Spotlight A+ Form 4 & 5 Chemistry KSSM

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Extra Features of This Book






CHAPTER
4 The Periodic Table of Elements

Important Learning Standard Page
4.1 The Development • Describe the historical development of the Periodic Table of Elements. 82
SMART SCOPE of Elements • Deduce the basic principle of arrangement of elements in the Periodic Table 85 CONCEPT MAP
of Periodic Table
of Elements.
4.2 The Arrangement
of in the Periodic • Describe briefly the modern Periodic Table of Elements. 85 86
• Generalise the relationship between the proton number and the position of
Table of Elements elements in the modern Periodic Table of Elements.
• Relate the inert nature of Group 18 to its stability. 88
4.3 Elements in • Generalise the changes in physical properties of elements when going down 89 Contents of the whole
Group 18.
Contains learning standard Group 18 • Describe briefly the uses of Group 18 elements in daily life. 89
EXAMPLE ©PAN ASIA PUBLICATIONS
• Generalise the physical changes of elements when going down Group 1. 90 topic are summarised in the
• Investigate through experiment the chemical properties of Group 1 elements with:
(LS) that need to be 4.4 Elements in ‒ Water ‒ Oxygen gas ‒ Chlorine 91 94 95
Group 1
• Generalise the changes in the reactivity of elements when going down Group 1.
• Reason out the physical and chemical properties of the other elements in Group 1.
• Generalise the changes in the physical properties of elements when going down 96 form of a concept map.
achieved in each topic. 4.5 Elements in • Summarise the chemical properties of Group 17 elements. 97 97
Group 17.
Group 17
• Generalise the changes in the reactivity of elements when going down Group 17.
Form • Predict the physical and chemical properties of the other elements in Group 17. 98
4
Chemistry Chapter 3 The Mole Concept, Formula and Chemical Equation • Describe the trends in physical properties of elements across Period 3. 98
4.6 Elements in • Conduct an experiment to observe changes in the properties of the oxides of 102
elements across Period 3.
Sulphur S 32 Period 3 Thus, the relative molecular mass of water 103
= 18
• Describe briefly the uses of semi-metals.
Chlorine Cl 35.5 3. Calculate the relative molecular mass or
• Determine the position of transition elements in the Periodic Table of Elements.
Potassium K 39 4.7 Transition relative formula mass 104 104
• Explain the special characteristics of a few transition elements with examples.
Calcium Ca 40 Elements (a) The relative molecular mass of a molecule 105
• List the uses of transition elements in industry
can be calculated by adding up the relative
Zinc Zn 65 atomic masses of all the atoms that are
Silver Ag 108 present in the molecule.
Lead Pb 207 Example: A molecule of carbon dioxide,
CO 2 , consists of 1 carbon atom and 2
• Alkali metal / Logam alkali • Metalloid / Metaloid Cleansing agent effectiveness
oxygen atoms.
• Amphoteric / Amfoterik • Monoatomic gas / Gas monoatomic in hard water and acidic water
CO
• Atomic radius / Jejari atom • Noble gas / Gas lengai
2
Measuring atomic mass • Chemical properties / Sifat kimia CO 2 = C 1 O 2 • Octet electron arrangement / Susunan elektron oktet Compare
One molecule of carbon
http://bit.ly/3706MgE • Diatomic molecules / Molekul dwiatom dioxide consists of 1 carbon Cleansing action
• Period / Kala
CHAP. • Duplet electron arrangement / Susunan elektron duplet • Periodic table / Jadual Berkala CHAP.
atom and 2 oxygen atoms.
3 • Elektronegativity / Keelektronegatifan • Physical properties / Sifat fizik 3 Study about
12 + 2(16) = 44
• Group / Kumpulan • Reactivity / Kereaktifan
• Transition element / Unsur peralihan
• Halogen / Halogen Therefore, the relative molecular mass of Discuss Soaps Detergents • Preservatives Usage
Why is the relative atomic mass of chlorine 35.5? • Inert / Lengai • Valence electron / Elektron valens about Traditional Types of • Antioxidants and
Natural chlorine exists in two isotopes, 35 Cl and carbon dioxide Uses in daily medicines additives • Flavourings effects
17
37 Cl. An ordinary sample of chlorine contains = RAM of carbon + 2(RAM of oxygen) • Colourings
17 = 12 + 2(16) life and Cleansing agents of using
approximately 75% chlorine-35 and 25% 80 RAM of C = 12 Medicines Food additives • Thickeners
chlorine-37. Therefore, the relative atomic mass is = 44 RAM of O = 16 misuse of • Stabilisers food
closer to 35 than 37. medicines Modern • Emulsifiers additives
RAM = 75 × 35 + 25 × 37 medicines Consumer Application of
100
100
= 35.5 How to determine the number of atoms in a Types of core and Industrial nanotechnology Graphene
Chemistry
in industry
Relative Molecular Mass, RMM molecule? • Water ingredients
The whole numbers in a chemical formula
1. The idea of relative atomic mass also can be represent the number of atoms of each element • Emulsifiers Cosmetics Application of Usage of
applied to compounds which may be molecules except “1” is not stated. • Preservatives Categorised into Fats and oils Green Technology sludge from
or ionic compounds. For example, in the formula of sulphuric acid, H 2 SO 4 • Thickeners in industrial waste waste water
• 2 means there are 2 atoms H,
2. The relative molecular mass of a molecule is the • 4 means there are 4 atoms O, • Moisturisers 1. Make-up cosmetics management treatment
2. Treatment cosmetics
Use of oils and
average mass of the molecule when compared • and 1 atom S even if it is not stated. • Colouring agents 3. Fragrances fats in daily life
• Fragrances
SPOTLIGHT PORTAL Relative Formula Mass, RFM Side effect of
H 2 SO 4 = H 2 S 1 O 4
with 1 of the mass of a carbon-12 atom.
12
Relative molecular mass of a molecule
=
The average mass of one molecule
For ionic compound, we use “relative formula
1 × the mass of one carbon-12 atom 1. We use “relative molecular mass” for molecules. cosmetic usage
12 mass”. 483
Example: Example: Form 4
Sodium oxide, Na 2 O is an ionic compound. Chemistry Chapter 6 Acid, Base and Salt
The mass of one molecule of water, H 2 O is 18 Therefore, the relative formula mass of sodium
Scan QR code to visit times greater than the 1 mass of a carbon-12 oxide 6.1 The Role of Water in Showing Acidic and Alkaline Properties
12
= 2(RAM of sodium) + RAM of oxygen
atom.
–– 1 of a carbon-12 atom = 2(23)+ 16 RAM of Na = 23
RAM of O = 16
O 12 = 62 Acids Hydroxonium ions, H 3 O + are the actual ions existing
in the aqueous solution that gives the acidic
websites or videos related to 18 1. When an acid is dissolved in water, hydrogen properties. To simplify explanation, we often use
H
H
atom in the molecule of acid is released as
hydrogen ion, H + .
You will learn about ionic compounds in Form 4 hydrogen ion, H + to represent hydroxonium ions,
2. Therefore, based on the Arrhenius theory, acid is
H 3 O + .
Chapter 5.
subtopics learnt. There are 3.1.1 Chemical substance that ionises in water to 6. Table 6.1 shows some examples of acids.
defined as follows:
Figure 3.3 One molecule of water is 18 times heavier
3.1.2
than 1 of a carbon-12 atom.
Table 6.1
48
12
produce hydrogen ions, H + .
Ions present in aqueous solution
Acid
videos for certain activities or 3. Hydrogen ions, H + cannot exist on their own. Hydrochloric HCl(aq) → H + (aq) + Cl – (aq)
With water, H + ions are pulled into water
acid
molecules, H 2 O, to form stable hydroxonium
ions, H 3 O + (aq). HCl(aq) → H+(aq) + Cl – (aq) Sulphuric H 2 SO 4 (aq) → 2H + (aq) + SO 4 (aq) INFORMATION GALLERY
2–
acid
experiments. 4. For example, when hydrogen chloride gas Nitric acid HNO 3 (aq) → H + (aq) + NO 3 (aq) –
is dissolved in water, molecules of hydrogen
chloride will ionise in water to produce hydrogen
H + (aq)
ions, H + and chloride ions, Cl – . Ethanoic acid CH 3 COOH(aq) CH 3 COO – (aq) +
H + (aq) + H 2 O(l) → H 3 O + (aq)
5. Hydrogen ions, H + will combine with the water Basicity of Acids
molecules, H 2 O to form stable hydroxonium Additional information
ions, H 3 O + . 1. Basicity of an acid refers to the number of
hydrogen ions, H + that can be produced by one
H H H H Form molecule of acid that ionises in water.
+
Cl
Chapter 1 Redox Equilibrium Chemistry O + + Cl – 5 2. Figure 6.2 shows the classfication of acids based related to the topic.
H
O
H on the basicity of acids.
CHAP. CHAP.
1 1.5 Figure 6.1 Formation of hydroxonium ion, H 3 O + 1
Aim: 3. The circuit is completed by connecting the
To investigate the effects of the type of electrode electrodes to the ammeter, batteries and switch
on the selection of ions to be reduced or oxidised as shown in Figure 1.27.
at the electrodes. Acids
Test tube
Problem statement: CHAP.
Do the types of electrodes affect the types of Copper(II) sulphate,
products formed during the electrolysis? 6 Carbon CuSO 4 solution
electrodes Monoprotic acid Diprotic acid Triprotic acid
Hypothesis: Form
When copper electrodes are used instead of carbon Ammeter A Switch 5
electrodes, the types of products formed at the Battery Chemistry Chapter 1 Redox Equilibrium
AktivitY / EXPERIMENT Variables: 4. The switch is turned on for 15 minutes Acid that produces two CHAP. 1. Figure 1.17 shows the of the cell while copper, Cu is the positive CHAP. 1
anode are different.
Daniell Cell
Acid that produces only one
Acid that produces three
Figure 1.27
hydrogen ion per molecule in
hydrogen ions per molecule 1
hydrogen ions per molecule Daniell cell. A Daniell
water.
in water.
(a) Manipulated: Types of electrodes
in water.
terminal of the cell.
(b) Responding: Types of products at the anode
Example:
Example:
voltaic cell.
(c) Fixed: Type of electrolyte, concentration of 5. All observations at the anode, cathode and cell or a zinc-copper cell is an example of a 5. Redox reaction that takes place in Daniell cell
electrolyte are recorded.
electrolyte 6. The gas produced at the anode is collected and 2– Example: 3– can be represented by an overall ionic equation:
H 2 SO 4 → 2H + + SO 4
HCl → H + + Cl –
–
HNO 3 → H + + NO 3
Material: tested with a glowing wooden splinter. H 2 C 2 O 4 2H + + C 2 O 4 2– H 3 PO 4 3H + + PO 4 A e – Voltmeter e –
0.1 mol dm –3 copper(II) sulphate, CuSO 4 solution, 7. Steps 1 to 6 are repeated using copper Anode Cathode 2e – are received for Negative terminal: Zn(s) → Zn 2+ (aq) + 2e –
Figure 6.2 Classification of acids based on the basicity of acid
sandpaper, wooden splinter, matches electrodes to replace the carbon electrodes by 2e – are released for (–) (+) every copper ion, Positive terminal: Cu 2+ (aq) + 2e – → Cu(s)
Overall ionic
using an apparatus set-up as shown in Figure
Complete activity or Apparatus: 1.28. every zinc, Zn atom Zn Cu oxidised equation: Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s)
Cu 2+ that is
that is oxidised
Batteries, carbon electrodes, copper electrodes,
connecting wires with crocodile clips, ammeter, Switch Battery 6. From the ionic equation, we can write cell
electrolytic cell, 50 cm 3 beaker, electronic balance, 146 A Ammeter e – Zn + B 6.1.1 6.1.2 C Cu 2+ notation for Daniell cell as the following:
switch, test tube
Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s)
experiment including results, Procedure: Copper, Cu e – Zn Zn 2+ Cu 2+ Cu e – 7. Standard cell potential, E 0 for Daniell cell can be
Salt bridge
1. Two carbon electrodes are cleaned with
sandpaper. electrodes Figure1.17 Daniel cell calculated using the following formula:
Copper(II) sulphate,
2. The copper(II) sulphate, CuSO 4 solution is
anode
cathode – E 0
E 0 = E 0
data analysis, discussion Observation: Figure 1.28 CuSO 4 solution Zn 2+ (aq) + 2e –  Zn(s) E 0 = –0.76 V with zinc, Zn as the anode and copper, Cu as the
cell
poured into an electrolytic cell until it is half full.
E 0 = +0.34 V
cathode.
Cu 2+ (aq) + 2e –  Cu(s)
Table 1.24 2. E 0 zinc is more negative. It indicates that zinc, Zn is Anode: Zn(s)  Zn 2+ (aq) + 2e – E 0 = –0.76 V
and conclusion to increase Electrode Anode Observation Cathode Electrolyte a stronger reducing agent. Thus, zinc, Zn plate Cathode: Cu 2+ (aq) + 2e –  Cu(s) E 0 = +0.34 V
is an anode where oxidation process occurs.
E 0 = E 0
cathode – E 0
copper is more positive. It indicates that the
cell
anode
= 1.10 V
Carbon Gas bubbles are released. A A brown solid is The blue colour of the 3. E 0 copper(II) ion, Cu 2+ is an oxidising agent. = (+0.34 V) – (–0.76 V)
students’ scientific skill. colourless gas that ignites the Chemistry Chapter 6 Acid, Base and Salt Copper plate is the cathode where reduction
Form
deposited on the cathode. solution become light
process occurs.
4
glowing wooden splinter is produced.
blue.
zinc which is more negative indicates that zinc, Zn
Copper Copper anode becomes thinner. A brown solid is The blue colour of A Zinc, Zn plate becomes thinner as zinc, E 0
deposited on the cathode. the solution remains Example 8
6.5 Concentration of Aqueous Solution Zn corrodes and dissolves in zinc sulphate, is more easily oxidised and acts as an anode.
unchanged. ZnSO 4 solution.
1. The concentration of a solution is a measurement Calculate the molarity of the following solutions: Zinc, Zn atom is oxidised to zinc ion, Zn 2+
Inference: that shows the quantity of solute dissolved in (a) 0.2 mol of solid calcium chloride, CaCl 2 in 500 by losing two electrons. 8. The E 0 obtained through the calculation is
cell
Conclusion:
The hypothesis is accepted. Electrolysis of copper(II)
1. Electrolysis using carbon electrodes one unit of volume of solution, normally in 1 cm 3 of distilled water. ) 2 is Zn(s) → Zn 2+ (aq) + 2e – actually the voltage produced in the Daniell cell
(b) 75.6
(a) Anode: Oxygen gas, O 2 is produced. 3 of solution. sulphate, CuSO 4 solution using carbon electrodes g of solid zinc nitrate, Zn(NO 3 based on the potential difference between two
dm
dissolved in water to make up 500 cm 3 of
(b) Cathode: Copper, Cu metal is produced. produces oxygen gas, O 2 and water, H 2 O at the B Potential difference between the two metal electrodes.
2. The more the solute in the solution, the higher
plates causes the flow of electrons from
solution.
anode and copper, Cu metal at the cathode.
2. Electrolysis using copper electrodes the concentration of the solution. [Relative atomic mass: N = 14, O = 16, the anode (zinc) to the cathode (copper) 9. The functions salt bridge of:
Electrolysis of copper(II) sulphate, CuSO 4 solution
3. The quantity of solute can be measured in gram
Zn = 65]
(a) Anode: Copper(II) ions, Cu 2+ are produced. using copper electrodes produces copper(II) ions, through wire. Therefore, electric current is (a) Complete the circuit by allowing the

or mole, thus the concentration of a solution can
Solution
(b) Cathode: Copper, Cu metal is produced. Cu 2+ at the anode and copper, Cu metal at the generated. movement of ions.
(b) Separate two different electrolytes
cathode.
be measured in the units of g dm –3 or mol dm –3 . (a) Molarity = 0.2 mol C Brown solid is deposited at the copper, Cu 10. Initially, oxidation of half-cell is neutral with
plate, making copper, Cu plate becomes
(a) Concentration in unit g dm –3 , is the mass of 0.5 dm 3 2−
1.4.3 333 = 0.4 mol thicker. Copper(II) ion, Cu 2+ is reduced zinc ions, Zn 2+ and sulphate ions, SO 4 in
solute found in 1 dm 3 solution. to the copper atom, Cu by gaining two the solution. When more and more zinc ions,
Concentration (g dm –3 ) (b) Number of moles electrons. Zn 2+ enter the solution, the solution becomes
Mass of solute (g)
mass
= Volume of solution (dm 3 ) = molar mass Cu 2+ (aq) + 2e ‒ → Cu(s) positively charged.
75.6 g
(b) Concentration in unit mol dm –3 , is the = 65 + 2 [14 + (3 × 16)] g mol −1 The intensity of blue solution decreases 11. Similarly, reduction of half-cell is neutral with 2−
number of moles of solute found in 1 = 0.4 mol dm –3 as the concentration of copper(II) ion, Cu 2+ copper(II) ions, Cu 2+ and sulphate ions, SO 4
dm 3 solution. This concentration is called decreases. in the solution. The solution will be negatively
molarity. Molarity = 0.4 mol 4. Oxidation process occurred at the anode and charged when more and more copper(II) ions,
Cu 2+ leave the solution and form copper atom, Cu.
0.5 dm 3
Molarity (mol dm –3 ) = 0.8 mol dm –3 reduction process at the cathode causes zinc 12. When two half-cells are charged, the voltaic
(anode) becomes relatively negative charge
= Number of moles of solute (mol) (electrons) as compared to copper (cathode). cell will not function. Therefore, a salt bridge is
Volume of solution (dm 3 ) 4. Both units of g dm –3 and mol dm –3 can be needed to connect the two half-cells.
Therefore, zinc, Zn is the negative terminal
Try Question 1 in Formative Zone 6.5 interchanged with one another as in the following
diagram. 318 1.3.1
Example 7 Molarity × molar mass Concentration
(mol dm –3 ) ÷ molar mass in g dm –3
Calculate the concentration of the following BRILLIANT TIPS
solutions in the unit of g dm –3 .
(a) 10 g of glucose is dissolved in 0.5 dm 3 of
water.
(b) 30 g of potassium hydroxide is dissolved in The unit for molarity is mol dm –3 or molar (M).
750 cm 3 of water. Mole is not the same as molar. Mole is the unit for
Example and complete CHAP. (c) 2 moles of sodium chloride in 10 dm 3 of measuring matter while molar is the number of
moles of solute in a given volume of solution.
distilled water. Given that the molar mass of
sodium chloride is 58.5 g mol –1 .
6 The volume of solution must be in dm 3 . Try Question 2 in Formative Zone 6.5 Useful tips for students
[1 dm 3 = 1000 cm 3 ]
solution to enchance Solution Example 9
10 g Calculate the concentration of 0.25 mol dm –3
(a) Concentration = 0.5 dm 3 750 cm 3 is sulphuric acid in the unit of g dm –3 . to solve problems in the
= 20 g dm –3
[Relative atomic mass: H = 1, O = 16, S = 32]
students’ understanding. (b) Concentration = 0.75 dm 3 converted to dm 3 Solution
30 g
= 750 cm 3
1000
= 40 g dm –3 = 0.75 dm 3 Molar mass H 2 SO 4 = 2(1) + 32 + 4(16) related subtopic.
= 98 g mol –1

(c) Mass of NaCl = Number of moles × molar mass Concentration = Molarity x molar mass
= 2 mol × 58.5 g mol –1 = 0.25 mol dm –3 × 98 g mol –1
= 117 g = 24.5 g dm –3
Concentration = 117 g
10 dm 3
= 11.7 g dm –3
164 6.5.1 6.5.2
ii

Tag OF ‘TRY QUESTION...
IN FORMATIVE ZONE...’ Form 5 Chemistry Chapter 2 Carbon Compound
2.3 Chemical Properties and 3. When the molecular size of an alkane increases,
Interconversion of Compounds molecules of alkanes burn with sootier flames. 5 Chemistry Chapter 2 Carbon Compound
it becomes more difficult to burn. Large Form
between Homologous Series This is because the number of carbon atoms per
Tag placed at the end of Chemical Properties of Alkane molecule increases as the molecular size of an 2.4
alkane increases. The percentage of carbon by
1. Alkanes are unreactive and do not react with mass in the alkane molecule also increases.
CHAP. most chemicals. 1. Name the following compound using the IUPAC nomenclature system. C2 CHAP.
examples to guide students in 2. Alkanes are saturated hydrocarbon Example 5 (a) H H H (b) H | 2
2
| | |
compounds. Each carbon atom in an alkane
molecule is already bonded to a maximum Explain why butane, C 4 H 10 burns with more sooty CHAP. H–C–C–C–H H–C–H | CHAP.
H H
|
|

number of atoms. The strong C-C and C-H
|
flame compared to ethane, C 2 H 6 .

