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Published by PENERBITAN PELANGI SDN BHD, 2023-04-05 22:35:15

FOCUS SPM Chemistry (2023)

WM: RM40.95
EM: RM42.95

Format 190mm X 260mm Extent : 528pg (25.26mm) confirmed (All 4C/70gsm) Status CRC Date 3/4

PELANGI BESTSELLER




SPM
CC038643



FORM KSSM FOCUS
4∙5
Chemistry SPM


Chemistry SPM





FOCUS SPM KSSM Form 4 • 5 – a complete and precise series of reference books
with special features to enhance students’ learning as a whole. FORM
This series covers the latest Kurikulum Standard Sekolah Menengah (KSSM) and
integrates Sijil Pelajaran Malaysia (SPM) requirements. (Dual Language Programme) 4∙5 KSSM
A great resource for every student indeed!


REVISION REINFORCEMENT EXTRA Chemistry • Lim Eng Wah

› i-Study SPM & ASSESSMENT FEATURES • Sim Ley Yee*
› Comprehensive Notes › SPM Practices › Examples • Francisca Lau*
› Concept Maps › SPM Model Paper › SPM Highlights • Low Swee Neo
› Activities & Experiments › Checkpoint › Digital Resources QR Codes • Chien Hui Siong*
› SPM Tips *Guru Cemerlang
› Complete Answers



TITLES IN THIS SERIES

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eBook here! • Ekonomi • Perniagaan

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W.M: RM40.95 / E.M: RM42.95
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CC038643
ISBN: 978-629-7537-63-4
Based on the
PELANGI LATEST SPM FORMAT


Format: 190mm X 260mm TPTV Focus SPM 2023 Chem BI version _pgi CRC












Chemistry SPM





FORM
4∙5
(Dual Language Programme) KSSM



• Lim Eng Wah
• Sim Ley Yee*
• Francisca Lau*
• Low Swee Neo
• Chien Hui Siong*
*Guru Cemerlang













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ISBN: 978-629-7537-63-4
eISBN: 978-629-7537-64-1 (eBook)


First Published 2023




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Exclusive Features of This Book





Theme 2: Organic Chemistry Form 5
Chapter Chemistry SPM Chapter 2 Carbon Compound
Chapter Focus 2 Carbon Compound
lists the learning CHAPTER FOCUS
objectives for an • Types of carbon compounds
• Homologous series
• Chemical properties and interconversion of
compounds between homologous series
overview of the chapter. • Isomers and naming based on IUPAC nomenclature

CHAPTER
2
i-Study SPM
Carbon atoms combine with one another to form
a C 60 carbon molecule.
What is the name of this C 60 carbon molecule? highlights the key
Access to
i-STUDY SPM concepts of each chapter
i-STUDY
to aid students in focused
PB 289 289
revision.
SPM Tips 7. The following example shows how to draw the Naming of Alkenes

structural formula for propane, C 3H 8 .
C C C
Step 1: Draw three carbon atoms in a row.

Chemistry SPM Chapter 2 Carbon Compound
to form a chain.

Step 2: Join the carbon atoms by single bonds
points out important Step 3: Add single bonds to each carbon atom presence and location of double bond between
C!C!C
1. The naming of alkenes follows the same IUPAC

rules used for naming alkanes. However, the
to ensure it forms four bonds.
C C C 2. The name of a straight-chain alkene consists
carbon atoms must be indicated.
tips for students to Step 4: Attach a hydrogen atom to each shows the longest Suffix
of two component parts:
Root name
H
H
H
‘empty’ bond of a carbon atom.
C
H
cabon chain with
double bonds
take note of. molecule, ensure that SPM Tips H H C C H H 3. Determining the root name: series and the position of
shows the homologous
double bond
Figure 2.7 Naming of alkenes
bonds • Identify the longest carbon chain that
When drawing the structural formula of an alkane
• each carbon atom has exactly four single bonds
contains the double bond.
chain.
a single bond
• carbon atoms are bonded to each other by single
• Count the number of carbon atoms in the
• each hydrogen atom is bonded to a carbon atom by
Alkenes 4. Determining the suffix:
but-, pent-, etc (Table 2.7).
• Name it appropriately as meth-, eth-, prop-,
SPM Highlights
homologous series.
1. Alkenes is another hydrocarbon.
• Use the suffix -ene to denote the alkene
2. The general formula for the alkene homologous
• Number the carbon atoms in the longest
provides exposure to
series is C nH 2n, where n = 2, 3, 4... and so on.
3. Alkenes are unsaturated hydrocarbons.
carbon atoms, C=C.
chain from the end of the nearest double
atoms the lowest possible number.
Every alkene has a double bond between
bond to give the double-bonded carbon
atom.
series. • Identify the location of the double bond by
4. The double bond between carbon atoms is the
SPM functional group of the alkene homologous the number of the first double-bonded carbon
suffix -ene.
Highlights • Place the number immediately in front of the
A C nH 2n the frequently-tested
320 B C n H 2n + 2 C C nH 2n + 1 OH • For molecules containing more than three
The general formula of the alkene homologous series is
questions that appear
D C n H 2n + 1COOH
Examiner’s Tip
bond must be specified.
carbon atoms, the position of the double
naming of an alkene.
5. Example 2.2 shows the steps involved in the
consist of only carbon and hydrogen.
Alkenes are unsaturated hydrocarbons which
Answer: A EXAMPLE 2.2 5 Form
an equivalent alkane (C n H 2n + 2 ) molecule.
An alkene molecule has 2 hydrogen atoms less than
H in the actual exam.
Name the alkene in Figure 2.9.
H H
H C H H H
C C
C C
C H
H H H
Figure 2.9 H
321
Activity / Chemistry SPM Chapter 4 The Periodic Table of Elements
Experiment
Problem statement: How does the reactivity of Group 1 elements with water and oxygen change
Experiment 4.1
helps students to Aim: To investigate the chemical properties of Group 1 elements with water and oxygen. Form
when going down the group?
A Reaction of alkali metals with water
master hands-on Hypothesis: When going down Group 1, reactivity of Group 1 elements with water increases. 4
(a) Manipulated variable: Type of alkali metal
Operational definition: An alkali metal that reacts more vigorously and rapidly with water is more
(b) Responding variable: Reactivity of alkali metal with water
scientific Variables: Materials: Small pieces of lithium, sodium, potassium, distilled water, filter paper, red litmus paper
(c) Constant variable: Size of alkali metal
reactive.
Form knowledge Apparatus: Glass trough, knife, forceps The practical work on the reaction of sodium or
Safety Precautions
potassium with water must be demonstrated by
the teacher only because these reactions are very
Safety Precautions
and skills. • Do not touch the extremely reactive alkali vigorous and may explode.
• Always wear safety goggles and gloves.
metals with your bare hands.
4
Procedure: Lithium Basin Distilled water
Figure 4.11 Reaction of alkali metals with water
1. A small piece of lithium is cut using a knife and forceps.
3. The lithium is dropped carefully onto the surface of water in the glass trough using forceps.
2. The paraffin oil on the surface of lithium is dried using filter paper.
6. Steps 1 to 5 are repeated by replacing lithium with sodium and potassium.
4. The observations are recorded.
5. The solution formed is tested with red litmus paper.
Table 4.3
Observation
Observation: Lithium moves slowly on the surface of water with a hissing sound. A colourless
Alkali metal solution is formed and turns red litmus paper to blue.
Lithium
71
Example
provides solution
Chemistry SPM Chapter 3 Thermochemistry
EXAMPLE 3.22
for example of
70 2.0 dm 3 of water at 32 °C.
Solution
questions in the
The fuel value of kerosene is 37 kJ g –1 . Calculate the mass of kerosene that must be burnt to boil
= mcθ
Calculate heat absorbed by water.
[Specific heat capacity of solution = 4.2 J g –1 °C –1 ; density of solution = 1 g cm –3 ]
Total heat absorbed by water, Q
= 2000 × 4.2 × 68
= 571 200 J
subtopics.
= 571.2 kJ
= (2000 cm 3 × 1 g cm –3 ) × (4.2 J g –1 o C –1 ) × (100 o C – 32 o C)
Calculate fuel value of kerosene
Fuel value of kerosene
Checkpoint = 571.2 kJ Heat released (kJ) Hence, 571.2 = 37
= mass of kerosene burnt (g)
m g
m
Given fuel value of kerosene = 37 kJ g –1
provides questions = 571.2 kJ kJ g –1 ∴ m = 571.2 = 15.4 g
m
37
to test students’ Q1 Hot pack is used to treat sport injuries. A hot pack
Checkpoint
chemicals is water.
contains two chemical substances. One of the
understanding of (b) What type of reaction occurs when water is (d) Explain the preparation of a hot pack with the
formula CuSO 4.
3.3
obtained?
(a) The second chemical substance has the
(c) How can the sodium ethanoate crystals be
Name this chemical substance.
(b).
crystals of sodium ethanoate.
mixed with the substance in (a)?
the subtopics and 5 Form Q2 Sodium ethanoate crystals, CH 3COONa, are Q3 Combustion of 1 mole of hydrogen gas releases
286 kJ of heat energy.
(c) Write a chemical to represent the reaction in
combustion of hydrogen gas.
(a) Write a thermochemical equation for the
gas.

obtained from a neutralisation reaction between
an acid X and an alkali Y. 100 cm 3 of 1.0 mol dm –3
(b) Calculate the mass of 1 mole of hydrogen
reinforce learning. (b) Calculate the number of moles of sodium (d) Calculate the volume of hydrogen gas
1.0 mol dm –3 alkali Y solution.
acid X solution is mixed with 100 cm 3 of
[Relative atomic mass: H = 1].
(a) Name acid X and alkali Y.
(c) What is the fuel value of hydrogen gas?
ethanoate produced.
[Specific heat capacity of water = 4.2 J g –1 °C –1 ;
(measured at room conditions) required
for combustion to raise the temperature of
100 cm 3 of water from 30 °C to 50 °C.
Density of water = 1 g cm –3 ; Molar volume of
gas = 24 dm 3 mol –1 at room conditions]
418
PB
ii ii
Exclusive Features.indd 2 03/04/2023 2:19 PM


Chemistry SPM Chapter 3 Thermochemistry
3.1 Heat Changes in Reactions is released by the reaction. Hence, this is an
exothermic reaction.
Digital Resources* 1. Thermochemistry studies the heat changes Magnesium Test tube
during physical processes and chemical
reactions.
help learners 2. All chemical substances have potential energy ribbon Dilute hydrochloric Heat Heat
stored in the form of chemical energy.
3. During a chemical reaction, the chemical
acid
energy in the reactants can be converted into
heat energy.
Heat
better comprehend 4. In a chemical reaction, heat energy is 5. Figure 3.3 shows some examples of exothermic
Figure 3.2 Example of an exothermic reaction
transferred to or from the surroundings. In
thermochemistry, these two types of reactions
reactions.
concepts and deepen are classified as: INFO Thermochemistry Combustion Combustion Reactions of reactive
• Exothermic reaction
metals with acids
• Endothermic reaction
of fuels
Reactions of
metal carbonates
of metals
learning. Exothermic Reactions Neutralisation Exothermic with acids Reactions of alkali
1. An exothermic reaction is a reaction that
releases heat to the surroundings. The heat
reactions
metals with water
reactions
released is a product of the reaction.
*Model 3D/Interactive 2. In an exothermic reaction: Respiration Rusting Contact Process Haber Process
(formation
Reactants → products + heat
of ammonia)
(formation of
of iron
Chart/Video Heat • chemical energy is Figure 3.3 Examples of exothermic reactions
sulphur dioxide)
converted into heat
energy
• heat energy is
Heat Heat transferred to the Endothermic Reactions
surroundings
• temperature of 1. An endothermic reaction is a reaction that
absorbs heat from the surroundings. Heat is
surroundings increases
• reaction mixture and required for the reaction to take place.
Heat container become hot
Figure 3.1 Exothermic reaction Reactants + heat → products
3. The surroundings do not involve in the 2. In an endothermic reaction: Form
reaction. The surroundings include: Heat • heat energy is
(a) container with reactants and their products Heat Heat converted into chemical 5
(b) solvent in which the reaction occurs energy
(c) air • heat energy is
absorbed from the
(d) thermometer Heat Heat surroundings
4. When a piece of magnesium ribbon is dropped • temperature of the
surroundings decreases
into a test tube containing dilute hydrochloric Heat Heat • reaction mixture and
acid, a reaction occurs and the test tube Heat container become cold
becomes hot. This phenomenon shows heat
Figure 3.4 Endothermic reaction

Chemistry SPM Chapter 3 The Mole Concept, Chemical Formula and Equation Concept Map
PB 375
SPM Practice CONCEPT MAP The mole concept, chemical Form summarises
formula and equation
provides sample Chemical formula 4 essential
questions to test Number of Stoichiometry Molecular formula concepts learnt in
students’ mastery particles ÷ NA × NA Number of moles Empirical formula the chapter using
of the chapter. × molar volume ÷ molar volume × molar mass ÷ molar mass concept map.
Volume of gas Mass (g) 3
Form
4
SPM Practice 4. Figure 1 shows a cooking gas
cylinder.
[Relative atomic mass: O = 16,
Mg = 24] 1 mole of propane
Objective Questions A 3 B 4 C 5 D 6 How many hydrogen atoms
gas, C 3 H 8
Choose the correct answer.
Figure 1
Relative Atomic Mass and
Relative Molecular Mass
[Avogadro’s constant: 6.02 ×
Mole Concept
1. Six atoms of element Y have 3. How many nitrate ions, are there in the gas cylinder?
NO 3 – are there in 2 mol of
3.2
3.1
the same mass with three
A 1 × 6.02 × 10 23
10 23 mol –1 ]
aluminium nitrate, Al(NO 3 ) 3?
tellurium atoms, Te. What is
B 3 × 6.02 × 10 23
[Avogadro’s constant: 6.02 ×
D 11 × 6.02 × 10 23
[Relative atomic mass: Te = 128]
10 23 mol –1 ]
Answers the relative atomic mass of A 1.204 × 10 24 mol –1 C 8 × 6.02 × 10 23 57
B 1.806 × 10 24 mol –1
Y?
C 32
C 3.010 × 10 24 mol –1
D 64
A 8
D 3.612 × 10 24 mol –1
B 16
have the equal mass with two
Enable students 2. How many oxygen atoms
magnesium atoms?
to compare their FORM 4 ANSWERS
Chapter 1 Chemistry SPM Answers
answers to the Q1 Chemistry is a branch of science Q3 He should remove his laboratory •
Introduction to Chemistry
mouth. A pipette should be used
Checkpoint
to suck up the solution/ liquid
1.1
chemical into the apparatus.
correct ones and Q2 Iron: It is used to make nails, screws body under the safety shower. At the • • manipulated variable.
Manipulated variable is
the factor that is purposely
coat immediately. If the acid wets his
that studies the composition,
changed in an experiment.
clothes, he should rinse his whole
structure, characteristics and
interaction of matter.
Responding variable is the
PB
factor that changes with the
same time, he should ask his friend
evaluate their level Q3 To carry out research and 1. D Objective Questions 1 (b) • experiment.
to report the incident to the teacher.
and nuts; Chlorine: It is used to treat
Fixed variable is the factor
SPM Practice
tap water; Calcium carbonate: It is
that is kept constant in an
manufacture cement.
used as construction material/ to
2. B
List out all the materials and
6. D
7. C
used.
8. C 3. C 4. B 5. A • Determine how to control
apparatus needed to be
development of food products/ To
of preparation. Checkpoint 1.2 1. (a) (i) Long hair that is not tied up • • Determine how to collect,
Subjective Questions
Section A
carry out analysis to check the
the manipulated and
quality of products/ To provide
section.
constant variables.
technical support to the marketing

