Suggested answers to in-text activities and...

New 21st Century Chemistry

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

Topic 6 Unit 24

In-text activities

Checkpoint (page 38)

1 a) P–H non-polar

δ+ δ–

b) H–F

δ– δ+

c) O–N

d) F–F non-polar

δ– δ+

e) N–H

f) C–I non-polar

2 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.

Checkpoint (page 44)Molecular formula

Electron diagram

Shape of molecule

Polar bond Polar molecule?

H2O V-shaped O–H yes

BeH2 linear Be–H no

CH2F2 tetrahedral C–F yes

Cl2O V-shaped Cl–O yes

Suggested answers to in-text activities and unit-end exercises 1 © Jing Kung. All rights reserved.Topic 6 Unit 24


2 a)

b) In trichloromethane, each C–Cl bond is polar.

The trichloromethane molecule is tetrahedral in shape.

The individual C–Cl bond dipole moments reinforce each other.

The molecule has a net dipole moment and it is polar.

When a positively charged rod is brought close to the jet of trichloromethane, negative

ends of the molecules are attracted towards the rod.

When a negatively rod is brought close to the jet of trichloromethane, positive ends of the

molecules are attracted towards the rod.


1 Substance Type of attractions

Instantaneousdipole-induced

dipole attractions

Permanentdipole-permanent

dipole attractionsa) Bromine ✔b) Liquid sulphur dioxide ✔ ✔c) Methane ✔

2 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.



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.

Discussion (page 60)

Without hydrogen bonding, the boiling point of H2O would be around –68 °C while that of HF

would be around –88 °C.

HF molecules can form an average of one hydrogen bond per molecule while H2O molecules can

form an average of two hydrogen bonds per molecule.

Hence the difference between the actual boiling point and estimated boiling point without hydrogen

bonding for H2O is much greater than that for HF.


1 a)Liquid Type of attractions

Instantaneousdipole-induced

dipole attractions

Permanentdipole-permanent

dipole attractions

Hydrogenbonds

a) ✔

b) ✔ ✔ ✔

c) ✔ ✔



b) The boiling point of a compound depends on the strength of its intermolecular attractions.

The boiling point of is the lowest.

Only weak instantaneous dipole-induced dipole attractions exist between the molecules.

The boiling point of is the highest.

Hydrogen bonds exist between the molecules.

2

(For the sake of clarity, interaction is not shown at every –OH group.)


1 (a) and (c)

2 a) Compound Type of attractions

Instantaneous dipole-induced

dipole attractions

Permanent dipole-permanent

dipole attractions

Hydrogen bonds

i) NH3 ✔ ✔ ✔ii) CH3OH ✔ ✔ ✔

iii)✔ ✔

iv) ✔



b)

Library Search & Presentation (page71)

Proteins

Biological functions of proteins

Proteins are large molecules that occur in every living organism. They are of many types and have

many biological functions. The following table lists some biological functions of proteins.Some biological functions of proteins

Type Function ExampleEnzymes catalyze biological processes pepsinHormones regulate body processes insulinStorage proteins store nutrients ferritinTransport proteins transport oxygen and other substances

through the bodyhemoglobin

Structural proteins form an organism’s structure collagenProtective proteins help fight infection antibodiesContractile proteins form muscles actin, myosinToxic proteins serve as a defense for the plant or animal snake venoms

Amino acids making up the body’s proteins

All proteins are made up of many amino acid units linked together in a long chain. Twenty different

amino acids are used to make the body’s proteins.

An amino acid contains both an amino group and a carboxyl group.

The following table lists some amino acids essential to living organisms.



Name Abbreviations Structure(shaded portion is the R group of the amino acid)

Alanine Ala

Aspartic acid Asp

Cysteine Cys

Glycine Gly

Histidine His

Serine Ser

Valine Val

Peptide link formation

Two amino acids can undergo a condensation reaction to form a dipeptide. A water molecule is

eliminated between the –NH2 group of one amino acid and the –COOH group of the other. The

amino acid units are linked by a peptide link.



