Focus SPM Chemistry samplebook

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CHEMISTRY üInfographics üComprehensive Notes üConcept Maps üActivities & Experiments üSPM Tips

REINFORCEMENT & ASSESSMENT

REVISION

üSPM Model Paper üComplete Answers

REINFORCEMENT

4∙5

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EXTRA FEATURES

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CHEMISTRY

CHEMISTRY

FOCUS SPM KSSM Form 4 • 5 – a complete and precise series of reference books with special features to enhance students’ learning as a whole.

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REVISION REVISI

üSPM Practices üCheckpoint

KSSM

SPM

4∙5

FOCUS

FORM

CC038642

SPM

PELANGI BESTSELLER

Chien Hui Siong Low Swee Neo Lim Eng Wah Salida Sani

NEW SPM ASSESSMENT FORMAT

2021

Format: 190mm X 260mm TP Focus SPM Chem BI 2022_pgi_2 IMP

CHEMISTRY

SPM

Chien Hui Siong Low Swee Neo Lim Eng Wah Salida Sani

© Penerbitan Pelangi Sdn. Bhd. 2022

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FORM

4∙5

KSSM

Exclusive Features of This Book Chapter Focus lists the learning objectives for an overview of the chapter.

r

te

ap

Ch

2

Chemistry

Carbon Co

mpound

SPM Chapte r 2 Carbon

Form 5 Compound

CHAPTER FOCUS • Types of carbon comp ounds • Homologo us series • Chemical properties and interconv compound ersion of s between homologou • Isomers s serie and naming based on IUPAC s nomenclat ure

Form 5

SPM Tips points out important tips for students to take note of.

Chemistry SPM

Oxidation

• Loss of oxygen n • Gain of hydroge s • Gain of electron • Decrease in oxidation number of them, try all er rememb to only. To get If you find it difficult definitions of oxidation opposite! the ering the rememb state reduction, simply the definition of • Gain of oxygen n • Loss of hydroge • • Increase in oxidation

solution s at a (d) Transfer of electron

289Loss of electrons

SPM Highlights

n?

Examiner’s Tip loss of oxygen, when there is a Reduction occurs or a decrease in gain of hydrogen gain of electrons, oxidation number.

reactions reactions. al reactions are redox 1. Not all chemic se reactions and double For example, acid-ba ation s (as in the precipit decomposition reaction ox reactions. method) are non-red Table 1.5 Detecting

Sodium Chem

istry

SPM

Chapter

Ek Exsp peer rim imen en 2.1

1 Red

ilibrium

iron(II) ions and

With iron ions Fe2+

a solution

noferrate(II) solution

ox Equ

3+ agent. e Fe2+ → Fe + +3 : +2 Oxidation number I) ions hand, changing iron(II e 4. On the other a reduction and therefor to iron(II) ions is agent. g requires a reducin 2+ – Fe3+ + e → Fe +2 : +3 number on Oxidati

the presence of

Reagent

Potassium hexacya

to Iron(III)

–

Answer: D

or ammoni hydroxide solution

distance

oxidation numbers: 1. Iron exhibits two 2+ ion, Fe (a) +2 as iron(II) 3+ ) ion, Fe (b) +3 as iron(III ions, containing iron(II) 2. An aqueous solution colour, whereas that 2+ is pale green in 3+ Fe yellow/ ) ions, Fe is containing iron(III in colour. yellowish-brown/brown ions is an ) ions to iron(III) g 3. Changing iron(II e requires an oxidisin oxidation and therefor

reductio cannot occur during Which of the following A Loss of oxygen s B Gain of electron n C Gain of hydroge number oxidation in D Increase

Types of redox

Equilibrium

2+

Iron(II) Ion, Fe Conversion of Versa 3+ Ion, Fe and Vice

number

Activity/ Experiment helps students to master hands-on scientific knowledge and skills.

Chapter 1 Redox

s are as of redox reaction 2. Four examples follows: ions to iron(III) ions (a) Changing of iron(II) and vice versa salt of metal from its (b) Displacement solution halide of halogen from its (c) Displacement

iron(III) ions

Observation

Green precipitate,

insoluble in excess

alkali

insoluble in excess

alkali

Fe3+

Brown precipitate,

Fe2+

Light blue precipita

Fe3+

Dark blue precipita

Fe2+

Dark blue precipita

SPM Highlights provides exposure to the frequentlytested questions that appear in the actual exam.

Form 5

Carbon atom SPM Tips C60 carbon s combine with one another of molecule. all the four ways What is the to form the summary of name of The following is this C60 carb and reduction. defining oxidation on molecule ?

Reduction

tion tion tion

253

Form

t 1.3 solution Aim: To Greenish-brown Fe3+ noferrate(III) solution study Potassium hexacya the Proble ion Pale red colourat m statem effects of typ Fe2+ es of ele ent: Do ctrodes ion es the Blood-red colourat solution Hypothe on the type of Fe3+ nium thiocyanate ele m/ammo sel ctr Potassiu sis: W ection electrod ode? hen ca of ion e affec s to be rbon t the where electrod discharg selectio as es are ed at the n of ion anode. when a co Variable use pper ele ele s to be d, ctrode hydroxide discharg ctrode. (a) Ma s: ions, is use nipula ed OH at the d, the (b) Re ted var copper – are dis spond iable : charged electrod ing (c) Co Types at ntrolled variable e, Cu : Pro of electrod is disch the anode, variab Mater ducts es arged les : ials: at the Type an formed at Apparat 0.1 mo the d conc l dm –3 us: entratio electrodes copper( n of the with cro Batteries, II) sulphate electroly carbo Proce codile n ele solution te dure: clips, , wood amme ctrodes, co 1. Tw en spl ter, tes pper ele o carbo int an t tube d sandp n electr and sw ctrodes, ele 2. An aper. itch. odes are ctroly electr tic ce cleaned sulphate olytic cell ll, conn with san is filled , CuSO ecting 3. Th wi dp wires th sol ap 4 e appa 0.1 mo er. ution ratus is until it l 4. Th is half dm –3 copper( set up e switc full. as sho II) through h is turne wn in Fig d the ure 1.2 electroly on to all 5. Th 3.

