6
258
8 linear
9 planar
10 tetrahedral
11 V-
13 lone pair-lone pair; lone pair-bond pair; bond pair-bond pair

15 T Electron diagram of a XeF
4
molecule:
F
F
Xe
F
F
(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
F
B
F
(Only electrons in the outermost shells are shown.)
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.
NCl
5
does not exist.
18 T Electron diagram of a carbon tetrachloride molecule:
Cl
Cl C Cl
Cl
(Only electrons in the outermost shells are shown.)
In a CCl
4
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 SF
4
molecule:
F
F
S
F
F
(Only electrons in the outermost shells are shown.)
The 5 pairs of electrons around the central S atom will adopt a trigonal bipyramidal arrangement.
259

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
F
S
120°
90°
F
I
F
F
F
120°
S
90°
F
II
Shape
I
II
Repulsions involving the lone pair
3 90° lone pair-bond pair repulsions (repulsions at angles greater than 90° can be ignored)
2 90° lond pair-bond pair repusions (repulsions at angles greater than 90° can be ignored)
∴ the SF
4
molecule has a seesaw shape.
Remark the repulsion is smaller for shape II, hence the electron pairs will adopt shape II
20 F Electron diagram of a SO
3
molecule:
O
O
S O
(Only electrons in the outermost shells are shown.)
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 PCl
4
+
ion:
+
Cl
Cl P
Cl
Cl
(Only electrons in the outermost shells are shown.)
In a PCl
4
+
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 PCl
4
+
ion has a tetrahedral shape.
+
Cl
Cl
P
Cl
Cl

22 F Electron diagram of a NO
3
–
ion:
–
O N O
O
(Only electrons in the outermost shells are shown.)
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°.
3
molecule has a trigonal planar shape.
NH
3
molecule has a trigonal pyramidal shape.
H
B
H H H
N
H
H
24 T Both molecules of BeF
2
and XeF
2
have a linear shape.
Molecule
BeF
2
Electron diagram
(Only electrons in the outermost shells are shown.)
F Be F
Shape
XeF
2
F
Xe
F
F — Be — F
F
Xe
F
25 T Bond angle: NH
4
+
> NH
3
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
NH
4
+
NH
3
Number of bond pairs
4
3
Number of lone pair
0
1
The electron pairs repel one another and stay as far apart as possible.
In the NH
4
+
ion, the furthest apart the four pairs of electrons can get is when they are arranged in a tetrahedral shape.
Thus the NH
4
+
ion has a tetrahedral shape. The H–N–H bond angles are 109.5°.
261

262
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the H–N–H bond angle in the NH
3
molecule is compressed to 107°.
26 F Buckminsterfullerene has a simple molecular structure. Van der Waals’ forces exist between the C
60 molecules.
27 C
28 A The sulphur atom has 6 electrons in its outermost shell and each oxygen atom has 6, i.e. a SO
2 molecule has a total of 18 electrons in the outermost shells.
Option
A
B
C
D
Electron diagram
(Only electrons in the outermost shells are shown.)
O S O
O
O
O S
S
S O
O
O
Remark
Correct
Incorrect as oxygen cannot form compounds with more than 8 electrons in the outermost shell of its atom.
Incorrect as there are 20 electrons in the outermost shells.
Incorrect 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 XeO
3 molecule has a total of 26 electrons in the outermost shells.
Option
A
Electron diagram
(Only electrons in the outermost shells are shown.)
O
O Xe O
Remark
Incorrect as there are 24 electrons in the outermost shells.
B O
O
Xe
O
Incorrect as there are 28 electrons in the outermost shells.
C
D
O
O Xe O
O
O Xe O
Incorrect as there are 32 electrons in the outermost shells.
Correct

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 H
3
O
+
ion carries one positive charge because it has lost 1 electron. That makes a total of 8 electrons in the outermost shells of the H
3
O
+
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 NH
2
–
ion carries one negative charge because it has gained 1 electron. That makes a total of 8 electrons in the outermost shell of the NH
2
–
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 SO
4
2–
ion carries two negative charges because it has gained 2 electrons. That makes a total of
32 electrons in the outermost shells of the SO
4
2–
ion.
Option
A
Electron diagram
(Only electrons in the outermost shells are shown.)
2–
O
O S
O
O
Remark
Incorrect as there are 28 electrons in the outermost shells.
B
C
D
2–
O
O S O
O
2–
O
O S O
O
O
O S
O
O
2–
Incorrect as there are 36 electrons in the outermost shells.
Correct
Incorrect as there are 40 electrons in the outermost shells.
33 A Electron diagram of a SiH
4
molecule:
H
H
Si H
(Only electrons in the outermost shells are shown.)
263

264
34 B Electron diagram of a NCl
3
molecule:
Cl N
Cl
Cl
(Only electrons in the outermost shells are shown.)
35 C Electron diagram of a SCl
2
molecule:
Cl S
Cl
(Only electrons in the outermost shells are shown.)
36 A Electron diagram of a PCl
4
+
ion:
+
Cl
Cl P
Cl
Cl
(Only electrons in the outermost shells are shown.)
37 A Electron diagram of a CH
3
+
ion:
+
H
H
C
H
(Only electrons in the outermost shells are shown.)
38 B
Option
A
B
C
Molecule
H
2
O
PH
3
PCl
5
Electron diagram
(Only electrons in the outermost shells are shown.)
H O
H
H P H
H
Cl
Cl Cl
P
Cl
Cl
D CH
2
Cl
2
H
H C Cl
Cl
Number of lone pair(s) of electrons in the outermost shell of the central atom
2
1
0
0

39 B
Molecule
XeF
2
XeF
4
Electron diagram
(Only electrons in the outermost shells are shown.)
F
Xe
F
F
F
Xe
F
F
Number of lone pairs of electrons in the outermost shell of the central atom
3
2
40 D Electron diagram of a BrF
5
molecule:
F
F
Br
F
F
F
(Only electrons in the outermost shells are shown.)
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.
BrF
5
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 SF
6
molecule has an octahedral shape. There are 6 electron pairs in the outermost shell of the central sulphur atom.
F
F F
S
F F
F
44 D Electron diagram of an OF
2
molecule:
F O
F
(Only electrons in the outermost shells are shown.)
265

266
An
2
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 OF
2
molecule is
V-shaped.
45 B Electron diagram of a BH
3
molecule:
H
H
B
H
(Only electrons in the outermost shells are shown.)
In a BH
3
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 BH
3
molecule has a trigonal planar shape.
46 D Electron diagram of a PF
3
molecule:
F P
F
F
(Only electrons in the outermost shells are shown.)
In a PF
3
molecule, there are one lone pair and three 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.
The shape of a molecule is determined only by the arrangement of atoms.
Thus the PF
3
molecule has a trigonal pyramidal shape.
47 B In an NH
3
molecule, there are one lone pair and three bond pairs of electrons in the outermost shell of the central nitrogen atom.
The shape that puts the four electron pairs the furthest apart is tetrahedral.
48 D Electron diagram of a XeF
4
molecule:
F
F
Xe
F
F
(Only electrons in the outermost shells are shown.)
In a XeF
4
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.

49 C Electron diagram of a ClF
3
molecule:
F Cl F
F
(Only electrons in the outermost shells are shown.)
In a ClF
3
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 CO
2
molecule:
O C O
(Only electrons in the outermost shells are shown.)
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 CO
2
molecule has a linear shape and the O–C–O bond angle is 180°.
52 B A H
2
O 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 H
2
O molecule is
V-shaped.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.
Thus in the H
2
O molecule, the two lone pairs will stay the furthest apart.
As a result, the H–O–H bond angle in the H
2
O molecule is compressed to 104.5°.
53 A Electron diagram of a C
2
H
4
molecule:
H
H
C C
H H
(Only electrons in the outermost shells are shown.)
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.
267

54 B
Option Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
A
B
H
2
S
BeCl
2
S H
H
Cl Be Cl
S
H
H
Cl Be Cl
C SCl
2
S Cl
Cl
S
Cl
Cl
D OF
2
O
F
F
O
F
F
55 A
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
SF
6
F
F
F
S
F
F
F
Three-dimensional structure
F
F
F
S
F
F
F
56 B
Ion
NH
4
+
Electron diagram
(Only electrons in the outermost shells are shown.)
+
H
H N H
H
Three-dimensional structure
H
H
N
H
H
+
268
57 B
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
XeF
4
F
F
Xe
F
F
F
F
Xe
F
F
Shape
V-shaped linear
V-shaped
V-shaped
Shape octahedral
Shape tetrahedral
Shape square planar

58 A Electron diagram of a CO
3
2–
ion:
2–
O
O
C
O
(Only electrons in the outermost shells are shown.)
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.
The shape that puts the electron pairs the furthest apart is trigonal planar.
Thus the CO
3
2–
ion has a trigonal planar shape.
59 C A H
2
O 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 H
2
O molecule is
V-shaped.
60 D
Option
A
Species
NO
3
–
Electron diagram
(Only electrons in the outermost shells are shown.)
–
O
O
N
O
Three-dimensional structure
O
O
N
O
–
Shape trigonal planar
B NCl
3
Cl N Cl
Cl trigonal pyramidal
C SF
4
F
F F
S
F
D SiF
4
F
F Si F
F
∴ the SiF
4
molecule has a tetrahedral shape.
Cl
N
Cl
Cl
F
F
F
S
F
F
F
Si
F
F seesaw tetrahedral
269

270
61 A
Option
A
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
F
BF
3
F
F
B
F
F
B
F
Shape trigonal planar
B NH
3
H N H
H
H
N
H
H trigonal pyramidal
C PH
3
H P H
H H
P
H
H trigonal pyramidal
D PCl
3
Cl P Cl
Cl
Cl
P
Cl
Cl trigonal pyramidal
∴ the BF
3
molecule has a shape different from the others.
62 C
Option
A
B
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
Cl
BCl
3
Cl
Cl
B
Cl
Cl
B
Cl
F
ClF
3
F Cl F
F
Cl F
F
Is the species planar?
yes yes
C NCl
3
Cl N Cl
Cl no
D XeF
4
F
F
Xe
F
F
Cl
N
Cl
Cl
F
F
Xe
F
F yes
∴ NCl
3
is NOT a planar species.

