5. Solution Guide to Supplementary Exercises

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Topic
6
Microscopic World II
Part A Unit-based exercise
Unit 23 Shapes of molecules
Fill in the blanks
1
a) four
b) tetrahedral
c) 109.5°
2
a) one; three
b) trigonal pyramidal
c) 107°
3
a) two; two
b) Vc) 104.5°
4
a) two
b) linear
c) 180°
5
a) three
b) trigonal planar
c) 120°
6
a) five
b) trigonal bipyramidal
7
a) six
b) octahedral
8
linear
9
planar
10 tetrahedral
11 V12 trigonal pyramidal
13 lone pair-lone pair; lone pair-bond pair; bond pair-bond pair
14 graphite; buckminsterfullerene
258
True or false
15 T Electron diagram of a XeF4 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
B
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.
There are less than 8 electrons in the outermost shell of the boron atom of a boron trifluoride
molecule. Thus the molecule does not obey the octet rule.
17 F
Nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its atom.
Thus NCl5 does not exist.
18 T Electron diagram of a carbon tetrachloride molecule:
Cl
Cl C Cl
Cl
(Only electrons in the outermost shells are shown.)
In a CCl4 molecule, there are four bond pairs of electrons in the outermost shell of the central carbon
atom.
The shape that puts the four electron pairs furthest apart is tetrahedral.
Thus the carbon tetrachloride molecule has a tetrahedral shape.
19 F
Electron diagram of a SF4 molecule:
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
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
F
120°
S
F
90°
F
120°
S
90°
F
F
I
Shape
II
Repulsions involving the lone pair
I
3 90° lone pair-bond pair repulsions (repulsions at
angles greater than 90° can be ignored)
II
2 90° lond pair-bond pair repusions (repulsions at
angles greater than 90° can be ignored)
Remark
the repulsion is smaller for shape II, hence
the electron pairs will adopt shape II
∴ the SF4 molecule has a seesaw shape.
20 F
Electron diagram of a SO3 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 PCl4+ ion:
+
Cl
Cl P Cl
Cl
(Only electrons in the outermost shells are shown.)
In a PCl4+ ion, there are four bond pairs of electrons in the outermost shell of the central phosphorus
atom.
The shape that puts the four electron pairs the furthest apart is tetrahedral.
Thus the PCl4+ ion has a tetrahedral shape.
+
Cl
P
Cl
Cl
Cl
260
22 F
–
Electron diagram of a NO3 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°.
23 F
A BH3 molecule has a trigonal planar shape.
An NH3 molecule has a trigonal pyramidal shape.
H
B
H
N
H
H
H
H
24 T Both molecules of BeF2 and XeF2 have a linear shape.
Electron diagram
(Only electrons in the
outermost shells are shown.)
Molecule
F Be F
BeF2
Shape
F — Be — F
F
XeF2
F
Xe
F
Xe
F
+
25 T Bond angle: NH4 > NH3
In the outermost electron shell of the central nitrogen atom in each of the species, the numbers of
bond pairs and lone pair of electrons are as follows:
Species
Number of bond pairs
Number of lone pair
NH4
4
0
NH3
3
1
+
The electron pairs repel one another and stay as far apart as possible.
In the NH4+ ion, the furthest apart the four pairs of electrons can get is when they are arranged in a
tetrahedral shape.
Thus the NH4+ ion has a tetrahedral shape. The H–N–H bond angles are 109.5°.
261
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the H–N–H bond angle in the NH3 molecule is compressed to 107°.
26 F
Buckminsterfullerene has a simple molecular structure. Van der Waals’ forces exist between the C60
molecules.
Multiple choice questions
27 C
28 A The sulphur atom has 6 electrons in its outermost shell and each oxygen atom has 6, i.e. a SO2
molecule has a total of 18 electrons in the outermost shells.
Option
Electron diagram
(Only electrons in the outermost
shells are shown.)
Remark
A
O
S
O
Correct
B
O
S
O
Incorrect as oxygen cannot form compounds with more
than 8 electrons in the outermost shell of its atom.
C
O
D
O
S
S
O
O
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 XeO3
molecule has a total of 26 electrons in the outermost shells.
Option
Electron diagram
(Only electrons in the outermost
shells are shown.)
O
A
O
Xe
O
O
B
O
Xe
O
O
C
O
D
262
Xe
O
O
O Xe O
Remark
Incorrect as there are 24 electrons in the outermost
shells.
Incorrect as there are 28 electrons in the outermost
shells.
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 H3O ion carries one positive charge because it has lost 1 electron. That makes a total of 8
electrons in the outermost shells of the H3O+ ion.
31 C The nitrogen atom has 5 electrons in its outermost shell and the two hydrogen atoms have 2 electrons,
i.e. a total of 7 electrons.
But the NH2– ion carries one negative charge because it has gained 1 electron. That makes a total of 8
electrons in the outermost shell of the NH2– ion.
32 C The sulphur atom has 6 electrons in its outermost shell and each of the four oxygen atoms has 6
electrons, i.e. a total of 30 electrons.
But the SO42– ion carries two negative charges because it has gained 2 electrons. That makes a total of
32 electrons in the outermost shells of the SO42– ion.
Option
Electron diagram
(Only electrons in the outermost
shells are shown.)
Remark
2–
O
A
O
S
Incorrect as there are 28 electrons in the outermost
shells.
O
O
2–
O
B
O
S
Incorrect as there are 36 electrons in the outermost
shells.
O
O
2–
O
C
O
S
O
Correct
O
2–
O
D
O
S
O
Incorrect as there are 40 electrons in the outermost
shells.
O
33 A Electron diagram of a SiH4 molecule:
H
H Si H
H
(Only electrons in the outermost shells are shown.)
263
34 B Electron diagram of a NCl3 molecule:
Cl N Cl
Cl
(Only electrons in the outermost shells are shown.)
35 C Electron diagram of a SCl2 molecule:
Cl S
Cl
(Only electrons in the outermost shells are shown.)
36 A Electron diagram of a PCl4+ ion:
+
Cl
Cl P Cl
Cl
(Only electrons in the outermost shells are shown.)
37 A Electron diagram of a CH3+ ion:
+
H
H
C
H
(Only electrons in the outermost shells are shown.)
38 B
Option
Molecule
A
H2O
B
PH3
Electron diagram
(Only electrons in the outermost
shells are shown.)
H O
Number of lone pair(s) of
electrons in the outermost shell
of the central atom
2
H
H
P
H
1
H
Cl
Cl
C
PCl5
Cl
P
Cl
0
Cl
H
D
CH2Cl2
H C Cl
Cl
264
0
39 B
Molecule
XeF2
Electron diagram
(Only electrons in the outermost
shells are shown.)
Number of lone pairs of
electrons in the outermost shell
of the central atom
Xe
3
F
F
F
F
Xe
XeF4
F
2
F
40 D Electron diagram of a BrF5 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.
BrF5 does not follow the octet rule as there are more than 8 electrons in the outermost shell of the
bromine atom.
41 A
42 C In a molecule with three bond pairs and one lone pair around the central atom, the shape that puts
the four electron pairs the furthest apart is tetrahedral.
The shape of a molecule is determined only by the arrangement of atoms.
Thus the molecule has a trigonal pyramidal shape.
43 B There are 6 electron pairs in the outermost shell of the central atom of a molecule with an octahedral
shape.
For example, a SF6 molecule has an octahedral shape. There are 6 electron pairs in the outermost shell
of the central sulphur atom.
F
F
F
S
F
F
F
44 D Electron diagram of an OF2 molecule:
F
O
F
(Only electrons in the outermost shells are shown.)
265
An OF2 molecule has two lone pairs and two bond pairs of electrons in the outermost shell of the
oxygen atom.
The four pairs of electrons will adopt a tetrahedral arrangement.
The shape of a molecule is determined only by the arrangement of atoms. Thus the OF2 molecule is
V-shaped.
45 B Electron diagram of a BH3 molecule:
H
H
B
H
(Only electrons in the outermost shells are shown.)
In a BH3 molecule, there are three bond pairs of electrons in the outermost shell of the central boron
atom.
The shape that puts the three electron pairs the furthest apart is trigonal planar.
Hence a BH3 molecule has a trigonal planar shape.
46 D Electron diagram of a PF3 molecule:
F
P
F
F
(Only electrons in the outermost shells are shown.)
In a PF3 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 PF3 molecule has a trigonal pyramidal shape.
47 B In an NH3 molecule, there are one lone pair and three bond pairs of electrons in the outermost shell of
the central nitrogen atom.
The shape that puts the four electron pairs the furthest apart is tetrahedral.
48 D Electron diagram of a XeF4 molecule:
F
F
Xe
F
F
(Only electrons in the outermost shells are shown.)
In a XeF4 molecule, there are six bond pairs of electrons in the outermost shell of the central xenon
atom.
The shape that puts the six electron pairs the furthest apart is octahedral.
266
49 C Electron diagram of a ClF3 molecule:
F
Cl
F
F
(Only electrons in the outermost shells are shown.)
In a ClF3 molecule, there are two lone pairs and three bond pairs of electrons in the outermost shell of
the central chlorine atom.
The shape that puts the five electron pairs the furthest apart is trigonal bipyramidal.
50 C
51 D Electron diagram of a CO2 molecule:
O C O
(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 CO2 molecule has a linear shape and the O–C–O bond angle is 180°.
52 B A H2O molecule has two lone pairs and two bond pairs of electrons in the outermost shell of the
oxygen atom.
The four pairs of electrons will adopt a tetrahedral arrangement.
The shape of a molecule is determined only by the arrangement of atoms. Thus the H2O 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 H2O molecule, the two lone pairs will stay the furthest apart.
As a result, the H–O–H bond angle in the H2O molecule is compressed to 104.5°.
53 A Electron diagram of a C2H4 molecule:
H
H
C
H
C
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.
The H–C–H bond angles are 120°.
267
54 B
Option
Molecule
A
H2S
Electron diagram
(Only electrons in the outermost
shells are shown.)
S
Three-dimensional
structure
Shape
S
V-shaped
H
H
H
H
B
BeCl2
C
SCl2
Cl Be Cl
Cl
Be
S Cl
Cl
V-shaped
S
Cl
linear
Cl
Cl
D
O
OF2
F
V-shaped
O
F
F
F
55 A
Molecule
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
F
F
F
SF6
F
F
S
F
Ion
F
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
Molecule
F
268
H
H
F
Three-dimensional structure
F
Xe
F
tetrahedral
N
H
Electron diagram
(Only electrons in the outermost
shells are shown.)
XeF4
Shape
+
H
H N H
H
57 B
F
F
H
NH4
octahedral
F
+
+
F
S
F
56 B
Shape
F
square planar
Xe
F
F
Shape
F
58 A Electron diagram of a CO32– 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 CO32– ion has a trigonal planar shape.
59 C A H2O molecule has two lone pairs and two bond pairs of electrons in the outermost shell of the
oxygen atom.
The four pairs of electrons will adopt a tetrahedral arrangement.
The shape of a molecule is determined only by the arrangement of atoms. Thus the H2O molecule is
V-shaped.
60 D
Option
Species
Electron diagram
(Only electrons in the outermost
shells are shown.)
–
O
A
NO3–
N
O
B
NCl3
Three-dimensional
structure
–
O
trigonal planar
N
O
O
Cl N Cl
Cl
Shape
O
trigonal pyramidal
N
Cl
Cl
Cl
F
C
SF4
F
F
F
F
S
seesaw
S
F
F
F
F
F
D
SiF4
F
Si
F
F
tetrahedral
Si
F
F
F
∴ the SiF4 molecule has a tetrahedral shape.
269
61 A
Option
A
Molecule
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional
structure
F
F
BF3
F
B
B
F
H N H
NH3
trigonal planar
B
F
H
Shape
F
trigonal pyramidal
N
H
H
H
C
H
PH3
P
H
H
trigonal pyramidal
P
H
H
H
D
Cl P Cl
PCl3
Cl
trigonal pyramidal
P
Cl
Cl
Cl
∴ the BF3 molecule has a shape different from the others.
62 C
Option
Molecule
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional
structure
Cl
Cl
A
BCl3
Cl
B
Is the species planar?
yes
B
Cl
Cl
Cl
F
B
ClF3
F
Cl
F
Cl
F
yes
F
F
C
NCl3
Cl N Cl
no
N
Cl
Cl
Cl
Cl
F
D
XeF4
F
∴ NCl3 is NOT a planar species.
270
F
F
F
yes
Xe
Xe
F
F
F
63 C Electron diagram of BCl3:
Cl B Cl
Cl
(Only electrons in the outermost shells are shown.)
In a BCl3 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 BCl3 molecule has a trigonal planar shape.
Electron diagram of a PH3 molecule:
H
P
H
H
(Only electrons in the outermost shells are shown.)
In a PH3 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 PH3 molecule has a trigonal pyramidal shape.
