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New 21st Century Chemistry
Suggested answers to in-text activities and unit-end exercises
Topic 6 Unit 23
In-text activities
Checkpoint (page 10)SiH4 H2S NF3 BCl3
Electron diagram
3-D structure
Shape ofmolecule
tetrahedralshape
V-shaped trigonal pyramidal
shape
trigonal planar shape
Checkpoint (page 15)
1 a)
b) Trigonal planar
2 a)
b) A phosphonium ion has a larger bond angle.
In the PH4+ ion, all the four electron pairs are bond pairs in the outermost shell of the
central nitrogen atom, while in the PH3 molecule, there are one lone pair and three bond
pairs.
The repulsion between a lone pair and a bond pair is stronger than that between a bond
pair and a bond pair.
∴ the bond pairs in PH3 are slightly compressed and the bond angles in PH3 are less than
those in PH4+.
Suggested answers to in-text activities and unit-end exercises 1 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
Checkpoint (page 17)
1 Electron diagram 3-D structure Shape of molecule
a)COCl2
trigonal planar shape
b)C2H2
linear shape
2 γ < β < α
When using the VSEPR theory, double bonds can be treated like single bonds.
Therefore carbon atom x has three pairs of electrons in its outermost shell.
The furthest apart the three pairs can get is at an angle of 120°, so the bond angle α is 120°.
Carbon atom y 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 angleβis 109.5°.
Oxygen atom z has two lone pairs and two bond pairs 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 bond angle γ is compressed to a value less than 109.5°.
Checkpoint (page 20)
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.
2 Graphite has a layered structure. Weak van der Waals’ forces exist between the layers.
The layers can easily slide over each other.
Hence graphite has a slippery feel and can be used as a lubricant.
3 Buckminsterfullerene has a simple molecular structure.
Weak van der Waals’ forces exist between the C60 molecules.
The molecules can slide over each other easily.
Hence buckminsterfullerene is soft and slippery.
Suggested answers to in-text activities and unit-end exercises 2 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
4 Balls of carbons would be very small and not linked to each other, appearing like powder.
Internet Search & Presentation (page 21)
Discovery of C60
The experiments that led to the discovery of C60 aimed at simulating in the laboratory the conditions
under which carbon nucleated in the atmosphere of a cool carbon-rich red giant star.
In 1985 researchers Harold Kroto, Robert Curl and Richard Smalley discovered that graphite,
vaporized by a laser in one billionth of a second, condensed to form a soot which contained clusters
of carbon atoms, of which C60 was the most common. Because of this work, Kroto, Smalley, and
Curl were awarded the Nobel Prize in Chemistry in 1996.
Applications of fullerenes
Antiviral activity
Compounds with antiviral activity are generally of great medical interest. Replication of the human
immunodeficiency virus (HIV) can be suppressed by several antiviral compounds, which are
effective in preventing or delaying the onset of acquired immunodeficiency syndrome (AIDS).
Fullerenes (C60) and their derivatives have potential antiviral activity, which has strong implications
on the treatment of HIV-infection. The antiviral activity of fullerene derivatives is based on several
chemical properties including their unique molecular architecture and antioxidant activity. It has
been shown that fullerenes derivatives can inhibit and make complex with HIV protease (HIV-P).
It has been shown that the fullerene can be accommodated inside the hydrophobic cavity present in
the enzyme and its location might prevent the interaction between the catalytic portions of HIV-P
and the virus substrates.
Computer designed accommodation of C60 in the HIV protease hydrophobic cavity
Suggested answers to in-text activities and unit-end exercises 3 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
Antioxidants and biopharmaceuticals
Fullerenes are powerful antioxidants, reacting readily and at a high rate with free radicals, which are
often the cause of cell damage or death. Fullerenes hold great promise in health and personal care
applications where prevention of oxidative cell damage or death is desirable, as well as in non-
physiological applications where oxidation and radical processes are destructive (food spoilage,
plastics deterioration, metal corrosion).
