5. solution guide to supplementary exercises - 2c

258

Topic 6

Part A Unit-based exercise

Unit 23 Shapes of molecules

Fill in the blanks

1 a) four

b) tetrahedral

c) 109.5°

2 a) one; three

b) trigonal pyramidal

c) 107°

3 a) two; two

b) V-

c) 104.5°

4 a) two

b) linear

c) 180°

5 a) three

b) trigonal planar

c) 120°

6 a) five

b) trigonal bipyramidal

7 a) six

b) octahedral

8 linear

9 planar

10 tetrahedral

11 V-

12 trigonal pyramidal

13 lone pair-lone pair; lone pair-bond pair; bond pair-bond pair

14 graphite; buckminsterfullerene

Microscopic World II

259

True or false

15 T Electron diagram of a XeF4 molecule:

XeFF

FF

(Only electrons in the outermost shells are shown.)

Thus there are 6 pairs of electrons in the outermost shell of the xenon atom.

16 T Electron diagram of a boron trifluoride molecule:

F

BF F


The octet rule suggests that atoms become stable by having eight electrons in their outermost shells (or two electrons in the case of some smaller atoms). An atom attains a stable electronic arrangement by sharing or transfer of electrons.

There are less than 8 electrons in the outermost shell of the boron atom of a boron trifluoride molecule. Thus the molecule does not obey the octet rule.

17 F Nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its atom.

Thus NCl5 does not exist.

18 T Electron diagram of a carbon tetrachloride molecule:

Cl C Cl

Cl

Cl


In a CCl4 molecule, there are four bond pairs of electrons in the outermost shell of the central carbon atom.

The shape that puts the four electron pairs furthest apart is tetrahedral.

Thus the carbon tetrachloride molecule has a tetrahedral shape.

19 F Electron diagram of a SF4 molecule:

SF F

FF


The 5 pairs of electrons around the central S atom will adopt a trigonal bipyramidal arrangement.

260

There are two different ways in which we may arrange the 4 bonding pairs and 1 lone pair into a trigonal bipyramid. The correct arrangement will be the one with the minimum repulsion.

F

F

I

FS90°90°

F120°

S

F120°F

FF

II

Shape Repulsions involving the lone pair Remark

I3 90° lone pair-bond pair repulsions (repulsions at angles greater than 90° can be ignored) the repulsion is smaller for shape II, hence

the electron pairs will adopt shape IIII

2 90° lond pair-bond pair repusions (repulsions at angles greater than 90° can be ignored)

∴ the SF4 molecule has a seesaw shape.

20 F Electron diagram of a SO3 molecule:

O

O

S O


When using the VSEPR theory, double bonds can be treated like single bonds. Therefore we can view the sulphur atom as having three pairs of electrons in its outermost shell.

The furthest apart the three pairs can get is at an angle of 120°.

Thus the sulphur trioxide molecule has a trigonal planar shape.

21 T Electron diagram of a PCl4+ ion:

+

Cl P

Cl

Cl

Cl


In a PCl4+ ion, there are four bond pairs of electrons in the outermost shell of the central phosphorus

atom.

The shape that puts the four electron pairs the furthest apart is tetrahedral.

Thus the PCl4+ ion has a tetrahedral shape.

+

ClCl Cl

P

Cl

261

22 F Electron diagram of a NO3– ion:

–

O N

O

O


When using the VSEPR theory, double bonds can be counted as single bonds. Therefore we can view the nitrogen atom as having 3 pairs of electrons in its outermost shell.

The electron pairs repel one another and stay as far apart as possible.

The shape that puts the electron pairs the furthest apart is trigonal planar.

Thus the bond angles are 120°.

23 F A BH3 molecule has a trigonal planar shape.

An NH3 molecule has a trigonal pyramidal shape.

B

H

H H HH

HN

24 T Both molecules of BeF2 and XeF2 have a linear shape.

MoleculeElectron diagram


Shape

BeF2 F Be F F — Be — F

XeF2Xe

FF

F

F

Xe

25 T Bond angle: NH4+ > NH3

In the outermost electron shell of the central nitrogen atom in each of the species, the numbers of bond pairs and lone pair of electrons are as follows:

Species Number of bond pairs Number of lone pair

NH4+ 4 0

NH3 3 1


In the NH4+ ion, the furthest apart the four pairs of electrons can get is when they are arranged in a

tetrahedral shape.

Thus the NH4+ ion has a tetrahedral shape. The H–N–H bond angles are 109.5°.

262

Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.

Thus the H–N–H bond angle in the NH3 molecule is compressed to 107°.

26 F Buckminsterfullerene has a simple molecular structure. Van der Waals’ forces exist between the C60 molecules.

Multiple choice questions

27 C

28 A The sulphur atom has 6 electrons in its outermost shell and each oxygen atom has 6, i.e. a SO2 molecule has a total of 18 electrons in the outermost shells.

Option

Electron diagram(Only electrons in the outermost

shells are shown.)Remark

A O S O Correct

B O S OIncorrect as oxygen cannot form compounds with more than 8 electrons in the outermost shell of its atom.

C O S OIncorrect as there are 20 electrons in the outermost shells.

D O S OIncorrect as there are 20 electrons in the outermost shells.

29 D The xenon atom has 8 electrons in its outermost shell and each oxygen atom has 6, i.e. a XeO3 molecule has a total of 26 electrons in the outermost shells.

OptionElectron diagram


Remark

AO Xe

O

O

Incorrect as there are 24 electrons in the outermost shells.

BO

XeO O


CO Xe

O

O


DO Xe

O

OCorrect

263

30 D The oxygen atom has 6 electrons in its outermost shell and the three hydrogen atoms have 3 electrons, i.e. a total of 9 electrons.

But the H3O+ ion carries one positive charge because it has lost 1 electron. That makes a total of 8

electrons in the outermost shells of the H3O+ ion.

31 C The nitrogen atom has 5 electrons in its outermost shell and the two hydrogen atoms have 2 electrons, i.e. a total of 7 electrons.

But the NH2– ion carries one negative charge because it has gained 1 electron. That makes a total of 8

electrons in the outermost shell of the NH2– ion.

32 C The sulphur atom has 6 electrons in its outermost shell and each of the four oxygen atoms has 6 electrons, i.e. a total of 30 electrons.

But the SO42– ion carries two negative charges because it has gained 2 electrons. That makes a total of

32 electrons in the outermost shells of the SO42– ion.

OptionElectron diagram


Remark

A O S

O

O

O

2–


B O S

O

O

O

2–


C O S

O

O

O

2–

Correct

D O S

O

O

O

2–


33 A Electron diagram of a SiH4 molecule:

SiH

H

H

H

H


264

34 B Electron diagram of a NCl3 molecule:

Cl N

Cl

Cl


35 C Electron diagram of a SCl2 molecule:

Cl S

Cl


36 A Electron diagram of a PCl4+ ion:

+

Cl P

Cl

Cl

Cl


37 A Electron diagram of a CH3+ ion:

+H

CH H


38 B

Option MoleculeElectron diagram


Number of lone pair(s) of electrons in the outermost shell

of the central atom

A H2OH O

H2

B PH3H P H

H1

C PCl5

Cl

PClCl

Cl Cl

0

D CH2Cl2 H C Cl

H

Cl

0

265

39 B



Number of lone pairs of electrons in the outermost shell

of the central atom

XeF2Xe

FF 3

XeF4 XeFF

FF2

40 D Electron diagram of a BrF5 molecule:

Br

FFF

FF


The octet rule suggests that atoms become stable by having eight electrons in their outermost shells (or two electrons in the case of some smaller atoms). An atom attains a stable electronic arrangement by sharing or transfer of electrons.

BrF5 does not follow the octet rule as there are more than 8 electrons in the outermost shell of the bromine atom.

41 A

42 C In a molecule with three bond pairs and one lone pair around the central atom, the shape that puts the four electron pairs the furthest apart is tetrahedral.

The shape of a molecule is determined only by the arrangement of atoms.

Thus the molecule has a trigonal pyramidal shape.

43 B There are 6 electron pairs in the outermost shell of the central atom of a molecule with an octahedral shape.

For example, a SF6 molecule has an octahedral shape. There are 6 electron pairs in the outermost shell of the central sulphur atom.

FF F

F FS

F

44 D Electron diagram of an OF2 molecule:

F O

F


266

An OF2 molecule has two lone pairs and two bond pairs of electrons in the outermost shell of the oxygen atom.

The four pairs of electrons will adopt a tetrahedral arrangement.

The shape of a molecule is determined only by the arrangement of atoms. Thus the OF2 molecule is V-shaped.

45 B Electron diagram of a BH3 molecule:

H

BH H


In a BH3 molecule, there are three bond pairs of electrons in the outermost shell of the central boron atom.

The shape that puts the three electron pairs the furthest apart is trigonal planar.

Hence a BH3 molecule has a trigonal planar shape.

46 D Electron diagram of a PF3 molecule:

F P

F

F


In a PF3 molecule, there are one lone pair and three bond pairs of electrons in the outermost shell of the central phosphorus atom.



Thus the PF3 molecule has a trigonal pyramidal shape.

47 B In an NH3 molecule, there are one lone pair and three bond pairs of electrons in the outermost shell of the central nitrogen atom.


48 D Electron diagram of a XeF4 molecule:

XeFF

FF


In a XeF4 molecule, there are six bond pairs of electrons in the outermost shell of the central xenon atom.

The shape that puts the six electron pairs the furthest apart is octahedral.

267

49 C Electron diagram of a ClF3 molecule:

F Cl F

F


In a ClF3 molecule, there are two lone pairs and three bond pairs of electrons in the outermost shell of the central chlorine atom.

The shape that puts the five electron pairs the furthest apart is trigonal bipyramidal.

50 C

51 D Electron diagram of a CO2 molecule:

O C O


When using the VSEPR theory, we can view the carbon atom as having 2 pairs of electrons in its outermost shell.

The two pairs must be at the opposite ends of a straight line in order to be as far apart as possible.

Thus the CO2 molecule has a linear shape and the O–C–O bond angle is 180°.

52 B A H2O molecule has two lone pairs and two bond pairs of electrons in the outermost shell of the oxygen atom.


The shape of a molecule is determined only by the arrangement of atoms. Thus the H2O molecule is V-shaped.


Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.

Thus in the H2O molecule, the two lone pairs will stay the furthest apart.

As a result, the H–O–H bond angle in the H2O molecule is compressed to 104.5°.

53 A Electron diagram of a C2H4 molecule:

C CH

H

H

H


When using the VSEPR theory, we can view each carbon atom as having three pairs of electrons in its outermost shell.

The overall arrangement of the three pairs of electrons around each carbon atom is trigonal planar.

Thus the ethene molecule has a planar structure.

The H–C–H bond angles are 120°.

268

54 B



Three-dimensional structure

Shape

A H2S S H

HH

HS V-shaped

B BeCl2 Cl Be Cl Be ClCl linear

C SCl2S Cl

Cl

ClCl

S V-shaped

D OF2

O

F

F

FF

O V-shaped

55 A



Three-dimensional structure Shape

SF6

F

S

FFF

FFF

F F

F FS

F

octahedral

56 B

IonElectron diagram



NH4+

+

H N H

H

HH

H HN

H+

tetrahedral

57 B




XeF4 XeFF

FF F F

F FXe square planar

269

58 A Electron diagram of a CO32– ion:

2–O

CO O


When using the VSEPR theory, double bonds can be counted as single bonds. Therefore we can view the carbon atom as having 3 pairs of electrons in its outermost shell.


Thus the CO32– ion has a trigonal planar shape.

59 C A H2O molecule has two lone pairs and two bond pairs of electrons in the outermost shell of the oxygen atom.


The shape of a molecule is determined only by the arrangement of atoms. Thus the H2O molecule is V-shaped.

60 D

Option SpeciesElectron diagram



Shape

A NO3–

O

NO O

–

N

O

O O

–

trigonal planar

B NCl3Cl N Cl

Cl ClCl

ClN trigonal pyramidal

C SF4 SFF

F F

FF

FS

F

seesaw

D SiF4 F Si F

F

FF

F FSi

F

tetrahedral

∴ the SiF4 molecule has a tetrahedral shape.

270

61 A




Shape

A BF3

F

BF F B

F

F F

trigonal planar

B NH3H N H

H HH

HN trigonal pyramidal

C PH3H P H

H HH

HP trigonal pyramidal

D PCl3Cl P Cl

Cl ClCl

ClP trigonal pyramidal

∴ the BF3 molecule has a shape different from the others.

62 C




Is the species planar?

A BCl3Cl

BCl Cl B

Cl

Cl Cl

yes

B ClF3F Cl F

FF

F

FCl yes

C NCl3Cl N Cl

ClCl

ClCl

N no

D XeF4 XeFF

FF F F

F FXe yes

∴ NCl3 is NOT a planar species.

271

63 C Electron diagram of BCl3:

Cl B Cl

Cl


In a BCl3 molecule, there are three bond pairs of electrons in the outermost shell of the central boron atom.



Hence a BCl3 molecule has a trigonal planar shape.

Electron diagram of a PH3 molecule:

H P H

H


In a PH3 molecule, there are one lone pair and three bond pairs of electrons in the outermost shell of the central phosphorus atom.



Thus the PH3 molecule has a trigonal pyramidal shape.

64 D (1) and (3) Electron diagram of a ClF3 molecule:

F Cl F

F


The 5 pairs of electrons around the central Cl atom will adopt a trigonal bipyramidal arrangement.

There are three different ways in which we may arrange the 3 bonding pairs and 2 lone pairs into a trigonal bipyramid. The correct shape is the one with the minimum repulsion.

F

FCl

F

I

F

FF

II

F

Cl

F

III

FCl90°90°

90°

90°

120°

90°

Shape Repulsion present Remark

I 90° lone pair-lone pair repulsion the molecule will not take up shape I

II6 90° lone pair-bond pair repulsions (repulsions at angles greater than 90° can be ignored) bond pair-bond pair repulsion is less

than lone pair-bond pair repulsion, hence shape III has the minimum repulsion

III

4 90° lone pair-bond pair repulsions + 2 90° bond pair-bond pair repulsions (repulsions at angles greater than 90° can be ignored)

∴ the ClF3 molecule is T-shaped.

