SUCCESS CHEMISTRY

Table 4.9 Some physical properties of the transition elements in the first series

Element Sc Ti V Cr Mn Fe Co Ni Cu Zn
0.138
Atomic radius (nm) 0.162 0.147 0.134 0.130 0.135 0.126 0.125 0.124 0.128 419
906
Melting point (°C) 1539 1668 1900 1875 1245 1536 1495 1453 1083 7.14
1.6
Boiling point (°C) 2730 3260 3450 2665 2150 3000 2900 2730 2959 2.5

Density (g cm–3) 3.0 4.51 6.1 7.19 7.43 7.86 8.9 8.9 8.96

Electronegativity 1.3 1.5 1.6 1.6 1.5 1.8 1.8 1.8 1.9

Hardness soft 6.0 7.0 8.5 6.0 4.0 5.0 4.0 3.0
(Mohs’ scale)

4 (Note: Hardness is measured using Mohs’ scale. Diamond, which is the hardest substance known, has a hardness of 10 on Mohs’ scale.)

Titanium alloy is corrosion Vanadium steel alloy is used in Titanium alloy
resistant and lightweight. It is making gears and crankshafts of is light and
used in orthopaedic and dental vehicles. have very
implants. high tensile
strength. It is
used to make
aircraft engines.
An engine of
a A380 Airbus
uses 11 tons
of titanium.

Uses of transition
metals

Nickel is used in many industrial Cobalt is alloyed with iron, nickel
and consumer products, including and other metals to make Alnico,
stainless steel, magnets, coinage an alloy of unusual magnetic
and special alloys. strength.

Chromium is used to make Iron is alloyed with carbon to Copper is used as electrical
stainless steel and for the make steel which is used in conductors and in piping.
electroplating of iron. making cars, ships and in building
industries.
Periodic Table of Elements
94

Special Properties of Transition Elements

Transition elements have variable oxidation numbers

1 Unlike elements in the main group of the Table 4.10 The oxidation numbers of transition
Periodic Table in which each has only one elements in their compounds
oxidation number, a transition element has
more than one oxidation number in its Compounds of transition Chemical Oxidation
compounds.
elements formula number
2 Oxidation number is the charge on the ion.
In other words, a transition element can form Chromium(III) chloride CrCl3 +3 4
ions with different charges. Potassium dichromate(VI) K2Cr2O7 +6
For example, magnesium (in Group 2), MnCl2
can form only Mg2+ ion, with an oxidation Manganese(II) chloride MnO2 +2
number of +2. Manganese(IV) oxide K2MnO4 +4
I ron, however, as a transition element, can Potassium manganate(VI) KMnO4 +6
form Fe2+ ion (oxidation number of +2) and Potassium manganate(VII) FeSO4 +7
Fe3+ ion (oxidation number of +3).
Iron(II) sulphate FeCl3 +2
3 Table 4.10 shows the different oxidation Iron(III) chloride NiSO4 +3
numbers of transition elements in their NiCl3
compounds. Nickel(II) sulphate Cu2O +2
Nickel(III) chloride CuO +3

Copper(I) oxide +1
Copper(II) oxide +2

Transition elements form coloured com­pounds

1 Unlike main group metal compounds which Name of ion of Formula of ion Colour of
transition of transition aqueous
are usually white, transition elements can element element solution
form compounds of different colours.
2 Unlike aqueous solutions of main group Manganese(II) ion Mn2+ Pink

compounds or ions which are usually Chromium(III) ion Cr3+ Green
colourless, aqueous solutions of transition
element compounds or their ions are Nickel(II) ion Ni2+ Green
coloured.
3 Table 4.11 shows the colours of some aqueous 4 Precious stones are coloured due to the
presence of comp­ ounds of transition elements.
solutions of ions of transition elements. Table 4.12 shows the transition elements
present which are responsible for the
Table 4.11 The colours of some aqueous solutions colours of some precious stones.
of ions of transition elements

Name of ion of Formula of ion Colour of Table 4.12 Examples of some precious stones and
transition of transition aqueous the transition elements which give
element element solution them their distinctive colours

Chromate ion CrO24– Yellow Precious Colour Transition element
Dichromate ion Cr2O72– Orange stone present
Permanganate ion MnO4– Purple
Iron(II) ion Fe2+ Green Ruby Red Chromium

Iron(III) ion Fe3+ Brown Sapphire Blue Iron and Titanium

Copper(II) ion Cu2+ Blue Emerald Green Chromium

Cobalt(II) ion Co2+ Pink Amethyst Purple Manganese

95 Periodic Table of Elements

Transition metals or their compounds have catalytic properties SPM

’08/P1/P2

1 A catalyst is a substance that speeds up the 3 Table 4.13 shows the uses of transition
rate of a reaction. A catalyst does not change elements or their compounds as catalysts in
industries.
chemically after a reaction. Catalysts are used

in almost all chemical manufacturing plants.
2 Many catalysts are transition elements or

their compounds.

Table 4.13 The uses of transition elements or their compounds as catalysts in industries

Transition element or its Industrial process which use the catalyst
compound

4 Fine iron powder, Fe Haber process in the manufacture of ammonia. Iron catalyses the
reaction between nitrogen and hydrogen gas to produce ammonia.

Vanadium(V) oxide, V2O5 N 2(g) + 3H2(g) ⎯⎯Fe → 2NH3(g)
Contact process in the manufacture of sulphuric acid. V2O5 catalyses the
SPM oxidation of sulphur dioxide to sulphur trioxide.

’10/P1 2 SO2(g) + O2(g) ⎯V2⎯O5→ 2SO3(g)

Nickel SO3 is used to manufacture sulphuric acid.
Manufacture of margarine. Nickel catalyses the hydrogenation of
Platinum unsaturated vegetable oil into saturated oil in the production of
margarine.

Ostwald process in the manufacture of nitric acid.

Transition elements can form complex ions

1 A complex ion is a polyatomic cation or anion Table 4.14 Formula
consisting of a central metal ion with other Complex ions
groups bonded to it. Tetraamminecopper(II) Cu(NH3)24+
Hexaamminechromium(III) Cr(NH3)36+
2 Table 4.14 shows some examples of complex Co(H2O)26+
ions formed by transition elements. Fe(CN)64–
Fe(CN)63–
Hexaaquocobalt(II)

Hexacyanoferrate(II)

Hexacyanoferrate(III)

Reaction of Aqueous Solutions of Transition 2 The ions of transition elements react with
Element Compounds with Sodium Hydroxide the hydroxide ions to form coloured metal
and Ammonia Solutions hydroxide precipitates.

1 The presence of ions of transition elements in 3 Table 4.15 gives some examples of the
a solution can be confirmed by using sodium
hydroxide solution or ammonia solution. reactions of aqueous solutions of ions of

transition elements with sodium hydroxide

solution and ammonia solution.

Periodic Table of Elements 96

Table 4.15

Ions Aqueous sodium hydroxide solution, NaOH(aq) Aqueous ammonia solution, NH3(aq)

Fe2+ Green precipitate of iron(II) hydroxide is Green precipitate of iron(II) hydroxide is formed.
formed. Precipitate is insoluble in excess NaOH Precipitate is insoluble in excess aqueous NH3
solution. solution.

Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s)
(from NaOH) (green precipitate) (from NH3) (green precipitate)

Fe3+ Brown precipitate of iron(III) hydroxide is Brown precipitate of iron(III) hydroxide is formed.
formed. Precipitate is insoluble in excess NaOH Precipitate is insoluble in excess aqueous NH3
solution. solution.

Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) 4
(from NaOH) (brown precipitate) (from NH3) (brown precipitate)

Cu2+ Blue precipitate of copper(II) hydroxide is Blue precipitate of copper(II) hydroxide is formed.
formed. Precipitate is insoluble in excess NaOH
solution. Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s)
(from NH3) (blue precipitate)
Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s)
(from NaOH) (blue precipitate) Precipitate is soluble iisn deaxrckesbs lNueHi3nsocolulotiuorn. to form
a complex ion which

Cu(OH)2(s) + 4NH3(aq) → Cu(NH3)24+(aq) + 2OH–(aq)
(dark blue solution)

6 ’03 Cu(NH3)24+ is a dark blue complex ion. Thus, Q is
copper(II) oxide.
Q is a compound. Study the flowchart below and Answer A
identify Q.

Solid Q

+ HNO3(aq) +NaOH(aq) Aqueous ammonia contains hydroxide ions:

Blue solution is formed Not soluble NH3 + H2O NH4+ + OH–

+ NaOH(aq) + excess ammonia

Blue precipitate solution Dark blue
solution

A Copper(II) oxide C Copper(II) sulphate Have high melting, Form Have more
boiling points and coloured than one
B Iron(III) oxide D Iron(II) sulphate densities compounds oxidation state

Comment

Q is a base, since it dissolves in nitric acid. Metal

oxides are bases. Since it forms a blue precipitate, Transition elements

Q must contain copper(II) ions. The above reactions

can be represented by the equations

CuO(s) + 2HNO3(aq) → Cu(NO3)2(aq) + H2O(l) The presence of Used as catalysts Form
their ions can be in chemical complex
Cu(NO3)2(aq) + 2NaOH(aq) → confirmed using reactions ions
2NaNO3(aq) + Cu(OH)2(s) NaOH(aq) or
(blue precipitate) NH3(aq)

Cu(OH)2(s) + 4NH3(aq) → Cu(NH3)24+(aq) + 2OH–(aq)

97 Periodic Table of Elements

4 4.6 in a table for easy studying. Scientists who
played prominent roles in the development of
1 You are given two blue aqueous solutions; one the Periodic Table were J. W. Dobereiner, John
containing Cu2+ ions and another containing a blue Newlands, Lothar Meyer, Dmitri Mendeleev
food colouring. Explain how you can differentiate and Henry Moseley.
between the two solutions.
Uses of the Elements and Compounds in
2 The body of U-2 spy plane is made of titanium Our Daily Life
alloy. Give three reasons why titanium alloy is
used? 1 An element is a substance that cannot be
broken down into simpler substances.
4.7 Appreciating the
Existence of Elements Table 4.17 shows the uses of some of these
and Their Compounds elements in daily life.

