New 21st Century Chemistry (2nd Edition)
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 24
Practice
P24.1 (page 28)
a)
non-polar
b)
c)
d)
non-polar
e)
f)
non-polar
P24.2 (page 34)
Molecular
formula
Shape of
molecule
Polar bond
Polar molecule?
H2O
V-shape or bent
shape
O–H
yes
HF
linear
H–F
yes
CH2F2
tetrahedral
C–F
yes
SF6
octahedral
S–F
no
Electron diagram
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 24
1
© Jing Kung. All rights reserved.
New 21st Century Chemistry (2nd Edition)
P24.3 (page 38)
The boiling point of a compound depends on the strength of its intermolecular attractions.
The intermolecular attractions in the carbon compounds are van der Waals’ forces.
The number of electrons in the molecule / the molecular mass increases from methane to propane.
Hence the strength of van der Waals’ forces also increases from methane to propane. This suggests
that the boiling points should increase from methane to propane in accordance with the data.
P24.4 (page 43)
1
a) Van der Waals’ forces
b) Hydrogen bonds and van der Waals’ forces
2
a)
b)
The boiling point of NH3 is higher than that of PH3.
The boiling point increases from PH3 to SbH3.
Consider an NH3 molecule. When a hydrogen atom is covalently bonded to a very
electronegative nitrogen atom, a very polar H–N covalent bond is formed. The hydrogen
atom has a strong partial positive charge.
The hydrogen atom is attracted strongly to a lone pair of electrons on the nitrogen atom of
another NH3 molecule. The resulting force of attraction is a hydrogen bond.
The boiling point of a compound depends on the strength of its intermolecular attractions.
Hydrogen bonds exist in NH3 while only van der Waals’ forces exist in PH3.
Hydrogen bonds are stronger than van der Waals’ forces. More heat is needed to separate
the NH3 molecules during boiling.
Thus, the boiling point of NH3 is higher than that of PH3.
The intermolecular attractions in PH3, AsH3 and SbH3 are van der Waals’ forces.
The number of electrons in one molecule increases from PH3 to SbH3. Hence the strength
of van der Waals’ forces also increases from PH3 to SbH3. This suggests that the boiling
points should increase from PH3 to SbH3, in accordance with the data.
P24.5 (page 52)
The boiling points of the compounds are in the order: A < C < D < B.
The boiling point of a compound depends on the strength of its intermolecular attractions.
Hydrogen bonds exist in B and D while only van der Waals’ forces exist in A and C.
Hydrogen bonds are stronger than van der Waals’ forces. More heat is needed to separate the
molecules of B and D during boiling.
Thus, the boiling points of B and D are higher than those of A and C.
Both B and D have a –OH group. Their boiling point difference is related to the difference in the
strength of van der Waals’ forces between molecules.
Molecule of B has a linear shape while that of D has a more spherical and compact shape. The
molecular shape of B allows greater surface contact between molecules.
The van der Waals’ forces in B are stronger.
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 24
2
© Jing Kung. All rights reserved.
New 21st Century Chemistry (2nd Edition)
Thus, the boiling point of B is higher than that of D.
One molecule of C contains more electrons than one molecule of A. Hence the strength of van der
Waals’ forces in C is higher than that in A.
Thus, the boiling point of C is higher than that of A.
Discussion (page 52)
The rate at which each steel ball drops depends on the viscosity of the alcohol. This is related to the
intermolecular attractions of the liquid.
Alcohol X has one, alcohol Y has two and alcohol Z has three –OH group(s) per molecule that can
take part in hydrogen bonding. Thus, the intermolecular attractions in alcohol Z is the strongest.
Among the three alcohols, the viscosity of alcohol Z is the highest because it has the strongest
intermolecular attractions and its molecules tend to become entangled due to the shape.
Thus, the ball drops most slowly in alcohol Z.
Unit Exercise (pages 56–62)
1
a) polar
b) non-polar
c) van der Waals’ forces
d) net dipole moment
2
e)
f)
van der Waals’ forces
F, O or N bonded to hydrogen
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.
b)
c)
3
CH4
a)
b)
c)
The electronegativity values of carbon and chlorine determine where the partial charges
are placed on the molecule.
Yes.
Each C–Cl bond is polar.
Because of its tetrahedral shape, the individual C–Cl bond dipole moments reinforce each
other.
Hence the whole molecule has a net dipole moment.
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 24
3
© Jing Kung. All rights reserved.
