14 Spotlight A+1 Form 4.5 Chemistry

Extra Features of This Book

CHAPTER The Periodic Table of Elements

4

SMART SCOPE 4.1 The Development Important Learning Standard Page CONCEPT MAP
of Periodic Table • Describe the historical development of the Periodic Table of Elements. 82
of Elements • Deduce the basic principle of arrangement of elements in the Periodic Table 85

4.2 The Arrangement of Elements. 85
of in the Periodic 86
Table of Elements • Describe briefly the modern Periodic Table of Elements.
• Generalise the relationship between the proton number and the position of

elements in the modern Periodic Table of Elements.

Contains learning standard 4.3 Elements in • Relate the inert nature of Group 18 to its stability. 88 Contents of the whole
(LS) that need to be Group 18 topic are summarised in the
achieved in each topic. • Generalise the changes in physical properties of elements when going down 89 form of a concept map.
4.4 Elements in Group 18. 89
Group 1 90
• Describe briefly the uses of Group 18 elements in daily life. 91
4.5 Elements in 94
Group 17 • Generalise the physical changes of elements when going down Group 1. 95

• Investigate through experiment the chemical properties of Group 1 elements with: 96
97
‒ Water ‒ Oxygen gas ‒ Chlorine 97

• 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
Group 17.

• Summarise the chemical properties of Group 17 elements.

• Generalise the changes in the reactivity of elements when going down Group 17.

Fo4rm • Predict the physical and chemical properties of the other elements in Group 17. 98
Chemistry Chapter 3 The Mole Concept, Formula and Chemical Equa• tiDonescribe the trends in physical properties of elements across Period 3. 98

4.6 Elements inThus, the rel•a teCivloeemndeumncttso alanec creouxslpsaePrriemrmieondat s3tso. oobsfervwea cthearnges in the properties of the oxides of 102
Period 3 103
Sulphur S 32 = 18
Chlorine Cl 35.5 • Describe briefly the uses of semi-metals.

3. Calculate the relative molecular mass or

Potassium K 39 4.7 Transition relative formu•laDmetearsmsine the position of transition elements in the Periodic Table of Elements. 104
Calcium Ca 40 Elements (a) The relativ• eEmxpolalienctuhlearspemciaasl schoafracatemrisotilcescouflae few transition elements with examples. 104
Zinc Zn 65 105
can be cal•cuLlaisttetdhebuyseasdodfitnragnsuitpiontheelemreelnatstiivneindustry
atomic masses of all the atoms that are
Silver Ag 108 present in the molecule.
Example: A molecule of carbon dioxide,
Lead Pb 207 ALmogfaooCmtexOryai2gkl,kenacloiantosmistss. of 1 carbon atom •anMd et2alloid / Metaloid
• Alkali metal / • Monoatomic gas / Gas monoatomic Cleansing agent effectiveness
• Amphoteric / • Noble gas / Gas lengai in hard water and acidic water
CO2• Atomic radius / Jejari atom
Measuring atomic mass • Chemical properties / MSiofaletkkuiml diwa iatomdOCiOnoxe2id=emCco1oOlen2csuisltes of ••ccaarrPObbeoocnnrtieotdel/eKctarloan arrangement / Susunan elektron oktet Compare
http://bit.ly/3706MgE • Diatomic molecules / of 1 Cleansing action
CHAP. • Duplet electron arrangement / Susunanatoemlekantrdo2nodxyugpenleatto•msP. eriodic table / Jadual Berkala
• Elektronegativity / Keelektronegatifan • Physical properties / Sifat fizik CHAP.
3 • Group / Kump1u2lan+ 2(16) = 44 • Reactivity / Kereaktifan
3 Study about

• Halogen / Halogen Therefore, the relative molecular m••asTVsraaloenfnsicteioenleecltermonen/tE/leUkntrsounr peralihan Discuss Soaps Detergents • Preservatives Usage
carbon dioxide valens about • Antioxidants and
Why is the relative atomic mass of chlorine 3•5.I5n?ert / Lengai Uses in daily Traditional Types of • Flavourings effects
35 Cl life and medicines additives • Colourings of using
Natural chlorine exists in two isotopes, 17 and = RAM of carbon + 2(RAM of oxygen) misuse of Modern Food additives • Thickeners food
medicines medicines • Stabilisers additives
37 Cl. An ordinary sample of chlorine contains 80 = 12 + 2(16) Cleansing agents Application of • Emulsifiers
17 25% = 44 nanotechnology
approximately 75% chlorine-35 and Consumer Graphene
chlorine-37. Therefore, the relative atomic mass is RAM of C = 12 Medicines and Industrial in industry
RAM of O = 16
closer to 35 than 37. Chemistry

RAM = 75 × 35 + 25 × 37
100 100
= 35.5
How to determine the number of atoms in a Types of core
ingredients
Relative Molecular Mass, RMM molecule? • Water
The whole numbers in a chemical formula
• Emulsifiers
1. The idea of relative atomic mass also can be represent the number of atoms of each element • Preservatives Cosmetics Fats and oils Application of Usage of
applied to compounds which may be molecules except “1” is not stated. • Thickeners Categorised into Use of oils and Green Technology sludge from
or ionic compounds. For example, in the formula of sulphuric acid, H2SO4 • Moisturisers fats in daily life in industrial waste waste water
2. The relative molecular mass of a molecule is the • 2 means there are 2 atoms H, • Colouring agents 1. Make-up cosmetics treatment
• 4 means there are 4 atoms O, • Fragrances 2. Treatment cosmetics management
• and 1 atom S even if it is not stated. 3. Fragrances
average mass of the molecule when compared
1 H2SO4 = H2S1O4
PORTALwith 12 of the mass of a carbon-12 atom.
SPOTLIGHT
Relative molecular mass of a molecule Relative Formula Mass, RFM
= The average mass of one molecule
1. We use “relative molecular mass” for molecules. Side effect of
For ionic compound, we use “relative formula cosmetic usage
1 the mass of one carbon-12 atom
12 × mass”. 483

Example: otnheanmtohleec1u12le of water, H2O is 18 Example: rNeala2tOiveis faonrmiouFnloa4ircmmcoamsspoofusnodCd.hiuemmistry Chapter 6 Acid, Base and Salt
The mass of mass of a carbon-12 Sodium oxide,
Scan QR code to visit times greater Therefore, the
atom. oxide 6.1
= 2(RAM of sodium) + RAM of oxygen The Role of Water in Showing Acidic and Alkaline Properties

O 1–12– of a carbon-12 atom = 2(23)+ 16 RAAcMidsof Na = 23 Hydroxonium ions, H3O+ are the actual ions existing
websites or videos relaHteH d to 18 = 62 1R.AMWohf eOn= a1n6 acid is dissolved in water, hydrogen in the aqueous solution that gives the acidic

atom in the molecule of acid is released as properties. To simplify explanation, we often use

You will learn about ionic compounds in FhToyhrdmerroe4fgoerne,iobnas,eHd+o. n the Arrhenius theory, acid is hydrogen ion, H+ to represent hydroxonium ions,
Chapter 5. 2. defined as follows: H3O+.
subtopics learnt. There areFigure 3.3Onemoleculeofwateris18timesheavier
48 than 1 of a carbon-12 atom. 3.1C.1hem3.i1c.a2l substance that ionises in water to 6. Table 6.1 shows some examples of acids.
12 produce hydrogen ions, H+. Table 6.1

videos for certain activities or 3. Hydrogen ions, H+ cannot exist on their own. Acid Ions present in aqueous solution
experiments.
With water, H+ ions are pulled into water Hydrochloric HCl(aq) → H+(aq) + Cl–(aq)
4. Fimoonorsle, cHeux3laOems+,(paHlqe,2)O. H,wChtoel(nafoqr)hm→ydrsHotag+be(nlaeq)hc+hydlCorroli–xd(oaeqn)iugmas acid
INFORMATION GALLERY
Sulphuric H2SO4(aq) → 2H+(aq) + SO42–(aq)
acid Additional information

is dissolved in water, molecules of hydrogen Nitric acid HNO3(aq) → H+(aq) + NO3–(aq)
chloride will ionise in water to produce hydrogen Ethanoic acid HCH+(3aCqO) OH(aq) CH3COO–(aq) +
ions, H+ and chloride ions, Cl–.
5. HydrogHen+(iaoqn)s+, HH+2Ow(ill)l →comHb3iOn+e(awqi)th the water
molecules, H2O to form stable hydroxonium Basicity of Acids
ions, H3O+. 1. Basicity of an acid refers to the number of

hydrogen ions, H+ that can be produced by one
molecule of acid that ionises in water.
5H H 2. Figure 6.2 shows the classfication of acids based
H H Form
related to the topic.CHAP.
ChapHter 1 CRledox+EquiliObrium ChemistryO+ + Cl– on the basicity of acids.
CHAP.
H

1 11.5 Figure 6.1 Formation of hydroxonium ion, H3O+

Aim: 3. The circuit is completed by connecting the Acids
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.

Problem statement: CHAP. Test tube Copper(II) sulphate, Diprotic acid Triprotic acid
Do the types of electrodes affect the types of Monoprotic acCiduSO4 solution Chemistry Chapter 1 Redox Equilibrium
6 Carbon
products formed during the electrolysis? electrodes

Hypothesis: Ammeter A Switch Fo5rm
When copper electrodes are used instead of carbon
AktivitY / EXPERIMENT electrodes, the types of products formed at the uTeTASe1tel.ltshhlee2selieentcc8peogttg.ssrrdbwooaassdliw1nhEHAwHytepcetiyxrcaCNtehasvrdaithtodelapmOrotaird→sotaopt.r3pihgueotag→lrae6uecHnrelrtn:BaeroseFp+npHtadiwcalure+ogtaao+otriasndeucentCd+rtrerdgpsuyoltetN–eeechhtrnwretehd1Oee-sf.opme.u32oao–oepa7ocrnnadnalo1leatFyeorde5csbindougdeomnselusenihe,psrinuoielencciuswnol6aietntnl.teecl2ghestrio.cCrnotdlcaedFeosdeiHAiEHhspgnisyxaacf22pudwaCSiinnbdcermOrdd2aeayorOtt4pthegi4→lroaee.tnn: p2i2oroHHfon+ad+scu++picdSeeCsOrs2bmtO4w2a–4oso2e–ledcounletheCHbAaPs.2ethievcae–ittr1HhiAEaDyyinsy.rxc3eaozdwaPiodifnmrrnFcvOxaeoacietoitidp4lhceegg,lelillirtaesuZlead.altne:snroi3dpCeeciraHrodeotc1no+afldeo.msl+1lurlepz7.cP–ZieenOnrssAAchmt4(n-3–ho–oc)orVwdoeloeepesclptumtCelhereaettehcr(Doe+dl)aleCniuisee–lalnce2Coeevxuelei–xd2l+ra.iaystrmheAecadoprtepDlicpeseeairnoviefioednlal,for CHAP.
anode are different. 14. of the cell while copper, Cu is the positive
terminal of the cell. 1
Variables: 5. 5. Redox reaction that takes place in Daniell cell
(a) Manipulated: Types of electrodes can be represented by an overall ionic equation:
(b) Responding: Types of products at the anode 6.
(c) Fixed: Type of electrolyte, concentration of
7.
electrolyte

Material: Negative terminal: Zn(s) → Zn2+(aq) + 2e–
0.1 mol dm–3 copper(II) sulphate, CuSO4 solution,
sandpaper, wooden splinter, matches Positive terminal: Cu2+(aq) + 2e– → Cu(s)
Overall ionic
Complete activity or Apparatus: equation: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Batteries, carbon electrodes, copper electrodes,
Switch Battery 6. From the ionic equation, we can write cell
connecting wires with crocodile clips, ammeter, notation for Daniell cell as the following:

electrolytic cell, 50 cm3 beaker, electronic balance, 146 e– B 6.1.1 6.C1.2 Cu2+
e– Zn+
experiment including results, switch,testtube A Ammeter Zn2+ Cu2+ e– Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Procedure: Copper, Cu Zn
1. Two carbon electrodes are cleaned with electrodes Cu 7. Standard cell potential, E0 for Daniell cell can be
Copper(II) sulphate, calculated using the following formula:
CuSO4 solution Salt bridge

sandpaper. Figure 1.28 Figure1.17 Daniel cell E0 = E0 – E0
cell cathode anode
data analysis, discussion 2. The copper(II) sulphate, CuSO4 solution is
poured into an electrolytic cell until it is half full. Zn2+(aq) + 2e–  Zn(s) E0 = –0.76 V with zinc, Zn as the anode and copper, Cu as the
Cu2+(aq) + 2e–  Cu(s) E0 = +0.34 V cathode.
Observation:

and conclusion to increase Table 1.24 2. aE0szitnrcoisngmeor rreendeugcaitnivgea. gIteinntd. iTcahtuess, that zinc, Zn is Anode: Zn(s)  Zn2+ (aq) + 2e– E0 = –0.76 V
students’ scientific skill. Observation zinc, Zn plate
is an anode where oxidation process occurs. Cathode: Cu2+(aq) + 2e–  Cu(s) E0 = +0.34 V
EXAMPLE Electrode E0 = E(+0c0at.h3o4de V– )E–0an(o–de0.76
Anode Cathode Electrolyte cEo0cpopppeerr(iIsI) more positive. It indicates that the cell =
3. ion, Cu2+ is an oxidising agent. V)

Carbon Gas bubbles are released. A A brown solid is The blue colour of the Copper plate is the cathode where reduction = 1.10 V

cgololowuinrlgesws ogoads ethnastpFilgoi4nrnmtietersisthperoduCcheedm. isdtreyposCitheadptoenr the cathode. solution become light process occurs.
6 Acid, Base anbdluSea.lt
A Zinc, Zn plate becomes thinner as zinc,
Copper Copper anode becomes 6th.i5nner. ConcAdeenbprtoorswaitnteidsooonlindotihsfeAcqatuheodoeu.sTthSheoeslboullutuietoiocnonloreumr oaifnEsxample 8 Zn corrodes and dissolves in zinc sulphate, iEs0zmincowrehiecahsiilsymoxoirdeisneedgaantidveacintsdiacsataens that zinc, Zn
anode.
In12f..erE((E(eaballn))ee) cccAACettrr:nnaootoohllddyyosseedii::sseOCuu: oxssCiipynnogpggpeepnccre(aoIgrIrp,)abCpisoo,eunnOrsme,e2leClieestucacp2ttl+rrriooosadd321rdpeue...erocspsedTtToTdtrhhodunmhhhead.ceeeet3euccqmouodcsouhfenn.noaosdcicrnotwe.eeltnonusitttftthryritaoehvaotntoeaEsCTupifisou.lonshlroousenqolinoelpnonmcuuddoglhocthtuaurefeleafyonutccttapelseohtieoynoaisscfe,sotpontayinhpsssCtldonuhooe:eouxotellsbruyiSfucfioegstsOtoesniocieiomlopos4oelninulpsnacpeus.tetac,aogpitrcorselmonae,uuendsortpdre,(,CireeIoaitImtsds)uOesnhsduasoep2iunuu.rlmlrlvnehaElsyopgemeicnildneghdrhitdgnuaehcaaanctlmitenwrn1ecertogaa,salertytCbcdesooiur.ts,hpSnoeHOpfe(C(e2ba4cOlarce))o(slacIcpdsZ[70coItRta)uomhp5.nlirt2uselolio.eauso36tl=datdmotritnoohe(eitleIos6vioenfIsgv.,)tn5eldeh.]doiesaofttmifisnololeomslwdilodaicalrwiictdmteaayrltaceozsitirufsn.o:mtchNmecn=hafilokto1relarl4oitd,ueweOp,i,nCgZ=5an0s1C(0oN6ll2,uOcitmn3iBo)235n0soi:0sZbPtptZfhhyloinaerntStoleceoOuan,ssngZ4tionhicsnadagolwuelaZdtuistwrioetn(efizsom(of.iesnnTe)rtic.hlhes→e)neecorcttexoeZfrlfiooondbntwri2ehse+st(ee,.wadoeqecflete)aontce+htlzteroh2iicndceetcer–ctowiu(nocrosnor,empfnZrpeotnetmrai2s)+l
8. aTchteuaEll0yceltlhoebvtoalitnaegde through the calculation is
produced in the Daniell cell
based on the potential difference between two
electrodes.