H H
answering related questions in bonds need a lot of energy to break. [Relative atomic mass: H = 1; C = 12] 2 H–C–H | H H–C–C–C–H 2
|

|
3. Alkanes undergo two types of reactions:
Solution
H H

(a) Combustion
|
(b) Substitution Percentange of carbon by mass in butane, C 4 H 10 H–C–H
H
Formative Zone. Combustion reaction = 4(12) + 10(1) × 100% = 82.76% 2. Draw the structural formula for each of the following organic compounds. C2
4(12)
1. Alkanes burn completely in excess oxygen, O 2 Percentage of carbon by mass in ethane, C 2 H 6 (a) 2-methylbutane
2(12)
(b) 2-methylpropene
to produce carbon dioxide, CO 2 and water, = 2(12) + 6(1) × 100% = 80%
H 2 O. The combustion of alkanes produces a lot 3. Name the following using the IUPAC nomenclature system.
of heat. Hence, alkanes are used as fuels. The percentage of carbon by mass of butane, (a) H H H (b) H H H H
Example: C 4 H 10 is higher. As a result, it burns with more | | | | | | |
CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) sooty flame. H–C–C=C–C–H H–C=C–C–C–C–H
| |
| | |

|
Methane H H H H H H
2C 2 H 6 (g) + 7O 2 (g) → 4CO 2 (g) + 6H 2 O(l) Substitution reaction 4. Draw the following isomers of pentene.
©PAN ASIA PUBLICATIONS
Ethane 1. A substitution reaction occurs when an atom (a) Pent-2-ene
Try Question 1 in Formative Zone 2.3
in a molecule.
2. Incomplete combustion occurs in limited or a group of atoms is substitute by other atoms (b) 2-methylbut-1-ene
(c) 3-methylbut-1-ene
supply of oxygen, O 2 . Alkanes burn in sooty 2. In this reaction, each hydrogen atom in an (d) 2-methylbut-2-ene FORMATIVE ZONE
flame and produce a mixture of carbon, C alkane molecule is substituted by a halogen 5. Name the following using the IUPAC nomenclature system. C2
particle (soot), carbon monoxide, CO gas, atom in the presence of sunlight or ultraviolet (a) H H (b) H
carbon dioxide, CO 2 gas and water, H 2 O. (UV) rays as the catalyst. | | |
Example: Example: H–C–CC–C–H | H–C–H H
|

2CH 4 (g) + 3O 2 (g) → 2CO(g) + 4H 2 O(l) Methane, CH 4 reacts with chlorine, Cl in the H H |
Methane 5 Form presence of sunlight or UV rays to produce four H–CC–C–C–H
CH 4 (g) + O 2 (g) → C(s) + 2H 2 O(l) products as shown in Figure 2.9. H H Questions to test
| |
Chemistry Chapter 4 Polymer

Ethane H
SPM Simulation HOTS Questions H – C – H 6. Draw the structural formula of pent-2-yne. C2
l
l 7. Name the following using the IUPAC nomenclature system. C2
H
1. Figure 1 shows the structural formula of
Methane
To balance a chemical equation: A Melting point (a) H H (c) H H H students' understanding
Step 1: compound Z which is used to make a pipe. B Density | | | | |
Cl 2 , UV ray
Check the number of carbon and hydrogen atoms C Empirical formula H–C–C–O–H H–C–C–C–O–H
H–C–H at the end of each
in the molecular formula of alkane. Balance the D Molecular formula | | H H | H H |
H Cl H Cl H Cl
C C C C C C
number of carbon atoms in CO 2 and the number of

Cl
hydrogen atoms in H 2 O by adding a number in front H l H Examiner’s comment: Cl l |
H H H H H H
of each formula. Compound Z H – C – Cl H – C – Cl H – C – Cl Cl – C – Cl (b) H

l Both substances have the empirical
l


Step 2: H l Cl molecular mass, so the density is higher H H H H
l formula of CH 2 . Substance S has bigger
l
l

Cl
| | | |
SPM SIMULATION Balance the number of oxygen atoms in O 2 . If the Chloromethane Dichloromethane Trichloromethane Tetrachloromethane 8. Draw the following isomers of butanol. C2 suptopic.

Cl
Check the number of oxygen atoms in CO 2 and H 2 O.
H–C–C–C–C–O–H
Figure 1
than R. With bigger molecular size, forces
| | | |

of attraction between molecules become
Which of the following is the structural formula
Figure 2.9 Products of substitution reaction of methane
number needed to add is a fraction, then multiply all
H H H H

stronger. Therefore, the melting point of
of the monomer of compound Z? C2
the numbers in front of the formulae by 2.
S is higher.
A
Answer: C
HOTS QUESTION 376 H Cl 3. Figure 3 shows the raincoat which is made (a) Butan-2-ol
2.3.1
C C
(b) 2-methylpropan-2-ol
H H n
from a synthetic polymer, polyvinyl chloride,
PVC.
B
H Cl
C C 404
CHAP. H H CHAP.
4 C 4
Provides complete solution H C C H
H Cl




H H
with examiner’s comment D H Cl
C C

H H SPM
to help students to answer Examiner’s comment: (a) State the name of the monomer for MODEL PAPER
Figure 3
Monomer of compound Z (polymer) must
have a carbon-carbon double bond, C = C. polyvinyl chloride, PVC. C1
Structural formula in Answer A is another (b) Draw the structural formula for the Paper 1
monomer. C2
HOTS questions. but not the monomer of compound Z. (c) State one reason why polyvinyl chloride, Instruction: Answer all questions. [40 marks]
simplified way to represent the polymer Z
PVC should not be disposed by open
Answer: B
burning? C3
2. Figure 2 shows the polymerisation process. 1. What is the meaning of nanotechnology? 5. The following statement refers to the
Examiner’s comment and answer: A The study of chemical bonds between metal characteristics of an element in the Periodic Table
H H H H (a) Vinyl chloride // Chloroethene atom and non-metal atom. of Elements.

C C C C B The manipulation of materials on an atomic


H H n (b) H Cl or molecular scale. • Brown colour and soft solid.
H H H C • Reacts with water to produce alkaline
R S H H C The study the importance of food additives solution.
in food processing industry and the
Form
• Burn in oxygen to produce a white solid.
SUMMATIVE ZONE 4 Chemistry Chapter 7 Rate of Reaction Figure 2 (c) Burning of polyvinyl chloride, PVC D The development and the application of Which element has the above characteristics?
evolution of food processing technology.
will release pollutant and acidic gas
Which of the following is similiar for substances
products or equipment, and a system to
such as hydrogen chloride gas, HC
R and S? C2
3. Diagram below shows a graph of volume of gas against time in a reaction. 60 cm 3 of hydrogen gas, H 2
is collected when excess magnesium powder, Mg reacts with 50 cm 3 of 0.1 mol dm –3 nitric acid, HNO 3 . to the atmosphere. conserve the environment. A D
2. Figure 1 shows two situations occur when B C
Volume of gas (cm 3 ) sunlight is shining on the glass X.
Question of various levels 474 90 6. Which statement explains the effective collision?
A The collision which takes place after a
60 Darker in reaction.
Answer (a) SPM MODEL PAPER sunlight B The collision which takes place before a
of thinking skill to evaluate 30 Glass X C The collision that causes a reaction.
reaction.
Figure 1
0 Time (s) What is the type of glass X? D The collision produces less activation
energy.
student’s understanding of A Fused glass 7. Figure 2 shows a glass cookware that usually
(a) On the same axes, sketch the curve that you would expect to obtain if 25 cm 3 of 0.2 mol dm –3

B Soda-lime glass
nitric acid, HNO 3 reacts with excess magnesium powder, Mg.
used in the kitchen.

(b) Explain your answer in 3(a).
C Borosilicate glass
D Photochromic glass
each topic. Examiner’s comment 1000 1000 3. The following statement refers to an element in
Number of moles of HNO 3 , n = 0.1(50) = 0.005 mol; Number of moles of HNO 3 , n = 0.2(25)
the Periodic Table of Elements.
= 0.005 mol
Number of moles of nitric acid, HNO 3 used in both experiments are the same, the number of • Located in Period 3 in the Periodic Table of
Elements
moles of hydrogen gas, H 2 produced will also be the same, which is 60 cm 3 . Concentration of nitric • Reacts with water to produce acidic SPM MODEL PAPER
solution and bleaching agent
acid, HNO 3 in the second experiment is higher. So, the rate of reaction is higher. That is why the Figure 2
gradient of the curve will be steeper. • Reacts with iron wool to produce a brown Which substance is added to the glass to make it
solid suitable for making the cookware?
A Lead(II) oxide, PbO
Which of the following shows the electron B Boron oxide, B 2 O 3
arrangement of the element? C Natrium carbonate, Na 2 CO 3
A 2.8.4 C 2.8.7 D Aluminium oxide, Al 2 O 3
4. Water molecule combines with hydrogen ion to SPM-oriented practice
D 2.8.8
B 2.8.5
Paper 1 8. Atom W has 4 neutrons and a nucleon number of
7. Which of the following is the correct symbol
1. Which of the following is a fast reaction? C1 form hydroxonium ion, H 3 O + . What is the type for atom W?
A Fermentation SPM Clone 3. Oxygen gas, O 2 is produced from the of the chemical bond formed? A 7 4 W C 3 4 W
C Metallic bond
B 7 W
4
B Photosynthesis decomposition of sodium chlorate(I), NaOCl A Dative bond D Hydrogen bondbased on latest SPM
D 7 3 W
C Formation of stalagmites in the presence of manganese(IV) oxide, MnO 2 . B Ionic bond
D Combustion of magnesium The following shows the chemical equation of the 524
SPM Clone 2. The reaction between sodium thiosulphate, reaction. MnO 2 format to assess students
Na 2 S 2 O 3 and sulphuric acid, H 2 SO 4 is represented 2NaOCl(aq) 2NaCl(aq) + O 2 (g)
by the following equation: Figure 1 shows the graph of volume of oxygen gas
Na 2 S 2 O 3 (aq) + H 2 SO 4 (aq) → Na 2 SO 4 (aq) + S(s) + against time.
CHAP.
CHAP. SO 2 (g) + H 2 O(l) Volume of oxygen gas (cm 3 ) on all the topics learnt in
7 Which of the following is the most suitable 7
method to determine the rate of reaction? C2
A Determine the change in temperature of the
solution with time. Form 4 & 5 textbooks.
B Determine the volume of water, H 2 O
produced with time.
C Determine the production of a fixed quantity 0 Time (s) Form
of sulphur, S precipitate with time. Figure 1 4
Chapter 4 The Periodic Table of Elements
D Determine the change in the concentration of Why the gradient of the curve decrease with Chemistry
sodium sulphate, Na 2 SO 4 with time. time? C2
264 Reinforcement & Assessment of Science Process Skill
Reinforcement exercise of science process skills in preparation for the Kertas Amali Bersepadu.
In this experiment you are required to investigate the properties of three types of oxides of elements in period
3 in the Periodic Table of Elements based on the reaction with alkali and acid solutions.
You are provided with the following materials:
• K1 = Oxide P
• K2 = Oxide Q
• K3 = Oxide R
• L1 = Sodium hydroxide solution, 2.0 mol dm –3
REINFORCEMENT & ASSESSMENT • L2 = Nitric acid, 2.0 mol dm –3 ANSWERS Complete Answers
https://bit.ly/3sn6L0o
Carry out the experiment according to the following instructions:
1. Put 2 spatula of substance K1 into two different test tubes.
2. Pour 5.0 cm 3 solution L1 into the first test tube. Form 4 (b) (i) Vinegar; salt; baking powder
OF SCIENCE PROCESS SKILL 3. Pour 5.0 cm 3 solution L2 into the second test tube. Chapter 1 Introduction to Chemistry (c) Chemist; doctor
(ii) Vinegar: preserves food

Salt: gives salty taste
4. Heat the mixture in each test tube slowly while stirring with a glass rod.
Baking powder: raises the dough
5. Record the solubility of substance K1 in each test tube.
1.1
CHAP.
can be poured directly into the sink. Concentrated
CHAP. 6. Repeat the experiment by replacing substance K1 with K2 and K3. 1. A field of science that studies the structures, properties, (d) Hydrogen peroxide waste with low concentration
hydrogen peroxide wastes need to be diluted with
4
4 Observation compositions and interactions between matter. water. Then, it is added with sodium sulphite for the
2. Herbicide; Hormone purpose of decomposition before being poured into the
3.
Oxide Reaction with nitric acid, HNO 3 Reaction with sodium With nanotechnology, sunscreens are no longer oily 2. (a) (i) Presence of water and oxygen
sink.
hydroxide solution, nowadays and are colourless when applied to the skin.
Questions for students to master the Oxide P (K1) NaOH 4. (a) Cosmetic consultant (b) Water and oxygen are needed for iron rusting.
(ii) Rusting of iron nail
(iii) Type of nail
(b) Dietitian
(c) Pathologist
(d) Veterinarian
(c)
Observation
science process skill (SPS) presented Oxide Q (K2) 1. A systematic scientific method used to solve science related Test tube A B Iron nail does not rust.
1.2
Iron nail rusts.
problems.
Oxide R (K3) 2. (a) When the temperature increases, the mass of salt C Iron nail does not rust.
in Science Practical Test. Scan the QR Table 1 (b) Manipulated variable: Temperature Section B (d) Oxygen and water must be present for the iron nail to
dissolved also increases.
rust.
Responding variable: Mass of salt dissolved
(**Complete Table 1 with reference to the Simulation Experiments and Sample Results given)

code below to get Tip dan Teknik Based on the experiment conducted: 1. • Do not eat, drink, chase or run in the laboratory. 3. (a) The scientific method is a systematic approach to solve ANSWER FORM 4 ANSWERS
1.3
problems in science.
(b)
1. State the following variables based on the experiment above:
Making an observation
• Do not pour the chemicals back to the reagent bottles.
(a) Manipulated:
• Do not point the mouth of the test tube at your face or
(b) Responding: 2. • Keep flammable substances away from the heat source. Making an inference
Prihatin Murid in answering Science (c) Control: 3. (a) To carry out experiment that involves the release of Identifying the problem
at other people.
toxic vapours, gases that can cause combustion or gases
with pungent smell.
2. Based on the results of the experiment, classify oxides P, Q and R into the table below according to their Making a hypothesis
respective properties.
(b) To remove dirt, oil, chemicals or microorganisms from
the hands.
Practical Test. Amphoteric Base Acid 4. (a) Kept in paraffin oil to prevent reaction between this Answers are provided. Scan
Identifying the variables
(c) To wash and clean the body if accident happens on
parts of the body. It is also used to put out fire at any
Controlling the variables
part of the body if there is fire.
Tip dan Teknik Table 2 5. Mercury poisoning is a phenomenon when a person is Planning an experiment
chemical with moisture, water and air.
(b) Kept in dark bottles to avoid the exposure of sunlight.
exposed to mercury in a certain amount. Two symptoms of QR code to get Complete
Collecting data
Interpreting data
Prihatin Murid mercury poisoning are vomiting and difficult in breathing. Making a conclusion
113
5. C Answers with explanations for
Paper 1
1. A 2. A 3. C 8. D 4. B 10. C Writing a report
7. B
9. B
6. B
Section C
https://bit.ly/349VB5H 11. B 12. C 13. A 14. C 15. D objective questions.
4. (a) (i) Ammonium nitrate / Urea

(ii) Occupation A: Pharmacist
Paper 2
Occupation B: Pathologist
Section A (b) • Inform the accident to the teacher immediately.
1. (a) Chemistry is defined as a field of science that • Make the spill area as a restricted area for students.
studies the structures, properties, compositions and • Try to stop the spill from spreading to other areas
interactions between matter. using sand to border it.
541
iii

CONTENTS




FORM 4
Revision Revision
date date

Theme 1 The Importance of Chemistry 4.5 Elements in Group 17 96
4.6 Elements in Period 3 98
©PAN ASIA PUBLICATIONS
Chapter 1 Introduction to Chemistry 1 4.7 Transition Elements 104
Summative Zone 107
1.1 Development in Chemistry
Field and Its Importance in Chapter 5 Chemical Bond 115
Daily Life 3 5.1 Basics of Compound
1.2 Scientific Investigation in Formation 117
Chemistry 6 5.2 Ionic Bond 118
1.3 Usage, Management and 5.3 Covalent Bond 122
Handling of Apparatus 5.4 Hydrogen Bond 126
and Materials 8 5.5 Dative Bond 129
Summative Zone 16 5.6 Metallic Bond 130
5.7 Properties of Ionic
Theme 2 Fundamentals of Chemistry Compounds and Covalent
Compounds 131
Chapter 2 Matter and the Atomic Structure 21 Summative Zone 137

2.1 Basic Concepts of Matter 23 Theme 3 Interaction between Matter
2.2 The Development of the
Atomic Model 29 Chapter 6 Acid, Base and Salt 143
2.3 Atomic Structure 31
2.4 Isotopes and Its Uses 34 6.1 The Role of Water in Showing
Summative Zone 38 Acidic and Alkaline
Properties 146
Chapter 3 The Mole Concept, Chemical Formula 6.2 pH Value 152
and Equation 45
6.3 Strength of Acids and
3.1 Relative Atomic Mass and Alkalis 155
Relative Molecular Mass 47 6.4 Chemical Properties of Acids
3.2 Mole Concept 50 and Alkalis 157
3.3 Chemical Formula 56 6.5 Concentration of Aqueous
3.4 Chemical Equation 66 Solution 164
Summative Zone 73 6.6 Standard Solution 166
6.7 Neutralisation 170
Chapter 4 The Periodic Table of Elements 80 6.8 Salts, Crystals and Their Uses

4.1 The Development of the in Daily Life 175
Periodic Table of Elements 82 6.9 Preparations of Salts 178
4.2 The Arrangement in the 6.10 Effect of Heat on Salts 190
Periodic Table of Elements 85 6.11 Qualitative Analysis 197
4.3 Elements in Group 18 88 Summative Zone 218
4.4 Elements in Group 1 90


iv

Revision Revision
date date
Chapter 7 Rate of Reaction 227 2.1 Types of Carbon
Compounds 359
7.1 Determining Rate of 2.2 Homologous Series 363
Reaction 229 2.3 Chemical Properties and
7.2 Factors Affecting Rate of Interconversion of
Reaction 241 Compounds between
7.3 Application of Factors that Homologous Series 376
©PAN ASIA PUBLICATIONS
Affect the Rate of Reaction 2.4 Isomers and Naming Based
in Daily Life 255 on IUPAC Nomenclature 394
7.4 Collision Theory 256 Summative Zone 406
Summative Zone 264
Theme 3 Heat
Theme 4 Industrial Chemistry
Chapter 3 Thermochemistry 416
Chapter 8 Manufactured Substances in 3.1 Heat Change in Reactions 418
Industry 273 3.2 Heat of Reaction 424