From a dipeptide, a tripeptide can be made by adding another amino acid molecule. Peptides that

contain many amino acid units are polypeptides. Proteins are polypeptides consisting of one or

more polypeptide chains.

Levels of protein structure

Chemists usually speak about four levels of structure when describing proteins.

Primary structure

The primary structure specifies the unique amino acid sequence of the polypeptide chain.

Secondary structure

Most polypeptide chains fold in such a way that the segments of the chain orient into regular

patterns, called secondary structures. There are two common kinds of patterns: the α-helix and the

β-pleated sheet.

In the α-helix, the polypeptide chain is coiled tightly in the fashion of a spring. The helix is

stabilized by hydrogen bonds between the N–H group of one amino acid unit and the C=O group on

the 4th amino acid unit away from it.

The following figure shows the α-helical secondary structure of keratin, a fibrous protein found in

wool, hair, fingernails and feathers.



The following figure shows the β-pleated sheet secondary structure found in fibroin, the fibrous

protein found in milk. A polypeptide chain doubles back on itself after a hairpin bend. The two

sections of the chain on either side of the bend line up in a parallel arrangement held together by

hydrogen bonds.

Tertiary structure

Secondary protein structures result primarily from hydrogen bonding between peptide links along

the protein backbone, but higher levels of structure result primarily from interactions of R groups in

the protein.



The tertiary structure of a protein is its three-dimensional shape that arises from further foldings of

its polypeptide chains, foldings superimposed on the coils of the α-helices.

Various forces are involved in stabilizing tertiary structures, including van der Waals’ forces, ionic

linkages, hydrogen bonds and disulphide bridges (refer to the following figure for details).

Quaternary structure

A protein molecule may be made up of more than one polypeptide chain. The overall arrangement

of the polypeptide chains is called the quaternary structure. A variety of interactions including

hydrogen bonding hold the various chains into a particular geometry.

There are two major categories of proteins with quaternary structure — fibrous and globular.

Collagen is a fibrous protein in tendons and muscles, consisting of intertwining polypeptide chains.



Globular proteins are mostly clumped into a shape of a ball. For example, the hemoglobin molecule

consists of four separate polypeptide chains or subunits. These subunits are held together by van der

Waals’ forces and ionic forces.

DNA

Functions of DNA

The nucleic acids are informational molecules because their primary structure contains a code or set

of directions by which they can duplicate themselves and guide the synthesis of proteins. There are

two types of nucleic acids which are polymers found in all living cells. Deoxyribonucleic acid

(DNA) is found mainly in the nucleus of the cell, while ribonucleic acid (RNA) is found mainly in

the cytoplasm of the cell.

Coded in an organism’s DNA is all the information that determines the nature of the organism and



all the directions that are needed for producing the thousands of different proteins required by the

organism.

Structure of nucleic acids

Nucleic acids are polymers made up of nucleotide units linked together to form a long chain. Each

nucleotide is composed of a nucleoside plus phosphoric acid, and each nucleoside is composed of a

sugar plus an amine base.

The sugar in DNA is 2-deoxyribose.

Four different cyclic amine bases occur in DNA: adenine, guanine, cytosine and thymine.

In DNA, the cyclic amine base is bonded to C1’ of the sugar, and the phosphoric acid is bonded to

the C5’ sugar position. The following figure shows the general structures of a nucleoside and a

nucleotide.



Nucleotides join together in nuclei acids by forming a phosphate ester bond between the phosphate

group at the 5’ end of one nucleotide and the hydroxyl group on the sugar component at the 3’ end

of another nucleotide.

This makes the nuclei acid a long unbranched chain with a backbone of sugar and phosphate units

with bases protruding from the chains at regular intervals.



The following diagram shows a segment of one DNA chain. See how the phosphate ester groups

link the 3’- and 5’- OH groups of the sugar units.