272

65 C




A CS2

S C S C SS

H2S

S H

HH

HS

B CH4

H C

H

H

HH

H HC

H

XeF4

XeFF

FF F

F

F

FXe

C NH4+

+

H N

H

H

H

+

HH H

N

H

SiCl4

Cl Si Cl

Cl

ClCl

Cl ClSi

Cl

D NH3

H N H

H HH

HN

NO3–

–

O N O

O

N

O

O O

–

∴ the NH4+ ion and SiCl4 molecule have a similar shape.

273

66 B




A BH3

H B H

H

B

H

H H

PH3

H P H

H HH

HP

B CO2O C O C OO

HCNH C N C NH

C CO32–

O

CO O

2– 2–

O OC

O

H3O+

H O H

H

+

HH

HO

+

D NH2–

H N H–

HH

N

–

XeF2

XeFF

F

Xe

F

∴ the CO2 and HCN molecules have a similar shape.

274

67 A




A PCl3

Cl P Cl

Cl ClCl

ClP

NCl3

Cl N Cl

Cl ClCl

ClN

B CF4

F C F

F

FF

F FC

F

SF4

S FF

F F

F

FF

FS

C SO2

O S O S

OO

CO2

O C O C OO

D PF5

F

P FF

F F F

FP

F

F

F

IF5

I

FFF

FFF

F F

F FI

∴ the PCl3 and NCl3 molecules have a similar shape.

275

68 D




A BF3

F

BF F B

F

F F

NF3

F N F

F FF

FN

B BCl3Cl

BCl Cl B

Cl

Cl Cl

PH3

H P H

H HH

HP

C BeCl2Cl Be Cl Be ClCl

SCl2

Cl S

Cl

ClCl

S

D CH3+

H C

H

H

+

C

H

H H

+

NO3–

O

NO O

–

N

O

O O

–

∴ the CH3+ ion and NO3

– ion have an identical geometry.

276

69 C H+

ammonia molecule ammonium ion

+

HH H

N

H

HH

HN

∴ there is a change of shape from trigonal pyramidal to tetrahedral.

70 A PF5 PF3

Electron diagrams of PF5 and PF3:

F

PFF

F FF P F

F


Arrangement of electron pairs in the outermost shell of the P atom in the molecules:

Thus the change in the three-dimensional arrangement of electron pairs in the outermost shell of the P atom is from trigonal bipyramidal to tetrahedral.

71 D




HCN H C NC

180°

NH

72 C Electron diagram of a CO32– ion:

O

CO O

2–





Thus the CO32– ion has a trigonal planar shape.

∴ the bond angle in a CO32– ion is 120°.

277

73 B




H2O H O

HH

HO

104.5°

NH3H N H

H HH

HN

107°

CH4 H C

H

H

HH

H HC

H

109.5°

CO2 O C O C OO

180°

∴ the correct order of bond angles in the molecules H2O, NH3, CH4 and CO2 is CO2 > CH4 > NH3 > H2O.

74 D




BeCl2 Cl Be ClBe

180°

ClCl

∴ the Cl–Be–Cl bond angle in a BeCl2 molecule is 180°.

75 B In a regular octahedral molecule, MX6, the number of X–M–X bonds at 180° is 3.

XX X

X X

M

X

180°

180°

180°

76 C

C

HH

H

C

Hα

When using the VSEPR theory, we can view each carbon atom as having three pairs of electrons in its outermost shell.

The overall arrangement of the three pairs of electrons around each carbon atom is trigonal planar.

Thus the H–C–H bond angle (i.e. α) is 120°.

278

77 C

C

HH

H

O H

β

Cx

Hα

H

Carbon atom x has four pairs of electrons in its outermost shell.

The furthest apart the pairs can get is when they are arranged in a tetrahedral shape. So, the bond angle α is 109°.

The oxygen atom has two lone pairs and two bond pairs of electrons in its outermost shell.

Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion,

while lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.

Thus the two lone pairs will stay the furthest apart. As a result, the C–O–H angle (i.e. β) is compressed to 105°.

78 A




A BF3

F

BF F

B

F

F F

120° 120°

120°

B CH4 H C H

H

HH

H HC

H

109.5°

C NH3H N H

H HH

HN

less than 109.5°

D PCl3Cl P Cl

Cl ClCl

ClP

less than 109.5°

∴ the Y–X–Y angle is the greatest in BF3.

79 CO N

O

O

–

When using the VSEPR theory, double bonds can be counted as single bonds. Therefore we can view the nitrogen atom as having 3 pairs of electrons in its outermost shell.

279



Thus the NO3– ion has a trigonal planar shape.

∴ the O–N–O bond angle is 120°.

80 B In the outermost electron shell of the central nitrogen atom in each of the species, the numbers of bond pairs and lone pair(s) of electrons are as follows:

Species Number of bond pairs Number of lone pair(s)

NH4+ 4 0

NH3 3 1

NH2– 2 2



tetrahedral shape.





Thus in the NH2– ion, the two lone pairs will stay the furthest apart. The H–N–H bond angle in the

NH2– ion is compressed to 104.5°.

∴ the order of the bond angles is NH4+ > NH3 > NH2

–.

81 C Consider the following change:

H2O H3O+

In the outermost electron shell of the oxygen atom in each of the species, the numbers of bond pairs and lone pair(s) of electrons are as follows:


H3O+ 3 1

H2O 2 2

In a H3O+ ion, the four pairs of electrons in the outermost shell of the oxygen atom will adopt a

tetrahedral arrangement.


Thus in a H2O molecule, the two lone pairs will stay the furthest apart.

As a result, the H–O–H bond angle in a H2O molecule is compressed to less than that in a H3O+ ion.

∴ when the H3O+ ion is formed from water, the bond angle increases slightly.

280

82 A Species

Electron diagram(Only electrons in the outermost shells are shown.)

(1) H3O+ H O

H

H+

(2) NH4+ H N

H

H

H

+

(3) SF6

F

S

FFF

FF

∴ both the H3O+ and NH4

+ ions have 8 electrons in the outermost shell of the central atom.

83 B Species


(1) CH4H C

H

H

H

(2) XeF2Xe FF

(3) NH2– H N H

–

∴ both the CH4 molecule and the NH2– ion have 8 electrons in the outermost shell of the central

atom. The electron pairs would adopt a tetrahedral arrangement.

84 D Species



(1) BeF2 F Be F Be FF

(2) CO2 O C O C OO

(3) HCN H C N C NH

∴ all the three species have a linear shape.

281

85 C

SpeciesElectron diagram



(1) BrF3

F Br F

F

F

F

FBr T-shaped

(2) NH3H N H

H HH

HN trigonal pyramidal

(3) PF3

F P F

F FF

FP trigonal pyramidal

∴ the NH3 and PF3 molecules have a trigonal pyramidal shape.

86 D




(1) Cl2OCl O

Cl

ClCl

O V-shaped

(2) NH2–

H N H–

HH

N

–

V-shaped

(3) SO2 OS

O S

OO

V-shaped

∴ all the three species are V-shaped.

282

87 B




(1) SiBr4 Br Si Br

Br

BrBr

Br BrSi

Br

tetrahedral

(2) SF4 SFF

F F

F

FF

FS seesaw

(3) NH4+ H N

H

H

H

+ +

HH H

N

H

tetrahedral

∴ the SiBr4 molecule and the NH4+ ion have a tetrahedral shape.

88 A




(1) boron trifluoride

F

BF F B

F

F F

trigonal planar

(2) ethene C CH

H

H

H

H

HC

HC

Hplanar

(3) nitrogen trichloride

Cl N Cl

Cl ClCl

ClN

trigonal pyramidal

∴ both boron trifluoride and ethene have a planar structure.

283

89 C




(1) BF3

F

BF F B

F

F F

trigonal planar

NH3

H N H

H HH

HN

trigonal pyramidal

(2) BeCl2Cl Be Cl Be ClCl linear

CO2

O C O C OO linear

(3) NH4+

H N

H

H

H

+ +

HH H

N

H

tetrahedral

CH4

H C

H

H

H

HH H

C

H

tetrahedral

∴ (2) the BeCl2 and CO2 molecules have a similar shape.

(3) the NH4+ ion and the CH4 molecule have a similar shape.

90 D




(1) CS2 S C S C SS

(2) PF5

F

PFF

F F

F

FF

FFP

(3) XeF4 XeFF

FF F

F

F

FXe

∴ all the three species have three atoms lying in a straight line.

284

91 B Three dimensional structure of PCl5:

Cl

ClCl

ClClP

120°

90°

∴ the bond angles 90° and 120° exist in a PCl5 molecule.

92 D




(1) boron trifluoride

F

BF F B

F

F F

trigonal planar

(2) carbonate ion

O

CO O

2–

C

O

O O

2–

trigonal planar

(3) nitrate ion

O

NO O

–

N

O

O O

–

trigonal planar

∴ the model could represent all the three species.

93 A (2) Van der Waals’ forces exist between molecules of buckminsterfullerene.

(3) Buckminsterfullerene has a simple molecular structure while graphite has a giant covalent structure.

Thus the melting point of graphite is higher than that of buckminsterfullerene.

94 C Nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its atom.

Thus nitrogen cannot form pentachloride.

95 B Electron diagram of a BF3 molecule:

F

BF F


There are less than 8 electrons in the outermost shell of the boron atom.

Thus the BF3 molecule does not conform to the octet rule.

285

Electron diagram of a NO2 molecule:

O N O


The NO2 molecule has an unpaired electron.

Thus the NO2 molecule does not conform to the octet rule.

96 C An NH3 molecule has one lone pair and three bond pairs of electrons in the outermost shell of the nitrogen atom.


The shape of a molecule is determined only by the arrangement of atoms. Thus the NH3 molecule has a trigonal pyramidal shape.

97 B Electron diagram of a methanal molecule:

OCH

H





Thus the methanal molecule has a trigonal planar shape.

So, the O–C–H bond angle is about 120°.

98 D Electron diagram of a CH3+ ion:

+H

CH H


There are 3 bond pairs of electrons in the outermost shell of the carbon atom.



Thus the CH3+ ion has a trigonal planar shape.

286

99 C




BCl3Cl

BCl Cl B

Cl

Cl Cl

trigonal planar

NCl3Cl N Cl

Cl ClCl

ClN trigonal pyramidal

∴ BCl3 has a planar structure but NCl3 does not.

100 B

IonElectron diagram



CO32–

O

CO O

2–

C

O

O O

2–

trigonal planar

NO3–

O

NO O

–

N

O

O O

–

trigonal planar

∴ both carbonate ion and nitrate ion have a trigonal planar shape.

101 C




PCl5

Cl

PClCl

Cl Cl

Cl

ClCl

ClClP trigonal bipyramidal

IF5I

FFF

FFF

F F

F FI square pyramidal

∴ molecules of PCl5 and IF5 have different shapes.

287

102 B Bond angle: NH3 > H2O

In the outermost electron shell of the central atom in each of the molecules, the numbers of bond pairs and lone pair(s) of electrons are as follows:

Molecule Number of bond pairs Number of lone pair(s)

NH3 3 1

H2O 2 2

In the NH3 molecule, the four pairs of electrons in the outermost shell of the nitrogen atom will adopt a tetrahedral arrangement.




Thus in the H2O molecule, the two lone pairs will stay the furthest apart.

As a result, the H–O–H bond angle in the H2O molecule is compressed to 104.5°.

103 D




Shape Bond angles

CF4 F C F

F

FF

F FC

F

tetrahedral 109.5°

SF4 SFF

F FF

F

FS

F

seesaw180°, 120°,

90°

∴ the bond angles in CF4 are 109.5° but those in SF4 are not.

104 C Graphite has a giant covalent structure. It is NOT soluble in carbon disulphide.

105 A

288

Unit 24 Bond polarity and intermolecular forces

Fill in the blanks

1 polar covalent

2 electronegativity

3 polar

4 fluorine

5 Dipole moment

6 a) polar

b) symmetrical

7 Instantaneous dipole-induced dipole

8 permanent dipole-permanent dipole; permanent dipole-induced dipole; instantaneous dipole-induced dipole

9 hydrogen; oxygen; nitrogen; fluorine

10 viscosity

True or false

11 T Fluorine is the most electronegative element.

12 F In general, electronegativity increases from left to right across a period, as the metallic character of elements decreases.

Both phosphorus and sulphur are Period 3 elements.

Sulphur is more electronegative than phosphorus.

13 T

14 T Within each group, electronegativity decreases with increasing atomic number and metallic character.

The electronegativity value of Cl is greater than that of I.

The electronegativity difference between H and Cl is greater than that between H and I.

Thus H–Cl bond is more polar than H–I bond.

15 F The electronegativity values of C and S are the same. Thus C=S bond is non-polar.

A carbon disulphide molecule is non-polar.

289

16 T The electronegativity value of nitrogen is greater than that of phosphorus.

Hence the bond pairs of electrons in NH3 will be attracted towards the nitrogen atom to a greater extent.

These bond pairs will repel each other to a greater extent.

Thus the H–N–H bond angle in NH3 is greater than the H–P–H bond angle in PH3.

17 F The electronegativity value of chlorine is greater than that of carbon. So each C–Cl bond is polar.

Due to the symmetry of the tetrahedral shape of tetrachloromethane, the four identical bond dipole moments cancel one another out exactly.

The molecule has no net dipole moment and it is non-polar.

∴ a stream of tetrachloromethane is NOT deflected by a charged rod.

18 F A phosphorous pentachloride molecule has a trigonal bipyramidal shape.

Cl

ClCl

ClClP

The electronegativity value of chlorine is greater than that of phosphorus. So each P–Cl bond is polar.

Due to the symmetry of the shape of the phosphorus pentachloride molecule, the individual bond dipole moments cancel one another out exactly.


∴ NO permanent dipole-permanent dipole attractions exist between phosphorus pentachloride molecules.

19 T Instantaneous dipole-induced dipole attractions exist between all atoms or molecules.

20 T




SO2O

SO S

OO

V-shaped

The electronegativity value of oxygen is greater than that of sulphur. So each S–O bond is polar.

The sulphur dioxide molecule has a V-shape. The individual S–O bond dipole moments reinforce each other.

The molecule has a net dipole moment and it is polar.

∴ both permanent dipole-permanent dipole attractions and instantaneous dipole-induced dipole attractions exist between sulphur dioxide molecules.

290

21 F Among the noble gases, the atomic number increases from helium to xenon and the number of electrons in one atom increases in the same order.

Hence the strength of van der Waals’ forces in noble gases also increases from helium to xenon.