1 In ancient times, gold, mercury, copper, 2 A compound is a substance made by
iron, sulphur, tin, antimony, lead, diamond, chemically combining atoms of two or more
and graphite were already discovered. Credit elements.
should be given to the scientists who, through
their persistent efforts, discovered and isolated 3 Many compounds have been synthesised by
other elements. Table 4.16 below shows the chemists to improve our standard of living.
scientists who isolated some of these elements. Many items in our home are made up of
elements. The shirt you are wearing may be
2 Then about a century (100 years) later, made of Terylene, your sofa may be made of
scientists discovered the subatomic particles of polyvinyl chloride, the cooking utensils in
atoms. Credit is given to J. J. Thomson, Ernest your kitchen are made of stainless steel, glass
Rutherford, James Chadwick and Niels Bohr or ceramics, part of the motor of your ceiling
in helping us understand the atomic structure fan is made of copper and the microprocessors
of atoms in elements. The atomic structure in the computer are made of silicon and
of atoms helps us understand how elements germanium. Elements are used to make
react to form compounds. vehicles, medicine and communication tools
like the handphone. Imagine our life without
3 As more elements are isolated, there is a need these elements. Coloured compounds of
to classify the elements. Elements with the transition elements are used as paint pigments
same chemical properties are grouped together and they make our homes more colourful.
Table 4.18 shows the uses of some of these
compounds.

Element Symbol Table 4.16 Isolated in
Hydrogen H Scientist 1766
Nitrogen N 1772
Oxygen O Henry Cavendish (British) 1774
Sodium Na Daniel Rutherford (British) 1807
Potassium K Carl Wilhelm Scheele (Swedish) 1807
Magnesium Mg Sir William Ramsay (British) 1808
Aluminium Al Sir Humphry Davy (British) 1825
Uranium U Sir Humphry Davy (British) 1841
Helium He Hans Christian Orsted 1895
Radium Ra Eugene Melchior Peligot (French) 1898
Sir William Ramsay (British)
Pierre Curie and Marie Curie (French)

Periodic Table of Elements 98

Element Formula Table 4.17 Uses of some elements 4
Hydrogen H2 Uses

Aluminium Al Used in the hydrogenation of unsaturated oil to produce margarine.
Used in the manufacture of ammonia.
Silicon Si Used in the production of duralumin alloy for use in the construction of
Sulphur S aeroplane bodies.
Chlorine Cl2 To make microchips
Iron Fe To make matches, fireworks and for the manufacture of sulphuric acid
Copper Cu To kill germs in drinking water
Cobalt-60 Co Production of steel
To make electrical wire, electrical motor, dynamo and coins (cupronickel alloy)
The gamma rays emitted from this radioactive element is used to kill cancer cells.

Table 4.18 Uses of some chemical compounds

Compound Chemical Formula Uses
Magnesium oxide
MgO In antacid drugs to treat gastric patients. Also used to treat acid
Tin fluoride poisonings.

Sodium bicarbonate SnF2 Is added to toothpaste.
Silver bromide Fluoride ions can strengthen the teeth.
Vinyl chloride
Ammonia NaHCO3 Used as a baking powder. Also used to treat acid burns.
Sodium hydroxide AgBr Used in the making of photographic films.

CH2CHCl To make PVC pipes, toys, raincoats and cushions
NH3 To manufacture fertiliser, nitric acid and explosives
NaOH To make soap

Preventing Wastage pollution because many of these items are
non-biodegradable.
1 Most of the elements needed to keep life 3 We can minimise wastage in the school
going are obtained from the Earth’s crust. A laboratory by practising the following:
number of these elements are obtained from (a) Weigh or use the correct amount of
the oceans and the air. After extraction, these
chemicals cannot be replaced and will be chemicals that is required to carry out an
depleted one day. So it is important that we experiment.
avoid wastage in using these elements. (b) Read the label of the chemicals carefully,
so that we do not take the wrong
2 Recycling of used materials is an alternative substance.
method. Used glass bottles, plastic bottles (c) Read the procedure of the experiment
and aluminium cans should be separated and plan the experiment carefully before
and thrown into different bins provided by carrying them out, so as to avoid the need
the government. This will also help to reduce to repeat the experiment.

99 Periodic Table of Elements

1 Chemical elements are classed into groups in the (b) HaloCgle2n(gs)re+acHt2Owi(tlh) → HCl(aq) + HOCl(aq)
alkali metals to form halide
Periodic Table, with elements in the same group
salts, e.g.
having the same chemical properties.
2 Group 1 elements are called alkali metals. 2Na(s) + hCol2t(igro) n→wo2oNlatCol(fso)rm
(c) Halogens react with brown
(a) They react with cold water to produce an
iron(III) halide salt, e.g.
alkaline solution and hydrogen gas, e.g.
5 The reactivity2Foef (sG)ro+up3C1l27(g)el→em2eFnetCiln3(csre)ases up
(b) 2Na(s) + w2Hith2Oh(la)lo→gen2NgaaOsHt(oaqf)or+mH2h(agl)ide
They react the group because the atomic radius decreases.

salts, e.g. Cfol2rm(g)m→eta2lNoaxCidl(ess),
2Na(s) + The smaller the atomic radius, the stronger the
(c) They burn in air to e.g.
4 electrostatic force of attraction between the nucleus

All oxides4Nofa(Gs)ro+upO21(l)m→eta2lsNac2aOn(sd)issolve in and the electron. Thus the elements higher in the

water to form alkaline solutions. group can accept electrons more easily.
6 Group 18 elements exist as monatoms because
reactivityNoaf2OG(ros)up+1He2lOem(l)en→ts 2NaOH(aq)
3 The increases down the they have attained the duplet or octet electronic

group because the valence electron is further from the configuration. They do not need to donate, accept

nucleus. The electrostatic force of attraction between or share electrons with other elements.
7 Transition elements are a block of elements
the nucleus and the valence electron becomes weaker.
between Group 2 and Group 13 in the Periodic
Thus the elements lower in the group can release their
Table. The characteristics of transition elements are
valence electrons more easily.
4 Group 17 elements are called halogens. (a) they form coloured compounds

(a) Halogens dissolve in water to form acidic (b) they have more than one oxidation number

solutions, e.g. (c) they catalyse chemical reactions

(d) they form complex ions

4

Multiple-choice Questions

4.1 Periodic Table of Elements 2 The structure of a compound 4 Atom of element X has proton
containing element Q and number of 19. Which of the
1 The diagram below shows the hydrogen is shown as follows. following statements are true
electron arrangement of atom Y. concerning X ?
H H I X has one valence electron.
x xo xo II The oxide of element X is
H ox Qxo xx xQo ox H soluble in water.
xx III X is in Period 4 of the
x Periodic Table.
IV X has the same chemical
Yxx x x H H properties as fluorine.
x x A I and III only
To which group of the Periodic B II and IV only
x C I, II and III only
xx Table does element Q belong? D I, III and IV only

A 14 C 16 5 An atom of element Y has a
nucleon number of 31 with 16
Which of the following is the B 15 D 17 neutrons.
position of atom Y in the
Periodic Table? 3 Which of the following elements
belong to the same group in the
A Group TC 52 Period Periodic Table?
B 3 2
7 P 9 Q 27 R 35 S T40
33 3 4 13 17
20

C 13 2 A P and R C P and S
B R and T D Q and T
D 13 3

Periodic Table of Elements 100

Which of the following statements 4.2 Group 18 Elements 14 Which of the following explains
is true concerning element Y? why sodium is more reactive
A Atom of element Y has 16 7 The symbols for four elements
are shown below. ’06 than lithium?
protons. A Sodium has more protons
B Atom of element Y has two 4 W 16 X 24 Y Z40 than lithium.
2 8 12 B Sodium has less valence
filled electron shells. 18 electrons than lithium.
C Atom of element Y has six C Sodium has a lower melting
Which of the following statements point than lithium. 4
valence electrons. is true? D Sodium can release its
D Element Y is in Group 15 of A Both elements W and Z are valence electron more easily.

the Periodic Table. monatomic. 15 An element X is burned in air.
B Element Z is more reactive The product of combustion is
6 An element X has a proton then dissolved in water.
number of 9. In which group than element X. The solution gives a pH value of
and period does X belong to in C Elements W and Y react to 14. The element X is
the Periodic Table? A potassium
form a compound with the B phosphorus
Group Period C aluminium
2 formula YW. D sulphur
A 16 2 D Elements X and Y react to
B 17 3 16 Which statement below is true
C 16 3 form a compound with the about sodium?
D 17 A It burns in air to form sodium
formula YX2. oxide of formula NaO which
is soluble in water.
8 Which of the inert gases below B It burns in air to form sodium
oxide of formula Na2O which
is used in a diver’s oxygen tank? is soluble in water.
C It burns in air to form sodium
A Neon C Helium oxide of formula NaO which
is insoluble in water.
B Argon D Krypton D It burns in air to form sodium
oxide of formula Na2O which
9 Which element, as represented by the symbols below, exists as is insoluble in water.
monatoms?
17 Which statement concerning the
A W B X C Y D Z ions of Group 1 elements is
correct?
10 The element which does not form compound with other elements is likely A Each contains more protons
than electrons.
to have a proton number of B Each contains more electrons
than protons.
’06 A 4 B 6 C 8 D 10 C Each has one electron in its
outer electron shell.
11 As we go down Group 18, III the boiling point increases. D Each contains the same
number of protons and
I the density increases. IV the atomic size decreases. electrons.
C I, II and III only
II the reactivity decreases. D I, III and IV only 18 As we go down Group 1,
A I and III only I the density increases
B II and IV only II the melting point increases
III the reactivity increases
12 Argon does not form compound with chlorine because IV the electropositivity decreases
A it has three filled electron shells. A I and II only
B it has low melting and boiling points. B I and III only
C it has eight electrons in the outermost electron shell. C II and III only
D the atoms have the same number of protons and neutrons in the D II and IV only

nucleus.

4.3 Group 1 Elements

13 Rubidium is below sodium in Group 1. Which statements below about

rubidium are correct?

’08 I It is more reactive than sodium.

II It burns in air to form rubidium oxide which is insoluble in water.

III It is produced from the electrolysis of molten rubidium chloride.

IV It reacts with cold water to form rubidium hydroxide and hydrogen.