Major pharmaceutical companies are exploring the use of fullerenes in controlling the neurological
damage of diseases which are a result of radical damage.
Fullerenes are known to behave like a ‘radical sponge’ as they can sponge-up and neutralize 20 or
more free radicals per fullerene molecule. They have shown performance 100 times more effective
than current leading antioxidants such as vitamin E. A company has launched a skin care cream
based on the C60 fullerene in Japan.
Organic solar cells
Solar cells are photovoltaic cells. Photovoltaics, as the word implies (photo = light, voltaic =
electricity), convert sunlight into electricity.
The following is a schematic diagram of a single layer organic solar cell.
Due to their relatively high electron affinity, the fullerenes serve as excellent electron acceptors. In
an organic solar cell, fullerene is used in conjunction with a polymer (electron donor). They are
blended and cast as the active layer to create what is known as a bulk heterojunction. The most
commonly used fullerene is C60.
Suggested answers to in-text activities and unit-end exercises 4 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
Efficiency of these cells is steadily marching upward. In 2009, a researcher suggested that practical
efficiency approaching 20% could be achieved using a polymer / fullerene bulk heterojunction
device.
Polymer additives
Much work has been done on the use of fullerenes as polymer additives to modify physical
properties and performance characteristics.
Catalysts
In 2009, researchers at Nanjing University in China showed that fullerenes could function
effectively as novel non-metal hydrogenation catalysts. Catalytic hydrogenation — used to refine
crude oil, synthesize ammonia, and produce bio-hydrocarbon fuels from fats and oils —
conventionally relies on transition metal catalysts.
Current catalysts and processes typically require high temperatures and pressures. The ability to
replace these catalysts with carbon-based substitutes operating under milder conditions could
reduce process costs, as well as environmental effects from metal pollution.
The discovery of C60 would be a starting point for more effective molecular-based non-metal
catalyst systems composed entirely of carbon.
STSE Connections (page 22)
1 Limitations would include: cost, lack of technology, unknown knowledge about nanotubes, etc.
2 Nano technology developed by Honda opens new potential in electronics
In 2009 Honda reported that microscopic carbon nanotubes might have the potential to
transport electricity faster and over greater distances with minimal loss of energy.
Researchers currently grow the carbon nanotubes on metal nanoparticles. When these tiny
carbon nanotubes exhibit metallic conductivity they possess extraordinary strength compared
to steel, have higher electrical properties than copper, are as efficient in conducting heat as
diamond and are as light as cotton.
Past research efforts to control the structural formation of carbon nanotubes with metallic
conductivity through conventional methodology resulted in a success rate of about 25 – 50 %.
Honda has achieved a success rate of 91% metallic conductivity.
Honda foresees the new technology affecting the production of batteries, cables, fuel cells and
solar cells, which could lead to more efficient vehicles. They also believe the findings could
lead to advancements in artificial muscles, robotics and electrodes for supercapacitors, etc.
Suggested answers to in-text activities and unit-end exercises 5 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
Car battery with carbon nanotube technology
Carbon nanotube battery (CNT battery) technology, introduced by Next Alternative Inc.,
allows cars to travel up to 380 miles per charge. These batteries recharge in ten minutes from a
regular electric outlet.
Micro Bubble Technology, Inc. (MBTI) developed the CNT battery technology. MBTI
developed a proprietary method of coating the anode and cathode, also modifying the
electrolyte with carbon nanotubes. The diminutive tubes hold 8 times as much energy as the
lead in lead-acid accumulators, and can hold a minimum of 2 times as much energy as
rechargeable lithium-based cells.
Next Alternative Inc. claims that CNT batteries are superior to lead-acid accumulators, lithium
cells and silicone cells powering electric cars today. Silicone-based cells perform better than
current lead-acid accumulators but do not allow electric cars to travel a long range and require
lengthy recharge times. Lithium-based cells are expensive to produce and have lengthy
recharge times.