The boiling point of an element depends on the strength of its intermolecular attractions.

Thus the boiling point of argon is higher than that of neon.

22 T The number of electrons in a HI molecule is greater than that in a HCl molecule.

Hence van der Waals’ forces between HI molecules are stronger than those between HCl molecules.

23 F The boiling point of H2S is higher than that of SiH4.

The boiling point of a compound depends on the strength of its intermolecular attractions.

H2S is a polar substance. There are permanent dipole-permanent dipole attractions and instantaneous dipole-induced dipole attractions between H2S molecules.

SiH4 is a non-polar substance. There are only instantaneous dipole-induced dipole attractions between SiH4 molecules.

More heat is needed to separate the H2S molecules during boiling.

24 F The electronegativity values of P and H are the same.

NO strong attraction exists between the hydrogen atom of one phosphine molecule (PH3) and the lone pair on the phosphorus atom of another phosphine molecule.

Thus NO hydrogen bond exists between phosphine molecules.

25 T Water has a high surface tension due to the hydrogen bonds between water molecules.

26 F The viscosity of a liquid is a measure of a liquid’s resistance to flow. The greater the viscosity, the more slowly the liquid flows.

27 T Ethanol is miscible with water (i.e. soluble in water in all proportions) because hydrogen bonds can form between ethanol molecules and water molecules. This helps the dissolving process.

CH3CH2

hydrogen bond

lone pair

Oδ–

Hδ+

Hδ+

Oδ–

Hδ+

291

28 T Propanone (H3C CH3C

O

) can form hydrogen bonds with water molecules.

H3C

Cδ+

Oδ–

H3C

Oδ–

HHδ+ δ+

hydrogen bond

lone pair

29 F A liquid with strong intermolecular forces has a higher viscosity than one with weak intermolecular forces.

The viscosity of ethanol is higher than that of tetrachloromethane because of its ability to form hydrogen bonds.

30 T Methoxymethane is fairly soluble in water because hydrogen bonds can form between methoxymethane molecules and water molecules.

Oδ–

HHδ+ δ+

hydrogen bond

lone pair

O

CH3H3C

δ–

Propane is non-polar. Propane does not mix with water due to the difference in the strength of intermolecular attractions between water molecules and those between propane molecules. Thus, propane is insoluble in water.


31 C Within each group, electronegativity decreases with increasing atomic number and metallic character.

i.e. the electronegativity of halogens decreases down the group.

The electronegativity difference between F and I is the greatest.

Thus the I–F bond is the most polar.

32 D C is the least electronegative among F, Cl, N and C.

Thus C–H bond is the least polar.

292

33 A Element W X Y Z

Atomic number 8 12 14 16

Name of element oxygen magnesium silicon sulphur

W (oxygen) is the most electronegative.

34 B Option Molecule Three-dimensional structure

A CO2 C OO

B NH3

HH

HN

C CCl4

ClCl Cl

C

Cl

D CH4

HH H

C

H

Option B — The electronegativity value of nitrogen is greater than that of hydrogen. So each N–H bond is polar.

The NH3 molecule has a trigonal pyramidal shape. The individual N–H bond dipole moments reinforce each other.


Option C — The electronegativity value of chlorine is greater than that of carbon. So each C–Cl bond is polar.

Due to the symmetry of the tetrahedral shape of the CCl4 molecule, the individual bond dipole moments cancel one another out exactly.


35 A Option A — The electronegativity value of fluorine is greater than that of hydrogen.

So the fluorine end has a partial negative charge while the hydrogen end has a partial positive charge.

Hence the HF molecule has a net dipole moment.

293

36 C Options A and C — The three-dimensional structures of CHCl3 and CHF3 are shown below:

ClCl Cl

C

H

net dipole moment

FF F

C

H

net dipole moment

The electronegativity difference between C and F is greater than that between C and Cl.

Thus C–F bond is more polar than C–Cl bond.

∴ CHF3 has a greater dipole moment than CHCl3.

Option D — The electronegativity value of chlorine is greater than that of carbon. So each C–Cl bond is polar.



37 C




SiF4 F Si F

F

FF

F FSi

F

tetrahedral

The electronegativity value of fluorine is greater than that of silicon. So each Si–F bond is polar.

Due to the symmetry of the tetrahedral shape of the SiF4 molecule, the individual bond dipole moments cancel one another out exactly.

Thus the SiF4 molecule has zero dipole moment.

38 B Options B and C — The electronegativity difference between H and Cl (2.1 and 3.0) is greater than that beween N and O (3.0 and 3.5).

∴ HCl has a greater net dipole moment than NO.

Option D —




SO3 O

O

S O S

O

O O

trigonal planar

294

The electronegativity value of oxygen is greater than that of sulphur. So each S–O bond is polar.

Due to the symmetry of the shape of the SO3 molecule, the individual bond dipole moments cancel one another out exactly.

Thus the SO3 molecule has no net dipole moment.

39 C Option C —




OF2

O F

F δ–

δ–

δ+

FF

O V-shaped

The electronegativity value of fluorine is greater than that of oxygen.

So the fluorine end has a partial negative charge while the oxygen end has a partial positive charge.

40 D Option Molecule Three-dimensional structure

A CHCl3

ClCl Cl

C

H

net dipole moment

δ–

δ–δ–

δ+

B HClnet dipole moment

Hδ+

Clδ–

C H2O

HH

O δ+

δ+

δ– net dipole moment

D CS2 C SS

Option D — The electronegativity values of C and S are the same.

Thus C=S bond is non-polar.

A carbon disulphide molecule is non-polar.

295

41 A Option Molecule Three-dimensional structure

A H2S

HH

Snet dipole moment

B CO2 no net dipole momentC OO

C CCl4

ClCl Cl

C

Cl

no net dipole moment

D BF3

F FB

F


∴ the H2S molecule has a net dipole moment and is polar.

42 D Option D — The electronegativity value of chlorine is greater than that of phosphorus. So each P–Cl bond is polar.

Due to the symmetry of the shape of the PCl5 molecule, the individual bond dipole moments cancel one another out exactly.

The PCl5 molecule has no net dipole moment and it is non-polar.

43 B Option Molecule Three-dimensional structure

A SiF4

FF F

Si

F


B SF4 net dipole moment

FF

FS

F

C BeF2 no net dipole momentBe FF

D BF3

F FB

F


∴ only the SF4 molecule has a net dipole moment.

296

44 C Option C — The electronegativity value of chlorine is greater than that of germanium. So each Ge–Cl bond is polar.

Due to the symmetry of the tetrahedral shape of the GeCl4 molecule, the individual bond dipole moments cancel one another out exactly.

The GeCl4 molecule has NO net dipole moment.

45 D Option D — The electronegativity value of fluorine is greater than that of beryllium. So each Be–F bond is polar.

As the polar Be–F bonds of the BeF2 molecule are arranged in a straight line, the two identical bond dipole moments cancel each other out exactly.

As a result, the BeF2 molecule has no net dipole moment.

So a BeF2 molecule is non-polar.

46 D Option D — The electronegativity value of chlorine is greater than that of carbon. So each C–Cl bond is polar.


The CCl4 molecule has no net dipole moment and it is non-polar.

47 C




A BeF2

F Be F no net dipole momentBe FF

NO2

O N O net dipole momentO O

N

B HClH Cl

net dipole moment

H Cl

CO2

O C O no net dipole momentC OO

C NO2

O N O net dipole momentO O

N

HClH Cl

net dipole moment

H Cl

297

D BeF2

F Be F no net dipole momentBe FF

CO2

O C O no net dipole momentC OO

∴ both NO2 and HCl molecules have a net dipole moment.

48 A Option Molecule I Molecule II Remark

A net dipole momentCH3

ClC

ClC

H3Cno net dipole moment

Cl

CH3

CCl

CH3C

the dipole moment in molecule 1 is greater than that in molecule 2

49 A The electronegativity value of nitrogen is greater than that of phosphorus.



Thus the H–N–H bond angle in NH3 is greater than the H–P–H bond angle in PH3.

Molecule Bond angle

PH3 93°

H2O 104.5°

NH3 107°

∴ the order of increasing bond angle is PH3 < H2O < NH3.

50 C Option C — Tetrachloromethane is a non-polar substance. There are only instantaneous dipole-induced dipole attractions between CCl4 molecules.

51 D

52 B Although H–Cl bond is polar, the attraction between the partially positively charged hydrogen atom and the lone pair on the chlorine atom of another HCl molecule is not strong.

The chlorine atom is quite large and the lone pairs are not very accessible to the hydrogen atom.

Thus hydrogen bonds do NOT exist in liquid HCl.

53 A

54 C In both ice and water, hydrogen bond is the strongest intermolecular attraction.

In ice, each water molecule forms four hydrogen bonds tetrahedrally.

The highly ordered structure leads to a very ‘open’ structure with large spaces in ice.

298

Thus ice has a lower density.

The ‘open’ structure collapses when ice melts to form liquid water.

The water molecules can pack more closely, so liquid water has a higher density.

55 B Diamond has a giant covalent structure.

56 A

57 C Option C — ICl is polar.

The electronegativity value of chlorine is greater than that of iodine.

So the chlorine end has a partial negative charge while the iodine end has a partial positive charge.

Hence the ICl molecule has a net dipole moment.

58 B The boiling point of an element depends on the strength of its intermolecular attractions.

The intermolecular attractions in the liquids are van der Waals’ forces.

Among CO2, H2, N2 and O2, a H2 molecule contains the smallest number of electrons.

Hence the van der Waals’ forces in liquid H2 is the weakest.

The least amount of heat is needed to separate the H2 molecules during boiling.

Hence liquid H2 is the most volatile.

59 B Option B — CH4, SiH4 and SnH4 are hydrides of Group IV elements.

The boiling point of a hydride depends on the strength of its intermolecular attractions.

The intermolecular attractions in the hydrides are van der Waals’ forces.

The number of electrons in one hydride molecule is in the order CH4 < SiH4 < SnH4.

Hence the strength of van der Waals’ forces in the hydrides is in the same order.

More heat, in their increasing order, is needed to separate the molecules during boiling,

and thus the order of the boiling points of the hydrides is CH4 < SiH4 < SnH4.

60 C Option C — The boiling point of H2S is higher than that of SiH4.


H2S is a polar substance. There are permanent dipole-permanent dipole attractions and instantaneous dipole-induced dipole attractions between H2S molecules.

SiH4 is a non-polar substance. There are only instantaneous dipole-induced dipole attractions between SiH4 molecules.


299

61 B The zig-zag polymeric structure in solid HF is due to hydrogen bonding:

F F F

F F F

H H HH H H

hydrogenbond

62 D In water, each molecule can be hydrogen-bonded to four other molecules.

H H

H H H

HHH

H H

O O

O

OO

63 D Option D — In a H C

H

H

C

O

H molecule, NO hydorgen atom is attached directly to a highly

electronegative atom (such as oxygen, nitrogen or fluorine). Thus the molecule would NOT form a hydrogen bond with another molecule of its own.

64 A The boiling point of a compound depends on the strength of its intermolecular attractions.

Hydrogen bonds exist in HF while only van der Waals’ forces exist in HCl, HBr and HI.

Hydrogen bonds are stronger than van der Waals’ forces.

More heat is needed to separate the HF molecules during boiling.

Thus the boiling point of HF is the highest.

The number of electrons in one molecule increases from HCl to HI.

Hence the strength of van der Waals’ forces also increases from HCl to HI.

Hence the boiling points of the hydrides increase from HCl to HI.

∴ HCl has the lowest boiling point.


Hydrogen bonds exist in H2O while only van der Waals’ forces exist in H2S.


More heat is needed to separate the H2O molecules during boiling.

Thus the boiling point of H2O is higher than that of H2S.

300

66 B The boiling point of a compound depends on the strength of its intermolecular attractions.

Both X and Y are non-polar. Relatively weak van der Waals’ forces exist in them.

Molecule of X is longer and somewhat spread-out whereas that of Y is more spherical and compact.

The shape of molecule of X allows greater surface contact between molecules.

The van der Waals’ forces in X are thus stronger.

More heat is needed to separate molecules of X during boiling.

Hence X has a higher boiling point than Y.

Hydrogen bonds exist in Z.

Hence Z has the highest boiling point.

67 D Option D — C2H5OH is miscible with water (i.e. soluble in water in all proportions) because it can form hydrogen bonds with water.

CH3CH2

hydrogen bond

lone pair

Oδ–

Hδ+

Hδ+

Oδ–

Hδ+

68 B In ice, hydrogen bond is the strongest intermolecular attraction.

Each water molecule forms four hydrogen bonds tetrahedrally.



∴ the melting of ice involves the cleavage of hydrogen bonds.


The compounds in Options A and C are non-polar. Relatively weak van der Waals’ forces exist in them.

The strength of the forces increases with the number of electrons in the molecule.

Hence the boiling point of the compound in Option C is higher than that of the compound in Option A.

Hydrogen bonds exist in the compounds in Options B and D.

Hence the compounds have higher boiling points.

∴ the boiling point of the compound in Option A is the lowest and thus the compound is the most volatile.

301

70 D The boiling point of a compound depends on the strength of its intermolecular attractions.

Relativity weak van der Waals’ forces exist in the compounds in Options A and B.

Hydrogen bonds exist in the compounds in Options C and D.

Hence the boiling points of compounds in Options C ad D are higher than those of compounds in Options A and B.

The intermolecular attractions in the compound in Option D is stronger as there are more electrons in one molecule of the compound.

∴ the compound in Option D has the highest boiling point.


Relatively weak van der Waals’ forces exist in H2 and N2.

The strength of the forces increases with the number of electrons in one molecule.

Hence the boiling point of N2 is higher than that of H2.

Hydrogen bonds exist in NH3.

Hence NH3 has the highest boiling point.

∴ the order of increasing boiling point is H2 < N2 < NH3.

72 D The viscosity of the compound in Option D is the highest.

All the compounds can form hydrogen bonds.

Each molecule of the compound in Option D has three –OH groups that can take part in hydrogen bonding while each molecule of other compounds has only one or two –OH groups. Each molecule of the compound in Option D can form more hydrogen bonds.

Furthermore, because of their shapes, molecules of the compound in Option D tend to become entangled rather than to slide past one another.

These factors contribute to the high viscosity of the compound in Option D.

73 D

74 B (2) The electronegativity value of fluorine is greater than that of boron. So each B–F bond is polar.

A BF3 molecular has a trigonal planar shape.