Determine how to measure
easily catches fire.
the responding variable.
(ii) Closed-up shoes will be
Q1 Make observations, make an
analyse and interpret data.
able to protect the feet from
(c) (i) Manipulated variable –
inference, identify the problem, make
water.
chemical spills or injury due
to glass apparatus falling
a hypothesis, identify the variables,
Mass of salt added to pure
onto the feet or floor.
control the variables, plan and carry
out an experiment, collect data,
Responding variable – Time
write a report. (iii) If the excess solution is (ii) The larger the mass of salt
interpret data, make a conclusion,
taken to freeze the water.
contaminated, the action
of pouring it back into
added to pure water, the longer
acid. acid, the higher the pH of the contaminate all the solution freeze.
its reagent bottle will
Q2 (a) The more water is added to the
time it takes for the water to
in the bottle. (iii) Volume of pure water used.
(b) Manipulated variable: Volume of (b) I will report the incident to the Type of container used.

water added to the acid
of acid Responding variable: pH value teacher immediately. Then, Type of freezer used.
(any two)
Fixed variable: Type and volume using gloves, I will dispose (d) (i) The larger the mass of salt
of the broken test tube into
a specific container that is
added to pure water, the
of acid, pH paper
Checkpoint
1.3 as instructed by the teacher. (ii) • The manipulated variable
higher is its density.
prepared and clean up the spill
chemicals (c) A fume chamber is a chamber
to pure water.
Q1 Goggles – to protect the eyes from
is the mass of salt added
where the air in it is always
being sucked out. A fume
Safety shower – to remove
chamber is used to carry
is added to pure water, for
body chemicals that the body comes in out activities that involve • So, different mass of salt
PB contact by showering all parts of the volatile, inflammable or and 5 g.
example 1 g, 2 g, 3 g, 4 g
victim’s body toxic chemicals so that the • So, he can measure the
• The responding variable is
Fire blanket – to put out the fire on a
the density of water.
vapour from the chemicals is
disposed of immediately from
Q2 A – Chemicals should not be sniffed
the fume chamber and does
initial mass and final mass
directly. Instead, the individual
and salt.
not contaminate the air in the
laboratory.
of each mixture of water
smell to his nose. Section B The density of the mixture
should use his hands to waft the
2. (a) • is calculated as follows:
B – The solution/ liquid chemical
Density =
should not be sucked using
Hypothesis is a statement
that links the manipulated
Mass of mixture (g)
Volume of mixture (cm 3 )
variable to the responding
variable. • The fixed variables are
salt used.
• So, pure water is used SPM Model Paper
pure water and type of
SPM Model Paper every time. Only one type prepares students
495
PAPER 1
One hour fifteen minutes
Answer all questions in this paper. First pair Second pair R and S for the SPM with
1. Figure 1 shows the arrangement of particles in three chemical III A 4. Figure 3 shows the standard the actual exam
S and T
P and Q
S and T
P and R
B
Q and R
R and T
substances at room temperature.
C
Q and S
D
Which of the following matches the arrangement of particles above? notation for atoms of element format.
II
X and M.
19 M
Figure 1
39
III
I
aluminium I aluminium mercury 16 8 X What is the formula of the
II
mercury
Figure 3
argon
aluminium
aluminium
mercury
combination of elements X
2. When 50 cm 3 of liquid P is mixed with 50 cm 3 of water, the total compound formed from the
argon
argon
A
argon
volume of the mixture is 98 cm 3 as shown in Figure 2.
B
mercury
and M?
C
Volume of
D
A MX
mixture
B M 2 X
C MX 2
D M 2X 2
liquid 50 cm 3 P + 50 cm 3 water 98 cm 3 5. Which of the following
represents the properties of
sodium chloride?
Which of the following explains the above observation? Substance point ( o C) Molten Aqueous
Electrical
Figure 2
A Liquid P reacts with water to form a solid substance.
conductivity
Melting
B Liquid P reacts with water to produce a volatile gas.
C Some particles of liquid P escape as a gas into the atmosphere.
D Particles of liquid P fill the spaces between water particles.
Weak
Weak
Good
3. Table 1 shows the subatomic particles of five different atoms. Number of 6 A B C D –100 –80 110 800 Weak Weak Insoluble Good
proton
Good
Table 1
Number of
neutron
electrons
Atom Number of 6 6 7 8 7 6 8
P 7 8 8
Q R 6 8 10
Which of the following two pairs are isotopes? 1
8
S
T
PB
iii iii
Exclusive Features.indd 3 03/04/2023 2:20 PM


Noble
18 gases 2 He Helium 4 10 Ne Neon 20 18 Ar Argon 40 36 Kr Krypton 84 54 Xe Xenon 131 86 Rn Radon 222 118 Og Oganesson 71 Lu Lutetium 175 103 Lr Lawrencium 262
Halogens
17 9 F Fluorine 19 17 Cl Chlorine 35.5 35 Br Bromine 80 53 I Iodine 127 85 At Astatine 210 117 Ts Tennessine 70 Yb Ytterbium 173 102 No Nobelium 259

16 8 O Oxygen 16 16 S Sulphur 32 34 Se Selenium 79 52 Te Tellurium 128 84 Po Polonium 209 116 Lv Livermorium (292) 69 Tm Thulium 169 101 Md Mendelevium 258

15 7 N Nitrogen 14 15 P Phosphorus 31 33 As Arsenic 75 51 Sb Antimony 122 83 Bi Bismuth 209 115 Mc Moscovium 68 Er Erbium 167 100 Fm Fermium 257
Periodic Table of Elements
14 6 C Carbon 12 14 Si Silicon 28 32 Ge Germanium 73 50 Sn Tin 119 82 Pb Lead 207 114 Fl Flerovium (289) 67 Ho Holmium 165 99 Es Einsteinium 252

13 Semi-metal Non-metal 5 B Boron 11 13 Al Aluminium 27 31 Ga Gallium 70 49 In Indium 115 81 Tl Thallium 204 113 Nh Nihonium 66 Dy Dysprosium 162.5 98 Cf Californium 251
Metal
12 30 Zn Zinc 65 48 Cd Cadmium 112 80 Hg Mercury 201 112 Cn Copernicium (285) 65 Tb Terbium 159 97 Bk Berkelium 247
Key:
11 29 Cu Copper 64 47 Ag Silver 108 79 Au Gold 197 111 Rg (272) 64 Gd Gadolinium 157 96 Cm Curium 247


10 28 Ni Nickel 59 46 Pd Palladium 106 78 Pt Platinum 195 110 Ds Darmstadtium Roentgenium (281) 63 Eu Europium 152 95 Am Americium 243
Group 9 Symbol of the element 27 Co Cobalt 59 45 Rh Rhodium 103 77 Ir Iridium 192 109 Mt Meitnerium (268) 62 Sm Samarium 150 94 Pu Plutonium 244
Proton number Relative atomic mass


8 26 Fe Iron 56 44 Ru Ruthenium 101 76 Os Osmium 190 108 Hs Hassium (265) 61 Pm Promethium 145 93 Np Neptunium 237


7 Transition metals 25 Mn Manganese 55 43 Tc Technetium 98 75 Re Rhenium 186 107 Bh Bohrium (264) 60 Nd Neodymium 144 92 U Uranium 238
1 H Hydrogen 1
6 24 Cr Chromium 52 42 Mo Molybdenum 96 74 W Tungsten 184 106 Sg Seaborgium (263) 59 Pr Praseodymium 141 91 Pa Protactinium 231

Name of the element
5 23 V Vanadium 51 41 Nb Niobium 93 73 Ta Tantalum 181 105 Db Dubnium (262) 58 Ce Cerium 140 90 Th Thorium 232

4 22 Ti Titanium 48 40 Zr Zirconium 91 72 Hf Hafnium 178.5 104 Rf Rutherfordium (261) 57 La Lanthanum 139 89 Ac Actinium 227

3 21 Sc Scandium 45 39 Y Yttrium 89 57 – 71 Lanthanides 89 – 103 Actinides
Alkaline earth metals Actinides
2 4 Be Beryllium 9 12 Mg Magnesium 24 20 Ca Calcium 40 38 Sr Strontium 88 56 Ba Barium 137 88 Ra Radium 226 Lanthanides
Alkali metals
1 1 H Hydrogen 1 3 Li Lithium 7 11 Na Sodium 23 19 K Potassium 39 37 Rb Rubidium 85.5 55 Cs Cesium 133 87 Fr Francium 223


1 2 3 4 Period 5 6 7










iv





Layout Perriodic Table.indd 4 03/04/2023 2:38 PM


CONTENTS






FORM 4
4.4 Elements in Group 1 69
Theme 1 The Importance of Chemistry
4.5 Elements in Group 17 76
Chapter 4.6 Elements in Period 3 84
1 Introduction to Chemistry 1 4.7 Transition Elements 90

1.1 Development in Chemistry Field and Its SPM Practice 4 92
Importance in Daily Life 2
Chapter
1.2 Scientific Investigation in Chemistry 4 5 Chemical Bond 94
1.3 Usage, Management and Handling of
Apparatus and Materials 5 5.1 Basics of Compound Formation 95
SPM Practice 1 8 5.2 Ionic Bond 98
5.3 Covalent Bond 103
5.4 Hydrogen Bond 107
Theme 2 Fundamentals of Chemistry
5.5 Dative Bond 111
Chapter Matter and the Atomic 5.6 Metallic Bond 113
2 Structure 11 5.7 Properties of Ionic Compounds and

2.1 Basic Concepts of Matter 12 Covalent Compounds 114
2.2 The Development of the Atomic SPM Practice 5 121
Model 18
2.3 Atomic Structure 20 Theme 3 Interaction between Matter
2.4 Isotopes and Their Uses 26
SPM Practice 2 30 Chapter
6 Acid, Base and Salt 123
Chapter The Mole Concept, Chemical
3 33 6.1 The Role of Water in Showing Acidic
Formula and Equation
and Alkaline Properties 124
3.1 Relative Atomic Mass and Relative 6.2 pH Value 130
Molecular Mass 34 6.3 Strength of Acids and Alkalis 133
3.2 Mole Concept 36 6.4 Chemical Properties of Acids and
3.3 Chemical Formula 42 Alkalis 135
3.4 Chemical Equation 52 6.5 Concentration of Aqueous Solution 142
SPM Practice 3 57 6.6 Standard Solution 145
6.7 Neutralisation 147
Chapter The Periodic Table of
4 60 6.8 Salts, Crystals and Their Uses in
Elements
Daily Life 154
4.1 The Development of the Periodic Table 6.9 Preparation of Salts 156
of Elements 61 6.10 Effect of Heat on Salts 170
4.2 The Arrangement in the Periodic 6.11 Qualitative Analysis 178
Table of Elements 63
4.3 Elements in Group 18 67 SPM Practice 6 189



v





Contents.indd 5 03/04/2023 1:59 PM


2.4 Isomers and Naming based on IUPAC
Chapter
7 Rate of Reaction 193 Nomenclature 353
SPM Practice 2 369
7.1 Determining Rate of Reaction 194
7.2 Factors Affecting Rate of Reactions 201
7.3 Application of Factors that Affect the Theme 3 Heat
Rate of Reaction in Daily Life 212
Chapter
7.4 Collision Theory 213 3 Thermochemistry 374
SPM Practice 7 219
3.1 Heat Change in Reactions 375

Theme 4 Industrial Chemistry 3.2 Heat of Reaction 381
3.3 Application of Exothermic and
Chapter Manufactured Substances in Endothermic Reactions in Daily Life 415
8 Industry 223 SPM Practice 3 419
8.1 Alloy and Its Importance 224
8.2 Composition of Glass and Its Uses 229 Theme 4 Technology in Chemistry
8.3 Composition of Ceramics and Its Uses 232
Chapter
8.4 Composite Materials and Its 4 Polymer 425
Importance 235
SPM Practice 8 239 4.1 Polymer 426
4.2 Natural Rubber 434
FORM 5 4.3 Synthetic Rubber 443
SPM Practice 4 446
Theme 1 Chemical Process
Chapter Consumer and Industrial
5 450
Chapter Chemistry
1 Redox Equilibrium 242
5.1 Oils and Fats 451
1.1 Oxidation and Reduction 243 5.2 Cleaning Agents 455
1.2 Standard Electrode Potential 264 5.3 Food Additives 466
1.3 Voltaic Cell 268 5.4 Medicines and Cosmetics 472
1.4 Electrolytic Cell 275 5.5 Application of Nanotechnology in
1.5 Extraction of Metal from Its Ore 292 Industry 479
1.6 Rusting 295 5.6 Application of Green Technology in
SPM Practice 1 305 Industrial Waste Management 484
SPM Practice 5 489

Theme 2 Organic Chemistry
ANSWERS 495
Chapter
2 Carbon Compound 313 SPM MODEL PAPER WITH ANSWERS
2.1 Types of Carbon Compounds 314
2.2 Homologous Series 318 https://qr.pelangibooks.com/?u=fTQ0fBqa
2.3 Chemical Properties and Interconversion
of Compounds between Homologous
Series 332



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Theme 2: Fundamentals of Chemistry Form 4

Chapter
2 Matter and the Atomic Structure













CHAPTER FOCUS Form

Form
• Basic Concepts of Matter 4
4
• The Development of the Atomic Model
• Atomic Structure
• Isotopes and Their Uses







































Matter is made up of tiny particles so small that it can only be seen through an
electron microscope. One of the particles is an atom. Scientists have discovered that
an atom has an inner structure. Each atom is built up from even smaller particles.
What are these particles?