63 C Electron diagram of BCl
3
:
Cl B Cl
Cl
(Only electrons in the outermost shells are shown.)
In a BCl
3
molecule, there are three bond pairs of electrons in the outermost shell of the central boron atom.
The electron pairs repel one another and stay as far apart as possible.
The shape that puts the three electron pairs the furthest apart is trigonal planar.
Hence a BCl
3
molecule has a trigonal planar shape.
Electron diagram of a PH
3
molecule:
H P H
H
(Only electrons in the outermost shells are shown.)
In a PH
3
molecule, there are one lone pair and three 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.
The shape of a molecule is determined only by the arrangement of atoms.
Thus the PH
3
molecule has a trigonal pyramidal shape.
64 D (1) and (3) Electron diagram of a ClF
3
molecule:
F Cl F
F
(Only electrons in the outermost shells are shown.)
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.
Shape
I
II
III
F
Cl
90°
F
90°
Cl
90°
F
F F
I
F
II
Repulsion present
90° lone pair-lone pair repulsion
6 90° lone pair-bond pair repulsions
(repulsions at angles greater than 90° can be ignored)
4 90° lone pair-bond pair repulsions +
2 90° bond pair-bond pair repulsions
(repulsions at angles greater than 90° can be ignored)
∴ the ClF
3
molecule is T-shaped.
120°
F
90°
Cl
90°
F
F
III
Remark the molecule will not take up shape I bond pair-bond pair repulsion is less than lone pair-bond pair repulsion, hence shape III has the minimum repulsion
271

272
65 C
Option
A
Species
CS
2
Electron diagram
(Only electrons in the outermost shells are shown.)
S C S
H
2
S
S H
H
B CH
4
H
H C
H
H
XeF
4
F
F
Xe
F
F
C NH
4
+
H
H N
H
H
+
SiCl
4
Cl
Cl Si Cl
Cl
D NH
3
H N H
H
NO
3
–
O
O N O
–
∴ the NH
4
+
ion and SiCl
4
molecule have a similar shape.
Three-dimensional structure
S C S
H
S
H
H
H
C
H
H
F
F
Xe
F
F
H
H
N
H
H
+
Cl
Cl
Si
Cl
Cl
H
N
H
H
O
O
N
O
–

66 B
Option
A
Species
BH
3
Electron diagram
(Only electrons in the outermost shells are shown.)
H
H B H
PH
3
H P H
H
B
C
CO
2
HCN
CO
3
2–
H
3
O
+
O C O
H C N
O
O
C
O
2–
H O H
H
+
D NH
2
–
H N H
–
XeF
2
F
Xe
F
∴ the CO
2
and HCN molecules have a similar shape.
Three-dimensional structure
H
H
B
H
H
P
H
H
O C O
H C N
O
O
C
O
2–
+
H
O
H
H
–
Xe
F
N
H
H
F
273

274
67 A
Option
A
Molecule
PCl
3
Electron diagram
(Only electrons in the outermost shells are shown.)
Cl P Cl
Cl
NCl
3
Cl N Cl
Cl
B CF
4
F
F C F
F
SF
4
F
F F
S
F
C SO
2
O S O
CO
2
O C O
D PF
5
F
F F
P
F
F
IF
5
F
F
I
F
F
F
∴ the PCl
3
and NCl
3
molecules have a similar shape.
Three-dimensional structure
Cl
P
Cl
Cl
Cl
N
Cl
Cl
F
F
C
F
F
F
F
F
S
F
O
S
O
O C O
F
F
F
F
F
P
F
F
I
F
F
F

68 D
Option
A
Species
BF
3
Electron diagram
(Only electrons in the outermost shells are shown.)
F
F
B
F
NF
3
F N F
F
B BCl
3
Cl
Cl
B
Cl
PH
3
H P H
H
C BeCl
2
Cl Be Cl
SCl
2
Cl S
Cl
D CH
3
+
H
H C H
+
NO
3
–
O
O
N
O
–
∴ the CH
3
+
ion and NO
3
–
ion have an identical geometry.
Three-dimensional structure
F
F
B
F
F
N
F
F
Cl
Cl
B
Cl
H
P
H
H
Cl Be Cl
H
S
Cl
Cl
H
C
H
+
O
O
N
O
–
275

69 C ammonia molecule
H
+
ammonium ion
H
+
H
N
H
H H
N
H
H
∴ there is a change of shape from trigonal pyramidal to tetrahedral.
70 A PF
5
PF
3
Electron diagrams of PF
5
and PF
3
:
F
F F
P
F
F
F P F
F
(Only electrons in the outermost shells are shown.)
Arrangement of electron pairs in the outermost shell of the P atom in the molecules:
USJHPOBM
CJQZSBNJEBM
UFUSBIFESBM
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
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
HCN H C N
H
180°
C N
72 C Electron diagram of a CO
3
2–
ion:
2–
O
O
C
O
(Only electrons in the outermost shells are shown.)
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.
The electron pairs repel one another and stay as far apart as possible.
276
The shape that puts the electron pairs the furthest apart is trigonal planar.
Thus the CO
3
2–
ion has a trigonal planar shape.
∴ the bond angle in a CO
3
2–
ion is 120°.

73 B
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
H O
H
Three-dimensional structure
H
2
O O
H
H
104.5°
NH
3
CH
4
H N H
H
H
H C
H
H
H
N
H
H
107°
H
H
C
H
H
109.5°
CO
2
O C O
O
180°
C O
∴ the correct order of bond angles in the molecules H
2
O, NH
3
, CH
4
and CO
2
is CO
2
> CH
4
> NH
3
>
H
2
O.
74 D
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
BeCl
2
Cl Be Cl
Cl
180°
Be Cl
∴ the Cl–Be–Cl bond angle in a BeCl
2
molecule is 180°.
75 B In a regular octahedral molecule, MX
6
, the number of X–M–X bonds at 180° is 3.
X 180°
X
M
X 180°
X
X
180°
X
76 C H
C
H
α
C
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 H–C–H bond angle (i.e. α ) is 120°.
277

278
77 C
H
H
α
Cx
H
H
C
β
O H
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
Option Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
A
B
BF
3
CH
4
F
F
B
F
H
H C H
H
F
120°
F
B
120°
120°
F
H
H
C
H
H
109.5°
C NH
3
H N H
H
H
N
H
H less than 109.5°
D PCl
3
Cl P Cl
Cl
Cl
P
Cl
Cl less than 109.5°
∴ the Y–X–Y angle is the greatest in BF
3
.
79 C –
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 NO
3
–
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
NH
4
+
NH
3
NH
2
–
Number of bond pairs
4
3
2
Number of lone pair(s)
0
1
2
The electron pairs repel one another and stay as far apart as possible.
In the NH
4
+ ion, the furthest apart the four pairs of electrons can get is when they are arranged in a tetrahedral shape.
Thus the NH
4
+
ion has a tetrahedral shape. The H–N–H bond angles are 109.5°.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the H–N–H bond angle in the NH
3
molecule is compressed to 107°.
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.
Thus in the NH
2
–
ion, the two lone pairs will stay the furthest apart. The H–N–H bond angle in the
NH
2
–
ion is compressed to 104.5°.
∴ the order of the bond angles is NH
4
+
> NH
3
> NH
2
–
.
81 C Consider the following change:
H
2
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:
Species
H
3
O
+
H
2
O
Number of bond pairs
3
2
Number of lone pair(s)
1
2
In a H
3
O
+
ion, the four pairs of electrons in the outermost shell of the oxygen atom will adopt a tetrahedral arrangement.
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.
Thus in a H
2
O molecule, the two lone pairs will stay the furthest apart.
As a result, the H–O–H bond angle in a H
2
O molecule is compressed to less than that in a H
3
O
+
ion.
∴ when the H
3
O
+
ion is formed from water, the bond angle increases slightly.
279

280
82 A
Species
(1) H
3
O
+
(2) NH
4
+
Electron diagram
(Only electrons in the outermost shells are shown.)
+
H O
H
H
H
H N
H
H
+
(3) SF
6
F
F
F
S
F
F
F
∴ both the H
3
O
+
and NH
4
+
ions have 8 electrons in the outermost shell of the central atom.
83 B
Species
(1) CH
4
Electron diagram
(Only electrons in the outermost shells are shown.)
H
H C
H
H
(2) XeF
2
(3) NH
2
–
F
Xe
F
H N H
–
∴ both the CH
4
molecule and the NH
2
–
ion have 8 electrons in the outermost shell of the central atom. The electron pairs would adopt a tetrahedral arrangement.
84 D
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
(1) BeF
2
F Be F F Be F
(2) CO
2
O C O O C O
H C N (3) HCN H C N
∴ all the three species have a linear shape.

85 C
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
(1) BrF
3
F Br F
F
Three-dimensional structure
F
Br
F
F
Shape
T-shaped
(2) NH
3
H N H
H
H
N
H
H
(3) PF
3
F P F
F
F
P
F
F
∴ the NH
3
and PF
3
molecules have a trigonal pyramidal shape.
86 D
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
(1) Cl
2
O
Cl O
Cl
(2) NH
2
–
H N H
–
O
Cl
Cl
–
N
H
H trigonal pyramidal trigonal pyramidal
Shape
V-shaped
V-shaped
(3) SO
2 O
S
O
O
S
O
V-shaped
∴ all the three species are V-shaped.
281

282
87 B
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
(1) SiBr
4
(2) SF
4
(3) NH
4
+
Br
Br Si Br
Br
F
F F
S
F
H
H N
H
H
+
Br
Br
Si
Br
Br
F
H
F
F
S
F
H
N
H
H
+
∴ the SiBr
4
molecule and the NH
4
+
ion have a tetrahedral shape.
88 A
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
(1) boron trifluoride
F
F
B
F
F
F
B
F
Shape tetrahedral seesaw tetrahedral
Shape trigonal planar
(2) ethene
H
H
C C
H
H
H
H
C C
H
H planar
(3) nitrogen trichloride
Cl N Cl
Cl
∴ both boron trifluoride and ethene have a planar structure.
Cl
N
Cl
Cl trigonal pyramidal

89 C
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
(1) BF
3
F
F
B
F
F
F
B
F
NH
3
H N H
H H
N
H
H
(2) BeCl
2
Cl Be Cl Cl Be Cl
CO
2
(3) NH
4
+
CH
4
O C O
H
H N
H
H
+
H
H C
H
H
O C O
H
H
N
H
H
+
H
H
C
H
H
∴ (2) the BeCl
2
and CO
2
molecules have a similar shape.
NH
4
+
ion and the CH
4
molecule have a similar shape.
90 D
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
(1) CS
2
S C S S C S
(2) PF
5 F
F F
P
F
F
F
F
F
P
F
F
(3) XeF
4
F
F
Xe
F
F F
F
Xe
F
F
∴ all the three species have three atoms lying in a straight line.
Shape trigonal planar trigonal pyramidal linear linear tetrahedral tetrahedral
283

284
91 B Three dimensional structure of PCl
5
:
Cl
Cl
Cl
P
90°
120°
Cl
Cl
∴ the bond angles 90° and 120° exist in a PCl
5
molecule.
92 D
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
(1) boron trifluoride
F
F
B
F
F
F
B
F
(2) carbonate ion
O
O
C
O
2–
O
O
C
O
2–
Shape trigonal planar trigonal planar
(3) nitrate ion
O
O
N
O
–
O
O
N
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 BF
3
molecule:
F
F
B
F
(Only electrons in the outermost shells are shown.)
There are less than 8 electrons in the outermost shell of the boron atom.
Thus the BF
3
molecule does not conform to the octet rule.