8.1 Alloy and Its Importance 275 3.3 Application of Endothermic
8.2 Composition of Glass and Its and Exothermic Reactions
Uses 279 in Daily Life 442
8.3 Composition of Ceramics Summative Zone 446
and Its Uses 281
8.4 Composite Materials and Its Theme 4 Technology in Chemistry
Importance 284
Summative Zone 288 Chapter 4 Polymer 453
4.1 Polymer 455
4.2 Natural Rubber 465
FORM 5 4.3 Synthetic Rubber 472
Summative Zone 475
Theme 1 Chemical Process
Chapter 5 Consumer and Industrial
Chapter 1 Redox Equilibrium 294 Chemistry 482
1.1 Oxidation and Reduction 296 5.1 Oils and Fats 484
1.2 Standard Electrode 5.2 Cleaning Agents 487
Potential 312 5.3 Food Additives 497
1.3 Voltaic Cell 315 5.4 Medicines and Cosmetics 502
1.4 Electrolytic Cell 321 5.5 Application of
1.5 Extraction of Metal from Nanotechnology
Its Ore 338 in Industry 509
1.6 Rusting 342 5.6 Application of Green
Summative Zone 352 Technology in Industrial
Waste Management 511
Theme 2 Organic Chemistry Summative Zone 517
SPM Model Paper 524
Chapter 2 Carbon Compound 357 Answers 541 – 578




v

18
2 He Helium 4 10 Ne Neon 20 18 Ar Argon 40 36 Kr Krypton 84 54 Xe Xenon 131 86 Rn Radon 118 Og Oganesson
17
9 F Fluorine 19 17 Cl Chlorine 35.5 35 Br Bromine 80 53 I Iodine 127 85 At Astatine 117 Ts Tennessine 71 Lu Lutetium 175 103 Lr Lawrencium
16
ABLE OF ELEMENTS
H ©PAN ASIA PUBLICATIONS
8 O Oxygen 16 16 S Sulphur 32 34 Se Selenium 79 52 Te Tellurium 128 84 Po Polonium 116 Lv Livermorium 70 Yb Ytterbium 173 102 No Nobelium
15
7 N Nitrogen 14 15 P Phosphorus 31 33 As Arsenic 75 51 Sb Antimony 122 83 Bi Bismuth 209 115 Mc Moscovium 69 Tm Thulium 169 101 Md Mendelevium
14
6 C Carbon 12 14 Si Silicon 28 32 Ge Germanium 73 50 Sn Stanum 119 82 Pb Plumbum 207 114 Fl Flerovium 68 Er Erbium 167 100 Fm Fermium Non-metal
13
5 B Boron 11 13 Al Aluminium 27 31 Ga Gallium 70 49 In Indium 115 81 Tl Thallium 204 113 Nh Nihonium 67 Ho Holmium 165 99 Es Einsteinium
12
30 Zn Zinc 65 48 Cd Cadmium 112 80 Hg Mercury 201 112 Cn Copernicium 66 Dy Dysprosium 162.5 98 Cf Californium

11
29 Cu Copper 64 47 Ag Argentum 108 79 Au Aurum 197 111 Rg Roentgenium 65 Tb Terbium 159 97 Bk Berkelium Semi metal
10
Proton number Name of the element 9 28 27 Ni Co Nickel Cobalt 59 59 46 45 Pd Rh Palladium Rhodium 106 103 78 77 Pt Ir Platinum Iridium 195 192 110 109 Ds Mt Darmstadtium Meitnerium 64 63 Gd Eu Gadolinium Europium 157 152 96 95 Cm Am Curium Americium




THE PERIODIC T
8
26 Fe Iron 56 44 Ru Ruthenium 101 76 Os Osmium 190 108 Hs Hassium 62 Sm Samarium 150 94 Pu Plutonium
Hydrogen 7 25 Mn Manganese 55 43 Tc Technetium 75 Re Rhenium 186 107 Bh Bohrium 61 Pm Promethium 93 Np Neptunium
1 H 1 Metal
6 24 Cr Chromium 52 42 Mo Molybdenum 96 74 W Tungsten 184 106 Sg Seaborgium 60 Nd 144 92 U Uranium 238
Symbol of the element Relative atomic mass 5 4 23 V Vanadium 51 41 Nb Niobium 93 73 Ta Tantalum 181 105 Db Dubnium 59 Pr Praseodymium Neodymium 141 91 Pa Protactinium 231 Legend:








3 22 Ti Titanium 48 40 Zr Zirconium 91 72 Hf Hafnium 178.5 104 Rf Rutherfordium 58 Ce Cerium 140 90 Th Thorium 232 (Source: International Union of Pure and Applied Chemistry, IUPAC)
21 Sc Scandium 45 39 Y Yttrium 89 57 – 71 Lantanides 89 – 103 Actinides 57 La Lanthanum 139 89 Ac Actinium

2
Be Beryllium 12 Mg Magnesium 24 20 Ca Calcium 40 38 Sr Strontium 88 56 Barium 137 88 Ra Radium
4 9 Ba
Group 1 1 Hydrogen 1 3 Li Lithium 7 11 Na Sodium 23 19 K Potassium 39 37 Rb Rubidium 85.5 55 Cs Caesium 133 87 Fr Francium Lanthanides series Actinides series



1 Period 2 3 4 5 6 7






vi

CHAPTER
5 Chemical Bond













©PAN ASIA PUBLICATIONS
5.1 Basics of Compound Important Learning Standard Page
Formation • Explain the basic formation of compounds. 117



5.2 Ionic Bond • Explain with examples the formation of an ionic bond. 118



• Explain with examples the formation of a covalent bond. 122
5.3 Covalent Bond
• Compare ionic and covalent bonds. 126



• Explain with examples the formation of a hydrogen bond. 126
5.4 Hydrogen Bond • Explain the effect of the hydrogen bond on the physical properties
of substances. 127



5.5 Dative Bond • Explain with examples the formation of a dative bond. 129



• Explain the formation of a metallic bond. 130
5.6 Metallic Bond
• Reason out the electrical conductivity of metal. 130


5.7 Properties of • Compare the properties of ionic and covalent compounds through 131
Ionic Compounds experiment.
and Covalent • Explain with examples the uses of ionic and covalent compounds
Compounds in daily life. 135









• Chemical bond / Ikatan kimia • Hydrogen bond / Ikatan hidrogen
• Covalent bond / Ikatan kovalen • Ionic compounds / Sebatian ion
• Covalent compounds / Sebatian kovalen • Ionic bond / Ikatan ion
• Dative bond / Ikatan datif • Metallic bond / Ikatan logam
• Delocalised / Dinyahsetempatkan • van der Waals attraction forces /
• Electron sea / Lautan elektron Daya tarikan van der Waals
• Giant molecular structure / Struktur molekul gergasi


115
115

Stable duplet Stable octet
electron electron Affect the solubility and boiling
arrangement arrangement point of covalent compounds
©PAN ASIA PUBLICATIONS
Attraction force between H
atom and F, O and N atoms in
Group 18 elements different molecules

The shared pair of
Compound Formation Hydrogen Bond electrons comes
from the same atom

Dative Bond
Ionic Bond Chemical Bond Delocalised
Metallic Bond electrons

Metal and Lattice structure
non-metal Covalent Bond of metal ions

Formation of Non-metal Sea of electrons
cations and non-metal


Formation of Sharing of
anions electrons

Electrostatic Van der Waals
attraction forces attraction forces


Ionic compounds Covalent compounds

High melting and Low melting and
boiling points boiling points


Normally can dissolve Normally can
in water but cannot dissolve in organic
dissolve in inorganic solvent but cannot
solvent dissolve in water

Can conduct electricity Cannot conduct
in molten state or electricity
aqueous state







116

Form
4
Chapter 5 Chemical Bond Chemistry
5.1 Basics of Compound Formation

1. Different elements that are chemically bonded Types of Chemical Bonds
together but not mixed together are called http://bit.ly/2OeUnAH
compounds.
Example:
• Water is a compound of hydrogen and oxygen 5. Chemical bonds are formed through the transfer
• Carbon dioxide is a compound of carbon and of electrons or share of electrons.
oxygen (a) During the transfer of electrons, a metal atom
©PAN ASIA PUBLICATIONS
• Rust is a compound of iron, oxygen and with one, two or three valence electrons will
hydrogen donate its valence electrons whereas a non-
2. Atoms form chemical bond to achieve a stable metal atom with five, six or seven valence
electron arrangement. electrons will receive the electrons.
3. Noble gases exist as monoatomic gases and Example:
are not chemically reactive because they have Sodium atom, Na with electron arrangement
achieved a stable duplet electron arrangement or of 2.8.1 will donate one valence electron
octet electron arrangement. to chlorine atom, Cl which has electron
(a) All Group 18 elements have eight valence arrangement of 2.8.7.
electrons except helium (helium has two
valence electrons).
Table 5.1 The electron arrangements of noble gases Na Cl
Noble gas Electron arrangement
Helium 2 Sodium atom, Na Chlorine atom, Cl
Transfer of electron from sodium
Neon 2.8 atom to chlorine atom
Argon 2.8.8 Figure 5.1 Transfer of one electron
from sodium atom, Na to chlorine atom, Cl
CHAP. Krypton 2.8.18.8 CHAP.
5 Xenon 2.8.18.18.8 5

Radon 2.8.18.32.18.8
By donating, gaining or sharing electrons, an atom
Try Questions 2 and 3 in Formative Zone 5.1 can change its electron arrangement.

(b) In sharing electrons, two or more non-metal
atoms will contribute their valence electrons
Atoms of Group 18 elements do not gain, lose or for sharing to achieve the duplet or octet
share electrons with atoms of other elements. electron arrangement.
Example:
(b) The electron arrangement of an atom where In water molecule, H O, each of two hydrogen
2
the valence shell is filled with eight valence atoms, H contributes one valence electron to
electrons is known as octet electron share with one oxygen atom, O.
arrangement.
(c) The electron arrangement of an atom with a
single shell filled with two valence electrons
is known as duplet electron arrangement. H O H
4. Atoms with octet or duplet electron arrangement
are very stable. Therefore, atoms of other elements
that have less than eight valence electrons tend Sharing of electrons between an
to achieve the stable octet or duplet electron oxygen atom and two hydrogen atoms
arrangement through the formation of chemical Figure 5.2 Sharing of electrons in water molecule, H O
bonds. 2

5.1.1 117

Form
4
Chemistry Chapter 5 Chemical Bond
6. There are two types of chemical bonds:
(a) Ionic bond Non-metal Covalent bond Non-metal
(b) Covalent bond
Figure 5.4 Sharing of electrons between atoms
of non-metal forms a covalent bond
Try Question 5 in Formative Zone 5.1
Chemical bonds are forces that hold atoms
together to make compounds or molecules through
the transfer of electrons or sharing of electrons
• Metals are placed on the left-hand side of the
©PAN ASIA PUBLICATIONS
Try Question 1 in Formative Zone 5.1 Periodic Table of Elements except hydrogen.
7. An ionic bond is formed when atoms are joined • Non-metals are placed on the right-hand side of
together by transferring electrons (between the Periodic Table of Elements.
atoms of metal and non-metal).
1 18
Ionic bond 1
Metal Non-metal 2 13 14 15 16 17
2 B
3 3 4 5 6 7 8 9 10 11 12 Si
Figure 5.3 Transferring of electrons between metal 4 GeAs
atoms and non-metal atoms forms ionic bonds 5 Sb Te
8. A covalent bond is formed when atoms are 6 Po
joined together by sharing electrons (between
atoms of non-metal). 7
Metal Non-metal
Figure 5.5 The positions of metals and non-metals
in the Periodic Table of Elements
5.1

1. What is chemical bond? C1
CHAP. 2. Why neon gas, Ne does not form a compound? C2 CHAP.
5 3. Is magnesium atom, Mg stable? Why? C2 5

4. State the electron arrangement of sodium atom, Na. How does sodium atom, Na become stable?
C3
5. Determine whether the following compounds are formed by ionic bond or covalent bond. C3
(a) Potassium chloride, KCl (d) Magnesium oxide, MgO
(b) Nitrogen dioxide, NO (e) Ammonia, NH
2 3
(c) Ethane, C H (f) Copper(II) oxide, CuO
2 6
5.2 Ionic Bond
2. A metal atom of Group 1 will donate one valence
Formation of Ions electron to form a cation with a charge of +1.
Example: Li → Li + e –
+
1. A cation is formed when a metal atom donates
its valence electrons to achieve the stable duplet Donates one +
or octet electron arrangement. electron
Li Li
X → X + ne –
n+
2.1 2
+
Lithium atom, Li Lithium ion, Li
There are more protons than electrons in a cation. Charge of 3 protons = +3 Charge of 3 protons = +3
In a neutral atom, the number of electrons is equal Charge of 3 electrons = –3 Charge of 2 electrons = –2
to the number of protons. Total charge = 0 Total charge = +1


118 5.1.1 5.2.1

Form
4
Chapter 5 Chemical Bond Chemistry
Role of Hydrogen Bonds in Daily Life
H δ+ Hydrogen
O δ– bond 1. Why our hair is stuck when get wet?
H δ+ H δ+ (a) Human hair consists between 65% – 95% of
O δ– protein. The protein molecules are attracted
Covalent to each other by hydrogen bond.
H δ+ bond (b) When the hair gets wet, the presence of water

H δ+ will break the hydrogen bonds between the
protein molecules.
O δ– (c) Protein molecules will form hydrogen
©PAN ASIA PUBLICATIONS
H δ+ bonds with water molecules and the water
molecules will form hydrogen bonds with
Oxygen atom is more electronegative than other hair protein molecules. As a result, the
hydrogen atom.
Figure 5.20 Formation of hydrogen bonds between hair sticks together.
water molecules, H O Dry hair
2
(b) Formation of hydrogen bond in ammonia, H
NH and hydrogen fluoride, HF O H O
3
H
Hydrogen bond
O H O
H δ+ H δ+ H
N δ– H δ+ N δ– O H O
H δ+ Protein molecules
H δ+ H δ+
Wet hair
Nitrogen atom is more electronegative than hydrogen Hydrogen bond
atom. Hydrogen bond is formed between a nitrogen
atom in an ammonia molecule with a hydrogen atom in H
another ammonia molecule. H O
H O
Figure 5.21 Formation of hydrogen bonds between O H Water
CHAP. ammonia molecules, NH 3 molecules CHAP.
5 (c) Pembentukan ikatan hidrogen dalam Protein molecules 5
hidrogen fluorida, HF Figure 5.23 Hydrogen bonds formed between the
hair protein molecules with water molecules,
Hydrogen bond H O when the hair gets wet
2
δ– F δ+ H 2. Why do we wet a finger to turn pages?
(a) The principal raw material for producing
paper is cellulose fibers.
H δ+ δ– F (b) Cellulose consists of hydroxyl (–OH) groups.
When a wet hand is used to turn pages, water
molecules will form hydrogen bond with
Fluorine atom is more electronegative than hydrogen –OH groups in cellulose in the paper. This
atom. Hydrogen bond formed between a fluorine atom
in a hydrogen fluoride molecule with a hydrogen atom makes the paper sticks to our fingers.
in another hydrogen fluoride molecule.

Figure 5.22 Formation of hydrogen bonds between Hydrogen bond
hydrogen fluoride molecules, HF H H
H O H
O H H O Water
O
O H molecules
H
O
Cellulose fiber surface
Hydrogen chloride, HCl does not form hydrogen
bond. This is because the atomic radius of chlorine
is very large and its less electronegative compared Figure 5.24 Hydrogen bond formed between cellulose
to N, O and F. in the paper with water molecule, H O at finger
2

5.4.1 5.4.2 127

Form
4
Chemistry Chapter 5 Chemical Bond
Effect of Hydrogen Bond on the Physical
Properties of Substances

1. Hydrogen bonds will affect the solubility and The oxygen atom is partially negatively charged,
boiling point of a covalent compound. and the hydrogen atom is partially positively
2. Boiling point: charged.
(a) The boiling point of ethane, C H is
2
6
–89 °C, whereas the boiling point of ethanol, 4. Covalent compounds such as ammonia, NH
3
and hydrogen fluoride, HF also can form
C H OH is 78 °C. hydrogen bonds with water molecules, H O.
2
5
(b) Molecules of ethane, C H are attracted by This explains why ammonia, NH and hydrogen
2
5©PAN ASIA PUBLICATIONS
2
6
weak Van der Waals attraction forces only. fluoride, HF can dissolve in water, H O and has
3
2
Molecules of ethanol, C H OH are attracted higher boiling point compared to other covalent
2
5
by weak Van der Waals attraction forces and compounds.
hydrogen bonds.
(c) More heat energy is required to overcome Hydrogen bond
the weak Van der Waals attraction forces and Hydrogen bond
to break the hydrogen bonds. H δ+ δ+
δ+ H O δ– H
van der Waals attraction forces δ–
N δ– H δ+ H δ+ O
H H H H δ+
δ+ H Water F δ– H
δ– δ+ δ– δ+ molecule, Water molecule, H O
H C C O H H C C O H δ+ H H O 2
2
Hydrogen fluoride, HF
H H H H
Ethanol Ammonia, NH 3
H H molecule,
Hydrogen bond Figure 5.27 Hydrogen bond between ammonia, NH and
C H OH
δ+ δ– 2 5 3
H O C C H hydrogen fluoride, HF with water molecules
H H Table 5.4 Comparison of boiling point between covalent
compounds with and without hydrogen bond
Ethanol molecule, C H OH
CHAP. 2 5 Covalent Covalent CHAP.
5 Figure 5.25 Molecules of ethanol are attracted by Van compound with compound 5
der Waals attraction forces and hydrogen bonds
with Van
3. Solubility: hydrogen bond Boiling der Waals Boiling
and Van der
point
point
(a) Ethane, C H is insoluble in water whereas Waals attraction attraction
6
2
ethanol, C H OH is highly soluble in water. forces force only
2
5
(b) This is because the molecules of ethanol,
C H OH can form hydrogen bonds with Ammonia, NH -33 0C Phosphine, -87 0C
2
5
water molecules, H O but molecules of 3 PH 3
2
ethane, C H cannot form hydrogen bonds Hydrogen Hydrogen
2
6
with water molecules, H O. fluoride, HF 19.5 0C chloride, HCl -85.05 0C
2
H H H δ+ Ethanol,
H δ+ 78 0C Ethane, C H 6 -89 0C
2
δ– δ+ δ– C H OH
H C C O H O 2 5
Water molecule,
H H H O
2
Hydrogen bond 5.4
Ethanol molecule,
C H OH 1. What is hydrogen bond? C1
H δ+
2 δ+
H 2. Explain why hydrogen fluoride, HF has a
O δ– higher boiling point than hydrogen chloride,
Figure 5.26 Hydrogen bonds between ethanol, HCI? C3
C H OH with water molecule, H O 3. Usually covalent compounds cannot dissolve
2
5
2
in water but ethanol, C H OH can dissolve in
2 5
water. Explain why. C4
128 5.4.2

Form
4
Chapter 5 Chemical Bond Chemistry


+ –
Bulb glows Battery Negative terminal Positive terminal
Positive + + + + +
metal ion
+ + + + +
– +
Metal rod + + + + +
Electron + + + + +
Metal
©PAN ASIA PUBLICATIONS
Figure 5.32 The electrical conductivity of metal
Try Question 3 in Formative Zone 5.6

5.6

1. What is meant by delocalised electrons? C1
2. Explain the formation of metallic bonds in magnesium, Mg. C2
3. Explain how aluminium, Al can conduct electricity. C2


5.7 Properties of Ionic Compounds Solubility in Water and Organic Solvents
and Covalent Compounds 1. Most ionic compounds are soluble in water but
are insoluble in organic solvents.
Electrical Conductivity 2. On the other hand, most covalent compounds are

1. In solid ionic compounds, ions are held together insoluble in water but can dissolve in organic
by strong electrostatic attraction forces in the solvents.
lattice structure. Try Question 2 in Formative Zone 5.7
CHAP. (a) Ions are in fixed positions and cannot move CHAP.
5 freely. 5
(b) Hence, ionic compounds in the solid state do
not conduct electricity. Water is a polar solvent. It has separation of charges.
2. In aqueous or molten state, ions are free to The positive ions are attracted to the negative ends
move. Therefore, the compounds can conduct of the water molecules (oxygen atom), while the
electricity. negative ions are attracted to the positive ends of
water molecules (hydrogen atom). The attraction
Ions are in fixed positions.
Ions are free to move. between the water molecules and the ions of the
They cannot move freely.
compounds are strong enough to overcome the
electrostatic attraction forces between the ions.
2δ – Na +
O
H H
δ+ δ+
Cl –


Solid state Aqueous or molten state
Figure 5.33 Electrical conductivity of ionic compounds Melting Point and Boiling Point
3. However, covalent compounds consist of neutral
molecules. They do not contain ions. Hence, 1. Ionic compounds consist of positive ions and
covalent compounds do not conduct electricity negative ions which are held together by strong
in all states. electrostatic attraction forces.