The three identical B–F bond dipole moments in a BF3 molecule cancel one another out exactly.

As a result, the BF3 molecule has no net dipole moment and is non-polar.

(3) The electronegativity value of oxygen is greater than that of sulphur. So each S–O bond is polar.

A sulphur dioxide molecule is V-shaped. The individual S–O bond dipole moments reinforce each other.

As a result, the SO2 molecule has a net dipole moment and is polar.

302

75 B (1) The electronegativity values of C and S are the same. Thus C=S bond is non-polar.

∴ the CS2 molecule is non-polar.

(2) The electronegativity value of S is greater than that of H. So each S–H bond is polar.

A H2S molecule is V-shaped. The individual S–H bond dipole moments reinforce each other.

Hence the molecule has a net dipole moment.

∴ the H2S molecule is polar.

(3) The electronegativity value of Cl is greater than that of P. So each P–Cl bond is polar.

Due to the symmetry of the trigonal bipyramidal shape of the PCl5 molecule, the individual bond dipole moments cancel one another out exactly.

The molecule has no net dipole moment.

∴ the PCl5 molecule is non-polar.

76 A (1) The electronegativity value of nitrogen is greater than that of hydrogen. So each N–H bond is polar.



∴ the NH3 molecule is polar.

(2) The electronegativity value of Cl is greater than that of S. So each S–Cl bond is polar.

A SCl2 molecule is V-shaped. The individual S–Cl bond dipole moments reinforce each other.


∴ the SCl2 molecule is polar.

(3) The electronegativity value of F is greater than that of Xe. So each Xe–F bond is polar.

Due to the symmetry of the square planar shape of the XeF4 molecule, the individual bond dipole moments cancel one another out exactly.


∴ the XeF4 molecule is non-polar.

77 A (3) Across the second period, the melting points of elements rise to Group IV and then fall to low values.

78 A Option A — The sulphur atom has 6 electrons in its outermost shell and each oxygen atom has 6, i.e. a SO3 molecule has a total of 24 electrons in the outermost shells.

303

79 B (1) Three-dimensional structure of a SO3 molecule:

S

O

O O

When using the VSEPR theory, double bonds can be counted as single bonds. Therefore we can view the sulphur atom as having 3 pairs of electrons in its outermost shell.



∴ a SO3 molecule has a trigonal planar shape.

(2) The electronegativity value of O is greater than that of S. So a SO3 molecule contains polar bonds.

(3) Due to the symmetry of the trigonal planar shape, the bond dipole moments cancel one another out exactly.

Hence the SO3 molecule has no net dipole moment and it is non-polar.

80 B Option B — The chlorine atom has 7 electrons in its outermost shell and each fluorine atom has 7, i.e. a ClF3 molecule has a total of 28 electrons in the outermost shells.

81 A (1) and (3) In a ClF3 molecule, the 5 pairs of electrons around the central Cl atom will adopt a trigonal bipyramidal arrangement.

There are three different ways in which we may arrange the 3 bonding pairs and 2 lone pairs into a trigonal bipyramid. The correct shape is the one with the minimum repulsion.

F

FCl

F

I

F

FF

II

F

Cl

F

III

FCl90°90°

90°

90°

120°

90°



II6 90° lone pair-bond pair repulsions (repulsions at angles greater than 90° can be ignored) bond pair-bond pair repulsion is less

than lone pair-bond pair repulsion, hence shape III has the minimum repulsion

III


∴ the ClF3 molecule is T-shaped.

(2) The electronegativity value of F is greater than that of Cl. So each Cl–F bond is polar.

304

82 A (1) In the molecule, the C–Cl bonds are on the same side of the ring and thus the bond dipole moments do not cancel each other out.

∴ the molecule is polar.

(2) In the molecule, the C–Cl bonds are on opposite sides of the planar carbon ring, so the individual bond dipole moments cancel each other out exactly.


∴ the molecule is non-polar.

(3) In the molecule, the C–Cl bonds are on opposite sides of the carbon-carbon double bond, so the individual bond dipole moments cancel each other out exactly.


∴ the molecule is non-polar.

83 A (1) Three-dimensional structure of trichloromethane:

ClCl Cl

C

H

net dipole moment

∴ the molecule is polar and thus the liquid is deflected by the negatively charged rod.

(2) and (3) Both molecules are non-polar.

∴ the liquids are NOT deflected by the negatively charged rod.

84 A (1) The electronegatvity value of Cl is higher than that of Si and Ge. So the Si–Cl and Ge–Cl bonds are polar.

(3) Due to the symmetry of the tetrahedral shape of the SiCl4 and GeCl4 molecules, the individual bond dipole moments cancel one another out exactly.

The molecules has no net dipole moment and they are non-polar.


Both SiCl4 and GeCl4 are non-polar. Relatively weak instantaneous dipole-induced dipole attractions exist in them.

The strength of the attractions increases with the number of electrons in one molecule.

Hence the boiling point of GeCl4 is higher than that of SiCl4.

85 C (1) The electronegativity value of nitrogen is greater than that of hydrogen. So each N–H bond is polar.



(2) The electronegativity value of N is greater than that of P.


These bond pairs will repel each other to a greater extent. Thus the H–N–H bond angle in NH3 is greater than the H–P–H bond angle in PH3.

305

(3) NH3 can form hydrogen bonds but PH3 cannot.

More heat is needed to separate the NH3 molecules during boiling.

Thus the boiling point of NH3 is higher than that of PH3.

86 A (1) Both X and Y have the same molecular formula, C5H12.

(3) The boiling point of a compound depends on the strength of its intermolecular attractions.

Molecule of Y is longer and somewhat spread-out whereas that of X is more spherical and compact.

The shape of molecule of Y allows greater surface contact between molecules.

The van der Waals’ forces in Y are thus stronger.

More heat is needed to separate molecules of Y during boiling.

Hence Y has a higher boiling point than X.

87 C (2) and (3) H C

H

H H

H

C O H and H C

H

H

C

O

O H molecules can form hydrogen bonds

with water molecules.

O

O H

H C

H

H

C

hydrogen bond

hydrogen bond

lone pair

δ+δ–

δ–

O

HHδ+

δ+

δ–

O H

Hδ+

δ+δ–

CH3CH2

hydrogen bond

lone pair

Oδ–

Hδ+

Hδ+

Oδ–

Hδ+

88 D

89 B (2) Due to the symmetry of the octahedral shape of a sulphur hexafluoride molecule, the six identical bond dipole moments cancel one another out exactly.


∴ instantaneous dipole-induced dipole attractions exist between sulphur hexafluoride molecules.

(3) NO permanent dipole-permanent dipole attractions exist between sulphur hexafluroide molecules.

306

90 D (2) and (3) In a methoxymethane molecule, the C–O bonds are polar.

H C

H

H

C

H

H

H net dipole momentO

The molecule has a net dipole moment and thus a methoxymethane molecule is polar.

∴ instantaneous dipole-induced dipole attractions and permanent dipole-permanent dipole attractions exist between methoxymethane molecules.

91 D (1) Hydrogen bonds exist between methanoic acid molecules.

C

O

H

O OH

H O

H C

hydrogen bond

hydrogen bondlone pair

δ+δ–

δ+

δ–

δ– δ–

92 C (3) Electron diagram of a HNO3 molecule:

H O N

O


The hydrogen atom has a strong positive charge because it is bonded to a highly electronegative oxygen atom.

The partially positively charged hydrogen atom can form hydrogen bond with a lone pair on the oxygen atom of another HNO3 molecule.

93 B In ice, hydrogen bond is the strongest intermolecular attraction.

Each water molecule forms four hydrogen bonds tetrahedrally.


94 B (2) There are NO hydrogen bonds in X and Y.


Both X and Y are non-polar. Relatively weak van der Waals’ forces exist in them.


Hence the boiling point of X is lower than that of Y.

307

95 A (1) Methanoic acid molecules can form hydrogen bonds with water molecules.

O

O H

H C

hydrogen bond

hydrogen bond

lone pair

δ+δ–

δ–

O H

Hδ+

δ+δ–

O

HHδ+

δ+

δ–

(2) Methoxymethane molecules can form hydrogen bonds with water molecules.

Oδ–

HHδ+ δ+

hydrogen bond

lone pair

Oδ–

CH3H3C

96 B (2) Intramolecular hydrogen bonds occur in

OH

NO2

.

O

H

N

O

O

key:

hydrogen bond

97 A (3) The boiling point of a compound depends on the strength of its intermolecular attractions.

Hydrogen bonds exist in ethane-1,2-diol while only van der Waals’ forces exist in propane.


More heat is needed to separate the ethane-1,2-diol molecules during boiling.

Thus the boiling point of ethane-1,2-diol is higher than that of propane.

308

98 A (1) Both water and trichloromethane are polar.

∴ streams of both liquids are deflected by a charged rod.

(2) Liquids that have strong intermolecular forces tend to have high surface tension.

Water has a high surface tension. The hydrogen-bonded molecules form an array across the water surface.

(3) The viscosity of a liquid depends on:

• the strength of attractive forces between molecules; and

• the tendency of molecules to become entangled with each other.

A liquid with strong intermolecular forces has a higher viscosity than one with weak intermolecular forces.

Water molecules can form hydrogen bonds but trichloromethane molecules cannot.

∴ the viscosity of water is higher than that of trichloromethane.

99 D (1) X is a non-polar compound while Y is a polar compound.


Hydrogen bonds exist in Y while only van der Waals’ forces exist in X.


More heat is needed to separate the molecules of Y during boiling.

Thus the boiling point of Y is higher than that of X.





Molecules of Y can form hydrogen bonds but molecules of X cannot.

∴ Y is more viscous than X.

100 B (2) The boiling point of a compound depends on the strength of its intermolecular attractions.

Van der Waals’ forces exist in X and Y.

The strength of forces increases with the number of electrons in one molecule.

Hence the boiling point of Y is higher than that of X.





Van der Waals’ forces in Y are stronger than those in X.

∴ Y is more viscous than X.

309

101 B The electronegativity of an element represents the power of an atom of that element to attract a bonding pair of electrons towards itself in a molecule.

102 A In N–H bond, the nitrogen atom gets a slightly negative charge because it has a greater share of the bonding electrons. The hydrogen atom becomes slightly positively charged because it has lost some of its share in the bonding electrons.

103 D In NH3, the bond pairs of electrons are attracted towards the nitrogen atom to a greater extent as nitrogen is more electronegative than hydrogen.

The bond pairs of electrons repel each other to a greater extent and thus the H–N–H bond angle is greater.

In NF3, the bond pairs of electrons are closer to the fluorine atom as fluorine is more electronegative than nitrogen.

The bond pairs of electrons repel each other to a less extent and thus the F–N–F bond angle is smaller.

The atomic size of fluorine is smaller than that of nitrogen.

104 B Within each group, electronegativity decreases with increasing atomic number and metallic character.

The electronegativity value of Cl is greater than that of I.

The electronegativity difference between H and Cl is greater than that between H and I.

Thus H–Cl bond is more polar than H–I bond.

105 C The electronegativity value of fluorine is greater than that of boron. So each B–F bond is polar.

A BF3 molecular has a trigonal planar shape. The three identical B–F bond dipole moments in a BF3 molecule cancel one another out exactly.

As a result, the BF3 molecule has no net dipole moment and is non-polar.

106 D Three dimensional shape of a SF4 molecule:

FF

FS

F

The electronegativity value of fluorine is greater than that of sulphur. So each S–F bond is polar.

The SF4 molecule has a seesaw shape. The individual S–F bond dipole moments reinforce each other.

The SF4 molecule has a net dipole moment.

107 C The electronegativity value of Cl is greater than that of Si. So each Si–Cl bond is polar.

Due to the symmetry of the tetrahedral shape of the SiCl4 molecule, the individual bond dipole moments cancel one another out exactly.


∴ NO permanent dipole-permanent dipole attractions exist between silicon tetrachloride molecules.

310

108 A Propanone molecule is polar.

net dipole momentC CC

O

H H

H H H H

δ–

δ+


Hydrogen sulphide is a gas at room temperature and pressure because the attractions between its molecules are weak.


Hydrogen bonds exist in HF while only van der Waals’ forces exist in HI.



Thus the boiling point of HF is higher than that of HI.

111 A NH3 molecules can form hydrogen bonds with water molecules.

Hδ+

δ+H

H

δ+

δ+

δ+

Nδ– δ–

H

O

Hhydrogen bond

lone pair

112 B Liquids that have strong intermolecular forces tend to have high surface tension.

Water has a high surface tension. The hydrogen-bonded molecules form an array across the water surface.

311

Part B Topic-based exercise


1 B Option B — The xenon atom has 8 electrons in its outermost shell and each fluorine atom has 7, i.e. a XeF4 molecule has a total of 36 electrons in the outermost shells.

2 A Option A — The sulphur atom has 6 electrons in its outermost shell and each fluroine atom has 7, i.e. a SF4 molecule has a total of 34 electrons in the outermost shells.

3 D Option A — An oxygen atom cannot have more than 8 electrons in its outermost shell.

Option B — A nitrogen atom cannot have more than 8 electrons in its outermost shell.

Opiton D — The nitrogen atom has 5 electrons in its outermost shell and each oxygen atom has 6 electrons. But the NO3

– ion carries one negative charge because it has gained 1 electron. Thus a NO3

– ion has a total of 24 electrons in the outermost shells.

4 D Option B — A carbon atom cannot have more than 8 electrons in its outermost shell.

Option D — The sulphur atom has 6 electrons in its outermost shell, the carbon atom has 4 and the nitrogen atom has 5. But the SCN– ion carries one negative charge because it has gained 1 electron. Thus a SCN– ion has a total of 16 electrons in the outermost shells.

5 A Option A — NCl5 does NOT exist because nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its atom.

6 C Option C — Consider a molecule with one lone pair and three bond pairs of electrons in the outermost shell of the central atom.


The four pairs of electrons in the molecule will adopt a tetrahedral arrangement.

The shape of a molecule is determined only by the arrangement of atoms. Thus, the molecule has a trigonal pyramidal shape.

7 C Option C — Electron diagram of a XeO3 molecule:

O Xe O

O


A XeO3 molecule has one lone pair and three bond pairs of electrons in the outermost shell of the Xe atom.



The shape of a molecule is determined only by the arrangement of atoms. Thus, the XeO3 molecule has a trigonal pyramidal shape.

