 8-1Check Point 8-1


8.8.22Dative Covalent Dative Covalent

BondsBonds


8.2 Dative Covalent Bonds (SB p.218)

Dative Covalent BondsDative Covalent Bonds

A dative covalent bond is formed by the overlapping of an empty orbital of an atom with an orbital occupied by a lone pair of electrons of another atom.

A dative covalent bond is formed by the overlapping of an empty orbital of an atom with 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 the dative covalent bond is known as the acceptor.

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


8.2 Dative Covalent Bonds (SB p.218 – 219)

A. NHA. NH33BFBF33 molecule molecule



B. Ammonium Ion (NHB. Ammonium Ion (NH44++))


AlCl3

Al: relative small atomic size; high I.E.’s required to become a cation of +3 charge.

Al: relative small atomic size; high I.E.’s required to become a cation of +3 charge.


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


Why doesn’t Al form ionic compounds with Cl?Why doesn’t Al form ionic compounds with Cl?(a dimer of AlCl3)


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



8.8.33Bond EnthalpiesBond Enthalpies


8.3 Bond Enthalpies (SB p.220)

Bond EnthalpyBond Enthalpy

Bond enthalpy is the energy associated with a chemical bond. When a chemical bond is broken or formed, a certain amount of energy is absorbed from or released to the surroundings.

Bond enthalpy is the energy associated with a chemical bond. When a chemical bond is broken or formed, a certain amount of energy is absorbed from or released to the surroundings.



• Example:

Combustion of methane



Standard enthalpy changes of combustion of the homologous series of alkanes and alkanols


e.g. H-H(g) 2H(g) H = +431 kJ mol-1ø


Bond Dissoication EnthalpyBond Dissoication Enthalpy

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

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


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)

4335)( 425)( 480)( 422)(

= +415.5 kJ mol-1

CH4(g) CH3(g) + H(g) H = +422 kJ mol-1

CH3(g) CH2(g) + H(g) H = +480 kJ mol-1

CH2(g) CH(g) + H(g) H = +425 kJ mol-1

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

øø

øø




Average Bond EnthalpiesAverage Bond Enthalpies

Average bond enthalpy is the average of the bond dissociation enthalpies required to break a particular chemical bond.

Average bond enthalpy is the average of the bond dissociation enthalpies required to break a particular chemical bond.


8.8.44Estimation of AveragEstimation of Averag

e Bond Enthalpies use Bond Enthalpies using Data from Enegeing Data from Enege

ticstics


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

A. Derived from the Enthalpy A. Derived from the Enthalpy Change of Atomization of a Change of Atomization of a CompoundCompound

Atomization of a compound means the breaking down of one mole of the gaseous compound into its constituent atoms in the gaseous state.

Atomization of a compound means the breaking down of one mole of the gaseous compound into its constituent atoms in the gaseous state.



• Example:

Atomization of methane

C(g) + 4H(g) H = +1 662 kJ mol-1ø

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 in strength.

The average bond enthalpy of C-H bonds

= ¼ x (+1 662) kJ mol-1 = +415.5 kJ mol-1



• Two ways to determine the enthalpy change of atomization of methane:

1. From successive bond dissociation enthalpies

2. From enthalpy cycle and Hess’s law


The enthalpy change of atomization of butane (C4H10) and pentane (C5H12) are +5165 kJ mol-1 and +6337 kJ mol-1 respectively. Find a values for the bond enthalpies of C-H and C-C based on the above data.