9. The functions salt bridge of:
(a) Complete the circuit by allowing the
movement of ions.
(b) Cathode: Copper, Cu metal is prodorucmedol.e, thus tChue2c+oantcetnhteraatinoondoef aansodluctoiopnpecar,nCu metaSlolauttiothne generated. (b) Separate two different electrolytes
be measured incatthheoudnei.ts of g dm–3 or mol dm–3. C Brown solid is deposited at the copper, Cu
(a) Molarity = 0.2 mol plate, making copper, Cu plate becomes 10. Initially, oxidation of half-cell is neutral with
(a) Concentration in unit g dm–3, is the mass of 0.5 dm3 thicker. Copper(II) ion, Cu2+ is reduced zinc ions, Zn2+ and sulphate ions, zSinOc42i−onisn,
1.4.3 solute found in 1 dm3 solution. 333 = 0.4 mol to the copper atom, Cu by gaining two the solution. When more and more
electrons.
Concentration (g dm–3) (b) Number of moles Cu2+(aq) + 2e‒ → Cu(s) Zn2+ enter the solution, the solution becomes
The intensity of blue solution decreases
= Mass of solute (g) = mass as the concentration of copper(II) ion, Cu2+ positively charged.
Volume of solution (dm3) molar mass
75.6 g 11. Similarly, reduction of half-cell is neutral with
(b) Concentration in unit mol dm–3, is the = + (3 × 16)] copper(II) ions, TChue2+soalnudtiosnulwphilaltebeionnesg, atSiOve4l2y−
number of moles of solute found in 1 65 + 2 [14 g mol−1 in the solution.
decreases.
dm3 solution. This concentration is called = 0.4 mol dm–3 4. Oxidation process occurred at the anode and charged when more and more copper(II) ions,
reduction process at the cathode causes zinc
molarity. Molarity = 0.4 mol Cu2+ leave the solution and form copper atom, Cu.
0.5 dm3
Molarity (mol dm–3) = 0.8 mol dm–3
Number of moles of solute (mol) (anode) becomes relatively negative charge 12. When two half-cells are charged, the voltaic
= Volume of solution (dm3) odnme–a3 naonthdermasoilndthme–f3olcT(laoehnlweecrinbetrfgeoonres,) as compared to copper (cathode). cell will not function. Therefore, a salt bridge is
4. Both units of g zinc, Zn is the negative terminal needed to connect the two half-cells.
interchanged with
Try Question 1 in Formative Zone 6.5 diagram.

Molarity × molar mass 318 1.3.1

Example 7 Concentration

(mol dm–3) ÷ molar mass in g dm–3 BRILLIANT TIPS

Calculate the concentration of the following

solutions in the unit of g dm–3.

(a) 10 g of glucose is dissolved in 0.5 dm3 of

Example and complete water. The unit for molarity is mol dm–3 or molar (M). Useful tips for students
solution to enchance Mole is not the same as molar. Mole is the unit for to solve problems in the
students’ understanding. (b) 30 g of potassium hydroxide is dissolved in measuring matter while molar is the number of related subtopic.
750 cm3 of water. moles of solute in a given volume of solution.

(c) 2 moles of sodium chloride in 10 dm3 of Try Question 2 in Formative Zone 6.5

distilled water. Given that the molar mass of Example 9
CHAP. sodium chloride is 58.5 g mol–1.
Calculate the concentration of 0.25 mol dm–3
6 The volume of solution must be in dm3. sulphuric acid in the unit of g dm–3.
[1 dm3 = 1000 cm3] [Relative atomic mass: H = 1, O = 16, S = 32]

Solution Solution
Molar mass H2SO4 = 2(1) + 32 + 4(16)
(a) Concentration = 10 g 750 cm3 is
0.5 dm3 = 98 g mol–1
= 20 g dm–3 converted to dm3 Concentration = Molarity x molar mass
750
(b) Concentration = 30 g = 1000 cm3 = 0.25 mol dm–3 × 98 g mol–1
0.75 dm3 = 0.75 dm3 = 24.5 g dm–3
= 40 g dm–3

(c) Mass of NaCl = Number of moles × molar mass
= 2 mol × 58.5 g mol–1

= 117 g

Concentration = 117 g
10 dm3
= 11.7 g dm–3

164 6.5.1 6.5.2

ii

Tag OF ‘TRY QUESTION...
IN FORMATIVE ZONE...’
Fo5rm Chemistry Chapter 2 Carbon Compound

2.3 Chemical Properties and 3. When the molecular size of an alkane increases,
it becomes more difficult to burn. flaLmaregseF. o5rm
Interconversion of Compounds molecules of alkanes burn with sootier
between Homologous Series
Tag placed at the end of This is because the number of carbon atoms per Chemistry Chapter 2 Carbon Compound
Chemical Properties of Alkane
1. Alkanes are unreactive and do not react with molecule increases as the molecular size of an
alkane increases. The percentage of carbon by 2.4

examples to guide students in2CHAP. mostchemicals. mass in the alkane molecule also increases. 1. Name the following compound using the IUPAC nomenclature system. C2 CHAP.
2. Alkanes are
saturated hydrocarbon Example 5 (a) H H H (b) H 2
compounds. Each carbon atom in an alkane | || |
molecule is already bonded to a maximum Explain why butane, C4H10 burns with more sooty
answering related questions in number of atoms. The strong C-C and C-H flame compared to ethane, C2H6. H–C–C–C–H H–C–H
bonds need a lot of energy to break. [Relative atomic mass: H = 1; C = 12] CHAP. || HH CH
3. Alkanes undergo two types of reactions: 2 HH ||
(a) Combustion Solution H–C–H H–C–C–C–H 2
| ||
H HH
H–C–H
(b) Substitution Percentange of carbon by mass in butane, C4H10
|
Formative Zone. Combustion reaction = 4(12) × 100% = 82.76% H
1. oHtAofl2kOhpae.nraoTet.dhsHuebceucenorcnmcea,bcrauoblsmkotainponlneedstoieoaflrxyeaildiukneas,enedxeCscaOesps2rsfouaodenxlusdyc.geewsnaa,telOort2, 4(12) + 10(1)
2. Draw the structural formula for each of the following organic compounds. C2
Percentage of carbon by mass in ethane, C2H6
2(12) (a) 2-methylbutane
= 2(12) + 6(1) × 100% = 80% (b) 2-methylpropene

3. Name the following using the IUPAC nomenclature system.
(a) H H H (b) H H H H
Example: The percentage of carbon by mass of butane, || | | | ||
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) C4H10 is higher. As a result, it burns with more
Methane sooty flame. H–C–C=C–C–H H– C = C – C – C– C – H
| || | ||
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) H HH H HH
Ethane
Substitution reaction 4. Draw the following isomers of pentene.
Try Question 1 in Formative Zone 2.3 1. A substitution reaction occurs when an atom (a) Pent-2-ene
or a group of atoms is substitute by other atoms
2. Incomplete combustion occurs in limited in a molecule. (b) 2-methylbut-1-ene
2. In this reaction, each hydrogen atom in an
supply aonfdoxpyrgoednu,ceO2a. Alkanes burn in sooty (c) 3-methylbut-1-ene FORMATIVE ZONE
flame mixture of carbon, C alkane molecule is substituted by a halogen (d) 2-methylbut-2-ene
atom in the presence of sunlight or ultraviolet
5. Name the following using the IUPAC nomenclature system. C2

particle (soot), carbon monoxide, CO gas, (a) H H (b) H

carbon dioxide, CO2 gas and water, H2O. (UV) rays as the catalyst. || |
H–C–CC–C–H H– C– H
Example: Example:
pMreetshenancee,oCf sHu4nlriegahcttos rwUitVh chlorine, Cl in the || H
MC2CHetHh4Fa(4ong5(re)gm)++O32O(g2)(→g) C→Ch(e2sm)Ci+Ost2(rygH)2+OC(4hlH)ap2Ote(rl4) Polymer rays to produce four HH |

products as shown in Figure 2.9. Questions to testH–CC–C–C–H
||
Ethane SPM Simulation HOTS Questions H HH
H – Cl – H
Hl 6. Draw the structural formula of pent-2-yne. C2 students' understanding
A MMeelthinagnepoint
7. Name the following using the IUPAC nomenclature system. C2
(a) H H (c) H H H
To balance a c1h. emFigicuarleequ1atisohno: ws the structural formula of || | ||
Step 1: compound Z which is used to make a pipe.
B Density H–C–C–O–H
Check the number of carbon and hydrogen atoms C EmpiricCal2l, fUoVrmrauyla || H–C–C–C–O–H
in the molecular formuHla Cofl aHlkaCnel . HBalCalnce the D Molecular formula HH || at the end of each
HH
number of carbon aHto2OHCmbsyCHinadCHCdOi2nCHagnadCHnthuHCemnbuemr ibnefrroonft H HExaminer’s coCml ment: Cl H–C–H
hydrogen atoms in Compound Z H – Cl – Cl DHich–lorCComtBfllmohl–ooearttmCnlhhealuncRelua.lsaWuTorHrbifictsmhh–tCloaaCCrbHnllsolmsci–2g,.eeCgtshsSealounrebhtmshaTteoevaCtrelnaeldccc–heeluCoCtllnlrhlaoSs–meriteCyshthialzaeinsesem, bhfpoiiggirrgihcceeeasrrl(b)
of each formula. Hl |
H
Step 2: Chloromethane H HH H
nCBuahlmeacnbkceterhnteheneeudnmeudbmeWtbor ehoarifdcoodhfxiooysxgfayetgfhnreeaanctftooaiotmllonFosm,iwgtishniuneirCngneOOim12s2a.utnIhlftdeitphHsleyt2rOaul.cl tural | || | suptopic.
SPM SIMULATION the numbers in froofntthoef mthoenfoormmeurlaoefbcyo2m. pound Z? C2 foFrimguurlae 2.9 Productosstforofanstugtrbeasrct.ittiTouhtnieorbneefrotewarece,teionthnemofommleeeclttuhinlaegnsepboeicnotmoef H–C–C–C–C–O–H
HOTS QUESTION | || |
H HH H

A H Cl S is higher. 8. Draw the following isomers of butanol. C2
Answer: C (a) Butan-2-ol
376 C C 3. the 2.3.1 is
H Hn Figure 3 shows raincoat which mad(eb) 2-methylpropan-2-ol

from a synthetic polymer, polyvinyl chloride,

B H Cl PVC.

CC 404

Provides complete solution 4CHAP. H H CHAP.
C H Cl
HCCH 4
HH

with examiner’s comment D H Cl SPM MODEL PAPER
CC
HH

to help students to answer Examiner’s comment: Figure 3 Paper 1
HOTS questions. Monomer of compound Z (polymer) must
have a carbon-carbon double bond, C = C. (a) State the name of the monomer for Instruction: Answer all questions. [40 marks]
Structural formula in Answer A is another polyvinyl chloride, PVC. C1
simplified way to represent the polymer Z 1. What is the meaning of nanotechnology? 5. The following statement refers to the
but not the monomer of compound Z. (b) Draw the structural formula for the A The study of chemical bonds between metal characteristics of an element in the Periodic Table
Answer: B monomer. C2 atom and non-metal atom. of Elements.
B The manipulation of materials on an atomic
2. Figure 2 shows the polymerisation process. (c) State one reason why polyvinyl chloride, or molecular scale. • Brown colour and soft solid.
PVC should not be disposed by open C The study the importance of food additives • Reacts with water to produce alkaline
burning? C3 in food processing industry and the
evolution of food processing technology. solution.
HH HH Examiner’s comment and answer: D The development and the application of • Burn in oxygen to produce a white solid.
products or equipment, and a system to Which element has the above characteristics?
CC CC (a) Vinyl chloride // Chloroethene conserve the environment.

H Hn (b) H Cl 2. Figure 1 shows two situations occur when
sunlight is shining on the glass X.
HH HC

Form RS HH

SUMMATIVE ZONE 4 Chemistry Chapter 7 Rate of Reaction Figure 2 (c) Burning of polyvinyl chloride, PVC
Diagram below shows a graph of will release pollutant and acidic gas
Which of the following is similiar for substances such as hydrogen chloride gas, HC
R and S? C2
3. volume of gas against time in a reaction. 60 cm3 of hydrogen gast,oHt2he atmosphere.
A D
is collected when excess magnesium powder, Mg reacts with 50 cm3 of 0.1 mol dm–3 nitric acid, HNO3. B C

Volume of gas (cm3)

Question of various levels SPM 49704 Time (s) Darker in 6. Which statement explains the effective collision?
of thinking skill to evaluate SPM sunlight A The collision which takes place after a
60 reaction.
SPM MODEL PAPERAnswer (a) Glass X B The collision which takes place before a
Figure 1 reaction.
30 C The collision that causes a reaction.
What is the type of glass X? D The collision produces less activation
0 A Fused glass energy.
B Soda-lime glass
student’s understanding of(a) On the same axes, sketch the curve that you would expect to obtain if 25 cm3 of 0.2 mol dm–3 C Borosilicate glass 7. Figure 2 shows a glass cookware that usually
nitric acid, HNO3 reacts with excess magnesium powder, Mg. D Photochromic glass used in the kitchen.
(b) Explain your answer in 3(a).

each topic. Examiner’s comment 3. The following statement refers to an element in
the Periodic Table of Elements.
Number of moles of HNO3, n = 0.1(50) = 0.005 mol; Number of moles of HNO3, n = 0.2(25)
= 0.005 mol 1000 1000
• Located in Period 3 in the Periodic Table of
Number of moles of nitric acid, HNO3 used in both experiments are the same, the number of Elements
moles of hydrogen gas, H2 produced will also be the same, which is 60 cm3. Concentration of nitric • Reacts with water to produce acidic SPM MODEL PAPERFigure2
acid, HNO3 in the second experiment is higher. So, the rate of reaction is higher. That is why the
gradient of the curve will be steeper. solution and bleaching agent Which substance is added to the glass to make it
• Reacts with iron wool to produce a brown suitable for making the cookware?
A Lead(II) oxide, PbO
solid

SPM-orientedWhich of the following shows the electron
arrangement of the element? B BNAolaurtrominuinmoixucimdaerb,ooBxin2dOaet3,e,ANl2aO2C3 O3
A 2.8.4 C 2.8.7 C
B 2.8.5 D 2.8.8 D practice

Paper 1 4. Water molecule combines with hydrogen ion to 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 basedofofrtmhehcyhdermoxicoanlibuomndiofno,rmHe3Od?+. What is the type for atom W?
A Fermentation Clone Oxygen gas, oOf2 is produced from the A 74 W on latestC 43W SPM
B Photosynthesis 3. decomposition sodium chlorate(I), NaOCl A Dative bond C Metallic bond B 74 W D 73 W
C Formation of stalagmites
D Combustion of magnesium iTnhethfeolplorwesienngcsehoofwms tahnegcahneemsei(cIaVl )eqouxaidtieo,nMonf tOh2e. B Ionic bond D Hydrogen bond

Clone The reaction between sodium thiosulphate, reaction. 524 format to assess students
2.
bNya2tSh2eOf3oallnodwisnuglpehquuraictioacni:d, H2SO4 is represented 2NaOCl(aq) MnO2 of2vNoalCuml(aeqo)f+oxOy2g(egn)
Figure 1 shows the graph gas
against time.
Na2S2O3(aq) + H2SO4(aq) → Ntah2SeSOOm24((oga)sqt)++sHuSi2t(Oasb)(ll+e) on all the topicsCHAP. learnt in
7CHAP. Which of the following is Volume of oxygen gas (cm3)
7
method to determine the rate of reaction? C2
A Determine the change in temperature of the
solution with time. Form 4 & 5 textbooks.
Determine the volume of water, H2O
B produced with time. Fo4rm

C Determine the production of a fixed quantity 0 Time (s)

of sulphur, S precipitate with time. Why the Figure 1 curve decreaCsehawptiethr 4 The Periodic Table of Elements Chemistry
D Determine the change in the concentration of gradient of the
sodium sulphate, Na2SO4 with time. time? C2
Reinforcement & Assessment of Science Process Skill
264

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
REINFORCEMENT & ASSESSMENT • K2 = Oxide Q ANSWERS Complete Answers
• K3 = Oxide R https://bit.ly/3sn6L0o
• L1 = Sodium hydroxide solution, 2.0 mol dm–3
• L2 = Nitric acid, 2.0 mol dm–3

Carry out the experiment according to the following instructions:
1. Put 2 spatula of substance K1 into two different test tubes.