64 D (1) and (3) Electron diagram of a ClF3 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.
F
F
90°
F
Cl
F
Cl
F
F
Shape
90°
Cl
F
F
II
Repulsion present
I
90° lone pair-lone pair repulsion
II
6 90° lone pair-bond pair repulsions
(repulsions at angles greater than 90°
can be ignored)
III
4 90° lone pair-bond pair repulsions +
2 90° bond pair-bond pair repulsions
(repulsions at angles greater than 90°
can be ignored)
∴ the ClF3 molecule is T-shaped.
120°
90°
F
I
90°
90°
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
65 C
Option
Species
A
CS2
Electron diagram
(Only electrons in the outermost
shells are shown.)
S
C
S
S
H
Three-dimensional structure
S
C
S
H2S
S
H
H
H
B
CH4
H
H
H C H
C
H
H
H
H
XeF4
F
F
F
F
F
Xe
F
C
Xe
NH4+
H
F
+
H
+
H N H
N
H
H
H
H
SiCl4
Cl
Cl
Cl Si Cl
Si
Cl
Cl
Cl
D
F
Cl
NH3
H N H
N
H
H
H
H
NO3–
O
–
O N O
∴ the NH4+ ion and SiCl4 molecule have a similar shape.
272
–
O
N
O
O
66 B
Option
Species
A
BH3
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
H
H
H
B
H
B
H
H
PH3
H
P
H
P
H
H
H
H
B
CO2
O C O
O
C
O
H C
H
C
N
HCN
C
N
CO32–
2–
O
O
C
2–
O
C
O
O
O
+
H3O
+
+
H O H
O
H
H
H
H
D
NH2–
–
–
N
H N H
H
H
XeF2
F
F
Xe
F
Xe
F
∴ the CO2 and HCN molecules have a similar shape.
273
67 A
Option
Molecule
A
PCl3
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
Cl P Cl
P
Cl
Cl
Cl
Cl
NCl3
Cl N Cl
Cl
N
Cl
Cl
Cl
B
CF4
F
F
F
C
F
C
F
F
F
F
SF4
F
F
S
S
F
F
F
F
F
F
C
SO2
O
S
O
S
O
O
CO2
O C O
D
O
PF5
F
P
F
F
F
F
P
F
F
IF5
F
F
F
I
F
F
F
∴ the PCl3 and NCl3 molecules have a similar shape.
274
O
F
F
F
C
F
F
I
F
F
68 D
Option
Species
A
BF3
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
F
F
F
B
B
F
F
F
NF3
F
N
F
N
F
F
F
F
B
BCl3
Cl
Cl
Cl
B
B
Cl
Cl
Cl
PH3
H
P
H
P
H
H
H
H
C
BeCl2
Cl Be Cl
Cl
Be
Cl
SCl2
Cl S
S
Cl
Cl
Cl
D
CH3+
+
C
H C H
H
NO3–
–
O
O
N
+
H
H
O
H
–
O
N
O
O
∴ the CH3+ ion and NO3– ion have an identical geometry.
275
69 C
ammonia molecule
H+
ammonium ion
+
H
N
N
H
H
H
H
H
H
∴ there is a change of shape from trigonal pyramidal to tetrahedral.
70 A PF5
PF3
Electron diagrams of PF5 and PF3:
F
F
P
F
F
F
P
F
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
Electron diagram
(Only electrons in the outermost
shells are shown.)
Molecule
Three-dimensional structure
180°
HCN
H C
N
H
C
N
72 C Electron diagram of a CO32– 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.
The shape that puts the electron pairs the furthest apart is trigonal planar.
Thus the CO32– ion has a trigonal planar shape.
276
∴ the bond angle in a CO32– ion is 120°.
73 B
Electron diagram
(Only electrons in the outermost
shells are shown.)
Molecule
Three-dimensional structure
H O
H2O
O
H
H
H
H N H
NH3
H
104.5°
N
H
H
H
H
H
CH4
C
H C H
H
107°
H
H
H
109.5°
180°
CO2
O C O
O
C
O
∴ the correct order of bond angles in the molecules H2O, NH3, CH4 and CO2 is CO2 > CH4 > NH3 >
H2O.
74 D
Molecule
Electron diagram
(Only electrons in the outermost
shells are shown.)
BeCl2
Cl Be Cl
Three-dimensional structure
180°
Cl
Be
Cl
∴ the Cl–Be–Cl bond angle in a BeCl2 molecule is 180°.
75 B In a regular octahedral molecule, MX6, the number of X–M–X bonds at 180° is 3.
X
X
180°
M
X
180°
X
180°
X
X
76 C
H
H
α
C
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
H
77 C
H
α
H
β
Cx
C
H
H
O
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
F
F
A
BF3
F
B
120°
F
F
CH4
H C H
NH3
C
H
H
H N H
H
PCl3
109.5°
N
H
H
H
D
F
120°
H
H
C
120°
H
H
B
B
less than 109.5°
Cl P Cl
Cl
P
Cl
Cl
Cl
less than 109.5°
∴ the Y–X–Y angle is the greatest in BF3.
–
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.
278
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 NO3 ion has a trigonal planar shape.
∴ the O–N–O bond angle is 120°.
80 B In the outermost electron shell of the central nitrogen atom in each of the species, the numbers of
bond pairs and lone pair(s) of electrons are as follows:
Species
Number of bond pairs
Number of lone pair(s)
NH4+
4
0
NH3
3
1
NH2–
2
2
The electron pairs repel one another and stay as far apart as possible.
In the NH4+ ion, the furthest apart the four pairs of electrons can get is when they are arranged in a
tetrahedral shape.
Thus the NH4+ ion has a tetrahedral shape. The H–N–H bond angles are 109.5°.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the H–N–H bond angle in the NH3 molecule is compressed to 107°.
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.
Thus in the NH2– ion, the two lone pairs will stay the furthest apart. The H–N–H bond angle in the
NH2– ion is compressed to 104.5°.
∴ the order of the bond angles is NH4+ > NH3 > NH2–.
81 C Consider the following change:
H2O
H3O
+
In the outermost electron shell of the oxygen atom in each of the species, the numbers of bond pairs
and lone pair(s) of electrons are as follows:
Species
Number of bond pairs
Number of lone pair(s)
H3O+
3
1
H2O
2
2
In a H3O+ ion, the four pairs of electrons in the outermost shell of the oxygen atom will adopt a
tetrahedral arrangement.
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.
Thus in a H2O molecule, the two lone pairs will stay the furthest apart.
As a result, the H–O–H bond angle in a H2O molecule is compressed to less than that in a H3O+ ion.
∴ when the H3O+ ion is formed from water, the bond angle increases slightly.
279
82 A
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
+
H O H
(1) H3O+
H
+
H
(2) NH4+
H N H
H
F
F
F
S
(3) SF6
F
F
F
+
∴ both the H3O and NH4+ ions have 8 electrons in the outermost shell of the central atom.
83 B
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
H
H C H
(1) CH4
H
(2) XeF2
F
Xe
F
–
(3) NH2
–
H N H
∴ both the CH4 molecule and the NH2– ion have 8 electrons in the outermost shell of the central
atom. The electron pairs would adopt a tetrahedral arrangement.
84 D
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
(1) BeF2
F Be F
F
Be
F
(2) CO2
O C O
O
C
O
(3) HCN
H C
H
C
N
N
∴ all the three species have a linear shape.
280
Three-dimensional structure
85 C
Species
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
Shape
F
F
(1) BrF3
Br
F
Br
F
F
T-shaped
F
H N H
(2) NH3
trigonal pyramidal
N
H
H
H
H
F
(3) PF3
P
F
trigonal pyramidal
P
F
F
F
F
∴ the NH3 and PF3 molecules have a trigonal pyramidal shape.
86 D
Species
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
Cl O
(1) Cl2O
V-shaped
O
Cl
Shape
Cl
Cl
–
–
(2) NH2
–
H N H
V-shaped
N
H
H
(3) SO2
S
O
V-shaped
S
O
O
O
∴ all the three species are V-shaped.
281
87 B
Species
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
Br
Br
(1) SiBr4
Shape
Si
Br Si Br
tetrahedral
Br
Br
Br
Br
F
F
F
(2) SF4
F
F
seesaw
S
S
F
F
F
+
H
(3) NH4
+
+
H
H N H
tetrahedral
N
H
H
H
H
∴ the SiBr4 molecule and the NH4+ ion have a tetrahedral shape.
88 A
Species
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
F
F
(1) boron
trifluoride
F
B
H
(2) ethene
C
(3) nitrogen
trichloride
F
H
H
H
H
Cl N Cl
Cl
F
H
C
C
H
trigonal planar
B
F
C
282
planar
H
N
Cl
Cl
Cl
∴ both boron trifluoride and ethene have a planar structure.
Shape
trigonal
pyramidal
89 C
Electron diagram
(Only electrons in the outermost
shells are shown.)
Species
(1)
Three-dimensional structure
BF3
F
F
F
Shape
B
trigonal planar
B
F
F
F
NH3
H N H
H
trigonal
pyramidal
N
H
H
H
(2)
BeCl2
Cl Be Cl
Cl
Be
Cl
linear
O C O
O
C
O
linear
CO2
(3)
NH4
+
+
H
+
H
H N H
tetrahedral
N
H
H
H
H
CH4
H
H
tetrahedral
C
H C H
H
H
H
H
∴ (2) the BeCl2 and CO2 molecules have a similar shape.
(3) the NH4+ ion and the CH4 molecule have a similar shape.
90 D
Species
(1) CS2
Electron diagram
(Only electrons in the outermost
shells are shown.)
S
C
F
P
F
F
S
P
F
F
F
F
F
Xe
(3) XeF4
C
F
F
F
S
F
F
F
(2) PF5
S
Three-dimensional structure
F
Xe
F
F
F
∴ all the three species have three atoms lying in a straight line.
283
91 B Three dimensional structure of PCl5:
Cl
Cl
90°
P
Cl
120°
Cl
Cl
∴ the bond angles 90° and 120° exist in a PCl5 molecule.
92 D
Species
Electron diagram
(Only electrons in the outermost
shells are shown.)
B
F
C
(3) nitrate ion
O
N
F
2–
F
2–
O
trigonal planar
C
O
O
–
O
trigonal planar
B
F
O
O
Shape
F
F
(1) boron
trifluoride
(2) carbonate
ion
Three-dimensional structure
O
O
–
O
trigonal planar
N
O
O
∴ the model could represent all the three species.
93 A (2) Van der Waals’ forces exist between molecules of buckminsterfullerene.
(3) Buckminsterfullerene has a simple molecular structure while graphite has a giant covalent structure.
Thus the melting point of graphite is higher than that of buckminsterfullerene.
94 C Nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its atom.
Thus nitrogen cannot form pentachloride.
95 B Electron diagram of a BF3 molecule:
F
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 BF3 molecule does not conform to the octet rule.
284
Electron diagram of a NO2 molecule:
O N O
(Only electrons in the outermost shells are shown.)
The NO2 molecule has an unpaired electron.
Thus the NO2 molecule does not conform to the octet rule.
96 C An NH3 molecule has one lone pair and three bond pairs of electrons in the outermost shell of the
nitrogen atom.
The four pairs of electrons will adopt a tetrahedral arrangement.
The shape of a molecule is determined only by the arrangement of atoms. Thus the NH3 molecule has
a trigonal pyramidal shape.
97 B Electron diagram of a methanal molecule:
H
C
O
H
(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 CH3+ 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 CH3+ ion has a trigonal planar shape.
285
99 C
Molecule
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
Cl
Cl
BCl3
B
Cl
NCl3
Shape
trigonal planar
B
Cl
Cl
Cl N Cl
Cl
trigonal pyramidal
N
Cl
Cl
Cl
Cl
∴ BCl3 has a planar structure but NCl3 does not.
100 B
Ion
Electron diagram
(Only electrons in the outermost
shells are shown.)
2–
O
CO32–
C
O
trigonal planar
C
O
–
O
–
O
–
N
O
trigonal planar
N
O
Shape
2–
O
O
O
NO3
Three-dimensional structure
O
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.)
Cl
P
Cl
Cl
Cl
F
IF5
I
F
F
∴ molecules of PCl5 and IF5 have different shapes.
286
P
Cl
trigonal bipyramidal
Cl
Cl
F
F
Shape
Cl
Cl
Cl
PCl5
Three-dimensional structure
F
F
F
I
F
square pyramidal
F
102 B Bond angle: NH3 > H2O
In the outermost electron shell of the central atom in each of the molecules, the numbers of bond
pairs and lone pair(s) of electrons are as follows:
Molecule
Number of bond pairs
Number of lone pair(s)
NH3
3
1
H2O
2
2
In the NH3 molecule, the four pairs of electrons in the outermost shell of the nitrogen atom will adopt
a tetrahedral arrangement.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the H–N–H bond angle in the NH3 molecule is compressed to 107°.