Access to
i-STUDY
i-STUDY SPM



11
PB 11




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Chemistry SPM Chapter 2 Matter and the Atomic Structure


2.1 Basic Concepts of Matter

1. Matter is anything that has mass and occupies space.
2. Matter is classified as element and compound.
3. An element is a substance that is made entirely from one type of atom.
4. A compound is a substance that consists of two or more elements chemically bonded together.
5. According to the Kinetic Theory of Matter, matter is made up of tiny discrete particles.
6. Particles consist of atoms, molecules or ions. Form

PARTICLE 4
Form
4
Atom
Ion Molecule
Smallest particle in an
element
Charged particle – positively-charged A group of atoms that are
or negatively-charged chemically bonded together

Figure 2.1 Classification of particles

7. Matter can be classified into two groups, that are elements and compounds.


Element MATTER Compound




Atom Molecule Molecule Ion











Gold, Au Oxygen gas, O 2 Water, H 2 O

Sodium chloride,
NaCl







Argon gas, Ar Sulphur, S 8 Carbon dioxide gas, CO 2

Figure 2.2 Classification of matter




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Chemistry SPM Chapter 2 Matter and the Atomic Structure

SPM Tips
• A molecule that consists of one type of atom only is a molecular element.
• A molecule that consists of two or more different types of atoms is a molecular compound.
• A compound that is made up of non-metal atoms normally consist of molecules.
• A compound that is formed from a combination of metal and non-metal atoms normally consists of ions.



States of Matter
1. Matter exists in three states, that are solid, liquid and gas.
2. The Kinetic Particle Theory is used to describe the three states of matter. Form
Form
State of Solid Liquid Gas 4
4
Matter
Arrangement
of particles





Particle Particles can only vibrate Particles can vibrate, rotate Particles can vibrate, rotate
motion and rotate about their fixed and move throughout the and move freely. The rate
positions. liquid. The particles collide of collision is higher than in
with one another. liquid.

Attractive Very strong attractive forces Particles are bonded by strong Very weak attractive forces
forces between particles. attractive forces but weaker between particles.
between than those in a solid.
particles
Energy Least energy content because High energy content due to Highest energy content due to
content of of restricted motion. particles able to move easily random motion of particles
particles
Shape and • Solids have a fixed volume • Liquids have a fixed volume. • Gases do not have a fixed
volume and shape. Liquids do not have a fixed shape and volume.
• Solids cannot be compressed. shape but take the volume • Gases can be easily
of the container. compressed.
• Liquids cannot be easily
compressed.
Figure 2.3 Arrangement and motion of particles in the three states of matter

Changes in States of Matter
1. Changes in states of matter are caused by heating or cooling.

Evaporation

• When a liquid is heated, the particles gain energy and move faster.
• Continued heating will supply more energy to the particles until the particles can overcome the attractive forces
between them and escape from the liquid surface.
• At this point, the liquid starts to become a gas.
• Evaporation takes place at all temperatures between the melting point and boiling point.








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Chemistry SPM Chapter 2 Matter and the Atomic Structure

Melting Boiling
• When a solid is heated, the particles gain kinetic energy • When a liquid is heated, the particles gain kinetic energy
and vibrate more rapidly. and move more rapidly.
• Continued heating will supply more energy to the • Continued heating will supply more energy to the
particles until the particles can overcome the attractive particles until the particles can overcome the attractive
forces between them and break away from each other. forces between them and escape.
• Particles move away from their fixed positions. • At this point, the liquid starts to become a gas.
• At this point, the solid starts to change to a liquid. • The temperature at which a liquid changes to a gas is
• The temperature at which a solid changes to a liquid is called the boiling point.
called the melting point.
Sublimation Form
Boiling /
Form
Melting Evaporation
Solid Cecair Gas 4
4
Freezing Condensation
Deposition

Freezing Condensation
• When a liquid is cooled, the particles lose kinetic energy • When a gas is cooled, the particles lose kinetic energy
and move more slowly. and move more slowly.
• Continued cooling will cause the particles to lose more • Continued cooling will cause the particles to lose more
energy and allow the formation of attractive forces energy and allow the formation of attractive forces
between particles to bond the particles in fixed positions. between particles to bond the particles more closely
• At this point, the liquid starts to become a solid. together.
• The temperature at which a liquid changes to a solid is • At this point, the gas starts to become a liquid.
called the freezing point. • The temperature at which a gas changes to a liquid is
called the condensation point.
• The condensation point is the same as the boiling point
of the liquid.
Sublimation and Deposition
• When a solid is heated, the particles gain kinetic energy and vibrate more rapidly.
• Continued heating will supply more energy to the particles until the particles can overcome the attractive forces
between them and break away from each other.
• Particles at the surface escape from their fixed positions, leading to a process called sublimation, in which the solid
turns directly into a gas.
• Conversely, when a gas loses energy, its particles slow down and eventually lose enough energy to form a solid, a
process called deposition.
• The temperature at which a solid changes directly to a gas (sublimation point) and the temperature at which a gas
changes directly to a solid (deposition point) are the same for a given substance under fixed conditions.
Figure 2.4 Changes in states of matter
Activity 2.1


Aim: To determine the melting and boiling points of naphthalene.
Materials: Naphthalene and tap water.
3
Apparatus: Boiling tube, 250 cm beaker, thermometer, tripod stand, retort stand with clamps, Bunsen
burner, stopwatch, conical flask and wire gauze.
Procedure:
A Heating naphthalene
1. A boiling tube is filled with naphthalene to a height of 3 cm and a thermometer is inserted into it.
2. The boiling tube is suspended in a beaker half-filled with water using a retort stand and clamp
as shown in Figure 2.5. The level of the naphthalene in the boiling tube must be below the water
level in the beaker.


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Chemistry SPM Chapter 2 Matter and the Atomic Structure


Thermometer

Boiling tube
Water
Naphthalene






Figure 2.5 Heating naphthalene Form
Form
3. The water is heated and the naphthalene is stirred slowly with a thermometer. 4
4
Safety Precautions
Naphthalene is flammable.

4. When the temperature of the naphthalene reaches 60°C, the stopwatch is started. The temperature
and state of naphthalene are recorded at half-a-minute intervals until the temperature reaches
90°C.
B Cooling naphthalene
1. The boiling tube in section A is removed from the water bath. The surface of the boiling tube is
dried and immediately placed into a conical flask, as shown in Figure 2.6. The naphthalene is
stirred continuously.
2. The temperature and state of the naphthalene is recorded at half-a-minute intervals until the
temperature falls to about 60°C.

Thermometer


Boiling tube

Naphthalene

Conical flask

Figure 2.6 Cooling naphthalene

Results:
Table 2.1
Heating naphthalene Cooling naphthalene
Time (min) Temperature (°C) State Time (min) Temperature (°C) State
0.0 61.0 Solid 0.0 87.0 Liquid
0.5 65.0 Solid 0.5 84.0 Liquid
1.0 67.0 Solid 1.0 83.0 Liquid
1.5 71.0 Solid 1.5 81.0 Liquid
2.0 74.0 Solid 2.0 80.0 Liquid and solid
2.5 76.0 Solid 2.5 80.0 Liquid and solid




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Chemistry SPM Chapter 2 Matter and the Atomic Structure

3.0 79.0 Solid 3.0 80.0 Liquid and solid
3.5 80.0 Solid and liquid 3.5 80.0 Liquid and solid
4.0 80.0 Solid and liquid 4.0 80.0 Liquid and solid

4.5 80.0 Solid and liquid 4.5 80.0 Liquid and solid
5.0 80.0 Solid and liquid 5.0 80.0 Liquid and solid
5.5 80.0 Solid and liquid 5.5 76.0 Solid
6.0 80.0 Solid and liquid 6.0 73.0 Solid
6.5 83.0 Liquid 6.5 69.0 Solid Form
Form
7.0 86.0 Liquid 7.0 66.0 Solid 4
4
7.5 89.0 Liquid 7.5 64.0 Solid

Discussion:
1. A graph of temperature against time was plotted for the heating of naphthalene, as shown in
Figure 2.7
Temperature (°C)

90 D
x
x
x
80 x x x x x x
x B C
x
x
70 x
x
x
60 x A



50 Time (min)
0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
Figure 2.7 Heating curve of naphthalene
2. Note that there is a specific section of the curve where there is no change in temperature with time
during the heating. At this temperature, both solid and liquid exist. This temperature is the melting
point of naphthalene. Hence, the melting point of naphthalene is 80 C.
o
(a) At point A, naphthalene exists as a solid.
(b) When the solid is heated, energy is absorbed. This causes the particles to gain kinetic energy
and vibrate more rapidly. The temperature increases from point A to point B.
(c) At point B, solid naphthalene starts to melt. During the melting process, the temperature of
naphthalene did not increase even though the heating is continuing. The temperature remains
constant because the heat energy absorbed by the particles is used to overcome the attractive
forces between particles so that the solid can change to a liquid. At this temperature, both solid
and liquid are present.
(d) At point C, all the solid naphthalene has melted.
(e) From point C to point D, the particles in the liquid naphthalene absorb heat energy and move
more rapidly. The temperature increases from point C to point D.




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Chemistry SPM Chapter 2 Matter and the Atomic Structure

3. The graph temperature against time was plotted for the cooling of naphthalene, as shown in
Figure 2.8.
Temperature (°C)


90
xE
x
x F G
80 x x x x x x x x
x
x Form
70
Form
x 4
4
x
x
H
60

50 Time (min)
0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
Figure 2.8 Cooling curve of naphthalene
4. Note that there is a specific section of the curve where the temperature remains constant with time
during the cooling. At this temperature, both solid and liquid exist. This temperature is the freezing
point of naphthalene. Hence, the freezing point of naphthalene is 80°C.
(a) At point E, naphthalene exists as a liquid.
(b) When the liquid cools, the particles in the liquid naphthalene lose kinetic energy. The particles
move more slowly when the temperature falls from point E to point F.
(c) At point F, liquid naphthalene starts of freeze. During the freezing process, the temperature of
naphthalene remains constant because the loss of heat to the surroundings is balanced by the
heat energy liberated when particles attract one another to form a solid. At this temperature,
both solid and liquid are present.
(d) At point G, all the liquid has frozen.
(e) From point G to point H, the particles in the solid naphthalene lose heat energy and vibrate
more slowly. The temperature falls from point G to point H.
5. During the heating of naphthalene,
(a) the water bath is used to replace direct heating with a Bunsen burner. This is to ensure the
naphthalene is heated uniformly. Furthermore, naphthalene is flammable.
(b) the naphthalene is stirred continuously to ensure uniform heating.
6. During the cooling of naphthalene, Temperature (°C)
(a) the boiling tube was placed inside a conical flask. This is to
ensure the cooling process is uniform to minimise loss of heat
energy to the surroundings. Boiling
point
(b) naphthalene is stirred continuously to prevent supercooling.
Supercooling is a situation where the temperature of a liquid Time (min)
being cooled falls below its normal freezing point without
appearance of a solid. Figure 2.9 Supercooling
7. A water bath is used in this experiment because the melting point of naphthalene is lower than
o
100 C, the maximum temperature achieved by the water bath. For solids with melting points
o
higher than 100 C, liquids with boiling points higher than water such as oil ought to be used.
Conclusion:
The melting point of naphthalene is the same as the freezing point of naphthalene, that is 80.0°C.





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Chemistry SPM Chapter 2 Matter and the Atomic Structure

SPM Highlights Checkpoint 2.1

Q1 Explain the change in terms of energy,
The following figure shows the cooling curve of movement and arrangement of particles when
substance X. a gas condenses to form a liquid.
Temperature (°C)
Q2 Figure 2.11 shows the heating curve of ice.
Temperature (°C)
Room
temperature
15 S
t 1 t 2 Time (min) Form
Figure 2.10 –3 0 Q R
Form
Which statement is correct about substance X? P 4
4
A X is a gas at room temperature
B Undergoes physical change at T°C Time (min)
C From t 1 to t 2 , X absorbs heat energy Figure 2.11
D From t 1 to t 2 , X exists as a liquid
(a) State the freezing point of ice.
Examiner’s tip (b) Why does the temperature remain
Substance X is a solid/liquid at room temperature. constant between Q and R even though
At the time interval t 1 to t 2 , X loses heat to the the heating is continued?
surroundings. Both liquid and solid/gas and liquid (c) Complete the curve in Figure 2.11 to
exist at time interval t 1 and t 2 . show the results that you would expect
Answer: B if the heating is continued until steam is
produced at 120°C.


2.2 The Development of the Atomic Model

1. The history of the atom started around 450 BC with the Greek philosopher Democritus.
2. He introduced the idea of atom from the Greek word ‘atomos’.
3. Table 2.2 shows the historical development of the atomic model.

Table 2.2 Historical development of atomic model
John Dalton’s atomic model • Atom is the smallest particle in an element.
(proposed by John Dalton in 1805) • All atoms of an element are the same.
• Different elements have different atoms.
• Atom cannot be divided, created nor destroyed.








Thomson’s atomic model • He discovered electrons in cathode rays.
(proposed by J. J. Thomson in 1897) • He proposed a “plum pudding” model.
• An atom is a positively charged sphere with electrons embedded
in it.
Negatively
charged electron
Positively
charged sphere






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Chemistry SPM Chapter 2 Matter and the Atomic Structure

Ernest Rutherford’s atomic model
(proposed by Ernest Rutherford in 1911) • He discovered the proton (positive particle) inside the atom.
• He proposed a nuclear atomic model.
Nucleus • The centre of the atom contains positive charge concentrated in
contains a very small region (called nucleus).
protons • Most of the space around the nucleus is empty.
Electrons • Electrons move randomly around the nucleus.
move
outside
the nucleus


Neils Bohr’s atomic model • He introduced the concept of shells to the atomic model. Form
(proposed by Neils Bohr in 1913) • Electrons orbit the nucleus in fixed shells.
Form
4
Electron 4
Nucleus
contains
protons
Shell

James Chadwick’s atomic model • He discovered the neutrons (neutral particles).
(proposed by James Chadwick in 1932)
• Nucleus is made up of protons and neutrons.
• Neutrons contribute nearly half the mass of an atom.
Electron • Electrons orbit the nucleus in fixed shells.
Nucleus
contains
protons
Shell



Subatomic Particles
1. An atom is made up of even smaller particles – subatomic particles. Subatomic
2. There are three types of subatomic particles – proton, neutron and electron. Particles
3. Figure 2.12 shows the positions of the subatomic particles in an atom. 3D Model
Subatomic Particles


Proton Neutron Electron
relative charge: +1 relative charge: 0 relative charge: –1
relative mass: 1 relative mass: 1 relative mass: 1/1840

Nucleus of atom: Electron:
• contains protons and neutrons • moves in the empty space outside the nucleus
• positively charged • moves in specific shells
• almost all the mass is concentrated here • negatively charged
• very small mass
SPM Tips • number of electrons same as number of
protons
Remember the charge of each subatomic particle:
Protons are positive
Neutrons are neutral Shell:
Hence electrons must be negative • electron pathway

Figure 2.12 The positions of the subatomic particles in an atom.