Electron diagram of a NO
2
molecule:
O N O
(Only electrons in the outermost shells are shown.)
NO
2
molecule has an unpaired electron.
Thus the NO
2
molecule does not conform to the octet rule.
96 C An NH
3
molecule has one lone pair and three bond pairs of electrons in the outermost shell of the nitrogen 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 NH
3
molecule has a trigonal pyramidal shape.
97 B Electron diagram of a methanal molecule:
H
H
C O
(Only electrons in the outermost shells are shown.)
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.
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 methanal molecule has a trigonal planar shape.
So, the O–C–H bond angle is about 120°.
98 D Electron diagram of a CH
3
+
ion:
+
H
H
C
H
(Only electrons in the outermost shells are shown.)
There are 3 bond pairs of electrons in the outermost shell of the carbon atom.
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 CH
3
+
ion has a trigonal planar shape.
285

286
99 C
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
BCl
3
Cl
Cl
B
Cl
Cl
Cl
B
Cl
Shape trigonal planar
NCl
3
Cl N Cl
Cl
Cl
N
Cl
Cl
∴ BCl
3
has a planar structure but NCl
3
does not.
100 B
Ion
CO
3
2–
Electron diagram
(Only electrons in the outermost shells are shown.)
2–
O
O
C
O
Three-dimensional structure
O
O
C
O
2–
NO
3
–
O
O
N
O
–
O
O
N
O
–
∴ both carbonate ion and nitrate ion have a trigonal planar shape.
101 C
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
PCl
5
IF
5
Cl
Cl Cl
P
Cl
Cl
F
F
F
I
F
F
Three-dimensional structure
Cl
Cl
P
Cl
Cl
F
F
I
F
Cl
F
F
∴ molecules of PCl
5
and IF
5
have different shapes.
trigonal pyramidal
Shape trigonal planar trigonal planar
Shape trigonal bipyramidal square pyramidal

102 B Bond angle: NH
3
> H
2
O
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
NH
3
H
2
O
Number of bond pairs
3
2
Number of lone pair(s)
1
2
In the NH
3
molecule, the four pairs of electrons in the outermost shell of the nitrogen atom will adopt a tetrahedral arrangement.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the H–N–H bond angle in the NH
3
molecule is compressed to 107°.
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.
Thus in the H
2
O molecule, the two lone pairs will stay the furthest apart.
As a result, the H–O–H bond angle in the H
2
O molecule is compressed to 104.5°.
103 D
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
CF
4
SF
4
F
F C F
F
F
F F
S
F
F
F
F
S
F
F
C
F
F
F
∴ the bond angles in CF
4
are 109.5° but those in SF
4
are not.
104 C Graphite has a giant covalent structure. It is NOT soluble in carbon disulphide.
Shape tetrahedral seesaw
Bond angles
109.5°
180°, 120°,
90°
105 A
287

2 electronegativity
3 polar
4 fluorine
288
8 permanent dipole-permanent dipole; permanent dipole-induced dipole; instantaneous dipole-induced dipole
9 hydrogen; oxygen; nitrogen; fluorine
10 viscosity
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.
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.

16 T The electronegativity value of nitrogen is greater than that of phosphorus.
Hence the bond pairs of electrons in NH
3
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 NH
3
is greater than the H–P–H bond angle in PH
3
.
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
Cl
P Cl
Cl
Cl
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.
The molecule has no net dipole moment and it is non-polar.
∴ 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
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure Shape
SO
2
O
S
O
S
V-shaped
O O
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.
289

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 H
2
S is higher than that of SiH
4
.
The boiling point of a compound depends on the strength of its intermolecular attractions.
H
2
S is a polar substance. There are permanent dipole-permanent dipole attractions and instantaneous dipole-induced dipole attractions between H
2
S molecules.
SiH
4
is a non-polar substance. There are only instantaneous dipole-induced dipole attractions between
SiH
4
molecules.
More heat is needed to separate the H
2
S 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 (PH
3
) 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.
CH
3
CH
2
δ +
H
δ –
O lone pair hydrogen bond
δ +
H
δ –
O
δ +
H

3
C
O
C CH
3
) can form hydrogen bonds with water molecules.
lone pair
H
3
C
δ +
C
H
3
C
δ –
O
δ +
H
δ –
O
δ +
H hydrogen bond
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.
δ +
H
δ –
O
δ +
H lone pair hydrogen bond
δ –
O
H
3
C CH
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.
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 least polar.
291

292
33 A
Element
Atomic number
Name of element
W
8 oxygen
X
12 magnesium
W (oxygen) is the most
electronegative.
34 B
Option Molecule
Y
14 silicon
Three-dimensional structure
A CO
2
O C O
Z
16 sulphur
B
C
NH
3
CCl
4
H
N
H
H
Cl
Cl
C
Cl
Cl
H
D CH
4
H
C
H
H
B electronegativity value of nitrogen is greater than that of hydrogen. So each N–H bond is polar.
The
3
molecule has a trigonal pyramidal shape. The individual N–H bond dipole moments reinforce each other.
The molecule has a net dipole moment and it is polar.
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 CCl
4
molecule, the individual bond dipole moments cancel one another out exactly.
The molecule has no net dipole moment and it is non-polar.
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.

36 C Options A and C — The three-dimensional structures of CHCl
3
and CHF
3
are shown below:
H H
Cl
C
Cl
Cl net dipole moment
F
C
F
F net dipole moment
The electronegativity difference between C and F is greater than that between C and Cl.
∴ CHF
3
has a greater dipole moment than CHCl
3
.
Option D — 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 CCl
4
molecule, the individual bond dipole moments cancel one another out exactly.
The molecule has no net dipole moment and it is non-polar.
37 C
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure Shape
F
SiF
4
F
F Si F
F
F
Si
F
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 SiF
4
molecule, the individual bond dipole moments cancel one another out exactly.
Thus the SiF
4
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.
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
O
SO
3
O
O
S O
O
S
O
Shape trigonal planar
293

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 SO
3
molecule, the individual bond dipole moments cancel one another out exactly.
Thus SO
3
molecule has no net dipole moment.
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure Shape
40 D
Option
OF
2
O F
F
δ +
O
δ –
F
δ –
F
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.
Molecule
A
B
CHCl
3
HCl
Three-dimensional structure
δ –
Cl
H
δ +
C
δ –
Cl
δ –
Cl net dipole moment
δ +
H net dipole moment
δ –
Cl
C H
2
O
δ –
O
δ +
H
δ +
H net dipole moment
S C S D CS
2
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.

41 A
Option Molecule Three-dimensional structure
A H
2
S
S
H
H net dipole moment
B
C
D
CO
2
CCl
4
BF
3
O C O no net dipole moment
Cl
Cl
C
Cl
Cl no net dipole moment
F
F
B
F no net dipole moment
∴ the H
2
S 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 PCl
5
molecule, the individual bond dipole moments cancel one another out exactly.
The
5
molecule has no net dipole moment and it is non-polar.
43 B
Option Molecule
A
B
SiF
4
SF
4
F
F
F
Three-dimensional structure
F
Si
F
F no net dipole moment
F
S
F net dipole moment
C BeF
2 F Be F no net dipole moment
D BF
3
F
F
B
F no net dipole moment
∴ only the SF
4
molecule has a net dipole moment.
295

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 GeCl
4
molecule, the individual bond dipole moments cancel one another out exactly.
The
4
molecule has NO net dipole moment.
bond is polar.
As the polar Be–F bonds of the BeF
2
molecule are arranged in a straight line, the two identical bond dipole moments cancel each other out exactly.
As a result, the BeF
2
molecule has no net dipole moment.
So BeF
2
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.
Due to the symmetry of the tetrahedral shape of the CCl
4
molecule, the individual bond dipole moments cancel one another out exactly.
The
4
molecule has no net dipole moment and it is non-polar.
47 C
Option Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
A BeF
2
F Be F
F Be F no net dipole moment
NO
2
O N O
B HCl
H Cl
O
N net dipole moment
O net dipole moment
H Cl
CO
2
O C O
O C O no net dipole moment
C NO
2
O N O
HCl
H Cl
O
N net dipole moment
O net dipole moment
H Cl

D BeF
2
F Be F
F Be F no net dipole moment
CO
2
O C O
O C O no net dipole moment
∴ both NO
2
and HCl molecules have a net dipole moment.
48 A
Option Molecule I Molecule II
A
Cl
H
3
C
C C
Cl
CH
3 net dipole moment
Cl CH
3
C C
H
3
C Cl no net dipole moment
Remark 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.
Hence the bond pairs of electrons in NH
3
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 NH
3
is greater than the H–P–H bond angle in PH
3
.
Molecule
PH
3
H
2
O
NH
3
Bond angle
93°
104.5°
107°
∴ the order of increasing bond angle is PH
3
< H
2
O < NH
3
.
50 C Option C — Tetrachloromethane is a non-polar substance. There are only instantaneous dipole-induced dipole attractions between CCl
4
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.
297

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.
CO
2
, H
2
, N
2
and O
2
, a H
2
molecule contains the smallest number of electrons.
Hence the van der Waals’ forces in liquid H
2
is the weakest .
The amount of heat is needed to separate the H
2
molecules during boiling.
Hence liquid H
2
is the most volatile.
B — CH
4
, SiH
4
and SnH
4
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 CH
4
< SiH
4
< SnH
4
.
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 CH
4
< SiH
4
< SnH
4
.
60 C Option C — The boiling point of H
2
S is higher than that of SiH
4
.
The boiling point of a compound depends on the strength of its intermolecular attractions.
H
2
S is a polar substance. There are permanent dipole-permanent dipole attractions and instantaneous dipole-induced dipole attractions between H
2
S molecules.
SiH
4
is a non-polar substance. There are only instantaneous dipole-induced dipole attractions between SiH
4
molecules.
More heat is needed to separate the H
2
S molecules during boiling.