Copper e obser ow ele te for (II) vat 15 sulp ctr ions at are rec hate icity to minutes solution orded the an . pass 6. Ste . ode, ca Ammet Carbon ps 1 to er A thode 5 are and ele electrod Obser repeated es ctroly vation te using Switch s: Figure copper Battery 1.23 Elec electrod Electr trolysis es to rep ode of cop solution per(II) lace the sulphat carbo e Carbo n electr n Ga odes.

Anode Obser s bubb vation les are colou relea rle glowing ss gas which sed. A Catho Copper de religh wood A brown ts a en spl The co int sol id is deposite pp produ is Electr the sol er electrod ced. olyte The int uti cathode d on the e ensity thinner. on. The an dissolves int . colou of the ode be o Th Discu r of blue comes e catho ssion: decrease the electr de beco thicke olyte 1. Th s. mes r. e aque The int ous hydro ensity gen ion solution of co of lou the copp 2. (a) r of s, H + an blue Durin remain the electr d hydro er(II) sulph g the ele oly s ate un xid te consi change e ions, The Cu 2+ ctroly d.

OH – tha sts of copp sis usi ng carbo selectiv ion is low er( t move er ely dis n freely. II) ions, Cu 2+ charged than the H + electrodes, , sulph (b) Th the Cu 2+ ion ate ion to form e SO 2– s, SO 2– ion copper in the electr the ele 4 ions and 4 , ochemi s and H + ion metal. ctroche OH – ion l Cu 2+ cal ser s (aq chemistry water. mical nt to boi ies. He move to the 3 Thermo series. s move to the ) + 2e – → Cu st be bur Chapter nce, the catho Hence, (s) that mu try SPM Cu 2+ ion de. the OH – anode. The (c) Th kerosene Chemis s are OH – ions are e inten mass of –3 selectiv ion is lower late the blue Cu 2+ sity of the 4O – –1 Calcu cm ] ely dis tha 272 3.22 ions de blue colou H (aq) → O n=1g kJ g .

E charged n the SO 2– EXAMPL (g) crease r of solutio ne is 37 4 to form ion in ose sity –1 s when of the electr 2 + 2H2 O(l) ker ; den of oxygen –1 + olyte more l value J g °C and decrease 4e – copper The fue3 water at 32 °C. ution = 4.2 is depo s sol of sited on because the acity of 2.0 dm co the ca heat cap thode ncentration [Specific . . of the by water

5

Example provides solution for example of questions in the subtopics.

Checkpoint provides questions to test students’ understanding of the subtopics and reinforce learning.

Solution heat absorbed o C) ter, Q e o C – 32 Calculat absorbed by wa –1 ) × (100 t –1 o Total hea Jg C –3 × (4.2 g cm ) = mcθ 3 cm × 1 –1 = (2000 4.2 × 68 kJ g × ne = 37 = 2000 of kerose J t value = 571 200 Given hea kJ osene = 571.2 571.2 = 37 ue of ker nce, m e heat val ne He lat lcu ose Ca 571.2 = 15.4 g ue of ker ) Heat val at released (kJ ∴ m = 37 (g) He ne burnt = mass of kerose 571.2 kJ = mg –1 3.3 571.2 kJ kJ g = tals be m ate crys

ethano sodium can the k with the (c) How d? a hot pac ation of obtaine prepar pack ate. s lain the . A hot ethano release (d) Exp sodium rt injuries One of the en gas tals of . hydrog treat spo crys of to ces d le stan use the of 1 mo al sub pack is the bustion energy. chemic ation for Q1 Hot equ has Com two t s al ce Q3 mic of hea er. contain substan . rmoche en 286 kJ als is wat mical te a the hydrogen gas le of hydrog chemic second che Wri of (a) 1 mo tion (a) The ula CuSO4. combus the mass of er is stance. form when wat mical sub culate this che ction occurs (b) Cal ?

H = 1].

Name of rea mass: gas rogen gas gas ce in (a)? reaction in at type e atomic t value of hyd en the substan (b) Wh [relativ hydrog with the n to represent d for is the hea volume of mixed uire at atio req Wh are of the (c) dition) te an equ ONa, culate perature (c) Wri room con (d) Cal n an , CH3CO the tem °C. red at (b). crystals ction betwee –3 raise 50 –1 (measu –1 °C ; rea anoate l dm tion to 30 °C to ium eth neutralisation cm3 of 1.0 mo 3 of combus3 water from water = 4.2 J g of me at cm Q2 Sod 100 d by a acity of –3 lar volu 100 cm obtaine alkali Y. ed with 100 c heat cap= 1 g cm ; mo and an mix –1 [Specifi 3 mol ] acid X solution is . of water solution density dition = 24 dm –3 alkali Y acid X ium Y. of sod alkali l dm room con moles 1.0 mo e acid X and ber of Nam num (a) the culate duced. (b) Cal ate pro ethano

oint

Checkp

Form

5

PM 21 4:40

05/03/20

394

ii

0b Exclusive Features Chemistry SPM.indd 2

03 SPM

dd 394

RY F5.in

CHEMIST

27/12/2021 11:49 AM

Chemistry SPM Chapter 3 Thermochemistry

• Energy level diagram:

Heat of Displacement 1. Heat change that occurs during a displacement reaction is called heat of displacement.

Energy Mg(s) + CuSO4(aq)

Heat of displacement is the heat change when one mole of a metal is displaced from its salt solution by a more electropositive metal.

QR Code Video/ Info is scanned with a smartphone to obtain related information or video.

ΔH = –352 kJ mol–1 MgSO4(aq) + Cu(s)

2. A more electropositive metal atom displaces an ion of a less electropositive atom. Examples of displacement reactions are given in Figure 3.19.

Figure 3.20 Energy level diagram for displacement reaction of metal

Displacement VIDEO

Magnesium sulphate solution (colourless)

5. If different metals are used to displace copper, the heats of displacement are different.

Example: Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s) ∆H = –352 kJ

Precipitate of copper metal (brown)

Figure 3.19 Displacement reaction of metal

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) ∆H = –210 kJ

When magnesium powder is added to copper(II) sulphate solution, the blue solution turns colourless. A brown solid is deposited at the bottom of the test tube.