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Chemistry SPM Chapter 2 Matter and the Atomic Structure

Checkpoint 2.2
Q1 (a) Hydrogen atom is said to be neutral. Explain this statement.
(b) The mass of an atom is concentrated at the nucleus. Explain.
Q2 John Dalton is the first scientist to propose the atomic model in 1805.
State the characteristics of the atom introduced by John Dalton.
Q3 (a) What particle is discovered by J.J. Thomson?
(b) How is the model proposed by Thomson different from Dalton’s model?
Q4 What particle is discovered by Ernest Rutherford and how did the discovery of this particle improve Thomson’s
model? Form
Q5 What idea is presented by the scientist Neils Bohr towards the atomic model in 1913?
Form
Q6 (a) What particle is discovered by James Chadwick? 4
4
(b) What is the contribution of this particle to the atomic model?

2.3 Atomic Structure



Proton Number and Nucleon Number
1. Atoms of different elements have different numbers of subatomic particles.
2. Proton number and nucleon number are used to determine the number of subatomic particles in an
atom.
neutron Nucleon number = number
contains are called total particles
+ Nucleons of protons + number of
Nucleus proton neutrons
of atom
Proton number = number
of protons
Figure 2.13 Proton number and nucleon number
3. Proton number can be used to identify an 6. Because an atom is neutral, the proton number
element. is also the number of electrons in an atom.
4. Proton number of an element is the number 7. Nucleon number of an element is the total
of protons in the nucleus of its atoms. number of protons and number of neutrons
5. Each element has its individual proton number. in an atom of the element.
Example 1: Oxygen atom has a proton number 8. The number of neutrons can be determined as
of 8: follows.
Hence, From the definition:
(i) all oxygen atoms have 8 protons
(ii) an atom which has 8 protons is an oxygen Nucleon = number of + number of
proton
number
neutron
atom
Example 2: Sodium atom has a proton number However, number of proton = proton number
of 11: Hence,
Hence,
(i) all sodium atoms have 11 protons number of = nucleon – proton
(ii) an atom which has 11 protons is a sodium neutron number number
atom
9. Nucleon number is also known as mass
number.





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Chemistry SPM Chapter 2 Matter and the Atomic Structure
10. The relative atomic mass of an atom is nearly
the same as its nucleon number. Nucleon EXAMPLE 2.2
number is sometimes used in calculations as Lithium has proton number 3 and nucleon
an estimate of the relative atomic mass. number 7. What is the number of protons,
electrons and neutrons present in a lithium
atom?
EXAMPLE 2.1
A chlorine atom has 17 protons and 18 neutrons. Solution
What is the proton number and nucleon number Number of protons = proton number = 3
of the chlorine atom? Atoms are neutral. Hence, number of electrons Form
Solution = number of protons = 3.
Form
Proton number = number of protons = 17. Number of neutrons = nucleon number – proton 4
4
Nucleon number = number of protons + number number = 7 – 3 = 4.
of neutrons = 17 + 18 = 35.


Number of Subatomic Particles in Ions
1. Ions are formed from atoms by gaining electrons or losing electrons.
Positive ion loses gains Negative ion
(cation) electrons ATOM electrons (anion)

2. Metal atoms lose electrons to form positive ions.
Example: Lithium atom (proton number, 3; nucleon number, 7)
loses 1 electron
Li Li

Lithium atom: Lithium ion:
*proton = 3 *proton = 3
*neutron = 7 – 3 *neutron = 7 – 3
= 4 = 4
*electron = 3 *electron = 3 – 1
= 2
Figure 2.14 Lithium atom loses one electron to form lithium ion
3. Non-metal atoms gain electrons to form negative ions.
Example: Fluorine atom (proton number, 9; nucleon number, 19)

gains 1 electron
F F

Fluorine atom: Fluoride ion:
*proton = 9 *proton = 9
*neutron = 19 – 9 *neutron = 19 – 9
= 10 = 10
*electron = 9 *electron = 9 + 1
= 10
Figure 2.15 Fluorine atom gains one electron to form fluoride ion
4. When forming an ion,
• the number of protons and number of neutrons remain the same.
• the formation of a positive ion causes the number of electrons to decrease.
• the formation of a negative ion causes the number of electrons to increase.





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Chemistry SPM Chapter 2 Matter and the Atomic Structure

Table 2.3 Elements and their symbols
EXAMPLE 2.3
Element Symbol Element Symbol
Iron atom has proton number 26 and nucleon
number 56. Iron atom loses 3 electrons to form Hydrogen H Sodium Na
an ion. What is the number of each subatomic Helium He Magnesium Mg
particle in the iron ion?
Lithium Li Aluminium Al
Solution
Number of protons = 26 Beryllium Be Silicon Si
Number of neutrons = 56 – 26 = 30 Boron B Phosphorus P
Number of electrons = 26 – 3 = 23 Form
Carbon C Sulphur S
Form
4
Nitrogen N Chlorine Cl 4
EXAMPLE 2.4 Oxygen O Argon Ar
Atom X has proton number 8 and nucleon Fluorine F Potassium K
number 18. The figure below shows an ion
formed from atom X. Neon Ne Calcium Ca
3. Note that:
X (a) each symbol consists of one or two letters.
(b) elements with symbols of two letters, the
first letter is always capitalised whereas
Figure 2.16 the second letter is always a lower case.
(a) For the ion formed, determine (c) some elements have symbols which
(i) number of protons and neutrons originate from Latin names.
(ii) number of electrons Example:
(b) Is the ion formed positively charged or Name of element: Sodium (Latin:
negatively charged? Natrium)
Explain your answer. Symbol of element: Na
Name of element: Potassium (Latin:
Solution Kalium)
(a) (i) Number of protons = 8 Symbol of element: K
Number of neutrons = 18 – 8 = 10
(ii) 10 4. The standard notation of an atom of any
(b) Negatively charged ion because the total element shows the proton number and nucleon
number of electrons increased from 8 to 10. number of the element.
Nucleon number A X Symbol of
Proton number Z element
Standard Notations of Elements Example: The standard notation for bromine
81
1. Each element is given a name and a symbol. atom is Br.
35
2. Table 2.3 shows the names and symbols of a This means the bromine atom has proton
few elements. number 35 and nucleon number 81.
5. Sometimes elements are represented by
using the nucleon number alone. As an
23
example, Na is represented as sodium-23.
11






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Chemistry SPM Chapter 2 Matter and the Atomic Structure

How to write electron arrangement for
EXAMPLE 2.5
an atom
A carbon atom has 6 protons and 7 neutrons. The following steps can be followed to write the
Write the standard notation of carbon atom. electron arrangement for an atom:
Step 1 Obtain the proton number of the atom.
Solution Step 2 Obtain the number of electrons of the atom
Symbol of carbon atom = C For neutral atoms, number of electrons is
Proton number, Z = 6 the same as number of protons.
Nucleon number, A = 6 + 7 = 13 Number of electrons = number of protons
13
Standard notation of carbon atom = C (proton number)
6 Form
Step 3 Arrange the electrons in the shells,
beginning with shell n = 1.
Form
EXAMPLE 2.6 Electrons are only filled in a new shell 4
4
when the previous shell has achieved its
A magnesium atom has 12 electrons and 12
neutrons. maximum.
Write the standard notation of magnesium atom. EXAMPLE 2.7
Solution What is the electron arrangement for sulphur atom?
Symbol of magnesium atom = Mg
Proton number, Z = number of electrons = 12
Nucleon number, A = 12 + 12 = 24 S
24
Standard notation of carbon atom = Mg
12
Figure 2.19 Electron arrangement for sulphur atom
Atomic Structure and Electron Solution
Arrangement Step 1: Proton number = 16
1. The structure of an atom shows the locations, Step 2: Number of electrons = 16
types and number of subatomic particles in Step 3: First shell (n = 1) = 2
an atom. Second shell (n = 2) = 8
Example: Structure of carbon atom Third shell (n = 3) = 6
Shells Electron arrangement for S atom
Nucleus 6 protons = 2.8.6
7 neutrons
Electrons
Figure 2.17 Structure of carbon atom EXAMPLE 2.8
2. Electrons are arranged in shells, starting from What is the electron arrangement for calcium atom?
the shell nearest to the nucleus.
3. Each shell can be filled with a specific number
of electrons. Ca
4. For atoms with proton numbers 1 to 20, the
maximum number of electrons that can be
placed into each shell is Figure 2.20 Electron arrangement for calcium atom
• 2 electrons for the first shell (n = 1) Solution
• 8 electrons for the second shell (n = 2) Step 1: Proton number = 20
• 8 electrons for the third shell (n = 3) Step 2: Number of electrons = 20
First shell: 2 electrons Step 3: First shell (n = 1) = 2
Second shell (n = 2) = 8
Second shell: 8 electrons
Third shell (n = 3) = 8
Third shell: 8 electrons/ Fourth shell (n = 4) = 2
maximum 18 electrons if
Nucleus proton number exceeds 20 Electron arrangement for Ca atom
Figure 2.18 Maximum numbers of electrons that can be = 2.8.8.2
placed into each shell for elements with proton numbers
22 1 to 20 23





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Chemistry SPM Chapter 2 Matter and the Atomic Structure

Valence Electrons
1. Valence electrons are electrons that are located in the outermost occupied electron shell of an atom.

Valence Valence electron
electron
Outermost occupied
electron shell
(valence shell)
Figure 2.21 Valence electrons are determined from the electron arrangement of an atom
2. Number of valence electrons in an atom can be determined from its electron arrangement.
Example: Argon atom (proton number: 18) Form
Electron arrangement: 2.8.8
Form
Number of valence electrons = 8 4
4
Table 2.4 Electron arrangements for elements with proton numbers 1 to 20 and number of corresponding valence electrons

Number of Electron Number of valence
Element Symbol Proton number
electrons arrangement electrons
Hydrogen 1 H 1 1 1 1
1
Helium 4 2 He 2 2 2 2
Lithium 7 Li 3 3 2.1 1
3
Beryllium 9 Be 4 4 2.2 2
4
Boron 11 B 5 5 2.3 3
5
Carbon 12 C 6 6 2.4 4
6
Nitrogen 14 N 7 7 2.5 5
7
Oxygen 16 O 8 8 2.6 6
8
Fluorine 19 F 9 9 2.7 7
9
Neon 20 Ne 10 10 2.8 8
10
Sodium 23 Na 11 11 2.8.1 1
11
Magnesium 24 Mg 12 12 2.8.2 2
12
Aluminium 27 Al 13 13 2.8.3 3
13
Silicon 28 Si 14 14 2.8.4 4
14
Phosphorus 31 P 15 15 2.8.5 5
15
Sulphur 32 S 16 16 2.8.6 6
16
Chlorine 35 Cl 17 17 2.8.7 7
17
Argon 40 Ar 18 18 2.8.8 8
18
Potassium 39 K 19 19 2.8.8.1 1
19
Calcium 40 Ca 20 20 2.8.8.2 2
20



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Chemistry SPM Chapter 2 Matter and the Atomic Structure


EXAMPLE 2.9
Determine the number of valence electrons for the following atoms.


P Q R S



Figure 2.22
Solution
Valence electrons are electrons located in the outermost occupied electron shell. Form
P atom: 5; Q atom: 6; R atom: 2; S atom: 8 4
Form
4

Checkpoint 2.3
Q1 Fluorine-19 atom has 10 neutrons.
(a) What is the proton number and nucleon number of this atom?
A
(b) Represent this atom in the form X.
Z
Q2 Atom X has electron arrangement as shown below.
29

X


Figure 2.23
(a) Write the electron arrangement for atom X.
(b) (i) What is the proton number of atom X?
(ii) Identify atom X.
(c) What is the relative atomic mass of atom X?
Q3 Data for five particles A, B, C, D and E are given below.

Particle Proton number Nucleon number Electron arrangement

A 3 7 2
B 17 37 2.8.8

C 12 24 2.8

D 8 16 2.8

E 18 40 2.8.8

(a) Which particle has number of protons equal to the number of neutrons?
(b) Which particle is a positive ion?
(c) Which particle is a negative ion?
(d) Which particle has eight valence electrons?









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Chemistry SPM Chapter 2 Matter and the Atomic Structure


2.4 Isotopes and Their Uses


What are isotopes?
1. The structure and standard notation for three atoms of hydrogen (proton number 1) are shown below.

1 H
1

Similarity: Difference: Similarity: 2 Differences: Form
Same number Different numbers Same proton H Different nucleon
of protons of neutrons number 1 numbers 4
Form
4
Proton 3 H
Neutron 1
Electron

2. Definition of isotopes:

Isotopes are atoms of an element which has the same number of protons but different number of
neutrons.

or

Isotopes are atoms of an element which has the same proton number but different nucleon numbers.


3. Table 2.5 shows examples of isotopes of a number of elements.
Table 2.5 Examples of isotopes of a number of elements

Proton Nucleon Number of Number of Number of
Element Isotope
number number proton electron neutron
Hydrogen Hydrogen-1 ( 1 H) 1 1 1 1 0
1
Hydrogen-2 ( 1 H) 1 2 1 1 1
2
Hydrogen-3 ( 1 H) 1 3 1 1 2
3
Oxygen Oxygen-16 ( 8 O) 8 16 8 8 8
16
Oxygen-17 ( 8 O) 8 17 8 8 9
17
Oxygen-18 ( 8 O) 8 18 8 8 10
18
Carbon Carbon-12 ( 6 C) 6 12 6 6 6
12
Carbon-13 ( 6 C) 6 13 6 6 7
13
Carbon-14 ( 6 C) 6 14 6 6 8
14
Chlorine Chlorine-35 ( 17 Cl) 17 35 17 17 18
35
Chlorine-37 ( 17Cl) 17 37 17 17 20
37
Bromine Bromine-79 ( 35 Br) 35 79 35 35 44
79
Bromine-81 ( 35 Br) 35 81 35 35 46
81



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Chemistry SPM Chapter 2 Matter and the Atomic Structure
4. Isotopes can be written in a number of forms.
Example: Uranium element, U has two isotopes.
First isotope: uranium-235 or U-235 or 235 U
92
Second isotope: uranium-238 or U-238 or 238 U
92
5. Isotopes are atoms of the same element. Hence, isotopes have the same chemical properties.
Physical properties such as density, melting point and boiling point are different because the numbers
of neutrons are not the same.