61 B The zig-zag polymeric structure in solid HF is due to hydrogen bonding: hydrogen bond
F F F
H H H
H H H
F F F
62 D In water, each molecule can be hydrogen-bonded to four other molecules.
H H
H
O O
H H
H
O
H H H H
O O
H
D C
O
C H molecule, NO hydorgen atom is attached directly to a highly
H 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.
65 A The boiling point of a compound depends on the strength of its intermolecular attractions.
Hydrogen bonds exist in H
2
O while only van der Waals’ forces exist in H
2
S.
Hydrogen bonds are stronger than van der Waals’ forces.
More heat is needed to separate the H
2
O molecules during boiling.
Thus the boiling point of H
2
O is higher than that of H
2
S.
299

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.
— C
2
H
5
OH is miscible with water (i.e. soluble in water in all proportions) because it can form hydrogen bonds with water.
CH
3
CH
2
δ +
H
δ –
O lone pair hydrogen bond
δ +
H
δ –
O
δ +
H
68 B In ice, hydrogen bond is the strongest intermolecular attraction.
Each water molecule forms four hydrogen bonds tetrahedrally.
The highly ordered structure leads to a very ‘open’ structure with large spaces in ice.
The ‘open’ structure collapses when ice melts to form liquid water.
∴ the melting of ice involves the cleavage of hydrogen bonds.
69 A The boiling point of a compound depends on the strength of its intermolecular attractions.
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.

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.
71 A The boiling point of a compound depends on the strength of its intermolecular attractions.
Relatively weak van der Waals’ forces exist in H
2
and N
2
.
The strength of the forces increases with the number of electrons in one molecule.
Hence the boiling point of N
2
is higher than that of H
2
.
Hydrogen bonds exist in NH
3
.
NH
3
has the highest boiling point.
∴ the order of increasing boiling point is H
2
< N
2
< NH
3
.
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 molecular has a trigonal planar shape.
The three identical B–F bond dipole moments in a BF
3
molecule cancel one another out exactly.
As a result, the BF
3
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 SO
2
molecule has a net dipole moment and is polar.
301

302
75 B (1) The electronegativity values of C and S are the same. Thus C=S bond is non-polar.
∴ the CS
2
molecule is non-polar.
(2) The electronegativity value of S is greater than that of H. So each S–H bond is polar.
Hence the molecule has a net dipole moment.
∴ the H
2
S 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 PCl
5
molecule, the individual bond dipole moments cancel one another out exactly.
The molecule has no net dipole moment.
∴ the PCl
5
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
3
molecule has a trigonal pyramidal shape. The individual N–H bond dipole moments reinforce each other.
Hence the molecule has a net dipole moment.
∴ the NH
3
molecule is polar.
(2) The electronegativity value of Cl is greater than that of S. So each S–Cl bond is polar.
SCl
2
molecule is V-shaped. The individual S–Cl bond dipole moments reinforce each other.
Hence the molecule has a net dipole moment.
∴ the SCl
2
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 XeF
4
molecule, the individual bond dipole moments cancel one another out exactly.
The molecule has no net dipole moment.
∴ the XeF
4
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 SO
3
molecule has a total of 24 electrons in the outermost shells.

79 B (1) Three-dimensional structure of a SO
3
molecule:
O
S
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.
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.
∴ a SO
3
molecule has a trigonal planar shape.
(2) The electronegativity value of O is greater than that of S. So a SO
3
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 SO
3
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 ClF
3
molecule has a total of 28 electrons in the outermost shells.
81 A (1) and (3) In a ClF
3
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
F
Cl
90°
I
F
F
F
90°
Cl
90°
F
II
120°
90°
F
Cl
90°
F
F
III
Shape
I
II
III
Repulsion present
90° lone pair-lone pair repulsion
6 90° lone pair-bond pair repulsions
(repulsions at angles greater than 90° can be ignored)
4 90° lone pair-bond pair repulsions +
2 90° bond pair-bond pair repulsions
(repulsions at angles greater than 90° can be ignored)
Remark the molecule will not take up shape I bond pair-bond pair repulsion is less than lone pair-bond pair repulsion, hence shape III has the minimum repulsion
∴ the ClF
3
molecule is T-shaped.
(2) The electronegativity value of F is greater than that of Cl. So each Cl–F bond is polar.
303

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 has no net dipole moment.
∴ 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 has no net dipole moment.
∴ the molecule is non-polar.
83 A (1) Three-dimensional structure of trichloromethane:
H
Cl
C
Cl
Cl 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 SiCl
4
and GeCl
4
molecules, the individual bond dipole moments cancel one another out exactly.
The molecules has no net dipole moment and they are non-polar.
The boiling point of a compound depends on the strength of its intermolecular attractions.
SiCl
4
and GeCl
4
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 GeCl
4
is higher than that of SiCl
4
.
85 C (1) The electronegativity value of nitrogen is greater than that of hydrogen. So each N–H bond is polar.
The
3
molecule has a trigonal pyramidal shape. The individual N–H bond dipole moments reinforce each other.
The molecule has a net dipole moment and it is polar.
(2) The electronegativity value of N is greater than that of P.
Hence the bond pairs of electrons in NH
3
will be attracted towards the nitrogen atom to a greater extent.
304
These bond pairs will repel each other to a greater extent. Thus the H–N–H bond angle in NH
3 is greater than the H–P–H bond angle in PH
3
.

NH
3
can form hydrogen bonds but PH
3
cannot.
More heat is needed to separate the NH
3
molecules during boiling.
Thus the boiling point of NH
3
is higher than that of PH
3
.
86 A (1) Both X and Y have the same molecular formula, C
5
H
12
.
(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.
H H
87 C (2) and (3) H C C O H and H
H
C
H H with water molecules.
H
O
C O H molecules can form hydrogen bonds
CH
3
CH
2
δ –
O
δ +
H lone pair
–
δ O
+
δ H
+
δ H hydrogen bond hydrogen bond
δ +
H
δ –
O
δ +
H lone pair
H
O
δ –
H C C
H
O
δ –
H hydrogen bond
δ +
δ
–
O
δ
+
H
δ
+
H
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.
The molecule has no net dipole moment and it is non-polar.
∴ instantaneous dipole-induced dipole attractions exist between sulphur hexafluoride molecules.
(3) NO permanent dipole-permanent dipole attractions exist between sulphur hexafluroide molecules.
305

306
90 D (2) and (3) In a methoxymethane molecule, the C–O bonds are polar.
H H
O
H C C H net dipole moment
H H
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.
H C
O
δ – lone pair hydrogen bond
δ +
H O
δ –
C H
O
δ –
H
δ + δ –
O hydrogen bond
92 C (3) Electron diagram of a HNO
3
molecule:
H O N
O
(Only electrons in the outermost shells are shown.)
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 HNO
3
molecule.
93 B In ice, hydrogen bond is the strongest intermolecular attraction.
Each water molecule forms four hydrogen bonds tetrahedrally.
The highly ordered structure leads to a very ‘open’ structure with large spaces in ice.
94 B (2) There are NO hydrogen bonds in X and Y.
(3) 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.
The strength of the forces increases with the number of electrons in one molecule.
Hence the boiling point of X is lower than that of Y.

95 A (1) Methanoic acid molecules can form hydrogen bonds with water molecules.
–
δ O
+
δ H
+
δ H hydrogen bond lone pair
O
δ –
H C
δ –
O H
δ + hydrogen bond
δ
–
O
δ
+
H
δ
+
H
(2) Methoxymethane molecules can form hydrogen bonds with water molecules.
H
3
C lone pair
δ –
O
CH
3
δ +
H
δ –
O
δ +
H hydrogen bond
96 B (2) Intramolecular hydrogen bonds occur in
O
N
O
H
O key: hydrogen bond
OH
NO
2
.
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.
Hydrogen bonds are stronger than van der Waals’ forces.
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.
307

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.
(2) The boiling point of a compound depends on the strength of its intermolecular attractions.
Hydrogen bonds exist in Y while only van der Waals’ forces exist in X.
Hydrogen bonds are stronger than van der Waals’ forces.
More heat is needed to separate the molecules of Y during boiling.
Thus the boiling point of Y is higher than that of X.
(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.
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.
(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.
308
Van der Waals’ forces in Y are stronger than those in X.
∴ Y is more viscous than X.

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 NH
3
, 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 , 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.
105 C The electronegativity value of fluorine is greater than that of boron. So each B–F bond is polar.
A molecular has a trigonal planar shape. The three identical B–F bond dipole moments in a BF
3 molecule cancel one another out exactly.
As a result, the BF
3
molecule has no net dipole moment and is non-polar.
106 D Three dimensional shape of a SF
4
molecule:
F
F
S
F
F
The electronegativity value of fluorine is greater than that of sulphur. So each S–F bond is polar.
The molecule has a seesaw shape. The individual S–F bond dipole moments reinforce each other.
The 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 SiCl
4
molecule, the individual bond dipole moments cancel one another out exactly.
The molecule has no net dipole moment and it is non-polar.
∴ NO permanent dipole-permanent dipole attractions exist between silicon tetrachloride molecules.
309

310
108 A Propanone molecule is polar.
O
δ –
H
C
C
δ +
C
H net dipole moment
H H H H
109 B The boiling point of a compound depends on the strength of its intermolecular attractions.
Hydrogen sulphide is a gas at room temperature and pressure because the attractions between its molecules are weak.
110 D 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 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 higher than that of HI.
111 A NH
3
molecules can form hydrogen bonds with water molecules.
δ +
H
δ +
H
δ +
H
δ –
N lone pair
δ +
H
δ –
O hydrogen bond
H
δ +
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.