312

8 C Option C — Electron diagram of a PCl4+ ion:

Cl P Cl

Cl

Cl

+


The four pairs of electrons in the outermost shell of the phosphorus atom will adopt a tetrahedral arrangement.

∴ a PCl4+ ion has a tetrahedral shape.

9 B Option B — Consider a molecule with one lone pair and five bond pairs of electrons in the outermost shell of the central atom.


The six pairs of electrons in the molecule will adopt an octahedral arrangement.

The shape of a molecule is determined only by the arrangement of atoms. Thus, the molecule has a square pyramidal shape.

10 D Option Species


A BF3

F

BF F

B CO32–

2–O

CO O

C NO3–

–O

NO O

D PCl3Cl P

Cl

Cl

∴ a PCl3 molecule has one lone pair of electrons in the outermosst shell of the central atom.

313

11 C




A C2H4 C CH

H

H

H H

HC

HC

Hplanar

B H3O+

+

H O H

H

+

HH

HO

trigonal pyramidal

C SF2

S F

F

FF

S V-shaped

D XeF2 XeFF

F

Xe

F

linear

∴ a SF2 molecule has a V-shape.

12 C




A BH3

H

BH H B

H

H H

trigonal planar

B CO32–

2–O

CO O C

O

O O

2–

trigonal planar

C PF3F P F

F FF

FP

trigonal pyramidal

D SO3O S O

OS

O

O O

trigonal planar

∴ a PF3 molecule has a trigonal pyramidal shape.

314

13 A




A I3– II I

–

I

I

I

–

linear

B O3 O O O O

OO

V-shaped

C OF2

O F

F

FF

O V-shaped

D SO2 OS

O S

OO

V-shaped

∴ an I3– ion has a linear shape.

14 C Option C — Electron diagram of a NCl3 molecule:

Cl N Cl

Cl


A nitrogen trichloride molecule has one lone pair and three bond pairs of electrons in the outermost shell of the nitrogen atom.



The shape of a molecule is determined only by the arrangement of atoms. Thus, the nitrogen trichloride molecule has a trigonal pyramidal shape.

Option D — When using the VSEPR theory, double bonds can be counted as single bonds. Therefore we

can view the carbon atom of a H C

O

H molecule as having 3 pairs of electrons in its outermost shell.



∴ the H C

O

H molecule is planar.

315

15 A




A BeCl2Cl Be Cl Be ClCl linear

XeF

XeFF

F

Xe

F

linear

B BF

F

BF F B

F

F F

trigonalplanar

PCl

Cl P Cl

Cl ClCl

ClP

trigonalpyramidal

C PF

F

PFF

F F

F

FF

FFP

trigonalbipyramidal

IF

I

FFF

FFF

F F

F FI square

pyramidal

D SiF4

F Si F

F

FF

F FSi

F

tetrahedral

SF4

SFF

F FF

F

FS

F

seesaw

∴ molecules of BeCl and XeF have an identical shape.

6 D Option D —




XeF4 XeFF

FF F F

F FXe

squareplanar

316

17 C




A PBr3

Br P Br

Br BrBr

BrP

trigonal pyramidal

PCl3

Cl P Cl

Cl ClCl

ClP

trigonal pyramidal

B NH3

H N H

H HH

HN

trigonal pyramidal

BF3

F

BF F B

F

F F

trigonal planar

C BF3

F

BF F B

F

F F

trigonal planar

CO32–

2–O

CO O C

O

O O

2–

trigonal planar

D ClF3

F Cl F

F

F

F

FCl T-shaped

PF3

F P F

F FF

FP

trigonal pyramidal

∴ the BF3 molecule and CO32– ion are trigonal planar.

317

18 A For a molecule with two lone pairs and three bond pairs of electrons in the outermost shell of the central atom, the shape that puts the five electron pairs the furthest apart is trigonal bipyramidal.

There are three different ways in which we may arrange the five electron pairs into a trigonal bipyramid. The correct shape is the one with the minimum repulsion.

X

XM

X

I

X

XX

II

X

M

X

III

XM90°90°

90°

90°

120°

90°



II6 90° lone pair-bond pair repulsions (repulsions at angles greater than 90° can be ignored) bond pair-bond pair repulsion is less than lone

pair-bond pair repulsion, hence shape III has the minimum repulsionIII


∴ the molecule is T-shaped.

19 A For a molecule with two lone pairs and four bond pairs of electrons in the outermost shell of the central atom, the shape that puts the six electron pairs the furthest apart is octahedral.

The most preferred shape of the molecule is square planar with respect to the atoms where the lone pairs of electrons are separated by 180°.

X

X

X

XM

20 B Option B — The xenon atom has 8 electrons in its outermost shell, the oxygen atom has 6 and each fluorine atom has 7, i.e. a XeOF4 molecule has a total of 42 electrons in the outermost shells.

Option C — Oxygen atom cannot have more than 8 electrons in its outermost shell.

318

21 C When using the VSEPR theory, double bonds can be treated like single bonds. Thus we can regard a XeOF4 molecule as having one lone pair and five bond pairs of electrons in the outermost shell of the xenon atom.



The shape of a molecule is determined only by the arrangement of atoms. Thus, the XeOF4 molecule has a square pyramidal shape.

OF F

F FXe

22 C

23 D Three-dimensional structure of a XeF4 molecule:

90°F

F

F

FXe

∴ the molecule contains 90° lone pair-bond pair repulsions.

24 C



Three-dimensional structure Bond angle

A H2O H O H

HH

O 104.5°

B CH3+

H C

H

H

+

C

H

H H

+

120°

C CO2 O C O C OO 180°

D SO2 OS

O S

OO

~120°

∴ a CO2 molecule has the greatest angle between two covalent bonds.

319

25 C

N

H

O

H

C N Hβ

αH

According to the VSEPR theory, electron pairs in the outermost shell of the central atom of a molecule repel one another and stay as far apart as possible.

When using the VSEPR theory, double bonds can be treated like single bonds. Therefore we can view the carbon atom as having three pairs of electrons in its outermost shell.


∴ α is about 120°.

The nitrogen atom has four pairs of electrons in its outermost shell.

The furthest apart the pairs can get is when they are arranged in a tetrahedral shape.

∴ β is about 109°.

26 B Option B — In an AlCl3 molecule, there are three bond pairs of electrons in the outermost shell of the central aluminium atom.



Hence an AlCl3 molecule has a trigonal planar shape.

In the adduct, the nitrogen atom in the NH3 molecule supplies a lone pair of electrons to the aluminium atom, forming a dative covalent bond.

Hence there are four bond pairs of electrons in the outermost shell of the aluminium atom in the adduct.

The furthest apart the four pairs of electrons can get is when they are arranged in a tetrahedral shape.

N

H

H Cl

Cl

+ Al ClH N

H

H

Cl

Cl

Al ClH

320

27 B Option Species Three-dimensional structure Bond angle(s)

A SiH4

HH H

Si

H

109.5°

NH3

HH

HN 107°

B SO2

S

OO

~120°

CS2 C SS 180°

CCS2 C SS 180°

SF4 FF

FS

F

90°, 120°, 180°

D H2O

HH

O 104.5°

H2S

HH

S 92.5°

Option D — The electronegativity value of oxygen is greater than that of sulphur.

Hence the bond pairs of electrons in H2O will be attracted towards the oxygen atom to a greater extent. These bond pairs will repel each other to a greater extent and the bond angle in H2O is greater than the bond angle in H2S.

∴ in Option B, the bond angle around the central atom in the second species (CS2) is larger than that in the first species (SO2).

321

28 C Molecule Three-dimensional structure Bond angle

BF3 B

F

F F

120°

PF3

FF

FP ~109°

ClF3

F

F

FCl ~90°

∴ the order of increasing bond angle is ClF3 < PF3 < BF3.

29 B

30 B Element W X Y Z

Atomic number 7 11 13 17

Name of element nitrogen sodium aluminium chlorine

X (sodium) is the least electronegative.

31 D Option D — Three-dimensional structure of a SiF4 molecule:

FF F

Si

F

The electronegativity value of fluorine is greater than that of Si. So each Si–F bond is polar.

Due to the symmetry of the tetrahedral shape of the SiF4 molecule, the individual bond dipole moments cancel one another out exactly.



Both CH4 and CCl4 are non-polar. Van der Waals’ forces exist in them.


Hence the boiling point of CCl4 is higher than that of CH4.

322


All the compounds are non-polar. Van der Waals’ exist in them.


Hence the boiling point of heptane (C7H16) is the highest.

34 D Option D — The boiling point of a compound depends on the strength of its intermolecular attractions.

Hydrogen bonds exist in H2O while only van der Waals’ forces exist in other Group VI hydrides.



Thus the boiling point of H2O is the highest.

The number of electrons in one hydride molecule increases from H2S to H2Te.

Hence the strength of van der Waals’ forces between the molecules also increases from H2S to H2Te.


and thus the boiling points of the hydrides increase from H2S to H2Te.

35 C The boiling point of a compound depends on the strength of its intermolecular attractions.

The intermolecular attractions in HCl, HBr and HI are van der Waals’ forces.

The number of electrons in one hydride molecule increases from HCl to HI.

Hence the strength of van der Waals’ forces between the molecules also increases from HCl to HI.


and thus the boiling points of the hydrides increase from HCl to HI.

36 A Option A — The boiling point of C2H5OH is higher than that of C2H5OC2H5. This is due to the presence of hydrogen bonds between molecules of C2H5OH.

37 D Option D — According to the VSEPR theory, electron pairs in the outermost shell of the central atom of a molecule repel one another and stay as far apart as possible.

When using the VSEPR theory, double bonds can be treated like single bonds. Therefore we can view the carbon atom as having three pairs of electrons in its outermost shell.


∴ the O–C–O bond angle is about 120°.

38 C Option C — The boiling point of X is higher than that of Y.


Hydrogen bonds exist in X while only van der Waals’ forces exist in Y.

323


More heat is needed to separate the molecules of X during boiling.

Thus the boiling point of X is higher than that of Y.

39 A (3) Electron diagram of a BH3 molecule:

H B H

H


40 B (2) BF3 molecule does NOT contain any unpaired electron.

(3) The octet rule suggests that atoms become stable by having eight electrons in their outermost shells (or two electrons in the case of some smaller atoms). An atom attains a stable electronic arrangement by sharing or transfer of electrons.

The NO molecule does NOT conform to the octet rule because it contains an unpaired electron.

The BF3 molecule does NOT conform to the octet rule because there are less than 8 electrons in the outermost shell of the boron atom.

41 B




(3) NO3–

–O

NO O N

O

O O

–

trigonal planar

∴ species X could NOT be a NO3– ion.

42 C




(1) CS2 S C S C SS linear

(2) Cl2OO Cl

Cl

ClCl

O V-shaped

(3) NH2–

–

H N H

HH

N

–

V-shaped

∴ species X could be Cl2O / NH2–.

324

43 D Species


(1) PCl5

Cl

PClCl

Cl Cl

(2) SF4 SFF

F F

(3) XeF2 XeFF

∴ all the three species have five electron pairs in the outermost shell of the central atom. The five electron pairs will adopt a trigonal bipyramidal arrangement.

44 B




(1) SF4 SFF

F FF

F

FS

F

seesaw

(2) SiF4 F Si F

F

FF

F FSi

F

tetrahedral

(3) XeF4 XeFF

FF F F

F FXe square planar

∴ the SiF4 and CH4 molecules have a tetrahedral shape.

325

45 B




(1) CO2

O C O C OO linear

C HC C

H

H

H

H H

HC

HC

Hplanar

H O

H O

HH

HO V-shaped

CSS C S C SS linear

HCNH C N C NH linear

XeF

XeFF

F

F

Xe linear

XeF

XeFF

F

F

Xe linear

OF

F O

F

FF

O V-shaped

SCl

Cl S

Cl

ClCl

S V-shaped

∴ the set CS, HCN and XeF contains only linear species.

326

46 A Species Three-dimensional structure

(1) CH3NH2

CH3

HH

N

(2) ClF3

F

F

FCl

(3) PF5

F

FF

FFP

∴ the shapes of the CH3NH2 and ClF3 molecules are influenced by the presence of lone pairs of electrons.

47 A Species Three-dimensional structure

(1) XeF4

F

F

F

FXe 90°

(2) PF5 F

F

P

F

90°F

F

(3) SiF4

109.5°FF F

Si

F

∴ the XeF4 and PF5 molecules contain 90° bond angles.

48 B Species Three-dimensional structure

(1) C2H4 CC

H

H

H

H

120°

120°

120°

(2) NF3

102°FF F

N

(3) PCl5

Cl

ClCl

ClClP

120°

∴ the C2H4 and PCl5 molecules contain 120° bond angles.

327

49 A Molecule



(1) BrF5Br

FFF

FFF

F F

F FBr

(2) ClF5

Cl

FFF

FFF

F F

F FCl

(3) PCl5

Cl

PClCl

Cl Cl

Cl

ClCl

ClClP

∴ the BrF5 and ClF5 molecules have the geometric arrangement shown.

50 B (1) H+

H2O H3O+

In the outermost electron shell of the central oxygen atom in each of the species, the numbers of bond pairs and lone pair(s) of electrons are as follows:


H3O+ 3 1

H2O 2 2


The furthest apart the four pairs of electrons can get is when they are arranged in a tetrahedral shape.


Thus the bond angle in the H3O+ ion is compressed to slightly less than 109.5°.


In the water molecule, the two lone pairs will stay the furthest apart.

As a result, the bond angle in the H2O molecule is further compressed.

∴ bond angle: H3O+ > H2O

(2) H2

C2H4 C2H6

328

Three-dimensional structures of the C2H4 and C2H6 molecules:

CC

H

H

H

H

120°

120°

120°

H HC

H

CH H

H

In a C2H6 molecule, each carbon atom has four pairs of electrons in its outermost shell. The overall arrangement of the four pairs of electrons around each carbon atom is tetrahedral. Thus, the bond angles are 109.5°.

∴ bond angle: C2H4 > C2H6.

(3) H+

NH3 NH4+

In the outermost electron shell of the central nitrogen atom in each of the species, the numbers of bond pairs and lone pair(s) of electrons are as follows:


NH4+ 4 0

NH3 3 1


In the NH4+ ion, the furthest apart the four pairs of electrons can get is when they are arranged in

a tetrahedral shape.