5.6.2 5.7.1 131

Form
4
Chemistry Chapter 5 Chemical Bond
Positive ion 5. Less heat energy is required to overcome the
Negative ion weak Van der Waals attraction forces. This
explains why covalent compounds have low
A lot of heat energy is melting and boiling points.
required to break the strong
electrostatic forces during 6. Covalent compounds usually exist as volatile
melting or boiling. liquids at room temperature.

Figure 5.34 Oppositely-charged ions in an ionic Try Question 1 in Formative Zone 5.7
compound are held together by strong
electrostatic attraction forces Table 5.5 Comparison between properties
of ionic and covalent compounds
2. A lot of heat energy is required to overcome
the strong electrostatic forces. This explains Properties Ionic Covalent
why ionic compounds have higher melting and compounds compounds
boiling points.
3. Ionic compounds are usually non-volatile Melting point High Low
solids at room temperature. and boiling
4. On the other hand, covalent compounds (with point
simple molecular structures) consist of neutral
molecules which are held together by Van der Electrical Conduct Do not conduct
Waals attraction forces. conductivity electricity in electricity in any
aqueous or states
Less heat energy is molten state
required to break the
weak Van der Waals
attraction forces Solubility Usually dissolve Usually dissolve in
during melting in water but do organic solvents such
Molecules or boiling.
not dissolve in as benzene but do not
Figure 5.35 Molecules in a covalent compound are held organic solvents dissolve in water
together by weak Van der Waals attraction forces


CHAP. ©PAN ASIA PUBLICATIONS CHAP.
5 5.1 5

Aim: Variables:
To compare the properties of ionic compounds (a) Manipulated: Type of compound
and covalent compounds. (b) Responding: Electrical conductivity
(c) Fixed: Carbon electrode
Problem statement:
What are the differences between properties of Procedure:
ionic compounds and covalent compounds?
Batteries Bulb Switch
Materials:
Magnesium chloride, MgCI , cyclohexane, C H
2 6 12
naphthalene, C H , distilled water, solid lead (II)
10 8 Carbon electrodes
bromide, PbBr
2
Lead(II) bromide, PbBr
Apparatus: 2
Test tubes, crucible, 10 cm measuring cylinder,
3
3
spatula, glass rod, 250 cm beaker, Bunsen burner,
wire gauze, tripod stand, carbon electrodes,
pipe clay triangle, battery, connecting wire with
crocodile clip, light bulb, switch Figure 5.36
A Electrical conductivity 1. A crucible is filled with lead(II) bromide, PbBr
2
Hypothesis: powder until it is half full.
Ionic compounds can conduct electricity in the 2. The apparatus is set up as shown in Figure
molten state but not in the solid state while 5.36.
covalent compounds do not conduct electricity in 3. The switch is turned on. The observation on
both states. the light bulb is recorded.

132 5.7.1

Form
4
Chapter 5 Chemical Bond Chemistry

4. The switch is turned off and the lead(II) 6. Steps 1 to 5 are repeated using naphthalene,
bromide, PbBr powder is then heated until it C H to replace lead(ll) bromide, PbBr .
2 10 8 2
melts completely. 7. The observation on the light bulb is recorded.
5. The switch is turned on again and the
observation is made on the light bulb.
Results:
Table 5.6
Observation
Compound Physical state Inference
on the light bulb
©PAN ASIA PUBLICATIONS
Lead(II) Solid The bulb does not light up. PbBr cannot conduct electricity in solid state
2
bromide, but can conduct electricity in molten state.
PbBr Molten The bulb lights up.
2
Naphthalene, Solid The bulb does not light up. C H cannot conduct electricity in solid and
10 8
C H molten states.
10 8
Molten The bulb does not light up.
Conclusion: Variables:
The hypothesis is accepted. Lead(II) bromide, (a) Manipulated: Type of compound, magnesium
PbBr is an ionic compound which able to conduct chloride, MgCl and naphthalene, C H
2 2 10 8
–
2+
electricity in molten state but not in solid state. (b) Responding: Solubility of Mg ions and Cl
Naphthalene, C H is a covalent compound that ions
10 8
cannot conduct electricity both in solid and molten (c) Fixed: Quantity of compound, volume of
states. solvent, temperature
Discussion: Procedure:
1. Lead(II) bromide, PbBr consists of lead(II) ions,
2
Pb and bromide ions, Br . In solid state, the
2+
–
oppositely-charged ions are held together by Magnesium
strong electrostatic forces of attraction in the Distilled chloride, Cyclohexane, C H 12
6
lattice structure. They are in fixed positions and water MgCl
CHAP. cannot move freely. As a result, solid lead(II) 2 CHAP.
5 bromide, PbBr cannot conduct electricity. Figure 5.37 5
2
2. When lead(II) bromide, PbBr is heated until 1. Half spatula of magnesium chloride, MgCl
2 2
it melts to become a liquid, the lead(II) crystals is placed in two test tubes separately.
ions, Pb and bromide ions, Br are free to 2. 5 cm of distilled water is added to the first
2+
–
3
move. Therefore, molten PbBr can conduct test tube and shaken.
2
electricity. 3. 5 cm of cyclohexane, C H is added to the
3
12
6
3. Naphthalene, C H only consists of neutral second test tube and shaken.
10 8
molecules. It does not contain ions. Thus, it 4. The solubility of magnesium chloride, MgCl in
cannot conduct electricity either in solid state both solvents are observed and recorded. 2
or molten state. 5. Steps 1 to 4 are repeated using solid
naphthalene, C H to replace magnesium
10 8
chloride, MgCl .
2
Ionic compounds which are soluble in water also Results:
can dissociate in water to produce free moving Table 5.7
ions.
Compound Solubility Solubility in
in water cyclohexane
B Solubility
Hypothesis: Magnesium Soluble Insoluble
chloride, MgCl
Most ionic compounds are soluble in water but 2
insoluble in organic solvents while most covalent
compounds are insoluble in water but can dissolve Naphthalene Insoluble Soluble
in organic solvents. C H 8
10
5.7.1 133

Form
4
Chemistry Chapter 5 Chemical Bond

Conclusion: 2. A spatula of naphthalene, C H powder is
10 8
The hypothesis is accepted. Magnesium chloride, placed in another test tube.
MgCl is an ionic compound which is soluble in 3. The two test tubes are heated in a beaker
2
water but insoluble in organic solvents. Napthalene, containing water as shown in Figure 5.38.
C H is a covalent compound which is insoluble in 4. The changes in the physical states of the
10 8
water but soluble in organic solvents. compounds are observed and recorded.
Discussion: Results:
1. Magnesium chloride, MgCl dissociates in Table 5.8
2
water to produce free moving magnesium ions
Mg and chloride ions, Cl . Compound Observation Inference
–
2+
+©PAN ASIA PUBLICATIONS
2. Naphthalene, C H consists of neutral
10 8 Magnesium No change Magnesium chloride,
molecules which cannot dissociate in water.
chloride, MgCl has a high
2
C Melting and boiling points MgCl melting point.
2
Hypothesis:
Ionic compounds have high melting and boiling Naphthalene, It melts Naphthalene, C H
8
10
points while covalent compounds have low melting C H 8 rapidly. has a low melting
10
and boiling points. point.
Variables: Conclusion:
(a) Manipulated: Type of compound, magnesium The hypothesis is accepted. Magnesium chloride,
chloride, MgCl and napthalene, C H MgCl is an ionic compound and naphthalene,
2 10 8 2
(b) Responding: Boiling point C H is a covalent compound. Magnesium
8
10
(c) Fixed: Quantity of compound chloride, MgCl has a higher melting point than
2
naphthalene, C H .
Procedure: 10 8
Discussion:
1. Magnesium chloride, MgCl consists of positive
2
2+
–
ions, Mg and negative ions, Cl . Mg ions
2+
Water, H O
–
2 and Cl ions are attracted together by strong
Magnesium
Naphthalene,
chloride, MgCl electrostatic forces. A lot of heat energy is
2 C H
10 8 required to overcome the strong electrostatic
CHAP. Heat attraction forces. Therefore, the melting point CHAP.
5 of magnesium chloride, MgCl , is high. 5
2
2. Naphthalene, C H consists of molecules which
10 8
Figure 5.38 are attracted by weak Van der Waals attraction
forces. As a result, less heat energy is required
1. A spatula of magnesium chloride, MgCl to overcome the weak Van der Waals attraction
2
powder is placed in a test tube. forces.
Structure of Ionic Compounds and Covalent Compounds
1. Ionic compounds consist of positive ions of
metals and negative ions of non-metals. The
positive ions and negative ions attract each other In the actual structure, ions are packed together
to form a rigid three-dimensional structure closely as shown in Figure 5.40. Figure 5.39 just
which is called lattice. Each ion in the lattice to make the structure easy to see.
structure is surrounded by other ions of opposite
charges.
Example:
Sodium chloride, NaCl consists of sodium ions,
Na and chloride ions, Cl .
–
+ Figure 5.40
Na
–
Cl 2. The arrangement shown in Figure 5.39 does not
exist separately on its own. It extends throughout
the crystal, involving millions of ions. This
structure is known as giant ionic lattice.
Figure 5.39 The lattice structure of NaCl
134 5.7.1

Form
4
Chapter 5 Chemical Bond Chemistry
4. Covalent compounds consist of molecules in 8. Giant molecular structures (or macromolecular
which structures can be classified as: structures) contain many hundreds of thousands
(a) Simple molecular structure. of atoms joined by strong covalent bonds to
(b) Giant molecular structure. form three-dimensional networks of atoms.
5. In a simple molecular structure, the There are no separate molecules and the weak
molecule consists of a few atoms. The atoms Van der Waals' forces do not exist.
in the molecule are joined together by strong Examples of giant molecular structures:
covalent bonds. Diamond, graphite, silicon(IV) oxide
6. However, the individual molecules are only
attracted to each other by weak Van der Waals Strong covalent bonds between
©PAN ASIA PUBLICATIONS
attraction forces. Only little heat is required to the carbon atoms in the three-
dimensional structure of diamond
overcome the attraction forces make it has low
melting and boiling point.
Examples of simple molecular structures:
Iodine, I , carbon dioxide, CO , water, H O
2
2
2
7. Simple molecule can exist as solid, liquid and
gas. Figure 5.42 A small part of the structure of diamond
I I 9. Compounds with giant molecules exist as solid.
Weak Van der Waals attraction
forces between the iodine
molecules, I 2
I I
Covalent compounds with giant molecular
Strong covalent bond between the
iodine atoms in the molecule. structure have very high melting point and boiling
point because the heat required to break the
Figure 5.41 The simple molecules of iodine covalent bond is high.



Uses of Ionic and Covalent Compound in Daily Life
CHAP. Table 5.9 Various uses of ionic and covalent compounds in daily life CHAP.
5 5
Field Example of ionic compounds Example of covalent compounds

Industry • Lithium iodide, LiI is used • Turpentine, C H is used as a solvent for paint.
10
18
to make pacemaker batteries, • Ether, (C H ) O is used to make inks and dyes.
2
5 2
which has a lifespan of 7 to 8
years.
Agriculture • Ammonium nitrate, NH NO • Bromoethane, C H Br and chloropicrin, CCl NO are
3
2
3
5
4
2
and pottasium chloride, KCl used as pesticides and insecticides.
are used to make fertiliser.
Medicine • Sodium bicarbonate, NaHCO • Paracetamol, C H NO is a pain reliever and a fever
9
8
3
2
reduces stomach acid. It reducer.
is used as an antacid to • Penicillin, C H N O S is an antibiotic. It is used
16
2
4
18
treat indigestion and upset to treat many different types of infection caused by
stomach. bacteria.
Domestic • Sodium chlorate, NaClO is • Glycerol, C H (OH) is a very effective moisturiser on
3
3
5
3
use used in the manufacture of skin. It absorbs water from air reducing dry and dull
bleaches and detergents. patches on your skin. Your skin gets soft, supple and
hydrated immediately after application.
5.7.1 5.7.2 135

Form
4
Chemistry Chapter 5 Chemical Bond

5.7

1. Sodium chloride, NaCl and hydrogen chloride, HCl are two compounds that consist of chlorine, Cl element.
The melting point of sodium chloride, NaCl is 808 °C but the melting point of hydrogen chloride, HCl is
–114 °C. Explain why the melting points of the two compounds are different. C3
2. Table 1 shows the proton number of elements P and Q.
Element Proton number
P 12
©PAN ASIA PUBLICATIONS
Q 17

Table 1
(a) What type of chemical bond is formed between: C2
(i) P and Q?
(ii) Q and Q?
(b) Explain why the compound formed by P and Q cannot conduct electricity in solid state but can
conduct electricity in aqueous or molten state. C3
(c) State whether the compound formed by Q and Q is an electrical conductor. Explain why. C3
(d) Between the compounds formed in (a)(i) and (a)(ii), determine which compound can dissolve in:
C3
(i) Water
(ii) Ethanol

SPM Simulation HOTS Questions



1 Which of the following particles contain 10 Examiner’s comment:
electrons? C4 • The properties above are for ionic
CHAP. [Proton number: Na = 11, Al = 13] compound. CHAP.
5 I Al III Na + • Lead(II) iodide and sodium chloride 5
II Na IV Al 3+ are ionic compounds but lead(II) iodide
A I and II C II and IV is insoluble in water.
B I and III D III and IV Answer: D

Examiner’s comment: 3. The electron arrangement of atom J is 2.8.2
• Al contains 13 electrons. and the electron arrangement of atom L is 2.7.
• Na contains 11 electrons.
+
• Na is the ion of Na which loses 1 electron. Elements J and L react to form a compound.
Thus, the number of electrons is 10. Write the chemical formula for the compound
• Al is the ion of Al which loses 3 electrons. formed and state two physical properties.
3+
Thus, the number of electrons is 10. C4
Answer: D
Examiner’s comment:
2 Compound Z has the following properties. J is a metal and L is a non-metal. An atom
• Soluble in water J will donate 2 electrons. Each of two atom
• Conduct electricity in aqueous or molten L will receive 1 electron from atom J. The
state ionic compound formed has the chemical
• Melting point = 800 °C formula JL .
2
Answer:
What is Z? C3 Chemical formula = JL 2
A Glucose The compound formed is soluble in water
B Naphthalene and has high melting point and boiling
C Lead(II) iodide point.
D Sodium chloride



136

Form
4
Chapter 5 Chemical Bond Chemistry







Paper 1


1. Which of the following pairs of physical
properties of sodium oxide, Na O is true? C2 C
2
©PAN ASIA PUBLICATIONS
Solubility in Electrical conductivity
water when molten D
A Soluble Conducting
B Soluble Not conducting
C Insoluble Not conducting
D Insoluble Conducting
6. Ethanol, C H OH is a covalent compound that
5
2
2. Which of the following substances is a covalent can dissolve in water. Which of the following
compound? C1 explain the solubility of ethanol, C H OH in
A Magnesium chloride, MgCl water? C2 2 5
B Calcium oxide, CaO A Ethanol molecules ionise in water.
C Lead(II) chloride, PbCl 2 B Ethanol molecules form hydrogen bonds
D Carbon dioxide, CO 2 with water molecules.
C Ethanol molecules form dative bonds with
3. The following shows information of a few water molecules.
elements. D Ethanol molecules donate electron to water
V: Sulphur molecules.
CHAP. W: Silver 7. Figure 1 shows the electron arrangement of an CHAP.
SPM Clone
5 X: Carbon ion M . Given that atom M has 18 neutrons. 5
–
Y: Chlorine
Z: Oxygen –
Which of the following elements can react to
form an ionic compound? C2 M
A V and X C X and Y
B W and Z D Y and Z
Figure 1
4. Which of the following compounds has the What is the nucleon number of atom M? C2
lowest melting point? C2 A 8 C 22
A Sodium chloride, NaCl B 10 D 35
B Sulphur dioxide, SO 2
SPM Clone
C Iron(III) oxide, Fe O 3 8. The electron arrangement of an atom J is 2.1 and
2
D Magnesium chloride, MgCl the electron arrangement of an atom T is 2.6.
2
Atom J reacts with atom T to form a compound.
5. Which of the following diagrams shows a Which of the following is true about the
compound that consists of double covalent reaction? C2
bonds? C1 A Atom T donates 6 electrons.
A B Atom J receives 1 electron.
C An ionic compound is formed.
D The chemical formula of the compound
B formed is JT .
6



137

Form
4
Chemistry Chapter 5 Chemical Bond
9. Element X and hydrogen, H combined to form 12. Table 2 shows the numbers of electrons in ions
compound H X. Element X and sodium, Na P , Q , R and T .
2–
+
–
2+
2
combined to form compound Na X. Which Number of Number of
2
of the following represents the electron Ion electrons neutrons
arrangement for a compound formed between P 2– 10 11
magnesium, Mg and element X? C2 Q + 10 12
A
R – 18 18
Mg X T 2+ 18 20
Table 2
©PAN ASIA PUBLICATIONS
Which of the following shows the correct
B nucleon number of the ion? C3
X Mg X Ion Nucleon number
A P 2– 20
C – 2+ – B Q + 22
X Mg X C R – 35
D T 2+ 38
D 2+ 2– 13. Figure 3 shows the electron arrangements of

+
2–
Mg X ions W and Z .
+ 2 –
W Z
10. What is the chemical bond shown in Figure 2?
C1
H F H F Figure 3
H N B F H N B F Calculate the difference between the numbers of
electrons in atom W and atom Z. C3
CHAP. H F H F A 2 C 4 CHAP.
Figure 2
5 A Covalent bond C Hydrogen bond B 3 D 5 5
B Dative bond D Ionic bond 14. Figure 4 shows the symbols of ion X and ion Y.
3+
SPM Clone X Y 2–
11. Table 1 shows the melting points and types of
particles of compounds P, Q, R and S. Figure 4
Calculate the relative formula mass of the
Melting compound formed when X reacts with Y.
Compound Type of particle
point ( C) Given that the relative atomic mass of X is 27
o
and the relative atomic mass of Y is 16. C3
P 70 Molecule A 113
Q 280 Ion B 102
R 650 Ion C 86
S 2 000 Atom
Table 1 15. An element reacts with chlorine, Cl to form a
Based on Table 1, which compound can conduct compound with the formula XCl. Which of the
following elements could be X?
electricity when it is heated to a temperature of I Sodium, Na III Hydrogen, H
320 °C? C3 II Calcium, Ca IV Aluminium, Al
A P A I and II
B Q B I and III
C R C II and IV
D S D III and IV



138

Form
4
Chapter 5 Chemical Bond Chemistry

Paper 2
Section A

SPM Clone [1 mark]
1. (a) (i) The electron arrangement of neon is 2.8. Why is this element very stable? C2
(ii) Name another element that has the same stability as neon. C1 [1 mark]
(b) Figure 1.1 shows an electron arrangement of magnesium oxide, MgO compound that is produced by the
2–
formation of an ionic bond between a magnesium ion, Mg and an oxide ion, O . C2
2+
2+ 2–
Mg O



Figure 1.1
(i) How are a magnesium ion, Mg and an oxide ion, O formed from their respective atoms? C1
2-
2+
[1 mark]
2-
(ii) What type of attractive forces exists between magnesium ions, Mg and oxide ions, O ? C2
2+
[1 mark]
(iii) What happens to the ions in this compound when it is heated until it melts completely? [1 mark]
(iv) Give one reason for your answer in (b)(iii). C2 [1 mark]
(c) Figure 1.2 shows the symbols of two elements, P and Q. The letters used do not represent the actual
symbols of the elements.