Base pairing in DNA: the Watson-Crick Model

According to the model, DNA consists of two polynucleotide strands coiled around each other in a

double helix. The two strands run in opposite directions and are held together by hydrogen bonds

between pairs of bases. Adenine (A) and thymine (T) form two strong hydrogen bonds to each

other, but not to guanine (G) or cytosine (C); G and C form three strong hydrogen bonds to each

other, but not to A or T.



The specific base pairing also means that the two chains of DNA are complementary. Wherever

adenine appears in one chain, thymine must appear opposite it in the other; wherever cytosine

appears in one chain, guanine must appear in the other (refer to the following diagram of the DNA

double helix showing complementary base pairing).

Replication of DNA

Just prior to cell division the double strand of DNA begins to unwind. Complementary strands are

formed along each chain. Each chain acts as a template for the formation of its complement. When

unwinding and duplication are complete, there are two identical DNA molecules where only one

had existed before. These two molecules can then be passed on, one to each daughter cell.



Unit-end exercises (pages 75-84)

Answers for the HKCEE (Paper 1) and HKALE questions are not provided.

1

2 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) δ– δ+ δ+ δ– δ– δ+ δ– δ+

C–H C=O O–H N–H

c) CH4



3 a)

b) The electronegativity values of carbon and chlorine determine where the partial charges

are placed on the molecule.

c) 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.

4 a)Substance Boiling point (K) Type(s) of intermolecular forces

Propane 229 instantaneous dipole-induced dipole attractionsMethanol 338 instantaneous dipole-induced dipole attractions,

permanent dipole-permanent dipole attractions, hydrogen bonds

b) The boiling point of a compound depends on the strength of its intermolecular attractions.

Hydrogen bonds exist between methanol molecules, in addition to permanent dipole-

permanent dipole attractions and instantaneous dipole-induced dipole attractions.

There are only weak instantaneous dipole-induced dipole attractions between propane

molecules.

Hence the boiling point of methanol is higher than that of propane.

5 The volatility of a halogen depends on the strength of its intermolecular attractions.

Van der Waals’ forces exist between halogen molecules.

The number of electrons in the molecule increases from chlorine to iodine. Hence the strength

of van der Waals’ forces also increases from chlorine to iodine.

Thus the trend in volatility of the three halogens is chlorine > bromine > iodine.

6 B Van der waals’ forces exist in methane and neon.

The stronger van der Waals’ forces in methane (due to greater molecular surface area

allowing greater contact between molecules) account for its higher boiling point as more

heat is required to separate its molecules during boiling.

7 C Although the H–Cl bond is polar, the chlorine atom is quite large and its lone pairs of

electrons are not very accessible to a hydrogen atom.

Hence there is no strong attraction between the hydrogen atom and the lone pair on the

chlorine atom of another HCl molecule.



A HCl molecule will not form a hydrogen bond with another HCl molecule.

8 D X is non-polar. It does not mix with water due to the difference in the strength of

intermolecular attractions between water molecules and those between molecules of X.

Thus, X is insoluble in water.

Y is soluble in water because hydrogen bonds can form between molecules of Y and

water molecules.

The water solubility of Z is higher than that of Y.

Each molecule of Z has two –OH groups that can take part in hydrogen bonding while

each molecule of Y has only one –OH group that can take part in hydrogen bonding.

9 D (1) In a BF3 molecule, each B–F bond is polar.

A BF3 molecule has a trigonal planar shape. The three identical bond dipole

moments cancel one another out exactly.

So a BF3 molecule is non-polar.

(2) In a CCl3Br molecule, all the bonds are polar.

The CCl3Br molecule is tetrahedral in shape. The individual bond dipole moments do

not cancel one another out exactly.

The molecule has a net dipole moment and it is polar.

(3) In a NF3 molecule, each N–F bond is polar.

A NF3 molecule has a trigonal pyramidal shape.



The individual N–F bond dipole moments reinforce each other.

The NF3 molecule has a net dipole moment and it is polar.

10 B (1) In a hexane molecule, each C–H bond can be regarded as non-polar. Thus hexane is

a non-polar liquid.