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.
Thus in the H2O molecule, the two lone pairs will stay the furthest apart.
As a result, the H–O–H bond angle in the H2O molecule is compressed to 104.5°.
103 D
Molecule
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional
structure
F
Bond angles
tetrahedral
109.5°
seesaw
180°, 120°,
90°
F
F
CF4
Shape
C
F
C
F
F
F
F
F
SF4
F
F
F
F
S
S
F
F
F
∴ the bond angles in CF4 are 109.5° but those in SF4 are not.
104 C Graphite has a giant covalent structure. It is NOT soluble in carbon disulphide.
105 A
287
Unit 24 Bond polarity and intermolecular forces
Fill in the blanks
1
polar covalent
2
electronegativity
3
polar
4
fluorine
5
Dipole moment
6
a) polar
b) symmetrical
7
Instantaneous dipole-induced dipole
8
permanent dipole-permanent dipole; permanent dipole-induced dipole; instantaneous dipole-induced dipole
9
hydrogen; oxygen; nitrogen; fluorine
10 viscosity
True or false
11 T Fluorine is the most electronegative element.
12 F
In general, electronegativity increases from left to right across a period, as the metallic character of
elements decreases.
Both phosphorus and sulphur are Period 3 elements.
Sulphur is more electronegative than phosphorus.
13 T
14 T Within each group, electronegativity decreases with increasing atomic number and metallic character.
The electronegativity value of Cl is greater than that of I.
The electronegativity difference between H and Cl is greater than that between H and I.
Thus H–Cl bond is more polar than H–I bond.
15 F
The electronegativity values of C and S are the same. Thus C=S bond is non-polar.
A carbon disulphide molecule is non-polar.
288
16 T The electronegativity value of nitrogen is greater than that of phosphorus.
Hence the bond pairs of electrons in NH3 will be attracted towards the nitrogen atom to a greater
extent.
These bond pairs will repel each other to a greater extent.
Thus the H–N–H bond angle in NH3 is greater than the H–P–H bond angle in PH3.
17 F
The electronegativity value of chlorine is greater than that of carbon. So each C–Cl bond is polar.
Due to the symmetry of the tetrahedral shape of tetrachloromethane, the four identical bond dipole
moments cancel one another out exactly.
The molecule has no net dipole moment and it is non-polar.
∴ a stream of tetrachloromethane is NOT deflected by a charged rod.
18 F
A phosphorous pentachloride molecule has a trigonal bipyramidal shape.
Cl
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
SO2
Electron diagram
(Only electrons in the outermost
shells are shown.)
S
O
Three-dimensional structure
Shape
S
V-shaped
O
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
21 F
Among the noble gases, the atomic number increases from helium to xenon and the number of
electrons in one atom increases in the same order.
Hence the strength of van der Waals’ forces in noble gases also increases from helium to xenon.
The boiling point of an element depends on the strength of its intermolecular attractions.
Thus the boiling point of argon is higher than that of neon.
22 T The number of electrons in a HI molecule is greater than that in a HCl molecule.
Hence van der Waals’ forces between HI molecules are stronger than those between HCl molecules.
23 F
The boiling point of H2S is higher than that of SiH4.
The boiling point of a compound depends on the strength of its intermolecular attractions.
H2S is a polar substance. There are permanent dipole-permanent dipole attractions and instantaneous
dipole-induced dipole attractions between H2S molecules.
SiH4 is a non-polar substance. There are only instantaneous dipole-induced dipole attractions between
SiH4 molecules.
More heat is needed to separate the H2S molecules during boiling.
24 F
The electronegativity values of P and H are the same.
NO strong attraction exists between the hydrogen atom of one phosphine molecule (PH3) and the lone
pair on the phosphorus atom of another phosphine molecule.
Thus NO hydrogen bond exists between phosphine molecules.
25 T Water has a high surface tension due to the hydrogen bonds between water molecules.
26 F
The viscosity of a liquid is a measure of a liquid’s resistance to flow. The greater the viscosity, the more
slowly the liquid flows.
27 T Ethanol is miscible with water (i.e. soluble in water in all proportions) because hydrogen bonds can
form between ethanol molecules and water molecules. This helps the dissolving process.
δ+
CH3CH2
H
δ–
O
lone pair
hydrogen
bond
δ+
H
δ+
H
290
δ–
O
O
28 T Propanone (H3C
C
CH3 ) can form hydrogen bonds with water molecules.
δ–
O
δ+
H3C
δ+
δ+
H
lone pair
H
hydrogen bond
δ–
C
O
H3C
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–
O
δ+
lone pair
δ+
H
H
hydrogen bond
δ–
O
H3C
CH3
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.
Multiple choice questions
31 C Within each group, electronegativity decreases with increasing atomic number and metallic character.
i.e. the electronegativity of halogens decreases down the group.
The electronegativity difference between F and I is the greatest.
Thus the I–F bond is the most polar.
32 D C is the least electronegative among F, Cl, N and C.
Thus C–H bond is the least polar.
291
33 A
Element
W
X
Y
Z
Atomic number
8
12
14
16
oxygen
magnesium
silicon
sulphur
Name of element
W (oxygen) is the most electronegative.
34 B
Option
Molecule
A
CO2
B
NH3
Three-dimensional structure
O
C
O
N
H
H
H
Cl
C
CCl4
C
Cl
Cl
Cl
H
D
CH4
C
H
H
H
Option B — The electronegativity value of nitrogen is greater than that of hydrogen. So each N–H
bond is polar.
The NH 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 CCl4 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.
292
36 C Options A and C — The three-dimensional structures of CHCl3 and CHF3 are shown below:
H
H
C
Cl
Cl
net dipole moment
C
net dipole moment
F
F
Cl
F
The electronegativity difference between C and F is greater than that between C
and Cl.
Thus C–F bond is more polar than C–Cl bond.
∴ CHF3 has a greater dipole moment than CHCl3.
Option D — The electronegativity value of chlorine is greater than that of carbon. So each C–Cl bond
is polar.
Due to the symmetry of the tetrahedral shape of the CCl4 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
F
F
SiF4
F
Si
Shape
F
tetrahedral
Si
F
F
F
F
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 SiF4 molecule has zero dipole moment.
38 B Options B and C — The electronegativity difference between H and Cl (2.1 and 3.0) is greater than
that beween N and O (3.0 and 3.5).
∴ HCl has a greater net dipole moment than NO.
Option D —
Molecule
SO3
Electron diagram
(Only electrons in the outermost Three-dimensional structure
shells are shown.)
O
O
O
S
Shape
O
S
O
O
trigonal
planar
293
The electronegativity value of oxygen is greater than that of sulphur. So each S–O bond is
polar.
Due to the symmetry of the shape of the SO3 molecule, the individual bond dipole
moments cancel one another out exactly.
Thus the SO3 molecule has no net dipole moment.
39 C Option C —
Molecule
Electron diagram
(Only electrons in the outermost Three-dimensional structure
shells are shown.)
O
OF2
F
δ+
O
F
δ–
δ–
Shape
V-shaped
F
F
The electronegativity value of fluorine is greater than that of oxygen.
So the fluorine end has a partial negative charge while the oxygen end has a partial
positive charge.
40 D
Option
Molecule
Three-dimensional structure
H
A
CHCl3
δ+
δ–
C
Cl
net dipole moment
δ–
δ–
Cl
Cl
net dipole moment
B
HCl
δ+
δ–
H
C
H2O
Cl
net dipole moment
δ–
O
δ+
δ+
H
H
D
CS2
S
C
S
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.
294
41 A
Option
Molecule
A
H2S
Three-dimensional structure
net dipole moment
S
H
H
B
CO2
O
O no net dipole moment
C
Cl
C
CCl4
no net dipole moment
C
Cl
Cl
Cl
F
D
BF3
no net dipole moment
B
F
F
∴ the H2S molecule has a net dipole moment and is polar.
42 D Option D — The electronegativity value of chlorine is greater than that of phosphorus. So each P–Cl
bond is polar.
Due to the symmetry of the shape of the PCl5 molecule, the individual bond dipole
moments cancel one another out exactly.
The PCl5 molecule has no net dipole moment and it is non-polar.
43 B
Option
Molecule
Three-dimensional structure
F
A
no net dipole moment
Si
SiF4
F
F
F
F
F
B
SF4
net dipole moment
S
F
F
C
BeF2
F
Be
F
no net dipole moment
F
D
BF3
no net dipole moment
B
F
F
∴ only the SF4 molecule has a net dipole moment.
295
44 C Option C — The electronegativity value of chlorine is greater than that of germanium. So each Ge–Cl
bond is polar.
Due to the symmetry of the tetrahedral shape of the GeCl4 molecule, the individual bond
dipole moments cancel one another out exactly.
The GeCl4 molecule has NO net dipole moment.
45 D Option D — The electronegativity value of fluorine is greater than that of beryllium. So each Be–F
bond is polar.
As the polar Be–F bonds of the BeF2 molecule are arranged in a straight line, the two
identical bond dipole moments cancel each other out exactly.
As a result, the BeF2 molecule has no net dipole moment.
So a BeF2 molecule is non-polar.
46 D Option D — The electronegativity value of chlorine is greater than that of carbon. So each C–Cl bond
is polar.
Due to the symmetry of the tetrahedral shape of the CCl4 molecule, the individual bond
dipole moments cancel one another out exactly.
The CCl4 molecule has no net dipole moment and it is non-polar.
47 C
Option
Molecule
A
BeF2
Electron diagram
(Only electrons in the outermost
shells are shown.)
F Be F
Three-dimensional structure
F
Be
F
no net dipole moment
NO2
O N O
B
net dipole moment
N
O
O
HCl
net dipole moment
H Cl
H
Cl
CO2
O C O
C
O
O no net dipole moment
C
NO2
O N O
HCl
H Cl
net dipole moment
N
O
O
net dipole moment
H
296
Cl
D
BeF2
F Be F
F
Be
F
O C O
O
C
O no net dipole moment
no net dipole moment
CO2
∴ both NO2 and HCl molecules have a net dipole moment.
48 A
Option
Molecule I
Cl
A
Molecule II
Cl
Cl
C
net dipole moment
C
H3C
CH3
Remark
CH3
C
C
H3C
Cl
no net dipole moment
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 NH3 will be attracted towards the nitrogen atom to a greater
extent.
These bond pairs will repel each other to a greater extent.
Thus the H–N–H bond angle in NH3 is greater than the H–P–H bond angle in PH3.
Molecule
Bond angle
PH3
93°
H2O
104.5°
NH3
107°
∴ the order of increasing bond angle is PH3 < H2O < NH3.
50 C Option C — Tetrachloromethane is a non-polar substance. There are only instantaneous dipole-induced
dipole attractions between CCl4 molecules.
51 D
52 B Although H–Cl bond is polar, the attraction between the partially positively charged hydrogen atom
and the lone pair on the chlorine atom of another HCl molecule is not strong.
The chlorine atom is quite large and the lone pairs are not very accessible to the hydrogen atom.
Thus hydrogen bonds do NOT exist in liquid HCl.
53 A
54 C In both ice and water, hydrogen bond is the strongest intermolecular attraction.
In ice, each water molecule forms four hydrogen bonds tetrahedrally.
The highly ordered structure leads to a very ‘open’ structure with large spaces in ice.
297
Thus ice has a lower density.
The ‘open’ structure collapses when ice melts to form liquid water.
The water molecules can pack more closely, so liquid water has a higher density.
55 B Diamond has a giant covalent structure.
56 A
57 C Option C — ICl is polar.
The electronegativity value of chlorine is greater than that of iodine.
So the chlorine end has a partial negative charge while the iodine end has a partial
positive charge.
Hence the ICl molecule has a net dipole moment.
58 B The boiling point of an element depends on the strength of its intermolecular attractions.
The intermolecular attractions in the liquids are van der Waals’ forces.
Among CO2, H2, N2 and O2, a H2 molecule contains the smallest number of electrons.
Hence the van der Waals’ forces in liquid H2 is the weakest.
The least amount of heat is needed to separate the H2 molecules during boiling.
Hence liquid H2 is the most volatile.
59 B Option B — CH4, SiH4 and SnH4 are hydrides of Group IV elements.
The boiling point of a hydride depends on the strength of its intermolecular attractions.
The intermolecular attractions in the hydrides are van der Waals’ forces.
The number of electrons in one hydride molecule is in the order CH4 < SiH4 < SnH4.
Hence the strength of van der Waals’ forces in the hydrides is in the same order.
More heat, in their increasing order, is needed to separate the molecules during boiling,
and thus the order of the boiling points of the hydrides is CH4 < SiH4 < SnH4.
60 C Option C — The boiling point of H2S is higher than that of SiH4.
The boiling point of a compound depends on the strength of its intermolecular
attractions.