23 35
11P 17Q

Figure 1.2
Element P reacts with element Q to form a compound.
CHAP. ©PAN ASIA PUBLICATIONS CHAP.
5 Draw one diagram to show the bonding formed between elements P and Q. C6 [2 marks] 5
SPM Clone
2. Table 2.1 shows the electrical conductivity and melting points of substances X, Y and Z.

Electrical conductivity in the state of
Substance Melting point (°C)
Solid Molten Aqueous

X No No No < – 60

Y No No No 70 – 90

Z No Yes Yes > 750

Table 2.1
(a) (i) State the types of particle and bonding of substance X. C1 [1 mark]
(ii) Explain why substance X has a melting point less than – 60 °C. C2 [1 mark]
(b) State how the bonds are formed in: C2
(i) Substance Y [1 mark]
(ii) Substance Z [1 mark]

(c) State why substance Z can conduct electricity in molten and aqueous states but not in solid state. C2
[1 mark]


139

Form
4
Chemistry Chapter 5 Chemical Bond
(d) Based on the information in Table 2.1, classify substances X, Y and Z according to their solubility in
water. C3
Solubility in water Soluble Insoluble
Substance
(i) (ii)

(iii)
Table 2.2
[3 marks]
3. Figure 3 shows the electron arrangements of five elements P, Q, R, S and T.








P Q R S T
Figure 3
(a) Which element can form metallic bond? C2 [1 mark]
(b) P reacts with Q.
(i) What type of chemical bond is formed in the reaction? C1 [1 mark]
(ii) Draw the electron arrangement of the compound formed. C3 [1 mark]
(iii) Predict whether the compound formed can conduct electricity or not. Explain
your answer. C3 [1 mark]

(c) Element R is burnt vigorously in gas S.
(i)
[1 mark]
©PAN ASIA PUBLICATIONS
What is the type of particle of compound formed in the reaction? C2
CHAP. (ii) Write the formula of the compound formed. C4 [1 mark] CHAP.
5 (d) State two differences of physical properties between compounds formed in (b) and (c). C4 5
[1 mark]
(e) Explain why element T exists as monoatomic gas at room temperature. C3 [1 mark]

Section B

SPM Clone
4. Figure 4 shows the chemical symbols which represent three elements, P, Q and R.

39 P 35 12 R
19 17Q 6
Figure 4
(a) (i) Write the electron arrangement of atoms P and R. C4 [2 marks]
(ii) State the number of neutrons in an atom of element Q. C2 [1 mark]
(iii) Write the symbol of an isotope of element R. C2 [1 mark]
(b) Element P reacts with element Q to form an ionic compound whereas element Q reacts with element R
to form a covalent compound. Explain how these ionic and covalent compounds are formed. C3
[8 marks]
(c) The ionic compound formed by the reaction between elements P and Q is able to conduct electricity
when it is dissolved in water. Describe how you could prove that this statement is correct. C6
[8 marks]




140

Form
4
Chapter 5 Chemical Bond Chemistry
Section C
SPM Clone
23
16
5. (a) J and L are symbols of two atoms.
11
8
Explain how atoms J and L can form ions and write the formulae of the ions formed. C4 [4 marks]
(b) Figure 5 shows the electron arrangement of a molecule PQ . These letters are not the actual symbols of
3
the elements.
Q P Q

Q
Figure 5
Based on Figure 5, write the electron arrangements of the atoms of element P and element Q. Explain the
position of element P in the Periodic Table of Elements. C2 [6 marks]
(c) Table 5 shows the electron arrangements of atoms W, X and Y. These letters are not the actual symbols
of the elements.

Element Electron arrangement
W 2.4
X 2.6
Y 2.8.2
Table 5
Using the information in Table 5, explain how two compounds can be formed by these elements based
on their electron arrangements. The two compounds should have different types of bond. C3
[10 marks]


Reinforcement & Assessment of Science Process Skill
CHAP. ©PAN ASIA PUBLICATIONS CHAP.
5 Reinforcement exercise of science process skills in preparation for Paper 3 (Practical Test). 5



In this experiment you are required to conduct an experiment to study the electrical conductivity of substances
M1 and M2 in solid and molten states.
You are provided with the following materials:
M1 = Solid compound X
M2 = Solid compound Y

Batteries Bulb
Switch

Carbon
electrodes
Solid M1






Figure 1
Carry out the experiment according to the following instructions:
1. Fill a crucible with M1 until half full.
2. Set up the apparatus as shown in Figure 1.
3. Turn on the switch and observe the brightness of the light bulb. Record the observation in Table 1.

141

Form
4
Chemistry Chapter 5 Chemical Bond
4. Turn off the switch and heat the M1 until completely melted.
5. Turn on the switch again and observe the brightness of the light bulb. Record the observation in Table 1.
6. Repeat steps 1 to 5 using M2. Record the observations in Table 1.
Substance Physical state Observation on the light bulb


Substance M1


Substance M2


Table 1
[**Refer to the Simulation experiments and sample results given to understand how to fill in Table 1]
Based on the experiment conducted:
1. State an inference about the electrical conductivity of compounds X and Y based on the observations of
this experiment.
2. State the type of particles in compounds X and Y.
3. What is the relationship between the type of particle and electrical conductivity?
4. Predict which compound has a higher melting point. Explain why.

Nota: Simulation experiments and sample results

Experiment Solid state Molten state




Carbon
CHAP. ©PAN ASIA PUBLICATIONS CHAP.
Carbon
electrodes
5 electrodes 5
I Solid M1 Molten M1









Carbon Carbon
electrodes electrodes
II Solid M2 Molten M2


















142

CHAPTER
1 Redox Equilibrium













Important Learning Standards Page
©PAN ASIA PUBLICATIONS
• Describe redox reactions. 296
1.1 Oxidation and
Reduction • Explain redox reaction based on the change in oxidation number. 301
• Investigate displacement reaction as a redox reaction. 306

1.2 Standard Electrode • Describe the standard electrode potential. 312
Potential • Determine oxidising agent and reducing agent based on their value 313
of standard electrode potentials.
1.3 Voltaic Cell • Explain redox reaction in voltaic cell. 315
• Describe electrolysis. 321
• Describe electrolysis of molten compound. 323
1.4 Electrolytic Cell • Explain factors that affect electrolysis of aqueous solution. 326
• Compare voltaic cell and electrolytic cell. 335
• Describe electroplating and purification of metal by electrolysis. 335
• Explain extraction of metal from its ore through electrolysis 338
1.5 Extraction of Metal process.
from Its Ore • Explain metal extraction from its ore through reduction process by 340
carbon.
• Describe metal corrosion process as redox reaction. 342
1.6 Rusting
• Prevent rusting. 345










• Active electrode / Elektrod aktif • Non-electrolyte / Bukan elektrolit
• Cast iron / Besi tuangan • Oxidation state / Keadaan pengoksidaan
• Chemical species / Spesies kimia • Oxidising agent / Agen pengoksidaan
• Corrosion / Kakisan • Polarisation / Pengutuban
• Daniell cell / Sel Daniell • Potential difference / Beza keupayaan
• Displacement of halogen / Penyesaran halogen • Purification of metal / Penulenan logam
• Displacement of metal / Penyesaran logam • Reducing agent / Agen penurunan
• Electrolyte / Elektrolit • Rusting / Pengaratan
• Electroplating of metal / Penyaduran logam • Slag / Sanga
• Extraction of metal / Pengekstrakan logam • Standard cell potential / Keupayaan sel piawai
• Froth floatation / Pengapungan berbuih • Standard electrode potential / Keupayaan elektrod
• Half-cell / Setengah sel piawai
• Half equation / Setengah persamaan • Voltaic cell / Sel kimia

294

Extraction of aluminium Extraction of iron



A less electropositive metal will increase the rate of rusting. A more electropositive metal will prevent iron from rusting. Rusting is the corrosion of iron. Metal is oxidised naturally with the release of electrons to form metal ions. Corrosion of metal is a redox reaction. Via electrolysis Extraction of metal Purification of metals Electroplating of metals Via reduction using carbon Products are affected by Types





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Corrosion of metal Application in industries Electrolytic cell Aqueous solution Concentration of ions in electrolyte










Decreases of oxidation number Gain of electron Gain of hydrogen Loss of oxygen Redox Equilibrium Electrical energy → chemical energy Molten compound E 0 value








Reduction Chemical energy → electrical energy Voltaic cell Daniell Simple cell cell











Oxidation Standard Electrode Potential




Increases of oxidation number Loss of electron Loss of hydrogen Gain of oxygen The more positive the E 0 value is, the easier for the chemical species to undergo reduction. The more negative the E 0 value is, the easier for the chemical species to undergo oxidation.












295

Form
5 Chemistry Chapter 1 Redox Equilibrium

CHAP. CHAP.
1 Activity 1.2 1


Aim:
To investigate the conversion of iron(II) ion, Fe to 4. The mixture is filtered into a test tube.
2+
3+
iron(III) ion, Fe and vice versa.
Materials:
–3
0.5 mol dm freshly prepared iron(II) sulphate, Test to confirm the presence of iron(II) ion, Fe .
2+
–3
FeSO solution, 0.5 mol dm iron(III) chloride,
4
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FeCl solution, bromine water, Br , zinc, Zn powder,
3 2
filter paper, 2.0 mol dm sodium hydroxide, NaOH 5. 0.2 mol dm of sodium hydroxide, NaOH
–3
–3
solution solution is added drop by drop until in excess.
The changes are recorded.
Apparatus:
Dropper, spatula, test tube holder, Bunsen burner, Observation:
filter funnel, test tube rack, measuring cylinder, Table 1.2
test tube
Activity Observation
Procedure: Iron(II) sulphate, The brown bromine water,
2+
A Conversion of iron(II) ion, Fe to iron(III) FeSO + bromine Br turns colourless.
ion, Fe 3+ 4 2
water, Br The pale green iron(II)
2
sulphate, FeSO solution
4
turns yellow.
Bromine Adding of sodium A brown precipitate is
water, Br
2
hydroxide, NaOH formed. It is insoluble in
solution excess sodium hydroxide,
Iron(II) sulphate, NaOH solution.
FeSO solution
4
Figure 1.8 Iron(III) chloride, The brown iron(III) chloride,
FeCl + zinc, Zn FeCl solution turns pale
3
3
1. 2 cm of freshly prepared iron(II) sulphate, powder green.
3
FeSO solution is measured and pour into a
4
test tube. Adding of sodium A dirty green precipitate
2. By using a dropper, bromine water, Br is hydroxide, NaOH is formed. It is insoluble in
2
added to the solution in the test tube drop by solution excess sodium hydroxide,
drop until no further changes are observed. NaOH solution.
3. The mixture is shaken well and warm gently.
Inference:
1. The pale green iron(II) sulphate, FeSO solution
4
2+
turns yellow because the iron(II) ions, Fe are
Test to confirm the presence of iron(III) ion, Fe . oxidised to iron(III) ions, Fe .
3+
3+
2. A brown precipitate is formed when tested
with sodium hydroxide, NaOH solution. This
–3
4. 0.2 mol dm of sodium hydroxide, NaOH confirms the presence of the iron(III) ions, Fe .
3+
solution is added drop by drop until in excess. 3. The brown iron(III) chloride, FeCl solution
The changes are recorded. 3 3+
turns pale green because the iron(III) ions, Fe
are reduced to iron(II) ions, Fe .
2+
B Conversion of iron(III) ion, Fe to iron(II)
3+
4. A dirty green precipitate is formed when
ion, Fe 2+
tested with sodium hydroxide, NaOH solution.
3
1. 2 cm of iron(III) chloride, FeCl solution is This confirms the presence of the iron(II) ions,
3
measured and pour into a test tube. Fe .
2+
2. Half spatula of zinc, Zn powder is added into
the solution in the test tube. Discussion:
2+
3. The mixture is shaken well and warm gently 1. Iron(II) ions, Fe are oxidised to iron(III) ions,
3+
until no further changes. Fe by bromine water, Br .
2
304 1.1.2

Form
5
Chapter 1 Redox Equilibrium Chemistry
CHAP. 2. Bromine water, Br acts as the oxidising agent, 5. Iron(III) ions, Fe are reduced to iron(II) ions, CHAP.
3+
1 whereas iron(II) ions, Fe act as the reducing Fe by zinc, Zn. 1
2
2+
2+
3+
agent. 6. Iron(III) ion, Fe act as the oxidising agent
3. An iron(II) ion, Fe loses an electron to form whereas zinc, Zn acts as the reducing agent.
2+
2+
iron(III) ion, Fe . Thus, iron(II) ion, Fe is 7. A zinc atom loses two electrons to form zinc
3+
oxidised to iron(III) ion, Fe . ion, Zn . Thus, zinc atom, Zn is oxidised to
3+
2+
2+
4. A bromine molecule, Br receives two zinc ion, Zn .
2
electrons from iron(II) ion, Fe to form 8. An iron(III) ion, Fe receives an electron from
3+
2+
2+
–
bromide ions, Br . Thus, bromine molecule, Br zinc, Zn to form iron(II) ion, Fe . Thus, iron(III)
2
2+
is reduced to bromide ions, Br . ion, Fe is reduced to iron(II) ion, Fe .
3+
–
©PAN ASIA PUBLICATIONS
Oxidation half equation: Oxidation half equation:
Fe (aq) → Fe (aq) + e – Zn(s) → Zn (aq) + 2e –
2+
3+
2+
Reduction half equation: Reduction half equation:
Br (aq) + 2e → 2Br (aq) 3+ – 2+
–
–
2 Fe (aq) + e → Fe (aq)
Overall ionic equation (Redox reaction): Overall ionic equation (Redox reaction):
2+
–
3+
2Fe (aq) + Br (aq) → 2Fe (aq) + 2Br (aq) 3+ 2+ 2+
2 2Fe (aq) + Zn(s) → 2Fe (aq) + Zn (aq)
Try Question 7 in Formative Zone 1.1
Name compound using the IUPAC nomenclature 5. For anions which contain metal that have
1. Many elements have just one oxidation number, variable oxidation numbers, the oxidation
but some elements such as transition metals, number of the metal ion are written in Roman
carbon, nitrogen and sulphur have more than numeral in brackets, immediately following the
one oxidation number. name of the anion.
2. To avoid confusion, Roman numeral (I, II, 6. Table 1.4 shows examples of metal element
III and etc) are included in naming some which has more than one oxidation number in
compound that consists of element which has anion.
more than one oxidation number. Table 1.4
3. For simple ionic compounds, the oxidation
number of a metal ion is written in Roman Chemical Oxidation IUPAC
numeral in brackets, immediately following the formula of Formula number nomenclature
name of the metal. compound of anion of metal of the
4. Table 1.3 shows examples of metal element elements compound
which has more than one oxidation number in
the ionic compound. K MnO 4 MnO 4 2– +6 Potassium
manganate(VI)
2
Table 1.3
Potassium
Chemical Oxidation IUPAC nomenclature KMnO 4 MnO 4 – +7 manganate(VII)
formula of number of of the compound
compound metal elements
K CrO CrO 2– +6 Potassium
Cu O +1 Copper(I) oxide 2 4 4 chromate(VI)
2
CuO +2 Copper(II) oxide Potassium
K Cr O 7 Cr O 7 2– +6 dichromate(VI)
2
2
2
PbCl 2 +2 Lead(II) chloride
7. For non-metal elements which have variable
PbCl +4 Lead(IV) chloride
4 oxidation numbers, the oxidation numbers of
the non-metal element are written in Roman
FeSO +2 Iron(II) sulphate
4 numeral in brackets, immediately following the
Fe (SO ) +3 Iron(III) sulphate name of the ion that contains the non-metal
2 4 3
element.
1.1.2 305

Form
5 Chemistry Chapter 1 Redox Equilibrium

CHAP. CHAP.
1 8. Zinc, Zn strip is placed into silver nitrate, (a) Oxidation half equation 1
Ag(NO ) solution. Silver, Ag and zinc nitrate,
3 2 (b) Reduction half equation
Zn(NO ) are formed. (c) Overall ionic equation
3 2
Zn(s) + 2AgNO (aq) → Ag(s) + Zn(NO ) (aq) (d) Oxidising agent
3 3 2 (e) Reducing agent
Based on the reaction equation, determine the
following: C3
1.2 Standard Electrode Potential potential of an unknown half-cell with a
standard electrode. Hydrogen electrode was
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1. The strength of an oxidising agent and a chosen as a standard electrode.
reducing agent depends on the standard
electrode potential, E . Platinum surface promotes oxidation of
0
2. Standard electrode potential, E 0 is a hydrogen molecule, H to hydrogen ions,
2
H or reduction of hydrogen ions, H to
+
+
measurement of the potential for equilibrium. hydrogen gas, H .
2
3. If a metal is immersed in a solution containing
the metal ions, the metal atoms tend to release Atoms of platinum, Pt electrode
electrons to form metal ions and enter the
solution. The electrons released accumulate on H (g) at e – H +
2
1 atm
the metal, making the metal negatively charged. H 2
4. In a short time, the metal is surrounded by Platinum, e – H +
Pt wire
positive ions. A small number of metal ions H (g) e – H +
2
will receive electrons and become metal atoms Platinum, Pt e – H 2
again. electrode e – H +
Acidic solution,
M(s)  M (aq) + ne – H 1.0 mol dm –3 Half equation of the
n+
+
hydrogen half-cell:
–
5. Thus, an equilibrium is achieved between metal 2H (aq) + 2e  H (g)
+
2
n+
M atoms and metal M ions in a half-cell. Figure 1.12 Standard hydrogen electrode half-cell
This condition results in a potential difference 10. Electrode potential of standard hydrogen
between the metal (electrode) and its solution electrode, E 0 is assigned as 0.00 V.
reference
(electrolyte). The potential difference is called as 11. The electrode potential in a half-cell when
electrode potential. compared relative to the standard hydrogen
6. Hydrogen gas, H can also be dissolved in electrode, SHE under standard conditions is
2
+
a solution to form hydrogen ions, H .The called standard electrode potential, E .
0
+
hydrogen ions, H in the solution will also 12. Standard conditions for the measurement of
receive electrons to become hydrogen gas, standard electrode potential of a half-cell:
H . An equilibrium is achieved between the (a) Concentration of ions in aqueous solutions
2
hydrogen gas, H and the hydrogen ions, H in is 1.0 mol dm .
+
-3
2
the solution. (b) Gas pressure of 1 atm or 101 kPa.
o
H (g)  2H (aq) + 2e – (c) Temperature at 25 C or 298 K.
+
2
(d) Platinum is used as an inert electrode.
7. Because hydrogen is in gaseous state, an inert
electrode, namely platinum, Pt is used as a Determine the value of the standard electrode
conductor so that it can come into contact with potential
hydrogen gas, H as shown in Figure 1.12. 1. Standard hydrogen electrode consists of a
2
8. Since, there is a potential difference between the platinum electrode immersed in a 1.0 mol
electrode that is in contact with hydrogen gas, dm of strong acid solution, H through which
−3
+
H and the solution containing hydrogen ions, hydrogen gas, H at a pressure of 1 atm is
2
H , the electrode potential for hydrogen half- bubbled. Thus, the reference half reaction is as
2
+
cell is produced. follows:
9. It is impossible to measure the absolute value of
–
+
an electrode potential in a half-cell. Therefore, 2H (aq; 1M) + 2e  H (g, 1 atm) E 0 reference = 0.00 V
2
scientists compare the value of electrode
312 1.2.1