∴ bond angle: NH4+ > NH3

51 C (1) The order of atomic size of the halogen is in the order Cl < Br < I.

52 D

53 A (3) Three-dimensional structure of a SF6 molecule:

F

FF

F

F

FS

The electronegativity value of F is greater than that of S. So each S–F bond is polar.

Due to the symmetry of the octahedral shape of the SF6 molecule, the individual bond dipole moments cancel one another out exactly.

The SF6 molecule has no net dipole moment and it is non-polar.

329

54 B Molecule Three-dimensional structure Polar bond Symmetrical shape? Polar molecule?

(1) BCl3 B

Cl

Cl Cl

δ+ δ–

B—Cl yes no

(2) Cl2O

ClCl

Oδ– δ+

O—Cl no yes

(3) SiCl4

ClCl Cl

Si

Cl

δ+ δ–

Si—Cl yes no

∴ the BCl3 and SiCl4 molecules are non-polar.

55 B (1) Consider a molecule with two lone pairs and four bond pairs of electrons in the outermost shell of the atom X.



The shape of a molecule is determined only by the arrangement of atoms. Thus, the molecule has a square planar shape.

∴ there are 12 electrons in the outermost shell of the central atom X.

(2) Suppose each X–Y bond is polar.

Due to the symmetry of the shape of the XY4 molecule, the individual bond dipole moments cancel one another out exactly.

The XY4 molecule has no net dipole moment and it is non-polar.

(3) The bond angles around the central atom X are 90°.

56 D (1) The 6 electron pairs in the outermost shell of the central atom X will adopt an octahedral arrangement.

(2) There are more than 8 electrons in the outermost shell of the central atom X.

∴ molecule XY6 does not conform to the octet rule.

57 D (2) and (3) Trichloromethane molecules are polar.

∴ both instantaneous dipole-induced dipole attractions and permanent dipole-permanent dipole attractions exist between trichloromethane molecules.

330

58 D (1) The shape of a methoxymethane is NOT linear.

O

CH3H3C

(2) The methoxymethane molecule is polar.

H3C

O

CH3

δ+ δ+

δ–

(3) The methoxymethane molecule can form hydrogen bonds with water molecules.

O

HH

hydrogen bondlone pairO

CH3H3C

δ+ δ+

δ–

δ–

59 C (1) The propanone molecule is polar.


O

H H

H H H H

δ–

δ+

(2) NO hydrogen bond exists between propanone molecules.

(3) The propanone molecule can form a hydrogen bond with a trichloromethane molecule.

OC

H3C

H3C

C

Clδ–

Clδ–

Clδ–

Hδ+

δ+ δ–

331

60 D (1) and (2) Three-dimensional structures of the NF3 and NH3 molecules:

FF

FN net dipole moment

HH

HN net dipole moment

Both the NF3 and NH3 molecules have a trigonal pyramidal shape.

Both the NF3 and NH3 molecules are polar.

(3) In NH3, the bond pairs of electrons are attracted towards the nitrogen atom to a greater extent as nitrogen is more electronegative than hydrogen.

The bond pairs of electrons repel each other to a greater extent and thus the H–N–H bond angle is greater.

In NF3, the bond pairs of electrons are closer to the fluorine atom as fluorine is more electronegative than nitrogen.

The bond pairs of electrons repel each other to a less extent and thus the F–N–F bond angle is smaller.

61 C (1) Although H–S bond is polar, the attraction between the partially positively charged hydrogen atom and the lone pair on the sulphur atom of another H2S molecule is not strong.

The sulphur atom is quite large and the lone pairs are not very accessible to the hydrogen atom.

∴ hydrogen bonds do NOT exist in H2S.


Hydrogen bonds exist in H2O while only van der Waals’ forces exist in H2S.



Thus the boiling point of H2O is higher than that of H2S.

(3) The electronegativity value of oxygen is greater than that of sulphur.

Hence the bond pairs of electrons in H2O will be attracted towards the oxygen atom to a greater extent.


Thus the H–O–H bond angle in H2O is greater than the H–S–H bond angle in H2S.

62 A (1) and (2) When using the VSEPR theory, double bonds can be counted as single bonds. Therefore we can view the carbon atom of a methanal molecule as having 3 pairs of electrons in its outermost shell.



∴ the methanal molecule has a planar structure and the H–C–O bond angle is approximately 120°.

332

(3) In a methanal molecule, NO hydrogen atom is attached directly to a highly electronegative atom (such as oxygen, nitrogen or fluorine). Thus NO hydrogen bond exists between the methanal molecules.

63 D (2) Hydrogen bond between H C

H

H

H

N H molecules:

hydrogen bond

lone pair

Nδ–

Hδ+ H

δ+H3C

δ+

Nδ–

Hδ+ H

δ+H3C

δ+

(3) Hydrogen bonds between H C

O

O H molecules:

C

O

H

O OH

H O

H C

hydrogen bond


δ+δ–

δ+

δ–

δ– δ–

64 A (1) Ethanol is miscible with water (i.e. soluble in water in all proportions) because hydrogen bonds can form between ethanol molecules and water molecules. This helps the dissolving process.

O

CH3CH2

hydrogen bond

lone pair

hydrogen bond

Oδ–

Hδ+

Hδ+

Oδ–

Hδ+

Hδ+

Hδ+

δ–

333

(2) Methoxymethane is fairly soluble in water because hydrogen bonds can form between methoxymethane molecules and water molecules.

O

HH

hydrogen bond

lone pairO

CH3H3C

δ+ δ+

δ–

δ–

However, the water solubility of methoxymethane is lower than that of ethanol since ethanol molecules are capable of forming more hydrogen bonds with water molecules.

(3) Propane is non-polar. Propane does not mix with water due to the difference in the strength of intermolecular attractions between water molecules and those between propane molecules.

Thus, propane is insoluble in water.

65 A (2) Both X and Y are soluble in water because they can form hydrogen bonds with water.

Hydrogen bond between molecule of Y and water molecule:

CH3CH2CH2

hydrogen bond

lone pair

Oδ–

Hδ+

Hδ+

Oδ–

Hδ+

(3) The boiling point of X is lower than that of Y.


The shape of molecule of Y is longer while that of X is more compact.

This allows greater surface contact between molecules of Y.

Van der Waals’ forces in Y are stronger than those in X.

More heat is needed to separate the molecules of Y during boiling.

66 B (1)no net dipole moment

Cl

HC

ClC

Hnet dipole moment

H

ClC

YX

ClC

H

In a molecule of X, both –Cl groups are on the same side of the carbon-carbon double bond. The molecule has a net dipole moment and it is polar.

In a molecule of Y, the bond dipole moments cancel one another out and there is no net dipole moment. The molecule is non-polar.

334

(2) The boiling point of X is higher than that of Y.


X is a polar compound. There are permanent dipole-permanent dipole attractions and instantaneous dipole-induced dipole attractions between molecules of X.

Y is a non-polar compound. There are only instantaneous dipole-induced dipole attractions between molecules of Y.


(3) Both X and Y CANNOT form hydrogen bond with another molecule of its own.

67 A (2) The boiling point of a compound depends on the strength of its intermolecular attractions.

Molecule of X is longer and somewhat spread-out whereas that of Y is more spherical and compact.

The shape of molecule of X allows greater surface contact between molecules.

The van der Waals’ forces in X are thus stronger.


Hence X has a higher boiling point than Y.

(3) Stronger van der Waals’ forces in X pull the molecules closer together.

So, the density of X is higher.

68 D (1) Both X and Y are soluble in water because they can form hydrogen bonds with water.

Hydrogen bond between molecule of X and water molecule:

O

O OH

H C

H

H

C

hydrogen bond

lone pair

δ+

Hδ+

Hδ+δ– δ–

δ–

Hydrogen bond between molecule of Y and water molecule:

Oδ–

HHδ+ δ+

hydrogen bond

lone pair

O

CH3H3C

δ+C

δ–

335


Hydrogen bonds exist in X while only van der Waals’ forces exist in Y.



Thus the boiling point of X is higher than that of Y.

(3) Stronger intermolecular attractions in X pull the molecules closer together.


69 A Electron diagram of a PCl5 molecule:

Cl

PClCl

Cl Cl

Phosphorus does not conform to the octet rule because it can form compounds with more than eight electrons in its outermost shell.

70 C A XeF4 molecule has two lone pairs and four bond pairs of electrons in the outermost shell of the xenon atom.



The shape of a molecule is determined only by the arrangement of atoms. Thus, the XeF4 molecule has a square planar shape.

F F

F FXe

71 B

IonElectron diagram



CO32–

2–O

CO O C

O

O O

2–

trigonal planar

NO3–

–O

NO O N

O

O O

–

trigonal planar

∴ CO32– ion and NO3

– ion are planar species.

336

72 B When using the VSEPR theory, double bonds can be counted as single bonds. Therefore we can view each carbon atom of an ethene molecule as having 3 pairs of electrons in its outermost shell.



Thus ethene has a planar structure.

73 B




BeF2 F Be F Be FF linear

XeF2Xe

FF

F

Xe

F

linear

∴ BeF2 and XeF2 are linear species.

74 B The electronegativity of oxygen is higher than that of sulphur.

Hence the bond pairs of electrons in H2O will be attracted towards the oxygen atom to a greater extent.


Thus the H–O–H bond angle in H2O is greater than the H–S–H bond angle in H2S.

75 C In buckminsterfullerene, the molecules are held by weak van der Waals’ forces. The molecules can easily slide over each other. Hence buckminsterfullerene is quite soft.

∴ buckminsterfullerene CANNOT replace diamond in cutting stones.

76 C The electronegativity value of Cl is greater than that of C. So each C–Cl bond is polar.

Due to the symmetry of the tetrahedral shape of the tetrachloromethane molecule, the individual bond dipole moments cancel one another out exactly.


77 A The propanone molecule is polar.


O

H H

H H H H

δ–

δ+

337

78 B Each S–O bond is polar.

A sulphur dioxide molecule is V-shaped. The individual S–O bond dipole moments reinforce each other.

Hence the SO2 molecule has a net dipole moment.

Each C=O bond is polar.

A carbon dioxide molecule is linear in shape.

The individual C=O bond dipole moments cancel each other out exactly.

Hence the CO2 molecule has no net dipole moment.

79 A A xenon atom contains more electrons than a argon atom.

Hence the van der Waals’ forces between xenon atoms are stronger than those between argon atoms.

More heat is needed to separate the xenon atoms during boiling.

Thus the boiling point of xenon is higher than that of argon.

80 C The boiling point of a compound depends on the strength of its intermolecular attractions.

The strongest type of intermolecular attractions in H2O is hydrogen bonds.

The strongest type of intermolecular attractions in H2Se is van der Waals’ forces.

The hydrogen bonds are stronger than van der Waals’ forces.

Hence more heat is needed to separate the H2O molecules during boiling.

Thus the boiling point of H2O is higher than that of H2Se.

81 D In a methoxymethane molecule, the C–O bonds are polar.

H C

H

H

C

H

H


The molecule has a net dipole moment and thus methoxymethane is polar.

Methoxymethane is fairly soluble in water because hydrogen bonds can form between methoxymethane molecules and water molecules.

O

HH

hydrogen bond

lone pairO

CH3H3C

δ+ δ+

δ–

δ–

338

82 B Propanone can form hydrogen bonds with trichloromethane.

OC

H3C

H3C

C

Clδ–

Clδ–

Clδ–

Hδ+

δ+ δ–

83 A Methanamine ( H

H H

H

C N H ) molecules can form hydrogen bonds with water molecules.

Hδ+

δ+H

H

δ+

δ+

δ+

Nδ– δ–

H3C

O

Hhydrogen bond

lone pair

84 D The viscosity of a liquid depends on:



An ethanol molecule contains more electrons than a methanol molecule.

Hence the strength of van der Waals’ forces in ethanol is higher than that in methanol.

Thus the viscosity of ethanol is higher than that of methanol.

Both methanol and ethanol molecules can form hydrogen bonds.

339

Short questions

85 Molecular model Geometry

(a) square pyramidal

(1)

(b) trigonal pyramidal

(1)

(c) square planar

(1)

(d) trigonal planar

(1)

(e) trigonal bipyramidal

(1)

(f) octahedral

(1)

340

86 Species Electron diagram Three-dimensional structure Shape

(a) BeCl2 Cl Be Cl

(1)

Be ClCl

(1)

linear

(1)

(b) HCN H C N

(1)

C NH

(1)

linear

(1)

(c) CS2 S C S

(1)

C SS

(1)

linear

(1)

(d) BF3

F

BF F

(1)

B

F

F F(1)

trigonal planar

(1)

(e) SiF4 F Si F

F

F(1) F

F FSi

F

(1)

tetrahedral

(1)

(f) PH3 H P H

H(1)

HH

HP

(1)

trigonal pyramidal

(1)

(g) OF2F O

F

(1) FF

O

(1)

V-shaped

(1)

(h) PCl5Cl

PClCl

Cl Cl

(1) Cl

ClCl

ClClP

(1)

trigonal bipyramidal

(1)

341

87 Species Electron diagram Three-dimensional structure Shape

(a) SCl2Cl S

Cl

(1)

ClCl

S

(1)

V-shaped

(1)

(b) ClF3

F Cl F

F

(1)

F

F

FCl

(1)

T-shaped

(1)

(c) ClF5Cl

FFF

FF

(1)

FF F

F FCl

(1)

square pyramidal

(1)

(d) ICl4– I

ClCl

ClCl–

(1)

Cl Cl

Cl ClI

–

(1)

square planar

(1)

(e) I3–

II I

–

(1)

I

I

I

–

(1)

linear

(1)

(f) SO42– O S O

O

O

2–

(1)

OO O

S

O2–

(1)

tetrahedral

(1)

88 a) δ+ δ–H—F (1)

b) δ+ δ–H—Cl (1)

c) δ+ δ–Cl—F (1)

d) δ– δ+N—H (1)

e) δ+ δ–Si—Cl (1)

342

89 Molecule Electron diagram Shape Polar bond

Symmetrical shape?

Polar molecule?