H2S is a polar substance. There are permanent dipole-permanent dipole attractions and
instantaneous dipole-induced dipole attractions between H2S molecules.
SiH 4 is a non-polar substance. There are only instantaneous dipole-induced dipole
attractions between SiH4 molecules.
More heat is needed to separate the H2S molecules during boiling.
298
61 B The zig-zag polymeric structure in solid HF is due to hydrogen bonding:
hydrogen
bond
F
H
H
F
H
H
F
F
H
F
H
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
63 D Option D — In a H
H
O
C
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 H2O while only van der Waals’ forces exist in H2S.
Hydrogen bonds are stronger than van der Waals’ forces.
More heat is needed to separate the H2O molecules during boiling.
Thus the boiling point of H2O is higher than that of H2S.
299
66 B The boiling point of a compound depends on the strength of its intermolecular attractions.
Both X and Y are non-polar. Relatively weak van der Waals’ forces exist in them.
Molecule of X is longer and somewhat spread-out whereas that of Y is more spherical and compact.
The shape of molecule of X allows greater surface contact between molecules.
The van der Waals’ forces in X are thus stronger.
More heat is needed to separate molecules of X during boiling.
Hence X has a higher boiling point than Y.
Hydrogen bonds exist in Z.
Hence Z has the highest boiling point.
67 D Option D — C2H5OH is miscible with water (i.e. soluble in water in all proportions) because it can form
hydrogen bonds with water.
δ+
CH3CH2
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.
300
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 H2 and N2.
The strength of the forces increases with the number of electrons in one molecule.
Hence the boiling point of N2 is higher than that of H2.
Hydrogen bonds exist in NH3.
Hence NH3 has the highest boiling point.
∴ the order of increasing boiling point is H2 < N2 < NH3.
72 D The viscosity of the compound in Option D is the highest.
All the compounds can form hydrogen bonds.
Each molecule of the compound in Option D has three –OH groups that can take part in hydrogen
bonding while each molecule of other compounds has only one or two –OH groups. Each molecule of
the compound in Option D can form more hydrogen bonds.
Furthermore, because of their shapes, molecules of the compound in Option D tend to become
entangled rather than to slide past one another.
These factors contribute to the high viscosity of the compound in Option D.
73 D
74 B (2) The electronegativity value of fluorine is greater than that of boron. So each B–F bond is polar.
A BF3 molecular has a trigonal planar shape.
The three identical B–F bond dipole moments in a BF3 molecule cancel one another out exactly.
As a result, the BF3 molecule has no net dipole moment and is non-polar.
(3) The electronegativity value of oxygen is greater than that of sulphur. So each S–O bond is polar.
A sulphur dioxide molecule is V-shaped. The individual S–O bond dipole moments reinforce each
other.
As a result, the SO2 molecule has a net dipole moment and is polar.
301
75 B (1) The electronegativity values of C and S are the same. Thus C=S bond is non-polar.
∴ the CS2 molecule is non-polar.
(2) The electronegativity value of S is greater than that of H. So each S–H bond is polar.
A H2S molecule is V-shaped. The individual S–H bond dipole moments reinforce each other.
Hence the molecule has a net dipole moment.
∴ the H2S molecule is polar.
(3) The electronegativity value of Cl is greater than that of P. So each P–Cl bond is polar.
Due to the symmetry of the trigonal bipyramidal shape of the PCl5 molecule, the individual bond
dipole moments cancel one another out exactly.
The molecule has no net dipole moment.
∴ the PCl5 molecule is non-polar.
76 A (1) The electronegativity value of nitrogen is greater than that of hydrogen. So each N–H bond is polar.
The NH3 molecule has a trigonal pyramidal shape. The individual N–H bond dipole moments
reinforce each other.
Hence the molecule has a net dipole moment.
∴ the NH3 molecule is polar.
(2) The electronegativity value of Cl is greater than that of S. So each S–Cl bond is polar.
A SCl2 molecule is V-shaped. The individual S–Cl bond dipole moments reinforce each other.
Hence the molecule has a net dipole moment.
∴ the SCl2 molecule is polar.
(3) The electronegativity value of F is greater than that of Xe. So each Xe–F bond is polar.
Due to the symmetry of the square planar shape of the XeF4 molecule, the individual bond dipole
moments cancel one another out exactly.
The molecule has no net dipole moment.
∴ the XeF4 molecule is non-polar.
77 A (3) Across the second period, the melting points of elements rise to Group IV and then fall to low
values.
78 A Option A — The sulphur atom has 6 electrons in its outermost shell and each oxygen atom has 6, i.e.
a SO3 molecule has a total of 24 electrons in the outermost shells.
302
79 B (1) Three-dimensional structure of a SO3 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 SO3 molecule has a trigonal planar shape.
(2) The electronegativity value of O is greater than that of S. So a SO3 molecule contains polar bonds.
(3) Due to the symmetry of the trigonal planar shape, the bond dipole moments cancel one another
out exactly.
Hence the SO3 molecule has no net dipole moment and it is non-polar.
80 B Option B — The chlorine atom has 7 electrons in its outermost shell and each fluorine atom has 7, i.e.
a ClF3 molecule has a total of 28 electrons in the outermost shells.
81 A (1) and (3) In a ClF3 molecule, the 5 pairs of electrons around the central Cl atom will adopt a
trigonal bipyramidal arrangement.
There are three different ways in which we may arrange the 3 bonding pairs and 2 lone
pairs into a trigonal bipyramid. The correct shape is the one with the minimum repulsion.
F
F
90°
F
Cl
F
Cl
F
F
Shape
120°
90°
Cl
F
90°
F
I
90°
90°
F
II
Repulsion present
I
90° lone pair-lone pair repulsion
II
6 90° lone pair-bond pair repulsions
(repulsions at angles greater than 90°
can be ignored)
III
4 90° lone pair-bond pair repulsions +
2 90° bond pair-bond pair repulsions
(repulsions at angles greater than 90°
can be ignored)
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
∴ the ClF3 molecule is T-shaped.
(2) The electronegativity value of F is greater than that of Cl. So each Cl–F bond is polar.
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
C
net dipole moment
Cl
Cl
Cl
∴ the molecule is polar and thus the liquid is deflected by the negatively charged rod.
(2) and (3) Both molecules are non-polar.
∴ the liquids are NOT deflected by the negatively charged rod.
84 A (1) The electronegatvity value of Cl is higher than that of Si and Ge. So the Si–Cl and Ge–Cl bonds
are polar.
(3) Due to the symmetry of the tetrahedral shape of the SiCl4 and GeCl4 molecules, the individual
bond dipole moments cancel one another out exactly.
The molecules has no net dipole moment and they are non-polar.
The boiling point of a compound depends on the strength of its intermolecular attractions.
Both SiCl4 and GeCl4 are non-polar. Relatively weak instantaneous dipole-induced dipole attractions
exist in them.
The strength of the attractions increases with the number of electrons in one molecule.
Hence the boiling point of GeCl4 is higher than that of SiCl4.
85 C (1) The electronegativity value of nitrogen is greater than that of hydrogen. So each N–H bond is polar.
The NH3 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 NH3 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 NH3 is
greater than the H–P–H bond angle in PH3.
(3) NH3 can form hydrogen bonds but PH3 cannot.
More heat is needed to separate the NH3 molecules during boiling.
Thus the boiling point of NH3 is higher than that of PH3.
86 A (1) Both X and Y have the same molecular formula, C5H12.
(3) The boiling point of a compound depends on the strength of its intermolecular attractions.
Molecule of Y is longer and somewhat spread-out whereas that of X is more spherical and
compact.
The shape of molecule of Y allows greater surface contact between molecules.
The van der Waals’ forces in Y are thus stronger.
More heat is needed to separate molecules of Y during boiling.
Hence Y has a higher boiling point than X.
87 C (2) and (3) H
H
H
C
C
H and H
O
H
O
C
C
O
H molecules can form hydrogen bonds
H
O
δ–
H
δ+
H
H
with water molecules.
hydrogen bond
δ+
δ+
CH3CH2
H
H
δ–
O
lone pair
lone pair
δ–
H
δ–
H
O
H
hydrogen bond
δ+
δ+
C
H
H
C
δ–
δ+
H
O
δ–
O
δ+
H
H
hydrogen
bond
O
δ+
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
90 D (2) and (3) In a methoxymethane molecule, the C–O bonds are polar.
H
H
O
H
C
C
H
H
H
net dipole moment
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.
lone pair
δ–
O
H
H
hydrogen bond
δ+
δ–
O
C
C
δ–
O
H
δ+
H
δ–
O
hydrogen bond
92 C (3) Electron diagram of a HNO3 molecule:
O
H O N
(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 HNO3 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.
306
O
δ–
H
δ+
95 A (1) Methanoic acid molecules can form hydrogen bonds with water molecules.
H
δ+
hydrogen bond
lone pair
δ–
O
H
δ+
hydrogen bond
H
δ+
δ–
H
O
δ–
O
δ+
C
H
(2) Methoxymethane molecules can form hydrogen bonds with water molecules.
δ–
O
δ+
lone pair
H
δ+
H
hydrogen bond
δ–
O
H3C
CH3
OH
NO2
96 B (2) Intramolecular hydrogen bonds occur in
.
O
N
O
O
H
key:
hydrogen bond
97 A (3) The boiling point of a compound depends on the strength of its intermolecular attractions.
Hydrogen bonds exist in ethane-1,2-diol while only van der Waals’ forces exist in propane.
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.
Van der Waals’ forces in Y are stronger than those in X.
∴ Y is more viscous than X.
308
101 B The electronegativity of an element represents the power of an atom of that element to attract a
bonding pair of electrons towards itself in a molecule.
102 A In N–H bond, the nitrogen atom gets a slightly negative charge because it has a greater share of the
bonding electrons. The hydrogen atom becomes slightly positively charged because it has lost some of
its share in the bonding electrons.
103 D In NH3, the bond pairs of electrons are attracted towards the nitrogen atom to a greater extent as
nitrogen is more electronegative than hydrogen.
The bond pairs of electrons repel each other to a greater extent and thus the H–N–H bond angle is
greater.
In NF3, the bond pairs of electrons are closer to the fluorine atom as fluorine is more electronegative
than nitrogen.
The bond pairs of electrons repel each other to a less extent and thus the F–N–F bond angle is smaller.
The atomic size of fluorine is smaller than that of nitrogen.
104 B Within each group, electronegativity decreases with increasing atomic number and metallic character.
The electronegativity value of Cl is greater than that of I.
The electronegativity difference between H and Cl is greater than that between H and I.
Thus H–Cl bond is more polar than H–I bond.
105 C The electronegativity value of fluorine is greater than that of boron. So each B–F bond is polar.
A BF3 molecular has a trigonal planar shape. The three identical B–F bond dipole moments in a BF3
molecule cancel one another out exactly.
As a result, the BF3 molecule has no net dipole moment and is non-polar.
106 D Three dimensional shape of a SF4 molecule:
F
F
S
F
F
The electronegativity value of fluorine is greater than that of sulphur. So each S–F bond is polar.
The SF4 molecule has a seesaw shape. The individual S–F bond dipole moments reinforce each other.
The SF4 molecule has a net dipole moment.
107 C The electronegativity value of Cl is greater than that of Si. So each Si–Cl bond is polar.
Due to the symmetry of the tetrahedral shape of the SiCl4 molecule, the individual bond dipole
moments cancel one another out exactly.
The molecule has no net dipole moment and it is non-polar.
∴ NO permanent dipole-permanent dipole attractions exist between silicon tetrachloride molecules.
309
108 A Propanone molecule is polar.
O
H
C
δ–
δ+
C
H
H
net dipole moment
C
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 NH3 molecules can form hydrogen bonds with water molecules.
δ+
lone pair
H
δ–
N
δ+
H
δ+
H
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.
310
Part B
Topic-based exercise
Multiple choice questions
1
B Option B — The xenon atom has 8 electrons in its outermost shell and each fluorine atom has 7, i.e.
a XeF4 molecule has a total of 36 electrons in the outermost shells.
2
A Option A — The sulphur atom has 6 electrons in its outermost shell and each fluroine atom has 7, i.e.
a SF4 molecule has a total of 34 electrons in the outermost shells.
3
D Option A — An oxygen atom cannot have more than 8 electrons in its outermost shell.
Option B — A nitrogen atom cannot have more than 8 electrons in its outermost shell.
Opiton D — The nitrogen atom has 5 electrons in its outermost shell and each oxygen atom has 6
electrons. But the NO3– ion carries one negative charge because it has gained 1 electron.
Thus a NO3– ion has a total of 24 electrons in the outermost shells.
4
D Option B — A carbon atom cannot have more than 8 electrons in its outermost shell.
Option D — The sulphur atom has 6 electrons in its outermost shell, the carbon atom has 4 and the
nitrogen atom has 5. But the SCN– ion carries one negative charge because it has gained
1 electron. Thus a SCN– ion has a total of 16 electrons in the outermost shells.