A stream of hexane would NOT be deflected by the charged rod.

(2) In a trichloromethane molecule, each C–Cl bond is polar.

The trichloromethane molecule is tetrahedral in shape. The individual C–Cl bond

dipole moments reinforce each other.

The molecule has a net dipole moment and thus trichloromethane is a polar liquid.

A stream of trichloromethane would be deflected by the charged rod.

(3) In a tetrachloromethane molecule, each C–Cl bond is polar.

A tetrachloromethane molecule has a tetrahedral shape. The four identical bond

dipole moments cancel one another out exactly.

As a result, the molecule has no net dipole moment. Thus tetrachloromethane is a

non-polar liquid.



A stream of tetrachloromethane would NOT be deflected by the charged rod.

11 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) A BF3 molecule has a trigonal planar shape.

The three identical bond dipole moments cancel one another out exactly.

As a result, the molecule has no net dipole moment.

12 a) In a propanone molecule, the C=O bond is polar.

The molecule has a net dipole moment and thus propanone is a polar molecule.

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.

13 a) PF3 PF5

b) PF3 – trigonal pyramidal shape

PF5 – trigonal bipyramidal shape

c) Each P–F bond is polar.

A PF3 molecule has a trigonal pyramidal shape. The individual polar P–F bond dipole

moments reinforce each other.

The PF3 molecule has a net dipole moment and it is polar.

d) NF5 do not exist.

Nitrogen cannot form compounds with more than 8 electrons in the outermost shell of its

atom.

14 —

15 a) i) Electron pairs repel one another and stay as far apart as possible.

A water molecule has two lone pairs and two bond pairs of electrons in the outermost

shell of the oxygen 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

water molecule is V-shaped.

ii) 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.



Therefore the two lone pairs will stay the furthest apart, and the separation between a

lone pair and a bond pair will be greater than that between two bond pairs.

As a result, the H–O–H bond angle in a water molecule is decreased to 104.5°.

b) i)

ii) The electronegativity values of boron and chlorine are different.

iii) A BCl3 molecule has a trigonal planar shape.

The three identical B–Cl bond dipole moments cancel one another out exactly.

As a result, the molecule has no net dipole moment. So the molecule is non-polar.

16 a)

b) The electronegativity value of hydrogen is greater than that of silicon. So in a SiH4

molecule, each Si–H bond is polar.

A SiH4 molecule has a tetrahedral shape. The four identical Si–H bond dipole moments

cancel one another out exactly.

As a result, the molecule has no net dipole moment. So the SiH4 molecule is non-polar.

c) The electronegativity value of sulphur is greater than that of hydrogen. So in a H2S

molecule, each S–H bond is polar.

Because of the V-shape of the H2S molecule, the individual S–H bond dipole moments

reinforce each other.

The H2S molecule has a net dipole moment and it is polar.


There are permanent dipole-permanent dipole attractions and instantaneous dipole-

induced dipole attractions between H2S molecules while there are only instantaneous

dipole-induced dipole attractions between SiH4 molecules.

Hence more heat is needed to separate the H2S molecules than the SiH4 molecules during

boiling.

17 a) The boiling point of a compound depends on the strength of its intermolecular attractions.

The intermolecular attractions in both chloroethane and 1-chloroprapane are van der

Waals’ forces.

A 1-chloropropane molecule contains more electrons than a chloroethane molecule.



So the van der Waals’ forces between 1-chloropropane molecules are stronger than that

between chloroethane molecules. More heat is need to separate the 1-chloropropane

molecules during boiling.

b) The shape of a 1-chloropropane molecule is more spread-out while that of a 2-

chloropropane molecule is more compact.

This allows greater surface contact between 1-chloropropane molecules than between 2-

chloropropane molecules.

Hence the van der Waals’ forces between 1-chloropropane molecules are greater than

those between 2-chloropropane molecules. More heat is needed to separate the 1-

chloropropane molecules during boiling.