Form
5
Chapter 1 Redox Equilibrium Chemistry
+
CHAP. (b) This means hydrogen ions, H undergo CHAP.
1 reduction to form hydrogen gas, H . 1
2
The standard reference half-cell is the standard Therefore hydrogen electrode acts as the
hydrogen electrode, SHE. cathode.
2H O (aq) + 2e → H (g) + 2H O(l)
–
+
3
2
2
2. An unknown standard electrode potential, or simplified as
E 0 unknown can be obtained by constructing a 2H (aq) + 2e → H (g)
–
+
2
voltaic cell consisting of a reference half-cell (c) On the other hand, zinc atoms, Zn are
and another unknown half-cell, to measure the oxidised to zinc ions, Zn . Hence, zinc
2+
electromotive force, e.m.f or commonly known electrode acts as the anode.
©PAN ASIA PUBLICATIONS
as standard cell potential, E 0 cell . Oxidation half reaction:
(a) If the hydrogen ion, H is reduced, the Zn(s) → Zn (aq) + 2e –
+
2+
reference half-cell acts as the cathode while (d) E 0 cell = E 0 cathode – E 0 anode
oxidation occurs at the unknown half-cell. 0.76 V = 0.00 V – E 0

zinc
E 0 = E 0 – E 0 E 0 = 0.00 V – 0.76 V
anode
cathode
cell = 0.00 V – E 0 unknown zinc = – 0.76 V
= – E 0 Zinc standard electrode potential:
unknown
Standard electrode potential of the half-cell: Zn (aq) + 2e  Zn(s) E 0 zinc = – 0.76 V
–
2+
E 0 unknown = –E 0 cell
(b) If the hydrogen gas, H is oxidised, the
2
reference half-cell acts as the anode while
reduction occurs at the unknown half-cell. Conventionally, standard electrode potential is
written in the form of standard reduction potential.
E 0 cell = E 0 cathode – E 0 anode
= E 0 unknown – 0.00 V
= E 0 unknown Oxidising Agents and Reducing Agents Based on
Standard electrode potential of the half-cell: The Value of Standard Electrode Potentials
0
E 0 = E 0 1. The value of standard electrode potential, E
unknown cell
Example: gives a direct measure of the ease of a chemical
Figure 1.13 shows a voltaic cell consisting of a species to be oxidised or reduced.
standard zinc half-cell and a standard hydrogen
half-cell which is prepared to determine the
standard electrode potential of a zinc. Chemical species can be atom, molecule,
Voltmeter monoatomic ion, polyatomic ion, or radical.
e – 0.76 V e –
2. We can compare the strength in oxidation and
Anode (–) Cathode (+) reduction of elements or ions based on the value
Zn H (g) of E for the half reactions involved.
0
2
(a) The more positive the E value is, the easier
0
H for the chemical species on the left side of the
2
e – Zn 2+ Pt 2H O half equation to undergo reduction.
2
0
e – 2H O + e – (b) The more negative the E value is, the easier
Zn 1 M Zn 2+ 1 M H O + 3
for the chemical species on the right side of
3
Salt bridge the half equation to undergo oxidation.
Figure 1.13 Example:
(a) There are a few observations that can be seen Figure 1.14 shows the standard electrode
in the voltaic cell: potentials of three ions.
(i) Gas bubbles are produced around the
platinum, Pt electrode. Zn (aq) + 2e ⇌ Zn (s) E = –0.76 V
0
–
2+
0
–
+
(ii) Zinc, Zn electrode becomes thinner. Strength as oxidising agent 2H (aq) + 2e ⇌ H (g) E = 0.00 V Strength as reducing agent
2
2+
–
0
(iii) Voltmeter reading shows 0.76 V. Cu (aq) + 2e ⇌ Cu (s) E = 0.34 V
Figure 1.14
1.2.1 1.2.2 313

Form
5 Chemistry Chapter 1 Redox Equilibrium

0
0
CHAP. (a) Referring to the E value, copper(II) ions, 4. In short, by considering the E value, we can CHAP.
1 Cu are most easily reduced compared to predict the followings: 1
2+
hydrogen ions, H . Zinc ions, Zn is the (a) Strength of the oxidising agent and the
2+
+
most difficult to be reduced. reducing agent.
(b) Thus, the strength of oxidising agents is in (b) Chemical species (atom, molecule or ion)
the order of Cu > H > Zn . that undergoes oxidation or reduction.
2+
2+
+
(c) Negative E value indicates that zinc atom,
0
Zn is more easily oxidised, followed by
hydrogen gas, H , and copper atom, Cu
2
being most difficult to be oxidised. Table of Standard Electrode
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(d) Thus, the strength of reducing agents are in Potential, E 0
the order of Zn > H > Cu. http://bit.ly/2NYV9lu
2
Example 3
The following are the E values of zinc and copper.
0
0
–
Zn (aq) + 2e  Zn(s) E = – 0.76 V
2+
0
2+
–
Cu (aq) + 2e  Cu(s) E = + 0.34 V
Based on the E values:
0
(a) Determine the substance that is oxidised and substance that is reduced.
(b) Identify the oxidising agent and reducing agent.
(c) Write the oxidation half equation and reduction half equation.
(d) Write the ionic equation for the redox reaction.
0
Solution The more negative the E value is, the easier for the chemical
species on the right of the half equation to undergo oxidation.
(a) Substance oxidised: Zinc, Zn Zn (aq) + 2e ⇌ Zn(s) E = – 0.76 V
–
0
2+
2+
Substance reduced: Copper(II) ion, Cu
(b) Oxidising agent: Copper(II) ion, Cu 2+ The more positive the E value is, the easier for the chemical
0
Reducing agent: Zinc atom, Zn species on the left of the half equation to undergo reduction.
–
0
2+
(c) Oxidation half equation: Zn(s) → Zn (aq) + 2e – Cu (aq) + 2e ⇌ Cu(s) E = + 0.34 V
2+
–
2+
Reduction half equation: Cu (aq) + 2e → Cu(s)
2+
2+
(d) Ionic equation: Zn(s) + Cu (aq) → Zn (aq) + Cu(s)
Example 4
A displacement reaction using copper, Cu and silver nitrate, AgNO solution
3
is carried out in the laboratory. Scan the QR code to observe the reaction Displacement
0
occurred. The following are the E values of silver and copper. reaction
–
0
Cu (aq) + 2e  Cu(s) E = + 0.34 V
2+
0
+
Ag (aq) + e  Ag(s) E = + 0.80 V https://bit.ly/3gyad3z
–
(a) State two observations that can be obtained in the displacement reaction.
(b) Based on the E values:
0
(i) Determine the oxidising agent and reducing agent.
(ii) Write the oxidation half equation and reduction half equation
(iii) Write the ionic equation for the redox reaction.
Solution
E value of silver is more positive. So,
0
(a) Silvery grey solid is deposited . Colourless solution turns blue silver ion, Ag is reduced.
+
+
(b) (i) Silver ion , Ag is reduced. So, it is an oxidising agent.
0
Copper atom, Cu is oxidised. So, it is a reducing agent. E value of copper is less positive.
2+
(ii) Oxidation half equation: Cu(s) → Cu (aq) + 2e – Hence, copper atom is oxidised.
–
+
Reduction half equation: Ag (aq) + e → Ag(s)
+
(iii) Ionic equation: Cu(s) + 2Ag (aq) → Cu (aq) + 2Ag(s)
2+
314 1.2.2

Form
5
Chapter 1 Redox Equilibrium Chemistry
CHAP. CHAP.
1 1.2 1


1. State the meaning of standard electrode potential, E . C1
0
2. What are the standard conditions for the measurement of standard electrode potential of a half-cell?
C1
0
3. Referring to the standard electrode potential, E given below, answer the following questions. C4
+
–
0
Ag (aq) + e  Ag(s) E = +0.80 V
–
Sn (aq) + 2e  Sn(s) E = –0.14 V
0
2+
©PAN ASIA PUBLICATIONS
(a) Determine the oxidising agent and substance to be oxidised.
(b) Write the oxidation half equation and reduction half equation.
(c) State two observations that can be obtained in the redox reaction.
(d) Construct an overall ionic equation to represent the reaction.
1.3 Voltaic Cell A Magnesium atom, Mg is oxidised to
2+
The Redox Reactions in Voltaic Cell magnesium ion, Mg by losing two
electrons. Thus, magnesium atom, Mg
1. A voltaic cell is an electrochemical cell that atom acts as reducing agent.
converts chemical energy to electrical energy. Oxidation half reaction:
2. A simple voltaic cell consists of two different Mg(p) Mg (ak) + 2e –
2+
metals that are connected to a bulb, voltmeter 2+
or galvanometer using connecting wires and Magnesium ion, Mg produced then
placed in an electrolyte. enters sulphuric acid, H SO . Therefore,
4
2
3. Potential difference between these two metals magnesium, Mg electrode becomes thinner.
causes the movement of electrons through B Electrons released from the magnesium,
external circuit (wire) and therefore, producing Mg electrode then moves through the wire
electric current. to copper, Cu electrode.
4. Figure 1.15 shows a simple voltaic cell that C Each hydrogen ion, H is reduced to a
+
uses magnesium, Mg and copper, Cu plates hydrogen atom, H by gaining one electron
as electrodes and sulphuric acid, H SO as on the surface of copper, Cu electrode.
2
4
+
electrolyte. Thus, hydrogen ion, H acts as oxidising
agent.
Current flow H + e → H
+
–
Wire
Magnesium, Mg B Copper, Cu
electrode electrode Two hydrogen atom, H then combine
e – e – together to form one molecule of hydrogen
Bulb
gas, H .
2
H + H → H
A C 2
Therefore, bubbles of hydrogen gas, H
e – 2
e – H + + e –
H 2 Hydrogen gas, are released at the surface of copper, Cu
H + + e –
Mg Mg 2+ H bubbles electrode.
2
Reduction half reaction:
+
–
Sulphuric acid, H SO 2H (ak) + 2e → H (g)
2 4 2
Figure 1.15 A simple voltaic cell
–
Mg(p) → Mg + 2e – 2H + 2e → H
+
2+
Mg (aq) + 2e ⇌ Mg(s) E = –2.38 V 2
2+
–
0
Cu (aq) + 2e ⇌ Cu(s) E = +0.34 V Electrons added to the Electrons added to the
–
2+
0
E 0 magnesium value is more negative. Therefore, right side of the equation left side of the equation
magnesium, Mg is a stronger reducing agent means loss of electrons. means gain of electrons.
compared to copper, Cu.
1.3.1 315

Form
5 Chemistry Chapter 1 Redox Equilibrium

CHAP. 5. Redox reaction occurred in the simple chemical 7. The relative charge on the electrode depends CHAP.
1 cell can be represented by the following overall on the source of electrons and the direction of 1
ionic equation: flow of electrons.
(a) Since the electrons released by the
Oxidation half equation: Mg(s) → Mg (aq) + 2e – magnesium atom, Mg enters the
2+
Reduction half equation: 2H (aq) + 2e → H (g) magnesium, Mg electrode while the
–
+
2 electrons that reach the copper, Cu
Overall ionic equation: Mg(s) + 2H (aq) → Mg (aq) + H (g) electrode are given to the hydrogen ion,
+
2+
2
H , magnesium, Mg electrode relatively
+
6. The area where oxidation occurs is called anode contains more negative charge (electrons)
©PAN ASIA PUBLICATIONS
and the area where reduction occurs is called compare to copper, Cu electrode.
cathode. Thus, the magnesium, Mg electrode (b) Thus, magnesium, Mg electrode is the
acts as an anode while the copper, Cu electrode negative terminal of the cell and the
acts as a cathode in this simple chemical cell. copper, Cu electrode is the positive
terminal of the cell.
U seful Acronym (c) The potential difference between
magnesium, Mg and copper, Cu causes the
transfer of electrons through the external
circuit and an electric current is produced.
(d) Hydrogen gas, H on the copper, Cu surface
2
results in polarisation and prevents
further reduction of hydrogen ions,
+
Red cat An ox H . Hence, simple chemical cell can only
Reduction; cathode Oxidation; anode function for a short period of time.
8. This problem was solved by John Frederic
Daniell, a British chemist in 1836 who invented
Daniell cell.


1.1

Aim: Procedure:
To show the production of electricity through redox 1. A copper, Cu plate and iron, Fe nail are cleaned
reaction in a simple voltaic cell. with sandpaper.
3
-3
Problem statement: 2. 150 cm of 1.0 mol dm sulphuric acid, H SO is
2
4
How can a simple voltaic cell produce electricity poured into a beaker.
through redox reaction?
3. Copper, Cu plate and iron nail are connected
to a voltmeter using connecting wires and
Hypothesis: dipped into the sulphuric acid, H SO as shown
When two different metals are dipped into an in Figure 1.16. 2 4
electrolyte and connected by wires, electricity is
produced.
Galvanometer
G
Variables:
(a) Manipulated: Pair of metal
(b) Responding: Generation of electricity
(c) Fixed: Type of electrolyte used Iron, Fe Copper,
nail Cu plate
Materials:
Zinc, Zn plate, copper, Cu plate, iron, Fe nail, 1.0
-3
mol dm sulphuric acid, H SO , sandpaper Sulphuric acid,
2 4 H SO 4
2
Apparatus: Figure 1.16
3
Connecting wires with crocodile clips, 250 cm
beaker, voltmeter
316 1.3.1

Form
5
Chapter 1 Redox Equilibrium Chemistry
CHAP. CHAP.
+
1 4. The voltmeter reading is recorded. All (e) Hydrogen ions, H are reduced to form 1
observations at the electrodes are recorded.
hydrogen gas, H by gaining electrons.
2
5. Steps 1 to 4 are repeated using zinc, Zn plate Reduction occurs at the copper, Cu plate.
+
and copper, Cu plate. Hydrogen ion, H acts as oxidising agent.
(f) Electrons move from iron, Fe nail to copper,
Results: Cu plate through wire. In other words,
Table 1.10 electrons flow from reducing agent to the
oxidising agent. Therefore, iron, Fe nail
Voltmeter
Pair of Observations at the is an anode while copper, Cu plate is the
reading cathode of the voltaic cell.
metals electrodes
©PAN ASIA PUBLICATIONS
(V) (g) Electrons that flow through the wire from
Iron, Fe 0.8 Iron, Fe nail becomes iron, Fe to copper, Cu produce electric
nail + thinner. current.
2. Zinc, Zn plate + copper, Cu plate:
copper, Gas bubbles are released (a) Oxidation half equation:
Cu plate around copper, Cu plate. 2+ –
Zn(s) → Zn (aq) + 2e
Zinc, Zn 1.1 Zinc, Zn plate becomes (b) Reduction half equation:
–
+
plate + thinner. 2H (aq) + 2e → H (g)
2
copper, Gas bubbles are released (c) Overall ionic equation:
Cu plate around copper, Cu plate. Zn(s) + 2H (aq) → Zn (aq) + H (g)
+
2+
2
(d) Zinc atom, Zn is oxidised to zinc ion, Zn
2+
Inference: by losing two electrons. Oxidation occurs
1. Iron, Fe nail and zinc, Zn plate dissolves and at the zinc, Zn plate and zinc, Zn acts as
corrodes to form iron(II) ion, Fe and zinc ion, reducing agent.
2+
2+
Zn respectively. (e) Hydrogen ions, H are reduced to form
+
2. Hydrogen gas, H is produced. hydrogen gas, H by gaining electrons.
2 2
Reduction occurs at the copper, Cu plate.
+
Conclusion: Hydrogen ion, H acts as oxidising agent.
Hypothesis is accepted. Two different metals that (f) Electrons move from zinc, Zn plate to
are dipped into an electrolyte and connected by copper, Cu plate through wire. In other
wire will produce electricity. words, electrons flow from reducing agent
to oxidising agent. Therefore, zinc, Zn plate
Discussion: is the anode while copper, Cu plate is the
1. Iron, Fe nail + copper, Cu plate: cathode of the voltaic cell.
(a) Oxidation half equation: (g) Electrons that flow through the wire from
Fe(s) → Fe (aq) + 2e – zinc, Zn to copper, Cu produce electric
2+
(b) Reduction half equation: current.
–
+
2H (aq) + 2e → H (g) 3. Reaction in the simple chemical cell is a redox
2
(c) Overall ionic equation: reaction because oxidation and reduction
2+
+
Fe(s) + 2H (aq) → Fe (aq) + H (g) occur at the same time.
2 4. If two copper, Cu electrodes are used, there
(d) Iron atom, Fe is oxidised to iron(II) ion, Fe is no potential difference between electrodes.
2+
by losing two electrons. Oxidation occurs When there is no electron flow, electric current
at the iron, Fe nail and iron, Fe acts as is not generated and voltmeter reading will
reducing agent.
show 0 V.