1 B Option B — The xenon atom has 8 electrons in its outermost shell and each fluorine atom has 7, i.e. a XeF
4
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 SF
4
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 NO
3
–
ion carries one negative charge because it has gained 1 electron.
Thus a NO
3
–
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.
Option —
5
does NOT exist because nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its atom.
Option Consider a molecule with one lone pair and three bond pairs of electrons in the outermost shell of the central atom.
The electron pairs repel one another and stay as far apart as possible.
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 XeO
3
molecule:
O
O Xe O
(Only electrons in the outermost shells are shown.)
A
3
molecule has one lone pair and three bond pairs of electrons in the outermost shell of the Xe atom.
The electron pairs repel one another and stay as far apart as possible.
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 XeO
3 molecule has a trigonal pyramidal shape.
311

8 C Option C — Electron diagram of a PCl
4
+
ion:
+
Cl
Cl P Cl
Cl
(Only electrons in the outermost shells are shown.)
The four pairs of electrons in the outermost shell of the phosphorus atom will adopt a tetrahedral arrangement.
∴ a PCl
4
+
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 electron pairs repel one another and stay as far apart as possible.
The six pairs of electrons in the molecule will adopt an octahedral arrangement.
10 D
Option
PDUBIFESBM
The shape of a molecule is determined only by the arrangement of atoms. Thus, the molecule has a square pyramidal shape.
Species
A
B
BF
3
CO
3
2–
Electron diagram
(Only electrons in the outermost shells are shown.)
F
F
B
F
2–
O
O
C
O
C NO
3
–
O
O
N
O
–
312
D PCl
3
Cl P
Cl
Cl
∴ a PCl
3
molecule has one lone pair of electrons in the outermosst shell of the central atom.

11 C
Option
A
B
Species
C
H
2
3
H
O
4
+
Electron diagram
(Only electrons in the outermost shells are shown.)
H
C C
H
H
H
Three-dimensional structure
H
H
C C
H
H
+
+
H O H
H H
O
H
H
Shape planar trigonal pyramidal
C
D
SF
2
XeF
2
S F
F
F
Xe
F
Xe
F
F
S
F
F
∴ a SF
2
molecule has a V-shape.
12 C
Option
A
B
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
BH
3
H
H
B
H
H
H
B
H
CO
3
2–
O
O
C
O
2–
O
O
C
O
2–
V-shaped linear
Shape trigonal planar trigonal planar
C PF
3
F P F
F trigonal pyramidal
D SO
3
O
O S O
∴ a PF
3
molecule has a trigonal pyramidal shape.
F
P
F
F
O
O
S
O trigonal planar
313

13 A
Option
A
Species
I
3
–
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
–
I I I
–
I
I
I
Shape linear
B O
3
O O O
O
O
O
V-shaped
C OF
2
O F
F
O
F
F
V-shaped
D SO
2
O
S
O S
V-shaped
O O
∴ an I
3
–
ion has a linear shape.
314
14 C Option C — Electron diagram of a NCl
3
molecule:
Cl N Cl
Cl
(Only electrons in the outermost shells are shown.)
A nitrogen trichloride molecule has one lone pair and three bond pairs of electrons in the outermost shell of the nitrogen atom.
The electron pairs repel one another and stay as far apart as possible.
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 nitrogen trichloride molecule has a trigonal pyramidal shape.
D When using the VSEPR theory, double bonds can be counted as single bonds. Therefore we
O can view the carbon atom of a H C its outermost shell.
H molecule as having 3 pairs of electrons in
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.
O
∴ the H C H molecule is planar.

15 A
Option
A
B
C
D
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
BeCl
2
Cl Be Cl
Three-dimensional structure
Cl Be Cl
XeF
F
Xe
F
F
Xe
F
BF
F
F
B
F
F
F
B
F
PCl
Cl P Cl
Cl
PF
IF
F
F
F
F F
P
I
F
F
F
F
F
Cl
F
F
F
P
I
F
F
F
P
Cl
Cl
F
F
F
F
SiF
4
SF
4
F
F
F Si F
F
F F
S
F
F
F
F
S
F
F
Si
F
F
F
∴ molecules of BeCl and XeF have an identical shape.
6 D Option D —
Molecule
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
XeF
4
F
F
Xe
F
F
F
F
Xe
F
F
Shape linear linear trigonal planar trigonal pyramidal trigonal bipyramidal square pyramidal tetrahedral seesaw
Shape square planar
315

316
17 C
Option
A
B
C
D
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
PBr
3
Br P Br
Br
Br
P
Br
Br
PCl
3
Cl P Cl
Cl
Cl
P
Cl
Cl
NH
3
H N H
H
H
N
H
H
BF
3
F
F
B
F
F
F
B
F
BF
3
F
F
B
F
F
F
B
F
CO
3
2–
2–
O
O
C
O
O
O
C
O
2–
ClF
3
F Cl
F
F
F
Cl
F
F
PF
3
F P F
F
F
P
F
F
∴ the BF
3
molecule and CO
3
2–
ion are trigonal planar.
Shape trigonal pyramidal trigonal pyramidal trigonal pyramidal trigonal planar trigonal planar trigonal planar
T-shaped trigonal pyramidal

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.
USJHPOBMCJQZSBNJEBM
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
X
M
90°
X
I
X
X
90°
M
90°
X
II
120°
90°
X
M
90°
X
X
III
Shape
I
II
III
Repulsion present
90° lone pair-lone pair repulsion
6 90° lone pair-bond pair repulsions (repulsions at angles greater than 90° can be ignored)
4 90° lone pair-bond pair repulsions + 2 90° bond pair-bond pair repulsions (repulsions at angles greater than 90° can be ignored)
Remark the molecule will not take up shape I bond pair-bond pair repulsion is less than lone pair-bond pair repulsion, hence shape III has the minimum repulsion
∴ 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.
PDUBIFESBM
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
M
X
X
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 XeOF
4
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.
317

21 C When using the VSEPR theory, double bonds can be treated like single bonds. Thus we can regard a
XeOF
4
molecule as having one lone pair and five bond pairs of electrons in the outermost shell of the xenon atom.
The electron pairs repel one another and stay as far apart as possible.
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 XeOF
4
molecule has a square pyramidal shape.
O
F F
Xe
F F
22 C
23 D Three-dimensional structure of a XeF
4
molecule:
F
F
Xe
90° F
F
∴ the molecule contains 90° lone pair-bond pair repulsions.
24 C
Option Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure Bond angle
A H
2
O H O H 104.5°
B CH
3
+
H
H C H
+
H
H
C
O
H
H
H
+
120°
C CO
2
O C O O C O 180°
318
D SO
2 O
S
O
O
∴ a CO
2
molecule has the greatest angle between two covalent bonds.
S
O
~120°

25 C
H N
H
O
α
C N H
β
H
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 furthest apart the three pairs can get is at an angle of 120°.
∴ 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°.
AlCl
3
molecule, there are three bond pairs of electrons in the outermost shell of the central aluminium atom.
The electron pairs repel one another and stay as far apart as possible.
The shape that puts the three electron pairs the furthest apart is trigonal planar.
Hence AlCl
3
molecule has a trigonal planar shape.
In the adduct, the nitrogen atom in the NH
3
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.
H Cl H Cl
H N + Al Cl H N Al Cl
H Cl H Cl
319

320
27 B
Option
A
Species
SiH
4
Three-dimensional structure
H
H
Si
H
H
Bond angle(s)
109.5°
NH
3
H
N
H
H
107°
B SO
2
CS
2
S
O
S
O
C S
~120°
180°
C
CS
2
SF
4
S
F
F
C
F
S
F
S 180°
90°, 120°, 180°
D H
2
O
O
H
H
104.5°
H
2
S
S
H
H
92.5°
Option D — The electronegativity value of oxygen is greater than that of sulphur.
Hence the bond pairs of electrons in H
2
O 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 H
2
O is greater than the bond angle in H
2
S.
∴ in Option B, the bond angle around the central atom in the second species (CS
2
) is larger than that in the first species (SO
2
).

28 C
Molecule
BF
3
Three-dimensional structure
F
F
B
F
Bond angle
120°
PF
3
F
P
F
F
F
Cl F ClF
3
F
∴ the order of increasing bond angle is ClF
3
< PF
3
< BF
3
.
~109°
~90°
29 B
30 B
Element
Atomic number
Name of element
W
7 nitrogen
X (sodium) is the least electronegative.
X
11 sodium
Y
13 aluminium
Z
17 chlorine
31 D Option D — Three-dimensional structure of a SiF
4
molecule:
F
Si
F F
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 SiF
4
molecule, the individual bond dipole moments cancel one another out exactly.
The molecule has no net dipole moment and it is non-polar.
32 D The boiling point of a compound depends on the strength of its intermolecular attractions.
CH
4
and CCl
4
are non-polar. Van der Waals’ forces exist in them.
The strength of the forces increases with the number of electrons in one molecule.
Hence the boiling point of CCl
4
is higher than that of CH
4
.
321

322
33 B The boiling point of a compound depends on the strength of its intermolecular attractions.
All the compounds are non-polar. Van der Waals’ exist in them.
The strength of the forces increases with the number of electrons in one molecule.
Hence the boiling point of heptane (C
7
H
16
) is the highest .
attractions.
Hydrogen bonds exist in H
2
O while only van der Waals’ forces exist in other Group VI hydrides.
Hydrogen bonds are stronger than van der Waals’ forces.
More heat is needed to separate the H
2
O molecules during boiling.
Thus the boiling point of H
2
O is the highest .
The number of electrons in one hydride molecule increases from H
2
S to H
2
Te.
Hence the strength of van der Waals’ forces between the molecules also increases from
H
2
S to H
2
Te.
More heat, in their increasing order, is needed to separate the molecules during boiling, and thus the boiling points of the hydrides increase from H
2
S to H
2
Te.
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.
More heat, in their increasing order, is needed to separate the molecules during boiling, and thus the boiling points of the hydrides increase from HCl to HI.
36 A Option A — The boiling point of C
2
H
5
OH is higher than that of C
2
H
5
OC
2
H
5
. This is due to the presence of hydrogen bonds between molecules of C
2
H
5
OH.
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 furthest apart the three pairs can get is at an angle of 120°.
∴ the O–C–O bond angle is about 120°.
38 C Option C — The boiling point of X is higher than that of Y.
The boiling point of a compound depends on the strength of its intermolecular attractions.
Hydrogen bonds exist in X while only van der Waals’ forces exist in Y.

Hydrogen bonds are stronger than van der Waals’ forces.
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 BH
3
molecule:
H B H
H
(Only electrons in the outermost shells are shown.)
40 B (2) BF
3
molecule does NOT contain any unpaired electron.
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 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
Species
(3) NO
3
–
Electron diagram
(Only electrons in the outermost shells are shown.)
–
O
O
N
O
Three-dimensional structure
O
O
N
O
–
Shape trigonal planar
∴ species X could NOT be a NO
3
–
ion.
42 C
Species
(1) CS
2
Electron diagram
(Only electrons in the outermost shells are shown.)
S C S
Three-dimensional structure
S C S
Shape linear
(2) Cl
2
O
O Cl
Cl
V-shaped
(3) NH
2
–
H N H
–
O
Cl
Cl
–
N
H
H
V-shaped
∴ species X could be Cl
2
O / NH
2
–
.
323

324
43 D
Species
(1) PCl
5
(2) SF
4
Electron diagram
(Only electrons in the outermost shells are shown.)
Cl
Cl Cl
P
Cl
Cl
F
F F
S
F
(3) XeF
2 F
Xe
F
∴ 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
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure Shape
(1) SF
4
(2) SiF
4
F
F F
S
F
F
F Si F
F
F
F
F
F
S
F
F
Si
F
F seesaw tetrahedral
(3) XeF
4
F
F
Xe
F
F
F
F
Xe
F
F square planar
∴ the SiF
4
and CH
4
molecules have a tetrahedral shape.