A I and III only C I, III and IV only

B II and IV only D I, II, III and IV

101 Periodic Table of Elements

19 The element X has a proton 23 Which of the statements below C Reacts with iron wool to
number of 19. Element X is not true concerning Group 17
A forms a basic oxide that is elements? form a compound with the
insoluble in water. A Bromine is a gas at room
B reacts with cold water to temperature. D formula FeX3. to form
form an acidic solution. B Chlorine is more reactive Reacts with sodium
C reacts with chlorine to form a than bromine.
compound with formula C As we go down Group 17, a compound with formula
D iXsCml2.ore reactive than an the density increases.
element with a proton D Chlorine is more NaX2.
number of 11. electronegative than bromine.
4 28 Which of the following
4.4 Group 17 Elements 24 As we go down Group 17,
I the reactivity decreases statements about the Periodic
20 Which of the following explains II electronegativity increases Table is true?
why chlorine is more reactive III the melting point increases A All inert gases have eight
than bromine? IV the solubility of the element
A Chlorine is less dense than in water decreases valence electrons.
bromine. A I, II and III only B All Group 1 elements are
B Chlorine has a lower boiling B I, III and IV only
point than bromine. C II, III and IV only equally reactive.
C Chlorine atom accepts an D I, II, III and IV C The atomic radius increases
electron more easily than
bromine atom. 25 Which statement about ions of in size across a period from
D Chlorine atom contains Group 17 elements is correct?
less protons than bromine A Each has eight valence left to right.
atom. electrons. D The electronegativity of
B Each contains more protons
21 Which of the following halogens than electrons. Group 17 elements increases
dissolve in water to produce C Each contains more protons
an acidic solution which can than neutrons. up the group.
decolourise the colour of litmus D Each has an odd number of
paper? electrons. 4.5 Elements in a Period
I Chlorine
II Bromine 26 Which of the following statements 29 The table shows the proton
III Iodine are true about chlorine and numbers of four elements.
IV Astatine bromine?
A I and II only I Both are gases at room Element P Q R S T
B I and III only temperature.
C II and IV only II They react with sodium to Proton 10 11 14 17 19
D III and IV only form soluble salts. number
III They react with heated iron
22 An element from Group X can wool to form iron(II) halides Arrange the atomic radii of the
dissolve in water to form an IV They dissolve in water to
acidic solution. form solutions with pH elements in increasing order.
This resulting solution reacts values of less than 7. A P, Q, R, S, T
with silver nitrate reagent to A I and III only B P, S, R, Q, T
form a white precipitate. B II and IV only C T, S, R, Q, P
In which group of the Periodic C I, II and IV only D T, Q, R, S, P
Table does the element belong D II, III and IV only
to? 30 Which of the following metallic
A 1 27 An element X has a proton
B 2 number of 35. Which statement oxides can react with both acid
C 16 about X is true?
D 17 A X forms a positive ion during ’06 and alkaline solutions?
chemical reactions.
B It belongs to Group 15 of I Aluminium oxide
the Periodic Table.
II Tin(II) oxide

III Lead(II) oxide

IV Copper(II) oxide
A I and III only
B II and IV only
C I, II and III only
D I, III and IV only

31 X, Y and Z are elements in the

same period of the Periodic

Table.

The oxide of X is acidic, the

oxide of Y is basic and the oxide

of Z is amphoteric.

Arrange the elements X, Y and Z

in increasing proton number.

A Z, X, Y C Y, X, Z

B X, Z, Y D Y, Z, X

Periodic Table of Elements 102

32 X, Y and Z are elements in Element XYZ Which of the following
Period 3 of the Periodic Table. statements below are true
The table shows the properties Proton number 11 13 17 about the elements in the
of the oxides of X, Y and Z Periodic Table?
when reacted with sodium Which of the following I Element W is inert.
hydroxide and nitric acid statements is true? II Element Y has more than
solutions. A X, Y and Z are all conductors
one oxidation state.
Metal Observation of electricity. III Element X reacts with cold
oxides B All the elements above are
X oxide Sodium Nitric acid water to form X oxide and
hydroxide solution made up of atoms. hydrogen.
Y oxide solution C The atomic radius decreases IV Element Z reacts with iron
to form a compound with
Z oxide Dissolves Dissolves in the order Z, Y, X. A fIIoramndulaIVFoenZl3y.
to form to form D The electronegativity B I, II and III only
colourless colourless C I, II and IV only
solution solution increases in the order X, Y, Z. D I, II, III and IV

4.6 Transition Elements 37 The properties of the transition 4
metals include
Dissolves No change 35 Which of the following elements
to form ’08 I they form white compounds.
colourless will form coloured compounds? II they act as catalysts in
solution chemical reactions.
I Cobalt III Aluminium III they have more than one
oxidation state.
II Nickel IV Manganese IV they form complex ions.
A I, II and III only
No change Dissolves A I, II and III only B I, II and IV only
to form C II, III and IV only
colourless B I, II and IV only D I, II, III and IV
solution
C II, III and IV only

D I, II, III and IV

What is the correct arrangement 36 The diagram shows part of a
in increasing proton number of Periodic Table.
the elements?
A X, Y, Z 38 The table shows the properties of four elements. Which element is most
B Y, X, Z likely to be a transition element?
C Z, Y, X
D Z, X, Y Element Melting point (°C) Density (g cm–3) Electrical conductivity

33 The graph shows the change A 360 6.8 Good
in a property as we go across B 620 3.1 Good
Period 3 of the Periodic Table. C 3500 3.5 Poor
D 3400 8.1 Good

39 An element X forms two oxides with the formula XO and X2O. Which of
the following statements is true about the element X ?

A It is a transition metal.

B It is a Group 1 element.

C It is a Group 17 element.

D The oxides of element X are amphoteric.

Which of the properties below 40 Which of the following shows correctly the colour of the ions of the
corresponds to the change as transition elements?
shown in the graph above?
A Atomic radius Colour
B Electropositivity
C Electrical conductivity Fe2+ Cr2O72– MnO4– Co2+
D Density A Brown Orange Purple Pink
B Green Orange Purple Pink
34 The table shows the proton C Brown Purple Orange Blue
numbers of three elements X, Y D Green Purple Orange Blue
and Z.

103 Periodic Table of Elements

Structured Questions

1 Bromine is an element of Group 17 in the Periodic 4 A list of elements represented by the letters with the
Table. nucleon numbers and proton numbers are given
below:
(a) What is the physical state of bromine at room

temperature? [1 mark] 11P, 126Q, 199R, 2173S, 3157T, U39

(b) Write an equation for the reaction between 19

bromine and water. [1 mark] (a) Choose two elements from the list above that
belong to the same group in the Periodic Table.
(c) Draw a labelled diagram for the apparatus that
[1 mark]
can be used to carry out a reaction between
(b) State the (i) group and (ii) period of element U.
bromine and iron wool. [2 marks] [2 marks]

(d) What is the number of valence electrons in (c) Give one use of the element P.

4 bromine? [1 mark] [1 mark]

(e) Iodine is below bromine in Group 17. Which of (d) Draw the atomic structure of element R.
[2 marks]
the two elements, iodine or bromine, is more

reactive? Explain your answer. [3 marks] (e) (i) Which of the elements in the list reacts with

2 Rubidium is placed below potassium in Group 1 in cold water to produce hydrogen gas?
the Periodic Table.
[1 mark]

(a) Give two physical properties of rubidium. (ii) Write a balanced chemical equation for the
[2 marks]
reaction in (i). [1 mark]

(b) How is rubidium stored in the laboratory? (f) Write the formula of the ion formed by the
[1 mark]
element U. [1 mark]

(c) Write the equations for the reactions of rubidium (g) (i) What is electronegativity?
(Rb) with (i) water and (ii) chlorine gas. (ii) Which is the more electronegative element

[2 marks] between R and T? Explain your answer.

(d) Explain why rubidium is more reactive than [3 marks]

potassium. [2 marks] 5 A list of the symbols of the transition elements in
Period 4 is given below:
(e) Write the formula of (i) rubidium nitrate and (ii)

rubidium sulphate. [2 marks] Ti, V, Cr, Mn, Fe, Co, Ni, Cu

(f) What is the colour of rubidium nitrate? [1 mark] (a) Give two physical properties of the transition

3 Diagram 1 shows a portion of the Periodic Table. elements. [2 marks]
The letters in the Periodic Table do not represent the
(b) What is the colour of the aqueous Cu2+ ion
’05 actual symbols of the elements.
solution? [1 mark]

(c) Name a transition element in the list which is

used as a catalyst in the

(i) Haber process to manufacture ammonia.

[1 mark]

Diagram 1 (ii) hydrogenation of unsaturated oil to make

(a) Choose two elements which are metals in the margarine. [1 mark]

Periodic Table above. [2 marks] (d) Other than forming coloured compounds and

(b) Write the formula of the ions formed by the having catalytic properties, name two other

elements (i) E and (ii) Q. [2 marks] properties of transition elements. [2 marks]

(c) Choose an element from the Periodic Table (e) Name a reagent that can be used to differentiate
above that can form a coloured compound.
between the ions of transition elements in the list
[1 mark]
above. [1 mark]

(d) Choose an element that exists as monatoms. 6 The symbols of the elements in Period 3 of the
Periodic Table are given below:
Give a use of this element. [2 marks]

(e) Which element from the Periodic Table above Na, Mg, Al, Si, P, S, Cl, Ar

can form an acidic oxide? [1 mark]

(f) Which is the more reactive element between E (a) Name an element that can conduct electricity in

and R? Explain your answer. [2 marks] Period 3. [1 mark]

Periodic Table of Elements 104

(b) How does the atomic radius change across the (f) (i) Name an acidic oxide of Period 3 that can

period? Explain your answer. [3 marks] dissolve in water. [1 mark]

(c) Name an element that exists as monatoms in (ii) Write an equation for the reaction that

Period 3. Explain your answer. [2 marks] takes place when the acidic oxide in (i) is

(d) Write the formulae of the oxides of the elements dissolved in water. [1 mark]

in Period 3. [3 marks] (g) The oxide of aluminium can react with both acid

(e) (i) Name a basic oxide of Period 3 that can and alkali solutions. What is the term given to

dissolve in water. [1 mark] such an oxide? [1 mark]

(ii) Write an equation for the reaction that takes (h) Give a use of the element silicon. [1 mark]

place when its basic oxide in (i) is dissolved

in water. [1 mark]

Essay Questions 4

1 (a) The reactivity of alkali metals increases down (b) Explain why Group 18 elements exist as
the group in the order:
monatoms. [5 marks]
lithium < sodium < potassium (c) Chlorine II Iron(III) chloride

Explain the statement given above. [7 marks] I

(b) Rubidium (Rb) is placed below potassium in Sodium chloride

Group 1 of the Periodic Table. Predict three The flowchart shows the conversion of chlorine to

physical properties and three chemical properties sodium chloride and iron(III) chloride.