OF SCIENCE PROCESS SKILL 2. Pour 5.0 cm3 solution L1 into the first test tube. Form 4 (b) (i) Vinegar; salt; baking powder
CHAP. 3. Pour 5.0 cm3 solution L2 into the second test tube. (ii) Vinegar: preserves food
4 Salt: gives salty taste
4. Heat the mixture in each test tube slowly while stirring with a glass rod. Chapter 1 Introduction to Chemistry Baking powder: raises the dough
Questions for students to master
5. Record the solubility of substance K1 in each test tube. 1.1 (c) Chemist; doctor
(d) Hydrogen peroxide waste with low concentration
6. Repeat the experiment by replacing substance K1 with K2 and K3. 41. A field of science that studCiHesAPt.he structures, properties,
Observation compositions and interactions between matter. can be poured directly into the sink. Concentrated
2. Herbicide; Hormone hydrogen peroxide wastes need to be diluted with
the Oxide Reaction with nitric acid, HNO3 Reaction with sod3i.umWith nanotechnology, sunscreens are no longer oily water. Then, it is added with sodium sulphite for the
Oxide P (K1) purpose of decomposition before being poured into the
hydroxide solution,nowadays and are colourless when applied to the skin. sink.
NaOH 4. (a) Cosmetic consultant 2. (a) (i) Presence of water and oxygen
(b) Dietitian (ii) Rusting of iron nail
(c) Pathologist (iii) Type of nail
(d) Veterinarian (b) Water and oxygen are needed for iron rusting.
(c)

science process skill (SPS) presented OxideQ(K2) Test tube Observation

1.2 A Iron nail does not rust.
1. A systematic scientific method used to solve science related
B Iron nail rusts.

Oxide R (K3) problems. C Iron nail does not rust. ANSWER FORM 4
2. (a) When the temperature increases, the mass of salt (d) Oxygen and water must be present for the iron nail to
in Science Practical Test. Scan the QR Table1
(**Complete Table 1 with reference to the Simulation Experiments and Sample Results given) dissolved also increases. rust.
(b) Manipulated variable: Temperature
Section B
Responding variable: Mass of salt dissolved

code below to get Tip dan Teknik Based on the experiment conducted: 1.3 3. (a) The scientific method is a systematic approach to solve
Prihatin Murid in answering Science 1. • Do not eat, drink, chase or run in the laboratory. problems in science.
Practical Test. 1. State the following variables based on the experiment above: (b) Making an observation ANSWERS
Tip dan Teknik (a) Manipulated: • Do not pour the chemicals back to the reagent bottles.
(b) Responding: 2. • Keep flammable substances away from the heat source. Making an inference
(c) Control: Identifying the problem
• Do not point the mouth of the test tube at your face or
at other people.

3. (a) To carry out experiment that involves the release of
toxic vapours, gases that can cause combustion or gases
Based on the results of the experiment, classify oxides P, and into the table below(abc) cwTooirtdhreipmnugonvgteeondtitrhstm,eoierilll,. chemicals Making a hypothesis
2. respective properties. Q R the hands. or microorganisms from 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
part of the body if there is fire.

4. (a) Kept in paraffin oil to prevent reaction between this
chemical with moisture, water and air.
Amphoteric Base Answers are provided. ScanAcid
Controlling the variables
Planning an experiment

Table 2 QR(b) Kept in dark bottles to avoid the exposure of sunlight. codeCollecting data to get Complete
Interpreting data
Prihatin Murid 5. Mercury poisoning is a phenomenon when a person is
exposed to mercury in a certain amount. Two symptoms of
mercury poisoning are vomiting and difficult in breathing.
Answers with explanations for113
Paper 1 2. A 3. C 4. B 5. C Making a conclusion
1. A Writing a report

7. B 9. B 10. C
12. C 14. C 15. D
https://bit.ly/349VB5H objective questions.6. B 8. D
13. A
11. B
Paper 2 Section C
Section A 4. (a) (i) Ammonium nitrate / Urea
1. (a) Chemistry is defined as a field of science that (ii) Occupation A: Pharmacist
Occupation B: Pathologist
(b) • Inform the accident to the teacher immediately.
• 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

Redvaistieon Redvaistieon

T heme 1 The Importance of Chemistry 4.5 Elements in Group 17 96
98
Chapter 1 Introduction to Chemistry 1 4.6 Elements in Period 3 104
107
1.1 Development in Chemistry 4.7 Transition Elements
Field and Its Importance in Summative Zone

Daily Life Chapter 5 Chemical Bond 115
1.2 Scientific Investigation in
Chemistry 3 5.1 Basics of Compound
1.3 Usage, Management and 6 Formation 117
Handling of Apparatus
and Materials 8 5.2 Ionic Bond 118
Summative Zone 16
5.3 Covalent Bond 122

5.4 Hydrogen Bond 126

5.5 Dative Bond 129

5.6 Metallic Bond 130

5.7 Properties of Ionic
Compounds and Covalent
T heme 2 Fundamentals of Chemistry

Compounds 131

Chapter 2 Matter and the Atomic Structure 21 Summative Zone 137

2.1 Basic Concepts of Matter 23 T heme 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

Chapter 3 The Mole Concept, Chemical Formula Properties 146

and Equation 45 6.2 pH Value 152

6.3 Strength of Acids and

3.1 Relative Atomic Mass and Alkalis 155
Relative Molecular Mass
3.2 Mole Concept 47 6.4 Chemical Properties of Acids
3.3 Chemical Formula 50
3.4 Chemical Equation 56 and Alkalis 157
Summative Zone 66
73 6.5 Concentration of Aqueous

Solution 164

6.6 Standard Solution 166

Chapter 4 The Periodic Table of Elements 80 6.7 Neutralisation 170

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

Redvaistieon Redvaistieon

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
Compounds between
Reaction 241
Homologous Series 376
7.3 Application of Factors that
2.4 Isomers and Naming Based
Affect the Rate of Reaction
on IUPAC Nomenclature 394
in Daily Life 255
Summative Zone 406
7.4 Collision Theory 256

Summative Zone 264 Theme 3 Heat

T heme 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
and Exothermic Reactions
8.2 Composition of Glass and Its
in Daily Life 442
Uses 279
Summative Zone 446
8.3 Composition of Ceramics

and Its Uses 281 T heme 4 Technology in Chemistry

8.4 Composite Materials and Its
Importance 284
288 Chapter 4 Polymer 453
Summative Zone

FORM 5 4.1 Polymer 455
T heme 1 Chemical Process 4.2 Natural Rubber 465
Chapter 1 Redox Equilibrium 4.3 Synthetic Rubber 472
Summative Zone 475

294 Chapter 5 Consumer and Industrial
Chemistry 482

1.1 Oxidation and Reduction 296 5.1 Oils and Fats 484
1.2 Standard Electrode
Potential 312 5.2 Cleaning Agents 487
1.3 Voltaic Cell 315
1.4 Electrolytic Cell 321 5.3 Food Additives 497
1.5 Extraction of Metal from
Its Ore 338 5.4 Medicines and Cosmetics 502
1.6 Rusting 342
Summative Zone 352 5.5 Application of

Nanotechnology

in Industry 509

5.6 Application of Green

Technology in Industrial

Waste Management 511

Theme 2 Organic Chemistry Summative Zone 517
Chapter 2 Carbon Compound
357 SPM Model Paper 524

Answers 541 – 578

v

vi THE PERIODIC TABLE OF ELEMENTS

Period Group 18
1

1 2 Symbol of the element 1 Proton number 13 14 15 16 17 2
Relative atomic mass Name of the element
H1 Hydrogen H 5 He

1 Hydrogen B Helium
1 4
34 Boron 6 7 8 9
11 10
2 Li Be C N O F
Ne
Lithium Beryllium Carbon Nitrogen Oxygen Fluorine
12 14 16 19 Neon
79 20

11 12 13 14 15 16 17 18

3 Na Mg 34 56 78 9 10 11 12 Al Si P S Cl Ar
Sodium Magnesium Aluminium Silicon Phosphorus Sulphur Chlorine Argon
23 24
27 28 31 32 35.5 40

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

4K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Potassium
39 Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton
40 45 48 51 52 55 56 59 59 64 65 70 73 75 79 80 84

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Argentum Cadmium Indium Stanum Antimony Tellurium Iodine Xenon
89 91 93 96 101 103 106 108 112 115 119 122 128 127 131
85.5 88

55 56 57 – 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
Lantanides
6 Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn
Caesium
133 Barium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Aurum Mercury Thallium Plumbum Bismuth Polonium Astatine Radon
137 178.5 181 184 186 190 192 195 197 201 204 207 209

87 88 89 – 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118

7 Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og
Actinides Rutherfordium Dubnium Seaborgium
Francium Radium Bohrium Hassium Meitnerium Darmstadtium Roentgenium Copernicium Nihonium Flerovium Moscovium Livermorium Tennessine Oganesson

57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

Lanthanides series La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Lanthanum
139 Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium
140 141 144 150 152 157 159 162.5 165 167 169 173 175
89
90 91 92 93 94 95 96 97 98 99 100 101 102 103
Actinides series Ac
Actinium Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium
232 231 238

Legend: Metal Semi metal Non-metal

(Source: International Union of Pure and Applied Chemistry, IUPAC)

CHAPTER Chemical Bond

5

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

5.4 Hydrogen Bond • Explain with examples the formation of a hydrogen bond. 126
127
5.5 Dative Bond • Explain the effect of the hydrogen bond on the physical properties
of substances.
5.6 Metallic Bond
• Explain with examples the formation of a dative bond. 129
5.7 Properties of
Ionic Compounds • Explain the formation of a metallic bond. 130
and Covalent • Reason out the electrical conductivity of metal. 130
Compounds
• Compare the properties of ionic and covalent compounds through 131
experiment. 135

• Explain with examples the uses of ionic and covalent compounds
in daily life.

• 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
• Giant molecular structure / Struktur molekul gergasi Daya tarikan van der Waals

115

Stable duplet Stable octet Affect the solubility and boiling
electron electron point of covalent compounds

arrangement arrangement

Group 18 elements Attraction force between H
atom and F, O and N atoms in

different molecules

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

Ionic Bond Chemical Bond Dative Bond Delocalised
Metallic Bond electrons
Metal and
non-metal Covalent Bond Lattice structure
Formation of of metal ions
Non-metal
cations and non-metal Sea of electrons
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
Normally can dissolve dissolve in organic
solvent but cannot
in water but cannot dissolve in water
dissolve in inorganic Cannot conduct
electricity
solvent
Can conduct electricity

in molten state or
aqueous state

116

Chapter 5 Chemical Bond Chemistry Fo4rm

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.
5. Chemical bonds are formed through the transfer
Example: of electrons or share of electrons.
• Water is a compound of hydrogen and oxygen (a) During the transfer of electrons, a metal atom
• Carbon dioxide is a compound of carbon and with one, two or three valence electrons will
donate its valence electrons whereas a non-
oxygen metal atom with five, six or seven valence
• Rust is a compound of iron, oxygen and electrons will receive the electrons.
Example:
hydrogen Sodium atom, Na with electron arrangement
2. Atoms form chemical bond to achieve a stable of 2.8.1 will donate one valence electron
to chlorine atom, Cl which has electron
electron arrangement. arrangement of 2.8.7.
3. Noble gases exist as monoatomic gases and

are not chemically reactive because they have
achieved a stable duplet electron arrangement or
octet electron arrangement.
(a) All Group 18 elements have eight valence

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

Neon 2.8 Transfer of electron from sodium
atom to chlorine atom

Argon 2.8.8 Figure 5.1 Transfer of one electron
from sodium atom, Na to chlorine atom, Cl
Krypton 2.8.18.8 CHAP.

Xenon 2.8.18.18.8 5

Radon 2.8.18.32.18.8 By donating, gaining or sharing electrons, an atom
can change its electron arrangement.
Try Questions 2 and 3 in Formative Zone 5.1
(b) In sharing electrons, two or more non-metal
Atoms of Group 18 elements do not gain, lose or atoms will contribute their valence electrons
share electrons with atoms of other elements. for sharing to achieve the duplet or octet
electron arrangement.
(b) The electron arrangement of an atom where
the valence shell is filled with eight valence Example:
electrons is known as octet electron Iantowmast,erHmcoolnetcruibleu,tHes2Oon, eeavcahloenf tcweoelheycdtrroongetno
arrangement.
share with one oxygen atom, O.
(c) The electron arrangement of an atom with a
single shell filled with two valence electrons H OH
is known as duplet electron arrangement.
Sharing of electrons between an
4. Atoms with octet or duplet electron arrangement oxygen atom and two hydrogen atoms
are very stable. Therefore, atoms of other elements Figure 5.2 Sharing of electrons in water molecule, H2O
that have less than eight valence electrons tend
to achieve the stable octet or duplet electron 117
arrangement through the formation of chemical
bonds.

5.1.1

Fo4rm Chemistry   Chapter 5 Chemical Bond

6. There are two types of chemical bonds: Non-metal Covalent bond Non-metal
(a) Ionic bond
(b) Covalent bond

Chemical bonds are forces that hold atoms Figure 5.4 Sharing of electrons between atoms
together to make compounds or molecules through of non-metal forms a covalent bond
the transfer of electrons or sharing of electrons
Try Question 5 in Formative Zone 5.1

Try Question 1 in Formative Zone 5.1 • Metals are placed on the left-hand side of the
Periodic Table of Elements except hydrogen.
7. Atongeitohneirc bboyndtriasnfosfremrreidngwheelnecattroomnss are joined
(between • Non-metals are placed on the right-hand side of
atoms of metal and non-metal). the Periodic Table of Elements.

Ionic bond 1 18

Metal Non-metal 12 13 14 15 16 17
2 B

Figure 5.3 Transferring of electrons between metal 3 3 4 5 6 7 8 9 10 11 12 Si
atoms and non-metal atoms forms ionic bonds
4 Ge As
8. Ajaotoinmceodsvoatlfoengnoettnhb-emornebdtayli)ss.haforrinmgedelewcthreonnsat(obmetsweaerne
5 Sb Te

6 Po

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

5 3. Is magnesium atom, Mg stable? Why? C2
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, NO2 (e) Ammonia, NH3
(c) Ethane, C2H6 (f) Copper(II) oxide, CuO

5.2 Ionic Bond 2. A metal atom of Group 1 will donate one valence
electron to form a cation with a charge of +1.
Formation of Ions Example: Li → Li+ + e–
1. A cation is formed when a metal atom donates
Donates one +
its valence electrons to achieve the stable duplet
or octet electron arrangement. electron
Li Li
X → Xn+ + ne–

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

Chapter 5 Chemical Bond Chemistry Fo4rm

Hδ+ Hydrogen Role of Hydrogen Bonds in Daily Life
Oδ– bond
Hδ+ 1. Why our hair is stuck when get wet?
Hδ+ Oδ– Covalent (a) Human hair consists between 65% – 95% of
bond protein. The protein molecules are attracted
Hδ+ Hδ+ to each other by hydrogen bond.
Oδ– (b) When the hair gets wet, the presence of water
will break the hydrogen bonds between the
Hδ+ protein molecules.
(c) Protein molecules will form hydrogen
Oxygen atom is more electronegative than bonds with water molecules and the water
hydrogen atom. molecules will form hydrogen bonds with
other hair protein molecules. As a result, the
Figure 5.20 Formation of hydrogen bonds between hair sticks together.