Suggested answers to in-text activities and unit-end exercises 6 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
Unit-end exercises (pages 170-179)
Answers for the HKCEE (Paper 1) and HKALE questions are not provided.
1
2 a) Trigonal pyramidal
b) Tetrahedral
c) Trigonal planar
d) Trigonal bipyramidal
e) Octahedral
3 Molecule Electron diagram Model
H2S (iii)
NH3 (i)
OCS (ii)
Suggested answers to in-text activities and unit-end exercises 7 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
4 a) Molecule Number of bond pairs in the
outermost shell of the central atomNumber of lone pairs in the outermost
shell of the central atom
BeCl2 2 0BF3 3 0CH4 4 0H2O 2 2PCl5 5 0SF6 6 0
b) & c)Molecule 3-D structure Shape of molecule
BeCl2 linear
BF3 trigonal planar
CH4 tetrahedral
H2O V-shaped
PCl5 trigonal bipyramidal
SF6 octahedral
5 a) Consider the electron diagrams of the molecules of CO2 and H2O.
Electron pairs repel one another and stay as far apart as possible.
Suggested answers to in-text activities and unit-end exercises 8 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
When using the VSEPR theory, double bonds can be counted as single bonds.
Therefore the carbon atom can be viewed as having two pairs of electrons in its outermost
shell.
The two pairs must be at opposite ends of a straight line.
Thus, carbon dioxide is a linear molecule.
A water molecule has two lone pairs and two bond pairs of electrons in the outermost
shell of the oxygen atom.
So, the four pairs of electrons in the water molecule will adopt a tetrahedral arrangement.
As the shape of a molecule is determined only by the arrangement of atoms, the water
molecule is V-shaped.
b) Electron pairs repel one another and stay as far apart as possible.
When using the VSEPR theory, double bonds can be counted as single bonds.
Therefore the carbon atom in the methanal molecule can be viewed as having three pairs
of electrons in its outermost shell.
The overall arrangement of the three pairs of electrons around the carbon atom is trigonal
planar.
Hence the methanal molecule has a trigonal planar structure.
c) Consider the electron diagram of OF2:
Oxygen cannot form compounds with more than eight electrons in the outermost shell of
its atom.
Hence oxygen forms OF2 only.
Sulphur can form some compounds with more than eight electrons in the outermost shell of
its atom.
Hence it can form SF4 and SF6 besides SF2.
6 BOption Molecule Electron diagram Shape of molecule
A H2O V-shaped
B HCN linear
C SO2 V-shaped
D Cl2O V-shaped
Suggested answers to in-text activities and unit-end exercises 9 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
7 AOption Species Electron diagram Shape of molecule
A BF3 trigonal planar
B H3O+ trigonal pyramidal
C NF3 trigonal pyramidal
D PCl3 trigonal pyramidal
8 BOption Species Electron diagram Shape of molecule
A BF3 trigonal planar
NH3 trigonal pyramidal
B CH4 tetrahedral
NH4+ tetrahedral
C BeCl2 linear
SO2 V-shaped
D H2O V-shaped
CO2 linear
Suggested answers to in-text activities and unit-end exercises 10 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
9 C
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 angleα is 109°.
Oxygen atom y has two lone pairs and two bond pairs 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 bond angle γ is compressed to a value less than 109°.
10 AOption Molecule 3-D structure
(1) H2O
NH3
CH4
(2) PH3
BF3
BeCl2
Suggested answers to in-text activities and unit-end exercises 11 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
Option Molecule 3-D structure(3) H2O
CO2
PH3
11 a)
b) i)
ii) Electron pairs repel one another and stay as far apart as possible.
In a boron trichloride molecule, there are three bond pairs of electrons in the
outermost shell of the central boron atom.
The furthest apart the three pairs can get is at an angle of 120°,
so the boron trichloride molecule is trigonal planar in shape.