of rubidium. [10 marks] Explain a method to carry out an experiment

(c) The nucleon number of sodium is 23 and its (i) to obtain sodium chloride from chlorine in

atom has 12 neutrons. The nucleon number of conversion I. [5 marks]

chlorine atom is 35 and it has 18 neutrons. Prove (ii) to obtain iron(III) chloride from chlorine in

that sodium and chlorine belong to the same conversion II. [5 marks]

period in the Periodic Table. [3 marks] 4 (a) Using a suitable period in the Periodic Table as

2 (a) Explain the following statements: an example, explain the trend in the properties of

(i) The reactivity of Group 17 elements the elements in terms of metals, non-metals and

’06 decreases down the group. [7 marks] semi-metals. [5 marks]

(ii) The atomic radius of Period 3 elements (b) Using a suitable period in the Periodic Table as

decreases across Period 3 from left to right. an example, explain why the electronegativity of

[5 marks] the elements increases across a period from left

(b) With suitable examples, discuss the four properties to right. [5 marks]
of transition elements in the Periodic Table.
(c) Write the formulae of all the oxides of the elements
[8 marks] in Period 3. Describe a suitable experiment to show
that the oxides change from basic to amphoteric
3 (a) Name three elements in Group 18 and state their and then to acidic across Period 3. [10 marks]

uses. [5 marks]

Experiments

1 The reactivity of a halogen in the reaction with
iron wool depends on its position in Group 17.
Diagram 1 shows the set-up of apparatus for an
experiment to determine the reactivity of halogens
in Group 17.

Diagram 1

105 Periodic Table of Elements

The experiment is carried out using bromine gas, chlorine gas and iodine vapour to react with
heated iron wool respectively. Observation of the experiment is shown in Table 1 below.

Halogen Observation
Bromine Hot iron wool glows moderately bright when bromine gas is passed over it.
Chlorine Hot iron wool glows brightly when chlorine gas is passed over it.
Iodine Hot iron wool glows dimly when iodine vapour is passed over it.

Table 1
(a) Complete Table 2 below based on the experiment.

4 Variables Action to be taken
(i) Manipulated variable : (i) The way to manipulate variable :

(ii) Responding variable : (ii) What to observe in the responding variable :

(iii) Constant variable : (iii) The way to maintain the constant variable :

Table 2 [6 marks]
(b) State one hypothesis for the experiment. [1 mark]
(c) Based on the observation, arrange bromine, chlorine and iodine in descending order of reactivity of [1 mark]
[1 mark]
halogens with iron wool. [1 mark]
(d) The proton numbers of the halogens increase in the order: chlorine < bromine < iodine.
[3 marks]
Make a conclusion regarding the positions of halogens in Group 17 in relation to their reactivities. [3 marks]
(e) Astatine is a halogen below iodine in Group 17. Predict the reactivity of astatine with iron wool. [3 marks]
[3 marks]
2 [3 marks]
Lithium, sodium and potassium are in Group 1 of the Periodic Table.

’08 The reactivity of Group 1 elements increases down the group from lithium to potassium.

You are required to design a laboratory experiment to prove the statements above. Your explanations
should include the following:
(a) Problem statement
(b) Hypothesis
(c) List of materials and apparatus
(d) Procedure
(e) Tabulation of data

Periodic Table of Elements 106

5CHAPTER FORM 4

THEME: Interaction between Chemicals

Chemical Bonds

SPM Topical Analysis

Year 2008 2009 2010 2011

Paper 1 2 31 2 31 2 31 2 3

Section ABC ABC ABC ABC

Number of questions 5 —13 – – – 3 —14 – – – 6 1 – – – 5 – – —21 –

ONCEPT MAP

CHEMICAL BONDS
To attain the stable electron arrangement of the noble gases

Ionic bonds Covalent bonds

Formed by transfer of electrons from Formed by sharing of electrons
metal atoms to non-metal atoms between non-metal atoms

Metal atoms donate electrons Non-metal atoms accept electrons Non-metal atoms share electrons

Example: CH4

Example: NaCl /
/

¶ /*/ /* /
5H *S
5H *S

/
/

Differences in physical properties

Melting point and boiling point Solubility Electrical conductivity

5.1 Formation of Compounds 9 Helium has only one electron shell filled with
2 electrons (a duplet electron arrangement).
Stability of Noble Gases All other noble gases have 8 electrons in
the valence shell. This is known as an octet
5 1 A compound is a chemical substance that is electron arrangement.
formed by combining two or more elements
10 A duplet electron arrangement (as in helium)
chemically in fixed proportions. or an octet electron arrangment (as in the
2 Almost all chemical substances exist as other noble gases) is very stable. As such,
atoms of noble gases do not donate, accept
compounds in nature. Examples of compounds or share electrons with other elements. Thus,
are water (H2O), carbon dioxide (CO2), table atoms of noble gases do not combine with
salt sodium chloride (NaCl) and minerals other elements or with itself. They exist as
monatoms.
such as metal silicates, metal oxides, metal
Conditions for the Formation of
carbonates and metal sulphides. Chemical Bonds
3 Only noble gases and a few minerals such as
1 Noble gases do not form chemical bonds
gold, diamond and platinum exist as pure because they have the stable duplet or octet
electron arrangement.
elements.
4 The tendency of elements to combine with 2 Atoms of elements from Group 1 to Group
17:
other elements to form compounds shows that (a) Have less than 8 valence electrons
compounds are more stable than elements. (b) Each atom will tend to donate, accept
5 The formation of compounds proves that or share electrons to achieve the stable
chemical bonds hold atoms of elements duplet or octet electron arrangement as
together. that of a noble gas
6 Noble gases are elements in Group 18 of the
Periodic Table. They are also known as inert 3 In the process of attaining the stable duplet or
octet electron arrangement, chemical bonds
gases which consists of helium, neon, argon, will form between atoms of these elements.

krypton, xenon and radon. 4 The two types of chemical bonds are
7 Noble gases exist as elements. They are very (a) ionic bonds (formed by transfers of
electrons), and
stable and are inert, which means they are (b) covalent bonds (formed by sharing of
non–reactive chemically. electrons).
8 The stability of noble gases is due to their electron
arrangements (or electronic configurations) as 5 In the formation of chemical bonds, only
valence electrons are involved in the donation,
shown in Table 5.1. acceptance or sharing of electrons. Electrons
in the inner shells are not involved.
Table 5.1 Electron arrangements of noble gases
6 The valence shell will then achieve an octet
Noble gas Symbol Electron electron arrangement or a duplet electron
arrangement arrangement (in the case where there is only
one electron shell).
Helium He 2
Neon is inert not just because it is a Group 18 element
Neon Ne 2.8 but because it has a stable octet electron arrangement
with eight electrons in the outermost shell.
Argon Ar 2.8.8

Krypton Kr 2.8.18.8

Xenon Xe 2.8.18.18.8

Radon Rn 2.8.18.32.18.8

Chemical Bonds 108

5.1 electrons respectively. In chemical reactions,
these metal atoms tend to donate all their
1 The proton numbers of neon and argon are 10 and valence electrons to achieve the stable duplet
18 respectively. Write the electron arrangements of or octet electron arrangement.
neon and argon. Explain why these two elements 6 A negative ion (or anion) is formed when an
exist as monatoms. atom accepts one or more electrons. The ion
formed is negatively-charged because there are
2 The electron arrangements of atoms P, Q and R more electrons than protons. For instance,
are given in the table below.

Atom Electron arrangement 2–
P 2.8.2
Q 2.8.7 accepts 2 10e
electrons 8p
8p
5
8e

R 2.8.8 Oxygen atom (O) Oxide ion (O2–)
(8p, 8e) (8p, 10e)
(a) Which atom is chemically inert? Explain your
answer. 7 Generally,
(a) non–metals usually form negative ions,
(b) Which atom will take part in chemical bonding? (b) charge of negative ion =
Explain your answer. number of electrons received by an atom

3 State two types of chemical bonds.

X + ne– → Xn–

8 Non–metal atoms from Groups 15, 16 and 17

5.2 Formation of Ionic Bonds SPM in the Periodic Table have 5, 6 and 7 valence
electrons respectively. In chemical reactions,
’08/P1

Formation of Ions non-metal atoms will accept electrons so
that the ion formed achieves the stable octet

1 Atoms are neutral because the number of electron arrangement.

protons is the same as the number of electrons.

2 An ion is formed when an atom donates or

receives one or more electrons. • The term electron arrangement is used interchangeably
3 A positive ion or cation is formed when an with electronic configuration.

atom donates one or more electrons. The • Donate electrons can also be explained as lose electrons
ion formed has less electrons than protons or release electrons.
and is positively-charged. For example,
• Accept electrons can also be explained as gain

+ electrons or receive electrons.
• The name of a metal ion is the same as the metal
11e donates an 10e
11p electron 11p atom. Examples: sodium ion (Na+), magnesium ion
(Mg2+), aluminium ion (Al3+).
• The name of the non-metal ion ends with -ide or

-ate (when oxygen is attached to the non-metal).

Examples: chloride ion (Cl–), sulphide ion (S2–),

Sodium atom (Na) Sodium ion (Na+) sulphate ion (SO42–), carbonate ion (CO32–).
(11p,11e) (11p,10e)

4 Generally, • The duplet or octet electron arrangement of noble
(a) metal atoms usually form positive ions, gases are very stable.
(b) charge of positive ion =
number of electrons released by an atom • Atoms form positive ions or negative ions so as to
attain the electron arrangement as that of the noble
M → Mn+ + ne– gases.

5 Metal atoms from Groups 1, 2 and 13 in
the Periodic Table have 1, 2 and 3 valence

109 Chemical Bonds

To prepare ionic compounds

Materials Magnesium ribbon, sodium, 2 The ignited sodium is placed in a gas jar filled with
Apparatus chlorine gas, iron wool and sodium chlorine gas. Any changes that occur are recorded.
hydroxide solution.

Tripod stand, clay pipe triangle,
Bunsen burner, crucible and lid,
sandpaper, gas jar, gas jar spoon,
combustion tube, filter funnel, retort
stand, clamp and beaker.