(b) water molecules, Hb2Oond in ammonia, Dry hair
Formation of hydrogen H
NH3 and hydrogen fluoride, HF O HO

Hydrogen bond H
O HO
Hδ+ Hδ+
Nδ– Hδ+ Nδ– H
O HO
Hδ+ Hδ+ Hδ+
Protein molecules

Nitrogen atom is more electronegative than hydrogen Wet hair
atom. Hydrogen bond is formed between a nitrogen Hydrogen bond
atom in an ammonia molecule with a hydrogen atom in
another ammonia molecule. H HO CHAP.
O HO Water
Figure 5.21 Formation of hydrogen bonds between molecules 5
ammonia molecules, NH3 H
Protein molecules
(c) Pembentukan ikatan hidrogen dalam
hidrogen fluorida, HF Figure 5.23 Hydrogen bonds formed between the
hair protein molecules with water molecules,
Hydrogen bond
H2O when the hair gets wet
δ–F δ+H 2. Why do we wet a finger to turn pages?

Hδ+ δ–F (a) The principal raw material for producing
paper is cellulose fibers.
Fluorine atom is more electronegative than hydrogen
atom. Hydrogen bond formed between a fluorine atom (b) Cellulose consists of hydroxyl (–OH) groups.
in a hydrogen fluoride molecule with a hydrogen atom When a wet hand is used to turn pages, water
in another hydrogen fluoride molecule. molecules will form hydrogen bond with
–OH groups in cellulose in the paper. This
Figure 5.22 Formation of hydrogen bonds between makes the paper sticks to our fingers.
hydrogen fluoride molecules, HF
Hydrogen bond

H H H H
O
O Water
O O H OH H molecules
H O

Hydrogen chloride, HCl does not form hydrogen Cellulose fiber surface
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, H2O at finger

5.4.1 5.4.2 127

Fo4rm 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
charged.
2. Boiling point:
(a) The °Cb,owilhinegreaps othinetbooilfingetphoainnet,ofCet2Hha6noils,
–89 4. Covalent compounds such as ammonia, fNorHm3
CM2oHle5OcuHlesis 78 °C. and hydrogen fluoride, HF also can
of ethane, hfThlyuihgdoihsrreoiedrgxeebp,nolaHiilnibFnsogcnwapdnhosyidnwaitsmisctoohmlmvoewpniaianatre,ewrNdaHmtteo3ro,oaleHtnhcd2ueOrlhecysaod,nvrdHaolghe2eOnanst.
(b) weak Van der Waals aCtt2rHa6ctaioren attracted by
forces only.
Molecules of deethraWnoala,lsCa2Httr5aOcHtioanrefoartcterascatnedd
by weak Van compounds.
hydrogen bonds.
(c) More heat energy is required to overcome
the weak Van der Waals attraction forces and Hydrogen bond Hydrogen bond

to break the hydrogen bonds. δ+H Hδ+ Hδ+
Nδ– Oδ– Oδ–
van der Waals attraction forces Hδ+ Hδ+
δ+H Fδ– Hδ+
HH HH δ+H Water
H C C Oδ– Hδ+ H C C Oδ– Hδ+ molecule, Water molecule, H2O
H2O
Hydrogen fluoride, HF
HH HH
Hydrogen bond
Ethanol Ammonia, NH3
molecule,
HH C2H5OH Figure 5.27 Hydrogen bond between ammonia, NH3 and
H hydrogen fluoride, HF with water molecules
Hδ+ Oδ– C C

HH Table 5.4 Comparison of boiling point between covalent
compounds with and without hydrogen bond
Ethanol molecule, C2H5OH
CHAP.
5 Figure 5.25 Molecules of ethanol are attracted by Van Covalent Covalent
der Waals attraction forces and hydrogen bonds compound with Boiling compound Boiling
hydrogen bond point with Van point
3. Solubility: der Waals
(a) Ethane, bCCe22cHHa6u5OsiseHintihsseohliumgbhollleyecisnuollewusabtloeefrinwethwheaarnteeoarl.s, and Van der attraction
(b) ethanol, Waals attraction force only
This is
forces

eCwwt2aihtHthaen5rOwe,aHmtCeor2clHemacn6oucllfeaeocnsr,unmloeHts,2hfOHoydr2mOrbo.ughteyndmrbooogleencnduslbeoswnidtohsf Ammonia, NH3 -33 0C Phosphine, -87 0C
Hydrogen 19.5 0C PH3 -85.05 0C
fluoride, HF
Hydrogen
chloride, HCl

HH Hδ+ Hδ+ Ethanol, 78 0C Ethane, C2H6 -89 0C
H C C Oδ– Hδ+ δ–O C2H5OH

Water molecule,
H2O
HH 5.4

Ethanol molecule, Hydrogen bond 1. What is hydrogen bond? C1
C2H5OH 2. Explain why hydrogen fluoride, HF has a
δ+H Hδ+
Oδ– higher boiling point than hydrogen chloride,
HCI? C3
Figure 5.26 Hydrogen bonds between ethanol, 3. Usually covalent compounds cannot dissolve
C2H5OH with water molecule, H2O in water but ethanol, C2H5OH can dissolve in
water. Explain why. C4
128
5.4.2

Chapter 5 Chemical Bond Chemistry Fo4rm

+–

Bulb glows Battery Negative terminal Positive terminal

Metal rod Positive +++++ +
metal ion +++++

– +++++
+++++
Electron Metal

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
Electrical Conductivity are insoluble in organic solvents.

1. In solid ionic compounds, ions are held together 2. On the other hand, most covalent compounds are
by strong electrostatic attraction forces in the insoluble in water but can dissolve in organic
lattice structure. solvents.
(a) Ions are in fixed positions and cannot move Try Question 2 in Formative Zone 5.7
freely.
(b) Hence, ionic compounds in the solid state do CHAP.
not conduct electricity.
5
2. In aqueous or molten state, ions are free to
move. Therefore, the compounds can conduct Water is a polar solvent. It has separation of charges.
electricity. The positive ions are attracted to the negative ends
of the water molecules (oxygen atom), while the
Ions are in fixed positions. Ions are free to move. negative ions are attracted to the positive ends of
They cannot move freely. water molecules (hydrogen atom). The attraction
between the water molecules and the ions of the
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
1. Ionic compounds consist of positive ions and
3. However, covalent compounds consist of neutral
molecules. They do not contain ions. Hence, negative ions which are held together by strong
covalent compounds do not conduct electricity electrostatic attraction forces.
in all states.

5.6.2 5.7.1 131

Fo4rm 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.
Try Question 1 in Formative Zone 5.7
Figure 5.34 Oppositely-charged ions in an ionic
compound are held together by strong Table 5.5 Comparison between properties
of ionic and covalent compounds
electrostatic attraction forces
2. A lot of heat energy is required to overcome Properties Ionic Covalent
compounds compounds
the strong electrostatic forces. This explains
why ionic compounds have higher melting and Melting point High Low
boiling points. and boiling
3. Ionic compounds are usually non-volatile point
solids at room temperature.
4. On the other hand, covalent compounds (with Electrical Conduct Do not conduct
simple molecular structures) consist of neutral conductivity electricity in electricity in any
molecules which are held together by Van der aqueous or states
Waals attraction forces. molten state

Less heat energy is Solubility Usually dissolve Usually dissolve in
required to break the in water but do organic solvents such
Molecules weak Van der Waals not dissolve in as benzene but do not
attraction forces organic solvents dissolve in water
during melting
or boiling.

Figure 5.35 Molecules in a covalent compound are held
together by weak Van der Waals attraction forces

CHAP. Variables:
(a) Manipulated: Type of compound
5 5.1 (b) Responding: Electrical conductivity
(c) Fixed: Carbon electrode
Aim: Procedure:
To compare the properties of ionic compounds
and covalent compounds. Batteries Bulb Switch

Problem statement: Carbon electrodes
What are the differences between properties of Lead(II) bromide, PbBr2
ionic compounds and covalent compounds?
Figure 5.36
Materials: 1. A crucible is filled with lead(II) bromide, PbBr2
Magnesium chloride, MgCI2, cyclohexane, C6H12
naphthalene, C10H8, distilled water, solid lead (II) powder until it is half full.
bromide, PbBr2 2. The apparatus is set up as shown in Figure

Apparatus: 5.36.
Test tubes, crucible, 10 cm3 measuring cylinder, 3. The switch is turned on. The observation on
spatula, glass rod, 250 cm3 beaker, Bunsen burner,
wire gauze, tripod stand, carbon electrodes, the light bulb is recorded.
pipe clay triangle, battery, connecting wire with
crocodile clip, light bulb, switch 5.7.1

A Electrical conductivity

Hypothesis:
Ionic compounds can conduct electricity in the
molten state but not in the solid state while
covalent compounds do not conduct electricity in
both states.

132

Chapter 5 Chemical Bond Chemistry Fo4rm

4. The switch is turned off and the lead(II) 6. Steps 1 to 5 are repeated using naphthalene,
bromide, PbBr2 powder is then heated until it C10H8 to replace lead(ll) bromide, PbBr2.
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

Compound Physical state Observation Inference
on the light bulb

Lead(II) Solid The bulb does not light up. PbBr2 cannot conduct electricity in solid state
bromide, Molten but can conduct electricity in molten state.
PbBr2
Solid The bulb lights up.
Naphthalene, Molten The bulb does not light up. C10H8 cannot conduct electricity in solid and
C10H8
molten states.
The bulb does not light up.

Conclusion: Variables:
The hypothesis is accepted. Lead(II) bromide, (a) Manipulated: Type of compound, magnesium
PbBr2 is an ionic compound which able to conduct
electricity in molten state but not in solid state. chloride, MgCl2 and naphthalene, C10H8
Naphthalene, C10H8 is a covalent compound that (b) Responding: Solubility of Mg2+ ions and Cl–
cannot conduct electricity both in solid and molten
states. ions
(c) Fixed: Quantity of compound, volume of
Discussion:
1. Lead(II) bromide, PbBr2 consists of lead(II) ions, solvent, temperature

Pb2+ and bromide ions, Br –. In solid state, the Procedure:
oppositely-charged ions are held together by
strong electrostatic forces of attraction in the Distilled Magnesium Cyclohexane, C6H12
lattice structure. They are in fixed positions and water chloride,
cannot move freely. As a result, solid lead(II) MgCl2
bromide, PbBr2 cannot conduct electricity. CHAP.
2. When lead(II) bromide, PbBr2 is heated until Figure 5.37
it melts to become a liquid, the lead(II) 5
ions, Pb2+ and bromide ions, Br – are free to
move. Therefore, molten PbBr2 can conduct 1. Half spatula of magnesium chloride, MgCl2
electricity. crystals is placed in two test tubes separately.
3. Naphthalene, C10H8 only consists of neutral
molecules. It does not contain ions. Thus, it 2. 5 cm3 of distilled water is added to the first
cannot conduct electricity either in solid state test tube and shaken.
or molten state.
3. 5 cm3 of cyclohexane, C6H12 is added to the
Ionic compounds which are soluble in water also second test tube and shaken.
can dissociate in water to produce free moving
ions. 4. The solubility of magnesium chloride, MgCl2 in
both solvents are observed and recorded.
B Solubility
Hypothesis: 5. Steps 1 to 4 are repeated using solid
Most ionic compounds are soluble in water but naphthalene, C10H8 to replace magnesium
insoluble in organic solvents while most covalent chloride, MgCl2.
compounds are insoluble in water but can dissolve
in organic solvents. Results:

Table 5.7

Compound Solubility Solubility in
in water cyclohexane

Magnesium Soluble Insoluble
chloride, MgCl2 Insoluble Soluble

Naphthalene
C10 H8

5.7.1 133

Fo4rm Chemistry   Chapter 5 Chemical Bond

Conclusion: 2. A spatula of naphthalene, C10H8 powder is
The hypothesis is accepted. Magnesium chloride, placed in another test tube.
MgCl2 is an ionic compound which is soluble in
water but insoluble in organic solvents. Napthalene, 3. The two test tubes are heated in a beaker
C10H8 is a covalent compound which is insoluble in containing water as shown in Figure 5.38.
water but soluble in organic solvents.
4. The changes in the physical states of the
Discussion: compounds are observed and recorded.
1. Magnesium chloride, MgCl2 dissociates in
Results:
water to produce free moving magnesium ions
Mg2+ and chloride ions, Cl–. Table 5.8
2. Naphthalene, C10H8 consists of neutral
molecules which cannot dissociate in water. Compound Observation Inference

C Melting and boiling points Magnesium No change Magnesium chloride,
Hypothesis: chloride, MgCl2 has a high
Ionic compounds have high melting and boiling MgCl2 melting point.
points while covalent compounds have low melting
and boiling points. Naphthalene, It melts Naphthalene, C10H8
has a low melting
Variables: C10H8 rapidly. point.
(a) Manipulated: Type of compound, magnesium
Conclusion:
chloride, MgCl2 and napthalene, C10H8 The hypothesis is accepted. Magnesium chloride,
(b) Responding: Boiling point MgCl2 is an ionic compound and naphthalene,
(c) Fixed: Quantity of compound C10H8 is a covalent compound. Magnesium
chloride, MgCl2 has a higher melting point than
Procedure: naphthalene, C10H8.

Magnesium Water, H2O Discussion:
chloride, MgCl2 Naphthalene, 1. Magnesium chloride, MgCl2 consists of positive
CHAP. C10H8
ions, Mg2+ and negative ions, Cl–. Mg2+ ions
5 Heat and Cl– ions are attracted together by strong
electrostatic forces. A lot of heat energy is
Figure 5.38 required to overcome the strong electrostatic
1. A spatula of magnesium chloride, MgCl2 attraction forces. Therefore, the melting point
of magnesium chloride, MgCl2, is high.
powder is placed in a test tube. 2. Naphthalene, C10H8 consists of molecules which
are attracted by weak Van der Waals attraction
forces. As a result, less heat energy is required
to overcome the weak Van der Waals attraction
forces.

Structure of Ionic Compounds and Covalent Compounds

1. Ionic compounds consist of positive ions of In the actual structure, ions are packed together
metals and negative ions of non-metals. The closely as shown in Figure 5.40. Figure 5.39 just
positive ions and negative ions attract each other to make the structure easy to see.
to form a rigid three-dimensional structure
which is called lattice. Each ion in the lattice Figure 5.40
structure is surrounded by other ions of opposite 2. The arrangement shown in Figure 5.39 does not
charges.
exist separately on its own. It extends throughout
Example: the crystal, involving millions of ions. This
Sodium chloride, NaCl consists of sodium ions, structure is known as giant ionic lattice.

Na+ and chloride ions, Cl– . 5.7.1

Na+
Cl–

Figure 5.39 The lattice structure of NaCl

134

Chapter 5 Chemical Bond Chemistry Fo4rm

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
covalent bonds. Examples of giant molecular structures:
6. However, the individual molecules are only Diamond, graphite, silicon(IV) oxide
attracted to each other by weak Van der Waals
attraction forces. Only little heat is required to Strong covalent bonds between
overcome the attraction forces make it has low the carbon atoms in the three-
melting and boiling point. dimensional structure of diamond

Examples of simple molecular structures:
Iodine, Im2,oclaercbuolen cdaionxiedxeis, tCaOs2,sowlaidte,r,liHqu2Oid
7. Simple and Figure 5.42 A small part of the structure of diamond
gas. 9. Compounds with giant molecules exist as solid.

II Weak Van der Waals attraction Covalent compounds with giant molecular
structure have very high melting point and boiling
forces between the iodine point because the heat required to break the
I I molecules, I2 covalent bond is high.

Strong covalent bond between the
iodine atoms in the molecule.

Figure 5.41 The simple molecules of iodine

Uses of Ionic and Covalent Compound in Daily Life CHAP.