45 B
Option
(1)
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Three-dimensional structure
CO
2
O C O O C O
C H
H
H
C C
H
H
H
H
C C
H
H
H O
H O
H
O
H
H
CS
S C S S C S
HCN
H C N H C N
XeF
XeF
F
F
Xe
Xe
F
F
F
Xe
F
F
Xe
F
OF
F O
F
O
F
F
SCl
Cl S
Cl
S
Cl
Cl
∴ the set CS, HCN and XeF contains only linear species.
Shape linear planar
V-shaped linear linear linear linear
V-shaped
V-shaped
325

46 A
Species Three-dimensional structure
(1) CH
3
NH
2
CH
3
N
H
H
F
(2) ClF
3
Cl F
F
F
F
(3) PF
5
P F
F
F
∴ the shapes of the CH
3
NH
2
and ClF
3
molecules are influenced by the presence of lone pairs of electrons.
47 A
Species
(1) XeF
4
(2) PF
5
Three-dimensional structure
F
F
Xe 90°
F
F
F
F
P
90°
F
F
F
F
(3) SiF
4
F
Si
F
F
109.5°
∴ the XeF
4
and PF
5
molecules contain 90° bond angles.
48 B
Species
(1) C
2
H
4
Three-dimensional structure
H
H
C
120°
C
120°
H
120°
H
(2) NF
3
F
N
F
F
102°
Cl
Cl
(3) PCl
5
Cl
P
120°
Cl
Cl
326
∴ the C
2
H
4
and PCl
5
molecules contain 120° bond angles.

49 A
Molecule
(1) BrF
5
Electron diagram
(Only electrons in the outermost shells are shown.)
F
F
Br
F
F
F
Three-dimensional structure
F
F F
Br
F F
(2) ClF
5
F
F
Cl
F
F
F
F
F
F
Cl
F
F
(3) PCl
5 Cl
Cl Cl
P
Cl
Cl
Cl
Cl
P
Cl
Cl
Cl
∴ the BrF
5
and ClF
5
molecules have the geometric arrangement shown.
50 B (1)
H
2
H
+
3
O
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:
Species
H
3
O
+
H
2
O
Number of bond pairs
3
2
Number of lone pair(s)
1
2
The electron pairs repel one another and stay as far apart as possible.
The furthest apart the four pairs of electrons can get is when they are arranged in a tetrahedral shape.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the bond angle in the H
3
O
+
ion is compressed to slightly less than 109.5°.
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.
In the water molecule, the two lone pairs will stay the furthest apart.
As a result, the bond angle in the H
2
O molecule is further compressed.
∴ bond angle: H
3
O
+
> H
2
O
(2)
C
2
H
4
H
2
C
2
H
6
327

328
Three-dimensional structures of the C
2
H
4
and C
2
H
6
molecules:
H H
H
H
C
H
C
120°
C
120°
120°
C
H H
H
H
H
In a C
2
H
6
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: C
2
H
4
> C
2
H
6
.
H +
NH
3
NH
4
+
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
NH
4
+
NH
3
Number of bond pairs
4
3
Number of lone pair(s)
0
1
The electron pairs repel one another and stay as far apart as possible.
In the NH
4
+
ion, the furthest apart the four pairs of electrons can get is when they are arranged in a tetrahedral shape.
Thus the NH
4
+ ion has a tetrahedral shape. The H–N–H bond angles are 109.5°.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the H–N–H bond angle in the NH
3
molecule is compressed to 107°.
∴ bond angle: NH
4
+
> NH
3
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 SF
6
molecule:
F
F F
S
F F
F
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 SF
6
molecule, the individual bond dipole moments cancel one another out exactly.
The molecule has no net dipole moment and it is non-polar.

54 B
Molecule
(1) BCl
3
Three-dimensional structure
Cl
Cl
B
Cl
Polar bond Symmetrical shape?
Polar molecule?
δ + δ –
B—Cl yes no
(2) Cl
2
O
(3) SiCl
4
O
Cl
Cl
Cl
Cl
Si
Cl
Cl
δ – δ +
O—Cl
δ + δ –
Si—Cl no yes yes no
∴ the BCl
3
and SiCl
4
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 electron pairs repel one another and stay as far apart as possible.
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 planar shape.
∴ there are 12 electrons in the outermost shell of the central atom X.
Due to the symmetry of the shape of the XY
4
molecule, the individual bond dipole moments cancel one another out exactly.
The 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 XY
6
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.
329

330
58 D (1) The shape of a methoxymethane is NOT linear.
O
H
3
C CH
3
(2) The methoxymethane molecule is polar.
δ –
O
H
3
δ +
C
δ +
CH
3
(3) The methoxymethane molecule can form hydrogen bonds with water molecules.
δ +
H
δ –
O
δ +
H
H
3
C
δ –
O
CH
3 hydrogen bond lone pair
59 C (1) The propanone molecule is polar.
O
δ –
H
C
C
δ +
C
H net dipole moment
H H H H
(2) NO hydrogen bond exists between propanone molecules.
(3) The propanone molecule can form a hydrogen bond with a trichloromethane molecule.
δ
+
H
δ
–
Cl
δ
–
Cl
C
δ
–
Cl
H
3
C
δ +
C
H
3
C
δ –
O

60 D (1) and (2) Three-dimensional structures of the NF
3
and NH
3
molecules:
F
N
F
F net dipole moment
H
N
H
H net dipole moment
Both the NF
3
and NH
3
molecules have a trigonal pyramidal shape.
Both the NF
3
and NH
3
molecules are polar.
(3) NH
3
, 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
3
, 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 H
2
S 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 H
2
S.
(2) The boiling point of a compound depends on the strength of its intermolecular attractions.
Hydrogen bonds exist in H
2
O while only van der Waals’ forces exist in H
2
S.
Hydrogen bonds are stronger than van der Waals’ forces.
More heat is needed to separate the H
2
O molecules during boiling.
Thus the boiling point of H
2
O is higher than that of H
2
S.
(3) The electronegativity value of oxygen is greater than that of sulphur.
Hence the bond pairs of electrons in H
2
O will be attracted towards the oxygen atom to a greater extent.
These bond pairs will repel each other to a greater extent.
Thus the H–O–H bond angle in H
2
O is greater than the H–S–H bond angle in H
2
S.
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 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.
∴ the methanal molecule has a planar structure and the H–C–O bond angle is approximately 120°.
331

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.
H
63 D (2) Hydrogen bond between H C
H
3
δ +
C lone pair
N
δ –
H
δ +
H
δ + hydrogen bond
H
N
H
H molecules:
H
3
δ +
C
N
δ –
H
δ +
H
δ +
O
(3) Hydrogen bonds between H C O H molecules:
H C
O
δ – lone pair hydrogen bond
δ +
H O
δ –
C H
O
δ –
H
δ + δ –
O hydrogen bond
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.
δ +
H hydrogen bond
δ –
O
δ +
H
CH
3
CH
2
δ +
H
δ –
O lone pair hydrogen bond
δ +
H
δ –
O
δ +
H

methoxymethane molecules and water molecules.
δ +
H
δ –
O
δ +
H
δ –
O hydrogen bond lone pair
H
3
C CH
3
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:
CH
3
CH
2
CH
2
δ +
H
δ –
O lone pair hydrogen bond
δ +
H
δ –
O
δ +
H
(3) The boiling point of X is lower than that of Y.
The boiling point of a compound depends on the strength of its intermolecular attractions.
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) Cl Cl Cl H
C C net dipole moment C C no net dipole moment
H H H Cl
X Y
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.
333

334
(2) The boiling point of X is higher than that of Y.
The boiling point of a compound depends on the strength of its intermolecular attractions.
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.
More heat is needed to separate the molecules of X during boiling.
(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.
More heat is needed to separate molecules of X during boiling.
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: lone pair
H
O
δ –
H C C
H
O
δ –
H hydrogen bond
δ +
δ +
H
δ –
O
δ +
H
Hydrogen bond between molecule of Y and water molecule: lone pair
H
3
C
δ –
O
C
δ +
CH
3
δ +
H
δ –
O
δ +
H hydrogen bond

(2) The boiling point of a compound depends on the strength of its intermolecular attractions.
Hydrogen bonds exist in X while only van der Waals’ forces exist in Y.
Hydrogen bonds are stronger than van der Waals’ forces.
More heat is needed to separate the molecules of X during boiling.
Thus the boiling point of X is higher than that of Y.
(3) Stronger intermolecular attractions in X pull the molecules closer together.
So, the density of X is higher.
69 A Electron diagram of a PCl
5
molecule:
Cl
Cl Cl
P
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 XeF
4
molecule has two lone pairs and four bond pairs of electrons in the outermost shell of the xenon atom.
The electron pairs repel one another and stay as far apart as possible.
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 XeF
4
molecule has a square planar shape.
F
F
Xe
F
F
71 B
Ion
CO
3
2–
Electron diagram
(Only electrons in the outermost shells are shown.)
2–
O
O
C
O
NO
3
–
O
O
N
O
–
∴ CO
3
2–
ion and NO
3
–
ion are planar species.
Three-dimensional structure
O
O
C
O
2–
O
O
N
O
–
Shape trigonal planar trigonal planar
335

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.
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 ethene has a planar structure.
73 B
Molecule
BeF
2
Electron diagram
(Only electrons in the outermost shells are shown.)
F Be F
Three-dimensional structure
F Be F
Shape linear
XeF
2 F
Xe
F
F
Xe
F linear
∴ BeF
2
and XeF
2
are linear species.
74 B The electronegativity of oxygen is higher than that of sulphur.
Hence the bond pairs of electrons in H
2
O will be attracted towards the oxygen atom to a greater extent.
These bond pairs will repel each other to a greater extent.
Thus the H–O–H bond angle in H
2
O is greater than the H–S–H bond angle in H
2
S.
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.
The molecule has no net dipole moment and it is non-polar.
77 A The propanone molecule is polar.
O
δ –
H
C
C
δ +
C
H net dipole moment
H H H H

A sulphur dioxide molecule is V-shaped. The individual S–O bond dipole moments reinforce each other.
Hence the SO
2
molecule has a net dipole moment.
A carbon dioxide molecule is linear in shape.
The individual C=O bond dipole moments cancel each other out exactly.
Hence the CO
2
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 H
2
O is hydrogen bonds.
The strongest type of intermolecular attractions in H
2
Se is van der Waals’ forces.
The hydrogen bonds are stronger than van der Waals’ forces.
Hence more heat is needed to separate the H
2
O molecules during boiling.
Thus the boiling point of H
2
O is higher than that of H
2
Se.
81 D In a methoxymethane molecule, the C–O bonds are polar.
H H
O
H C C H net dipole moment
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.
δ +
H
δ –
O
δ +
H
H
3
C
δ –
O
CH
3 hydrogen bond lone pair
337

338
82 B Propanone can form hydrogen bonds with trichloromethane.
δ
+
H
δ
–
Cl
δ
–
Cl
C
δ
–
Cl
H
3
C
δ +
C
H
3
C
δ –
O
H
C N H ) molecules can form hydrogen bonds with water molecules.
H H
H
3
δ +
C
δ +
H
δ +
H
δ –
N lone pair
δ +
H
δ –
O hydrogen bond
H
δ +
84 D 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.
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.