Activity 5.1 (A) Preparation of magnesium oxide Figure 5.2 Preparation of sodium chloride
5
Procedure (C) Preparation of iron(III) chloride

1 A 5 cm length of magnesium ribbon is cleaned Procedure
with a piece of sandpaper.
1 A little iron wool is placed inside a combustion
2 The magnesium ribbon is placed in the crucible. tube.
3 The magnesium ribbon is heated strongly. Any
2 The end of the combustion tube is connected to
changes that occur are recorded. a filter funnel inverted into a beaker with some
sodium hydroxide solution.

3 The iron wool is heated strongly until it glows.
4 Chlorine gas is passed through the iron wool

while being heated. Any changes that occur are
recorded.

Figure 5.1 Preparation of magnesium oxide

(B) Preparation of sodium chloride

Procedure

1 A small piece of sodium metal is placed in a gas
jar spoon and is heated carefully until it begins
to ignite.

Figure 5.3 Preparation of iron(III) chloride

Results

Method Observation Inference
The white powder formed is
Heating of magnesium in air • The magnesium ribbon burns with a magnesium oxide
bright flame
The white powder formed is
• White powder is formed sodium chloride

Burning of sodium in chlorine • Sodium burns with a bright yellow flame The brown powder formed is
gas • The yellowish-green colour of chlorine iron (III) chloride

gas is decolourised.
• White fumes are produced and deposited

as white powder

Heating of iron in chlorine gas • The iron wool continues to glow brightly
in the chlorine gas

• A brown powder is formed

Chemical Bonds 110

Discussion 7 Sodium hydroxide solution is used to absorb the 5
1 When magnesium is heated, magnesium excess chlorine gas. Besides sodium hydroxide,
soda lime can also be used.
atom combines with oxygen in the air to form
magnesium oxide. 8 A filter funnel is used to prevent the sodium
hydroxide solution from being suctioned back
2Mg(s) + O2(g) → 2MgO(s) into the combustion tube as chlorine gas is very
soluble in water.
2 Magnesium oxide is an ionic compound that
contains Mg2+ ions and O2– ions. Conclusion

3 When sodium burns in chlorine gas, the white 1 Generally, the reaction between metals and non-
powder formed is sodium chloride. metals produces ionic compounds.

2Na(s) + Cl2(g) → 2NaCl(s) 2 Ionic compounds such as magnesium oxide,
sodium chloride and iron(III) chloride can be
4 Sodium chloride is an ionic compound that prepared by direct combination of the metal and
contains Na+ ions and Cl– ions. non-metal elements.

5 When iron burns in chlorine gas, the brown Metal Non-metal Ionic compound
powder formed is iron(III) chloride.
Magnesium + oxygen → magnesium oxide
2Fe(s) + 3Cl2(g) → 2FeCl3(s)
Sodium + chlorine → sodium chloride
6 Iron(III) chloride is an ionic compound that
contains Fe3+ ions and Cl– ions. Iron + chlorine → iron(III) chloride

Formation of Ionic Bonds SPM

’09/P1

1 Metals from Groups 1, 2 or 13 react with (c) non–metal atoms receive electrons to
non–metals from Groups 15, 16 or 17 in the form negative ions (anions).
Periodic Table to form ionic compounds.
(d) positive ions and negative ions are then
2 In the formation of an ionic bond,
attracted to each other by the strong
(a) electrons are transferred from a metal electrostatic force of attraction. The bond

atom to a non–metal atom. formed between ions of opposite charges
(b) metal atoms donate valence electrons to is known as ionic bond or electrovalent
bond.
form positive ions (cations).

Formation of ionic bond in sodium chloride SPM
’06/P2

1 A sodium atom with an electron Cl + e– ⎯→ Cl–
arrangement of 2.8.1 achieves stability after 2.8.7 2.8.8
it donates one valence electron to form a
sodium ion, Na+. The electron arrangement 3 Sodium ions, Na+ and chloride ions, Cl–
of the sodium ion, Na+, is 2.8, with an octet with opposite charges are attracted to
of valence electrons. each other by the electrostatic force of
attraction. This force of attraction is called
Na ⎯→ Na+ + e– the ionic bond.
2.8.1 2.8

2 A chlorine atom with an electron
arrangement of 2.8.7 achieves stability after
it accepts one electron from a sodium atom
to form a chloride ion, Cl­–. The electron

arrangement of the chloride ion, Cl–, is

2.8.8, with an octet of valence electrons.

111 Chemical Bonds

Formation of ionic bond in magnesium oxide SPM

’09/P1

1 A magnesium atom with the electron arrangement O + 2e– ⎯→ O2–
of 2.8.2 achieves the stable octet electron
arrangement when it donates two valence 2.6 2.8
electrons to form a magnesium ion, Mg2+.
3 The electrostatic force of attraction that exists
Mg ⎯→ Mg2+ + 2e– between the oppositely-charged magnesium
2.8.2 2.8 ions, Mg2+ and oxide ions, O2– forms the
ionic bond.

5 2 An oxygen atom with an electron arrangement
of 2.6 achieves the stable octet electron
arrangement when it accepts two electrons to
form an oxide ion, O2–.

Formation of ionic bond in potassium oxide

1 A potassium atom with an electron arrangement electrons from two potassium atoms to
of 2.8.8.1 achieves the stable electron form an oxide ion, O2–.
arrangement of an octet when it donates one 3 Hence each of the two potassium atoms donates
valence electron to form a potassium ion, K+. one valence electron to be accepted by one
oxygen atom. Ionic bonds are formed between
K ⎯→ K+ + e– the two potassium ions and the oxide ion.
2.8.8.1 2.8.8

2 An oxygen atom with an electron arrangement
of 2.6 achieves the stable octet electron
arrangement when it accepts two valence

How to Predict the Formula of an Ionic Compound

When drawing the electron 1 Metal atoms will donate their valence electrons to achieve
arrangements to show the formation the stable duplet or octet electron arrangement of the noble
of ionic bonds, gases.
• do not overlap the outermost
2 Non-metal atoms will accept electrons in order to achieve
electron shells of atoms. the stable octet electron arrangement of the inert gases.
• the outermost shells of all ions
3 For cations Mb+ and anion Xa–, the formula of an ionic
must have eight electrons except compound formed between them is written as
Li+ and H+.
• use ‘dots‘ or ‘crosses‘ to represent MaXb
the electrons from different atoms.
• show the charge of the ions clearly Number of electrons Number of electrons
outside the brackets of the ions. that will be received by that will be donated by
X (or charge of X ion) M (or charge of M ion)
The chemical formula of sodium
chloride, NaCl, tells us that 1 mol 4 The overall positive charge of the cation must be equal to the
of sodium ions combines with 1 overall negative charge of the anion in an ionic compound.
mol of chloride ions and it is not 1 Hence the formula of an ionic compound formed between
molecule of sodium chloride. them can also be derived as

Chemical Bonds aMb++ bXa– → MaXb

112

1 2

Atoms in element J and element Q have proton Element M is an element from Group 13 and element
numbers 12 and 17 respectively. Explain what type X is an element from Group 16 in the Periodic
of bond will form between J and Q. Predict the Table. What is the formula of the compound formed
formula of the compound formed. between element M and element X?

Solution Solution
An atom of J, with a proton number of 12 and
an electron arrangement of 2.8.2, will achieve the M from Group 13 has three valence electrons and
stable octet electron arrangement by donating two will form M 3+ ion. X from Group 16 will receive
electrons to form a J2+ ion. two electrons to form X 2– ion. The formula of the
compound formed is

J ⎯⎯→ J2+ + 2e– M2X3 5
2.8.2 2.8
Charge of X 2– ion Charge of M 3+ ion

An atom of element Q with proton number 17 and 5.2
an electron arrangement of 2.8.7 will achieve the
stable octet electron arrangement by accepting one 1 Give the formulae of the ions formed by the
electron to form a Q– ion. following elements:
(a) Calcium
Q + e– ⎯⎯→ Q– (b) Phosphorus
(c) Sulphur
2.8.7 2.8.8 (d) Potassium
(e) Nitrogen
Two Q atoms will accept one electron each from the
two electrons donated by every J atom. Ionic bonds [Proton number: N, 7; P, 15; S, 16; K, 19; Ca, 20]
are formed between J 2+ ions and Q – ions to produce
a compound with formula JQ2. 2 The proton numbers of element P and element Q
are 9 and 20 respectively.
1 ’02 (a) Write the equations that show the transfer of
electrons and the formulae of the ions formed
R reacts with S to form an ionic compound with by atom P and atom Q respectively.
(b) Draw the valence electron arrangement to
the formula R2S3. Which of the following electron show the formation of a compound formed
arrangements of atoms R and atoms S is true? between P and Q.

Electron Electron 3 The table shows the groups of four elements in
arrangement arrangement the Periodic Table represented by the letters E, F, G
and H.
of atom R of atom S

A 2.8.2 2.8.3

B 2.8.3 2.8.2 Group 1 2 16 17
Element E F GH
C 2.8.2 2.5

D 2.8.3 2.6

Comments (a) Write the formulae of the ions formed by the
Since an ionic compound is formed, R must be elements E, F, G and H.
a metal and S must be a non-metal. The formula
of R2S3 shows that R will donate three electrons (b) What is the formula of the compound formed
whereas S will accept two electrons to achieve the
octet electron arrangement. Hence, R has three from
valence electrons and S has six valence electrons. (i) E and G ?
(ii) E and H ?
Answer D (iii) F and G ?
(iv) F and H ?

113 Chemical Bonds

5.3 Formation of Covalent 3 A Lewis structure is a diagram which shows
Bonds only the valence electrons of the atoms
represented by dots. The Lewis structure of
1 A covalent bond is a bond that is formed from the hydrogen molecule is
SPM the sharing of valence electrons between
’08/P1 non-metal atoms to achieve the stable octet H• + H• → H •• H (or H — H)

or a duplet electron arrangement. Formation of a chlorine molecule, Cl2
2 Non-metals from Groups 15, 16 or 17 in the
1 A chlorine atom with an electron arrangement
Periodic Table react with other non-metals of of 2.8.7, needs to share one electron to achieve
the same group or different groups to form the stable octet electron arrangement of 2.8.8.
covalent compounds.
3 Hydrogen is a non-metal element. It can form5
covalent bonds with other non-metal atoms
from Groups 14, 15, 16 and 17. Examples are ••
CH4, NH3, H2S and HCl. ••
4 Molecules with covalent bonds can be formed ••
from: ••
(a) Atoms of the same element ••
••
Examples: Hydrogen (H2), chlorine (Cl2), ••
oxygen (O2) and nitrogen (N2). ••
(b) Atoms of different elements 2 Each chlorine atom contributes one valence
Examples: Water (H2O), ammonia (NH3), electron to be shared. The sharing of one pair of
carbon dioxide (CO2), methane (CH4), electrons results in the formation of one single
tetrachloromethane (CCl4). covalent bond between two chlorine atoms.
5 The types of covalent bond formed depend
on the number of pairs of electrons shared 3 The Lewis structure showing the formation
between two atoms. There are three types of of chlorine molecule is
covalent bonds.
(a) Single bond: One pair of electrons • Cl • + • Cl • → • Cl • Cl •
shared between two atoms. • • • • •
(b) Double bond: Two pairs of electrons
shared between two atoms. Formation of methane molecule, CH4
(c) Triple bond: Three pairs of electrons
shared between two atoms. 1 A carbon atom with an electron arrangement
of 2.4 needs to share four electrons to achieve
Formation of Single Covalent Bonds the stable octet electron arrangement of 2.8.