5
A Option A — NCl 5 does NOT exist because nitrogen cannot form compounds with more than 8
electrons in the outermost shell of its atom.
6
C Option C — Consider a molecule with one lone pair and three bond pairs of electrons in the
outermost shell of the central atom.
The 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 XeO3 molecule:
O
O Xe O
(Only electrons in the outermost shells are shown.)
A XeO3 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 XeO3
molecule has a trigonal pyramidal shape.
311
8
C Option C — Electron diagram of a PCl4+ 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 PCl4+ ion has a tetrahedral shape.
9
B Option B — Consider a molecule with one lone pair and five bond pairs of electrons in the outermost
shell of the central atom.
The 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.
PDUBIFESBM
The shape of a molecule is determined only by the arrangement of atoms. Thus, the
molecule has a square pyramidal shape.
10 D
Option
Species
A
BF3
Electron diagram
(Only electrons in the outermost shells are shown.)
F
F
B
F
2–
O
B
CO32–
O
C
O
–
O
C
NO3–
O
D
PCl3
N
O
Cl P Cl
Cl
∴ a PCl3 molecule has one lone pair of electrons in the outermosst shell of the central atom.
312
11 C
Option
Species
A
C2H4
Electron diagram
(Only electrons in the outermost
shells are shown.)
H
C
C
H
H
Shape
H
H
H
C
Three-dimensional structure
C
planar
H
H
+
+
B
H O H
+
H3O
trigonal
pyramidal
O
H
H
H
H
C
S
SF2
F
V-shaped
S
F
F
F
F
D
XeF2
Xe
F
Xe
F
linear
F
∴ a SF2 molecule has a V-shape.
12 C
Option
A
Species
Electron diagram
(Only electrons in the outermost
shells are shown.)
BH3
H
H
H
B
B
H
2–
CO3
O
C
PF3
F
C
P
O
F
H
2–
O
trigonal
planar
C
O
F
Shape
trigonal
planar
H
2–
O
B
Three-dimensional structure
O
P
F
F
trigonal
pyramidal
F
O
D
SO3
O
O
S
O
S
O
O
trigonal
planar
∴ a PF3 molecule has a trigonal pyramidal shape.
313
13 A
Option
Species
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
A
I3
–
I
I
–
I
–
Shape
I
I
linear
I
B
O3
O
O O
O
C
O
OF2
V-shaped
O
F
O
V-shaped
O
F
F
F
D
S
SO2
O
V-shaped
S
O
O
O
∴ an I3– ion has a linear shape.
14 C Option C — Electron diagram of a NCl3 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.
Option 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
its outermost shell.
C
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
314
∴ the H
C
H molecule is planar.
15 A
Option
Molecule
A
BeCl2
Electron diagram
(Only electrons in the outermost
shells are shown.)
Cl Be Cl
Three-dimensional structure
Cl
Be
XeF
Cl
Shape
linear
F
Xe
F
linear
Xe
F
F
B
BF
F
F
F
B
trigonal
planar
B
F
F
F
PCl
Cl P Cl
P
Cl
Cl
Cl
trigonal
pyramidal
Cl
C
PF
P
F
F
F
F
F
P
F
F
F
trigonal
bipyramidal
F
IF
F
F
F
F
I
F
SiF4
F
F
square
pyramidal
F
F
F
F
I
F
F
D
F
Si
F
tetrahedral
Si
F
F
F
F
SF4
F
F
F
F
F
S
S
F
seesaw
F
F
∴ molecules of BeCl and XeF have an identical shape.
6
D Option D —
Molecule
Electron diagram
(Only electrons in the outermost Three-dimensional structure
shells are shown.)
F
F
F
Xe
XeF4
F
F
Xe
F
F
F
Shape
square
planar
315
17 C
Option
Species
A
PBr3
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
Shape
P
Br
trigonal
pyramidal
Cl
trigonal
pyramidal
Br P Br
Br
Br
Br
PCl3
Cl P Cl
P
Cl
Cl
Cl
B
NH3
H N H
trigonal
pyramidal
N
H
H
H
H
BF3
F
F
B
F
C
F
F
BF3
CO3
B
D
F
F
2–
O
C
trigonal
planar
B
2–
O
F
F
F
F
trigonal
planar
B
F
2–
O
trigonal
planar
C
O
O
ClF3
O
F
F
Cl
F
Cl
F
F
T-shaped
F
PF3
F
P
F
F
P
F
F
F
∴ the BF3 molecule and CO32– ion are trigonal planar.
316
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
X
90°
M
M
X
90°
90°
X
120°
90°
M
X
90°
X
X
X
II
I
Shape
III
Repulsion present
I
90° lone pair-lone pair repulsion
II
6 90° lone pair-bond pair repulsions (repulsions
at angles greater than 90° can be ignored)
III
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 XeOF4 molecule has a total of 42 electrons in the outermost
shells.
Option C — Oxygen atom cannot have more than 8 electrons in its outermost shell.
317
21 C When using the VSEPR theory, double bonds can be treated like single bonds. Thus we can regard a
XeOF4 molecule as having one lone pair and five bond pairs of electrons in the outermost shell of the
xenon atom.
The 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 XeOF4 molecule
has a square pyramidal shape.
O
F
F
Xe
F
F
22 C
23 D Three-dimensional structure of a XeF4 molecule:
F
90°
F
Xe
F
F
∴ the molecule contains 90° lone pair-bond pair repulsions.
24 C
Option
Species
A
H2O
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
Bond angle
O
104.5°
H O H
H
H
B
CH3
H
+
CO2
D
SO2
O C O
S
O
O
H
C
O
O
∴ a CO2 molecule has the greatest angle between two covalent bonds.
180°
~120°
S
O
318
120°
C
H C H
H
C
+
H
+
O
O
25 C
α
H
N
C
H
N
β
H
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°.
26 B Option B — In an AlCl3 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 an AlCl3 molecule has a trigonal planar shape.
In the adduct, the nitrogen atom in the NH3 molecule supplies a lone pair of electrons to
the aluminium atom, forming a dative covalent bond.
Hence there are four bond pairs of electrons in the outermost shell of the aluminium
atom in the adduct.
The furthest apart the four pairs of electrons can get is when they are arranged in a
tetrahedral shape.
Cl
H
H
N
H
+
Al
Cl
Cl
H
H
Cl
N
Al
H
Cl
Cl
319
27 B
Option
Species
Three-dimensional structure
A
SiH4
H
Si
Bond angle(s)
109.5°
H
H
H
NH3
107°
N
H
H
H
B
SO2
~120°
S
O
C
O
CS2
S
C
S
180°
CS2
S
C
S
180°
SF4
F
F
90°, 120°, 180°
S
F
F
D
H2O
104.5°
O
H
H
H2S
92.5°
S
H
H
Option D — The electronegativity value of oxygen is greater than that of sulphur.
Hence the bond pairs of electrons in H2O will be attracted towards the oxygen atom to
a greater extent. These bond pairs will repel each other to a greater extent and the bond
angle in H2O is greater than the bond angle in H2S.
∴ in Option B, the bond angle around the central atom in the second species (CS2) is larger than that
in the first species (SO2).
320
28 C
Molecule
Three-dimensional structure
Bond angle
F
BF3
120°
B
F
F
PF3
~109°
P
F
F
F
F
ClF3
Cl
~90°
F
F
∴ the order of increasing bond angle is ClF3 < PF3 < BF3.
29 B
30 B
Element
W
X
Y
Z
Atomic number
7
11
13
17
nitrogen
sodium
aluminium
chlorine
Name of element
X (sodium) is the least electronegative.
31 D Option D — Three-dimensional structure of a SiF4 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 SiF4 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.
Both CH4 and CCl4 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 CCl4 is higher than that of CH4.
321
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 (C7H16) is the highest.
34 D Option D — The boiling point of a compound depends on the strength of its intermolecular
attractions.
Hydrogen bonds exist in H2O while only van der Waals’ forces exist in other Group VI
hydrides.
Hydrogen bonds are stronger than van der Waals’ forces.
More heat is needed to separate the H2O molecules during boiling.
Thus the boiling point of H2O is the highest.
The number of electrons in one hydride molecule increases from H2S to H2Te.
Hence the strength of van der Waals’ forces between the molecules also increases from
H2S to H2Te.
More heat, in their increasing order, is needed to separate the molecules during boiling,
and thus the boiling points of the hydrides increase from H2S to H2Te.
35 C The boiling point of a compound depends on the strength of its intermolecular attractions.
The intermolecular attractions in HCl, HBr and HI are van der Waals’ forces.
The number of electrons in one hydride molecule increases from HCl to HI.
Hence the strength of van der Waals’ forces between the molecules also increases from HCl to HI.
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 C2H5OH is higher than that of C2H5OC2H5. This is due to the presence
of hydrogen bonds between molecules of C2H5OH.
37 D Option D — According to the VSEPR theory, electron pairs in the outermost shell of the central atom
of a molecule repel one another and stay as far apart as possible.
When using the VSEPR theory, double bonds can be treated like single bonds. Therefore
we can view the carbon atom as having three pairs of electrons in its outermost shell.
The 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.
322
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 BH3 molecule:
H
B
H
H
(Only electrons in the outermost shells are shown.)
40 B (2) BF3 molecule does NOT contain any unpaired electron.
(3) The octet rule suggests that atoms become stable by having eight electrons in their outermost shells (or
two electrons in the case of some smaller atoms). An atom attains a stable electronic arrangement
by sharing or transfer of electrons.
The NO molecule does NOT conform to the octet rule because it contains an unpaired electron.
The BF3 molecule does NOT conform to the octet rule because there are less than 8 electrons in
the outermost shell of the boron atom.
41 B
Species
Electron diagram
(Only electrons in the outermost
shells are shown.)
–
O
(3) NO3–
O
N
Three-dimensional structure
–
O
trigonal planar
N
O
O
Shape
O
∴ species X could NOT be a NO3– ion.
42 C
Species
Electron diagram
(Only electrons in the outermost
shells are shown.)
S
(1) CS2
(2) Cl2O
C
S
Three-dimensional structure
S
O Cl
C
S
linear
V-shaped
O
Cl
Shape
Cl
Cl
–
–
(3) NH2
–
H N H
V-shaped
N
H
H
∴ species X could be Cl2O / NH2–.
323
43 D
Species
Electron diagram
(Only electrons in the outermost shells are shown.)
Cl
Cl
(1) PCl5
P
Cl
Cl
Cl
F
F
(2) SF4
F
(3) XeF2
F
S
Xe
F
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
F
(1) SF4
F
F
F
F
S
F
seesaw
S
F
F
F
F
(2) SiF4
F
Si
F
F
F
tetrahedral
F
F
F
F
Xe
(3) XeF4
F
square planar
Xe
F
∴ the SiF4 and CH4 molecules have a tetrahedral shape.
324
Si
F
F
F
F
45 B
Option
Species
(1)
CO2
Electron diagram
(Only electrons in the outermost
shells are shown.)
Three-dimensional structure
O
O C O
CH
H
H
C
C
H
H
linear
H
C
C
H
O
Shape
C
planar
H
H
HO
H O
V-shaped
O
H
H
H
CS
C
S
S
S
C
S
linear
N
H
C
N
linear
HCN
H C
XeF
F
Xe
F
F
linear
Xe
F
XeF
F
Xe
F
F
linear
Xe
F
OF
F
O
F
V-shaped
O
F
F
SCl
Cl S
Cl
V-shaped
S
Cl
Cl
∴ the set
CS, HCN and XeF contains only linear species.
325
46 A
Species
Three-dimensional structure
(1) CH3NH2
N
H
CH3
H
F
(2) ClF3
Cl
F
F
F
F
(3) PF5
P
F
F
F
∴ the shapes of the CH3NH2 and ClF3 molecules are influenced by the presence of lone pairs of electrons.
47 A
Species
Three-dimensional structure
F
F
Xe
(1) XeF4
90°
F
F
F
F
90°
(2) PF5
F
P
F
F
F
(3) SiF4
Si
F
F
F
109.5°
∴ the XeF4 and PF5 molecules contain 90° bond angles.
48 B
Species
Three-dimensional structure
H
H
120°
C
(1) C2H4
C
120°
120°
H
H
(2) NF3
N
F
F
F
102°
Cl
Cl
(3) PCl5
P
Cl
120°
Cl
Cl
326
∴ the C2H4 and PCl5 molecules contain 120° bond angles.
49 A
Electron diagram
(Only electrons in the outermost shells are shown.)
Molecule
Three-dimensional structure
F
F
F
F
Br
(1) BrF5
F
F
F
F
F
F
F
F
(3) PCl5
F
Cl
F
F
Cl
Cl
P
F
Cl
F
Cl
F
F
Cl
(2) ClF5
F
Br
Cl
P
Cl
Cl
Cl
Cl
Cl
∴ the BrF5 and ClF5 molecules have the geometric arrangement shown.