CO2 O C O linearδ+ δ–

C—O yes no

(a) NH3 H N H

H(1)

trigonal pyramidal

(1)

δ– δ+

N—H

(1)

no

(1)

yes

(1)

(b) HBr H Br

(1)

linear

(1)

δ+ δ–

H—Br

(1)

no

(1)

yes

(1)

(c) BeCl2 Cl Be Cl

(1)

linear

(1)

δ+ δ–

Be—Cl

(1)

yes

(1)

no

(1)

(d) SO2 OS

O

(1)

V-shaped

(1)

δ+ δ–

S—O

(1)

no

(1)

yes

(1)

(e) PF5

F

P FF

F F

(1)

trigonal bipyramidal

(1)

δ+ δ–

P—F

(1)

yes

(1)

no

(1)

(f) SiCl4 Cl Si Cl

Cl

Cl(1)

tetrahedral

(1)

δ+ δ–

Si—Cl

(1)

yes

(1)

no

(1)

(g) XeF4Xe

FF

FF

(1)

square planar

(1)

δ+ δ–

Xe—F

(1)

yes

(1)

no

(1)

(h) SF6

F

S

FFF

FF

(1)

octahedral

(1)

δ+ δ–

S—F

(1)

yes

(1)

no

(1)

343

Structured questions

90 a)

Liquid Structure

Type of attractions

Instantaneous dipole-induced

dipole attractions

Permanent dipole-permanent dipole

attractions

Hydrogen bonds

(i) Sulphur dioxideO

S

O✔

(0.5)✔

(0.5)

(ii) Tetrachloromethane C

Cl

Cl

Cl

Cl ✔

(0.5)

(iii) Methoxymethane C

H

H

H

O HC

H

H

✔

(0.5)

✔

(0.5)

(iv) PropanoneH3C C

O

CH3

✔

(0.5)✔

(0.5)

(v) Ethanal C

H

H

H

C

O

H ✔

(0.5)

✔

(0.5)

(vi) Ethanoic acid C

H

H

H

C H

O

O ✔

(0.5)

✔

(0.5)

✔

(0.5)

b)

C

O

C

H

H

H

O OH

H O

H C

H

H

C

hydrogen bond


δ+δ–

δ+

δ–

δ– δ–

(Students need to show only one hydrogen bond between the two molecules, 1 mark for showing the hydrogen bond between the lone pair of oxygen and hydrogen; 1 mark for showing the partial charges and other lone pairs of electrons) (2)

344

c)

O

HH

hydrogen bond

lone pairO

CH3H3C

δ+ δ+

δ–

δ–

(1 mark for showing the hydrogen bond between the lone pair of oxygen and hydrogen; 1 mark for showing the partial charges and other lone pairs of electrons) (2)

91 a) In a BH3 molecule, there are three bond pairs of electrons in the outermost shell of the central boron atom. (0.5)

The electron pairs repel one another and stay as far apart as possible. (0.5)

The shape that puts the three electron pairs furthest apart is trigonal planar.

Hence a BH3 molecule has a trigonal planar shape. (0.5)

In an NH3 molecule, there are one lone pair and three bond pairs of electrons in the outermost shell of the central nitrogen atom. (0.5)

The shape that puts the four electron pairs furthest apart is tetrahedral. (0.5)


Thus the NH3 molecule has a trigonal pyramidal shape. (0.5)

b) Each S–O bond is polar. (0.5)

A sulphur dioxide molecule is V-shaped. (0.5)

The individual S–O bond dipole moments reinforce each other. (0.5)


Each C=O bond is polar. (0.5)

A carbon dioxide molecule is linear in shape. (0.5)

The individual C=O bond dipole moments cancel each other out exactly. (0.5)

Hence the molecule has no net dipole moment.

c) The boiling point of an element depends on the strength of its intermolecular attractions. (1)

The intermolecular attractions in halogens are van der Waals’ forces. (1)

The number of electrons in the halogen molecule increases down the group.

Hence the strength of van der Waals’ forces between halogen molecules also increases down the group. (1)

More heat is needed to separate the molecules during boiling,

and thus the boiling points of the halogens increase down the group.

345

d) In both ice and water, hydrogen bond is the strongest intermolecular attraction. (1)

In ice, each water molecule forms four hydrogen bonds tetrahedrally. (1)

The highly ordered structure leads to a very ‘open’ structure with large spaces in ice. (1)

Thus ice has a lower density.


The water molecules can pack more closely, (1)

so liquid water has a higher density.

e) In H3PO4, hydrogen bond is the strongest intermolecular attraction. (1)

Each H3PO4 molecule has three –OH groups that can take part in hydrogen bonding with other H3PO4 molecules. (1)

Concentrated H3PO4 has a high viscosity due to its strong intermolecular attractions. (1)

92 a)

CH

H

H

HH

methane (1)

NH H

HH

ammonia (1)

OH

HH

water (1)

b) Bond angle: CH4 > NH3 > H2O (1)

In the outermost electron shell of the central atom in each of the molecules, the numbers of bond pairs and lone pair(s) of electrons are as follows:

Molecule Number of bond pairs Number of lone pair(s)

CH4 4 0

NH3 3 1

H2O 2 2


In the methane molecule, the furthest apart the four pairs of electrons can get is when they are arranged in a tetrahedral shape. (0.5)

Thus the methane molecule has a tetrahedral shape. The H–C–H bond angles are 109.5°. (0.5)

Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion. (0.5)

Thus the H–N–H bond angle in the ammonia molecule is compressed to 107°. (0.5)

Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion. (0.5)

In the water molecule, the two lone pairs will stay the furthest apart.

As a result, the H–O–H bond angle in the water molecule is compressed to 104.5°. (0.5)

346

c) The ’octet rule’ is commonly used to account for the formation of chemical bonds.

i) The octet rule suggests that atoms become stable by having eight electrons in their outermost shells (or two electrons in the case of some smaller atoms). An atom attains a stable electronic arrangement by sharing or transfer of electrons. (1)

ii) Any one of the following:

• PCl5 / SF6 (1)

There are more than 8 electrons in the outermost shell of the phosphorus / sulphur atom. (1)

• BeCl2 / BF3 (1)

There are less than 8 electrons in the outermost shell of the beryllium / boron atom. (1)

93 a) α = 120° (1)

β = 104.5° – 109.5° (1)

b) According to the VSEPR theory, electron pairs in the outermost shell of the central atom of a molecule repel one another and stay as far apart as possible. (1)

When using the VSEPR theory, double bonds can be treated like single bonds. Therefore the carbon atom has three pairs of electrons in its outermost shell. (1)


The oxygen atom has four pairs of electrons / two lone pairs and two bond pairs in its outermost shell. (1)

The furthest apart the pairs can get is when they are arranged in a tetrahedral shape.

94 a) Species Diagram Shape

NH2–(g)

HH

N

–

(1)

V-shaped

(1)

NH3(g)H

HH

N

(1)

trigonal pyramidal

(1)

NH4+(g)

HH H

N

H+

(1)

tetrahedral

(1)

347

b) Bond angle: NH4+(g) > NH3(g) > NH2

–(g) (1)

In the outermost electron shell of the central nitrogen atom in each of the species, the numbers of bond pairs and lone pair(s) of electrons are as follows:


NH4+(g) 4 0

NH3(g) 3 1

NH2–(g) 2 2



tetrahedral shape. (0.5)

Thus the NH4+ ion has a tetrahedral shape. The H–N–H bond angles are 109.5°. (0.5)


Thus the H–N–H bond angle in the NH3 molecule is compressed to 107°. (0.5)

Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion. (0.5)

In the NH2– ion, the two lone pairs will stay the furthest apart. The H–N–H bond angle in the NH2

– ion is compressed to 104.5°. (0.5)

95 a)

C

Cl

H

F

Cl

δ–

δ+ δ–

δ–

(1)

b) Fluorine and chlorine are more electronegative than carbon. (1)

The fluorine and chlorine atoms have a greater share of the bonding electrons. (1)

c)

ClH Cl

C

F

(1)

d) The individual bond dipole moments do not cancel one another out exactly. (0.5)

The molecule has a net dipole moment (0.5)

and the molecule is polar. (1)

e) Permanent dipole-permanent dipole attractions (1)

348

96 a)

B

F

F F (1)

BrI

(1)

b) i) IBr is polar. (0.5)

The electronegativity value of bromine is greater than that of iodine. (0.5)

So the bromine end has a partial negative charge while the iodine end has a partial positive charge. (0.5)

Hence the molecule has a net dipole moment. (0.5)

ii) BF3 is non-polar. (0.5)

The electronegativity value of fluorine is greater than that of boron. So each B–F bond is polar. (0.5)

The three identical B–F bond dipole moments in a BF3 molecule cancel one another out exactly. (0.5)

As a result, the molecule has no net dipole moment. (0.5)

97 a) Hydride Three-dimensional structure Shape

SiH4

HH H

Si

H

(1)

tetrahedral

(1)

PH3

HH

HP

(1)

trigonal pyramidal

(1)

H2S

HH

S

(1)

V-shaped

(1)

b) The boiling point of H2S is higher than that of SiH4. (1)

The boiling point of a compound depends on the strength of its intermolecular attractions. (1)

H2S is a polar substance. There are permanent dipole-permanent dipole attractions and instantaneous dipole-induced dipole attractions between H2S molecules. (1)

SiH4 is a non-polar substance. There are only instantaneous dipole-induced dipole attractions between SiH4 molecules. (1)


349

c) The electronegativity of nitrogen is higher than that of phosphorus. (1)

Hence the bond pairs of electrons in NH3 will be attracted towards the nitrogen atom to a greater extent. (1)

These bond pairs will repel each other to a greater extent (1)

and the H–N–H bond angle in NH3 is greater than the H–P–H bond angle in PH3.

d) i) B F

F

F (1)

Trigonal planar (1)

ii) H–S–H bond angle: 107° (1)

The electron pairs in the outermost shell of the sulphur atom repel one another and stay as far apart as possible. (0.5)

The four pairs of electrons adopt a tetrahedral arrangement. (0.5)


The three bond pairs around the sulphur atom is slightly compressed together, (0.5)

making the bond angle less than 109.5°.

98 a) i) 2–

orO

CO O

2–O

CO O

(1)

ii) O C O C OOor

(1)

b) i) The carbonate ion has a trigonal planar shape. (0.5)

When using the VSEPR theory, double bonds can be counted as single bonds. (0.5)

Therefore we can view the carbon atom as having 3 pairs of electrons in its outermost shell. (0.5)

The electron pairs repel one another and stay as far apart as possible. (0.5)

The shape that puts the electron pairs furthest apart is trigonal planar.

ii) The carbon dioxide molecule has a linear shape. (0.5)

When using the VSEPR theory, we can view the carbon atom as having 2 pairs of electrons in its outermost shell. (0.5)

The two pairs must be at the opposite ends of a straight line in order to be as far apart as possible.

350

c) A carbon dioxide molecule is non-polar. (0.5)

Each C=O bond is polar. (0.5)

As the polar C=O bonds of the molecule are arranged in a straight line, the two identical bond dipole moments cancel each other out exactly. (0.5)

As a result, the molecule has no net dipole moment. (0.5)

So a carbon dioxide molecule is non-polar.

99 a) Allotropes are two (or more) forms of the same element (1)

in which the atoms or molecules are arranged in different ways. (1)

b) Molecules of buckminsterfullerene are held together by van der Waals’ forces. The same forces held molecules of carbon disulphide together. (1)

Hence molecules of buckminsterfullerene and carbon disulphide mix together easily. (1)

c) Buckminsterfullerene is insoluble in water. (1)

The intermolecular forces between water molecules are strong hydrogen bonds. (1)

The weak intermolecular forces between buckminsterfullerene and water are not strong enough to overcome the hydrogen bonds. (1)

Hence molecules of buckminsterfullerene and water do not mix easily.

d) Diamond is harder than buckminsterfullerene. (1)

In buckminsterfullerene, the molecules are held by weak van der Waals’ forces. The molecules can easily slide over each other. Hence buckminsterfullerene is quite soft. (1)

In diamond, each carbon atom is bonded to other carbon atoms by strong covalent bonds. Relative motion of the atoms is restricted. Hence diamond is very hard. (1)

100 a) i) A slightly positive charge. (1)

ii) The electronegativity of hydrogen is lower than that of nitrogen. (1)

b) i) A BF3 molecule has a trigonal planar shape. (1)

In a BF3 molecule, there are three bond pairs of electrons in the outermost shell of the central boron atom. (1)

The electron pairs repel one another and stay as far apart as possible. (1)

The shape that puts the electron pairs furthest apart is trigonal planar.

ii) An NH3 molecule has a trigonal pyramidal shape. (1)

An NH3 molecule has one lone pair and three bond pairs of electrons in the outermost shell of the nitrogen atom. (1)

351

The four pairs of electrons will adopt a tetrahedral arrangement. (1)

The shape of a molecule is determined only by the arrangement of atoms. Thus the NH3 molecule has a trigonal pyramidal shape.

c) 107° (1)

Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion. (1)

The three bond pairs are thus slightly compressed together. (1)

This results in the H–N–H bond angles being 107° instead of 109.5°.

d) i) The nitrogen atom in the NH3 molecule supplies a lone pair of electrons to the boron atom, (1)

forming a dative covalent bond. (1)

ii) H

HH

F

FF

B

N

(1 mark for sharing the covalent bond between N and B atoms; 1 mark for sharing the tetrahedral arrangement of atoms around N and B) (2)

e) i)

HH

N

–

(1)

ii) An NH2– ion has two lone pairs and two bond pairs of electrons in the outermost shell of the

nitrogen atom.

Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion, (1)

while lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.

The two lone pairs will stay the furthest apart. As a result, the H–N–H angle in the NH2– ion is

compressed to a value smaller than that in the NH3 molecule. (1)

101 a) i)

Cl

PClCl

Cl Cl

(1)

ii)

Cl

ClCl

ClClP

(1)

Trigonal bipyramidal shape (1)

352

iii) 90° (1)

120° / 180° (1)

iv) A phosphorus pentachloride molecule is non-polar. (0.5)

Each P–Cl bond is polar. (0.5)

Due to the symmetry of the shape of the molecule,

the individual bond dipole moments cancel one another out exactly. (0.5)

The molecule has no net dipole moment (0.5)

and it is non-polar.

b) +

(1)

shape of [PCl4]+:

ClCl Cl

P

Cl–

(1)

shape of [PCl6]–:

ClCl Cl

Cl ClP

Cl

c) i) Cl P

Cl

Cl

(1)

ii)

ClCl Cl

P

(1)

Trigonal pyramidal shape (1)

iii) A phosphorus trichloride molecule has one lone pair and three bond pairs of electrons in the outermost shell of the phosphorus atom.