B. Derived from the Enthalpy B. Derived from the Enthalpy Changes of Atomization of Two Changes of Atomization of Two CompoundsCompounds

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


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), we obtain 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 B. Derived from the Enthalpy Changes of Atomization of Two Changes of Atomization of Two CompoundsCompounds

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


8.8.55Use of Average Use of Average

Bond Enthalpies Bond Enthalpies to Estimate the to Estimate the

Enthalpy Changes Enthalpy Changes of Reactionsof Reactions


Sum of bond enthalpies of products

Enthalpy change

of reaction =

Sum of bond enthalpies of reactants

-

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

Reaction between ethene and Reaction between ethene and hydrogenhydrogen



Enthalpy level diagram for the Enthalpy level diagram for the reaction between ethene and reaction between ethene and hydrogenhydrogen



Reaction between methane and Reaction between methane and oxygenoxygen



8.8.66Bond Enthalpies, Bond Enthalpies,

Bond Lengths and Bond Lengths and Covalent RadiiCovalent Radii


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

A. Bond Enthalpies as an Indication A. Bond Enthalpies as an Indication of the Strength of Covalent Bondsof the Strength 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)


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 electron diffraction, X-ray diffraction or spectroscopic techniques



Any conclusion for the relationship between bond length & bond enthalpy?

Any conclusion for the relationship between bond length & bond enthalpy?

Usually a longer bond length corresponds to a lower value of bond enthalpy (weaker bond).

Usually a longer bond length corresponds to a lower value of bond enthalpy (weaker bond).

BondBond length

(nm)Bond enthalpy

(kJ mol-1)

H-H

Cl-Cl

Br-Br

I-I

H-F

H-Cl

H-Br

H-I

0.074

0.199

0.228

0.266

0.092

0.127

0.141

0.161

436

242

193

151

565

431

364

299

C. Relationship between Bond C. Relationship between Bond Lengths and Bond EnthalpiesLengths and Bond Enthalpies



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 F22


Explain why the bond enthalpy of F-F is smaller than that

of Cl-Cl even though the bond length of F-F is the shortes

t among the halogens.


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

between the non-bonding pairs of electrons on the fluorine

atoms weaken the F-F bond.

FF

Non-bonding e-

/ lone pair of e-




D. Covalent RadiiD. Covalent Radii

• Half the internuclear distance between two atoms in a covalently bonded molecule

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



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



The covalent radii (in nm) of some elements


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

Bond length of a covalent bond A-B

=Covalent radius of atom A

+Covalent radius of atom B



BondCalculated bond

length (nm)Experimentally determined

bond length (nm)

C-O

C-F

C-Cl

C-Br

C-C

H-Cl

C-H

N-Cl

0.150

0.149

0.176

0.191

0.154

0.136

0.114

0.173

0.143

0.138

0.177

0.193

0.154

0.128

0.109

0.174

By what technique can the bond lengths be determined experimentally?

By what technique can the bond lengths be determined experimentally?

Similarelectronegativity

8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.229 – 230)Calculated and experimentally Calculated and experimentally determined bond lengthdetermined bond length


BondCalculated bond

length (nm)Experimentally determined

bond length (nm)

C-O

C-F

C-Cl

C-Br

C-C

H-Cl

C-H

N-Cl

0.150

0.149

0.176

0.191

0.154

0.136

0.114

0.173

0.143

0.138

0.177

0.193

0.154

0.128

0.109

0.174

8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.229 – 230)Calculated and experimentally Calculated and experimentally determined bond lengthdetermined bond length

Quite differentelectronegativity



8.8.77Shapes of Covalent Shapes of Covalent

Molecules and PolyaMolecules and Polyatomic Ionstomic Ions


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

Shapes of Covalent Molecules and PolyShapes of Covalent Molecules and Polyatomic Ionsatomic 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



Valence shell electron pair Valence shell electron pair repulsion theoryrepulsion theory

Electron pairs in the valence shell of the central atom of a molecule will stay as far apart as possible to minimize the electrostatic repulsion between electron pairs in the valence shell. The electron pairs are oriented with the maximum separation in space so as to minimize the coulombic repulsion of electron clouds.

Electron pairs in the valence shell of the central atom of a molecule will stay as far apart as possible to minimize the electrostatic repulsion between electron pairs in the valence shell. The electron pairs are oriented with the maximum separation in space so as to minimize the coulombic repulsion of electron clouds.