New 21st Century Chemistry (2nd Edition)
4
C
5
C
6
A
7
B
A tetrachloromethane molecule has a tetrahedral shape. Due to the symmetry of the
tetrahedral shape, the four identical C–Cl bond dipole moments cancel one another out
exactly. As a result, the molecule has no net dipole moment and it is non-polar.
CO2 and CCl4 molecules contain polar bonds. However, they have symmetrical shapes.
The bond dipole moments cancel one another out exactly. Thus, these molecules are
non-polar.
8
A
1-chloropropane molecules are polar.
Thus, a stream of 1-chloropropane would be affected by an electric field.
9
C
10
D
11
C
12
D
13
D
Molecules of Y and Z can form hydrogen bonds with water molecules while those of X
cannot. Thus, the water solubilities of Y and Z are higher than that of X.
Each molecule of Z has two –OH groups while each molecule of Y has only one.
Molecules of Z can form more hydrogen bonds with water molecules. Thus, the water
solubility of Z is higher than that of Y.
The order of increasing water solubility is: X < Y < Z
14
B
(2) The electronegativity of Group I elements decreases with increasing atomic number.
15
A
(1) Both alcohols have a –OH group. Their boiling point difference is due to the
difference in the strength of van der Waals’ forces between molecules.
The CH3C(CH3)2CH3 molecule is spherical and compact. The molecular surface area
available for coming into contact with neighbouring molecules is the smallest. Thus, the
van der Waals’ forces between CH3C(CH3)2CH3 molecules are weakest.
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 24
4
© Jing Kung. All rights reserved.
New 21st Century Chemistry (2nd Edition)
Molecule of X has a linear shape while that of Y has a more spherical and compact
shape. The molecular shape of X allows greater surface contact between molecules.
The van der Waals’ forces in X are stronger. Thus, the boiling point of X is higher
than that of Y.
(2) The van der Waals’ forces in X are stronger, pulling the molecules close together.
Thus, the density of X is higher than that of Y.
16
a)
b)
17
a)
i)
Octahedral
ii)
Molecule of BF3
ii)
BF3 has a symmetrical shape.
The three identical bond dipole moments cancel one another out exactly.
Molecule of NH3
i)
ii)
A hydrogen sulphide molecule has two lone pairs and two bond pairs of electrons in
the outermost shell of the sulphur atom.
These electron pairs repel to get as far apart as possible.
So, the four electron pairs will adopt a tetrahedral arrangement.
The shape of a molecule is determined only by the arrangement of atoms. Thus, the
hydrogen sulphide molecule is V-shape or bent shape.
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 24
5
© Jing Kung. All rights reserved.
New 21st Century Chemistry (2nd Edition)
b)
The H–S bonds are slightly polar.
As the hydrogen sulphide molecule is not symmetrical in shape, the individual bond
dipole moments reinforce each other.
The molecule has a net dipole moment and it is polar.
18
a)
b)
i)
The jet of propanone is deflected by the positively charged rod.
The negative ends of the molecules are attracted towards the rod.
ii)
The jet of propanone is deflected by the negatively charged rod.
The positive ends of the molecules are attracted towards the rod.
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 24
6
© Jing Kung. All rights reserved.
New 21st Century Chemistry (2nd Edition)
19
The boiling point of an element depends on the strength of its intermolecular attractions.
The intermolecular attractions in halogens are van der Waals’ forces.
The number of electrons in the halogen molecule increases from fluorine to iodine. Hence the
strength of van der Waals’ forces between halogen molecules also increases from fluorine to
iodine.
More heat is needed to separate the molecules during boiling, and thus the boiling points
increase from fluorine to iodine.
20
a)
b)
The boiling point of a compound depends on the strength of its intermolecular attractions.
The intermolecular attractions in both compounds X and Y are van der Waals’ forces.
A molecule of Y contains more electrons than a molecule of X.
So the van der Waals’ forces between molecules of Y are stronger than that between
molecules of X. More heat is need to separate the molecules of Y during boiling.
The shape of a molecule of Y is more spread-out while that of a molecule of Z is more
compact.
This allows greater surface contact between molecules of Y than between molecules of Z.
Hence the van der Waals’ forces between molecules of Y are greater than those between
molecules of Z. More heat is needed to separate the molecules of Y during boiling.
21
Answers for the HKDSE question are not provided.