6. The further the position of the displacing metal from the metal being displaced in the electrochemical series, the higher the value of the heat of displacement.

3. The more electropositive magnesium atom releases electrons.

Mg(s) → Mg2+(aq) + 2e– (grey)

7. Figure 3.21 compares the heat of displacement of copper by zinc.

(colourless)

Copper(II) ion which is less electropositive readily accepts the electrons.

Cu2+(aq) + 2e– → Cu(s)

Magnesium Reactivity of metal increases

(blue)

(brown)

The overall equation for displacement of copper by magnesium:

ΔH = –352 kJ

Zinc ΔH = –210 kJ Copper

Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)

Form 5

Magnesium

Copper(II) sulphate (blue)

Figure 3.21 Comparing heats of displacement of copper by magnesium and zinc

4. Thermochemical equation for the displacement reaction of copper by magnesium is given below.

SPM

Heat of displacement of copper by magnesiumObje ctive Ques is higher compared to the heat of displacement Choo tions of copper by zinc. The displacement of copper se the corre ct an swer. by magnesium releases more heat compared to 1.1 Ox idatio zinc. n an d

Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s) ∆H = –352 kJ • Displacement of copper by magnesium is an exothermic reaction.

• Heat of displacement of copper by magnesium is 352 kJ. • 352 kJ is given out when 1 mole of copper is displaced by magnesium.

Practi

ce

Chem

istry

03 SPM CHEMISTRY F5.indd 365

)

Salt brid Figure

CHEMIS

TRY

F5.indd

1 PAPER s minute r fifteen One hou paper. ns in this all questio

ge

B C

Which of A B C D

1

t of ngemen

arra ches the II argon m aluminiu

wing mat

the follo

I m aluminiu argon argon mercury

mercury argon

particle

D

s above?

III

mercury mercury m aluminiu m aluminiu

P and Q P and R Q and R Q and S

16 8

S and T S and T R and T

39

X

19 Figure

l

water

+ 2 Figure ? ervation above obs stance. lains the solid sub gas. wing exp form a . the follo water to a volatile osphere Which of ts with the atm produce id P reac with water to a gas into particles.

A Liqu ts ape as water id P reac of liquid P esc between B Liqu s ces icle atoms. the spa e part different C Som s of liquid P fill s of five icle ticle ic part D Par of subatom Number Table 1 ws the of le 1 sho proton 3. Tab Number of 6 neutron Number s 7 6 electron Atom 6 7 6 P 8 8 7 Q 8 8 6 R 10 8 S s?

8 isotope T pairs are wing two the follo Which of

M

3

ula of the the form from the What is ed nd form ents X compou tion of elem combina and M? A MX B M2X C MX2 D M2X2 wing s of of the follo 5. Which nts the propertie represe ? chloride sodium Electrical y Melting conductivit nt eous ce poi ten Aqu Substan (o C) Mol Weak Weak –100 Good A Weak –80 e Insolubl B Weak 110 Good C d Goo 800 D

Chem

istry

05/03/2021

Y F5.indd

CHEMISTR

SPM

Chapt

er 2 Ca

CONCE PT M AP

450

06 KM SPM

PM

Second pair R and S

standard ws the ent re 3 sho s of elem 4. Figu for atom notation X and M.

the tota 3 of water, 50 cm 2. ed with in Figure P is mix 3 3 of liquid as shown me of Volu 50 cm is 98 cm 2. When of the mixture3 mixture3 3 50 cm 50 cm volume 98 cm liquid P

)

21 5:00

A

II

SPM Practice provides ample questions to test students’ mastery of the chapter.

Y 2+(aq

3

05/03/20

First pair

chemical

III Figure

ilibrium

285

Answer

I

dox Equ

285

er del Pap SPM Mo

part

er 1 Re

Form

X 2+(aq

01 SPM

ment of arrange ws the ture. re 1 sho room tempera 1. Figu ces at substan

Chapt

5

SPM Model Paper prepares students for the SPM with the actual exam format.

three icles in

SPM

1

5. Wh at is the ox of ch idatio rom Redu dichro ium in po n numb 1. Wh ction er mate( tassiu ich A +2 10.

VI), K m Which redox of the follow 2 Cr O 2 7?

B +3 ing is reduc the follow A Ne reaction? a ing e C +5 Mn 2+(aq manganes can B Es utralisatio D +6 n terific at 29 ) to mang e(II) ion, C Co ation 8 K? anes 6. Wh e me D Do mbustion at is tal [Elec uble tro bromi the chan deco 2. Wh Mn is de poten ne ox ge of mpos tia ich sta when idatio –1 ition A P + .18 V at l of Mn 2+ n BrO – temen oxida / BrO3 298 K] ion tur number /P t defin tio B – ion ns E° A Ga n? + es Q to ? = A +1 C R / Q E° +0.80 V B Lo in of electr = –0.15 B –1 → +5 2 / R– 365 ons ss of D S E° = C Ga V oxygen C +1 → –5 2 / S– +0.54 E° = D Inc ining hydro D +5 → –5 –1.37 V ge rease → +1 V 1.3 Vo oxida n 3. Th 7. Wh tion nu ltaic e fol ich Cell mber half-e lowing is chan of the follow an ox quati 11. Fig ge on. idatio ure A HB s is a red ing 05/03/2021 4:59 PM n uction? r → Br 2I –(aq up of 2 shows B the se )→I the ap SO 2– 2 expe t2 (aq) C Cr 3 → SO 2– Base rimen paratus for + 2e – d 2– 4 t. an D Mn2 O7 → Cr of the on the eq O 2– uatio O4 – → n, wh A Ele following Mn 2+ 4 8. Wh is ich ctron at is s are true? iod of ma the oxida Magne dona B Ele ine. ng ted by tion nu electro sium with a anese in ctron mber de a for co iodine s are ac mu mpou A 0 Iron la Mn cepte C Ele . nd 3 (PO d by electro B +1 ctron 4 )2 ? de s are iod Glacia C +2 D Ele ide ions. donated by acid l ethanoic D +3 Figure ctron 2 iodide s are ac The bu cepte ions.

