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New 21st Century Chemistry Suggested answers to in-text activities and unit-end exercises Topic 6 Unit 23 I n-text activities Checkpoint (page 10) SiH 4 H 2 S NF 3 BCl 3 Electron diagram 3-D structure Shape of molecule tetrahedral shape 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 PH 4 + ion, all the four electron pairs are bond pairs Suggested answers to in-text activities and unit-end exercises 1 © Jing Kung. All rights reserved. Topic 6 Unit 23

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Page 1: Suggested answers to in-text activities and exercisesintranet.stmgss.edu.hk/.../Chemistry/CHEM_TB_ANS/Sugg…  · Web viewResearchers currently grow the carbon nanotubes on metal

New 21st Century Chemistry

Suggested answers to in-text activities and unit-end exercises

Topic 6 Unit 23

I n-text activities Checkpoint (page 10)

SiH4 H2S NF3 BCl3

Electron diagram

3-D structure

Shape ofmolecule

tetrahedralshape

V-shapedtrigonal

pyramidalshape

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

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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.

Suggested answers to in-text activities and unit-end exercises 2 © Jing Kung. All rights reserved.Topic 6 Unit 23

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New 21st Century Chemistry

Hence buckminsterfullerene is soft and slippery.

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 fullerenesAntiviral activityCompounds 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.

Suggested answers to in-text activities and unit-end exercises 3 © Jing Kung. All rights reserved.Topic 6 Unit 23

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New 21st Century Chemistry

Computer designed accommodation of C60 in the HIV protease hydrophobic cavity

Antioxidants and biopharmaceuticalsFullerenes 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 cellsSolar 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

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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 additivesMuch work has been done on the use of fullerenes as polymer additives to modify physical properties and performance characteristics.

CatalystsIn 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 electronicsIn 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

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New 21st Century Chemistry

Car battery with carbon nanotube technologyCarbon 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.

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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 pyramidalb) Tetrahedralc) Trigonal planard) Trigonal bipyramidale) Octahedral

3 Molecule Electron diagram Model

H2S (iii)

NH3 (i)

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OCS (ii)

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 atomBeCl2 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

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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.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.

Suggested answers to in-text activities and unit-end exercises 9 © Jing Kung. All rights reserved.Topic 6 Unit 23

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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

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

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Option Species Electron diagram Shape of moleculeNH3 trigonal pyramidal

B

CH4 tetrahedral

NH4+ tetrahedral

CBeCl2 linear

SO2 V-shaped

DH2O V-shaped

CO2 linear

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 A

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Option Molecule 3-D structure

(1)

H2O

NH3

CH4

(2)

PH3

BF3

BeCl2

(3)

H2O

CO2

PH3

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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.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

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Molecule Electron diagram 3-D structure Shape of molecule

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.

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.

Suggested answers to in-text activities and unit-end exercises 14 © Jing Kung. All rights reserved.Topic 6 Unit 23

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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)

b)

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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°.

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.

Suggested answers to in-text activities and unit-end exercises 16 © Jing Kung. All rights reserved.Topic 6 Unit 23

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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 17 © Jing Kung. All rights reserved.Topic 6 Unit 23