Table 5.9 Various uses of ionic and covalent compounds in daily life 5

Field Example of ionic compounds Example of covalent compounds

Industry • Lithium iodide, LiI is used • ETuthrepre,n(tCin2He,5)C2O10His18uisseudsetod as a solvent for paint.
to make pacemaker batteries, • make inks and dyes.
which has a lifespan of 7 to 8
years.

Agriculture • aAnmdmpootntiausmiumnitcrhaltoer,iNdeH, 4KNCOl 3 • Busreodmaosetpheasntiec,idCe2sHa5nBdr and chloropicrin, CCl3NO2 are
are used to make fertiliser. insecticides.

Medicine • Sodium sbtiocmarabcohnaatceid, N. IatHCO3 • Paracetamol, C8H9NO2 is a pain reliever and a fever
reduces reducer.
is used as an antacid to • tPoentriecailtlimn,aCny16Hdi1f8fNer2eOn4tStiyspaens antibiotic. It is used
treat indigestion and upset of infection caused by
stomach. bacteria.

Domestic • Sodium thcehlmoraanteu,faNcatuCrleOo3 fis • sGkliync.eIrtoal,bCso3Hrb5s(OwHat)e3rifsraomverayir effective moisturiser on
use used in 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

Fo4rm 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
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
C3
(c) State whether the compound formed by Q and Q is an electrical conductor. Explain why.

(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
5CHAP. [Proton number: Na = 11, Al = 13]
compound.
I Al III Na+ • Lead(II) iodide and sodium chloride
IV Al3+
II Na are ionic compounds but lead(II) iodide
is insoluble in water.
A I and II C II and IV Answer: D

B I and III D III and IV 3. The electron arrangement of atom J is 2.8.2
and the electron arrangement of atom L is 2.7.
Examiner’s comment: Elements J and L react to form a compound.
• Al contains 13 electrons.
• Na contains 11 electrons. Write the chemical formula for the compound
• Na+ is the ion of Na which loses 1 electron. formed and state two physical properties.
Thus, the number of electrons is 10. C4
• Al 3+ is the ion of Al which loses 3 electrons.
Thus, the number of electrons is 10.
Answer: D

2 Compound Z has the following properties. Examiner’s comment:
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
formula JL 2.
• Melting point = 800 °C 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

Chapter 5 Chemical Bond Chemistry Fo4rm

Paper 1

1. Which of the following pairs of physical
properties of sodium oxide, Na2O is true? C2 C

Solubility in Electrical conductivity
D
water when molten

A Soluble Conducting

B Soluble Not conducting

C Insoluble Not conducting

D Insoluble Conducting

2. Which of the following substances is a covalent 6. Ecatnhandoisls,oClv2eH5iOn Hwaitsera. covalent compound that
Which of the following
compound? C1 explain the solubility of ethanol, C2H5OH in
A Magnesium chloride, MgCl water? C2
B Calcium oxide, CaO
CLeaardb(oInI)dciholxoirdied,eC, POb2Cl2 A Ethanol molecules ionise in water.
C B Ethanol molecules form hydrogen bonds
D
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
molecules.
V: Sulphur
W: Silver SPM SPMClone Figure 1 shows the electron arrangement of an CHAP.
X: Carbon ion M –. Given that atom M has 18 neutrons.
Y: Chlorine 7. 5
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

4. Which of the following compounds has the Figure 1
What is the nucleon number of atom M? C2
lowest melting point? C2 A 8 C 22
A Sodium chloride, NaCl
MISruoalnpg(hnIueIsIr)iudomixoixdcihde,leoF,reSi2dOOe2,3 MgCl2 B 10 D 35
B
C Clone The electron arrangement of an atom J is 2.1 and
D the electron arrangement of an atom T is 2.6.
8.

5. Which of the following diagrams shows a Atom J reacts with atom T to form a compound.
compound that consists of double covalent Which of the following is true about the
bonds? C1 reaction? C2
A A Atom T donates 6 electrons.
B Atom J receives 1 electron.
C An ionic compound is formed.
D The chemical formula of the compound
B formed is JT6.

137

Fo4rm Chemistry   Chapter 5 Chemical Bond

9. Element X and hydrogen, H combined to form 12. Table 2 shows the numbers of electrons in ions
occoof mmpbthioneuenddfotlHolo2wXfoi.nrmgElemcroeepmnrtpesoXeunntasdndNthsaeo2Xd.iuelmWec,htriNocnha P 2–, Q +, R – and T 2+.
arrangement for a compound formed between
magnesium, Mg and element X? C2 Ion Number of Number of
A electrons neutrons

P2– 10 11
12
Q + 10 18
20
R – 18

Mg X T  2+ 18

B Table 2
Which of the following shows the correct
X Mg X
nucleon number of the ion? C3

C – 2+ – Ion Nucleon number
A P2– 20
X Mg X B Q+ 22
C R – 35
D T 2+ 38

D 2+ 2– 13. Figure 3 shows the electron arrangements of
ions W + and Z  2–.
Mg X
+ 2–

10. What is the chemical bond shown in Figure 2? WZ

C1

HF HF Figure 3
Calculate the difference between the numbers of
HN B F HNB F electrons in atom W and atom Z. C3

CHAP. H F HF A 2 C 4

5 Figure 2 B 3 D 5
A Covalent bond C Hydrogen bond
B Dative bond D Ionic bond 14. Figure 4 shows the symbols of ion X and ion Y.

SPMClone Table 1 shows the melting points and types of X  3+ Y 2–
particles of compounds P, Q, R and S.
11. Figure 4
Calculate the relative formula mass of the
Compound Melting Type of particle
point (oC) compound formed when X reacts with Y.
Given that the relative atomic mass of X is 27
P 70 Molecule and the relative atomic mass of Y is 16. C3
A 113
Q 280 Ion B 102
C 86
R 650 Ion

S 2 000 Atom 15. An element reacts with chlorine, Cl to form a

Table 1 compound with the formula XCl. Which of the
Based on Table 1, which compound can conduct following elements could be X?
I Sodium, Na III Hydrogen, H
electricity when it is heated to a temperature of II Calcium, Ca IV Aluminium, Al
320 °C? C3 A I and II
A P B I and III
B Q C II and IV
C R D III and IV
D S

138

Chapter 5 Chemical Bond Chemistry Fo4rm

Paper 2

Section A

SPM SPMC lone (a) (i) The electron arrangement of neon is 2.8. Why is this element very stable? C2 [1 mark]
1. (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
formation of an ionic bond between a magnesium ion, Mg2+ and an oxide ion, O2–. C2

2+ 2–

Mg O

Figure 1.1

(i) How are a magnesium ion, Mg2+ and an oxide ion, O2- formed from their respective atoms? C1
[1 mark]
(ii) What type of attractive forces exists between magnesium ions, Mg2+ and oxide ions, O2-? C2
[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.

P Q23 35

11 17

Figure 1.2 C6 [2 marks] CHAP.
Element P reacts with element Q to form a compound.
Draw one diagram to show the bonding formed between elements P and Q. 5

Clone Table 2.1 shows the electrical conductivity and melting points of substances X, Y and Z.

2.

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 [1 mark]
(i) Substance Y [1 mark]
(ii) Substance Z

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

139

Fo4rm 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) [3 marks]

Table 2.2


3. Figure 3 shows the electron arrangements of five elements P, Q, R, S and T.

PQ R S T

Figure 3 [1 mark]
(a) Which element can form metallic bond? C2
(b) P reacts with Q. [1 mark]
(i) What type of chemical bond is formed in the reaction? C1
(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) What is the type of particle of compound formed in the reaction? C2 [1 mark]
CHAP. (ii) Write the formula of the compound formed. C4 [1 mark]
5 (d) State two differences of physical properties between compounds formed in (b) and (c). C4
[1 mark]
(e) Explain why element T exists as monoatomic gas at room temperature. C3 [1 mark]

Section B

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

P39 Q35 R12

19 17 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

Chapter 5 Chemical Bond Chemistry Fo4rm

Clone Section C

SPM 5. (a) J23 and L16  are symbols of two atoms.
8
11

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 PQ3. These letters are not the actual symbols of
the elements.

Q PQ

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.

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

Fo4rm 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

CHAP. Carbon Carbon
electrodes electrodes
5 Solid M1 Molten M1
I

Carbon Carbon
electrodes electrodes

II Solid M2 Molten M2

142

CHAPTER Redox Equilibrium

1

1.1 Oxidation and Important Learning Standards Page
Reduction • Describe redox reactions. 296
• Explain redox reaction based on the change in oxidation number. 301
1.2 Standard Electrode • Investigate displacement reaction as a redox reaction. 306
Potential
• Describe the standard electrode potential. 312
1.3 Voltaic Cell • Determine oxidising agent and reducing agent based on their value 313

1.4 Electrolytic Cell of standard electrode potentials. 315
321
1.5 Extraction of Metal • Explain redox reaction in voltaic cell. 323
from Its Ore 326
• Describe electrolysis. 335
1.6 Rusting • Describe electrolysis of molten compound. 335
• Explain factors that affect electrolysis of aqueous solution.
• Compare voltaic cell and electrolytic cell. 338
• Describe electroplating and purification of metal by electrolysis.
340
• Explain extraction of metal from its ore through electrolysis
process. 342
345
• Explain metal extraction from its ore through reduction process by
carbon.

• Describe metal corrosion process as redox reaction.
• Prevent rusting.

• 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
• Half equation / Setengah persamaan piawai
294 • Voltaic cell / Sel kimia

Increases of Decreases of A less electropositive metal will increase
oxidation number oxidation number the rate of rusting.
Loss of electron Gain of electron A more electropositive metal will
Loss of hydrogen Gain of hydrogen prevent iron from rusting.
Rusting is the corrosion of iron.
Gain of oxygen Loss of oxygen Metal is oxidised naturally with the
release of electrons to form metal ions.

Oxidation Reduction Corrosion of metal is a redox reaction.
Corrosion of metal

Via electrolysis Extraction of
aluminium
Standard Electrode Potential Redox Equilibrium Application Extraction of metal Extraction

Chemical energy → Electrical energy → in industries of iron
electrical energy chemical energy
The more positive the Purification of metals
E0 value is, the easier for
the chemical species to Electrolytic cell Electroplating of metals
undergo reduction.
Voltaic cell

Via reduction
using carbon
Molten compound Aqueous solution
The more negative the
E0 value is, the easier for Simple Daniell Products are affected by

the chemical species to cell cell
undergo oxidation.
E0 value Concentration of Types of electrodes
295 ions in electrolyte

Fo5rm Chemistry   Chapter 1 Redox Equilibrium

CHAP.

1 Activity 1.2

Aim: 4. The mixture is filtered into a test tube.
To investigate the conversion of iron(II) ion, Fe2+ to
iron(III) ion, Fe3+ and vice versa.

Materials: Test to confirm the presence of iron(II) ion, Fe2+.
0.5 mol dm–3 freshly prepared iron(II) sulphate,
FeSO4 solution, 0.5 mol dm–3 iron(III) chloride, 5. 0.2 mol dm–3 of sodium hydroxide, NaOH
FeCl3 solution, bromine water, Br2, zinc, Zn powder, solution is added drop by drop until in excess.
filter paper, 2.0 mol dm–3 sodium hydroxide, NaOH
The changes are recorded.
solution

Apparatus: Observation:
Dropper, spatula, test tube holder, Bunsen burner,
Table 1.2
filter funnel, test tube rack, measuring cylinder,
Activity Observation
test tube

Procedure: Iron(II) sulphate, The brown bromine water,
A Conversion of iron(II) ion, Fe2+ to iron(III) FeSO4 + bromine Br2 turns colourless.
water, Br2 The pale green iron(II)
ion, Fe3+ sulphate, FeSO4 solution
turns yellow.

Bromine Adding of sodium A brown precipitate is
water, Br2 hydroxide, NaOH formed. It is insoluble in
solution
Iron(II) sulphate, excess sodium hydroxide,
FeSO4 solution
NaOH solution.
Figure 1.8
1. 2 cm3 of freshly prepared iron(II) sulphate, Iron(III) chloride, The brown iron(III) chloride,
FeCl3 + zinc, Zn FeCl3 solution turns pale
FeSO4 solution is measured and pour into a powder green.
test tube.
2. By using a dropper, bromine water, Br2 is Adding of sodium A dirty green precipitate
added to the solution in the test tube drop by hydroxide, NaOH is formed. It is insoluble in
drop until no further changes are observed. solution
3. The mixture is shaken well and warm gently. excess sodium hydroxide,

NaOH solution.

Test to confirm the presence of iron(III) ion, Fe3+. Inference:
1. The pale green iron(II) sulphate, FeSO4 solution
4. 0.2 mol dm–3 of sodium hydroxide, NaOH
solution is added drop by drop until in excess. turns yellow because the iron(II) ions, Fe2+ are
The changes are recorded. oxidised to iron(III) ions, Fe3+.
2. A brown precipitate is formed when tested
B Conversion of iron(III) ion, Fe3+ to iron(II) with sodium hydroxide, NaOH solution. This
ion, Fe2+ confirms the presence of the iron(III) ions, Fe3+.
3. The brown iron(III) chloride, FeCl3 solution
1. 2 cm3 of iron(III) chloride, FeCl3 solution is turns pale green because the iron(III) ions, Fe3+
measured and pour into a test tube. are reduced to iron(II) ions, Fe2+.
4. A dirty green precipitate is formed when
2. Half spatula of zinc, Zn powder is added into tested with sodium hydroxide, NaOH solution.
the solution in the test tube. This confirms the presence of the iron(II) ions,
Fe2+.
3. The mixture is shaken well and warm gently
until no further changes. D iscussion:
1. Iron(II) ions, Fe2+ are oxidised to iron(III) ions,

Fe3+ by bromine water, Br2.

304 1.1.2

Chapter 1 Redox Equilibrium Chemistry Fo5rm

2. Bromine water, Br2 acts as the oxidising agent, 5. Iron(III) ions, Fe3+ are reduced to iron(II) ions, CHAP.
whereas iron(II) ions, Fe2+ act as the reducing Fe2+ by zinc, Zn.
agent. 1
6. Iron(III) ion, Fe3+ act as the oxidising agent
3. An iron(II) ion, Fe2+ loses an electron to form whereas zinc, Zn acts as the reducing agent.
iron(III) ion, Fe3+. Thus, iron(II) ion, Fe2+ is
oxidised to iron(III) ion, Fe3+. 7. A zinc atom loses two electrons to form zinc
ion, Zn2+. Thus, zinc atom, Zn is oxidised to
4. A bromine molecule, Br2 receives two zinc ion, Zn2+.
electrons from iron(II) ion, Fe2+ to form
bromide ions, Br–. Thus, bromine molecule, Br2 8. An iron(III) ion, Fe3+ receives an electron from
is reduced to bromide ions, Br–. zinc, Zn to form iron(II) ion, Fe2+. Thus, iron(III)
ion, Fe3+ is reduced to iron(II) ion, Fe2+.
Oxidation half equation:
Fe2+(aq) → Fe3+(aq) + e– Oxidation half equation:
Zn(s) → Zn2+(aq) + 2e–
Reduction half equation: Reduction half equation:
Br2(aq) + 2e– → 2Br–(aq) Fe3+(aq) + e– → Fe2+(aq)
Overall ionic equation (Redox reaction):
Overall ionic equation (Redox reaction): 2Fe3+(aq) + Zn(s) → 2Fe2+(aq) + Zn2+(aq)
2Fe2+(aq) + Br2(aq) → 2Fe3+(aq) + 2Br–(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
number of the metal ion are written in Roman
but some elements such as transition metals, numeral in brackets, immediately following the
carbon, nitrogen and sulphur have more than name of the anion.
one oxidation number.
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 Formula Oxidation IUPAC
numeral in brackets, immediately following the formula of of anion number nomenclature
name of the metal. compound of metal
4. Table 1.3 shows examples of metal element elements of the
which has more than one oxidation number in compound
the ionic compound.
K2MnO4 MnO42– +6 Potassium
Table 1.3 manganate(VI)

Chemical Oxidation IUPAC nomenclature KMnO4 MnO4– +7 Potassium
formula of number of of the compound manganate(VII)
compound metal elements
K2CrO4 CrO42– +6 Potassium
Cu2O +1 Copper(I) oxide chromate(VI)
CuO +2 Copper(II) oxide
K2Cr2O7 Cr2O72– +6 Potassium
dichromate(VI)
PbCl2 +2 Lead(II) chloride
PbCl4 +4 Lead(IV) chloride 7. For non-metal elements which have variable
FeSO4 +2 Iron(II) sulphate oxidation numbers, the oxidation numbers of
Fe2(SO4)3 +3 Iron(III) sulphate the non-metal element are written in Roman
numeral in brackets, immediately following the
name of the ion that contains the non-metal
element.