5. Th d by A wa lb lights up e fol 1.2 Stand ter is ard Ele adde when reacti lowing sh gla Poten d to the ows a on. ctrod tial B the cial ethan e redox oic ac magn 9. Fig 2H2 S(g id. esium ure is rep ) + SO ele laced hydro 1 shows 2 (g) → ele with an ctrode the gen ele C me ctrode. iron 3S(s) ctrod standard What + 2H thylbe e. is the 2 O(l) nzen to the of su e is oxida lph acid. glacial eth added comp ur in each tion numb an ound er D the oic Hydro s in the of its position gen reacti gas Acid magn s of the In H on? es 2S solutio In SO iron ele ium and A n Electro 2 –2 In S interc ctrodes are de hang –2 B Figure 12. ed.

Figure 1 +4 Which 0 –2 C the ap 3 shows statem of the follow the pa 0 0 ratus se cell. the ele ents is co ing of a ch t-up of –2 D rrect emica –2 +4 abou A Ele ctrode? l t +4 B Pr ctrode us e0 ed is essu V Metal is 10 re of hydro nickel.

X egen ga C Te 1 kPa. mpera s Metal solution ture of ac id is 27 Y 3 K.

rbon

Comp

ound

5:03 PM

Sourc es: • pe tro • na leum tur • coal al gas

450

Hydro

Uses : • fer tilis • fue ers l • synth • solve etic polym ers • me nts dicine • foo s d addit ives

CARB COMP ON OUND S

carbo

ns

Satur

ated

Homo logous series

Unsa

No hydro n carbo ns

turate

d

Homo logous series

Alkan es Cn H 2n+

Alcoh

ol

2

& & !C !C ! & &

Concept Map summarises the essential concepts learnt in the chapter.

Single

Alken es Cn H 2n

C"

bond

Doubl

C

e bon

Chem ica • comb l Propert ies: • subs ustion titutio n

Form 5 CHEM

ISTRY

F5.indd

Alkyn es Cn H 2n–

!C

d

Chem ica • comb l Propert ies: • addit ustion ion – hydro – halog gen – hid en rog – oxida en halid tion wit e KM h • addit nO4 ion polym erisa tion

2

#C

Triple

bond

!

Homo logous series

Cn H

2n+1

No hydro n carbo ns

Ester

Homo logous series

Homo logous series

OH Cn H Cn H

2n+2

Chem ica • comb l Propert ies: • oxida ustion tion • de hydra tion

2n+1

CO C

COOH

2

m

H2m+

1

Phys ical Pr • fru opert ity ies: • colou smell • ins rless liquid olu • solub ble in wa ter le solve in organ ic nts • les s de water nse than

Chem ical Pr • rea ction operties: – base with – me tal ca rbona – rea • ester ctive me te tal ific reacti ation on

344

02 SPM

Homo logous series

344

05/03/20

21 5:14

PM

iii

0b Exclusive Features Chemistry SPM.indd 3

27/12/2021 11:49 AM

iv

0c Periodic Table of Elements.indd 4

27/12/2021 11:49 AM

Period

7

6

5

4

3

2

1

2

5

23

V

24

27

Co

28

Ni

29

Cu

13

30

B

5

Non-metals

Semi-metals

Metals

Zn

Key:

12

20

39

88 89 – 103 Actinides

57 – 71 Lanthanides

Yttrium 89

Actinides

Lanthanides

Radium 226

Ra

Fr

Francium 223

87

Barium 137

Cesium 133

Ba

56

55

Cs

Strontium 88

Rubidium 85.5

Y

38

Sr

Rb

37

Scandium 45

Ca

Calcium 40

Potassium 39

19

K

Mg

Magnesium 24

Sodium 23

Na

106

60

Actinium 227

Thorium 232

Th

90

89

Ac

Cerium 140

Lanthanum 139

Protactinium 231

Pa

91

Uranium 238

U

92

Praseodymium Neodymium 144 141

Nd

59

58

Pr

Ce

57

La

107

Bh

Bohrium (264)

Seaborgium (263)

Sg

105

Re

Rhenium 186

75

Technetium 98

43

Tc

Dubnium (262)

Db

25

Mn

Manganese 55

Rutherfordium (261)

Rf

104

W

74

Molybdenum 96

42

Mo

Tungsten 184

Ta

Cr

Chromium 52

Tantalum 181

Hf

73

Niobium 93

41

Nb

Vanadium 51

Hafnium 178.5

72

Zirconium 91

40

Zr

Titanium 48

26

Os

108

Neptunium 237

Np

93

Promethium 145

61

Pm

Hassium (265)

Hs

Osmium 190

76

Ruthenium 101

44

Ru

Iron 56

Fe

Ir

109

Pt

110

Ds

Platinum 195

78

Palladium 106

46

Pd

Nickel 59

111

Rg

Gold 197

Au

79

Silver 108

47

Ag

Copper 64

Hg

112

Cn

Mercury 201

80

Cadmium 112

48

Cd

Zinc 65

Plutonium 244

Pu

94

Samarium 150

62

Sm

Americium 243

Am

95

Europium 152

63

Eu

Curium 247

Cm

96

Gadolinium 157

64

Gd

Berkelium 247

Bk

97

Terbium 159

65

Tb

Meitnerium Darmstadtium Roentgenium Copernicium (272) (268) (281) (285)

Mt

Iridium 192

77

Rhodium 103

45

Rh

Cobalt 59

31

Tl

113

Californium 251

Cf

98

Dysprosium 162.5

66

Dy

Nihonium

Nh

Thallium 204

81

Indium 115

49

In

Gallium 70

Ga

Aluminium 27

Al

13

22

Relative atomic mass

11

Boron 11

Ti

10

12

21

9

Proton number Symbol of the element

8

Transition metals

7

Beryllium 9

Sc

H

1

Hydrogen 1

6

11

Be

4

4

Name of the element

3

Lithium 7

Li

3

Hydrogen 1

H

1

Alkali Alkaline metals earth metals

1

Group

Si

32

114

Einsteinium 252

Es

99

Holmium 165

67

Ho

Flerovium (289)