85
Molecular model
(a)
(d)
(e)
(b)
(c)
(f)
Geometry square pyramidal
(1) trigonal pyramidal
(1) square planar
(1) trigonal planar
(1) trigonal bipyramidal
(1) octahedral
(1)
339

340
86
Species
(a) BeCl
2
(b) HCN
(c) CS
2
(d) BF
3
(e) SiF
4
(f) PH
3
(g) OF
2
(h) PCl
5
Electron diagram
Cl Be Cl
H
S
F
H
C
C
F
B
F
F
F Si F
F
P
H
N
S
H
(1)
(1)
(1)
(1)
(1)
(1)
Three-dimensional structure
Cl Be Cl
(1)
H C N
(1)
S C S
(1)
F
F
B
F
F
F
Si
F
F
(1)
(1)
H
P
H
H
(1)
F O
F
Cl
Cl Cl
P
Cl
Cl
(1)
(1)
Cl
Cl
P
Cl
Cl
O
F
F
Cl
(1)
(1)
Shape linear
(1) linear
(1) linear
(1) trigonal planar
(1) tetrahedral
(1) trigonal pyramidal
(1)
V-shaped
(1) trigonal bipyramidal
(1)

87
Species
(a) SCl
2
(b) ClF
3
(c) ClF
5
(d) ICl
4
–
(e) I
3
–
(f) SO
4
2–
88 a)
δ + δ –
H—F
δ + δ –
b)
δ + δ –
c)
δ – δ +
d)
δ + δ –
e)
Electron diagram
Cl S
Cl
(1)
F Cl F
F
(1)
F
F
Cl
F
F
F
(1)
Cl
Cl
I
Cl
Cl
–
(1)
–
I I I
(1)
O
O S O
O
2–
(1)
Three-dimensional structure
S
Cl
Cl
(1)
F
Cl
F
F
(1)
F
F
F
Cl
F
F
(1)
–
Cl
Cl
I
Cl
Cl
(1)
–
I
I
I
(1)
O
O
S
O
O
2–
(1)
Shape
V-shaped
T-shaped square pyramidal square planar linear tetrahedral
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
341

342
89
Molecule
CO
2
(a) NH
3
(b) HBr
(c) BeCl
2
(d) SO
2
(e) PF
5
(f) SiCl
4
(g) XeF
4
(h) SF
6
Electron diagram
O C O
Shape linear
Polar bond
Symmetrical shape?
Polar molecule?
δ + δ –
C—O yes no
H N
H
H Br
H
(1)
(1) trigonal pyramidal linear
(1)
(1)
Cl Be Cl linear
(1) (1)
O
F
S
F F
P
F
O
F
V-shaped
(1) (1)
(1) trigonal bipyramidal
(1)
Cl
Cl Si Cl
Cl
(1) tetrahedral
(1)
F
F
Xe
F
F
(1) square planar
(1)
F
F
F
S
F
F
F
(1) octahedral
(1)
δ – δ +
N—H
(1)
δ + δ –
H—Br
(1)
δ + δ –
Si—Cl
(1)
δ + δ –
Xe—F
(1)
δ + δ –
S—F
(1)
δ + δ –
Be—Cl
(1)
δ + δ –
S—O
(1)
δ + δ –
P—F
(1) no no yes no
(1)
(1)
(1)
(1) yes
(1) yes yes yes
(1)
(1)
(1) yes
(1) yes
(1) no
(1) yes
(1) no no no no
(1)
(1)
(1)
(1)

90 a)
Liquid Structure
(ii) Tetrachloromethane
(iii) Methoxymethane
(iv) Propanone
(v) Ethanal
(vi) Ethanoic acid
S
O
Cl
Cl
C
Cl
H
O
Cl
H
H C
H
O C
H
O
H
3
C C
H O
CH
3
H
H
H
H
C
H
O
C C
H
C
O
H
H
Type of attractions
Instantaneous dipole-induced dipole attractions
Permanent dipolepermanent dipole attractions
Hydrogen bonds
✔ ✔
(0.5) (0.5)
✔
✔
✔
✔
✔
(0.5)
(0.5)
(0.5)
(0.5)
(0.5)
✔
✔
✔
✔
(0.5)
(0.5)
(0.5)
(0.5)
✔
(0.5)
b)
H
O
δ –
H C C
H
O
δ –
H hydrogen bond
δ + lone pair hydrogen bond
δ +
H O
δ – H
C C
δ –
O
H
H
(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)
343

c)
δ +
H
δ –
O
δ +
H
δ –
O hydrogen bond lone pair
H
3
C CH
3
(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 BH
3
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.
The shape that puts the three electron pairs furthest apart is trigonal planar.
(0.5)
Hence a BH
3
molecule has a trigonal planar shape. (0.5)
In an NH
3
molecule, there are one lone pair and three bond pairs of electrons in the outermost shell of the central nitrogen atom. (0.5)
(0.5) The shape that puts the four electron pairs furthest apart is tetrahedral.
The shape of a molecule is determined only by the arrangement of atoms.
Thus the NH
3
molecule has a trigonal pyramidal shape. (0.5)
(0.5)
A sulphur dioxide molecule is V-shaped.
The individual S–O bond dipole moments reinforce each other.
Hence the molecule has a net dipole moment.
(0.5)
(0.5)
A carbon dioxide molecule is linear in shape.
The individual C=O bond dipole moments cancel each other out exactly.
Hence the molecule has no net dipole moment.
c) The boiling point of an element depends on the strength of its intermolecular attractions.
The intermolecular attractions in halogens are van der Waals’ forces.
The number of electrons in the halogen molecule increases down the group.
(0.5)
(0.5)
(0.5)
(1)
(1)
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,
344 and thus the boiling points of the halogens increase down the group.

d) 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.
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.
(1)
e) H
3
PO
4
, hydrogen bond is the strongest intermolecular attraction. (1)
PO
4
molecule has three –OH groups that can take part in hydrogen bonding with other H
3
PO
4 molecules. (1)
4
has a high viscosity due to its strong intermolecular attractions. (1)
(1)
(1)
(1)
92 a)
H
H
C H methane (1)
H N H ammonia (1)
H O water (1) b) Bond angle: CH
4
> NH
3
> H
2
O (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
CH
4
NH
3
H
2
O
Number of bond pairs
4
3
2
Number of lone pair(s)
0
1
2
The electron pairs repel one another and stay as far apart as possible.
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°.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the H–N–H bond angle in the ammonia molecule is compressed to 107°.
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.
(0.5)
(0.5)
(0.5)
(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)
345

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:
PCl
5
/ SF
6
(1)
There are more than 8 electrons in the outermost shell of the phosphorus / sulphur atom. (1)
BeCl
2
/ BF
3
There are less than 8 electrons in the outermost shell of the beryllium / boron atom.
(1)
(1)
93 a) α = 120°
β = 104.5° – 109.5°
(1)
(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 furthest apart the three pairs can get is at an angle of 120°.
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
–
NH
2
–
(g) N
H
H
V-shaped
(1) (1)
NH
3
(g) trigonal pyramidal
NH
4
+
(g)
H
N
H
H
H
H
N
H
H
+
(1) tetrahedral
(1)
(1) (1)

b) Bond angle: NH
4
+
(g) > NH
3
(g) > NH
2
–
(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:
Species
NH
4
+
(g)
NH
3
(g)
NH
2
– (g)
Number of bond pairs
4
3
2
Number of lone pair(s)
0
1
2
The electron pairs repel one another and stay as far apart as possible.
In the NH
4
+
ion, the furthest apart the four pairs of electrons can get is when they are arranged in a tetrahedral shape. (0.5)
Thus the NH
4
+
ion has a tetrahedral shape. The H–N–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 NH
3
molecule is compressed to 107°. (0.5)
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion. (0.5)
In the NH
2
–
ion, the two lone pairs will stay the furthest apart. The H–N–H bond angle in the NH
2
–
ion is compressed to 104.5°. (0.5)
95 a)
H
F
δ –
C
δ +
Cl
δ –
Cl
δ – b) Fluorine and chlorine are more electronegative than carbon.
The fluorine and chlorine atoms have a greater share of the bonding electrons.
c) F
H
C
Cl
Cl d) The individual bond dipole moments do not cancel one another out exactly.
The molecule has a net dipole moment and the molecule is polar. e) Permanent dipole-permanent dipole attractions
(1)
(0.5)
(0.5)
(1)
(1)
(1)
(1)
(1)
347

96 a)
I Br
F
F
B
F (1) (1) b) i) IBr is polar.
The electronegativity value of bromine is greater than that of iodine.
(0.5)
(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)
BF
3
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 BF
3
molecule cancel one another out exactly. (0.5)
As a result, the molecule has no net dipole moment. (0.5)
97 a)
Hydride Shape
SiH
4
Three-dimensional structure
H
H
Si
H
H
(1) tetrahedral
(1)
PH
3
H
P
H
H trigonal pyramidal
(1) (1)
348
H
2
S S
H
H
V-shaped
(1) (1) b) The boiling point of H
2
S is higher than that of SiH
4
. (1)
The boiling point of a compound depends on the strength of its intermolecular attractions. (1)
H
2
S is a polar substance. There are permanent dipole-permanent dipole attractions and instantaneous dipole-induced dipole attractions between H
2
S molecules. (1)
SiH
4
is a non-polar substance. There are only instantaneous dipole-induced dipole attractions between
SiH
4
molecules. (1)
More heat is needed to separate the H
2
S molecules during boiling.