Differences Similarities

• Different nucleon numbers • Same proton number Form
Form
• Different number of neutrons Isotopes of • Same number of protons 4
• Different physical properties an element • Same number of electrons
4
because of different number of • Same number of valence electrons
neutrons • Same chemical properties due to the
same number of electrons
Calculating relative atomic masses for elements with isotopes
1. For elements with isotopes, the relative atomic mass of the element is the average mass of all its
isotopes.
2. Follow the following steps to calculate the relative atomic mass of an element.
Step 1 Identify all the known isotopes.
Step 2 Identify the abundance of each isotope.
Step 3 Calculate relative atomic mass by using the formula:

(% isotope 1 × relative isotopic mass of isotope 1) + (% isotope 2 ×
relative isotopic mass of isotope 2) + (% isotope 3 × relative isotopic
Relative atomic mass = mass of isotope 3) + ...
100
Note: Natural abundance is the percentage of isotopes found in a naturally occurring sample of the
element.


EXAMPLE 2.10 EXAMPLE 2.11
Chlorine has two isotopes with the following Bromine has two isotopes with the following
abundances. abundances.
Chlorine-35, 75% and chlorine-37, 25% Bromine-79, 50% and bromine-81, 50%
Calculate the relative atomic mass of chlorine. Calculate the relative atomic mass of bromine.

Solution Solution
Step 1 Isotopes: Cl-35 and Cl-37 Step 1 Isotopes: Br-79 and Br-81
Step 2 Abundance : 75% Cl-35 and 25% Cl-37 Step 2 Abundance: 50% Br-79 and 50% Br-81
Step 3 R.A.M chlorine Step 3 R.A.M. bromine
(% isotope Cl-35 × relative isotopic mass Cl-35) (% isotope Br-79 × relative isotopic mass
= + (% isotope Cl-37 × relative isotopic mass Cl-37) Br-79) + (% isotope Br-81 × relative isotopic
100 = mass Br-81)
(75 × 35) + (25 × 37) 100
= 100 (50 × 79) + (50 × 81)
= 35.5 = 100
= 80

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Chemistry SPM Chapter 2 Matter and the Atomic Structure

Uses of Isotopes in Daily Life
1. There two types of isotopes, radioactive isotopes and non-radioactive isotopes.
2. Radioactive isotopes or radioisotopes decay to give out dangerous radiation.
3. However, radioisotopes have important uses if handled properly. Radioisotopes are used in medicine,
industry, agriculture and research.



Agriculture
• The uptake and metabolism of phosphorus by plants can be investigated using
fertilisers that contain phosphorus-32. Form

Form
A solution containing A detector is used to 4
4
phosphorus-32 is injected detect the movement of
into the root system of the phosphorus-32 throughout
plant. the plant.

• Radioactive tracer research use carbon-14 to help understand photosynthesis and
synthesis of proteins.


Power Source
• Uranium-235 is a fuel
commonly used in nuclear
Medicine power stations.
• Cobalt-60 which emits gamma rays • Uranium-235 undergoes
is used in chemotherapy to treat nuclear fission to release
cancer. a large amount of heat
• Superficial cancers such as skin USES OF energy.
cancer can be treated with weak RADIOISOTOPES
radiation from phosphorus-32 or
strontium-90.
• Pacemakers containing Archaeology
plutonium-238 are used to control • Carbon-14 is used to
the heartbeats of patients with heart estimate the age of
problems. bones, wood or fossils by
• Iodine-131 is used in the treatment measuring the percentage
of thyroid. of carbon-14 it contains.
• Carbon-14 dating only
works for organic materials
that were once alive and
absorbed carbon during
their lifetime.



Industry
• Sodium-24 can be used to detect leakages in gas or oil pipes and ventilation
systems.
• Gamma radiation from cobalt-60 is passed through food to destroy bacteria which
causes food to go bad without changing the quality and texture of the food.
• Radiation from kyptron-85 can be used to control the thickness of plastic sheets
in industry.

Figure 2.24 Uses of radioisotopes





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Chemistry SPM Chapter 2 Matter and the Atomic Structure


Checkpoint 2.4

Q1 Lead-206 and lead-208 are two isotopes of lead. Number of subatomic particles for lead-206 is given below.
Particle Number
Electron 82
Neutron 124
Proton 82
(a) Explain the term isotopes.
(b) In the table below, give the number of subatomic particles for lead-208. Form
Form
Particle Number
4
Electron 4
Neutron
Proton

Q2 The abundances of three isotopes of iron: Fe-54, 5.85% Fe-56, 91.75% Fe-57, 2.40%
Calculate the relative atomic mass of iron.
Q3 Sulphur has two main isotopes: sulphur-32 and sulphur-34. Data from experiments showed that the relative
atomic mass of sulphur is 32.1. What is the abundance of both isotopes of sulphur?
Q4 State one use for the following isotopes.
(a) Phosphorus-32
(b) Iodine-131
(c) Cobalt-60
(d) Sodium-24








CONCEPT MAP

MATTER AND ATOMIC STRUCTURE



Arrangement of Atomic structure Electron structure Isotope
particles


Solid Gas Proton Electron Examples Uses of isotopes
and natural – Archaeology
Liquid Neutron abundance – Agriculture
– Medical
– Power source
First shell: 2 electrons
– Industry
Second shell: 8 electrons
Third shell: 8 electrons
maximum 18 electrons if
Nucleus proton number exceeds 20




29
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Chemistry SPM Chapter 2 Matter and the Atomic Structure

SPM Practice 2




Objective Questions 5. A liquid is heated until it 2.2 The Development of the
Choose the correct answer. becomes a gas. The gas is Atomic Model
then cooled until it becomes
2.1 Basic Concepts of Matter a solid. Figure 2 shows a 7. Which scientist discovered
graph of temperature against the electron?
time for the change that A J. J. Thomson
1. Which substance is made up occurred. B Ernest Rutherford
of atoms? Form
A Iron Temperature (°C) C Niels Bohr
Form
B Nitrogen D James Chadwick 4
4
C Water B
D Sodium chloride A C 8. Figure 4 shows a model of
an atom.
2. A gas jar filled with air is D Shell
placed above another gas
jar filled with bromine. After a Electron
few minutes, a uniform colour Time (min) Nucleus
is seen in both jars. What Figure 2 containing
process has taken place? At which section is freezing protons
A Sublimation taking place? Figure 4
B Diffusion
C Evaporation Which statement is correct
D Condensation 6. Figure 3 shows the cooling about the atomic model?
curve for a sample of pure A Constructed based on the
3. Which statement is correct vapour. existence of neutron.
about the Kinetic Theory of B Proposed by James
Matter? Temperature (°C) Chadwick.
A The particles in a liquid C Shows the nucleus is
are closely packed and neutral.
orderly. D Shows electrons move
B The particles in a solid in shells around the
move randomly. nucleus.
C The attractive forces t Time (min)
between gas particles are 9. Which of the following is
stronger than those in a Figure 3 correct about the charge of
liquid. an electron, neutron and
D The energy content of a Which statement is correct proton?
gas is higher than that of about the arrangement of Electron Neutron Proton
a solid.
particles at time t? HOTS A Negative Positive Neutral
4. Figure 1 shows a change in A All the molecules move
state. randomly. B Negative Neutral Positive
B All the molecules vibrate
Process X and rotate about fixed C Positive Neutral Negative
Ice Water positions. D Positive Negative Neutral
Figure 1 C Some molecules vibrate
and rotate about fixed
Which of the following is positions and the rest 10. Which scientist described an
process X? move freely. atom as a sphere that cannot
A Melting D Some molecules move in be divided?
B Freezing an orderly manner and A Bohr
C Boiling the rest move randomly. B Chadwick
D Condensation C Dalton
D Rutherford



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Chemistry SPM Chapter 2 Matter and the Atomic Structure
11. Which of the following is 16. A fluorine atom 9 F has the A G and L C L and M
19
correct about the atomic same number of neutrons as B J and L D J and M
model proposed by Ernest an atom of
3
Rutherford? A 10 Ne C 20 Ca 22. Helium has two isotopes, 2 He
40
12
4
A Atom is positively 23 32 and 2 He. Which statements
charged. B 11 Na D 16 S are correct?
B Atom is very tiny. 17. Which of the following shows I 3 2 He has two protons.
C Protons are found in the the electron arrangement of II 4 2 He has four electrons.
nucleus. atom 7 N? III He has one neutron.
15
3
D Almost all the mass of an A 2.5 C 2.8.5 4 2
atom is concentrated at B 2.6 D 2.8.6 IV 2 He has nucleon number
the nucleus. 2. Form
18. Atom Y has 18 protons and A I and II
Form
12. In Chadwick’s atomic model, 22 neutrons in its nucleus. B I and III
he stated that What is the number of C II and III 4
4
A electrons move in shells. valence electrons in atom Y? D III and IV
B protons and neutrons A 2 C 5
are concentrated at the B 4 D 8 23. Uranium has two isotopes
92 U and 92 U. The lighter
centre of the atom. 19. An atom of element X has 235 238
C nucleus of an atom is nucleon number 14 and 8 isotope has
positively charged. neutrons. Which statement A 146 neutrons.
D neutrons and protons are is correct about an atom of B 235 electrons.
of the same size. C a total of 238 protons and
element X?
A The atom has 7 protons. neutrons.
2.3 Atomic Structure B Its electron arrangement D three neutrons less than
is 2.4. the heavier isotope.
C The atom has 3 valence
13. Atom X has proton number 5 electrons. 24. Which isotope is used by
and nucleon number 11. doctors to detect and treat
14
31
What is the number of 20. Atom 7 X and atom 15 Y have thyroid disease?
protons, electrons and same number of A C-14
neutrons in atom X? A protons. B Co-60
B neutrons. C I-131
Proton Electron Neutron C electrons. D valence D U-238
A 5 5 11 electrons.
B 5 5 6 25. Which pairing of isotopes and
their uses is correct?
C 5 6 6 2.4 Isotopes and Their Uses A Iodine-131; to make
D 6 6 5 atomic battery
21. Table 1 shows the proton B Carbon-14; to measure
number, nucleon number and the density of concrete
14. The nucleus of an atom
contains number of electrons for four C Cobalt-60; to treat cancer
A only electrons. particles. D Uranium-238; dating of
B only neutrons. Proton Nucleon Number of organic artifacts
C both protons and Particle Number Number electrons
neutrons. 26. Isotopes cobalt-60 and
D both protons and G 15 31 15 iodine-131 are used in
electrons. A dating of geological
J 16 32 18 formations.
15. An atom of chlorine-37 has L 15 32 15 B measuring and detection
17 electrons. What is the in industries.
number of neutrons in an M 18 40 18 C medical procedures.
atom of chlorine-37? Table 1 D generating nuclear
A 17 energy.
B 20 Which pair of particles
C 37 are isotopes of the same
D 54 element?



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Chemistry SPM Chapter 2 Matter and the Atomic Structure

Subjective Questions

Section A
1. Temperature (ºC) Figure 1 shows a graph of temperature against time for vapour
X as it is cooled until it becomes a solid.
T P
1 (a) Name the process of change in state of matter when vapour
T Q R X is cooled. [1 mark]
2
(b) At what temperature vapour X will change to a solid?
T [1 mark]
3
S
(c) Draw the arrangement of the particles at Form
(i) temperature T 1 HOTS
Time (min)
Form
0 (ii) temperature T 3 [2 marks]
Figure 1 4
4
(d) Explain why temperature remains constant at T 2? [2 marks]
(e) (i) State how the particles X move between R and S.
(ii) Explain your answer in 1(e)(i). [2 marks]
(f) Draw a graph of temperature against time which you will
obtain if you heat X from T 3 to T 1 . [2 marks]
Section B
2. (a) (i) Give the definition of isotope. [1 mark]
(ii) Name one element that has more than one isotope. [1 mark]
(iii) Based on your answer in 2(a)(ii), compare and differentiate the isotopes. [6 marks]
(iv) Isotopes are used in medicine, industry, archaeology and agriculture. State one use of isotope in
each field. [4 marks]
(b) Figure 2 shows changes in states of matter.
I II
Ice Water Steam

Figure 2
(i) Name process I. [1 mark]
(ii) State the change of kinetic energy of particles in process II. [1 mark]

(c) Matter is made up of tiny discrete particles.
Plan a laboratory experiment to prove the statement.
Include in your answer the following:
• Chemical substances required
• Experimental procedure
• Observation
• Conclusion
[6 marks]

Section C
3. Figure 3 shows the arrangement of particles of substance J
at two different temperatures. HOTS
(a) Compare and differentiate the two arrangements from
the following aspect:
• Particle arrangement
• Movement of particles
• Attraction force between particles
• Energy content of particles
At temperature T 1 At temperature T 2
Figure 3 [8 marks]
(b) Solid G has a melting point of 120 °C. Describe a laboratory experiment to determine the melting point
of G. Your answer should show how the melting point is determined. [12 marks]


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Theme 4: Technology in Chemistry Form 5

Chapter Chemistry SPM Chapter 4 Polymer
4 Polymer



















































CHAPTER FOCUS According to a news report from Bernama on April

2020, the Deputy Minister of Plantation Industries and
• Polymer Commodities, Willie Mongin said, Malaysia being the largest
• Natural Rubber latex glove producer in the world, is expecting the total Form
• Synthetic Rubber export for this year to be 225 billion units with the value
of more than RM20 billion. He said that this is due to
Malaysia being the main producer of latex gloves to the 5
health sector that is currently facing the critical situation
following the COVID-19 pandemic. Are latex gloves made
from natural rubber or synthetic rubber?





Access to
i-STUDY
i-STUDY SPM



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Chemistry SPM Chapter 4 Polymer
polymers and are found in animals and plants
4.1 Polymer from our surroundings.

3. Synthetic polymers are man-made polymers
1. A polymer is a long chain molecule that is through chemical reactions in laboratories or
made from a combination of many repeating factories.
basic units known as monomers.
4. Table 4.1 shows a few examples of natural
2. Polymerisation reaction is a monomer polymers, their monomers and sources.
combination reaction to produce a long chain
molecule known as a polymer. Table 4.1 Examples of natural polymers and their
monomers
Polymerisation
Natural
polymer Monomer Source
Polymer
Monomer Starch Glucose Plant
Figure 4.1 Polymerisation reaction Protein Amino acid Plant or
animal
3. A polymer is a macromolecule made from Fats Glycerol and Plant or
thousands of repeating basic units. Hence, the fatty acids animal
polymer has a very large relative molecular Natural rubber Isoprene Plant
mass.
Cellulose Glucose Plant
4. The physical and chemical properties of a Deoxyribonucleic Nucleotide Plant or
polymer are different from its monomers. acid (DNA) animal
5. Polymers can be classified into several groups 5. Continuous research and development has
based on their sources, properties and type of enabled scientists to produce synthetic
polymerisation reaction to produce them. polymers using the similar monomers found
in the natural polymers.
Source of Polymer
6. Plastic is the general name of synthetic
1. Polymers can be classified into natural polymer.
polymers or synthetic polymers.
7. Table 4.2 shows the examples of synthetic
2. Natural polymers are naturally available
polymers and their monomers.