1.1.2 305

Fo5rm Chemistry   Chapter 1 Redox Equilibrium

CHAP. (a) Oxidation half equation
(b) Reduction half equation
1 8. Zinc, Zn strip is placed into silver nitrate, (c) Overall ionic equation
Ag(NO3)2 solution. Silver, Ag and zinc nitrate, (d) Oxidising agent
Zn(NO3)2 are formed. (e) Reducing agent
Zn(s) + 2AgNO3(aq) → Ag(s) + Zn(NO3)2(aq)

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
1. The strength of an oxidising agent and a chosen as a standard electrode.
reducing agent depends on the standard
electrode potential, E0. Platinum surface promotes oxidation of
hydrogen molecule, H2 to hydrogen ions,
2. Standard electrode potential, E0 is a H+ or reduction of hydrogen ions, H+ to
measurement of the potential for equilibrium. hydrogen gas, H2.

3. If a metal is immersed in a solution containing Atoms of platinum, Pt electrode
the metal ions, the metal atoms tend to release
electrons to form metal ions and enter the H2(g) at e– H+
solution. The electrons released accumulate on 1 atm e–
the metal, making the metal negatively charged. e– H2
Platinum, e–
4. In a short time, the metal is surrounded by Pt wire H+
positive ions. A small number of metal ions H2(g) H+
will receive electrons and become metal atoms Platinum, Pt
again. electrode H2
M(s)  Mn+ (aq) + ne–
H+

Acidic solution, Half equation of the
H+ 1.0 mol dm–3 hydrogen half-cell:
2H+(aq) + 2e–  H2(g)
5. Thus, an equilibrium is achieved between metal
M atoms and metal Mn+ ions in a half-cell.
6. Electrode potential is defined as the potential Figure 1.12 Standard hydrogen electrode half-cell

difference produced when an equilibrium is 11. Electrode potential of standard hydrogen
eTlheectreoledcet,roEd0reeferepncoe tiesnatsisailgnined as 0.00 V.
achieved between metal M and the aqueous 12. a half-cell when
solution containing metal ions, Mn+ in a half-cell.
Hydrogen gas, fHor2mcanhydarlsoogenbe dissolved in compared relative to the standard hydrogen
7. a solution to ions, H+.The electrode, SHE under standard conditions is

hydrogen ions, H+ in the solution will also called standard electrode potential, E0.
13. Standard conditions for the measurement of
receive electrons to become hydrogen gas, standard electrode potential of a half-cell:
tHhhy2ed. rsoAoglnuetnieoqgnua.sil,ibHr2iuamnd is achieved between the
the hydrogen ions, H+ in (a) Concentration of ions in aqueous solution is
1.0 mol dm-3.
(b) Gas pressure of 1 atm or 101 kPa.
(c) Temperature at 25 oC or 298 K.
H2(g)  2H+ (aq) + 2e– (d) Platinum is used as an inert electrode.

8. 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
hydrogen gas, aHp2 oatsesnhtoiawl ndiifnferFeingucereb1et.1w2e.en potential
Since, there is
9. the 1. Standard hydrogen electrode consists of a
platinum electrode immersed in a 1.0 mol
electrode that is in contact with hydrogen gas, dm−3 of strong acid solution, H+ through which
HH2+, and the solution containing hydrogen ions,
the electrode potential for hydrogen half- hydrogen gas, tHhe2 at a pressure of 1 atm is
cell is produced. bubbled. Thus, reference half reaction is as
follows:
10. It is impossible to measure the absolute value of
an electrode potential in a half-cell. Therefore, 2H+ (aq; 1M) + 2e–  H2 (g, 1 atm) E0 = 0.00 V
reference
scientists compare the value of electrode

312 1.2.1

Chapter 1 Redox Equilibrium Chemistry Fo5rm

(b) This means hydrogen ions, H+ undergo CHAP.
reduction to form hydrogen gas, Hth2e.
The standard reference half-cell is the standard Therefore hydrogen electrode acts as 1
hydrogen electrode, SHE.
cathode.
2H3O+(aq) + 2e– → H2(g) + 2H2O(l)
or simplified as
2. An unknown standard electrode potential,
vEo0ulntkaniocwnceclal ncobnesistoibntgainofeda by constructing a 2H+(aq) + 2e– →zinHc 2(agto)ms,
reference half-cell (c) On the other hand, Zn are
and another unknown half-cell, to measure the oxidised to zinc ions, Zn2+. Hence, zinc
electromotive force, e.m.f or commonly known electrode acts as the anode.
as standard cell potential, EH0+celli.s
(a) If the hydrogen ion, reduced, the Oxidation half reaction:
Zn(s) → Zn2+(aq) + 2e–
reference half-cell acts as the cathode while (d) E0 cell = E0 – E0
oxidation occurs at the unknown half-cell. cathode anode
0.76 V = 0.00 V – E00.7zin6c
E0cell = E0–.00Eca0t0huoVndken–o–wnEE00uannokndeown E0 zinc = 0.00 V – V
= = – 0.76 V
= Standard electrode potential of zinc:

Standard electrode potential of the half-cell: Zn2+ (aq) + 2e–  Zn(s) E0 = – 0.76 V
(b) EIf0unthkneownh=yd–rEo0gceelln gas, zinc
reference half-cell acts
Has2 is oxidised, the Conventionally, standard electrode potential is
the anode while written in the form of standard reduction potential.
reduction occurs at the unknown half-cell.
E0cell = EEE000cuuanntkkhnnooodwwe nn––E00a.n0o0deV Oxidising Agents and Reducing Agents Based on
= The Value of Standard Electrode Potentials
= 1. The value of standard electrode potential, E0

Standard electrode potential of the half-cell: gives a direct measure of the ease of a chemical
E0 = E0 species to be oxidised or reduced.
unknown cell

Example:
Figure 1.13 shows a voltaic cell consisting of a
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 e– monoatomic ion, polyatomic ion, or radical.
e– 0.76 V
2. We can compare the strength in oxidation and
Anode (–) Cathode (+) reduction of elements or ions based on the value
Zn H2(g) of E0 for the half reactions involved.
(a) The more positive the E0 value is, the easier
e– Zn2+ 1 M Zn2+ Pt H2 e– for the chemical species on the left side of the
e– 1 M H3O+ 2H2O half equation to undergo reduction.
(b) The more negative the E0 value is, the easier
Zn 2H3O+ for the chemical species on the right side of
the half equation to undergo oxidation.
Salt bridge Example:

Figure 1.13 Figure 1.14 shows the standard electrode
potentials of three ions.
(a) There are a few observations that can be seen
in the voltaic cell: Strength as
(i) Gas bubbles are produced around the oxidising
platinum, Pt electrode.
(ii) Zinc, Zn electrode becomes thinner. agent
(iii) Voltmeter reading shows 0.76 V. Strength as
reducing

agent
Zn2+ (aq) + 2e– ⇌ Zn (s) E0 = –0.76 V
C2Hu2++((aaqq))++22ee––⇌⇌HC2u(g(s)) E0 = 0.00 V
E0 = 0.34 V

Figure 1.14

1.2.1 1.2.2 313

Fo5rm Chemistry   Chapter 1 Redox Equilibrium

CHAP. (a) Referring to the E0 value, copper(II) ions, 4. In short, by considering the E0 value, we can
1 Cu2+ are most easily reduced compared to predict the followings:
hydrogen ions, H+. Zinc ions, Zn2+ is the (a) Strength of the oxidising agent and the
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 Cu2+ > H+ > Zn2+. that undergoes oxidation or reduction.
(c) Negative E0 value indicates that zinc atom,
Zn is more easily oxidised, followed by
hbyeidnrgogmenostgdasif,fiHcu2l,t and copper atom, Cu
to be oxidised. Table of Standard Electrode
(d) Thus, the strength of reducing agents are in Potential, E0
the order of Zn > H2 > Cu. http://bit.ly/2NYV9lu

Example 3

The following are the E0 values of zinc and copper.

Zn2+(aq) + 2e–  Zn(s)  E0 = – 0.76 V
Cu2+(aq) + 2e–  Cu(s)  E0 = + 0.34 V

Based on the E0 values:
(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.

Solution The more negative the E0 value is, the easier for the chemical

(a) Substance oxidised: Zinc, Zn species on the right of the half equation to undergo oxidation.
Substance reduced: Copper(II) ion, Cu2+ Zn2+(aq) + 2e– ⇌ Zn(s) E0 = – 0.76 V

(b) Oxidising agent: Copper(II) ion, Cu2+ The more positive the E0 value is, the easier for the chemical

Reducing agent: Zinc atom, Zn species on the left of the half equation to undergo reduction.

(c) Oxidation half equation: Zn(s) → Zn2+(aq) + 2e– Cu2+(aq) + 2e– ⇌ Cu(s) E0 = + 0.34 V

Reduction half equation: Cu2+(aq) + 2e– → Cu(s)
(d) Ionic equation: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Example 4

A displacement reaction using copper, Cu and silver nitrate, AgNO3 solution Displacement
is carried out in the laboratory. Scan the QR code to observe the reaction reaction
occurred. The following are the E0 value of silver ion and copper ion. https://bit.ly/3gyad3z

Cu2+(aq) + 2e–  Cu(s)  E0 = + 0.34 V
Ag+(aq) + e–  Ag(s)   E0 = + 0.80 V

(a) State two observations that can be obtained in the displacement reaction.
(b) Based on the E0 values:
(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 E0 value of silver is more positive. So,
silver ion, Ag+ is reduced.
(a) Silvery grey solid is deposited . Colourless solution turns blue
(b) (i) Silver ion , Ag+ is reduced. So, it is an oxidising agent. E0 value of copper is less positive.
Copper atom, Cu is oxidised. So, it is a reducing agent. Hence, copper atom is oxidised.
(ii) Oxidation half equation: Cu(s) → Cu2+(aq) + 2e–
Reduction half equation: Ag+(aq) + e– → Ag(s)
(iii) Ionic equation: Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

314 1.2.2

Chapter 1 Redox Equilibrium Chemistry Fo5rm

CHAP.

1.2 1

1. State the meaning of standard electrode potential, E0. C1

2. What are the standard conditions for the measurement of standard electrode potential of a half-cell?
C1

3. Referring to the standard electrode potential, E0 given below, answer the following questions. C4

Ag+(aq) + e–  Ag(s) E0 = +0.80 V
Sn2+(aq) + 2e–  Sn(s) E0 = –0.14 V

(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
magnesium ion, Mg2+ by losing two
The Redox Reactions in Voltaic Cell electrons. Thus, magnesium atom, Mg
atom acts as reducing agent.
1. A voltaic cell is an electrochemical cell that
converts chemical energy to electrical energy.
2. A simple voltaic cell consists of two different Oxidation half reaction:

metals that are connected to a bulb, voltmeter Mg(p) Mg2+(ak) + 2e–

or galvanometer using connecting wires and Magnesium ion, Mg2+ produced then
placed in an electrolyte. emnatgernsessiuumlp,hMurgiceleacctirdo,deHb2SeOco4m. eTshtehrienfnoreer,.
3. Potential difference between these two metals
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, H2SO4 as on the surface of copper, Cu electrode.
electrolyte. Thus, hydrogen ion, H+ acts as oxidising
agent.
Magnesium, Mg Current flow Copper, Cu H+ + e– → H
electrode B Wire electrode
Two hydrogen atom, H then combine
e– e– together to form one molecule of hydrogen
gas, H2.
Bulb

A C H + H → H2
Therefore, bubbles of hydrogen gas, CHu2
e– H+ + e– are released at the surface of copper,
e– H+ + e– H2 Hydrogen gas,
H2 bubbles
Mg Mg2+ electrode.

Reduction half reaction:

Sulphuric acid, H2SO4 2H+(ak) + 2e– → H2(g)
Figure 1.15 A simple voltaic cell

Mg2+(aq) + 2e– ⇌ Mg(s) E0 = –2.38 V Mg(p) → Mg2+ + 2e– 2H+ + 2e– → H2
Cu2+(aq) + 2e– ⇌ Cu(s) E0 = +0.34 V
mE0amgagnneesisuimuvma,luMe gisismaosrteronneggaetrivreed. Tuhcienrgefaogreen, t Electrons added to the Electrons added to the
compared to copper, Cu. right side of the equation left side of the equation
1.3.1 means loss of electrons. means gain of electrons.

315

Fo5rm Chemistry   Chapter 1 Redox Equilibrium

1CHAP. 5. Redox reaction occurred in the simple chemical 7. The relative charge on the electrode depends
cell can be represented by the following overall on the source of electrons and the direction of
ionic equation: flow of electrons.
(a) Since the E0 value of the magnesium is
Oxidation half equation: Mg(s) → Mg2+(aq) + 2e– more negative, magnesium atom, Mg tends

Reduction half equation: 2H+(aq) + 2e– → H2(g) to release the electrons. Electrons released
Overall ionic equation: Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) enters the magnesium, Mg electrode.
(b) Thus, magnesium, Mg acts as the negative
terminal because magnesium electrode
6. The area where oxidation occurs is called anode relatively contains more negative charges
and the area where reduction occurs is called (electrons) compare to copper, Cu
cathode. Thus, the magnesium, Mg electrode
acts as an anode while the copper, Cu electrode electrode.
acts as a cathode in this simple chemical cell. (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.
Useful Acronym (d) Electrons that reach the copper, Cu

electrode are given to the hydrogen ion,
H+. Copper acts as the positive terminal
because copper, Cu electrode relatively
loses negative charges (electrons).
(e) Hreysudlrtosgeningasp, oHla2 roinsatthioencopapnedr, Cu surface
Red cat An ox prevents

Reduction; cathode Oxidation; anode further reduction of hydrogen ions,
H+. Hence, simple chemical cell can only
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.

Activity 1.5 3. Copper, Cu plate and iron nail are connected
to a voltmeter using connecting wires and
Aim: dipped into the sulphuric acid, H2SO4 as shown
To show the production of electricity through redox in Figure 1.16.
reaction in a simple voltaic cell.
Galvanometer
Materials: G
Zinc, Zn plate, copper, Cu plate, iron, Fe nail, 1.0
mol dm-3 sulphuric acid, H2SO4, sandpaper Iron, Fe Copper,
nail Cu plate
Apparatus:
Connecting wires with crocodile clips, 250 cm3 Sulphuric acid,
beaker, voltmeter H2SO4

Procedure: Figure 1.16
1. A copper, Cu plate and iron, Fe nail are cleaned 4. The voltmeter reading and the observations at

with sandpaper. the electrodes are recorded.

2. 150 cm3 of 1.0 mol dm-3 sulphuric acid, H2SO4 is
poured into a beaker.

316 1.3.1

Chapter 1 Redox Equilibrium Chemistry Fo5rm

Results: 2. Reduction half equation: CHAP.
2H+(aq) + 2e– → H2(g)
Table 1.10 1
3. Overall ionic equation:
Voltmeter Observation Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g)
reading
Pair of (V) Negative Positive 4. Iron atom, Fe is oxidised to iron(II) ion, Fe2+ by
metals terminal terminal losing two electrons. Oxidation occurs at the
(anode) (cathode) iron, Fe nail and iron, Fe acts as reducing agent.

