Page 2 - Spotlight A+ Form 4 & 5 Chemistry KSSM

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