Table 4.2 Examples of synthetic polymers and their monomers
Synthetic polymer Monomer Characteristics Use
Polyethene Ethene • Durable • Plastic packaging
(Polythene / Polyethylene) • Strong • Plastic bottles
• Plastic bag
Polypropene Propene • Durable • Textile
Form
(Polypropylene) • Toys
• Plastic bottle
5
Polyvinyl chloride, PVC Chloroethene • Hard • Electrical insulator
(Polychloroethene) • Strong • Water pipes
• Artificial leather
Polystyrene Styrene • Light • Food packaging
(Polyphenylethene / (Phenylethene / • Heat insulator • Toys
Polyphenylethylene) Phenylethylene) • Heat insulators
Perspex Methyl 2-methylpropenoate • Hard • Signboard
(Polymethyl methacrylate) (Methyl methacrylate) • High tensile strength • Window glazing
• Rigid • Furniture



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Chemistry SPM Chapter 4 Polymer

Types of Polymers
1. Table 4.3 shows types of polymers, their properties and examples.
Table 4.3 Types of polymers, their properties and examples
Type of polymers Properties Examples
Thermoplastic polymers (a) Thermoplastic polymers are polymers that can be • Polyethene
remoulded and recycled many times. • Polypropene
(b) When heated, thermoplastic polymer chains slide • Polyvinyl chloride
against one another to make it soft and melts. (PVC)
(c) When cooled, thermoplastic polymer chains stick to • Nylon
each other and solidify to become hardened. • Polystyrene
(d) Thermoplastic polymers have low heat resistance, • Perspex
elastic, light and can be bent easily. • Polyester

Thermosetting polymers (a) Thermosetting polymers can be only cast once. • Melamine
They are normally disintegrated, decomposed due to • Bakelite
chemical degradation and burnt when heated to very • Epoxy
high temperatures. Hence, they cannot be remoulded
and recycled after heating.
(b) The presence of cross-links between polymer chains
makes the polymer chains difficult to slide against one
another and difficult to melt.
Cross-link
(c) Thermosetting polymers are heat resistant, hard, long-
lasting and can be moulded once.
Elastomer polymers Elastomer polymers possess high elasticity properties, can • Natural rubber
be stretched repeatedly or subject to tension and return to • Styrene-
Cross-link
their original shape and size when released. butadiene rubber
(SBR)
• Polyurethane
• Polybutadiene






Polymerisation Reaction 4. The addition polymerisation of ethene is shown
1. There are two types of polymerisation in Figure 4.2.
reactions: H H H H H H
(a) Addition polymerisation + C C + C C + C C +
(b) Condensation polymerisation H H H H H H
Ethene molecules
Addition Polymerisation Addition
polymerisation Form
1. The monomers for addition polymerisation
have double covalent bonds between two H H H H H H 5
carbon atoms, C=C. C C C C C C
2. During the addition polymerisation, the double H H H H H H
covalent bond between two carbon atoms is Polyethene
broken to form a single covalent bond with the H H H H
carbon atom from the adjacent monomer. The n C C C C
monomers are added to the molecule chain to
form a polymer. H H H H n
Ethene Polyethene
3. Addition polymerisation normally occurs at Figure 4.2 Addition polymerisation of ethene
high pressure of 1000 atm and temperature of
200 °C.

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Chemistry SPM Chapter 4 Polymer
5. Table 4.4 shows the examples of chemical equations of addition polymerisation.

Table 4.4 Examples of chemical equations of addition polymerisation
Monomer Polymer Chemical equation of addition polymerisation

Ethene Polyethene H H H H
(Polythene / Polyethylene)
n C C C C
H H H H n
Ethene Polyethene
(Polythene/ Polyethylene)

Propene Polypropene H CH 3 H CH 3
(Polypropylene)
n C C C C
H H H H n
Propene Polypropene
(Polypropylene)


Chloroethene Polyvinyl chloride, PVC H CI H CI
(Polychloroethene)
n C C C C
H H H H n
Chloroethene Polyvinyl chloride, PVC
(Polychloroethene)

Styrene Polystyrene
(Phenylethene / (Polyphenylethene / H H
Phenylethylene) Polyphenylethylene)
n C C C C
H H H H n
Styrene Polystyrene
(Phenylethene/ (Polyphenylethene/
Phenylethylene) Polyphenylethylene)


Methyl 2-methylpropenoate Perspex H CH H CH
(Methyl methacrylate) (Polymethyl methacrylate) 3 3
n C C C C
H CO CH 3 H CO CH 3 n
2
2
Methyl 2-methylpropenoate Perspex
(Methyl methacrylate) (Polymethyl methancrylate)

Condensation Polymerisation 3. Examples of polymers that are produced
Form
1. Condensation polymerisation involves from condensation polymerisation are
5
a combination of at least two different terylene (polyester), nylon (polyamide) and
monomers and the removal of small polyurethane.
molecules as a by-product such as water, H 2 O, 4. Table 4.5 shows the examples of chemical
hydrogen chloride, HCl or ammonia, NH 3 . equations of condensation polymerisation.
2. The monomers involved in condensation
polymerisation consist of two or more
functional groups.





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Chemistry SPM Chapter 4 Polymer

Table 4.5 Examples of chemical equations of condensation polymerisation
Polymer Monomers Chemical equation of condensation polymerisation
Terylene • 1,2-ethanediol H H O H H H O O
• Terephthalic acid n O C C O + n C C O C C O C C + n H O
H H HO OH 2
H H H H n
1,2-ethanediol Terephthalic acid Terylene Water
Nylon • 1,6-hexanediamine H H H H H H H O H H H H H H H H O
• Decanedioyl n H n C C C C C C C C C C
dichloride H N C C C C C C N + CI
H H H H H H H CI H H H H H H H H
1,6-hexanediamine Decanedioyl dichloride


H H H H H H H H O H H H H H H H H O
N C C C C C C N C C C C C C C C C C + n HCI
H H H H H H H H H H H H H H n
Nylon Hydrogen
chloride
Polyurethane • Methylene diphenyl H H H
diisocyanate n O C N C N C O C C OH
• 1,2-ethanediol + n HO
H H H
Methylene diphenyl diisocyanate 1,2-ethanediol

O H O H H
C N C N C O C C O
H H H H H n
Polyurethane


Activity 4.1


Aim: To synthesise nylon and to study the properties of the nylon produced.
Materials: 1,6-hexanediamine, C 6 H 16 N 2 , decanedioyl dichloride, C 10 H 16 C 12 O 2 , hexane, C 6 H 14 , sodium
–3
hydroxide powder, NaOH, distilled water, ethanol, C 2 H 5 OH, 5 mol dm hydrochloric acid,
–3
HCl, 5 mol dm sodium hydroxide solution, NaOH and acetone, (CH 3 ) 2 CO
Apparatus: Beaker, measuring cylinder, glass rod, forceps, meter rule, test tube, test tube rack, scissors
Procedure:
1. 50 cm of hexane, C 6 H 14 is measured and poured into a beaker. Form
3
2. 2 cm of decanedioyl dichloride, C 10 H 16 C 12 O 2 is added into hexane, C 6 H 14 in the beaker. The
3
mixture is stirred and labelled as Solution A. 5
3. 50 cm of distilled water is measured and poured into another beaker.
3
3
4. 3 cm of 1,6-hexanedianime, C 6 H 16 N 2 and 1 g of sodium hydroxide powder, NaOH are added into
50 cm of the distilled water in the beaker. The mixture is stirred and labelled as Solution B.
3
5. By using a glass rod, Solution B is poured slowly through the wall of the beaker to the surface of
Solution A to avoid mixing.
6. The layer formed between the two solutions is pinched using forceps and pulled out slowly then
coiled around the glass rod as shown in Figure 4.3.





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Chemistry SPM Chapter 4 Polymer




Glass rod
Nylon thread


Solution B
(1,6-hexanediamine, C H N 2
16
6
and sodium hydroxide,
NaOH in distilled water)
Nylon produced
between the
two solutions Solution A
(Decanedioyl dichloride,
C H CI O in hexane, C H )
2
16
2
14
10
6
Figure 4.3

7. The glass rod is winded slowly to pull out the nylon until no more nylon is produced.
8. The nylon produced is rinsed with ethanol, C 2 H 5 OH on the glass rod, followed by distilled water
and dried.
9. When the nylon is dry, the colour of the nylon is observed and its length is measured and recorded.
10. The nylon threads are cut into a length of 15 cm and are pulled at both ends.
11. Whether the nylon thread breaks or not is observed and recorded.
12. Three nylon thread strands with a length of 3 cm are cut and prepared.
13. The nylon thread is immersed into three test tubes separately containing 5 cm of 5 mol dm
–3
3
concentrated hydrochloric acid, HCl, 5 cm of 5 mol dm sodium hydroxide solution, NaOH and
–3
3
5 cm of acetone, (CH 3 ) 2 CO respectively.
3
14. Solubility of nylon thread in acid, alkali and organic solvent is observed and recorded.
Results:
Length of nylon thread formed 3.6 m
Colour of nylon thread formed White
Condition of nylon thread after stretching Nylon thread does not break
Nylon thread immersed in concentrated hydrochloric Nylon thread dissolves
acid, HCl
Nylon thread immersed in sodium hydroxide solution, No change. Nylon thread does not dissolve
NaOH
Nylon immersed in acetone, (CH 3 ) 2 CO (organic Nylon thread dissolves very slowly
solvent)
Discussion:
1. Nylon is produced through condensation polymerisation between 1,6-hexanediamine, C 6 H 16 N 2
and decanedioyl dichloride, C 10 H 16 C 12 O 2 .
Form
5
O O O O
n H N (CH ) NH + n CI C (CH ) 2 8 C CI NH (CH ) 2 6 NH C (CH ) 2 8 C + n HCI
2
2 6
2
n
1,6-hexanediamine Decanedioyl dichloride Nylon Hydrogen chloride
2. Nylon is a co-polymer because it is made of two different monomers.
3. Decanedioyl dichloride, C 10 H 16 C 12 O 2 dissolves in hexane, C 6 H 14 while 1,6-hexanediamine,
C 6 H 16 N 2 dissolves in water.
4. When the two solutions get in touch with each other, decanedioyl dichloride, C 10 H 16 C 12 O 2 and
1,6-hexanediamine, C 6 H 16 N 2 react in between the layers of the solution.



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Chemistry SPM Chapter 4 Polymer


5. Nylon thread is pulled out from the polymer film that forms in between the surface of the solutions.
6. When the polymer film is pulled out, the new polymer film forms immediately. Hence, the nylon
formation happens continuously until all reactants react completely.
7. The nylon formed is white. Nylon can be dipped into dyeing solutions to produce almost all
colours. Nylon is elastic and very strong, which is stronger than polyester fibre. Nylon is a
flammable synthetic fibre and dissolves in concentrated acids and organic solvents. Nylon is non-
toxic and does not cause irritation to the skin, therefore it is used to make fabrics and important in
textile industry.
Discussion:
Nylon is produced from condensation polymerisation between decanedioyl dichloride, C 10 H 16 C 12 O 2
and 1,6-hexanediamine, C 6 H 16 N 2 . The hydrogen chloride molecule, HCl is the by-product of the
condensation polymerisation reaction. The nylon produced is white, elastic, flammable, non-toxic and
very strong.


The Use of Polymer in Daily Life 3. Characteristics of synthetic polymers:
1. Synthetic polymers are widely used in various (a) Very stable, chemically inert and
aspects in daily life because they have the unreactive
properties that cannot be found in natural (b) Resistant to high heat
polymers. Normally, the synthetic polymers do (c) Good heat insulator
not corrode or decay easily with the presence (d) Strong and hard
of oxygen, water, sunlight and other chemicals. (e) Lightweight and low density
2. One advantage of synthetic polymers is they (f) Non-flammable
can be designed for specific purpose or specific (g) Easy to shape
use with their specific properties. Combination (h) Easy to colour
of various properties in a specific designed (i) Relatively cheap
synthetic polymer causing it to be very useful (j) More durable
in certain purposes. 4. Table 4.6 shows the uses of synthetic polymers
in daily life.

Table 4.6 Uses of synthetic polymers in daily life
Field Synthetic Polymer Properties Uses
Medical • Polyethene or • Light To make protheses, devices installed to
polyethylene (PE) • Strong replace the missing part of the body due to
• Polypropene or accidents or illnesses e.g. artificial limb
polypropylene (PP) Form









Polypropene or Chemically inert and Polypropylene is a cost-effective medical- 5
Polypropylene resistant to many grade plastic material such as syringes,
(PP) solvents blood collection tubes and personal
protective equipment (PPE)










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Chemistry SPM Chapter 4 Polymer

Polyurethane (PU) • High tensile strength To make the coating for cardiac pacemaker
• Resistant to water, oil leads, coating for breast implants and
and grease catheters (tube that is inserted to bladder)
• Resistant to mucor and other general tubing.
and fungus
Polyethylene • Very durable To make artificial arteries and cardiovascular
terephthalate (PET) • Lightweight implants such as heart valve implant
• Resistant to impact
and shatter
Packaging • Polyethene or • Lightweight To make plastic bags, bottles and food
polyethylene (PE) • Non-toxic packaging
• Polypropene or • High cracking
polypropylene (PP) resistance







Coating Teflon or • Non-stick To make the coating on non-stick cookware
polytetrafluoroethylene • Chemically inert
(PTFE) • Impermeable






Safety Bakelite Good heat and To make cookware handles and electrical
electrical insulators switches









Kevlar • Strong To make firemen uniforms, bulletproof vests,
• Heat resistance soldiers’ helmets









Form
Textile Nylon • Elastic To make threads, socks, clothes, swimwear,
5
• Very strong conveyer belts, seat belts, fishnets, fishing
• Resilient lines and parachutes.
• Dry quickly
• Damage resistant
to oil and many
chemicals









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Chemistry SPM Chapter 4 Polymer

Terylene • Elastic, To make clothes, raincoats, sailclothes,
• Creasing resistance fleece jackets and parachutes
• Dry quickly








Building Acrylic polymer • High heat resistance To make roof waterproofing
• Sunlight resistance
• Water resistance









Polymers and the Environment 6. Air pollution
1. Diverse and useful properties found in synthetic Open burning of synthetic polymers releases
polymers increase their demands and uses over pungent, acidic and toxic gases such as
the years. This is because synthetic polymers hydrogen chloride gas, HCl, hydrogen cyanide
have the properties of long-lasting, lightweight gas, HCN and methane gas, CH 4 into the air.
and can be shaped and moulded easily. Acidic gas dissolves in rainwater to produce
2. The main raw material for making synthetic acid rain. On the other hand, methane gas, CH 4
polymers is non-renewable and diminishing may cause serious greenhouse effect while
petroleum. hydrogen chloride gas, HCl and hydrogen
3. The stable, durable and non-biodegradable cyanide gas, HCN may cause respiratory
problems to humans.
properties of synthetic polymers make them
take a very long time to disintegrate and cause 7. Although synthetic polymers bring adverse
serious land, water and air pollution. effects on the environment, we can still use
them wisely and responsibly.
4. Land pollution
Landfills are filled with non-biodegradable 8. The following are several methods to overcome
synthetic polymer products e.g. plastic bags the pollution problems by using more
and plastic bottles. When plastic products are sustainable synthetic polymers:
not disposed of properly, they will become the (a) Adding additives to synthetic polymers
breeding grounds for mosquitoes and cause to enable plastics to decompose naturally Form
the spreading of dengue disease. On the other by bacteria (biodegradable) or decompose
hand, they will also lead to the blockage of by light (photodegradable) to reduce 5
drainage system and cause flash floods. pollution.
5. Water pollution (b) Recycling to ensure that synthetic
polymers do not end up in landfills.
Plastic waste contaminates lakes, rivers and (c) Reusing plastic products to reduce the
oceans. This causes death of aquatic life, as plastic dumping issue in landfills. It will
well as the introduction of microplastics be safer for wildlife.
into the food chain. Eventually, it will affect (d) Proper way of disposing synthetic
humans health. polymers.