Fl

Lead 207

Pb

82

Tin 119

50

Sn

Germanium 73

Ge

Silicon 28

14

Carbon 12

C

6

14

P

33

Bi

115

Fermium 257

Fm

100

Erbium 167

68

Er

Moscovium

Mc

Bismuth 209

83

Antimony 122

51

Sb

Arsenic 75

As

Phosphorus 31

15

Nitrogen 14

N

7

15

S

34

Po

116

17

18

Cl

35

At

117

Nobelium 259

No

102

Ytterbium 173

70

Yb

Tennessine

Ts

Astatine 210

85

Iodine 127

I

53

Bromine 80

Br

Chlorine 35.5

17

Fluorine 19

F

9

18

36

86

118

Lawrencium 262

103

Lr

Lutetium 175

71

Lu

Oganesson

Og

Radon 222

Rn

Xenon 131

54

Xe

Krypton 84

Kr

Argon 40

Ar

Neon 20

10

Ne

Helium 4

2

He

Halogens Noble gases

Mendelevium 258

Md

101

Thulium 169

69

Tm

Livermorium (292)

Lv

Polonium 209

84

Tellurium 128

52

Te

Selenium 79

Se

Sulphur 32

16

Oxygen 16

O

8

16

PERIODIC TABLE OF ELEMENTS

CONTENTS Form 4

4.4 Elements in Group 1 4.5 Elements in Group 17 4.6 Elements in Period 3 4.7 Transition Elements SPM Practice 4

ter

ap

Ch

1

Introduction to Chemistry

1

1.1

Development in Chemistry Field and Its Importance in Daily Life 2 1.2 Scientific Investigation in Chemistry 4 1.3 Usage, Management and Handling of Apparatus and Materials 5 SPM Practice 1 8 ter

ap

Ch

2

Matter and the Atomic Structure

2.1 2.2

Basic Concepts of Matter The Development of the Atomic Model 2.3 Atomic Structure 2.4 Isotopes and Their Uses SPM Practice 2 ter

ap

Ch

3

Relative Atomic Mass and Relative Molecular Mass 3.2 Mole Concept 3.3 Chemical Formula 3.4 Chemical Equations SPM Practice 3 ter

ap

4 4.1 4.2 4.3

5

Basics of Compound Formation Ionic Bond Covalent Bond Hydrogen Bond Dative Bond Metallic Bond Properties of Ionic Compounds and Covalent Compounds SPM Practice 5

11 12 18 20 26 30

The Periodic Table of Elements

33 34 36 42 52 57

60

The Development of the Periodic Table of Elements 61 The Arrangement in the Periodic Table of Elements 63 Elements in Group 18 67

94

Chemical Bond

5.1 5.2 5.3 5.4 5.5 5.6 5.7

95 98 103 107 111 113 114 120

ter

ap

3.1

Ch

ter

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Ch

Ch

The Mole Concept, Chemical Formula and Equation

69 76 84 90 92

6

Acid, Base and Salt

123

6.1

The Role of Water in Showing Acidic and Alkaline Properties 6.2 pH Value 6.3 Strength of Acids and Alkalis 6.4 Chemical Properties of Acids and Alkalis 6.5 Concentrations of Aqueous Solution 6.6 Standard Solution 6.7 Neutralisation 6.8 Salts, Crystals and Their Uses in Daily Life 6.9 Preparation of Salts 6.10 Effect of Heat on Salts 6.11 Qualitative Analysis SPM Practice 6

124 130 133 135 142 145 147 154 156 170 178 189

v

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ter

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7

Rate of Reaction

7.1 7.2 7.3

Determining Rate of Reaction Factors Affecting Rate of Reactions Application of Factors that Affect the Rate of Reaction in Daily Life 7.4 Collision Theory SPM Practice 7 ter

p ha

C

8.4

193

Alloy and Its Importance Composition of Glass and Its Uses Composition of Ceramics and Its Uses Composite Materials and Their Uses SPM Practice 8

1

Redox Equilibrium

242

1.1 Oxidation and Reduction 1.2 Standard Electrode Potential 1.3 Voltaic Cell 1.4 Electrolytic Cell 1.5 Extraction of Metal from Its Ore 1.6 Rusting SPM Practice 1

243 260 261 265 276 278 285

er

pt

a Ch

2

Carbon Compound

289

350 351 357 391 395

ter

ap

Ch

4

Polymer Chemistry

4.1 Polymer 4.2 Natural Rubber 4.3 Synthetic Rubber SPM Practice 4

232

ter

C

Heat Change in Reactions Heat of Reaction Application of Exothermic and Endothermic Reactions in Daily Life SPM Practice 3

224 229

235 239

Thermochemistry

3.1 3.2 3.3

212 213 220

Form 5 p ha

3

194 201

Manufactured Substances in Industry 223

8 8.1 8.2 8.3

ter

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Ch

er

pt

a Ch

5

Consumer and Industrial Chemistry

5.1 5.2 5.3 5.4 5.5

Oils and Fats Cleaning Agents Food Additives Medicines and Cosmetics Application of Nanotechnology in Industry 5.6 Application of Green Technology in Industrial Waste Management SPM Practice 5

401 402 407 414 416

419 420 423 432 436 441 443 446

SPM MODEL PAPER

450

ANSWERS

464

2.1 2.2 2.3

Types of Carbon Compound 290 Homologous Series 294 Chemical Properties and Interconversion of Compounds between Homologous Series 308 2.4 Isomers and Naming based on IUPAC Nomenclature 329 SPM Practice 2 345 vi

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r

e pt

a

Ch

Form 4

3

The Mole Concept, Chemical Formula and Equation OCUS

CHAPTER F

Relative Atomic Mass and Relative Molecular Mass Mole Concept Chemical Formula Chemical Equation

Form 4

• • • •

In 1814, Jacob Berzelius (1779 – 1848), a Swedish chemist, had developed a system that is used still today to represent chemicals. The only difference is that superscript numbers are used in the system while subscript numbers are used in the modern chemical formulae. For example, he used the H2O formula for water while modern chemistry uses the H2O formula.

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Chemistry SPM Chapter 3 The Mole Concept, Chemical Formula and Equation

3.1 Relative Atomic Mass and Relative Molecular Mass Relative Atomic Mass, RAM 1. An atom has a very small mass and thus, is difficult to measure. Therefore, chemists compare the mass of an atom with that of a standard atom.