A. Molecules and Polyatomic Ions without LA. Molecules and Polyatomic Ions without Lone Pair Electrons on the Central Atomone Pair Electrons on the Central Atom• Examples:

1. Beryllium Chloride (BeCl2) Molecule

2. Boron Trifluoride (BF3) Molecule

3. Methane (CH4) Molecule

4. Ammonium Ion (NH4+)

5. Phosphorus Pentachloride (PCl5) Molecule

6. Sulphur Hexafluoride (SF6) Molecule


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

BeClCl

Electronic Diagram

Shape in word Linear

Bond angle= angle between 2 bonds

Bond angle= angle between 2 bonds

Shape in Diagram



2. Boron Trifluoride Molecule (BF2. Boron Trifluoride Molecule (BF33))

Shape in word

Trigonal planar

B

F F

F


Electronic Diagram Shape in Diagram


3. Methane (CH3. Methane (CH44) Molecule) Molecule

Tetrahedral

CH H

H

H


Shape in word



Tetrahedral

NH H

H

H


Shape in word


4. Ammonium Molecule 4. Ammonium Molecule (NH(NH44

++))


Trigonal bipyramidal



5. Phosphorus Pentachloride (PCl5. Phosphorus Pentachloride (PCl55) Molecule) Molecule

P

Shape in word

Cl

Cl

Cl Cl

Cl


Octahedral



6. Sulphur Hexafluoride (SF6. Sulphur Hexafluoride (SF66) Molecule) Molecule

Shape in word

S

F

F

F

F

F

F



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

• The valence shell electron pair repulsion theory states

Electrostatic repulsion decreases in the following order:

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



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

1. Ammonia (NH3) Molecule

2. Water (H2O) Molecule

3. Amide Ion (NH2–)


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


V-shaped


> bp-bp repulsion


> bp-bp repulsion



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

Shape in word

H HO


V-shaped


> bp-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)

Example 8-7AExample 8-7A Example 8-7BExample 8-7B



8.8.88Multiple BondsMultiple Bonds


8.8 Multiple Bonds (SB p.238)

Single BondsSingle Bonds

• A covalent bond with two shared electrons

Multiple BondsMultiple Bonds

• Some atoms share more than two electrons in a bond

e.g. double bond, triple bond


Bond Bond orderBond length

(nm)

Bond enthalpy

(kJ mol-1)

C – C

C = C

C C

1

2

3

0.154

0.134

0.120

+348

+612

+837

N – N

N = N

N N

1

2

3

0.146

0.120

0.110

+163

+409

+944

C – O

C = O

1

2

0.143

0.122

+360

+743

Comparison of bond lengths and bond Comparison of bond lengths and bond enthalpies between single and multiple enthalpies between single and multiple bondsbonds




Effect of Multiple Bonding on Shapes of Effect of Multiple Bonding on Shapes of MoleculesMolecules

• Predict the shapes of molecules or polyatomic ions with multiple bonds

• Examples:

1. Ethene (CH2 = CH2) Molecule

2. Ethyne (CH CH) Molecule

3. Carbon Dioxide (CO2) Molecule

4. Sulphur Dioxide (SO2) Molecule


Trigonal planar

C C

H

HH

H



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

Shape in word


Linear



2. Ethyne (CH 2. Ethyne (CH CH) Molecule CH) Molecule

C C HH

Shape in word


Linear



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

Shape in word

CO O


V-shaped



4. Sulphur dioxide (SO4. Sulphur dioxide (SO22) Molecule) Molecule

Shape in word

SO O



8.8.99Covalent CrystalsCovalent Crystals


8.9 Covalent Crystals (SB p.240)

Covalent CrystalsCovalent Crystals

• May have simple molecular structures or giant covalent structures



Substances with Simple Molecular Substances with Simple Molecular StructuresStructures

• Consist of discrete molecules held together by weak intermolecular forces

• Atoms in a molecule are held together by strong covalent bonds

• Examples: H2 , O2 , H2O, CO2, I2



Substances with Giant Covalent Substances 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)



1. Diamond1. Diamond

• Each C atom is covalently bonded to 4 other C atoms to form a three-dimensional network

• The C – C bonding pattern accounts for the high m.p., stability and extreme hardness