Formation of a hydrogen molecule, H2 2 A hydrogen atom needs to share one
electron to achieve the stable duplet electron
1 A hydrogen atom with an electron arrangement arrangement.
of 1, needs to share one electron to achieve
the stable duplet electron arrangement. 3 One carbon atom contributes four valence
electrons to be shared with four hydrogen
2 Each hydrogen atom will contribute one atoms respectively to form four single covalent
valence electron to be shared between two bonds in the methane molecule, CH4.
hydrogen atoms. One pair of electrons
shared between two hydrogen atoms form a
single covalent bond.

Chemical Bonds 114

4 The Lewis structure for the formation of Formation of ammonia molecule, NH3
methane molecule is
1 A nitrogen atom with an electron
HH arrangement of 2.5 needs to share three
• •• electrons to achieve the stable octet electron
⏐ arrangement of 2.8.

4H • + •C• → H • C • H or H ⎯ C ⎯H 2 A hydrogen atom needs to share one
• • electron to achieve the stable duplet electron
• •• arrangement.
⏐
H 3 One nitrogen atom contributes three valence
H electrons to be shared with three hydrogen
atoms (which contributes one electron
Formation of tetrachloromethane SPM respectively) to form three single covalent
’10/P1, bonds in the ammonia molecule, NH3.
(carbon tetrachloride), CCl4 ’11/P1

1 A carbon atom with an electron 5
arrangement of 2.4 needs to share four
4 The Lewis structure showing the formation
electrons to achieve the stable octet electron of ammonia molecule, NH3 is

arrangement of 2.8. 3H • + • • •
2 A chlorine atom with an electron • • •
N H • N ••H or H ⎯ N⎯H
arrangement of 2.8.7 needs to share one • H⏐
••
electron to achieve the stable octet electron

arrangement of 2.8.8.
3 Four chlorine atoms contribute one valence

electron each to be shared with one carbon
atom to form four single covalent bonds in

the tetrachloromethane molecule, CCl4.

• →
•

•

H

Formation of water molecule, H2O

1 An oxygen atom with an electron arrangement
of 2.6 needs to share two electrons to achieve
the stable octet electron arrangement of 2.8.

2 A hydrogen atom needs to share one
electron to achieve the stable duplet electron
arrangement.

3 One oxygen atom contributes two valence
electrons to be shared with two hydrogen
atoms (which contributes one electron
respectively) to form two single covalent
bonds in the water molecule, H2O.

4 The Lewis structure of the formation of 4 The Lewis structure showing the formation
tetrachloromethane molecule is

•• • Cl • • • Cl • of water molecule, H2O is
•• ••• • •• ⏐ •
•• • •••
•• • ••• •
•• •
•• •
•

4 • Cl • + •C• → • Cl • C • Cl • or • Cl ⎯ C ⎯ Cl • HH
• • • • • • •
⏐ • •• ⏐
• Cl • • • Cl • 2H • + O• • H O• • or H O⎯ •
• • •• • → ••• • •• •
• •• •
•
•
•

115 Chemical Bonds

Formation of Double Covalent Bonds Formation of carbon dioxide molecule, SPM
CO2 ’06/P2
Formation of oxygen molecule, O2 ’08/P1

1 An oxygen atom with an electron 1 A carbon atom with an electron arrangement
arrangement of 2.6 needs to share two of 2.4 needs to share four electrons to achieve
electrons to achieve the stable octet electron
arrangement of 2.8. the stable octet electron arrangement of 2.8.
2 An oxygen atom with an electron arrangement
2 Each oxygen atom will contribute two
valence electrons to be shared between two of 2.6 needs to share two electrons to achieve
oxygen atoms.
the stable octet electron arrangement of 2.8.
3 The sharing of two pairs of electrons between 3 One carbon atom contributes four valence
two atoms results in the formation of a double
covalent bond in an oxygen molecule, O2. electrons to be shared with two oxygen atoms
5
(which contribute two electrons each) to form
•• two double bonds in the carbon dioxide
••
•• molecule, CO2 as shown in the diagram below.
••
4 The Lewis structure below shows the formation
•
• of a carbon dioxide molecule, CO2.

•• •• •• •• ••
2O • + C• • O C O•••• or O == C == O
•• • → •• • •
•• • • •• •• ••

4 The Lewis structure showing the formation Formation of Triple Covalent Bonds
of oxygen molecule is

•• •• •• Formation of nitrogen molecule, N2 SPM
O• + • O O O•• or O O• •
• → == • ’10/P1
• •• • • •• •

In the formation of a covalent bond 1 Each nitrogen atom with an electron
• an atom that requires one more electron to achieve arrangement of 2.5 will contribute three

the stable electron arrangement as in the noble gases valence electrons to be shared between two
will form one single covalent bond. For example:
hydrogen and chlorine form single covalent bonds in nitrogen atoms so as to achieve the stable
H – H; Cl – Cl; H – Cl.
• an atom (such as oxygen and sulphur) that requires octet electron arrangement of 2.8.
two electrons to achieve the stable electron arrangement 2 The sharing of three pairs of electrons
will form two single covalent bonds or one double
covalent bond. For example: results in the formation of a triple covalent
bond in the nitrogen molecule, N2.

H – O – H ; O=O 3 The Lewis structure for nitrogen molecule,

2 single covalent bonds 1 double covalent bond N2 is shown below.
per oxygen atom per oxygen atom • • ••
• • ••
••
•N• + • N• → N N• • or • N ≡ N •
• • • • • •

Chemical Bonds 116

How to Predict the Formula of a Covalent 2 ’03
Compound
The electron arrangement of atom P is 2.8.6 and
1 Covalent compounds are formed from the atom Q has four valence electrons. What is the formula 5
non-metal elements (from Group 14 to of the compound formed between P and Q?
Group 17).
ABC QQQ2PPP42
2 The number of electrons required by a non- D Q4P
metal atom to achieve the stable octet electron
Comments
arrangement as that of a noble gas is known P, with 6 valence electrons, will need to share 2
as the valency. electrons in order to achieve the stable octet electron
3 The valency of a non-metal is (8–x) where x arrangement. Q, with 4 valence electrons, will need
is the number of valence electrons. Table 5.2 to share 4 electrons to achieve the octet electron
arrangement. Hence the formula of the compound is
shows the valency of non-metal elements.
Q2P4 or QP2
Table 5.2

Element Number of valence Valency
electrons (x) (8–x)

Group 14 4 4

Group 15 5 3 2 electrons to be shared 4 electrons to be shared
with Q by P with P by Q

Group 16 6 2 Answer A

Group 17 7 1

4 If a non-metal element M, with a valency of 3
SPM x, combines with another non-metal element,
’09/P1 N, with a valency of y, the formula of the Element X is from Group 14, element Y is from
Group 16 and element Z is from Group 17 of
covalent compound formed will be MyNx. the Periodic Table. What is the formula of the
compound formed between
MyNx (a) element X and element Y?
(b) element X and element Z?
valency of N valency of M
Solution
5 Table 5.3 below shows the formulae of covalent (a) An atom of element X from Group 14 with 4
compounds formed between elements from
different groups. valence electrons needs to share 4 electrons in
order to achieve the octet electron arrangement.
Table 5.3 An atom of element Y from Group 16 has 6
valence electrons and needs to share 2 electrons in
Non-metal Non-metal Formula order to achieve the octet electron arrangement.
element, element, N of covalent Examples The formula of the compound formed is
compound
M X2Y4 or XY2

Group 14 H or Group 17 MN4 CH4, CCl4, 2 electrons required by Y to 4 electrons required by X to
(valency 4) (valency 1) SiCl4 be shared to achieve the octet be shared to achieve the octet
M2N4 electron arrangement electron arrangement
Group 14 Group 16 or MN2 CO2, SiO2
(valency 4) (valency 2) (b) An atom of element Z from Group 17 requires 1
MN3 NH3, PH3, electron to be shared, hence the formula of the
Group 15 H or Group 17 PCl3 compound is
(valency 3) (valency 1) MN2 X1Z4 or XZ4
H2O, H2S,
Group 16 H or Group 17 SCl2 Z shares 1 electron with X X shares 4 electrons with Z
(valency 2) (valency 1)

117 Chemical Bonds

4 The electron arrangement of an atom of Q is 2.8.7.
Both J and Q are non-metals and they will form a
The proton numbers of element J and element Q covalent compound.
are 16 and 17 respectively. Explain the type of bond An atom J needs to share 2 electrons and an atom
and the formula of the compound formed between Q needs to share 1 electron in order to achieve the
J and Q. stable octet electron arrangement. The formula of the
compound formed is JQ2.
Solution
The electron arrangement of an atom of J is 2.8.6.

Guidelines in drawing the electron arrangement of a covalent compound formed by atom M and atom N.

5 Step 1 Step 2

Determine the number of valence electrons Draw the positions of atom M and atoms
and the valency of atom M and atom N in the formula with atoms N (more
N, then determine the formula of the number of atoms) surrounding atom M.
compound. Example:
Example
Atom M has 5 valence electrons and valency
is 3. Atom N has 7 electrons and valency is
1. Formula of compound is M1N3.
[Proton number: M, 7; N, 9]

Step 4 Step 3

Determine the number of electrons to Draw the valence electron shells overlapping
be shared (= 8 – x) where x = number of between the two types of atoms.
valence electrons.
Draw the electrons to be shared as dot or
crosses at the
overlapped area.
Example:

Step 5 Step 6

Draw the balance of the valence electrons Draw the inner electron shells for atom M
not shared (x – number of electrons shared) and atoms N and the electrons as dots and
on the shell outside the overlapped area. crosses.