12 a)
b)
c) Electron pairs repel one another and stay as far apart as possible.
A phosphine molecule has one lone pair and three bond pairs of electrons in the outermost
shell of the phosphorus atom.
The four pairs of electrons in the molecule will adopt a tetrahedral arrangement.
Suggested answers to in-text activities and unit-end exercises 12 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
As the shape of a molecule is determined only by the arrangement of atoms, thus the
phosphine molecule has a trigonal pyramidal shape.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the bond pairs are slightly compressed and the H–P–H angle is less than 109.5°.
13 Molecule Electron diagram 3-D structure Shape of molecule
CF4 tetrahedral
NF3 trigonal pyramidal
PF5 trigonal bipyramidal
SF6 octahedral
14 a) Tetrahedral shape; 109.5°
b) Consider the electron diagram of NH2-:
Electron pairs repel one another and stay as far apart as possible.
There are two bond pairs and two lone pairs of electrons in the outermost shell of the
central nitrogen atom.
The pairs of electrons will adopt a tetrahedral arrangement.
Since the shape of a species is determined only by the arrangement of atoms, the amide
ion is V-shaped.
Suggested answers to in-text activities and unit-end exercises 13 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
15 a)
b) Electron pairs repel one another and stay as far apart as possible.
A hydrazine molecule has one lone pair and three bond pairs of electrons in the outermost
shell of each nitrogen atom.
The four pairs of electrons around each nitrogen atom will adopt a tetrahedral
arrangement.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the bond pairs are slightly compressed and the H–N–H angle is about 107°.
16 Molecule 3-D structure
SCl2
BF3
ICl4+
XeO3
17 a)
Suggested answers to in-text activities and unit-end exercises 14 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
b)
18 a)
b)
19 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 0NH3 3 1H2O 2 2
Electron pairs repel one another and stay as far apart as possible.
The furthest apart four pairs of electrons can get is when they adopt a tetrahedral arrangement.
Thus, the methane molecule has a tetrahedral shape. The H–C–H bond angles are 109.5°.
Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
Thus the H–N–H bond angle is compressed to 107°.
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.
As a result, the H–O–H bond angle in a water molecule is decreased to 104.5°.
Suggested answers to in-text activities and unit-end exercises 15 © Jing Kung. All rights reserved.Topic 6 Unit 23
New 21st Century Chemistry
20
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 109.5°.
When using the VSEPR theory, double bonds can be treated like single bonds.
Therefore carbon atom y has three pairs of electrons in its outermost shell.
The furthest apart the three pairs can get is at an angle of 120°,
so the bond angle β is 120°.
Carbon atom z can be viewed as having two pairs of electrons in its outermost shell.
The two pairs must be at opposite ends of a straight line in order to be as far apart as possible.
So, the bond angle γ is 180°.
21 α = 107°
β= 120°
22 —
23 —
24 a) In diamond, each carbon atom is covalently bonded to four other carbon atoms in the
form of a tetrahedron.
Hence diamond has a giant structure consisting of a network of covalent bonds.
b) i) In graphite, the layers of carbon atoms are held by weak van der Waals’ forces.
The layers can easily slide over each other. Hence graphite is quite soft.
ii) Graphite has a layered structure. Within each layer, each carbon atom uses three
outermost shell electrons in forming covalent bonds with three other atoms.
The remaining electron is delocalized between the layers of carbon atoms.
Graphite can conduct electricity due to the presence of delocalized electrons.
c) The melting point of diamond is higher than that of buckminsterfullerene.
Diamond has a giant covalent structure. The carbon atoms are held together by strong
covalent bonds.
There are weak van der Waals’ forces between the buckminsterfullerene molecules.
More heat is needed to break the strong covalent bonds between atoms in diamond. Hence
diamond has a higher melting point.
25 —
Suggested answers to in-text activities and unit-end exercises 16 © Jing Kung. All rights reserved.Topic 6 Unit 23