2. The mass of an atom compared to a standard atom is called relative atomic mass. Relative atomic mass has no unit as it is just a comparison value.

Form 4

At first, hydrogen atom which is the lightest atom was used as the standard atom.

• The mass of hydrogen atom is taken as 1 unit.

• Problems that arose: – The relative atomic masses of most elements cannot be determined as they do not easily combine with hydrogen.

– Hydrogen gas is difficult to handle.

In 1850, oxygen atom was used as the standard atom.

• The relative atomic masses of most elements can be determined as most of them can combine with oxygen.

• Problems arose when three oxygen isotopes were found.

– Chemists used natural oxygen which contains all the three isotopes as the standard.

– Physicists used oxygen-16 isotope as the standard.

Figure 3.1 The history of choosing the standard atom in determining relative atomic masses

In 1961, carbon-12 was accepted as the single standard atom internationally.

• The mass of carbon-12 atom is taken as 12 units.

• Carbon-12 was chosen due to a few reasons: – Carbon-12 is the most abundant carbon isotope (almost 98%). So, the mass of 12 units given to a carbon-12 atom is accurate.

– Carbon-12 is easy to handle in laboratories as it exists as a solid at room temperature.

– Many elements can combine with carbon-12.

– The mass spectrometer at that time was already using carbon-12 as the standard.

3. Based on the carbon-12 scale, the relative atomic mass of an element is defined as the average mass 1 of the mass of a carbon-12 atom. of an atom of the element compared to 12 Relative atomic mass of an element =

Average mass of one atom of the element 1 × mass of one carbon-12 atom 12

4. For example, the relative atomic mass of lithium is 7. This means that the mass of one lithium atom 1 is 7 times the mass of one carbon-12 atom.

12

SPM Highlights

1 The average mass of a molybdenum atom is 96 times larger than of the mass of a carbon-12 atom. What is the 12 relative atomic mass of molybdenum?

A 8 C 48 B 12 D 96 Examiner’s tip Students need to remember the definition of ‘relative atomic mass’ (refer to the definition above). Based on the definition, the relative mass of molybdenum is 96. Note that the relative mass of one carbon-12 atom is taken as 1 exactly 12 units. So, the phrase, of the mass of carbon-12’ equals to 1 unit.

12 Answer: D

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Chemistry SPM Chapter 3 The Mole Concept, Chemical Formula and Equation

EXAMPLE

3.1

How many beryllium atoms have the same mass as two silver atoms?

[Relative atomic mass: Be, 9; Ag, 108] Solution The number of beryllium atoms = n The mass of n beryllium atoms = Mass of 2 silver atoms So, n × RAM of Be = 2 RAM of Ag n × 9 = 2 × 108 2 × 108 n= 9 = 24 atoms EXAMPLE

3.2

How many times is the mass of a zinc atom larger than the mass of a helium atom?

[Relative atomic mass: He, 4; Zn, 65] Solution Mass of a zinc atom RAM of Zn = Mass of a helium atom RAM of He 65 = 4 = 16.25 times Relative Molecular Mass, RMM 1. Based on carbon-12 scale, relative molecular mass of a substance is defined as the average mass of a molecule of the substance compared 1 of the mass of a carbon-12 atom. to 12 Relative molecular = mass of a substance

Average mass of one molecule of the substance 1 mass of one × 12 carbon-12 atom

2. For example, the relative molecular mass of carbon dioxide is 44. This means that one molecule of carbon dioxide is 44 times 1 of 12 the mass of a carbon-12 atom.

3. One molecule is made up of a few atoms.

Therefore, we can calculate the relative molecular mass of a substance by adding up the relative atomic masses of all the atoms that form the molecule. To do so, you need to know the formula of the substance. See Figure 3.2 and Table 3.1.

Carbon dioxide molecule, CO2

RAM for O = 16

O

C

O

RAM for O = 16

RAM for C = 12

RMM for CO2 = 16 + 12 + 16 = 44

Figure 3.2 Calculating the relative molecular mass of carbon dioxide

Form 4

5. We can solve various problems by comparing the relative atomic masses of elements. See Examples 3.1 and 3.2.

Table 3.1 Calculation of relative molecular mass of several substances

Molecular substance Hydrogen gas Oxygen gas Water

Formula of substance

Relative molecular mass, RMM

H2

2(1) = 2

O2 H 2O

2(16) = 32 2(1) + 16 = 18

Ammonia

NH3

14 + 3(1) = 17

Ethanol

C2H5OH

2(12) + 5(1) + 16 + 1 = 46

[Relative atomic mass: H, 1; C, 12; N, 14; O, 16] Relative Formula Mass, RFM 1. The term ‘relative molecular mass’ is only used for substances made up of molecules. For substances made up of ions, we use the term ‘relative formula mass’.

2. In the same manner, we can calculate the relative formula mass of an ionic substance by adding up the relative atomic mass of all the atoms of elements shown in the formula. See Table 3.2.

Table 3.2 Calculation of relative formula mass of a few substances

Ionic substance Sodium chloride Aluminium oxide Calcium hydroxide Magnesium nitrate Copper(II) sulphate hydrate

Formula of substance NaCl Al2O3

Relative formula mass, RFM 23 + 35.5 = 58.5 2(27) + 3(16) = 102

Ca (OH)2

40 + 2[16+ 1]= 74

Mg (NO3)2

24 + 2[14 + 3(16)] = 148

CuSO4.5H2O

64 + 32 + 4(16) + 5[2(1) + 16] = 250

[Relative atomic mass : H,1; N, 14; O, 16; Na, 23, Mg, 24; S, 32; Cl, 35.5; Ca, 40; Cu, 64]

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Chemistry SPM Chapter 3 The Mole Concept, Chemical Formula and Equation

3. Using the knowledge about relative mass, we can solve various calculation problems. See Example 3.3.

EXAMPLE

3.3

Form 4

The relative formula mass of X2SO4 is 142.