c) The electronegativity of nitrogen is higher than that of phosphorus. (1)
Hence the bond pairs of electrons in NH
3
will be attracted towards the nitrogen atom to a greater extent. (1)
(1) These bond pairs will repel each other to a greater extent and the H–N–H bond angle in NH
3
is greater than the H–P–H bond angle in PH
3
.
F
B F
F (1)
(1)
(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)
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
The three bond pairs around the sulphur atom is slightly compressed together, making the bond angle less than 109.5°.
(0.5)
(0.5)
98 a) i) 2–
O
O
C
O or ii)
O C O or b) i) The carbonate ion has a trigonal planar shape.
O
O
O
C
O
C O
2–
(1)
The electron pairs repel one another and stay as far apart as possible.
The shape that puts the electron pairs furthest apart is trigonal planar.
(1)
(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)
(0.5) 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.
349

350 c) A carbon dioxide molecule is non-polar.
Each C=O bond is polar.
(0.5)
(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.
So a carbon dioxide molecule is non-polar.
(0.5)
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. c) Buckminsterfullerene is insoluble in water.
(1)
(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)
A molecule has a trigonal planar shape. (1)
BF
3
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.
NH
3
molecule has a trigonal pyramidal shape. (1)
NH
3
molecule has one lone pair and three bond pairs of electrons in the outermost shell of the nitrogen atom. (1)

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 NH
3
molecule has a trigonal pyramidal shape.
(1)
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
The three bond pairs are thus slightly compressed together.
This results in the H–N–H bond angles being 107° instead of 109.5°.
(1)
(1) d) i) The nitrogen atom in the NH
3
molecule supplies a lone pair of electrons to the boron atom, forming a dative covalent bond. ii)
H
H
N
H
(1)
(1)
F
B
F
F
(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)
–
N
H
H
(1)
NH
2
– 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 NH
2
–
ion is compressed to a value smaller than that in the NH
3
molecule. (1)
101 a) i)
Cl
Cl Cl
P
Cl
Cl
(1) ii)
Cl
Cl
P
Cl
Cl
Cl
(1)
(1)
351

352 iv) A phosphorus pentachloride molecule is non-polar.
(1)
(1)
(0.5)
(0.5)
Due to the symmetry of the shape of the molecule, the individual bond dipole moments cancel one another out exactly.
The molecule has no net dipole moment and it is non-polar.
b) shape of [PCl
4
]
+
:
Cl
+
Cl
P
Cl
Cl
(1) shape of [PCl
6
]
–
:
Cl
Cl
P
Cl
Cl
Cl
Cl
–
(1)
(0.5)
(0.5)
Cl P
Cl
Cl ii)
(1)
Cl
P
Cl
Cl
(1)
(1) iii) 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.
(1)
Nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its atom. (1)

102 a)
Molecule Electron diagram Three-dimensional structure
SF
2
F S
F
S
F
F
(1) (1)
SF
4
F
F F
S
F
(1)
SF
6
F
F
F
S
F
F
F
(1) b) Fluorine is more electronegative than sulphur.
The fluorine atom has a greater share of the bonding electrons.
SF
6
is non-polar.
Due to the symmetry of the octahedral shape, the six identical bond dipole moments cancel one another out exactly.
The molecule has no net dipole moment and it is non-polar.
d) Consider the electron diagram of OF
2
:
F O
F
F
F
F
S
F
F
F
F
S
F
F
F
(1)
Oxygen cannot form compounds with more than 8 electrons in the outermost shell of its atom. (0.5)
Hence oxygen forms OF
2
only.
Sulphur can form some compounds with more than 8 electrons in the outermost shell of its atom. (0.5)
Hence it can form SF
4
and SF
6
besides SF
2
.
(1)
(1)
(1)
(0.5)
(0.5)
(0.5)
(0.5)
103 a)
F N F
F
(1)
(1)
(1)
353

354
c)
3
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
3
molecule has a trigonal pyramidal shape. The individual N–F bond dipole moments reinforce each other. (0.5)
The molecule has a net dipole moment and it is polar.
(0.5)
(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.
Hydrogen bonds exist in HF
(1)
(1)
(1) while only van der Waals’ forces exist in HCl.
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 higher than that of HCl.
ii) A HI molecule contains more electrons than a HBr molecule.
Hence the strength of van der Waals’ forces in HI is higher than that in HBr.
More heat is needed to separate the HI molecules during boiling.
Thus the boiling point of HI is higher than that of HBr.
(1)
(1)
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.
The fluorine atom has a greater share of the bonding electrons.
(1)
(1)
NH
3 iii) Nitrogen is the least electronegative among nitrogen, oxygen and fluorine. /
Nitrogen and hydrogen have the smallest electronegativity difference.
(1)
(1)

c) i) (1) instantaneous dipole-induced dipole attractions ii) (1) There is a large difference in electronegativity between H and F.
In HF, the hydrogen atom has a stronger positive charge. atom of another molecule is known as a hydrogen bond.
(2) lone pair
δ –
F
δ +
H
δ –
F
δ +
H hydrogen bond
(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 boiling point of a compound depends on the strength of its intermolecular attractions. (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. so each H
2
O molecule can take part in hydrogen bonding to twice the extent.
More heat is needed to separate the water molecules during boiling.
(1)
The oxygen atom in a water molecule has two lone pairs and there are two hydrogen atoms in each molecule,
(1)
106 a) A covalent bond in which the electrons are unequally shared between the two bonded atoms. b) Fluorine is more electronegative than carbon.
The fluorine atom has a greater share of the bonding electrons.
(1)
(1)
(1)
(1) ii) iii) F
H
C C
F
H (1)
F
H
C C
H
F
F
F
C C
H
H (1)
(1)
355
(1)
(1)
(1)
(1)
(1)
(1)

107 a)
XBUFS XBUFS
356 m
(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.
b)
δ +
H
δ –
O
δ +
H lone pair
δ –
O
δ +
H
δ +
H
δ –
O
δ +
H
δ +
H hydrogen bond
(0.5)
δ –
O
δ +
H
δ +
H
(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)
+
H
O
H
H
(1)
(1)

108 a) A pair of equal and opposite charges separated by a distance is called a dipole.
Dipole moment is the product of the charge and the distance between the charges.
(1)
(1) b) i) Permanent dipole-permanent dipole attraction is the attraction between molecules which have permanent dipoles. (1) atom and the lone pair on another electronegative atom (usually oxygen, nitrogen or fluorine).
(1)
(1)
c)
δ +
H
δ +
H
δ +
H lone pair
δ –
N
δ +
H
δ –
O hydrogen bond
H
δ +
(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.
The van der Waals’ forces in X are thus stronger.
(1)
(1)
More heat is needed to separate molecules of X during boiling.
Hence X has a higher boiling point than Z.
357

358 b) Y has a structure in between the linear structure of X and the spherical structure of Z.
The van der Waals’ forces in Y are probably in between those in X and Z.
Hence the boiling point of Y is probably in between those of X and Z. c) Stronger van der Waals’ forces in X pull the molecules closer together.
So, the density of X is higher.
d) X is insoluble in water.
X is non-polar.
(1)
(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)
(1)
(1)
(1)
(1)
110 a) i) Van der Waals’ forces / instantaneous dipole-induced dipole attractions ii) They have the same number of electrons per molecule.
C–C–O bond angle: 120° ii) Permanent dipole-permanent dipole attractions c) Propanone molecules can form hydrogen bonds with water molecules, but butane molecules cannot.
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1) iii) lone pair
C
3
H
7
δ –
O
δ +
H
δ –
O
δ +
H
C
3
H
7 hydrogen bond
(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) e) The boiling point of a compound depends on the strength of its intermolecular attractions.
The shape of a propan-1-ol molecule is longer.
(1)
(1)
This allows greater surface contact between molecules.
Van der Waals’ forces in propan-1-ol are stronger than those in propan-2-ol.
More heat is needed to separate the propan-1-ol molecules during boiling.
(1)
(1)

111 a) Boiling point increases in the following order:
D < C < A < B
The boiling point of a compound depends on the strength of its intermolecular attractions.
(1)
(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)
CH
3
CH
2
δ +
H
δ –
O lone pair hydrogen bond
δ +
H
δ –
O
δ +
H
(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) c) The viscosity of compound Y is higher than that of compound X.
Both compounds X and Y can form hydrogen bonds.
(1)
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)
359

360
O N O
(1)
(1) d) Any one of the following:
PCl
5
/ SF
6
There are more than 8 electrons in the outermost shell of the phosphorus / sulphur atom.
(1)
(1)
BeCl
2
/ BF
3
There are less than 8 electrons in the outermost shell of the beryllium / boron atom.
(1)
(1)
4NO
2
(g) + O
2
(g) + 2H
2
4HNO
3
(aq) (1)
O
S
O ii)
(1)
S
O O iii) Sulphur dioxide is polar.
(1)
(0.5)
(0.5) each other.
The molecule has a net dipole moment and it is polar.
113 a) Any two of the following:
(0.5)
(0.5)
(1)
(1)
(1) • High potency / effectiveness as an anaesthetic
b)
N N O or N N O c) In an ethoxyethane molecule, the C–O bonds are polar.
H H H H
O
H C C C C H net dipole moment
H H H H
(1)
(1)

The molecule has a net dipole moment and thus ethoxyethane is a polar molecule.
(1) 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.
δ +
H
δ –
O
δ +
H lone pair hydrogen bond
δ –
O
H H
H C C H
C H H C
H H H H
(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) ii) Each hydrogen atom of a water molecule has a partial positive charge.
A hydrogen bond forms between the hydrogen atom of a water molecule
(1)
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)
The individual C–Cl bond dipole moments reinforce each other.
The molecule has a net dipole moment and it is polar.
(0.5)
(0.5)
(0.5)
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)
361

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.
The instantaneous dipole can affect the electron distributions in neighbouring molecules and induce similar dipoles in these neighbours.
Attractions exist between the instantaneous dipole and the induced dipoles.
(0.5)
(0.5)
(0.5)
(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.
Hence the strength of van der Waals’ forces also increases down the group.
(0.5)
(0.5)
The boiling point of an element depends on the strength of its intermolecular attractions.
Hence the boiling points of halogens increase down the group.
(0.5)
(0.5)
(3 marks for organization and presentation)