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


A diamond crystal



The structure of diamond




2. Graphite2. Graphite

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

• Weak van der Waals’ forces hold the layers togetherDelocalized e- free to move within layers

• Properties: soft and slippery (used as pencil ‘lead’), conductor



Graphite



The structure of graphite


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)

Hardness

Melting point (C)

Colour

Electrical conductivity

3.51

10 (hardest)

3 827

Colourless

None

2.27

< 1 (very soft)

3 652 (sublime)

Shiny black

High

Comparison of the properties of Comparison of the properties of diamond and graphitediamond and graphite




3. Quartz3. Quartz

• Each Si atom is bonded tetrahedrally to 4 neighbouring O atoms

• Each O atom is bonded to 2 Si atoms, one at the centre of each of two adjacent tetrahedral

• Gives rise to a tetrahedral diamondlike structure in quartz



Quartz



The structure of quartz


The END


(a) How many lone pair and bond pair electrons are present in NH3 and H2O molecules 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



(b) The electronic configuration of nitrogen is 1s22s22px12py12pz1. Its outermost shell electrons are filled in the second quantum shell. There are no lowlying d orbitals available for nitrogen to expand octet. It has a maximum of three half-filled p orbitals to form three bonds, i.e. NCl3.

(b) Nitrogen can only form one chloride, NCl3, while phosphorus can form two chlorides, PCl3 and PCl5. Explain briefly.

Answer




(b) The ground state electronic configuration of phosphorus is 1s22s22p6

3s23px13py13pz1. It has three half-filled p obitals for bond formation. Thus, three chlorine atoms can form bonds with it to give PCl3.

After promoting one 3s electron to the low-lying d orbitals, the excited state electronic configuration of phosphorus becomes 1s22s22p63s13p

x13py13pz13d1. It now has five half-filled orbitals available for bond form

ation. Therefore, five chloride atoms can form bonds with it to give PCl5.


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


Back


(a) Draw a “dot and cross” diagram for the product formed in the reaction between an ammonia molecule and a hydrogen chloride molecule. Answer


(a)


(b) There is a dative covalent bond present in a HNO3 molecule. Draw a “dot and cross” diagram of the molecule.

Answer


(b)


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

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

Answer


Back


Why do two atoms bond together? How does covalent bond strength compare with ionic bond

strength?

Back

They are of similar strength. For example, the lattice enthalpy of NaCl is 771 kJ mol–1 while the H–H bond enthaly is 436 kJ mol–1. It is a misconception that ionic bond must be stronger (or weaker) than covalent bond.

Answer



(a) Referring to Table 8-2 on page 222, calculate the enthalpy change for the following reactions and state whether the reactions are endothermic or exothermic.(i) Reaction between nitrogen and hydrogen.

N2(g) + 3H2(g) → 2NH3( 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


(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.



(a) (ii) H – H + Cl – Cl → 2H – Cl


= E(H – H) + E(Cl – Cl)

= (+436 + 242) kJ mol–1

= +678 kJ mol–1

(a) (ii) Reaction between hydrogen and chlorine.

H2(g) + Cl2(g) → 2HCl(g) Answer

Back



(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




(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


(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




(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


(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




(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


(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




(b) Calculate the enthalpy change for the reaction

CH4(g) + H2O(g) → CO(g) + 3H2(g)

using the following bond enthalpies.

E(C – H in CH4) = +435 kJ mol–1

E(C O in CO) = +1 078 kJ mol–1

E(H – H in H2) = +436 kJ mol–1

E(H – O in H2O) = +464 kJ mol–1


Answer


(b) CH4(g) + H2O → CO(g) + 3H2(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


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 the size of an atom is determined by its electron cloud which has a diffuse shape. In different compounds, the electron cloud of a given atom may vary slightly due to the different internal environment (i.e. the atom that is bonded to).

Answer



(a) Predict the approximate bond lengths of

Si – H, P – H, S – H and H – Cl from the following data:

(Hint: Assume that covalent radii are additive.)