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Chemistry SPM Chapter 4 Polymer

Checkpoint 4.1 4.2 Natural Rubber
Q1 Certain beds are covered by a material that
contains polymer A which consists of a basic
unit. The structure of polymer A is shown in 1. Natural rubber is a natural polymer found in
Figure 4.4. latex from rubber trees (Hevea brasiliensis).
H H H H H H
2. Latex is a colloidal white fluid obtained when
C C C C C C a rubber tree bark is tapped.
CI H CI H H H
Figure 4.4 3. Rubber particles disperse separately in water
(a) What is meant by polymer? to form a colloid.
(b) State the IUPAC name of the basic unit that 4. The IUPAC name of isoprene, which
forms polymer A.
(c) Draw the structural formula of the monomer is the monomer of natural rubber, is
involved. 2-methylbut-1,3-diene.
(d) Name the type of polymerisation reaction H CH H H
3
used to produce polymer A.
C
C
H
H
C
C
(e) State one similarity and one difference Isoprene or 2-methylbut-1,3-diene
between the structural formula of polymer
A and its basic unit. Figure 4.5 Isoprene
(f) Name the acidic gas released when 5. Thousands of isoprene units undergo addition
polymer A is disposed of by burning.
Q2 Plastic bottles produced from a synthetic polymerisation to form polyisoprene or natural
polymer can be used to make various products. rubber.
Improper disposal of these products can cause H CH H H H CH H H
3
3
pollution.
(a) Suggest a product that can be made by n C C C C C C C C
using plastic bottles. H H H H n
(b) Justify the use of synthetic polymers. Isoprene Polyisoprene or natural rubber
Figure 4.6 Polymerisation of isoprene
Characteristics of Natural Rubber
1. Rubber polymers are natural elastomeric polymers.
2. The characteristics of natural rubber are shown in Table 4.7.
Table 4.7 Characteristics of natural rubber
Characteristics Descriptions
Soft Natural rubber exists as a soft white solid at room conditions.
Elastic Natural rubber can be stretched and returns to its original shape and size when released.
Reasons:
(a) In normal conditions, rubber polymer chains are folded until they are tangled.
(b) When natural rubber is stretched, the rubber becomes elongated and thinner because rubber
polymer chains are straightened. When the stretching force is released, the rubber polymer
chains return to their original states.
Form
5 cm 5 cm
5
7 cm
4 cm Streched 3 cm Released 4 cm


Rubber polymer Rubber polymer Rubber polymer chains
chains tangled chains straightened return to their original states







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Chemistry SPM Chapter 4 Polymer

Low heat • At a temperature higher than 50 °C, natural rubber becomes soft, melts and sticky. It will
resistance decompose when heated above 200 °C.
• When the natural rubber is cooled to a temperature lower than 10 °C, it becomes hard and brittle
like plastics.
Electrical Natural rubber does not conduct electricity.
insulator Reasons:
(a) Rubber polymer chains exist as neutral molecules.
(b) There are no free-moving ions or electrons to carry charges.
Not resistant to Natural rubber is easily oxidised.
oxidation Reasons:
The double covalent bond between carbon atoms, C=C in rubber polymer chains can be reacted
with oxygen, O 2 and ozone, O 3 in the atmosphere in the presence of ultraviolet light. Hence, the
natural rubber will be decomposed.
Reactive to Natural rubber reacts easily with organic solvents, acids and alkalis.
chemicals
Does not (a) Natural rubber is impermeable to water.
dissolve (b) Natural rubber dissolves in organic solvents such as methylbenzene, petrol, and kerosene.
in water Reasons:
(Waterproof) (a) Rubber polymer chain is made of a hydrocarbon chain. Intermolecular forces (van der Waals
forces) between rubber polymer chains are different from hydrogen bonding between water
molecules. The rubber polymer chain does not form hydrogen bonds with water molecules.
Hence, natural rubber does not dissolve in water.
(b) Intermolecular forces between rubber polymer chains and intermolecular forces in organic
solvents are the same (i.e. van der Waals forces). Hence, natural rubber dissolves in organic
solvents.

Uses of Natural Rubber (c) Rubber blocks and rubber balls which
1. Natural rubber can be manufactured for various are used in the buildings near the railway
products. Its limitations can be overcome by tracks or in the earthquake areas (as
adding certain additives to obtain the desired earthquake isolators) to absorb vibration
characteristics. Coagulation of Latex
2. Natural rubber is used to make: 1. Latex is a milky fluid obtained from tapped
(a) Elastic products such as tyres, balloons, rubber tree bark. It is a colloidal suspension
rubber bands and engineering components of rubber particles in water.
(b) Gloves, rubber tubes and rubber covers 2. Rubber particle is made of rubber polymer
which are impermeable to water and other chains surrounded by a layer of protein
liquids membrane.
(c) Electrical insulators for cables and
electrical appliances 3. The negatively-charged protein membrane
(d) High friction materials such as rubber causes rubber particles to repel each other. Form
door stoppers, floor mats and shoe soles This prevents rubber particle fusion and hence
(e) the mixture of rubberised bitumen and prevents latex coagulation. 5
latex cement which is used to pave the Repulsion Negative charge
surface of roads and concrete of bridges.
3. Vulcanised rubber which has better
characteristics (more elastic, more resistant to
heat and oxidation) is used to make:
(a) Rubber tubes and gloves Rubber polymer Protein membrane
(b) Vehicles tyres. Carbon is added to make Figure 4.7 Repulsion between the negatively-charged
the vulcanised rubber more durable and protein membranes of rubber particles
maintain its elasticity for a longer duration


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Chemistry SPM Chapter 4 Polymer

Experiment 4.1


Aim: To study the coagulation process in latex and method to prevent latex coagulation
Problem statement: Does acid cause latex to coagulate while alkali prevents latex coagulation?
Hypothesis: The presence of acid causes latex to coagulate while the presence of alkali prevents latex
from coagulating

Variables:
(a) Manipulating variable: Ethanoic acid, CH 3 COOH and ammonia solution, NH 3
(b) Responding variable: Coagulation of latex
(c) Fixed variable: Volume of latex

–3 –3
Materials: Latex, 1 mol dm ethanoic acid, CH 3 COOH and 1 mol dm ammonia solution, NH 3
Apparatus: Beaker, measuring cylinder, dropper, glass rod
Procedure:
1. Three beakers are labelled A, B and C respectively.
3
2. 20 cm of latex is measured using a measuring cylinder and poured into each beaker.
3. 2 cm ethanoic acid, CH 3 COOH is measured using another measuring cylinder and poured into
3
beaker A while stirring continuously with a glass rod.
4. Step 3 is repeated using ammonia solution, NH 3 for beaker B to replace ethanoic acid, CH 3 COOH
while beaker C is set as a control without adding any solution to latex.
5. All the three beakers are left overnight. The observations are recorded.
Observations:

Beaker Mixture Observation

A Latex + ethanoic acid, A white solid lump is formed quickly. Latex coagulates
CH 3 COOH quickly.

B Latex + ammonia solution, No change. Latex does not coagulate.
NH 3

C Latex A white solid lump is formed slowly. Latex coagulates
slowly.

Discussion:
1. Ethanoic acid, CH 3 COOH ionises in water to produce hydrogen ions, H . Hydrogen ions, H
+
+
Form
neutralise the negative charges on the protein membrane of rubber particles causing latex to
coagulate.
5
2. Ammonia solution, NH 3 neutralises lactic acid produced by bacteria on the protein membrane.
The negatively-charged protein membrane of rubber particles are remained to prevent latex from
coagulating.
3. When latex is exposed to the air, bacteria attack the protein membrane slowly to produce lactic
acid and coagulate the latex.
Conclusion:
The presence of acid causes latex to coagulate while the presence of alkali prevents latex from
coagulating. Hypothesis is accepted.




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Chemistry SPM Chapter 4 Polymer

Coagulation Process of Latex 4. The rubber polymer chains combine causing
1. When acid such as methanoic acid (formic the formation of white lumps and latex
acid) is added, coagulation of latex occurs coagulates.
faster. 5. Latex coagulation can occur naturally by
leaving the latex exposed to the air.
2. Hydrogen ions, H from acid neutralise
+
negative charges on the protein membrane. 6. Bacteria in the air attack the protein membrane
to produce lactic acid. Hydrogen ions, H
+
3. The neutralised rubber particles collide with
each other causing the protein membranes to from the acid neutralise the negatively-charged
break. protein membrane of rubber particles causing
latex to coagulate.
Neutralised
rubber particles Latex Coagulation Prevention
H + H + H + H + H + H + H + + Coagulation 1. Alkaline solution such as ammonia solution,
H + H + H + H Process of NH 3 is added into latex to prevent coagulation.
H + + H + Latex
H 3D Model 2. Hydroxide ions, OH from alkali neutralise

H + + + H + H +
H H hydrogen ions, H from acid produced by
+
Collide bacteria.
Protein membrane
ruptures 3. The negatively-charged protein membrane of
rubber particles remained. The rubber particles
Rubber polymers will continue to repel each other and cannot
coagulate
combine to prevent latex coagulation.
Figure 4.8 Coagulation of latex

Activity 4.2


Aim: To produce a product from latex
Materials: Latex, ethanoic acid, CH 3 COOH and distilled water
3
Apparatus: 250 cm beaker, 250 cm measuring cylinder, 10 cm measuring cylinder and glass rod
3
3
Procedure:
1. 200 cm of latex is measured using a 250 cm measuring cylinder and poured into a beaker.
3
3
2. 5 cm of ethanoic acid, CH 3 COOH is measured using a 10 cm measuring cylinder and poured
3
3
into the beaker.
3. The mixture is stirred using a glass rod.
4. The glass rod is left in the mixture for 1 minute to allow a layer of latex to form on the glass rod. Form
5. The glass rod is taken out from the mixture and the latex layer is allowed to dry.
6. The glass rod is dipped into the mixture again to allow a thicker layer of latex to form. 5
7. Steps 5 and 6 are repeated until 1 cm thickness of latex layer is obtained.
8. The glass rod is removed from the mixture and left to dry for 2 days.
9. The dry latex layer is stripped from the glass rod to get a rubber tube.
10. The rubber tube is washed with distilled water and dried in the air.

Observations:
1. White solid lump is formed when ethanoic acid, CH 3 COOH is added to the latex.
2. Rubber tube is formed when the dry latex layer is stripped from the glass rod.



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Chemistry SPM Chapter 4 Polymer

Discussion:
1. The rubber tube produced is washed with distilled water to remove the excess ethanoic acid,
CH 3 COOH on it.
2. Besides rubber tube, rubber eraser and rubber glove can be produced using this method.
3. The rubber tube produced cannot be stored for a very long time because the product of natural
rubber (non-vulcanised rubber) is easily oxidised.
4. The rubber tube produced is elastic, soft and has low heat resistance.
Conclusion:
Rubber tube can be produced from latex. Hypothesis is accepted.


SPM Highlights Vulcanisation of Rubber
1. Natural rubber which is soft and easily oxidised
Glove is made of latex in liquid form. Which substance has limitations for certain applications.
can be used to prevent coagulation of latex?
A Ethanol 2. The characteristics of natural rubber can be
B Methanoic acid improved through a vulcanisation process.
C Sodium chloride
D Ammonia 3. Vulcanisation is a process of producing better
quality rubber that is more elastic through
Examiner’s Tip
Alkali, ammonia, NH 3 ionises in water to produce the production of cross-links between rubber
negative ions, OH . OH ions neutralise H ions from polymer chains.
+


acid produced by bacteria. The negatively-charged 4. Vulcanisation can be carried out by:
protein membrane of rubber particles remained. The
rubber particles will continue to repel each other. (a) Heating a mixture of 1-3% of sulphur by
Latex coagulation can be prevented. mass and natural rubber in the industry
Answer: D (b) Immersing natural rubber into disulphur
dichloride solution, S 2 Cl 2 in the laboratory
5. During vulcanisation process, some of the double covalent bonds between carbon atoms, C=C in
rubber molecules will react with sulphur or other substances to produce sulphur cross-links as shown in
Figure 4.9. Normally, there are one to four sulphur atoms in a sulphur cross-link.