50 B (1)
H2O
H+
H3O+
In the outermost electron shell of the central oxygen atom in each of the species, the numbers of
bond pairs and lone pair(s) of electrons are as follows:
Species
Number of bond pairs
Number of lone pair(s)
H3O+
3
1
H2O
2
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 H3O+ 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 H2O molecule is further compressed.
∴ bond angle: H3O+ > H2O
(2)
C2H4
H2
C2H6
327
Three-dimensional structures of the C2H4 and C2H6 molecules:
H
H
H
C
120°
C
C
H
H
120°
C
120°
H
H
H
H
H
In a C2H6 molecule, each carbon atom has four pairs of electrons in its outermost shell. The overall
arrangement of the four pairs of electrons around each carbon atom is tetrahedral. Thus, the bond
angles are 109.5°.
∴ bond angle: C2H4 > C2H6.
(3)
H+
NH3
NH4+
In the outermost electron shell of the central nitrogen atom in each of the species, the numbers of
bond pairs and lone pair(s) of electrons are as follows:
Species
Number of bond pairs
Number of lone pair(s)
+
4
4
0
NH3
3
1
NH
The electron pairs repel one another and stay as far apart as possible.
In the NH4+ ion, the furthest apart the four pairs of electrons can get is when they are arranged in
a tetrahedral shape.
Thus the NH4+ ion has a tetrahedral shape. The H–N–H bond angles are 109.5°.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the H–N–H bond angle in the NH3 molecule is compressed to 107°.
∴ bond angle: NH4+ > NH3
51 C (1) The order of atomic size of the halogen is in the order Cl < Br < I.
52 D
53 A (3) Three-dimensional structure of a SF6 molecule:
F
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 SF6 molecule, the individual bond dipole
moments cancel one another out exactly.
The SF6 molecule has no net dipole moment and it is non-polar.
328
54 B
Molecule
Three-dimensional structure
Polar bond
Symmetrical shape?
Polar molecule?
yes
no
no
yes
yes
no
Cl
δ+
(1) BCl3
Cl
Cl
δ–
(2) Cl2O
δ–
B—Cl
B
O
Cl
δ+
O—Cl
Cl
Cl
(3) SiCl4
δ+
Si
Cl
Cl
δ–
Si—Cl
Cl
∴ the BCl3 and SiCl4 molecules are non-polar.
55 B (1) Consider a molecule with two lone pairs and four bond pairs of electrons in the outermost shell of
the atom X.
The 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.
(2) Suppose each X–Y bond is polar.
Due to the symmetry of the shape of the XY4 molecule, the individual bond dipole moments cancel
one another out exactly.
The XY4 molecule has no net dipole moment and it is non-polar.
(3) The bond angles around the central atom X are 90°.
56 D (1) The 6 electron pairs in the outermost shell of the central atom X will adopt an octahedral
arrangement.
(2) There are more than 8 electrons in the outermost shell of the central atom X.
∴ molecule XY6 does not conform to the octet rule.
57 D (2) and (3) Trichloromethane molecules are polar.
∴ both instantaneous dipole-induced dipole attractions and permanent dipole-permanent
dipole attractions exist between trichloromethane molecules.
329
58 D (1) The shape of a methoxymethane is NOT linear.
O
H3C
CH3
(2) The methoxymethane molecule is polar.
δ–
O
δ+
δ+
H3C
CH3
(3) The methoxymethane molecule can form hydrogen bonds with water molecules.
δ–
O
δ+
δ+
H
H
hydrogen bond
lone pair
δ–
O
H3C
CH3
59 C (1) The propanone molecule is polar.
O
H
C
δ–
δ+
C
H
H
net dipole moment
C
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–
Cl –
δ
C
δ+
H3 C
δ+
C
H3C
330
δ–
O
H
Cl
δ–
Cl
60 D (1) and (2) Three-dimensional structures of the NF3 and NH3 molecules:
N
N
net dipole moment
F
F
net dipole moment
H
H
F
H
Both the NF3 and NH3 molecules have a trigonal pyramidal shape.
Both the NF3 and NH3 molecules are polar.
(3) In NH3, the bond pairs of electrons are attracted towards the nitrogen atom to a greater extent as
nitrogen is more electronegative than hydrogen.
The bond pairs of electrons repel each other to a greater extent and thus the H–N–H bond angle is
greater.
In NF 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 H2S molecule is not strong.
The sulphur atom is quite large and the lone pairs are not very accessible to the hydrogen atom.
∴ hydrogen bonds do NOT exist in H2S.
(2) The boiling point of a compound depends on the strength of its intermolecular attractions.
Hydrogen bonds exist in H2O while only van der Waals’ forces exist in H2S.
Hydrogen bonds are stronger than van der Waals’ forces.
More heat is needed to separate the H2O molecules during boiling.
Thus the boiling point of H2O is higher than that of H2S.
(3) The electronegativity value of oxygen is greater than that of sulphur.
Hence the bond pairs of electrons in H2O will be attracted towards the oxygen atom to a greater
extent.
These bond pairs will repel each other to a greater extent.
Thus the H–O–H bond angle in H2O is greater than the H–S–H bond angle in H2S.
62 A (1) and (2) When using the VSEPR theory, double bonds can be counted as single bonds. Therefore
we can view the carbon atom of a methanal molecule as having 3 pairs of electrons in its
outermost shell.
The 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
(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
lone pair
C
N
H
H
H molecules:
δ–
N
δ+
H3C
δ+
δ+
H
H
hydrogen bond
δ–
N
δ+
H3C
δ+
δ+
H
H
O
(3) Hydrogen bonds between H
lone pair
δ–
O
H
H
C
δ–
O
C
δ–
H
H molecules:
hydrogen bond
δ+
C
O
O
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
δ–
O
hydrogen
bond
δ+
H
δ+
CH3CH2
H
δ–
O
lone pair
hydrogen
bond
δ+
H
δ+
H
332
δ–
O
(2) Methoxymethane is fairly soluble in water because hydrogen bonds can form between
methoxymethane molecules and water molecules.
δ–
O
δ+
δ+
H
H
hydrogen bond
δ–
lone pair
O
H3C
CH3
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:
δ+
CH3CH2CH2
H
δ–
O
lone pair
δ+
hydrogen
bond
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
C
C
H
H
X
H
C
net dipole moment
no net dipole moment
C
H
Cl
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
(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
δ–
H
O
δ+
H
C
H
C
δ–
O
H
H
δ–
δ+
O
δ+
H
hydrogen bond
Hydrogen bond between molecule of Y and water molecule:
δ–
O
δ+
lone pair
H
δ+
H
hydrogen bond
δ–
O
C
H3C
334
δ+
CH3
(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 PCl5 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 XeF4 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 XeF4 molecule has
a square planar shape.
F
F
Xe
F
F
71 B
Ion
Electron diagram
(Only electrons in the outermost
shells are shown.)
2–
O
2–
CO3
O
C
O
N
O
–
O
trigonal planar
O
–
O
trigonal planar
N
O
Shape
2–
O
C
O
O
NO3–
Three-dimensional structure
O
∴ CO32– ion and NO3– ion are planar species.
335
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
Electron diagram
(Only electrons in the outermost
shells are shown.)
BeF2
F Be F
Three-dimensional structure
F
Be
F
Shape
linear
F
XeF2
F
Xe
F
Xe
linear
F
∴ BeF2 and XeF2 are linear species.
74 B The electronegativity of oxygen is higher than that of sulphur.
Hence the bond pairs of electrons in H2O will be attracted towards the oxygen atom to a greater
extent.
These bond pairs will repel each other to a greater extent.
Thus the H–O–H bond angle in H2O is greater than the H–S–H bond angle in H2S.
75 C In buckminsterfullerene, the molecules are held by weak van der Waals’ forces. The molecules can
easily slide over each other. Hence buckminsterfullerene is quite soft.
∴ buckminsterfullerene CANNOT replace diamond in cutting stones.
76 C The electronegativity value of Cl is greater than that of C. So each C–Cl bond is polar.
Due to the symmetry of the tetrahedral shape of the tetrachloromethane molecule, the individual bond
dipole moments cancel one another out exactly.
The molecule has no net dipole moment and it is non-polar.
77 A The propanone molecule is polar.
O
H
C
δ–
δ+
C
H
336
H
net dipole moment
C
H
H
H
78 B Each S–O bond is polar.
A sulphur dioxide molecule is V-shaped. The individual S–O bond dipole moments reinforce each other.
Hence the SO2 molecule has a net dipole moment.
Each C=O bond is polar.
A carbon dioxide molecule is linear in shape.
The individual C=O bond dipole moments cancel each other out exactly.
Hence the CO2 molecule has no net dipole moment.
79 A A xenon atom contains more electrons than a argon atom.
Hence the van der Waals’ forces between xenon atoms are stronger than those between argon atoms.
More heat is needed to separate the xenon atoms during boiling.
Thus the boiling point of xenon is higher than that of argon.
80 C The boiling point of a compound depends on the strength of its intermolecular attractions.
The strongest type of intermolecular attractions in H2O is hydrogen bonds.
The strongest type of intermolecular attractions in H2Se is van der Waals’ forces.
The hydrogen bonds are stronger than van der Waals’ forces.
Hence more heat is needed to separate the H2O molecules during boiling.
Thus the boiling point of H2O is higher than that of H2Se.
81 D In a methoxymethane molecule, the C–O bonds are polar.
H
H
O
H
C
C
H
H
H
net dipole moment
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–
O
δ+
δ+
H
H
hydrogen bond
δ–
lone pair
O
H3C
CH3
337
82 B Propanone can form hydrogen bonds with trichloromethane.
δ–
Cl –
δ
C
δ+
H 3C
δ+
C
Cl
δ–
Cl
H
δ–
O
H 3C
H
83 A Methanamine ( H
δ+
δ–
H
N
δ+
H
N
H
H
H ) molecules can form hydrogen bonds with water molecules.
lone pair
H3C
δ+
C
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.
338
Short questions
85
Molecular model
(a)
Geometry
square pyramidal
(1)
(b)
trigonal pyramidal
(1)
(c)
square planar
(1)
(d)
trigonal planar
(1)
(e)
trigonal bipyramidal
(1)
(f)
octahedral
(1)
339
86
Species
Electron diagram
(a) BeCl2
Cl Be Cl
Three-dimensional structure
Cl
Be
Cl
(1)
(b) HCN
H C
C
N
S
C
S
S
C
B
F
F
(1)
Si
Si
F
P
F
H
trigonal pyramidal
H
O
Cl
V-shaped
F
(1)
Cl
P
Cl
Cl
trigonal bipyramidal
Cl
Cl
(1)
340
(1)
Cl
Cl
P
(1)
F
(1)
(h) PCl5
(1)
O
F
Cl
(1)
H
(1)
F
(1)
P
H
H
(g) OF2
tetrahedral
F
F
(1)
H
(1)
F
F
(f) PH3
(1)
trigonal planar
B
F
F
F
linear
F
(1)
(e) SiF4
(1)
(1)
F
F
linear
S
(1)
(d) BF3
(1)
(1)
(1)
(c) CS2
linear
(1)
H
N
Shape
Cl
(1)
(1)
87
Species
Electron diagram
Three-dimensional structure
Shape
S
V-shaped
Cl S
(a) SCl2
Cl
Cl
Cl
(1)
(1)
(1)
F
F
(b) ClF3
Cl
F
Cl
F
F
T-shaped
F
(1)
F
F
(1)
F
F
F
Cl
(c) ClF5
(1)
F
Cl
F
F
F
square pyramidal
F
(1)
(1)
(1)
–
–
(d) ICl
–
4
Cl
Cl
Cl
I
Cl
Cl
I
square planar
Cl
Cl
Cl
(1)
(1)
(e) I3
I
I
–
I
–
–
I
I
(1)
linear
I
(1)
(1)
2–
(f) SO42–
O
S
O
δ+
δ–
δ+
δ–
b) H—Cl
c) Cl—F
δ– δ+
d) N—H
δ+
δ–
e) Si—Cl
O
O
(1)
δ+ δ–
tetrahedral
S
O
O
88 a) H—F
2–
O
O
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
341
89
Molecule
Electron diagram
Shape
Polar bond
CO2
O C O
linear
C—O
H N H
trigonal
pyramidal
N—H
(a) NH3
δ+ δ–
Symmetrical
shape?