Iron, Fe 0.8 Iron, Fe nail Gas bubbles 5. Hydrogen ions, H+ are reduced to form hydrogen
gas, H2 by gaining electrons. Reduction occurs
nail + becomes are released at the copper, Cu plate. Hydrogen ion, H+ acts
as oxidising agent.
copper, thinner. around
6. Electrons move from iron, Fe nail to copper, Cu
Cu plate copper, Cu plate through wire. In other words, electrons
flow from reducing agent to the oxidising
plate. agent. Therefore, iron, Fe nail is an anode while
copper, Cu plate is the cathode of the voltaic
Inference: cell.
1. Iron, Fe nail dissolves and corrodes to form
7. Electrons that flow through the wire from iron,
iron(II) ion, Fe2+. Fe to copper, Cu produce electric current.
2. Hydrogen gas, H2 is produced.
8. Reaction in the simple chemical cell is a redox
Conclusion: reaction because oxidation and reduction
Hypothesis is accepted. Two different metals that occur at the same time.
are dipped into an electrolyte and connected by
wire will produce electricity. 9. If two copper, Cu electrodes are used, there
is no potential difference between electrodes.
Discussion: When there is no electron flow, electric current
1. Oxidation half equation: is not generated and voltmeter reading will
Fe(s) → Fe2+(aq) + 2e– show 0 V.

Daniell Cell 3. cEo0cpopppeerr(iIsI) more positive. It indicates that the
1. Figure 1.17 shows a Daniell cell. A Daniell cell ion, Cu2+ is an oxidising agent.
Copper plate is the cathode where reduction
or a zinc-copper cell is an example of a voltaic process occurs.
cell.
A Zinc, Zn plate becomes thinner as zinc,
A Voltmeter e– 2e– are received for Zn corrodes and dissolves in zinc sulphate,
e– every copper ion,
Cu2+ that is ZZinnScOa4tosomlu, tZionni.s oxidised to zinc ion, Zn2+
2e– are released for Anode Cathode reduced by losing two electrons.
every zinc atom, Zn Zn (–) (+)
SO42– Na+ Zn(s) → Zn2+(aq) + 2e–
that is oxidised Cu B Potential difference between the two metal

e– B C Cu2+ plates causes the flow of electrons from
Zn2+ the anode (zinc) to the cathode (copper)
Zn2+ Cu2+ e– through wire. Therefore, electric current is
e– generated.
Zn Cu C Brown solid is deposited at the copper, Cu
plate, making copper, Cu plate becomes
Salt bridge, Na2SO4 thicker. Copper(II) ion, Cu2+ is reduced
Figure1.17 Daniel cell to the copper atom, Cu by gaining two
electrons.
Zn2+(aq) + 2e–  Zn(s) E0 = –0.76 V Cu2+(aq) + 2e‒ → Cu(s)
Cu2+(aq) + 2e–  Cu(s) E0 = +0.34 V The intensity of blue solution decreases
as the concentration of copper(II) ion, Cu2+
2. EZ0nzincisisa more negative. It indicates that zinc, decreases.
stronger reducing agent. Thus, zinc,
Zn plate is an anode where oxidation process
occurs.

1.3.1 317

Fo5rm Chemistry   Chapter 1 Redox Equilibrium

CHAP. 4. Oxidation process occurred at the anode and 8. aTchteuaEll0yceltlhoebvtoalitnaegde through the calculation is
produced in the Daniell cell
1 reduction process at the cathode causes zinc based on the potential difference between two
(anode) becomes relatively negative charge electrodes.
(electrons) as compared to copper (cathode).
Therefore, zinc, Zn is the negative terminal of 9. The functions of salt bridge:
the cell while copper, Cu is the positive terminal (a) Complete the circuit by allowing the
of the cell. movement of ions.
5. Redox reaction that takes place in Daniell cell (b) Separate two different electrolytes
can be represented by an overall ionic equation: 10. Initially, oxidation of half-cell is neutral with
zinc ions, Zn2+ and sulphate ions, zSinOc42i−onisn,
Negative terminal: Zn(s) → Zn2+(aq) + 2e– the solution. When more and more
Positive terminal: Cu2+(aq) + 2e– → Cu(s)
Overall ionic Zn2+ enter the solution, the solution becomes
equation: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) positively charged.
11. Similarly, reduction of half-cell is neutral with
copper(II) ions, TChue2+soalnudtiosnulwphilaltebeionnesg, atSiOve4l2y−
6. From the ionic equation, we can write cell in the solution.
notation for Daniell cell as the following: charged when more and more copper(II) ions,

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Cu2+ leave the solution to form copper atom, Cu.

7. Standard cell potential, E0 for Daniell cell can be 12. When two half-cells are charged, the voltaic
calculated using the following formula: cell will not function. Therefore, a salt bridge is
needed to connect the two half-cells.
E0 = E0 – E0 13. Based on Figure 1.17, the sodium ions, Na+ from
cell cathode anode

With zinc, Zn as the anode and copper, Cu as the salt bridge move to copper(II) sulphate,
the cathode.
hCmwauhosSevmrOee4oatsroseosulzpuilopntishcoitanistveuweliphochhnicaasht,regSi,esO.Zl4a2n−cSkfOroo4mf spotohluseittisiovanelt charge
Anode: Zn2+(aq) + 2e–  Zn(s) E0 = –0.76 V bridge
Cathode: Cu2+(aq) + 2e–  Cu(s) E0 = +0.34 V which
E 0cell == (E+0c0at.h3o4de V– )E–0an(o–de0.76 V)
= 1.10 V 14. Therefore, the solutions in both half-cells are
maintained in a neutral state and the voltaic cell
continues to function well.

Eis0zmincowrehiecahsiilsymoxoirdeisneedgaantidveacintsdiacsataens that zinc, Zn
anode.

1.1 Variables:
(a) Manipulated: Different pairs of metals
Aim: (b) Responding: Voltage of the cell
To determine the voltage of a voltaic cell using (c) Fixed: Volume and concentration of electrolytes
different pairs of metals.
Materials:
Problem statement: Copper, Cu plate, iron, Fe nail, zinc, Zn plate,
How do different pairs of metals in electrolyte affect magnesium, Mg ribbon, sandpaper, 1.0 mol dm-3
the voltage of a voltaic cell? iron(II) sulphate, Fe2SO4 solution, 1.0 mol dm-3
copper(II) sulphate, CuSO4 solution, 1.0 mol
Hypothesis: dm-3 zinc sulphate, ZnSO4 solution, 1.0 mol dm-3
Pair of metals with greater difference in the value magnesium sulphate, MgSO4 solution
of standard electrode potential, E0 will produce
greater voltage. 1.3.1

318

Chapter 1 Redox Equilibrium Chemistry Fo5rm

Apparatus: CHAP.
Voltmeter, 250 cm3 beaker, connecting wires with crocodile clips, porous pot
1
Procedure:
1. All metal plates are cleaned with sandpaper.
2. A porous pot is filled with zinc sulphate, ZnSO4 solution until it is two-third full.
3. A zinc, Zn plate is dipped into zinc sulphate, ZnSO4 solution.
4. A beaker is filled with copper(II) sulphate, CuSO4 solution until it is half full.
5. The copper, Cu plate is dipped into copper(II) sulphate, CuSO4 solution.
6. The porous pot is placed into the beaker.
7. The circuit is completed by connecting the metals to a voltmeter as shown in Figure 1.18.

Voltmeter
V

Copper, Zinc,
Cu plate Zn plate

Copper(II) Porous pot
sulphate, that contains
CuSO4 solution zinc sulphate,
ZnSO4 solution

Figure 1.18
8. The reading of the voltmeter is recorded.
9. Steps 2 to 8 are repeated by replacing zinc, Zn plate and zinc sulphate, ZnSO4 solution in the porous pot

with;

(a) iron nail, Fe and iron(II) sulphate, FeSO4 solution.
(b) magnesium, Mg ribbon and magnesium sulphate, MgSO4 solution.

Result:

Table 1.11

Set Pair of metals Voltage (V) Observations at the electrodes

Negative (anode) Positive (catnode)

I Zinc, Zn plate + 1.0 Zinc, Zn plate Brown solid is deposited on
copper, Cu plate becomes thinner. the copper, Cu plate.

II Iron, Fe nail + 0.7 Iron, Fe nail Brown solid is deposited on
copper, Cu plate becomes thinner. the copper, Cu plate.

III Magnesium, Mg Magnesium, Mg Brown solid is deposited on
ribbon + copper, 2.7 ribbon becomes the copper, Cu plate.
Cu plate
thinner.

Inference: Discussion:
1. Zinc, Zn, iron, Fe and magnesium, Mg corrode 1. Below are the value of the standard electrode

and dissolve into their salt solutions. potential, E0 of metals used in the experiemnt.
2. Copper, Cu metal is deposited on the copper,
Mg2+(aq) + 2e–  Mg(s) E0 = –2.38 V
Cu plate.
Zn2+(aq) + 2e–  Z(s) E0 = – 0.76 V
Conclusion:
The hypothesis is accepted. Pairs of metal with Fe2+(aq) + 2e–  Fe(s) E0 = – 0.44 V
greater difference in the value of standard electrode
potential, E0 will produce greater voltage. 2H+(aq) + 2e–  H2(g) E0 = 0.00 V

Cu2+(aq) + 2e–  Cu(s) E0 = +0.34 V

1.3.1 319

Chapter 1 Redox Equilibrium Chemistry Fo5rm

iron, Fe and are accepted by positive ions CHAP.
such as hydrogen ions, H+ or oxygen, O2 1
and water, H2O in the electrolyte.
Magnesium, Mg and iron, Fe form oxide coating
which is porous , not tightly packed and weak.
This oxide coating is easily peeled off and cannot
protect the metals from corrosion.

Electrochemical corrosion Cars rust more rapidly in a country where salt is
used to melt the snow and ice on the road.
1. Electrochemical corrosion is the process Dissolved salts in the water droplet greatly
by which a metal is corroded through loss of increase the conductivity of the electrolyte. Thus
electron to form a cation with the presence of increasing the rate of corrosion.
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
(–) (+) Electrolyte electropositivity compared to magnesium, Mg
Mg Fe and iron, Fe.

e– Mg e–
e– Fe e– The rate of

Figure 1.38 Electrochemical transferring of
(a) Figure 1.38 shows that magnesium, Mg is in series electrons from Mg to
Cu is higher.
contact with iron, Fe and the two metals are
put into an electrolyte. Cu
(b) Magnesium, Mg is more electropositive
than iron, Fe. Thus, magnesium, Mg is Figure 1.39 The difference in electropositivity
corroded (loss of electrons) and iron, Fe is between magnesium, ferum and copper
prevented from corrosion.
The greater the difference in electropositivity of
Mg(s) → Mg2+(aq) + 2e– two metals, the higher the rate of corrosion of the
more electropositive metal.
(c) Magnesium, Mg acts as a negative terminal,
whereas iron, Fe acts as a positive terminal. Try Question 1 in Formative Zone 1.6
Electrons flow from magnesium, Mg to

Activity 1.8 A Burning of copper, Cu strip

Aim: Procedure:
To study the corrosion of copper and iron. 1. Use a pair of tongs to hold a copper, Cu strip.

Materials: 2. Heat the copper, Cu strip using a Bunsen
Copper, Cu strip, iron, Fe wool, oxygen gas, O2 burner for a few minutes as shown in Figure
1.40.
Apparatus:
Bunsen burner, combustion spoon, gas jar with 3. All observations are recorded.
lids, a pair of tongs

1.6.1 343

Fo5rm Chemistry   Chapter 1 Redox Equilibrium

CHAP. Results:

1 Copper, Cu Table 1.27
strip
Tongs Bunsen burner A Black layer formed on the brown copper strip.

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, Fe2O3 is formed.

Procedure: Discussion:
1. A small amount of iron, Fe wool is placed in a 1. Copper, Cu is oxidised to copper(II) oxide,

combustion spoon. CuO which is black in colour.
2. The iron, Fe wool is heated until it starts to
2Cu(s) + O2(g) → 2CuO(s)
burn. Cu → Cu2+ + 2e–
3. The burning iron, Fe wool is quickly lowered
Copper atom, Cu releases two electrons
into a gas jar filled with oxygen gas, O2 as to form copper(II) ion, Cu2+. Copper, Cu
shown in Figure 1.41. undergoes corrosion.
4. All observations are recorded.
2. Iron, Fe is oxidised to iron(III) oxide, FeO
Combustion which is reddish-brown in colour.
spoon
Gas jar 4Fe(s) + 3O2(g) → 2Fe2O3(s)
Fe → Fe3+ + 3e–
Oxygen gas, O2
Iron, Fe wool Iron atom, Fe releases three electrons to form
iron(III) ion, Fe3+. Iron, Fe undergoes corrosion.
Figure 1.41

• 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

faonrdmosxaygreedn-,bOro2 winnthheydprraetseednmceeotaflwoaxtideer,,H(F2eO2Oo3r.xmHo2Ois)t,ucroe.mWmhoennlyiroknn,oFwencaosrrroudset.s, it

• 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 Formation of
finally to blue green. This oxide laanyderp, irsocteaclltesdthpeatuinnad,eCrulyCinOg3.Ccoup(OpHer)2lafiyrmerlsyfraodmhefruersthtoer patina on outdoor
the outer surface of the copper
corrosion. This is the reason why copper, Cu is used on roofs, gutter work and outdoor sculpture
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
ammonia compounds.

344 1.6.1

Chapter 1 Redox Equilibrium Chemistry Fo5rm

Iron Rusting as a Redox Reaction (e) Redox reaction equation: CHAP.
Anode: 2Fe(s) → 2Fe2+(aq) + 4e–
1. Rusting is a metal corrosion that occurs on iron Cathode: O2(g) + 2H2O(l) + 4e– → 4OH–(aq) 1
due to redox reaction. Rusting of iron requires 2Fe(s) + O2(g) + 2H2O(l) → 2Fe(OH)2(s)
the presence oxygen (air) and water.
Combination of 2Fe2+ + 4OH–
2. Figure 1.42 shows the mechanism of rusting
of iron. The surface of the iron and a water (f) Ion(II) hydroxide, byFe(oOxHyg)e2 n formed is
droplet constitute a simple chemical cell in further oxidised to form
which different regions of the surface of the iron hydrated irsiooronlin(dII(IsI)uIbIo)sxtiaodnxeci,deFee,k2OnFoe3.w2xOHn3.2xOaHs. 2rOusits,
act as anode (negative terminal) and cathode (g) Hydrated
(positive terminal) while the water droplet act a brown
as the electrolyte.

Water droplet Rust (Fe2)3.xH2O)

whereby the value of x varies.