CH CH CH
3 3 3
CH C CH CH CH C CH CH CH C CH CH
2 2 2 2 2 2
+
CH C CH CH CH C CH CH CH C CH CH
2 2 2 2 2 2
CH 3 CH 3 CH 3
+
Sulphur (S)
Vulcanisation
CH 3 CH 3 CH 3
Form
CH 2 C CH CH 2 CH 2 C CH CH 2 CH 2 C CH CH 2
5
S S S S Sulphur
cross-link
S S S S
CH C CH CH CH C CH CH CH C CH CH
2 2 2 2 2 2
CH CH CH
3 3 3
Figure 4.9 Formation of sulphur cross-link during vulcanisation







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Chemistry SPM Chapter 4 Polymer

Activity 4.3


Aim: To prepare vulcanised rubber
Materials: Latex, ethanoic acid, CH 3 COOH and disulphur dichloride, S 2 Cl 2 in methylbenzene, C 7 H 8
solution

Apparatus: Beaker, dropper, glass rod, retort stand, white tile, bulldog clip, knife and measuring
cylinder
Procedure:

Latex + ethanoic acid,
CH COOH
3

Glass rod
White tile
White tile

Rubber
sheet

Figure 4.10
1. 25 cm of latex is measured using a measuring cylinder and poured into a beaker.
3
2. 5 drops of ethanoic acid, CH 3 COOH are added into the latex. The mixture is stirred using a glass
rod.
3. The latex paste is poured onto a white tile to spread over using the glass rod.
4. The rubber sheet on the white tile is left to dry for 2 days.
5. The dried rubber sheet is cut on the white tile into two smaller pieces of the same size using a
knife.
6. One of the rubber pieces is dipped into disulphur dichloride solution, S 2 Cl 2 for 5 minutes.
7. The rubber pieces are dried by hanging them on a retort stand using bulldog clips.
8. Both rubber pieces are held and pressed.
9. All of the observations are recorded.
Observations:
Type of rubber Observation
Natural rubber piece Soft and less elastic
Hard and more elastic
Rubber piece dipped in disulphur dichloride solution, S 2 Cl 2
Discussion: Form
1. Ethanoic acid, CH 3 COOH is added to coagulate the latex.
2. Latex needs to be spread on the white tile to obtain an even thickness of rubber sheet. 5
3. When the rubber piece becomes harder and more elastic after dipping into disulphur dichloride
solution, S 2 Cl 2 , it indicates that a vulcanised rubber is successfully produced.
Conclusion:
Vulcanised rubber which is harder and more elastic is prepared by dipping the rubber piece into
disulphur dichloride solution, S 2 Cl 2 .









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Chemistry SPM Chapter 4 Polymer

Alternative Vulcanisation Method 2. A few alternative vulcanisation methods to
1. Vulcanisation with sulphur cannot be used for produce a sulphur-free vulcanised rubber:
certain types of rubber, for example, synthetic (a) Using metal oxides
rubber which does not contain double covalent (b) Using peroxides
bonds between carbon atoms, C=C. (c) Irradiation
3. The sulphur-free vulcanised rubber is more environmentally friendly.
Experiment 4.2


Aim: To investigate the elasticity of unvulcanised rubber and vulcanised rubber
Problem statement: Is vulcanised rubber more elastic than unvulcanised rubber?
Hypothesis: Vulcanised rubber is more elastic than unvulcanised rubber
Variables: Vulcanised rubber is more elastic than unvulcanised rubber
(a) Manipulating variable: Unvulcanised rubber strip and vulcanised rubber strip
(b) Responding variable: The elongation of rubber strips when the weight is removed
(c) Fixed variable: Mass of weight, size of rubber strips
Materials: Unvulcanised rubber strip and vulcanised rubber strip
Apparatus: Retort stand with clamp, bulldog clip, 50 g weight, meter ruler, knife and hook
Procedure:


Bulldog clip Bulldog clip
Vulcanised Unvulcanised
rubber strip rubber strip
Weight Weight


Figure 4.11
1. Unvulcanised rubber strip and vulcanised rubber strip are hung using bulldog clips as shown in
Figure 4.11.
2. The initial length, L 1 of both rubber strips is measured and recorded.
3. A 50 g weight is hung on each rubber strip and each final length, L 2 is measured and recorded.
4. The weights are removed and the length of each rubber strip, L 3 is measured and recorded.
Results::
Type of rubber strip Unvulcanised rubber Vulcanised rubber
Initial length, L 1 (cm) 10.0 10.0
Form
Final length when 50 g weight is hung, L 2 (cm) 12.7 11.5
5
Length when weight is removed, L 3 (cm) 10.5 10.0
Discussion:
1. Unvulcanised rubber strip shows the highest elongation.
2. Vulcanised rubber is more elastic because it returns to its original length after the 50 g weight is
removed.
3. It is predicted that the unvulcanised rubber strip will snap first if the mass of weight continues to
increase because unvulcanised rubber is weaker than vulcanised rubber.
Conclusion:
Vulcanised rubber is more elastic than unvulcanised rubber. Hypothesis is accepted.

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Chemistry SPM Chapter 4 Polymer

Properties of Vulcanised Rubber
1. Table 4.8 shows the comparison of properties between unvulcanised rubber and vulcanised rubber.
Table 4.8 Comparison of properties between unvulcanised rubber and vulcanised rubber

Unvulcanised rubber Properties Vulcanised rubber
Less elastic Elasticity More elastic
5 cm 5 cm
4 cm 4 cm S S S
S
Stretched
Stretched
7 cm
Unvulcanised rubber 7 cm Vulcanised rubber
3 cm strip is elongated and S S strip is elongated and
becomes thinner. 3 cm S
S becomes thinner.
Stretching is Stretching is
released released
6 cm Unvulcanised rubber 5 cm
strip cannot return to its Vulcanised rubber strip
original length and returns to its original
4 cm size because it was S S S length and size because
stretched over its 4 cm S its elasticity limit is not
elasticity limit. over.
Reason: Reason:
Weak Van der Waals forces between rubber Strong covalent bond in sulphur cross-links
polymer chains enable them to easily slide over between rubber polymer chains still enables
each other when the rubber is stretched but them to slide over each other when the rubber is
slightly difficult to return to its original shape and stretched and easier to return to its original shape
size after the rubber is released because it has and size after the rubber is released because it is
exceeded the elasticity limit. still in its elasticity limit.
Elasticity limit of unvulcanised rubber is low. Elasticity limit of vulcanised rubber is higher.

Weaker and softer Strength Stronger and harder
Reason: and Reason:
Weak van der Waals forces between rubber hardness Strong covalent bonds in sulphur cross-links
polymer chains are easier to overcome. between rubber polymer chains are more difficult
to break.

Low heat resistance (Easy to melt) Resistance High heat resistance (Hard to melt)
Reason: to heat Reason:
Weak van der Waals forces between rubber Strong covalent bond in sulphur cross-link
polymer chains need low heat energy to between rubber polymer chains needs high heat
overcome. Hence, unvulcanised rubber melts energy to break. Hence, vulcanised rubber will Form
easily when it is heated. melt at high temperatures. 5

Easily oxidised by oxygen, O 2 and ozone, O 3 Resistance Not easily oxidised by oxygen, O 2 and ozone, O 3
to
C C C C C C C C C C C C
oxidation
C C C C S S
S S
C C C C C C C C C C
Rubber polymer chains in unvulcanised rubber S S S S
(The number of C C is bigger) C C C C C C
Rubber polymer chains in vulcanised rubber
(The number of C C is smaller)





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Chemistry SPM Chapter 4 Polymer

Reason:
The presence of a lot of double covalent bonds Reason:
between carbon atoms, C=C in the rubber polymer The presence of fewer double covalent bonds
chains enables oxidation to occur more easily. between carbon atoms, C=C in the rubber polymer
chains causes oxidation more difficult to occur.
Insoluble in water Solubility Insoluble in water
Reason: in water Reason:
Rubber polymer chains are made of hydrophobic Rubber molecules are made of hydrophobic
hydrocarbon. hydrocarbon and sulphur.
Intermolecular attractive forces (van der Waals Intermolecular attractive forces (van der Waals
forces) between rubber polymer chains are forces) between rubber molecules are different
different from intermolecular hydrogen bonding from intermolecular hydrogen bonding between
between water molecules. water molecules.
Rubber polymer chains do not form hydrogen Rubber molecules do not form hydrogen bonds
bonds with water molecules. with water molecules.
Dissolves in organic solvent with higher solubility Solubility Dissolves in organic solvent with lower solubility
Reason: in organic Reason:
Intermolecular attractive forces (Van der Waals solvent Intermolecular attractive forces (Van der Waals
forces) between rubber polymer chains and forces) between rubber polymer chains and
intermolecular attractive forces between organic intermolecular attractive forces between organic
solvent molecules are the same. solvent molecules are the same.
Smaller molecular size of unvulcanised rubber Molecular size of vulcanised rubber is bigger.
causes the unvulcanised rubber to dissolve more Spaces between organic solvent molecules are
easily in organic solvent. limited. It is more difficult for bigger size vulcanised
rubber molecules to squeeze into the limited
spaces between organic solvent molecules.

Checkpoint 4.2
Q1 Figure 4.12 shows part of the structure of natural rubber polymer.
CH 3 CH 3 CH 3
CH 2 C CH CH 2 CH 2 C CH CH 2 CH 2 C CH CH 2
Figure 4.12
(a) Draw the structural formula of monomer of natural rubber and name the monomer using IUPAC nomenclature.
(b) State the type of reaction to form natural rubber.
(c) Write a balanced chemical equation for formation of natural rubber.
(d) Explain briefly how latex coagulation occurs naturally.
(e) Suggest a method to prevent latex coagulation.
Q2 Soft and easily oxidised natural rubber is unsuitable for certain applications. After improvement, the structure of
natural rubber changes to the structure as shown in Figure 4.13.
CH 3 CH 3 CH 3
CH C CH CH CH C CH CH CH C CH CH
2 2 2 2 2 2
S S S S
Form
S S S S
5
CH C CH CH CH C CH CH CH C CH CH
2 2 2 2 2 2
CH 3 CH 3 CH 3
Figure 4.13
(a) State a method to improve the properties of natural rubber. Describe briefly how to carry out the method you
mentioned in the laboratory.
(b) State one similarity and two differences in terms of properties between vulcanised rubber and unvulcanised
rubber.
(c) Explain why the vulcanised rubber is more elastic than the unvulcanised rubber.
(d) Suggest two alternative methods to overcome sulphur vulcanisation process. What is the advantage of the
product formed via the suggested methods?



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Chemistry SPM Chapter 4 Polymer

(a) Elastic
4.3 Synthetic Rubber (b) Hard
(c) Resistant to heat
(d) Resistant to chemicals such as acids and
1. Synthetic rubber is a synthetic elastomer alkalis
polymer which is elastic in nature. It will (e) Resistant to oxidation
return to its original shape and length when (f) Resistant to solvent (water and oil)
the stretching force is removed. (g) Good electrical insulator
2. The molecular structure of synthetic rubber is Uses of Synthetic Rubber
similar to natural rubber. Hence, the products 1. Nowadays, synthetic rubber is the main choice
made of synthetic rubber have natural rubber for manufacturing industries to produce various
characteristics, such as high elasticity. useful items due to its unique characteristics,
the advantage of mass production capacity and
3. Most synthetic rubber is produced through it does not reliant on the weather or diseases
addition polymerisation reaction of various like rubber trees.
petroleum-based monomers.
2. Table 4.9 shows the types of synthetic rubber,
4. The general characteristics of synthetic characteristics and uses.
polymers:
Table 4.9 Types of synthetic rubber, characteristics and their uses
Types of synthetic rubber Characteristics Uses
Neoprene (polychloroprene) • Elastic To make gloves
Addition • Resistant to oxidation To make bearing pads used in
polymerisation CH C CH CH and corrosion bridges as a safety mechanism
n CH 2 C CH CH 2 2 2 and conveyor belts
CI CI n • Resistant to high heat To make fire seals and firemen
Chloroprene Neoprene
2-chlorobuta-1,3-diene (Polychloroprene) (Not easily flammable) suspenders
• Resistant to solvent To make petrol rubber hoses
• High tensile strength For outdoor applications such
as gaskets and seals
Styrene-butadiene rubber (SBR) • Resistant to corrosion Used for surfaces corrosion
protection in metallic tanks in
H H H H H Addition H H H H H photochemical industries
polymerisation
n C C + n C C C C C C C C C C • Less elastic and Used in light application
H H H H H H H H n resistant to high heat automotive tyres and parts
• Abrasion resistance To make conveyor belts, rubber
Styrene Buta-1,3-diene Styrene-butadiene rubber (SBR)
(Not easily wear and gaskets and shoe soles
tear)
• Can be easily To make shoe soles
vulcanised Form
Silicone rubber • High temperature To make aircraft components,
resistance automotive components and 5
CH 3 CH 3 CH 3 cooking utensils
H C Si O Si O Si CH 3 • Low temperature To make sealants
3
CH CH n CH resistance
3 3 3
• Inert chemically To make medical implants,
(Chemicals resistance) catheter and prosthesis
• Non-toxic To make baby care products








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Chemistry SPM Chapter 4 Polymer

Thiokol rubber • Chemical solvent and To make chemical solvent and
oil resistance oil storage tank coatings, petrol
S S pipes and sealants
CH CH 2 S S
2 n
Nitrile rubber • Chemical solvent and To make gloves, synthetic
oil resistance leather, gaskets, oil seals and
engine hoses
CH 2 CH CH CH 2 CH 2 CH
n C N m


Use of Rubber and the Environment 5. The disposal of rubber material waste can be
1. Natural rubber is biodegradable and can be reduced by:
decomposed biologically in a short time (a) Recycle discarded tyres into fuel to
by microorganisms. Hence, natural rubber generate electricity
waste does not retain for a long time in the (b) Recycle the rubber material waste
environment. Example: Turning old tyres into innovative
items such as planters and stools
2. However, the use of natural rubber is limited (c) Reuse the old tyres by sticking a new
due to its less heat resistance and less resistant layer of rubber on them as jollop tyres
to chemical solvents. This phenomenon causes
manufacturing industries to use synthetic Checkpoint 4.3
rubber in making various products.
Q1 (a) What is the meaning of synthetic rubber?
3. Most of the synthetic rubber products are (b) List down three general characteristics of
synthetic rubber.
non-biodegradable and take a very long time (c) State two types of synthetic rubber and their
to decompose. Unsustainable use of rubber uses.
materials and difficulty in synthetic rubber Q2
disposal lead to environmental pollution. For “Shoe soles can be manufactured by using
natural rubber or synthetic rubber.”
example, large quantities of synthetic rubber
vehicle tyres need to be decomposed. (a) Name the type of synthetic rubber that can
be used to manufacture shoe soles.
4. Improper disposal of synthetic rubber waste (b) State two characteristics of natural rubber
shoe soles that are different from those
causes blockage of drainage system and pollute produced by using synthetic rubber.
rivers and lakes. Open burning of synthetic Q3 “Tyre made of styrene-butadiene rubber (SBR)
rubber releases poisonous gases such as carbon is one of synthetic rubber products which is
monoxide, CO and hydrogen cyanide, HCN difficult to decompose naturally”.
causes lungs to become swollen and blocked Justify the use of SBR tyres.
as well as air pollution.



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Chemistry SPM Chapter 4 Polymer








Environmental impacts and remedial measures Environmental Uses impact






Elastomer Synthetic rubber Characteristics




Uses in daily life Synthetic polymer Types Thermoset Types







Condensation polymerisation Thermoplastic




POLYMER Polymerisation Environmental impact



Addition polymerisation Vulcanised rubber Uses






Types Alternative vulcanisation Characteristics Latex coagulation prevention
Vulcanisation Coagulation of latex






Definition Natural polymer Example Natural rubber Uses Latex coagulation process Form

CONCEPT MAP Characteristics
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