Polar molecule?
yes
no
no
yes
δ– δ+
H
(1)
(1)
linear
linear
(1)
(d) SO2
S
O
(e) PF5
F
yes
(1)
(1)
yes
(1)
no
(1)
(1)
δ+ δ–
V-shaped
O
F
P
(1)
δ–
Be—Cl
(1)
(1)
F
(1)
δ+
Cl Be Cl
no
H—Br
(1)
(1)
(c) BeCl2
(1)
δ+ δ–
H Br
(b) HBr
(1)
(1)
trigonal
bipyramidal
F
no
S—O
(1)
yes
(1)
(1)
δ+ δ–
yes
P—F
no
F
(1)
(1)
(1)
(1)
(1)
Cl
(f) SiCl4
δ+
Cl Si Cl
tetrahedral
δ–
yes
Si—Cl
no
Cl
(1)
(1)
F
(g) XeF4
F
δ+
Xe
F
(1)
square planar
F
(1)
(1)
(1)
δ–
yes
Xe—F
(1)
(1)
no
(1)
(1)
F
F
(h) SF6
F
F
octahedral
yes
S—F
no
F
F
342
δ+ δ–
S
(1)
(1)
(1)
(1)
(1)
Structured questions
90 a)
Type of attractions
Liquid
Structure
Instantaneous
dipole-induced
dipole attractions
Permanent dipolepermanent dipole
attractions
✔
✔
Hydrogen
bonds
S
(i)
Sulphur dioxide
O
O
(0.5)
(0.5)
Cl
Cl
(ii) Tetrachloromethane
C
✔
Cl
Cl
(0.5)
H
H
(iii) Methoxymethane
H
C
O
C
H
✔
H
✔
H
(0.5)
(0.5)
O
✔
(iv) Propanone
H3C
H
(v) Ethanal
C
✔
CH3
H
O
C
C
(0.5)
✔
H
✔
H
H
(vi) Ethanoic acid
(0.5)
H
O
C
C
O
✔
H
(0.5)
b)
lone pair
δ–
H
C
O
δ–
O
H
(0.5)
(0.5)
H
O
C
δ+
✔
hydrogen bond
δ+
C
δ–
H
H
(0.5)
✔
H
H
(0.5)
C
H
δ–
O
H
hydrogen bond
(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–
O
δ+
δ+
H
H
hydrogen bond
δ–
lone pair
O
H3C
CH3
(1 mark for showing the hydrogen bond between the lone pair of oxygen and hydrogen; 1 mark for
showing the partial charges and other lone pairs of electrons)
(2)
91 a) In a BH3 molecule, there are three bond pairs of electrons in the outermost shell of the central boron
atom.
(0.5)
The electron pairs repel one another and stay as far apart as possible.
(0.5)
The shape that puts the three electron pairs furthest apart is trigonal planar.
Hence a BH3 molecule has a trigonal planar shape.
(0.5)
In an NH3 molecule, there are one lone pair and three bond pairs of electrons in the outermost shell of
the central nitrogen atom.
(0.5)
The shape that puts the four electron pairs furthest apart is tetrahedral.
(0.5)
The shape of a molecule is determined only by the arrangement of atoms.
Thus the NH3 molecule has a trigonal pyramidal shape.
b) Each S–O bond is polar.
(0.5)
(0.5)
A sulphur dioxide molecule is V-shaped.
(0.5)
The individual S–O bond dipole moments reinforce each other.
(0.5)
Hence the molecule has a net dipole moment.
Each C=O bond is polar.
(0.5)
A carbon dioxide molecule is linear in shape.
(0.5)
The individual C=O bond dipole moments cancel each other out exactly.
(0.5)
Hence the molecule has no net dipole moment.
c) The boiling point of an element depends on the strength of its intermolecular attractions.
The intermolecular attractions in halogens are van der Waals’ forces.
(1)
(1)
The number of electrons in the halogen molecule increases down the group.
Hence the strength of van der Waals’ forces between halogen molecules also increases down the
group.
(1)
More heat is needed to separate the molecules during boiling,
and thus the boiling points of the halogens increase down the group.
344
d) In both ice and water, hydrogen bond is the strongest intermolecular attraction.
(1)
In ice, each water molecule forms four hydrogen bonds tetrahedrally.
(1)
The highly ordered structure leads to a very ‘open’ structure with large spaces in ice.
(1)
Thus ice has a lower density.
The ‘open’ structure collapses when ice melts to form liquid water.
The water molecules can pack more closely,
(1)
so liquid water has a higher density.
e) In H3PO4, hydrogen bond is the strongest intermolecular attraction.
(1)
Each H3PO4 molecule has three –OH groups that can take part in hydrogen bonding with other H3PO4
molecules.
(1)
Concentrated H3PO4 has a high viscosity due to its strong intermolecular attractions.
92 a)
(1)
H
H C H
H
methane
H N H
H
ammonia
(1)
H O
(1)
H
water
(1)
b) Bond angle: CH4 > NH3 > H2O
(1)
In the outermost electron shell of the central atom in each of the molecules, the numbers of bond
pairs and lone pair(s) of electrons are as follows:
Molecule
Number of bond pairs
Number of lone pair(s)
CH4
4
0
NH3
3
1
H2O
2
2
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°.
(0.5)
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
(0.5)
Thus the H–N–H bond angle in the ammonia molecule is compressed to 107°.
(0.5)
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.
(0.5)
In the water molecule, the two lone pairs will stay the furthest apart.
As a result, the H–O–H bond angle in the water molecule is compressed to 104.5°.
(0.5)
345
c) The ’octet rule’ is commonly used to account for the formation of chemical bonds.
i) The octet rule suggests that atoms become stable by having eight electrons in their outermost shells (or
two electrons in the case of some smaller atoms). An atom attains a stable electronic arrangement
by sharing or transfer of electrons.
(1)
ii) Any one of the following:
• PCl5 / SF6
(1)
There are more than 8 electrons in the outermost shell of the phosphorus / sulphur atom.
• BeCl2 / BF3
(1)
(1)
There are less than 8 electrons in the outermost shell of the beryllium / boron atom.
93 a) α = 120°
(1)
(1)
β = 104.5° – 109.5°
(1)
b) According to the VSEPR theory, electron pairs in the outermost shell of the central atom of a molecule
repel one another and stay as far apart as possible.
(1)
When using the VSEPR theory, double bonds can be treated like single bonds. Therefore the carbon
atom has three pairs of electrons in its outermost shell.
(1)
The 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
–
–
V-shaped
N
NH2 (g)
H
H
(1)
NH3(g)
(1)
trigonal pyramidal
N
H
H
H
(1)
+
H
NH4+(g)
(1)
tetrahedral
N
H
H
H
(1)
346
(1)
b) Bond angle: NH4+(g) > NH3(g) > NH2–(g)
(1)
In the outermost electron shell of the central nitrogen atom in each of the species, the numbers of
bond pairs and lone pair(s) of electrons are as follows:
Species
Number of bond pairs
Number of lone pair(s)
+
4
NH (g)
4
0
NH3(g)
3
1
–
2
2
2
NH (g)
The electron pairs repel one another and stay as far apart as possible.
+
In the NH4 ion, the furthest apart the four pairs of electrons can get is when they are arranged in a
tetrahedral shape.
(0.5)
Thus the NH4+ 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 NH3 molecule is compressed to 107°.
(0.5)
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion.
(0.5)
In the NH2– ion, the two lone pairs will stay the furthest apart. The H–N–H bond angle in the NH2– ion
is compressed to 104.5°.
(0.5)
95 a)
F
H
C
δ–
δ+
Cl
δ–
Cl δ–
(1)
b) Fluorine and chlorine are more electronegative than carbon.
The fluorine and chlorine atoms have a greater share of the bonding electrons.
c)
(1)
(1)
F
C
Cl
H
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)
347
96 a)
F
I
Br
B
(1)
F
F
(1)
b) i) IBr is polar.
(0.5)
The electronegativity value of bromine is greater than that of iodine.
(0.5)
So the bromine end has a partial negative charge while the iodine end has a partial positive
charge.
(0.5)
Hence the molecule has a net dipole moment.
(0.5)
ii) BF3 is non-polar.
(0.5)
The electronegativity value of fluorine is greater than that of boron. So each B–F bond is polar. (0.5)
The three identical B–F bond dipole moments in a BF3 molecule cancel one another out exactly. (0.5)
As a result, the molecule has no net dipole moment.
97 a)
Hydride
(0.5)
Three-dimensional structure
Shape
H
tetrahedral
Si
SiH4
H
H
H
PH3
(1)
(1)
trigonal pyramidal
P
H
H
H
(1)
H2S
(1)
V-shaped
S
H
H
(1)
b) The boiling point of H2S is higher than that of SiH4.
The boiling point of a compound depends on the strength of its intermolecular attractions.
(1)
(1)
(1)
H2S is a polar substance. There are permanent dipole-permanent dipole attractions and instantaneous
dipole-induced dipole attractions between H2S molecules.
(1)
SiH4 is a non-polar substance. There are only instantaneous dipole-induced dipole attractions between
SiH4 molecules.
(1)
More heat is needed to separate the H2S molecules during boiling.
348
c) The electronegativity of nitrogen is higher than that of phosphorus.
(1)
Hence the bond pairs of electrons in NH3 will be attracted towards the nitrogen atom to a greater
extent.
(1)
These bond pairs will repel each other to a greater extent
(1)
and the H–N–H bond angle in NH3 is greater than the H–P–H bond angle in PH3.
d) i)
F
B
F
(1)
F
Trigonal planar
(1)
ii) H–S–H bond angle: 107°
(1)
The electron pairs in the outermost shell of the sulphur atom repel one another and stay as far apart
as possible.
(0.5)
The four pairs of electrons adopt a tetrahedral arrangement.
(0.5)
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
(0.5)
The three bond pairs around the sulphur atom is slightly compressed together,
(0.5)
making the bond angle less than 109.5°.
98 a) i)
2–
2–
O
O
O
C
or
O
C
O
O
(1)
ii)
O C O
or
O
C
O
(1)
b) i) The carbonate ion has a trigonal planar shape.
(0.5)
When using the VSEPR theory, double bonds can be counted as single bonds.
(0.5)
Therefore we can view the carbon atom as having 3 pairs of electrons in its outermost shell.
(0.5)
The electron pairs repel one another and stay as far apart as possible.
(0.5)
The shape that puts the electron pairs furthest apart is trigonal planar.
ii) The carbon dioxide molecule has a linear shape.
(0.5)
When using the VSEPR theory, we can view the carbon atom as having 2 pairs of electrons in its
outermost shell.
(0.5)
The two pairs must be at the opposite ends of a straight line in order to be as far apart as
possible.
349
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.
(0.5)
So a carbon dioxide molecule is non-polar.
99 a) Allotropes are two (or more) forms of the same element
in which the atoms or molecules are arranged in different ways.
(1)
(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.
The intermolecular forces between water molecules are strong hydrogen bonds.
(1)
(1)
(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.
ii) The electronegativity of hydrogen is lower than that of nitrogen.
b) i) A BF3 molecule has a trigonal planar shape.
(1)
(1)
(1)
In a BF3 molecule, there are three bond pairs of electrons in the outermost shell of the central
boron atom.
(1)
The electron pairs repel one another and stay as far apart as possible.
(1)
The shape that puts the electron pairs furthest apart is trigonal planar.
ii) An NH3 molecule has a trigonal pyramidal shape.
(1)
An NH3 molecule has one lone pair and three bond pairs of electrons in the outermost shell of the
nitrogen atom.
(1)
350
The four pairs of electrons will adopt a tetrahedral arrangement.
(1)
The shape of a molecule is determined only by the arrangement of atoms. Thus the NH3 molecule
has a trigonal pyramidal shape.
c) 107°
(1)
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
(1)
The three bond pairs are thus slightly compressed together.
(1)
This results in the H–N–H bond angles being 107° instead of 109.5°.
d) i) The nitrogen atom in the NH3 molecule supplies a lone pair of electrons to the boron atom,
forming a dative covalent bond.
(1)
(1)
ii)
H
H
H
N
B
F
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)
e) i)
–
N
H
H
(1)
ii) An NH2– ion has two lone pairs and two bond pairs of electrons in the outermost shell of the
nitrogen atom.
Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion,
(1)
while lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
The two lone pairs will stay the furthest apart. As a result, the H–N–H angle in the NH2– ion is
compressed to a value smaller than that in the NH3 molecule.
(1)
101 a) i)
Cl
Cl
Cl
P
Cl
Cl
ii)
(1)
Cl
Cl
P
Cl
Cl
Cl
Trigonal bipyramidal shape
(1)
(1)
351
iii) 90°
(1)
120° / 180°
(1)
iv) A phosphorus pentachloride molecule is non-polar.
(0.5)
Each P–Cl bond is polar.
(0.5)
Due to the symmetry of the shape of the molecule,
the individual bond dipole moments cancel one another out exactly.
(0.5)
The molecule has no net dipole moment
(0.5)
and it is non-polar.
+
b)
–
shape of [PCl4] :
shape of [PCl6] :
+
Cl
–
Cl
Cl
Cl
P
P
Cl
Cl
Cl
Cl
Cl
Cl
(1)
c) i)
(1)
Cl P Cl
Cl
ii)
(1)
P
Cl
Cl
Cl
Trigonal pyramidal shape
(1)
(1)
iii) A phosphorus trichloride molecule has one lone pair and three bond pairs of electrons in the
outermost shell of the phosphorus atom.