Bond Bond length (nm)

H – H

Si – Si

P – P (P4)

S – S (S4)

Cl – Cl

0.074

0.235

0.221

0.207

0.199

Answer



(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

20.221

nm2

0.074nm

20.207

nm2

0.199nm

20.074


(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 close to that calculated in (a), describe the relationship between bond length and bond enthalpy.


Bond Bond enthalpies (kJ mol–1)

Si – H

P – H

S – H

Cl – H

+318

+322

+338

+431

Answer



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

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(a) Explain why a molecule of CCl4 is tetrahedral, but a molecule of NCl3 is trigonal pyramidal in shape.

(a) In a CCl4 molecule, there are four bond pairs of electrons on the central carbon atom. The bond pairs have to stay as far away as possible. They take up the shape of a tetrahedron and thus the molecule is tetrahedral in shape. The four electron pairs in a NCl3 molecule take up the shape of a tetrahedron as well. However, one of the electron pairs is a lone pair and the other three are bond pairs. The shape of a NCl3 molecule is thus trigonal pyramidal.

Answer



(b) Deduce the shape of a molecule of BCl3.

(b) A BCl3 molecule has six outermost shell electrons around the central boron atom, forming three bond pairs. The shape of the BCl3 molecule is thus trigonal planar.

Answer



(c) Draw the structures of molecules of XeF2, XeF4 and XeF6 where Xe is a noble gas element with eight electrons in its outermost shell.

7.5 Ionic Radii (SB p.208)

Answer(c)

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The following data refer to the molecules NH3, H2O and HF.


Molecule Bond length (nm) Bond angle

NH3 0.101 107

H2O 0.096 104.5

HF 0.092 –


(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 bond lengths 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 atomic sizes decrease across a period owing to the increase in effective nuclear charge, the bond lengths of the three molecules decrease accordingly.


(b) Explain why the bond angle of H2O is less than that of NH3.


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

The central oxygen atom in H2O has two lone pairs and two bond pairs of electrons while the central nitrogen atom in NH3 has one lone pair and three bond pairs of electrons.

The electrostatic repulsion between electron pairs decreases in this order:

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

Thus, the bond angle of H2O is less than that of NH3.

Answer


(c) Match the following bond enthalpies to the bonds in the above three molecules:

+562 kJ mol–l, +388 kJ mol–l, +463 kJ mol–l


Answer(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 bond length and increase in polarity of bonds.

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What are the shapes of a H2S molecule and a H3O+ ion?

Explain their shapes in terms of the valence shell electron pair repulsion theory.

Answer


H2S molecule is V-shaped. In H2S molecule, there are two bond pairs and two lone pairs of electrons in the outermost shell of the central sulphur atom. All three types of electrostatic repulsion (lone pair – lone pair, lone pair – bond pair, bond pair – bond pair) are present. The two lone pairs will stay 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 H2S molecule is about 104.5 instead of 109.5 in tetrahedron.



H3O+ ion has a trigonal pyramidal shape. In H3O+ ion, the central oxygen atom forms two covalent bonds with two hydrogen atoms respectively. Also, one dative covalent bond is formed between the oxygen atom and the remaining hydrogen ion. We can regard the central oxygen atom has three bond pairs and one lone pair of electrons. According to the valence shell electron pair repulsion theory, the lone pair will stay further away from the three bond pairs. The three bond pairs are in turn compressed closer together. Thus, the H – O – H bond angles in the H3O+ ion are about 107 instead of 109.5 in tetrahedron.

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(a) Does sulphur obey the octet rule in forming a SO2 molecule? Explain your answer.

Answer(a) In the formation of SO2 molecule, sulphur does not obey the octet rule

because sulphur has 10 electrons in its outermost shell.



(b) Draw a “dot and cross” diagram of the hydrogen cyanide molecule (HCN). Describe and 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 electrostatic repulsion, the two electron clouds of the central carbon atom are separated at a maximum with bond angles of 180.

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New Way Chemistry for Hong Kong A-Level Book 11 Covalent Bonding 8.1Formation of Covalent Bonds 8.2Dative Covalent Bonds 8.3Bond Enthalpies 8.4Estimation