Calculate the relative atomic mass of element X. [Relative atomic mass: O, 16; S, 32] Solution RAM of X = x Given RFM of X2SO4 = 142 2(x) + 32 + 4(16) = 142 2x + 96 = 142 2x = 142 – 96 = 46 46 = 23 x= 2 RAM of X is 23.

3.1

Checkpoint

Q1 How many oxygen atoms have the same mass as three copper atoms?

[Relative atomic mass: C, 12; Cu, 64]

Q2 The mass of a sulphur atom is 8 times more than the mass of a helium atom. What is the relative atomic mass of sulphur?

[Relative atomic mass: He, 4] Q3 Calculate the relative molecular mass of the following substances: (a) Phosphorus, P4 (b) Carbon monoxide gas, CO (c) Sucrose, C12H22O11 (d) Benzoic acid, C6H5COOH [Relative atomic mass: H, 1; C, 12; O, 16; P, 31] Q4 Calculate the relative formula mass of the following substances: (a) Sodium oxide, Na2O (b) Zinc nitrate, Zn(NO3)2 (c) Potassium thiosulphate, K2S2O3 (d) Hydrated sodium carbonate, Na2CO3.10H2O [Relative atomic mass: H,1; C,12; N,14; O, 16; Na, 23; S, 32; K, 39; Zn, 65] Q5 Element Z forms a chloride salt with the formula ZCI2 and relative formula mass of 95. What is the relative atomic mass of element Z? [Relative atomic mass: Cl, 35.5]

3.2 Mole Concept 1. Chemists use the ‘mole’ unit to represent the quantity of a substance.

One mole of substance is defined as the quantity of substance containing the same number of atoms contained in 12 g of carbon-12, which is 6.02 × 1023 particles

Avogadro’s constant, NA is defined as the number of particles.

(NA = 6.02 × 1023 mol-1)

2. The number of particles per mole, which is 6.02 × 1023 is determined experimentally and is known as Avogadro’s constant or Avogadro’s number.

3. The way of using the mole unit is exactly the same as that of using the dozen unit. See Figure 3.3.

1 dozen of pencils = 12 units of pencils Contains 6.02 x 1023 molecules of water

2 dozen of pencils = 24 units of pencils Contains 2 x 6.02 x 1023 molecules of water

Figure 3.3 The mole unit is used in the same way as the dozen unit

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Chemistry SPM Chapter 3 The Mole Concept, Chemical Formula and Equation

4. The particles referred depends on the type of substances (Table 3.3).

Table 3.3 Type of substances and type of particles

Example of substances

Atomic substance (consists of atoms)

Molecular substance (consists of molecules)

• All metal elements • Most non-metal elements such as • All noble gases hydrogen (H2), oxygen (O2), nitrogen • Some non-metal elements (N2), fluorine (F2), chlorine (Cl2), such as carbon and silicon bromine (Br2), iodine (I2), sulphur (S8) and phosphorus (P4).

• All covalent compounds

The meaning 1 mole of atomic substance of 1 mole of has 6.02 × 1023 atoms. For example, 1 mole of zinc substance contains 6.02 × 1023 zinc atoms.

1 mole of molecular substance has 6.02 × 1023 molecules. For example, 1 mole of oxygen gas contains 6.02 × 1023 O2 molecules.

Ionic substance (consists of ions) All ionic compounds such as sodium chloride and zinc oxide

1 mole of ionic substance has 6.02 × 1023 formula units. For example, 1 mole of sodium chloride contains 6.02 × 1023 NaCl units.

Form 4

Type of substance

5. We can calculate the number of particles in a fraction of mole of substance by using the Avogadro’s constant, NA as follows. (NA: 6.02 × 1023 mol–1) Number of moles EXAMPLE

× NA ÷ NA

Number of particles

3.4

Calculate the number of atoms in: 1. 0.5 mole of copper.

2. 2 moles of hydrogen gas.

Solution 1. Copper is an atomic substance.

So, 1 mole of copper has 6.02 × 1023 of copper atoms.

The number atoms in 0.5 mole of copper = Number of mole × NA = 0.5 × 6.02 × 1023 atoms = 3.01 × 1023 atoms 2. Hydrogen gas is a molecular substance.

So, 1 mole of hydrogen gas has 6.02 × 1023 H2 molecules.

The number of H2 molecules in 2 moles of hydrogen gas = number of mole × NA = 2 × 6.02 × 1023 H2 molecules Based on the formula H2, each H2 molecule is made up of 2 H atoms.

Therefore, the number of H atoms in 2 moles of hydrogen gas = 2 × 2 × 6.02 × 1023 atoms = 24.08 × 1023 atoms = 2.408 × 1024 atoms

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Chemistry SPM Chapter 3 The Mole Concept, Chemical Formula and Equation EXAMPLE

Table 3.3 Molar mass of a few substances

3.5

How many moles of carbon dioxide contain 1.806 × 1023 CO2 molecules?

Form 4

Solution The number of moles of carbon dioxide = Number of particles ÷ NA 1.806 × 1023 = 6.02 × 1023 = 0.3 mol EXAMPLE

3.6

0.5 mole of magnesium chloride, MgCl2 are dissolved in a beaker of water. How many ions are present in the beaker?

Solution Magnesium chloride, MgCl2 is an ionic substance. So, 1 mole of magnesium chloride has 6.02 × 1023 MgCl2 units.

The number of MgCl2 units in 0.5 mole of magnesium chloride = number of moles × NA = 0.5 × 6.02 × 1023 units Each MgCl2 unit is made up of three ions, which are one magnesium ion and two chloride ions. So, the number of ions in the beaker = 3 × number of MgCl2 = 3 × 0.5 × 6.02 × 1023 ions = 9.03 × 1023 ions Number of moles and mass of substance 1. Chemists measure the number of moles of substance by just weighing its mass. Mass of substance and number of moles are linked by molar mass as shown below.

Number of moles

× molar mass ÷ molar mass

Mass in grams

2. Molar mass of a substance is the mass of one mole of the substance in grams. The unit for molar mass is gram per mole (g mol–1).