The electron pairs repel one another and stay as far apart as possible.
(1)
The four pairs of electrons in the molecule will adopt a tetrahedral arrangement.
(1)
The shape of a molecule is determined only by the arrangement of atoms. Thus, the phosphorus
trichloride molecule has a trigonal pyramidal shape.
d) Not agree
Nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its atom.
352
(1)
(1)
102 a)
Molecule
Electron diagram
F
SF2
Three-dimensional structure
S
S
F
F
F
(1)
(1)
F
SF4
F
F
F
F
S
S
F
F
F
(1)
F
F
F
F
F
S
SF6
(1)
F
S
F
F
F
F
(1)
F
F
b) Fluorine is more electronegative than sulphur.
(1)
(1)
The fluorine atom has a greater share of the bonding electrons.
c) SF6 is non-polar.
(1)
(0.5)
Due to the symmetry of the octahedral shape,
(0.5)
the six identical bond dipole moments cancel one another out exactly.
(0.5)
The molecule has no net dipole moment
(0.5)
and it is non-polar.
d) Consider the electron diagram of OF2:
F
O
F
Oxygen cannot form compounds with more than 8 electrons in the outermost shell of its atom.
(0.5)
Hence oxygen forms OF2 only.
Sulphur can form some compounds with more than 8 electrons in the outermost shell of its atom. (0.5)
Hence it can form SF4 and SF6 besides SF2.
103 a)
F
N
F
F
(1)
b) i) Tetrahedral
ii) Trigonal pyramidal
(1)
(1)
353
c) NF3 is polar.
(0.5)
The electronegativity value of fluorine is greater than that of nitrogen. So each N–F bond is polar. (0.5)
The NF3 molecule has a trigonal pyramidal shape. The individual N–F bond dipole moments reinforce
each other.
(0.5)
The molecule has a net dipole moment
(0.5)
and it is polar.
d) Not agree.
Nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its atom.
(1)
(1)
104 a) Dipole moment of a diatomic molecule is the product of the charge and the distance between the
charges.
(1)
b) The electronegativity of fluroine is much higher than that of iodine.
(1)
For HF and HI, the effect of electronegativity difference outweighs the effect of difference in the
distance between the charges.
(1)
c) i) The boiling point of a compound depends on the strength of its intermolecular attractions.
(1)
Hydrogen bonds exist in HF
(1)
while only van der Waals’ forces exist in HCl.
(1)
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.
(1)
(1)
More heat is needed to separate the HI molecules during boiling.
Thus the boiling point of HI is higher than that of HBr.
105 a) The electronegativity of an element represents the power of an atom of that element to attract a
bonding pair of electrons towards itself in a molecule.
(1)
b) i) Fluorine is more electronegative than hydrogen.
The fluorine atom has a greater share of the bonding electrons.
ii) NH3
(1)
(1)
(1)
iii) Nitrogen is the least electronegative among nitrogen, oxygen and fluorine. /
Nitrogen and hydrogen have the smallest electronegativity difference.
354
(1)
c) i) (1) instantaneous dipole-induced dipole attractions
(1)
(2) permanent dipole-permanent dipole attractions
(1)
(3) hydrogen bonds
(1)
ii) (1) There is a large difference in electronegativity between H and F.
(1)
In HF, the hydrogen atom has a stronger positive charge.
(1)
The attraction between the hydrogen atom of one molecule and a lone pair on the fluorine
atom of another molecule is known as a hydrogen bond.
(1)
(2)
δ–
F
lone pair
δ+
H
δ+
H
hydrogen bond
δ–
F
(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.
(1)
The oxygen atom in a water molecule has two lone pairs and there are two hydrogen atoms in each
molecule,
so each H2O molecule can take part in hydrogen bonding to twice the extent.
(1)
More heat is needed to separate the water molecules during boiling.
106 a) A covalent bond in which the electrons are unequally shared between the two bonded atoms.
b) Fluorine is more electronegative than carbon.
(1)
The fluorine atom has a greater share of the bonding electrons.
(1)
c) i) B
(1)
ii)
F
H
C
H
iii)
(1)
F
H
F
(1)
F
F
C
C
H
C
C
H
(1)
F
C
H
(1)
355
107 a)
XBUFS
XBUFS
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)
(0.5)
δ+
H
δ–
δ–
O
δ+
H
O
δ+
H
δ+
lone pair
H
δ+
H
hydrogen bond
δ–
O
δ+
H
δ–
O
δ+
δ+
H
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)
+
c) i)
O
H
H
H
ii) 107°
356
(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)
ii) Hydrogen bonding is the attraction between the hydrogen atom attached to an electronegative
atom
(1)
and the lone pair on another electronegative atom (usually oxygen, nitrogen or fluorine).
c)
δ+
H
lone pair
δ–
H
δ+
N
δ+
δ+
H
(1)
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.
The strongest type of intermolecular attractions in water is hydrogen bonds.
(1)
(1)
The strongest type of intermolecular attractions in hydrogen sulphide is permanent dipole-permanent
dipole attractions / van der Waals’ forces.
(1)
The hydrogen bonds are stronger than permanent dipole-permanent dipole attractions / van der Waals’
forces.
(1)
Hence more heat is needed to separate the water molecules during boiling.
e) In a molecule of X, both –Cl groups are on the same side of the carbon-carbon double bond. The
molecule has a net dipole moment and it is polar.
(1)
These molecules are held together by permanent dipole-permanent dipole attractions.
(1)
In a molecule of Y, the bond dipole moments cancel one another out and there is no net dipole
moment.
(1)
These molecules are held together by instantaneous dipole-induced dipole attractions.
(1)
Less heat is required to separate the molecules of Y during boiling. Hence Y has a lower boiling point
than X.
109 a) The boiling point of a compound depends on the strength of its intermolecular attractions.
(1)
Molecule of X is longer and somewhat spread-out whereas that of Z is more spherical and compact. (1)
The shape of molecule of X allows greater surface contact between molecules.
(1)
The van der Waals’ forces in X are thus stronger.
(1)
More heat is needed to separate molecules of X during boiling.
Hence X has a higher boiling point than Z.
357
b) Y has a structure in between the linear structure of X and the spherical structure of Z.
(1)
The van der Waals’ forces in Y are probably in between those in X and Z.
(1)
Hence the boiling point of Y is probably in between those of X and Z.
(1)
c) Stronger van der Waals’ forces in X pull the molecules closer together.
(1)
So, the density of X is higher.
d) X is insoluble in water.
(1)
X is non-polar.
(1)
X does not mix with water due to the difference in the strength of intermolecular attractions between
water molecules and those between molecules of X.
(1)
110 a) i) Van der Waals’ forces / instantaneous dipole-induced dipole attractions
ii) They have the same number of electrons per molecule.
b) i) H–C–H bond angle: 109.5°
(1)
(1)
(1)
C–C–O bond angle: 120°
(1)
ii) Permanent dipole-permanent dipole attractions
(1)
c) Propanone molecules can form hydrogen bonds with water molecules,
(1)
but butane molecules cannot.
d) i) C–O–H bond angle: 104.5°
(1)
ii) Hydrogen bonds
(1)
iii)
δ+
δ–
lone pair
H
O
C3H7
δ+
C3H7
H
δ–
hydrogen bond
O
(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)
This allows greater surface contact between molecules.
(1)
Van der Waals’ forces in propan-1-ol are stronger than those in propan-2-ol.
(1)
More heat is needed to separate the propan-1-ol molecules during boiling.
358
(1)
111 a) Boiling point increases in the following order:
D < C < A < B
(1)
The boiling point of a compound depends on the strength of its intermolecular attractions.
(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)
δ+
CH3CH2
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.
(1)
Both compounds X and Y can form hydrogen bonds.
Each molecule of Y has three –OH groups that can take part in hydrogen bonding while each molecule
of X has only two –OH groups. Each molecule of Y can form more hydrogen bonds.
(1)
Furthermore, because of their shapes, the molecules of Y tend to become entangled rather than to
slide past one another.
(1)
These factors contribute to the high viscosity of compound Y.
112 a) Carbon monoxide
(1)
b) The octet rule suggests that atoms become stable by having eight electrons in their outermost shells (or
two electrons in the case of some smaller atoms). An atom attains a stable electronic arrangement by
sharing or transfer of electrons.
(1)
359
c) i)
O N O
(1)
ii) Dative covalent bond
(1)
d) Any one of the following:
• PCl5 / SF6
(1)
There are more than 8 electrons in the outermost shell of the phosphorus / sulphur atom.
• BeCl2 / BF3
(1)
(1)
There are less than 8 electrons in the outermost shell of the beryllium / boron atom.
e) 4NO2(g) + O2(g) + 2H2O(l)
4HNO3(aq)
f) i)
(1)
(1)
S
O
O
(1)
ii)
S
O
O
(1)
iii) Sulphur dioxide is polar.
(0.5)
Each S–O bond is polar.
(0.5)
The sulphur dioxide molecule has a V-shape. The individual S–O bond dipole moments reinforce
each other.
(0.5)
The molecule has a net dipole moment
(0.5)
and it is polar.
113 a) Any two of the following:
• Non-toxic
(1)
• Non-flammable / non-explosive
(1)
• High potency / effectiveness as an anaesthetic
(1)
b)
N
N O
or
N
N
O
(1)
c) In an ethoxyethane molecule, the C–O bonds are polar.
H
H
H
H
O
H
360
C
C
C
C
H
H
H
H
H
net dipole moment
(1)
The molecule has a net dipole moment
(1)
and thus ethoxyethane is a polar molecule.
d) The boiling point of a compound depends on the strength of its intermolecular attractions.
(1)
Only relatively weak instantaneous dipole-induced dipole attractions and permanent dipole-permanent
dipole attractions exist in ethoxyethane.
(1)
Not too much heat is needed to separate the molecules during boiling.
e) i)
δ–
O
δ+
H
lone pair
δ+
H
hydrogen bond
δ–
H
H
C
H
C
H
H
H
C
H
H
C
H
H
O
(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.
(1)
A hydrogen bond forms between the hydrogen atom of a water molecule
and the lone pair of electrons on the highly electronegative oxygen atom of an ethoxyethane
molecule.
(1)
f) A trichloromethane molecule is polar.
(0.5)
Each C–Cl bond is polar.
(0.5)
The individual C–Cl bond dipole moments reinforce each other.
(0.5)
The molecule has a net dipole moment
(0.5)
and it is polar.
114 The shape of a molecule can be predicted by using what is called the valence-shell electron-pair repulsion
(VSEPR) theory.
(1)
The basis of the VSEPR theory is that electron pairs in the valence shell (i.e. outermost electron shell)
of the central atom of a molecule repel one another and stay as far apart as possible, thus causing the
molecule to assume a specific shape.
(1)
The VSEPR theory states that the repulsion decreases in the following order:
lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion
(1)
361
In a beryllium chloride molecule,
there are two bond pairs of electrons in the outermost shell of the central beryllium atom.
(0.5)
The electron pairs must be at opposite ends of a straight line in order to be as far apart as possible.
Hence the molecule has a linear shape.
(0.5)
In a boron trifluoride molecule, there are three bond pairs of electrons in the outermost shell of the
central boron atom.
(0.5)
The shape that put the three electron pairs furthest apart is trigonal planar. Hence the molecule has a
trigonal planar shape.
(0.5)
In a methane molecule, there are four bond pairs of electrons in the outermost shell of the central carbon
atom.
(0.5)
The shape that puts the four electron pairs furthest apart is tetrahedral. Hence the molecule has a
tetrahedral shape.
(0.5)
(3 marks for organization and presentation)
115 Halogens exist as diatomic molecules. Averaged over time, the distribution of electrons throughout a
halogen molecule is symmetrical.
(0.5)
The electrons in a halogen molecule are in constant motion. At a particular instant, there may be more
electrons at one end of the molecule than at the other;
(0.5)
instantaneously the molecule has developed a dipole, i.e. an instantaneous dipole.
(0.5)
The instantaneous dipole can affect the electron distributions in neighbouring molecules
(0.5)
and induce similar dipoles in these neighbours.
(0.5)
Attractions exist between the instantaneous dipole and the induced dipoles.
(0.5)
A large molecule has many electrons. The electron distribution in the molecule can be distorted easily. (0.5)
This leads to an increased chance for the occurrence of instantaneous dipoles and hence stronger van der
Waals’ forces between molecules.
(0.5)
The number of electrons in a halogen molecule increases down the group.
(0.5)
Hence the strength of van der Waals’ forces also increases down the group.
(0.5)
The boiling point of an element depends on the strength of its intermolecular attractions.
(0.5)
Hence the boiling points of halogens increase down the group.
(0.5)
(3 marks for organization and presentation)
362
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