The electron pairs repel one another and stay as far apart as possible. (1)

The four pairs of electrons in the molecule will adopt a tetrahedral arrangement. (1)

The shape of a molecule is determined only by the arrangement of atoms. Thus, the phosphorus trichloride molecule has a trigonal pyramidal shape.

d) Not agree (1)

Nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its atom. (1)

353

102 a) Molecule Electron diagram Three-dimensional structure

SF2F S

F

(1)F

FS

(1)

SF4 SFF

F F

(1)

FF

FS

F(1)

SF6

F

S

FFF

FF

(1)

FF F

F FS

F(1)

b) Fluorine is more electronegative than sulphur. (1)

The fluorine atom has a greater share of the bonding electrons. (1)

c) SF6 is non-polar. (0.5)

Due to the symmetry of the octahedral shape, (0.5)

the six identical bond dipole moments cancel one another out exactly. (0.5)

The molecule has no net dipole moment (0.5)

and it is non-polar.

d) Consider the electron diagram of OF2:

F O

F

Oxygen cannot form compounds with more than 8 electrons in the outermost shell of its atom. (0.5)

Hence oxygen forms OF2 only.

Sulphur can form some compounds with more than 8 electrons in the outermost shell of its atom. (0.5)

Hence it can form SF4 and SF6 besides SF2.

103 a) F N F

F (1)

b) i) Tetrahedral (1)

ii) Trigonal pyramidal (1)

354

c) NF3 is polar. (0.5)

The electronegativity value of fluorine is greater than that of nitrogen. So each N–F bond is polar. (0.5)

The NF3 molecule has a trigonal pyramidal shape. The individual N–F bond dipole moments reinforce each other. (0.5)


and it is polar.

d) Not agree. (1)

Nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its atom. (1)

104 a) Dipole moment of a diatomic molecule is the product of the charge and the distance between the charges. (1)

b) The electronegativity of fluroine is much higher than that of iodine. (1)

For HF and HI, the effect of electronegativity difference outweighs the effect of difference in the distance between the charges. (1)

c) i) The boiling point of a compound depends on the strength of its intermolecular attractions. (1)

Hydrogen bonds exist in HF (1)

while only van der Waals’ forces exist in HCl. (1)



Thus the boiling point of HF is higher than that of HCl.

ii) A HI molecule contains more electrons than a HBr molecule. (1)

Hence the strength of van der Waals’ forces in HI is higher than that in HBr. (1)

More heat is needed to separate the HI molecules during boiling.

Thus the boiling point of HI is higher than that of HBr.

105 a) The electronegativity of an element represents the power of an atom of that element to attract a bonding pair of electrons towards itself in a molecule. (1)

b) i) Fluorine is more electronegative than hydrogen. (1)


ii) NH3 (1)

iii) Nitrogen is the least electronegative among nitrogen, oxygen and fluorine. /

Nitrogen and hydrogen have the smallest electronegativity difference. (1)

355

c) i) (1) instantaneous dipole-induced dipole attractions (1)

(2) permanent dipole-permanent dipole attractions (1)

(3) hydrogen bonds (1)

ii) (1) There is a large difference in electronegativity between H and F. (1)

In HF, the hydrogen atom has a stronger positive charge. (1)

The attraction between the hydrogen atom of one molecule and a lone pair on the fluorine atom of another molecule is known as a hydrogen bond. (1)

(2)

Hδ+

Fδ–

F

H

δ–

δ+

hydrogen bond

lone pair

(1 mark for showing the hydrogen bond between the lone pair of fluorine and hydrogen; 1 mark for showing the partial charges and other lone pairs of electrons) (2)

d) The boiling point of water is higher than that of hydrogen fluoride. (1)


The fluorine atom in a hydrogen fluoride molecule has three lone pairs of electrons but there is only one hydrogen atom in each molecule,

hence each HF molecule can only use one lone pair to form a hydrogen bond on average. (1)

The oxygen atom in a water molecule has two lone pairs and there are two hydrogen atoms in each molecule,

so each H2O molecule can take part in hydrogen bonding to twice the extent. (1)

More heat is needed to separate the water molecules during boiling.

106 a) A covalent bond in which the electrons are unequally shared between the two bonded atoms. (1)

b) Fluorine is more electronegative than carbon. (1)


c) i) B (1)

ii)

F

HC

FC

H (1)

iii)

H (1)

FC

FC

H H (1)

HC

FC

F

356

107 a)

(1) (1)

ii) Because of its V-shape, individual bond dipole moments in a water molecule reinforce each other. (0.5)

The water molecule has a net dipole moment. (0.5)

The jet of water is deflected by the positively charged rod.

Negative ends of the molecules are attracted towards the rod. (0.5)

The jet of water is also deflected by the negatively charged rod.

Positive ends of the molecules are attracted towards the rod. (0.5)

b)

Hδ+

Hδ+

Hδ+

Hδ+

H

Oδ–

δ+Hδ+

hydrogen bondOδ–

Hδ+

Hδ+

Oδ–

Oδ–

lone pair

(2 marks for showing correct hydrogen bonds, i.e. the middle water molecule can form hydrogen bonds tetrahedrally; 1 mark for showing δ+ on H and δ– on O across at least one hydrogen bond) (3)

c) i) +

HH H

O

(1)

ii) 107° (1)

357

108 a) A pair of equal and opposite charges separated by a distance is called a dipole. (1)

Dipole moment is the product of the charge and the distance between the charges. (1)

b) i) Permanent dipole-permanent dipole attraction is the attraction between molecules which have permanent dipoles. (1)

ii) Hydrogen bonding is the attraction between the hydrogen atom attached to an electronegative atom (1)

and the lone pair on another electronegative atom (usually oxygen, nitrogen or fluorine). (1)

c)

Hδ+

δ+H

H

δ+

δ+

δ+

Nδ– δ–

H

O

Hhydrogen bond

lone pair

(1 mark for showing the hydrogen bond between the lone pair of nitrogen and hydrogen; 1 mark for showing the partial charges and other lone pairs of electrons) (2)

d) The boiling point of a compound depends on the strength of its intermolecular attractions. (1)

The strongest type of intermolecular attractions in water is hydrogen bonds. (1)

The strongest type of intermolecular attractions in hydrogen sulphide is permanent dipole-permanent dipole attractions / van der Waals’ forces. (1)

The hydrogen bonds are stronger than permanent dipole-permanent dipole attractions / van der Waals’ forces. (1)

Hence more heat is needed to separate the water molecules during boiling.

e) In a molecule of X, both –Cl groups are on the same side of the carbon-carbon double bond. The molecule has a net dipole moment and it is polar. (1)

These molecules are held together by permanent dipole-permanent dipole attractions. (1)

In a molecule of Y, the bond dipole moments cancel one another out and there is no net dipole moment. (1)

These molecules are held together by instantaneous dipole-induced dipole attractions. (1)

Less heat is required to separate the molecules of Y during boiling. Hence Y has a lower boiling point than X.

109 a) The boiling point of a compound depends on the strength of its intermolecular attractions. (1)

Molecule of X is longer and somewhat spread-out whereas that of Z is more spherical and compact. (1)

The shape of molecule of X allows greater surface contact between molecules. (1)

The van der Waals’ forces in X are thus stronger. (1)


Hence X has a higher boiling point than Z.

358

b) Y has a structure in between the linear structure of X and the spherical structure of Z. (1)

The van der Waals’ forces in Y are probably in between those in X and Z. (1)

Hence the boiling point of Y is probably in between those of X and Z. (1)

c) Stronger van der Waals’ forces in X pull the molecules closer together. (1)


d) X is insoluble in water. (1)

X is non-polar. (1)

X does not mix with water due to the difference in the strength of intermolecular attractions between water molecules and those between molecules of X. (1)

110 a) i) Van der Waals’ forces / instantaneous dipole-induced dipole attractions (1)

ii) They have the same number of electrons per molecule. (1)

b) i) H–C–H bond angle: 109.5° (1)

C–C–O bond angle: 120° (1)

ii) Permanent dipole-permanent dipole attractions (1)

c) Propanone molecules can form hydrogen bonds with water molecules, (1)

but butane molecules cannot.

d) i) C–O–H bond angle: 104.5° (1)

ii) Hydrogen bonds (1)

iii)

Oδ–

Hδ+

C3H7

hydrogen bond

lone pair

Oδ–

Hδ+

C3H7


e) The boiling point of a compound depends on the strength of its intermolecular attractions. (1)

The shape of a propan-1-ol molecule is longer. (1)

This allows greater surface contact between molecules. (1)

Van der Waals’ forces in propan-1-ol are stronger than those in propan-2-ol. (1)

More heat is needed to separate the propan-1-ol molecules during boiling.

359

111 a) Boiling point increases in the following order:

D < C < A < B (1)


Both compounds C and D are non-polar. Relatively weak instantaneous dipole-induced dipole attractions exist in them. (1)

The strength of the attractions increases with the number of electrons in the molecule. (1)

Hence the boiling point of compound C is higher than that of compound D.

Hydrogen bonds exist in compound B. (1)

Hence compound B has the highest boiling point.

Compound A has a net dipole moment. Permanent dipole-permanent dipole attractions exist in it. (1)

These attractions are stronger than those in compounds C and D but weaker than those in compound B.

Hence the boiling point of compound A is higher than those of compounds C and D but lower than that of compound B.

b)CH3CH2

hydrogen bond

lone pair

Oδ–

Hδ+

Hδ+

Oδ–

Hδ+


c) The viscosity of compound Y is higher than that of compound X. (1)

Both compounds X and Y can form hydrogen bonds.

Each molecule of Y has three –OH groups that can take part in hydrogen bonding while each molecule of X has only two –OH groups. Each molecule of Y can form more hydrogen bonds. (1)

Furthermore, because of their shapes, the molecules of Y tend to become entangled rather than to slide past one another. (1)

These factors contribute to the high viscosity of compound Y.

112 a) Carbon monoxide (1)

b) The octet rule suggests that atoms become stable by having eight electrons in their outermost shells (or two electrons in the case of some smaller atoms). An atom attains a stable electronic arrangement by sharing or transfer of electrons. (1)

360

c) i) O N O

(1)

ii) Dative covalent bond (1)

d) Any one of the following:

• PCl5 / SF6 (1)

There are more than 8 electrons in the outermost shell of the phosphorus / sulphur atom. (1)

• BeCl2 / BF3 (1)

There are less than 8 electrons in the outermost shell of the beryllium / boron atom. (1)

e) 4NO2(g) + O2(g) + 2H2O(l) 4HNO3(aq) (1)

f) i) O

SO

(1)

ii)

S

OO (1)

iii) Sulphur dioxide is polar. (0.5)

Each S–O bond is polar. (0.5)

The sulphur dioxide molecule has a V-shape. The individual S–O bond dipole moments reinforce each other. (0.5)


and it is polar.

113 a) Any two of the following:

• Non-toxic (1)

• Non-flammable / non-explosive (1)

• High potency / effectiveness as an anaesthetic (1)

b)N N O or N N O

(1)

c) In an ethoxyethane molecule, the C–O bonds are polar. (1)

H C

H

H

C

H

H

C

H

H

C

H

H


361

The molecule has a net dipole moment (1)

and thus ethoxyethane is a polar molecule.

d) The boiling point of a compound depends on the strength of its intermolecular attractions. (1)

Only relatively weak instantaneous dipole-induced dipole attractions and permanent dipole-permanent dipole attractions exist in ethoxyethane. (1)

Not too much heat is needed to separate the molecules during boiling.

e) i)

H

C

H

H C

H

H

Oδ–

HHδ+ δ+

hydrogen bond

lone pair

H

C

H

H

H

C

H

Oδ–


ii) Each hydrogen atom of a water molecule has a partial positive charge. (1)

A hydrogen bond forms between the hydrogen atom of a water molecule

and the lone pair of electrons on the highly electronegative oxygen atom of an ethoxyethane molecule. (1)

f) A trichloromethane molecule is polar. (0.5)

Each C–Cl bond is polar. (0.5)

The individual C–Cl bond dipole moments reinforce each other. (0.5)


and it is polar.

114 The shape of a molecule can be predicted by using what is called the valence-shell electron-pair repulsion (VSEPR) theory. (1)

The basis of the VSEPR theory is that electron pairs in the valence shell (i.e. outermost electron shell) of the central atom of a molecule repel one another and stay as far apart as possible, thus causing the molecule to assume a specific shape. (1)

The VSEPR theory states that the repulsion decreases in the following order:

lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion (1)

362

In a beryllium chloride molecule,

there are two bond pairs of electrons in the outermost shell of the central beryllium atom. (0.5)

The electron pairs must be at opposite ends of a straight line in order to be as far apart as possible. Hence the molecule has a linear shape. (0.5)

In a boron trifluoride molecule, there are three bond pairs of electrons in the outermost shell of the central boron atom. (0.5)

The shape that put the three electron pairs furthest apart is trigonal planar. Hence the molecule has a trigonal planar shape. (0.5)

In a methane molecule, there are four bond pairs of electrons in the outermost shell of the central carbon atom. (0.5)

The shape that puts the four electron pairs furthest apart is tetrahedral. Hence the molecule has a tetrahedral shape. (0.5)

(3 marks for organization and presentation)

115 Halogens exist as diatomic molecules. Averaged over time, the distribution of electrons throughout a halogen molecule is symmetrical. (0.5)

The electrons in a halogen molecule are in constant motion. At a particular instant, there may be more electrons at one end of the molecule than at the other; (0.5)

instantaneously the molecule has developed a dipole, i.e. an instantaneous dipole. (0.5)

The instantaneous dipole can affect the electron distributions in neighbouring molecules (0.5)

and induce similar dipoles in these neighbours. (0.5)

Attractions exist between the instantaneous dipole and the induced dipoles. (0.5)

A large molecule has many electrons. The electron distribution in the molecule can be distorted easily. (0.5)

This leads to an increased chance for the occurrence of instantaneous dipoles and hence stronger van der Waals’ forces between molecules. (0.5)

The number of electrons in a halogen molecule increases down the group. (0.5)

Hence the strength of van der Waals’ forces also increases down the group. (0.5)

The boiling point of an element depends on the strength of its intermolecular attractions. (0.5)

Hence the boiling points of halogens increase down the group. (0.5)

(3 marks for organization and presentation)

5. solution guide to supplementary exercises - 2c

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