N has 7 valence
electrons:
1 shared,
remainder 6
unshared.

Chemical Bonds 118

How to predict the type of chemical bonds formed and the formula of the compound formed
from two elements, X and Y.

Step 1 Step 2

Write the electron arrangement to • If one of the elements (say X) has 1, 2 or 3
determine the number of valence valence electrons and the other element (say
electrons of elements X and Y. Y) has 5, 6 or 7 valence electrons then X and Y
form an ionic bond.

• If both elements have 4, 5, 6 or 7 valence
electrons, or one of the elements is hydrogen,
X and Y form a covalent bond.

5

Step 3

Determine the number of electrons X or Y needs to donate/
accept/share to achieve the stable octet electron arrangement.

Ionic Bonding Covalent Bonding
Formula is Xa Yb. Formula is Xa Yb.

Number of Number of Number of electrons Number of electrons
electrons that atom electrons that that atom Y needs that atom X needs
Y needs to accept atom X needs to to share to achieve to share to
to achieve an donate to achieve an octet electron achieve an octet
octet electron an octet electron arrangement electron
arrangement arrangement arrangement

1 Elements with 1, 2 or 3 valence electrons can only form ionic bonds with non-metals (elements with 5, 6 or 7

valence electrons).
2 Elements with 4 valence electrons can only form covalent bonds.
3 Elements with 5, 6 or 7 valence electrons can form both ionic and covalent bonds.

119 Chemical Bonds

3 ’04 5.3

The diagram shows the electron arrangement of a 1 Write the Lewis structures for the following
compound formed between atoms X and Y.
compounds:
Which of the following
statements is true about the (a) Hydrogen chloride, HCl
compound?
(b) HWyadteror,gHe2nOcyanide, HCN
A The compound is formed (c)
by ionic bonds.
(d) P[Phroostopnhonruums tbriecrh:lHor,id1e; ,CP, C6l;3N, 7; P, 15; Cl, 17]
B The compound is formed
by electron transfer. 2 The table shows four elements in different groups
of the Periodic Table represented by the letters V, X,
C Atom X is a metal and atom Y is a non-metal. Y and Z.
D It is a covalent compound.
5 Group 14 15 16 17
Comments
The diagram shows the overlap of the outermost Element V X Y Z
electron shells of atoms X and Y where the sharing
of electrons is represented by dots and crosses. The (a) Write the formula of the compound formed
diagram represents a covalent compound formed by between
the sharing of electrons between non-metal atoms.
(i) V and Z
Answer D (ii) V and Y
(iii) X and Z
Comparison between Ionic Bonding and
Covalent Bonding (b) Draw the Lewis structures for the compounds
formed in (a).
Ionic bonding Covalent bonding
(c) What type of compounds are formed in (a)?
Similarity Atoms achieve the stable (duplet or
octet) electron arrangement after the 3 The proton numbers of elements P and Q are 6
formation of bonds. and 9 respectively. Draw the electron diagrams for
the formation of compounds between
(a) Q and hydrogen
(b) Q and P
(c) Q and Q

Difference Involves the Involves the CFCs (chlorofluorocarbon) are covalent compounds
transfer of sharing of that consist of chlorine, fluorine and carbon atoms
electrons from electrons between bonded by covalent bonds. Two examples of CFCs
metal atoms to non-metal atoms. are CF2Cl2 and CFCl3. CFCs are colourless, odourless
non-metal atoms. and non-toxic gases with low boiling points. CFCs are
used in aerosol and as refrigerants in air-conditioners.
Positively- No charged ions However, since 1992, the usage of CFC was banned
charged ions are formed. because it was believed to deplete the ozone layer
and negatively- Molecules are which protects the earth from getting too much
charged ions are formed. harmful ultraviolet rays from the sun.
formed.
• Metals only form ionic bonding.
Strong Van der • Non-metals form both ionic and covalent bondings.
electrostatic force Waals forces
of attraction holds of attraction
the oppositely- exist between
charged ions the covalent
together. molecules.

Chemical Bonds 120

5.4 The Properties of SPM 8 The differences between the physical properties
SPM of ionic compounds and covalent compounds
’08/P1 ’09/P1 are shown in Table 5.4.

Ionic Compounds and
Covalent Compounds
Table 5.4 Comparison of properties of ionic and
Properties of Ionic Compounds and covalent compounds

Covalent Compounds SPM Ionic Covalent
compound compound
’10/P2
(simple
1 The physical properties of ionic and covalent Properties molecules)

compounds are different because of the (a) Melting point High Low
different types of bonds formed and the and boiling
difference in the structures of the compounds. point 5
2 An ionic compound is formed when ions
of opposite charges are held by the strong (b) Volatility Non-volatile Volatile (can
electrostatic force of attraction. change to
3 The ions of ionic compounds are arranged vapour when
heated)
in an orderly and compact manner and form
(c) Solubility Usually Usually
large ionic structures. soluble in soluble in
4 Figure 5.4 shows the arrangement of ions in water and organic
polar solvents solvents such
a three-dimensional network forming a large but insoluble as benzene
structure in an ionic compound. in organic but insoluble
solvents in water

(d) Electrical Conducts Does not
conductivity electricity in conduct
the molten electricity in
Figure 5.4 Three-dimensional network of ions state or any state
aqueous
5 Most covalent compounds consist of simple solution
molecules.
Most ionic compounds are soluble in water. However
6 The covalent bond within a molecule is strong there are some ionic compounds that are insoluble in
but the intermolecular forces (van der Waals water. For example, lead(II) bromide, lead(II) chloride
forces of attraction) are weak. and calcium carbonate are insoluble in water. There
are also some covalent compounds that are polar and
7 Figure 5.5 shows the intermolecular forces are soluble in water. For example, ethanol, ethanoic
(van der Waals forces of attraction) between acid and sugar are soluble in water.
methane molecules and the intramolecular
bonds (covalent bonds) in the methane
molecules. The dotted line represents the weak
intermolecular forces (van der Waals forces).

Figure 5.5 Types of bonds in methane molecules • The low melting and boiling points of covalent
compounds are due to the weak intermolecular
forces between covalent molecules.

• They are not due to the strength of covalent bonds
inside the molecules.

121 Chemical Bonds

To study the physical properties of ionic and covalent SPM
compounds
’06/P2

(A) To investigate the melting point and volatility 3 Steps 1 and 2 are repeated with naphthalene
of compounds and hexane respectively to replace magnesium
chloride.
Apparatus Crucible, spatula, tripod stand, wire
gauze and Bunsen burner. Figure 5.6 To investigate the melting point of a
compound
Materials Magnesium chloride, naphthalene
and hexane.

Procedure

Activity 4.2 1 A spatula of magnesium chloride solid is placed
5 in a crucible.

2 Magnesium chloride solid is heated slowly
at first and then strongly. The change in the
physical state of the compound is recorded.

Results

Compound Observation Inference

Magnesium chloride Physical state Action of heat High melting point
Naphthalene Low melting point
Hexane Solid No change Low melting point and very volatile

Solid Melts rapidly

Liquid Vaporise rapidly

Conclusion Results

1 Magnesium chloride is an ionic compound and Compound Solubility
has high melting and boiling points. Naphthalene
and hexane are covalent compounds and they Magnesium In water In organic solvent
have low melting and boiling points. chloride
Naphthalene Soluble Insoluble
2 Ionic compounds are non-volatile whereas Hexane
covalent compounds are volatile. Insoluble Soluble
Insoluble Soluble
(B) To investigate the solubility of compounds

Apparatus Boiling tubes and spatula

Materials Magnesium chloride, naphthalene, Conclusion
hexane, water and cyclohexane.
1 Magnesium chloride is an ionic compound and is
Procedure soluble in water but insoluble in organic solvents.

1 A boiling tube is filled with 5 cm3 of distilled 2 Naphthalene and hexane are covalent compounds
water. and are insoluble in water but soluble in organic
solvents.
2 Half a spatula of magnesium chloride solid is
3 Hexane and water form two layers of liquid in a
added to the distilled water and shaken. The test tube. This shows hexane is insoluble in water.

solubility of magnesium chloride in water is noted. (C) To investigate the electrical conductivity of
3 Steps 1 to 2 are repeated with naphthalene and compounds

hexane to replace magnesium chloride. Apparatus Crucible, spatula, graphite rods,
4 Another boiling tube is filled with 5 cm3 of batteries, light bulb, switch,
connecting wires, tripod stand, clay
cyclohexane (an organic solvent). triangle and Bunsen burner.
5 Half a spatula of magnesium chloride is added
Materials Lead(II) bromide, magnesium chloride,
to the cyclohexane and shaken. The solubility of sugar and naphthalene.

magnesium chloride in cyclohexane is noted.
6 Steps 4 and 5 are repeated with naphthalene and

hexane in turn to replace magnesium chloride.

Chemical Bonds 122

Procedure SPM

’04/P2

1 Three spatulas of lead(II) bromide solid is placed in a crucible.

2 Two graphite rods are dipped in the lead(II) bromide solid and

the circuit is completed by connecti­ng to the batteries and switch.

3 The switch is turned on and the bulb is checked if it lights up.

4 Lead(II) bromide is heated strongly until it melts. The switch is

turned on again to check if the bulb lights up.

5 Steps 1 to 4 are repeated using naphthalene to replace lead(II)

bromide.