22
a)
i)
Type of bond:
Explanation:
ii)
Liquid ethanol:
Liquid propane:
Covalent bond
One pair of outermost shell electrons are shared between the C
and H atoms.
Hydrogen bonds
Van der Waals’ forces
b)
23
a)
b)
Hydrogen bonds between ethanol molecules were broken.
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 24
7
© Jing Kung. All rights reserved.
New 21st Century Chemistry (2nd Edition)
c)
i)
Amount of heat taken in by ethanol
ii)
= 8.0 g × 2.41 J g–1 K–1 × 3.8 K
= 73.3 J
Amount of heat taken in by cyclohexane
= 16.0 g × 1.83 J g–1 K–1 × 3.8 K
= 111 J
iii) Number of moles of ethanol used =
8.0 g
46.0 g mol–1
= 0.174 mol
(73.3+ 111) J
0.174 mol–1
= 1.06 kJ mol–1
iv) Strength of hydrogen bond between ethanol molecules =
24
a)
b)
25
a)
b)
The molecules in ice are held further apart than in liquid water.
i) Hydrogen bonds
ii) Van der Waals’ forces
iii) Hydrogen bonds are stronger than van der Waals’ forces.
More heat is needed to separate the molecules during melting of ice.
The boiling point of a compound depends on the strength of its intermolecular attractions.
Hydrogen bonds exist in propan-1-ol while only van der Waals’ forces exist in propene.
Hydrogen bonds are stronger than van der Waals’ forces. More heat is needed to separate
the propan-1-ol molecules during boiling.
Thus, the boiling point of propan-1-ol is higher than that of propene.
26
a)
The boiling point of a compound depends on the strength of its intermolecular attractions.
A HI molecule contains more electrons than a HBr molecule. Hence the strength of van
der Waals’ forces in HI is higher than that in HBr.
More heat is needed to separate the HI molecules during boiling.
Thus, the boiling point of HI is higher than that of HBr.
b)
Hydrogen bonds exist in HF while only van der Waals’ forces exist in HCl.
Hydrogen bonds are stronger than van der Waals’ forces.
More heat is needed to separate the HF molecules during boiling.
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 24
8
© Jing Kung. All rights reserved.
New 21st Century Chemistry (2nd Edition)
Thus, the boiling point of HF is higher than that of HCl.
27
a)
b)
Hydrogen bonds exist between ethanol molecules in addition to van der Waals’ forces.
There are only van der Waals’ forces between tetrachloromethane molecules.
A liquid with strong intermolecular forces has a higher viscosity than one with weak
intermolecular forces.
Hence the viscosity of ethanol is higher than that of tetrachloromethane.
Both ethanol and glycerol molecules can form hydrogen bonds.
Each glycerol molecule has three –OH groups that can take part in hydrogen bonding
while each ethanol moleucle has only one –OH group.
Each glycol molecule can form more hydrogen bonds.
Furthermore, because of their shape, the glycerol molecules tend to become entangled
rather than to slide past one another. These factors contribute to the high viscosity of
glycerol.
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 24
9
© Jing Kung. All rights reserved.
New 21st Century Chemistry (2nd Edition)
c)
A hydrogen bond requires both a lone pair of electrons on an atom of N, O or F, and a
hydrogen atom.
In H2O, there are two hydrogen atoms in each molecule and two lone pairs of electrons on
the oxygen atom, allowing the formation of two hydrogen bonds.
In HF, there is one hydrogen atom in each molecule and three lone pairs of electrons on
the fluorine atom. This means each molecule can only form one hydrogen bond.
Thus, more heat is required to separate the H2O molecules during boiling.
27
a)
b)
Hydrogen bonds exist between ethanol molecules in addition to permanent
dipole-permanent dipole attractions and instantaneous dipole-induced dipole attractions.
There are only instantaneous dipole-induced dipole attractions between
tetrachloromethane molecules.
A liquid with strong intermolecular forces has a higher viscosity than one with weak
intermolecular forces.
Hence the viscosity of ethanol is higher than that of tetrachloromethane.
Both ethanol and glycerol molecules can form hydrogen bonds.
Each glycerol molecule has three –OH groups that can take part in hydrogen bonding
while each ethanol moleucle has only one –OH group.
Each glycol molecule can form more hydrogen bonds.
Furthermore, because of their shape, the glycerol molecules tend to become entangled
rather than to slide past one another. These factors contribute to the high viscosity of
glycerol.
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 24
10
© Jing Kung. All rights reserved.