1.3.1 317

Form
5 Chemistry Chapter 1 Redox Equilibrium

CHAP. Daniell Cell CHAP.
1 1. Figure 1.17 shows a Daniell cell. A Daniell cell of the cell while copper, Cu is the positive 1

or a zinc-copper cell is an example of a voltaic terminal of the cell.
cell. 5. Redox reaction that takes place in Daniell cell
can be represented by an overall ionic equation:
A Voltmeter
e – e – 2+ –
2e are received for Negative terminal: Zn(s) → Zn (aq) + 2e
–
Anode Cathode
2+
–
–
2e are released for (–) (+) every copper ion, Positive terminal: Cu (aq) + 2e → Cu(s)
2+
every zinc atom, Zn – Cu that is Overall ionic
2+
2+
Zn SO 4 Na + Cu Zn(s) + Cu (aq) → Zn (aq) + Cu(s)
©PAN ASIA PUBLICATIONS
that is oxidised reduced equation:
6. From the ionic equation, we can write cell
e – B C Cu 2+ notation for Daniell cell as the following:
Zn 2+
2+
2+
e – e – Zn(s) | Zn (aq) || Cu (aq) | Cu(s)
Zn Zn 2+ Cu 2+ Cu
7. Standard cell potential, E for Daniell cell can be
0
Salt bridge, Na SO
2 4 calculated using the following formula:
Figure1.17 Daniel cell
E 0 cell = E 0 cathode – E 0 anode
Zn (aq) + 2e  Zn(s) E = –0.76 V With zinc, Zn as the anode and copper, Cu as
–
0
2+
Cu (aq) + 2e  Cu(s) E = +0.34 V the cathode.
0
–
2+
2. E 0 zinc is more negative. It indicates that zinc, Zn is Anode: Zn (aq) + 2e  Zn(s) E = –0.76 V
0
–

2+
a stronger reducing agent. Thus, zinc, Zn plate Cathode: Cu (aq) + 2e  Cu(s) E = +0.34 V
0
–
2+
is an anode where oxidation process occurs. E 0 = E 0 – E 0
cathode
anode
3. E 0 copper is more positive. It indicates that the cell = (+0.34 V) – (–0.76 V)
copper(II) ion, Cu is an oxidising agent. = 1.10 V
2+
Copper plate is the cathode where reduction
process occurs.
A Zinc, Zn plate becomes thinner as zinc, 0
which is more negative indicates that zinc, Zn
Zn corrodes and dissolves in zinc sulphate, E is more easily oxidised and acts as an anode.
zinc
ZnSO solution.
4
Zinc atom, Zn is oxidised to zinc ion, Zn 8. The E 0 obtained through the calculation is
2+
by losing two electrons. actually the voltage produced in the Daniell cell
cell
Zn(s) → Zn (aq) + 2e –
2+
B Potential difference between the two metal based on the potential difference between two
plates causes the flow of electrons from electrodes.
the anode (zinc) to the cathode (copper) 9. The functions of salt bridge:
through wire. Therefore, electric current is (a) Complete the circuit by allowing the
generated. movement of ions.
C Brown solid is deposited at the copper, Cu (b) Separate two different electrolytes
plate, making copper, Cu plate becomes 10. Initially, oxidation of half-cell is neutral with
2+
2−
thicker. Copper(II) ion, Cu is reduced zinc ions, Zn and sulphate ions, SO in
2+
4
to the copper atom, Cu by gaining two the solution. When more and more zinc ions,
electrons. Zn enter the solution, the solution becomes
2+
Cu (aq) + 2e → Cu(s) positively charged.
2+
‒
The intensity of blue solution decreases 11. Similarly, reduction of half-cell is neutral with
as the concentration of copper(II) ion, Cu copper(II) ions, Cu and sulphate ions, SO
2+
2−
2+
4
decreases. in the solution. The solution will be negatively
4. Oxidation process occurred at the anode and charged when more and more copper(II) ions,
reduction process at the cathode causes zinc Cu leave the solution to form copper atom, Cu.
2+
(anode) becomes relatively negative charge 12. When two half-cells are charged, the voltaic
(electrons) as compared to copper (cathode). cell will not function. Therefore, a salt bridge is
Therefore, zinc, Zn is the negative terminal needed to connect the two half-cells.
318 1.3.1

Form
5
Chapter 1 Redox Equilibrium Chemistry
+
CHAP. 13. Based on Figure 1.17, the sodium ions, Na from move to zinc sulphate, ZnSO solution which CHAP.
4
1 the salt bridge move to copper(II) sulphate, has more positive charge. 1
CuSO solution which is lack of positive charge 14. Therefore, the solutions in both half-cells are
4
whereas sulphate ions, SO from the salt bridge maintained in a neutral state and the voltaic cell
2−
4
continues to function well.


1.2
©PAN ASIA PUBLICATIONS
Aim: 2. A porous pot is filled with zinc sulphate, ZnSO
4
To determine the voltage produced from two solution until it is two-third full.
different pairs of metals in a chemical cell. 3. A zinc, Zn plate is dipped into zinc sulphate,
ZnSO solution.
4
Problem statement: 4. A beaker is filled with copper(II) sulphate,
How do different pairs of metals in chemical cells CuSO solution until it is half full.
4
affect the voltage produced? 5. The copper, Cu plate is dipped into copper(II)
sulphate, CuSO solution.
4
Hypothesis: 6. The porous pot is placed into the beaker.
Pair of metals with greater difference of standard 7. The circuit is completed by connecting the
electrode potential value will produce greater metals to a voltmeter as shown in Figure 1.18.
voltage.
Voltmeter
Variables: V
(a) Manipulated: Different pairs of metals
(b) Responding: Voltage reading of the cell
(c) Fixed: Copper, Cu electrode, volume and Copper, Zinc,
Cu plate Zn plate
concentration of electrolytes
Copper(II) Porous pot
Materials:
sulphate, that contains
Copper, Cu plate, iron, Fe nail, zinc, Zn plate, CuSO solution zinc sulphate,
magnesium, Mg ribbon, sandpaper, 1.0 mol dm 4 ZnSO solution
-3
-3
iron(II) sulphate, Fe SO solution, 1.0 mol dm 4
2 4 Figure 1.18
copper(II) sulphate, CuSO solution, 1.0 mol
4
-3
dm zinc sulphate, ZnSO solution, 1.0 mol dm 8. The reading of the voltmeter is recorded.
-3
4
magnesium sulphate, MgSO solution 9. Steps 2 to 8 are repeated by replacing zinc, Zn
4
plate and zinc sulphate, ZnSO solution in the
4
Apparatus: porous pot with;
Voltmeter, 250 cm beaker, connecting wires with (a) iron nail and iron(II) sulphate, FeSO
3
4
crocodile clips, porous pot solution.
(b) magnesium, Mg ribbon and magnesium
Procedure: sulphate, MgSO solution.
4
1. All metal plates are cleaned with sandpaper.
Result:
Table 1.11
Set Pair of metals Voltage (V) Observations at the electrodes
I Zinc, Zn plate + 1.1 Zinc, Zn plate becomes thinner.
copper, Cu plate Brown solid is deposited on the copper, Cu plate.
II Iron, Fe nail + 0.7 Iron, Fe nail becomes thinner.
copper, Cu plate Brown solid is deposited on the copper, Cu plate.
III Magnesium, Mg ribbon + 2.7 Magnesium, Mg ribbon becomes thinner.
copper, Cu plate Brown solid is deposited on the copper, Cu plate.



1.3.1 319

Form
5
Chapter 1 Redox Equilibrium Chemistry
CHAP. iron, Fe and are accepted by positive ions CHAP.
1 such as hydrogen ions, H or oxygen, O
2 1
+
Magnesium, Mg and iron, Fe form oxide coating and water, H O in the electrolyte.
2
which is porous , not tightly packed and weak.
This oxide coating is easily peeled off and cannot
protect the metals from corrosion.
Cars rust more rapidly in a country where salt is
used to melt the snow and ice on the road.
Electrochemical corrosion Dissolved salts in the water droplet greatly
1. Electrochemical corrosion is the process increase the conductivity of the electrolyte. Thus
Materials: ©PAN ASIA PUBLICATIONS
by which a metal is corroded through loss of increasing the rate of corrosion.
electron to form a cation with the presence of
electrolyte when the metal is in contact with 2. The rate of corrosion of magnesium, Mg
other less electropositive metal. increases when magnesium, Mg is contacted
Example: with copper, Cu. This is because magnesium,
Mg and copper, Cu have a great difference in
electropositivity compared to magnesium, Mg
(–) (+) Electrolyte and iron, Fe.
Mg Fe

e –
Mg
e – e –
Fe e – The rate of
Electrochemical
transferring of
Figure 1.38 series electrons from Mg to
Cu is higher.
(a) Figure 1.38 shows that magnesium, Mg is in
contact with iron, Fe and the two metals are Cu
put into an electrolyte.
(b) Magnesium, Mg is more electropositive Figure 1.39 The difference in electropositivity
than iron, Fe. Thus, magnesium, Mg is between magnesium, ferum and copper
corroded (loss of electrons) and iron, Fe is
prevented from corrosion.

Mg(s) → Mg (aq) + 2e –
2+
The greater the difference in electropositivity of
(c) Magnesium, Mg acts as a negative terminal, two metals, the higher the rate of corrosion of the
more electropositive metal.
whereas iron, Fe acts as a positive terminal.
Electrons flow from magnesium, Mg to
Try Question 1 in Formative Zone 1.6





Activity 1.7


Aim: A Burning of copper, Cu strip
To study the corrosion of copper and iron.
Procedure:
1. Use a pair of tongs to hold a copper, Cu strip.
Copper, Cu strip, iron, Fe wool, oxygen gas, O
2 2. Heat the copper, Cu strip using a Bunsen
burner for a few minutes as shown in Figure
Apparatus: 1.40.
Bunsen burner, combustion spoon, gas jar with
lids, a pair of tongs 3. All observations are recorded.



1.6.1 343

Form
5 Chemistry Chapter 1 Redox Equilibrium

CHAP. Results: CHAP.
1 Copper, Cu Table 1.27 1
strip
Tongs A Black layer formed on the brown copper strip.
Bunsen burner
B The grey iron wool turns into a reddish-brown
solid.

Figure 1.40 Inference:
1. Copper(II) oxide, CuO is formed.
B Burning of iron, Fe wool 2. Iron(II) oxide, Fe O is formed.
©PAN ASIA PUBLICATIONS
2 3
Procedure:
1. A small amount of iron, Fe wool is placed in a Discussion:
combustion spoon. 1. Copper, Cu is oxidised to copper(II) oxide,
2. The iron, Fe wool is heated until it starts to CuO which is black in colour.
burn. 2Cu(s) + O (g) → 2CuO(s)
3. The burning iron, Fe wool is quickly lowered 2
2+
into a gas jar filled with oxygen gas, O as Cu → Cu + 2e –
2
shown in Figure 1.41. Copper atom, Cu releases two electrons
4. All observations are recorded. to form copper(II) ion, Cu . Copper, Cu
2+
undergoes corrosion.
Combustion
spoon
2. Iron, Fe is oxidised to iron(III) oxide, FeO
Gas jar
which is reddish-brown in colour.
Oxygen gas, O
4Fe(s) + 3O (g) → 2Fe O (s)
2
Iron, Fe wool 2 2 3
Fe → Fe + 3e –
3+
Iron atom, Fe releases three electrons to form
Figure 1.41 iron(III) ion, Fe . Iron, Fe undergoes corrosion.
3+


• Corrosion of copper, Cu and iron, Fe
Corrosion of copper, Cu and iron, Fe will occur at moist open atmosphere. Iron, Fe is more electropositive than
copper, Cu. Thus, iron, Fe corrodes easily compared to copper, Cu.
• Rusting of iron
Rusting is the common term for corrosion of iron, Fe. Rusting is the reaction of iron, Fe
and oxygen, O in the presence of water, H O or moisture. When iron, Fe corrodes, it
2
2
forms a red-brown hydrated metal oxide, (Fe O .xH O), commonly known as rust.
2 3 2
• Corrosion Corrosion of
Corrosion of copper, Cu occur to the materials made of copper or copper alloys. When metal chains
exposed to the atmosphere, copper oxidises, causing the bright copper surfaces to
tarnish. After a few years, this tarnish gradually changes to dark brown or black, and
finally to blue green. This oxide layer, is called patina, CuCO .Cu(OH) firmly adheres to
2
3
the outer surface of the copper and protects the underlying copper layers from further
corrosion. This is the reason why copper, Cu is used on roofs, gutter work and outdoor
sculptures. Corrosion of copper, Cu occurs at negligible rates in polluted air, water
and deaerated non-oxidising acids. However, it is susceptible to more rapid attack in
oxidising acids, oxidising heavy metal salts, sulphur, ammonia and some sulphur and Formation of
ammonia compounds. patina on outdoor
sculpture





344 1.6.1

Form
5
Chapter 1 Redox Equilibrium Chemistry
CHAP. Iron Rusting as a Redox Reaction (e) Redox reaction equation: CHAP.
1 1. Rusting is a metal corrosion that occurs on iron Anode: 2Fe(s) → 2Fe (aq) + 4e – 1
2+
due to redox reaction. Rusting of iron requires Cathode: O (g) + 2H O(l) + 4e → 4OH (aq)
–
–
2
2
the presence oxygen (air) and water. 2Fe(s) + O (g) + 2H O(l) → 2Fe(OH) (s)
2. Figure 1.42 shows the mechanism of rusting 2 2 2
of iron. The surface of the iron and a water
droplet constitute a simple chemical cell in Combination of 2Fe + 4OH –
2+
which different regions of the surface of the iron
act as anode (negative terminal) and cathode (f) Ion(II) hydroxide, Fe(OH) formed is
2
©PAN ASIA PUBLICATIONS
(positive terminal) while the water droplet act further oxidised by oxygen to form
as the electrolyte. hydrated iron(III) oxide, Fe O .xH O.
3
2
2
(g) Hydrated iron (III) oxide, Fe O .xH O is
3
2
2
Water droplet Rust (Fe ) .xH O) a brown solid substance known as rust,
2 3 2
whereby the value of x varies.
Oxidation
OH – Fe(OH) (s) Fe O . xH O(s)
O O 2 2 3 2
2 2
Fe 2+
e –
Cathode (+) Anode (–) Cathode (+) • Iron(II) hydroxide, Fe(OH) is first oxidised to
2
iron(III) hydroxide, Fe(OH) .
Iron metal 3
• Iron(III) hydroxide, Fe(OH) is then decomposes to
3
Figure 1.42 Mechanism of rusting of iron hydrated iron(III) oxide, Fe O xH O.
2
3
2
3. In the mechanism of rusting of iron:
(a) Iron surface in the centre of a water droplet
acts as the anode (negative terminal). Iron
surface at the edge of the water droplet Other explanation for the rusting of iron:
2+
acts as the cathode (positive terminal). Anode: Fe(s) → Fe (aq) + 2e – –
+
The concentration of oxygen gas, O that Cathode: O (g) + 4H (aq) + 4e → 2H O(l)
2
2
2
dissolves at the edge of the water droplet is The reaction of atmospheric CO with water forms H +
higher than at the centre. and HCO 3− 2
(b) At the anode, iron atoms lose electrons and The Fe ions produced in the initial reaction are
2+
undergoes oxidation to form iron(II) ions, then oxidised by atmospheric oxygen to produce
2+
Fe . the insoluble hydrated oxide containing Fe , as
3+
represented in the following equation:
Oxidation half equation: 4Fe (aq) + O (g) + (2+4x)H O(l) →
2+
Fe(s) → Fe (aq) + 2e – 2 2 2Fe O .xH O + 4H (aq)
2+
+
2
3
2
Ferum(II) ions, Fe dissolve in water.
2+
(c) Electrons flow through the iron metal to the
edge of the water droplet (cathode) and are
received by oxygen and water molecules to Preventing Rusting
form hydroxide ions, OH . 1. When iron is in contact with a more
–
electropositive metal, rusting of iron is
Reduction half equation: prevented. The more electropositive metal will
O (g) + 2H O(l) + 4e → 4OH (aq)
–
−
2 2 corrode.
(d) The iron(II) ions, Fe combine with Example:
2+
hydroxide ions, OH to form iron(II) If iron, Fe nail is wrapped with a magnesium,
−
hydroxide, Fe(OH) . Mg ribbon and put into a test tube filled with
2 water, magnesium, Mg will corrode and iron, Fe
Fe (aq) + 2OH (aq) → Fe(OH) (s) nail will be prevented from rusting.
2+
−
2
1.6.1 1.6.2 345

Form
5 Chemistry Chapter 1 Redox Equilibrium

CHAP. Electrochemical series strip and put into a test tube filled with water, CHAP.
1 iron, Fe will corrode while copper, Cu will be 1
Iron, Fe nail prevented from rusting.
Mg
Electrochemical series
e –
Magnesium,
Mg ribbon Fe Iron,
Figure 1.43 Fe nail Fe
(a) Magnesium, Mg is more electropositive Copper, e –
than iron, Fe. Cu strip Cu
(b) Magnesium, Mg is corroded to form
©PAN ASIA PUBLICATIONS
magnesium ions, Mg . Figure 1.44
2+
(a) Iron, Fe is more electropositive than
Mg(s) → Mg (aq) + 2e – copper, Cu. Iron, Fe is corroded to form
2+
2+
(c) Iron(II) ions, Fe are not present. Thus, iron(II) ions, Fe .
2+
iron, Fe is prevented from rusting. Fe(s) → Fe (aq) + 2e –
2+
2. When iron is in contact with a less
electropositive metal, rusting of iron is speeded (b) The presence of copper, Cu increases the
up. rate of the formation of iron(II) ions, Fe .
2+
Example: As a result, the rusting of iron, Fe is speeded
If iron, Fe nail is wrapped with a copper, Cu
up.


1.8

Aim: are cleaned with sandpaper.
To investigate the effects of other metals on the 2. One clean iron, Fe nail is placed in test tube A.
rusting of iron. 3. The other four iron nails are coiled with
magnesium, Mg ribbon, copper, Cu strip,
Problem statement: zinc, Zn strip and tin, Sn strip respectively and
How do different types of metals in contact with
placed in test tube B, C, D and E as shown in
iron affect the rusting of iron?
Figure 1.45.
Hypothesis: A B C
When a more electropositive metal comes in
contact with an iron, the metal will prevent the
rusting of iron. When a less electropositive metal Iron nail Iron nail + Iron nail +
comes in contact with an iron, the metal will speed magnesium zinc
up the rusting of iron.
Variables:
(a) Manipulated: Different types of metals
(b) Responding: The presence of blue colouration/ Agar solution + potassium hexacyanoferrate(III)
Rusting of iron + phenolphthalein
(c) Fixed: Iron nails, hot agar solution D E
Materials:
Iron, Fe nails, magnesium, Mg ribbon, copper, Cu
strip, zinc, Zn strip, tin, Sn strip, hot agar solution, Iron nail + Iron nail +
copper
tin
potassium hexacyanoferrate(III), K Fe(CN) solution,
3 6
phenolphthalein indicator, sandpaper
Apparatus:
Test tube, test tube rack
Agar solution + potassium hexacyanoferrate(III)
Procedure: + phenolphthalein
1. Five iron, Fe nails, magnesium, Mg ribbon, Figure 1.45
copper, Cu strip, zinc, Zn strip and tin, Sn strip
346 1.6.2

Form
5
Chapter 1 Redox Equilibrium Chemistry
CHAP. CHAP.
1 4. A hot agar mixture which contains of potassium hexacyanoferrate(III) solution and phenolphthalein are 1
poured into each test tube to completely immerse all the iron nails.
5. The test tubes are placed in a test tube rack and leave aside for two days.
6. Changes of the colour of the solution in each test tube are observed and recorded.
Observation:
A B C D E



©PAN ASIA PUBLICATIONS
Iron nail + Iron nail + Iron nail + Iron nail +
Iron nail
only magnesium zinc copper tin







Agar solution + potassium hexacyanoferrate(III) solution + phenolphthalein
Table 1.28