18 —

19 a) 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.

b) 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.

20 a) i) Hydrogen bond

ii)

b) Ammonia molecules can form hydrogen bonds with water molecules while phosphine

molecules cannot.



21 —

22 a) i) The boiling point of a compound depends on the strength of its intermolecular

attractions.

Only weak van der Waals’ forces exist between methane molecules while there are

hydrogen bonds and van der Waals’ forces between ammonia molecules.

More heat is needed to separate the ammonia molecules than the methane molecules

during

boiling. Hence the boiling point of ammonia is higher than that of methane.

ii) The electronegativity value of oxygen is greater than that of nitrogen, making the O–

H bonds more polar than the N–H bonds. Hence the hydrogen bonds in water are

stronger than those in ammonia.

Furthermore, the oxygen atom in each water molecule has 2 lone pairs of electrons

for hydrogen bonding while the nitrogen atom in each ammonia molecule has only 1

lone pair of electrons for hydrogen bonding. Hence each water molecule can take

part in hydrogen bonding to twice the extent.

More heat is needed to separate the water molecules than the ammonia molecules

during boiling.

Hence the boiling point of water is higher than that of ammonia.

iii) The intermolecular attraction in both methane and silane are van der Waals’ forces.

A silane molecule contains more electrons than a methane molecule.

So the van der Waals’ forces between silane molecules are stronger than those

between methane molecules.

More heat is needed to separate the silane molecules than the methane molecules

during boiling.

Hence the boiling point of silane is higher than that of methane.

b) i) Hydrogen bonding exists in hydrogen fluoride. Hence its boiling point should be

high.

As fluorine is more electronegative than nitrogen, the hydrogen bonds in hydrogen

fluoride should be stronger than those in ammonia.

Hence the boiling point of hydrogen fluoride should be higher than that of ammonia.

There is 1 hydrogen atom per hydrogen fluoride molecule for hydrogen bonding

while there are 2 hydrogen atoms per water molecule for hydrogen bonding. Thus

the hydrogen bonding in hydrogen fluoride should be less extensive.

Hence the boiling point of hydrogen fluoride should be lower than that of water.

ii) The intermolecular attractions in both silane and germanium hydride are van der

Waals’ forces.

A germanium hydride molecule contains more electrons than a silane molecule.



So the van der Waals’ forces between germanium hydride molecules should be

stronger than those between silane molecules.

More heat should be needed to separate the germanium hydride molecules than the

silane molecules during boiling. Hence the boiling point of germanium hydride

should be higher than that of silane.

23 a)

b) A large dipole between a hydrogen atom and a highly electronegative oxygen atom

A lone pair of electrons on another oxygen atom, with which the partially positively

charged hydrogen atom can line up

c) Permanent dipole-permanent dipole attractions

d) i) The boiling point of a compound depends on the strength of its intermolecular

attractions.

Only van der Waals’ forces exist between chloromethane molecules

while there are hydrogen bonds and van der Waals’ forces between water molecules.

More heat is needed to separate the water molecules than the chloromethane

molecules during boiling.

ii) The intermolecular attractions in both chloromethane and bromomethane are van der

Waals’ forces.

A bromomethane molecule contains more electrons than a chloromethane molecule.

So the van der Waals’ forces between bromomethane molecules are stronger than

those between chloromethane molecules.

More heat is needed to separate the bromomethane molecules than the

chloromethane molecules during boiling.

24 a) Evaporation is a process that takes in heat. It can cause a temperature decrease in the

surroundings.

Other factors being equal, the higher the rate at which molecules leave the liquid surface,

the more extreme the temperature in the surroundings drops.

As the temperature change for pentane is greater than that for butan-1-ol, it can be

deduced that pentane evaporates faster than butan-1-ol.

b) Only weak van der Waals’ forces exist between pentane molecules

while there are hydrogen bonds and van der Waals’ forces between butan-1-ol molecules.

Thus pentane molecules can break away from the rest of the liquid more easily.


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