OH– Fe(OH)2(s) Oxidation Fe2O3. xH2O(s)
O2 O2

Fe2+

e–

Cathode (+) Anode (–) Cathode (+) • IIihrrryooodnnnr(((aIIIIIItI)I))ehdhhyyydiddrrorrooonxxx(iIidiIddIe)ee,o,,FxFFeieed(((OeOO,HHHF))e)2332.iOsisf3tixrhHset2nOo.xdiedcisoemdptooses to
•
Iron metal

Figure 1.42 Mechanism of rusting of iron Other explanation for the rusting of iron:
3. In the mechanism of rusting of iron: Anode: Fe(s) → Fe2+(aq) + 2e–
(a) Iron surface in the centre of a water droplet Cathode: O2(g) + 4H+(aq) + 4e– → 2H2O(l)

acts as the anode (negative terminal). Iron The reaction of atmospheric CO2 with water forms H+
surface at the edge of the water droplet and HCO3−
acts as the cathode (positive terminal).
dTihsesoclvoenscaetntthraetieodngeoof fotxhyegwenatgearsd, rOop2letht aist The Fe2+ ions produced in the initial reaction are
higher than at the centre. then oxidised by atmospheric oxygen to produce
(b) At the anode, iron atoms lose electrons and the insoluble hydrated oxide containing Fe3+, as
undergoes oxidation to form iron(II) ions, represented in the following equation:
Fe2+. 4Fe2+(aq) + O2(g) + (2+4x)H2O(2l)F→e2O3.xH2O + 4H+(aq)

Oxidation half equation: Preventing Rusting
Fe(s) → Fe2+(aq) + 2e–
Ferum(II) ions, Fe2+ dissolve in water. 1. When iron is in contact with a more
(c) Electrons flow through the iron metal to the electropositive metal, rusting of iron is
edge of the water droplet (cathode) and are prevented. The more electropositive metal will
received by oxygen and water molecules to corrode.
form hydroxide ions, OH–.
Example:
Reduction half equation: If iron, Fe nail is wrapped with a magnesium,
O2(g) + 2H2O(l) + 4e– → 4OH−(aq)
(d) The iron(II) ions, Fe2+ combine with Mg ribbon and put into a test tube filled with
water, magnesium, Mg will corrode and iron, Fe
hydroxide ions, OH− to form iron(II) nail will be prevented from rusting.
hydroxide, Fe(OH)2.
Fe2+(aq) + 2OH−(aq) → Fe(OH)2(s)

1.6.1 1.6.2 345

Fo5rm Chemistry   Chapter 1 Redox Equilibrium

CHAP. Electrochemical series strip and put into a test tube filled with water,
Mg iron, Fe will corrode while copper, Cu will be
1 prevented from rusting.
Iron, Fe nail
Electrochemical series
e–
Magnesium,    Fe Iron,
Mg ribbon Fe nail
Copper, Fe
Figure 1.43 Cu strip
e–
(a) Magnesium, Mg is more electropositive   Cu
than iron, Fe.
Figure 1.44
(b) Magnesium, Mg is corroded to form (a) Iron, Fe is more electropositive than
magnesium ions, Mg2+.
copper, Cu. Iron, Fe is corroded to form
Mg(s) → Mg2+(aq) + 2e– iron(II) ions, Fe2+.

(c) Iron(II) ions, Fe2+ are not present. Thus, Fe(s) → Fe2+(aq) + 2e–
iron, Fe is prevented from rusting.
(b) The presence of copper, Cu increases the
2. When iron is in contact with a less rate of the formation of iron(II) ions, Fe2+.
electropositive metal, rusting of iron is speeded As a result, the rusting of iron, Fe is speeded
up. up.

Example:
If iron, Fe nail is wrapped with a copper, Cu

1.7 are cleaned with sandpaper.
2. One clean iron, Fe nail is placed in test tube A.
Aim: 3. The other four iron nails are coiled with
To investigate the effects of other metals on the
rusting of iron. 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:
When a more electropositive metal comes in AB C
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.
Agar solution + potassium hexacyanoferrate(III)
Variables: + phenolphthalein
(a) Manipulated: Different types of metals
(b) Responding: The presence of blue colouration/ DE

Rusting of iron Iron nail + Iron nail +
(c) Fixed: Iron nails, hot agar solution copper tin

Materials: Agar solution + potassium hexacyanoferrate(III)
Iron, Fe nails, magnesium, Mg ribbon, copper, Cu + phenolphthalein
strip, zinc, Zn strip, tin, Sn strip, hot agar solution, Figure 1.45
potassium hexacyanoferrate(III), K3Fe(CN)6 solution,
phenolphthalein indicator, sandpaper 1.6.2

Apparatus:
Test tube, test tube rack

Procedure:
1. Five iron, Fe nails, magnesium, Mg ribbon,

copper, Cu strip, zinc, Zn strip and tin, Sn strip

346

Chapter 1 Redox Equilibrium Chemistry Fo5rm

4. A hot agar mixture which contains of potassium hexacyanoferrate(III) solution and phenolphthalein are CHAP.
poured into each test tube to completely immerse all the iron nails.
1
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 BC DE

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 Inference

tube metals blue colour pink colour

A Fe only Low None Iron(II) ions, Fe2+ presence. Iron nail undergoes
rusting.

B Fe + Mg None Very high Iron(II) ions, Fe2+ are not presence. No rusting occurs.
Very high concentration of hydroxide ions, OH–.

C Fe + Zn None High Iron(II) ions, Fe2+ are not presence. No rusting occurs.
Hydroxide ions, OH– are present.

D Fe + Cu Very high None Very high concentration of iron(II) ions, Fe2+. Iron nail
rust at the highest rate.

E Fe + Sn High None Very high concentration of iron(II) ions, Fe2+. Iron nail
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
electropositive metals prevent the rusting of iron. is transparent and it slows down the diffusion
The less electropositive metals increase the rate of
rusting of iron. process. Agar solution can also be used to trap

Discussion: if gas bubbles produced in the reaction.
1. Potassium hexacyanoferrate(III), K3Fe(CN)6 4. Test tube A:
(a) Test tube A is used as a control to compare
solution is added to detect the presence
of iron(II) ions, Fe2+. When iron(II) ions, Fe2+ the effect of other metals in contact on the
presence, a dark blue colour formed. The
more iron(II) ions, Fe2+ formed, the higher the rusting of iron.
intensity of the dark blue colour formed.
2. Phenolphthalein is added to detect the (b) In the presence of water and oxygen, iron
presence of hydroxide ions, OH–. The presence
of hydroxide ions, OH– increases the alkalinity nail rusts a little. The iron is oxidised to
of the solution and gives pink colour to the
solution. iron(II) ions, Fe2+.
3. Agar solution is used to enable us to see the Fe(s) → Fe2+(aq) + 2e–

The presence of iron(II) ions, Fe2+ give

a dark blue colour with potassium

hexacyanoferrate(III), K3Fe(CN)6 solution.
(c) The oxygen in the solution together with

the water is reduced to hydroxide ions,

OH–. O2(g) + 2H2O(l) + 4e– → 4OH–(aq)



1.6.2 347

Fo5rm Chemistry   Chapter 1 Redox Equilibrium

1CHAP. (v) Write the ionic equation to represent the reaction that take place in test tube P. C3 [1 mark]
(b) 1,1,1-trichloromethane is added to test tubes P and Q. The mixture in both test tubes are shaken gently.
(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
LO2N.E Figure 2.1 shows the electrolysis of potassium chloride, KCl solution using carbon electrodes.
PMC PMC
PM CS Hydrogen Chlorine gas
S gas, H2 1.0 mol dm–3
Carbon potassium chloride,
S electrode P KCl solution
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:

2OKCHl+222+++Oe42+–He–2+e+–K42 eC– Hl– 2 +2H2O2OH – EEEE0000 = +1.36 V
= +1.23 V
= –0.83 V
= –2.93 V

(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]

LONE V

(c) Figure 2.2 shows a voltaic cell. Given that E0 value Metal Q Copper,
of Q is more positive compared to copper, Cu. Solution R Cu metal
(i) State the positive and negative terminal of Copper(II) nitrate,
the cell. C2 Cu(NO3)2 solution
(ii) Suggest metal Q and suitable solution to be
solution R. C3
[4 marks]

Figure 2.2

Section C Voltmeter
L O3N.E (a) Two different metals are placed in a lime. The circuit is
Copper,
completed by connecting the metals to a voltmeter as shown Cu coin
in Figure 3.1. What happens to the iron nail? Write a half Lime
equation for the reaction. C3 [2 marks] Figure 3.1

Iron,
Fe nail

354

Chapter 1 Redox Equilibrium Chemistry Fo5rm

(b) Figure 3.2 shows two types of cells, A and B. CHAP.

V 1

Silver, Zinc, Silver,
Ag plates Zn plate Ag plate

Cell A Silver Cell B
nitrate,
sAogluNtiOon3

Figure 3.2
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

Fo5rm Chemistry   Chapter 1 Redox Equilibrium

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

II – Strip M1 + Solution L2

I II – Strip M2 + Solution L1

Table 1
[**Refer to the Simulation experiments and sample results to understand how to fill in Table 1.]

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

Strip M1 Colourless Shiny grey Colourless
solution L1 solid solution
Experiment II

Strip M1 Colourless Brown Colourless
solution L2 solid solution
Experiment III

Colourless Shiny grey Blue solution
solution L1 solid
Strip M2

356

SPM MODEL PAPER

Paper 1

Answer all questions. [40 marks]

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
or molecular scale. • Brown colour and soft solid.
C The study the importance of food additives • Reacts with water to produce alkaline
in food processing industry and the
evolution of food processing technology. solution.
D The development and the application of • Burn in oxygen to produce a white solid.
products or equipment, and a system to Which element has the above characteristics?
conserve the environment.
A D
2. Figure 1 shows two situations occur when B C
sunlight is shining on the glass X.

SPM MODEL PAPER Darker in 6. Which statement explains the effective collision?
sunlight A The collision which takes place after a
reaction.
Glass X B The collision which takes place before a
Figure 1 reaction.
C The collision that causes a reaction.
What is the type of glass X? D The collision produces less activation
A Fused glass energy.
B Soda-lime glass
C Borosilicate glass 7. Figure 2 shows a glass cookware that usually
D Photochromic glass used in the kitchen.

3. The following statement refers to an element in Figure 2
the Periodic Table of Elements.
• Located in Period 3 of the Periodic Table of Which substance is added to the glass to make it
Elements suitable for making the cookware?
• Reacts with water to produce acidic A Lead(II) oxide, PbO
solution and bleaching agent
• Reacts with iron wool to produce a brown
solid

Which of the following shows the electron B ANBolaurtrominuinmoixucimdaerb,ooBxin2dOaet3,e,ANl2aO2C3 O3
arrangement of the element? C
A 2.8.4 C 2.8.7 D
B 2.8.5 D 2.8.8
8. Atom W has 4 neutrons and a nucleon number of
4. iiWsonatht,eeHrt+ymptooeleofocfurtmlhee, hHcyhd2eOrmoxcicooanmliubbimonnedisofnwo,ritmHh3eOhdy?+d. rWogheant 7. Which of the following is the correct symbol
for atom W?
A Dative bond C Metallic bond A 74 W C 34 W
B 74 W D 73 W
B Ionic bond D Hydrogen bond

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 Fe2O3 C
BC
compound. Fe2O3
IV Conduct electricity in solid state. C Fe CO2
Fe
A I and III D CO2
B I and IV
C II and III 20. The equation below shows the reaction between
D III and IV excess zinc carbonate, ZnCO3 with dilute
17. Figure 6 shows the reaction between calcium hydrochloric acid, HCl.
CcaHrb3ConOaOteH, .CaCO3 and glacial ethanoic acid,
ZnCO3(s) + 2HCl(aq) → CaCl2(aq) + COH2(2gO) (+l)
Glacial ethanoic acid,
CH3COOH

Limewater The changes of the quantity of reactants and
products are recorded with time until the
reaction is completed. Which graph shows the
correct changes?
A Mass of zinc carbonate, ZnCO3 (g)

SPM MODEL PAPER Calcium carbonate, CaCO3

Figure 6
No changes are observed after the reaction. What
should be done in order to make the limewater
cloudy? Time (s)
A Heat the mixture.
B Add water to the mixture. B Concentration of calcium chloride, CaCl2
C Substitute calcium carbonate, CaCO3 with
magnesium, Mg powder. solution (mol dm–3)

D Change ccaalrcbiounmatcea, rCbaoCnaOte3,pCowaCdOer3. chips to
calcium

18. Bronze is an alloy which contains copper and tin Time (s)
atoms. In an activity, Shahriza found that bronze
is harder than pure copper. Which statement C Concentration of hydrochloric acid, HCl (mol dm–3)
explains the situation above?
A Tin atom makes strong bonds between the Time (s)
pure copper atom.
B Tin atom fills in all the empty spaces D Volume of carbon dioxide, CO2 (cm3)
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.

19. The following equation represents a redox
reaction.
2Fe2O3 + 3C → 4Fe + 3CO2

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
Magnesium, Silver, Ag plate precipitate
Mg plate
Nitric naictirda,teH, ANgON3 Ofo3llsoowluetdiownith
Sodium chloride, silver

NaCl solution White
Figure 11 precipitate

Which half equations represent the reactions at Figure 13
the positive terminal and the negative terminal of
the cell?

Positive terminal Negative terminal Which of the following is most likely to be
solution J?
A 2H+ + 2e− → H2 Ag → Ag+ + e− A Iron(II) csisouhudlllpoipdhrhieada,tteeFe,,e,FFlP2eebCSSOlO2 4 4
B Ag+ + e− → Ag Mg → Mg2+ + 2e− B Lead(II)
C Iron(II)
C 2H+ + e− → H2 Mg → Mg2+ + 2e− D Iron(II)

D Na+ + e− → Na 4OH− → O2 + 2H+2O4e−

33. Figure 12 shows the conversion oaf nedtheenvee,nCtu2aHll4y, 36. Figure 14 shows the results of an experiment to
into ethanoic acid, CH3COOH determine the heat of combustion of butanol,
compound Y. C4H9OH.

Wind shield Thermometer SPM MODEL PAPER

Process X Oxidation Copper can
Water (250 cm3)
Ethene, Ethanol, Compound
C2H4 C2H5OH Y Spirit lamp
Butanol, C4H9OH
Figure 12 Wooden block
Which of the following is process X and

compound Y?

Process X Compound Y Figure 14
A Hydration
B Esterification Ethanoic acid, 1.11 tghoefwbauttearnforlo,mC42H54OºCHtoisTcoºmC.pDleetetelyrmbuinrentthtoe
C Addition CH3COOH heat
D Oxidation Ethanoic acid, T ºC.
CH3COOH [Specific heat capacity of water: 4.2 J g–1 ºC–1;
Ethyl ethanoate, relative molecular mass of butanol: 74; heat of
CH3COOC2H5 combustion of butanol: – 2 450 kJ mol–1]
Ethyl ethanoate, A 60.0 °C
CH3COOC2H5 B 42.0 °C
C 35.0 °C
D 30.0 °C
34. The following equation represents a combustion
of methane, CH4. 37. The following statements describe the properties
of particles of a substance at room temperature.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
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?
[Relative atomic mass: H=1; C=12; O=16] • The particles vibrate, rotate and move to
the whole region.
A 5.6 g
B 12.6 g • The particles always collide with each
C 15.4 g other.
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?

Substances Melting Boiling Solution X Solution Y Solution Z
point (°C) point (°C)
AW A Sodium Soap Lime juice
BX 120 2.4 hydroxide solution
CY –36 27
DZ 12 40 B Lime juice Soap Potassium
–79 –5 solution hydroxide

C Ammonia Sodium Lime juice
solution hydroxide
38. Figure 15 shows the apparatus set up to study the
sreoalicdtiiosnproofdcuhcleodriante,thCele2 nadndofitrhoen,reFaec.tiAonb. rown D Potassium Lime juice Soap
hydroxide solution
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
Cl2 rusting of iron.

Combustion tube Set Experiment Observation
Heat

Figure 15 I Low intensity of
Determine the mass of the brown solid formed. blue spot
[Relative atomic mass: Cl = 35.5; Fe = 56]
SPM MODEL PAPER Iron nail
A 12.1875 g
B 13.7215 g Metal P
C 16.25 g Jelly solution + potassium
D 24.375 g hexacyanoferrate(III) solution

39. Table 6.1 shows the colour changes for three II High intensity of
types of indicators. blue spot

Indicator Colour Colour in pH Iron nail
in pH 2 12 solution
solution Metal Q
Jelly solution + potassium
Phenolphthalein Colourless Pink hexacyanoferrate(III) solution

Methyl orange Red Yellow III No changes

Universal Red Dark blue
indicator

Iron nail
Table 6.1 Metal R
Table 6.2 shows the colours of three indicators in Jelly solution + potassium
hexacyanoferrate(III) solution
solutions X, Y dan Z. Table 7

Solution Which of the following is the correct sequence,
in ascending order of metals P, Q and R in terms
Solution X + Solution Y + Solution Z + of tendency to form ions?
phenolphthalein methyl universal A P, Q, R
orange indicator B R, P, Q
Colour Pink C Q, P, R
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.

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. SPM MODEL PAPER

CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 O

CH2 CH2 CH2 CH2 CH2 CH2 CH2 C
O–

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: SZMoinadgciuncmehslionurimtidraestu,eZl,pNnhCaaNtl2eO,soM3lusgotSiloOunt4iosonlution
B:
C:

(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 [2 marks]
Table 3 531