3. The value of molar mass is the same as the relative mass of the substance.

Substance Magnesium, Mg Atomic substance Zinc, Zn Hydrogen gas, Molecular H2 substance Methane, CH4

Relative mass

Molar mass (g mol-1)

RAM = 24 RAM = 65

24 65

RMM = 2(1) = 2

2

RMM = 12 + 4(1) = 16 Zinc bromide, RFM = 65 + 2(80) ZnBr2 = 225 Ionic substance Sodium sulphate, RFM = 2(23) + 32 Na2SO4 + 4(16) = 142

16 225 142

[Relative atomic mass : H,1; C,12; O,16; Na, 23; Mg, 24; S, 32; Zn, 65; Br, 80]

Figure 3.4 Molar mass of magnesium is 24 g mol-1 EXAMPLE

3.7

What is the mass of 1.2 moles of sodium hydroxide, NaOH? [Relative atomic mass: H, 1; O, 16; Na, 23] Solution RFM of sodium hydroxide, NaOH = 23 + 16 + 1 = 40 So, the molar mass of sodium hydroxide = 40 g mol-1 The mass of 1.2 moles of sodium hydroxide = number of moles × molar mass = 1.2 × 40 = 48 g EXAMPLE

3.8

How many moles of molecules are there in 450 g of water, H2O? [Relative atomic mass: H, 1; O, 16] Solution RMM of water, H2O = 2(1) + 16 = 18 So, the molar mass of water = 18 g mol–1 The number of moles of molecules in 450 g of water = mass ÷ molar mass 450 = 25 mol = 18

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Chemistry SPM Chapter 3 The Mole Concept, Chemical Formula and Equation

SPM Tips

SPM Highlights

Examiner’s tip Students need to know the molar mass and the mass of each substance, including glucose.

Substance 1 mole C6H12O6

Mass (g) = number of moles × molar mass 6(12) + 12(1) + 6(16) = 180

3 moles C3H7OH

3 × [3(12) + 7(1) + 16 + 1] = 3 × 60 = 180

4 moles C3H6

4 × [3(12) + 6(1)] = 4 × 42 = 168

6 moles C2H6

6 × [2(12) + 6(1)] = 6 × 30 = 180

10 moles H2O

10 × [2(1) + 16] = 10 × 18 = 180

Answer: B

4. Substances with the same number of moles have equal number of particles even though the particles are different in size and mass.

12 g 0.5 mole of magnesium

28 g 0.5 mole of iron

Figure 3.5 Both blocks of magnesium and iron have 0.5 x 6.022 × 1023 atoms [Relative atomic mass: Mg, 24; Fe, 56] EXAMPLE

3.9

What is the mass of gold that has the same number of atoms as 4 g of oxygen?

[Relative atomic mass: O, 16; Au, 197] Solution The number of oxygen atoms, O = mass ÷ molar mass = 4 ÷ 16 = 0.25 mol 0.25 mole of gold has the same number of atoms as 0.25 mole of oxygen. So, the mass of gold = number of moles × molar mass = 0.25 × 197 = 49.25 g

For molecular elements such as oxygen, make sure you read the question carefully to understand what is required in the question. Use RAM to calculate the number of atoms of oxygen, O. Use RMM to calculate the number of molecules of oxygen, O2.

5. We can compare the number of particles in substances by just comparing the number of moles of the substances. See Example 3.10.

EXAMPLE

3.10

How many times the number of atoms in 7 g of nitrogen gas is larger than that in 7 g of iron?

[Relative atomic mass: N, 14; Fe, 56]

Form 4

Which substance has a different mass with 1 mole of glucose, C6H12O6?

[Relative atomic mass: H,1; C, 12; O, 16] A 3 moles of propanol, C3H7OH B 4 moles of propane, C3H6 C 6 moles of ethane, C2H6 D 10 moles of water, H2O

Solution We only need to compare the number of moles of both samples.

The number of moles of nitrogen atoms = mass ÷ molar mass = 7 ÷ 14 = 0.5 mol The number of moles of iron atoms = mass ÷ molar mass = 7 ÷ 56 = 0.125 mol 0.5 Number of moles of nitrogen atoms = 0.125 = 4 Number of moles of iron atoms Therefore, 7 g of nitrogen gas has 4 times more number of atoms than that in 7 g of iron.

Number of moles and volume of gases 1. The number of moles of a gas can also be determined through its volume. This is not true of a solid or liquid substance.

2. All gases with the same volume under the same pressure and temperature, have the same number of particles.

3. The volume of 1 mole of gas is known as molar volume.

• Molar volume of any gas is 22.4 dm3 mol–1 at STP or 24 dm3 mol–1 at room conditions.

• STP refers to the standard temperature of 0 ºC and the pressure of 1 atm.

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Introduction to Chemistry

I N FO G R Chapter 1

Form

S IC H AP

4

Careers in Chemistry

Engineer

Doktor

Forensic scientist

Pharmacist

Researcher

Chemical-based Industries Palm oil industry

Cosmetic industry

Detergents

Cosmetic products

0a Infografik SPM Chemistry.indd 1

Petroleum industry

Cooking oils

27/12/2021 11:46 AM

Matter and the Atomic Structure

I N FO G R

S IC H AP

Chapter 2

Form

4

What are the characteristics of matter?

Matter exists in three states, i.e. solid, liquid and gas.

Solid

Liquid

Gas

Characteristics of Solid, Liquid and Gas State of matter

Solid

Liquid

Gas

Particles are loosely packed and disorderly.

Particles are separated very far apart from each other and distributed randomly.

Particles can vibrate, rotate and flow.

Particles are constantly colliding with each other.

Particles can vibrate, rotate and move freely.

Particles are constantly colliding with each other.

1) Diagram of arrangement of particles

2) Arrangement of particles

Particles are closely packed and orderly.

3) Movement of Particles vibrate and particles rotate about their fixed positions.

4) Forces of attraction between particles

Very strong forces of attraction between particles.

Forces of attraction between particles are weaker compared to those in solid.

Very weak forces of attraction between particles.

5) Kinetic energy of particles

Very low energy.

Moderate energy.

Highest energy.

0a Infografik SPM Chemistry.indd 2

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Format 190mm X 260mm Extent : 504pg (21.78 mm) = 1.08 mm (24pg70gsm) + 19.2 mm (480pg 60gsm) + 1.5mm Status CRC Date 14/1

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