Test Pair of Intensity of the Intensity of the
tube metals blue colour pink colour Inference
A Fe only Low None Iron(II) ions, Fe presence. Iron nail undergoes
2+
rusting.
B Fe + Mg None Very high Iron(II) ions, Fe are not presence. No rusting occurs.
2+
–
Very high concentration of hydroxide ions, OH .
C Fe + Zn None High Iron(II) ions, Fe are not presence. No rusting occurs.
2+
–
Hydroxide ions, OH are present.
D Fe + Cu Very high None Very high concentration of iron(II) ions, Fe . Iron nail
2+
rust at the highest rate.
E Fe + Sn High None Very high concentration of iron(II) ions, Fe . Iron nail
2+
undergoes rusting faster than iron nail in test tube A.
Conclusion: blue and pink colour in the solution because it
The hypothesis is accepted. The more is transparent and it slows down the diffusion
electropositive metals prevent the rusting of iron. process. Agar solution can also be used to trap
The less electropositive metals increase the rate of if gas bubbles produced in the reaction.
rusting of iron. 4. Test tube A:
(a) Test tube A is used as a control to compare
Discussion: the effect of other metals in contact on the
1. Potassium hexacyanoferrate(III), K Fe(CN)
3 6 rusting of iron.
solution is added to detect the presence (b) In the presence of water and oxygen, iron
2+
of iron(II) ions, Fe . When iron(II) ions, Fe nail rusts a little. The iron is oxidised to
2+
presence, a dark blue colour formed. The 2+
2+
more iron(II) ions, Fe formed, the higher the iron(II) ions, Fe . 2+ –
Fe(s) → Fe (aq) + 2e
intensity of the dark blue colour formed. The presence of iron(II) ions, Fe give
2+
2. Phenolphthalein is added to detect the a dark blue colour with potassium
presence of hydroxide ions, OH . The presence hexacyanoferrate(III), K Fe(CN) solution.
–
–
of hydroxide ions, OH increases the alkalinity 3 6
(c) The oxygen in the solution together with
of the solution and gives pink colour to the the water is reduced to hydroxide ions,
solution. OH .
–
3. Agar solution is used to enable us to see the
–
O (g) + 2H O(l) + 4e → 4OH (aq)
–
2 2
1.6.2 347

Form
5 Chemistry Chapter 1 Redox Equilibrium

CHAP. (v) Write the ionic equation to represent the reaction that take place in test tube P. C3 [1 mark] CHAP.
1 (b) 1,1,1-trichloromethane is added to test tubes P and Q. The mixture in both test tubes are shaken gently. 1

(i) Predict the colour of 1,1,1-trichloromethane in both test tubes. C4 [1 mark]
(ii) Explain the formation of the colour of 1,1,1-trichloromethane in test tube Q. C3 [1 mark]


Section B

SPM CLONE
2. Figure 2.1 shows the electrolysis of potassium chloride, KCl solution using carbon electrodes.
©PAN ASIA PUBLICATIONS
Hydrogen Chlorine gas
gas, H
2 1.0 mol dm –3
potassium chloride,
KCl solution
Carbon
electrode P Carbon
electrode Q




Figure 2.1
(a) State the factors that determine the products formed at electrode P and electrode Q. C2 [2 marks]
Given that:
Cl + 2e  2Cl – E = +1.36 V
–
0
2
O + 4H + 4e  2H O E = +1.23 V
+
–
0
2
2
2H O + 2e  H + 2OH – E = –0.83 V
–
0
2
2
K + e  K E = –2.93 V
–
0
+
(b) Explain the reactions at electrodes P and Q. Include the following in your explanation: C4
(i) List of cation and anion of electrolyte attracted to each of electrodes, P and Q. [2 marks]
(ii) Names of the chemical species oxidised and reduced at each electrode. [2 marks]
(iii) The reason why the chemical species are chosen to be oxidised or reduced. [4 marks]
(iv) Half equations for each reaction. [2 marks]
SPM CLONE 0
(c) Figure 2.2 shows a voltaic cell. Given that E value
of Q is more positive compared to copper, Cu. V
(i) State the positive and negative terminal of
the cell. C2 Metal Q Copper,
(ii) Suggest metal Q and suitable solution to be Cu metal
solution R. C3 Solution R Copper(II) nitrate,
Cu(NO ) solution
[4 marks] 3 2
Figure 2.2
Section C
SPM CLONE
3. (a) Two different metals are placed in a lime. The circuit is
completed by connecting the metals to a voltmeter as shown Voltmeter
in Figure 3.1. What happens to the iron nail? Write a half
equation for the reaction. C3 [2 marks]
Copper,
Iron, Cu coin
Fe nail Lime
Figure 3.1
354

Form
5
Chapter 1 Redox Equilibrium Chemistry
CHAP. (b) Figure 3.2 shows two types of cells, A and B. CHAP.
1 1
V
Zinc,
Silver, Zn plate Silver,
Ag plates Ag plate
Silver
nitrate,
AgNO 3
solution
Cell A Cell B
Figure 3.2
©PAN ASIA PUBLICATIONS
Compare and contrast between cells A and B. Include in your answer the observations and half equations
for the reactions at the electrode in both cells. K4 [8 marks]
(c) A student intends to purify an impure iron bar. Design a laboratory experiment to purify the iron bar.
Your answer should consist of the following: K3
(i) Procedures of the experiment
(ii) Labelled diagram showing the set up of apparatus
(iii) Observations
(iv) Half equations involved in the reaction
[10 marks]




Reinforcement & Assessment of Science Process Skill

Reinforcement exercise of science process skills in preparation for Paper 3 (Practical test).
In this experiment you are required to determine the position of the three metals P, W and T in the
Electrochemical series by displacement reaction.
You are provided with the following materials:
M1 = Strip of metal P
M2 = Strip of metal T
L1 = Solution of ion W
L2 = Solution of ion T
Carry out the experiment according to the following instructions:
Experiment I:
1. Clean M1 (strip metal P) with sandpaper.
2. Pour solution L1 into a test tube until half full.
3. Put M1 into solution L1 in the test tube.
4. Record the colour change of L1 solution and any solid deposit in the test tube.
Experiment II:
1. Clean M1 (strip metal P) with sandpaper.
2. Pour solution L2 into a test tube until half full.
3. Put M1 into solution L2 in the test tube.
4. Record the colour change of L2 solution and any solid deposit in the test tube.

Experiment III:
1. Clean M2 (strip metal T) with sandpaper.
2. Pour solution L1 into a test tube until half full.
3. Put M2 into solution L1 in the test tube.
4. Record the colour change of L1 solution and any solid deposit in the test tube.

355

Form
5 Chemistry Chapter 1 Redox Equilibrium

CHAP. Experiment Observation CHAP.
1 1
I – Strip M1 + Solution L1

II – Strip M1 + Solution L2
III – Strip M2 + Solution L1
Table 1
[**Refer to the Simulation experiments and sample results to understand how to fill in Table 1.]
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Based on experiments conducted:
1. State an inference based on the observations in Experiment I.
2. State the name of the ion that gives blue colour to the solution and the shiny grey solid formed in
Experiment III.
3. Compare the strength of electropositivity of metal P and metal T based on Experiment II.
4. Compare the strength of electropositivity of metal T and metal W based on Experiment III.
5. Arrange metals P, T and W in ascending order of electropositivity in the electrochemical series.


Note: Simulation experiments and sample results

Experiment I








Colourless Shiny grey Colourless
Strip M1 solution L1 solid solution



Experiment II








Colourless Colourless
Strip M1 solution L2 Brown solution
solid

Experiment III









Colourless Shiny grey Blue solution
Strip M2 solution L1 solid




356

SPM MODEL PAPER






Paper 1
[40 marks]
Answer all questions.

1. What is the meaning of nanotechnology? 5. The following statement refers to the
A The study of chemical bonds between metal characteristics of an element in the Periodic Table
atom and non-metal atom. of Elements.
B The manipulation of materials on an atomic • Brown colour and soft solid.
or molecular scale.
C The study the importance of food additives • Reacts with water to produce alkaline
solution.
in food processing industry and the • Burn in oxygen to produce a white solid.
evolution of food processing technology.
D The development and the application of Which element has the above characteristics?
products or equipment, and a system to
conserve the environment. A D
2. Figure 1 shows two situations occur when B C
sunlight is shining on the glass X. 6. Which statement explains the effective collision?

SPM MODEL PAPER ©PAN ASIA PUBLICATIONS
A The collision which takes place after a
reaction.
Darker in
sunlight
B The collision which takes place before a
reaction.
C The collision that causes a reaction.
Glass X
energy.
What is the type of glass X?
A Fused glass Figure 1 D The collision produces less activation
B Soda-lime glass 7. Figure 2 shows a glass cookware that usually
C Borosilicate glass used in the kitchen.
D Photochromic glass
3. The following statement refers to an element in
the Periodic Table of Elements.
• Located in Period 3 of the Periodic Table of
Elements
• Reacts with water to produce acidic Figure 2
solution and bleaching agent
• Reacts with iron wool to produce a brown Which substance is added to the glass to make it
solid suitable for making the cookware?
A Lead(II) oxide, PbO
Which of the following shows the electron B Boron oxide, B O
3
2
arrangement of the element? C Natrium carbonate, Na CO
A 2.8.4 C 2.8.7 D Aluminium oxide, Al O 2 3
B 2.8.5 D 2.8.8 2 3
8. Atom W has 4 neutrons and a nucleon number of
4. Water molecule, H O combines with hydrogen 7. Which of the following is the correct symbol
2
ion, H to form hydroxonium ion, H O . What for atom W?
+
+
3
is the type of the chemical bond formed? A 7 4 W C 3 4 W
A Dative bond C Metallic bond B 4 W D 7 W
B Ionic bond D Hydrogen bond 7 3
524

Which of the following are correct about the Which of the following are the oxidising agent
compound? and reducing agent in the reaction?
I Low melting point Oxidising agent Reducing agent
II Exist as liquid at room temperature.
III Double covalent bond formed in the A Fe O 3 C
2
compound. B C Fe O
2
IV Conduct electricity in solid state. C Fe CO 3
A I and III 2
B I and IV D CO 2 Fe
C II and III
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D III and IV 20. The equation below shows the reaction between
17. Figure 6 shows the reaction between calcium excess zinc carbonate, ZnCO with dilute
3
carbonate, CaCO and glacial ethanoic acid, hydrochloric acid, HCl.
3
CH COOH. ZnCO (s) + 2HCl(aq) → CaCl (aq) + CO (g) +
3
2
Glacial ethanoic acid, 3 2 H O(l)
CH COOH 2
3
The changes of the quantity of reactants and
products are recorded with time until the
reaction is completed. Which graph shows the
correct changes?
Limewater
A Mass of zinc carbonate, ZnCO (g)
3
SPM MODEL PAPER No changes are observed after the reaction. What Time (s)
Calcium carbonate, CaCO
3
Figure 6
should be done in order to make the limewater
cloudy?
A Heat the mixture.
B Add water to the mixture.
2
C Substitute calcium carbonate, CaCO with
–3
solution (mol dm )
magnesium, Mg powder. 3 B Concentration of calcium chloride, CaCl
D Change calcium carbonate, CaCO chips to
3
calcium carbonate, CaCO powder.
3
18. Bronze is an alloy which contains copper and tin
atoms. In an activity, Shahriza found that bronze Time (s)
is harder than pure copper. Which statement
explains the situation above? C Concentration of hydrochloric acid, HCl (mol dm )
–3
A Tin atom makes strong bonds between the
pure copper atom.
B Tin atom fills in all the empty spaces
between pure copper atom.
C Tin atom compresses the arrangement of Time (s)
atom in pure copper.
D Tin atom reduced the layer of pure copper
atoms from sliding. D Volume of carbon dioxide, CO (cm )
3
2
19. The following equation represents a redox
reaction.
2Fe O + 3C → 4Fe + 3CO
2 3 2
Time (s)



526

32. Figure 11 shows a voltaic cell. 35. Figure 13 shows a series of tests carried out on
solution J.
V
Solution NaOH Green
J precipitate
Magnesium, Silver, Ag plate
Mg plate
Nitric acid, HNO followed with
3
silver nitrate, AgNO solution
Sodium chloride, 3
NaCl solution
Figure 11 White
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Which half equations represent the reactions at precipitate
the positive terminal and the negative terminal of
the cell? Figure 13

Positive terminal Negative terminal Which of the following is most likely to be
+
A 2H + 2e → H Ag → Ag + e − solution J?
−
+
2 A Iron(II) sulphate, FeSO 4
B Ag + e → Ag Mg → Mg + 2e − B Lead(II) sulphate, PbSO 4
2+
−
+
2+
−
+
C 2H + e → H 2 Mg → Mg + 2e − C Iron(II) chloride, FeCl 2
D 4OH → O + 2H O D Iron(II) iodide, Fel 2
−
Na + e → Na 2 2
+
−
+ 4e − 36. Figure 14 shows the results of an experiment to
determine the heat of combustion of butanol,
33. Figure 12 shows the conversion of ethene, C H , C H OH.
2
4
into ethanoic acid, CH COOH and eventually 4 9
3
compound Y. Wind shield Thermometer
Process X Oxidation
Copper can
Ethene, Ethanol, Compound SPM MODEL PAPER
3
C H C H OH Y Water (250 cm )
2 4 2 5
Figure 12 Spirit lamp
Which of the following is process X and Butanol, C H OH
4
9
compound Y? Wooden block
Process X Compound Y Figure 14
A Ethanoic acid,
Hydration 1.11 g of butanol, C H OH is completely burnt to
4
4
CH COOH heat the water from 25 ºC to T ºC. Determine the
3
B Ethanoic acid, T ºC.
Esterification
CH COOH [Specific heat capacity of water: 4.2 J g ºC ;
–1
–1
3
C Ethyl ethanoate, relative molecular mass of butanol: 74; heat of
Addition combustion of butanol: – 2 450 kJ mol ]
–1
CH COOC H
3 2 5 A 60.0 °C
D Ethyl ethanoate,
Oxidation B 42.0 °C
CH COOC H
3 2 5 C 35.0 °C
34. The following equation represents a combustion D 30.0 °C
of methane, CH . 37. The following statements describe the properties
4
of particles of a substance at room temperature.
CH (g) + 2O (g) → CO (g) + 2H O(l)
2
2
4
2
What is the mass of products formed when 5.6 g • The arrangement of the particles is not in
the orderly manner.
of methane is burnt completely? • The particles vibrate, rotate and move to
[Relative atomic mass: H=1; C=12; O=16] the whole region.
A 5.6 g • The particles always collide with each
B 12.6 g other.
C 15.4 g
D 28.0 g
529

Which substance has the above properties at Which of the following substances match the
room temperature? observation in Table 6.2?

Melting Boiling Solution X Solution Y Solution Z
Substances
point (°C) point (°C) A Sodium Soap
A W 120 2.4 hydroxide solution Lime juice
B X –36 27 B Lime juice Soap Potassium
C Y 12 40 solution hydroxide
D Z –79 –5 C Ammonia Sodium Lime juice
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solution hydroxide
38. Figure 15 shows the apparatus set up to study the D Potassium Soap
reaction of chlorine, Cl and iron, Fe. A brown hydroxide Lime juice solution
2
solid is produced at the end of the reaction.
8.4 g of iron wool (ferum) Soda lime
40. Table 7 shows three experiments carried out
Chlorine gas, to investigate the effect of other metals on the
Cl rusting of iron.
2
Combustion tube
Heat Set Experiment Observation
Figure 15 I Low intensity of
Determine the mass of the brown solid formed. blue spot
[Relative atomic mass: Cl = 35.5; Fe = 56] Iron nail
A 12.1875 g
SPM MODEL PAPER 39. Table 6.1 shows the colour changes for three II Jelly solution + potassium High intensity of
B 13.7215 g
Metal P
C 16.25 g
D 24.375 g
hexacyanoferrate(III) solution
types of indicators.


in pH 2
Indicator Colour Colour in pH Iron nail blue spot
12 solution
solution
Metal Q
Phenolphthalein Colourless Pink
Jelly solution + potassium
Methyl orange Red Yellow hexacyanoferrate(III) solution
Universal Red Dark blue III No changes
indicator

Table 6.1 Iron nail
Table 6.2 shows the colours of three indicators in Metal R
solutions X, Y dan Z.
Solution Jelly solution + potassium
hexacyanoferrate(III) solution
Table 7
Which of the following is the correct sequence,
in ascending order of metals P, Q and R in terms
of tendency to form ions?
Solution X + Solution Y + Solution Z + A P, Q, R
phenolphthalein methyl universal B R, P, Q
orange indicator
C Q, P, R
Colour Pink Red Dark blue D Q, R, P

Table 6.2
530

Paper 2
Section A
[60 marks]

Answer all questions.
1. Figure 1 shows the manufactured substances in industries.


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Reinforced concrete Bronze Laboratory glassware
Figure 1
(a) Reinforced concrete is produced by reinforce steel to the concrete. Between the reinforce steel and
concrete, which one acts as the matrix? [1 mark]
(b) Bronze is a type of alloy. Name two elements in the bronze. [2 marks]
(c) State one reason for using borosilicate glass to make laboratory glassware. [1 mark]

2. Soap dissolves in water to form soap anion. Soap anion consists of two parts as shown in Figure 2.
CH CH CH CH CH CH CH CH O
3 2 2 2 2 2 2 2
CH CH CH CH CH CH CH C
2 2 2 2 2 2 2
O – SPM MODEL PAPER
Part P Part Q
Figure 2
(a) Name part P and part Q. [2 marks]
(b) The following are three types of solutions, A, B and C used to dissolve a soap.
A: Magnesium sulphate, MgSO solution
4
B: Sodium nitrate, NaNO solution
3
C: Zinc chloride, ZnCl solution
2
(i) Choose a solution in which the cleansing action of the soap is not effective. [1 mark]
(ii) Explain your answer in (b)(i). [2 marks]
3. Figure 3.1 shows the atomic structure of a carbon atom.




W


Figure 3.1
(a) Based on Figure 3.1, complete Table 3.
Subatomic particle Relative mass Relative charge
W

Table 3
[2 marks]
531

(iii) Sodium chloride, NaCl cannot conduct electricity in solid state but it can conduct electricity in
aqueous state. Explain why. [2 marks]
(b) Figure 7.2 shows the symbol of two elements, N and H. Element N reacts with element H to form a
compound.
14 N 1 H
7 1
Figure 7.2
(i) State the type of particle in the compound formed. [1 mark]
(ii) Draw a diagram to show the electron arrangement in the compound formed between atoms N
and H. [2 marks]
(iii) The molecule formed in (b)(ii) can combine with hydrogen ion, H to form ammonium ion,
+
NH . State the type of chemical bond and explain briefly how the chemical bond is formed.
+
4
[2 marks]
8. (a) Figure 8.1 shows three test tubes contain dilute H X acid, HY acid and dry glacial ethanoic acid,
2
CH COOH. H X acid and HY acid are strong acids.
3 2
P Q R

0.1 mol dm 0.1 mol dm Anhydrous
–3
–3
dilute H X acid dilute HY acid ethanoic acid,
2
CH COOH
3
Calcium carbonate, Calcium carbonate, Calcium carbonate,
CaCO
CaCO
CaCO
SPM MODEL PAPER ©PAN ASIA PUBLICATIONS
3
3
3
Figure 8.1
There are no gas bubbles produced in test tube R. Suggest a method that can make calcium
(i)
carbonate, CaCO reacts with ethanoic acid, CH COOH to produce gas bubbles.
[1 mark]
3
3
(ii) Table 8 shows the pH value of aqueous solution of ethanoic acid, CH COOH and dilute HY acid
3
of the same molarity.
Type of acid
0.1 mol dm
Molarity Ethanoic acid Dilute HY acid
–3
0.1 mol dm
–3
pH value 4.0 1.0
Table 8
Explain why the pH value of aqueous solution of ethanoic acid, CH COOH is higher than the
3
pH value of dilute HY acid. [2 marks]
(iii) Figure 8.2 shows the apparatus set up for the titration of 20 cm of 0.5 mol dm sodium
–3
3
hydroxide, NaOH solution with H X acid of an unknown concentration. Phenolphthalein is
2
used as the indicator.
H X acid
2
Sodium hydroxide, NaOH
solution + phenolphthalein
Figure 8.2
The chemical equation for the reaction is as the following.
H X(aq) + 2NaOH(aq) → Na X (aq) + 2H O(l)
2 2 2
3
If 25 cm of H X acid is required to neutralise the sodium hydroxide, NaOH solution in the
2
conical flask, calculate the concentration of the H X acid solution in mol dm . [3 marks]
–3
2
534