6 Three spatulas of magnesium chloride solid is placed in a crucible.

7 The switch is turned on to check if the bulb lights up.

8 Water is added to the magnesium chloride. The mixture is stirred Figure 5.7 To investigate the
electrical conductivity of
with a glass rod until all the magnesium solid dissolves in water. compounds 5

9 The switch is turned on again to check if the bulb lights up. Inference
Conducts electricity in the liquid but
10 Steps 6 to 9 are repeated using sugar to replace magnesium not in the solid state

chloride. Does not conduct electricity in any
state
Results
Conducts electricity in aqueous state
Compound State of compound Observation but not in the solid state

Lead(II) bromide Solid Light bulb does not Does not conduct electricity in any
light up state
Naphthalene Molten
Solid Light bulb lights up

Molten Light bulb does not
light up
Magnesium chloride Solid
Light bulb does not
Sugar Aqueous solution light up
Solid
Light bulb does not
Aqueous solution light up

Light bulb lights up

Light bulb does not
light up

Light bulb does not
light up

Conclusion an electric current flows through the liquid or
aqueous solution.
1 Ionic compounds such as lead(II) bromide and 2 Lead(II) bromide has a lower melting point than
magnesium chloride do not conduct electricity magnesium chloride. Therefore, lead(II) bromide
in the solid state but conduct electricity in the is more suitable for use in the investigation of the
molten state or in aqueous solution due to the electrical conductivity of ionic compounds in the
presence of mobile ions. liquid state compared to magnesium chloride.
3 Lead(II) bromide is insoluble in water whereas
2 Covalent compounds such as naphthalene and magnesium chloride is soluble in water. Therefore,
sugar do not conduct electricity in any state due magnesium chloride is more suitable for the
to the absence of mobile ions. investigation of the electrical conductivity of
ionic compound in an aqueous solution.
Discussion 4 Naphthalene does not dissolve in water. Some
covalent compounds such as sugar are soluble in
1 Besides light bulbs, galvanometers and ammeters water but they still do not conduct electricity.
can be used to determine the electrical conductivity.
The needle of a galvanometer will deflect if

123 Chemical Bonds

Explanation for the Differences between consist of a covalent network of molecules
the Physical Properties of Ionic and forming a giant molecular structure. Some
Covalent Compounds examples are carbohydrates, proteins and
silicon dioxide. Strong covalent bonds hold
5 Melting points, boiling points and volatility the atoms together in these giant molecules.
A lot of heat energy is required to overcome
1 Ionic compounds have higher melting and the strong covalent bonds of these giant
boiling points than covalent compounds. molecules.
This is because the electrostatic force of
attraction between oppositely charged Solubility
ions is very strong. A lot of heat energy is
required to overcome this strong force of 1 Ionic compounds are soluble in water
attraction. because they can form bonds with water
molecules but are insoluble in organic
2 There are two types of covalent solvents.
compounds:
(i) Simple molecules 2 Covalent compounds are soluble in
(ii) Giant molecules organic solvents because they can form
bonds with organic solvent molecules but
3 Below are some examples of simple are insoluble in water.
molecules and giant molecules. Most simple
molecules exist as gases or liquids while 3 Some covalent molecules can dissolve in
giant molecules exist as solids. water because they form hydrogen bonds
with water. Some examples of these covalent
Simple covalent Giant covalent compounds are sugar, ethanol, acetone and
molecules molecules carboxylic acids.

Carbon dioxide, CO2 Graphite 4 Most giant covalent molecules are in­soluble
Tetrachloromethane, CCl4 Diamond in both water and organic solvents.
Ammonia, NH3 Silicon
Hydrogen gas, H2 Silicon dioxide Electrical conductivity
Oxygen gas, O2 Carbohydrates
Water, H2O Protein 1 All ionic compounds can conduct electricity
in the molten state and in aqueous solution
4 Melting and boiling points of covalent because the charged ions can move freely.
compounds of simple molecules such as Ionic compounds in the solid state do not
tetrachloromethane, naphthalene, chlorine, conduct electricity because the ions are held
bromine and iodine are low because by the strong electrostatic force of attraction
the intermolecular forces of attraction in the lattice structure and are not free to
between molecules (van der Waals forces move.
of attraction) are very weak. Little heat
energy is required to overcome the weak 2 Covalent compounds do not conduct
electricity in any state because they consist
intermolecular forces. of molecules. There are no freely moving
5 The low boiling points of the covalent charged particles.

compounds cause the covalent compounds 3 A few covalent compounds can dissociate
into ions when dissolved in water. So, they
to be volatile (easily changed to the can conduct electricity in aqueous solution.
Some examples are ammonia (NH3),
vapour state when heated). Hence covalent hydrogen chloride (HCl) and sulphur
dioxide (SO2).
compounds should be kept far away from a
4 Electrical conductivity in the molten or
heat source. liquid state is the best physical property to
6 Some covalent compounds have high differentiate between ionic compounds
and covalent compounds.
melting and boiling points because they

Chemical Bonds 124

1 The flowchart below helps us to determine the type of chemicals.

Chemical
Does it conduct electricity in the solid state?

Yes No
Metal Ionic compound or covalent compound

Does it conduct electricity in the liquid state?

Yes No 5
Ionic compound Covalent compound

Does it have high melting and boiling points?

Yes No
Giant covalent compound Simple covalent compound

Does it dissolve in water?

Yes No
Polar covalent compound Non-polar covalent compound

4 ’07

The table shows electrical condutivity and melting points of three substances X, Y, and Z.

Substance Solid Electrical conductivity Aqueous Melting point (°C)
No Molten No
X No No No –117
Y No No Yes 80
Z Yes 801

(a) State the types of structure and bonding of network of ions held by ionic bonds.
substances X, Y and Z. (b) Substance X consists of simple covalent molecules

(b) Explain why substance X has a low melting point. held by weak intermolecular forces of attraction
(c) Show how the bonds are formed in substance Y that requires little energy to overcome them.
(c) The bonds in substance Y are formed by the
and substance Z. sharing of valence electrons to achieve the
(d) Based on the information in the table, predict the stable octet electron arrangement. The bonds in
substance Z are formed by the transfer of electrons
solubility of substances X, Y and Z in water. to achieve the stable octet electron arrangement.
Solution (d) Substances X and Y are insoluble in water.
(a) Substances X and Y are covalent molecules with Substance Z is soluble in water.

covalent bonds.
Substance Z is an ionic compound with a giant

125 Chemical Bonds

Uses of Covalent Compounds as Organic • Covalent bonds formed between atoms are very
Solvents strong. When a covalent compound is heated, heat
energy is used to overcome the weak intermolecular
1 Organic solvents are liquids consisting of forces of attraction which is the van der Waals forces
simple covalent compounds.
of attraction. The covalent bonds are not broken.
2 Examples of organic solvents are ethanol,
• Some covalent compounds are polar (examples: ethanol,
ether, acetone, benzene, turpentine, petrol and
hydrogen chloride, ammonia) and can dissolve in water.
tetrachloromethane.
3 Organic solvents are used • Some ionic compounds are insoluble in water

(a) to clean or remove stains that cannot be (examples: Al2O3, PbCl2).

5 cleaned by water.
(b) to extract organic compounds.
Ionic Covalent compound
(c) as solvents for drugs in medicine or compound

compounds used in cosmetics.
4 Some common useful organic solvents and

their uses are given in Table 5.5.

Table 5.5 Uses of some organic solvents Property Simple Giant

Organic solvents Uses molecule molecule

Petrol, kerosene and To remove grease and (a) Melting and High Low High
boiling points
turpentine paint stains

Ethanol, acetone To prepare medicated (b) Volatility Non-volatile Very volatile Non-volatile
solution, perfume and
ink (c) Solubility Soluble in Soluble Insoluble
in both
water, insoluble in organic water and
organic
in organic solvents, solvents

Acetone, turpentine To dissolve varnish solvents insoluble
compounds such as
shellacs, lacquer and in water
paints
(d) Electrical Conducts Does not conduct electricity
conductivity electricity in in any state
liquid and
Ether To extract organic in aqueous
compounds solution

5.4 2 The table shows the number of valence electrons of
three elements represented by the letters Q, R and S.
Aluminium oxide, silicon dioxide, bromine,
ethanol, ammonium nitrate, barium sulphate, Element QR S
naphthalene, tetrachloromethane, hydrogen
iodide, copper(II) chloride No. of valence electrons 2 4 7

1 Based on the chemicals in the list above, choose Choose two elements which can combine to form a
(a) two compounds with high melting points. compound that
(b) two compounds which can conduct electricity in (a) has low melting and boiling points,
the aqueous state. (b) conducts electricity in the liquid state,
(c) one compound which dissolves in water but (c) is soluble in water.
cannot conduct electricity.
(d) one covalent compound with a high melting point. 3 Explain clearly why carbon dioxide is a gas whereas
(e) two compounds which can be used as organic magnesium chloride is a solid at room temperature.
solvents. Which compound can conduct electricity in the
(f) one compound which cannot dissolve in water liquid state? Explain your answer.
but can conduct electricity in the molten state.

Chemical Bonds 126

1 Inert gases are non-reactive because they have either (b) Double bond: Sharing of two pairs of 5
a stable duplet or octet electron arrangement. electrons between two atoms.

2 Inert gases do not release, accept or share (c) Triple bond: Sharing of three pairs of
electrons with other elements. electrons between two atoms.

3 Atoms of elements from Group 1 to Group 17 have 14 For a covalent compound formed by a non-metal
less than eight electrons. These elements will form element P combined with another non-metal
chemical bonds. element, Q, the formula of the covalent compound
formed will be PYQX. where
4 Ionic bonds are formed between atoms of metals
and atoms of non-metals. PYQX

5 Metal atoms from Groups 1, 2 and 13 will release Number of electrons Number of electrons
their valence electrons to achieve the stable duplet required by Q to required by P to achieve
or octet electron arrangement. achieve duplet or octet duplet or octet electron
electron arrangement arrangement
6 A positive ion (cation) is formed when an atom
releases electrons. 1 5 The differences in physical properties between ionic
compounds and covalent compounds:
7 Non-metal atoms from Groups 15, 16 and 17
will accept electrons to achieve the stable octet Property Ionic Covalent
electron arrangement. compound compound

8 A negative ion (anion) is formed when an atom Melting point High (simple
receives electrons. and boiling molecule)
point Non-volatile
9 The positive ions and negative ions are attracted Volatility Low
to each other by strong electrostatic force of Solubility Soluble
attraction in ionic bonds. in water, Volatile
Electrical insoluble
1 0 The total positive charge of the cation must be equal conductivity in organic Soluble
to the total negative charge of the anion in an ionic solvents in organic
compound. solvents,
Conducts insoluble in
1 1 For an ionic compound consisting of cations Mb+ electricity water
and anions Xa–, the formula of the ionic compound in molten
formed between them is written as form and Does not
in aqueous conduct
MaXb solution electricity in
any state
Number of electrons Number of electrons
that will be received by that will be released by
X (or charge of X ion) M (or charge of M ion)

12 A covalent bond is formed between atoms of non-

metals.
1 3 Non-metal atoms share valence electrons to achieve

the stable duplet or octet electron arrangement

in covalent compounds. There are three types of

covalent bond.
(a) Single bond: Sharing of one pair of electrons

between two atoms.

127 Chemical Bonds

5

Multiple-choice Questions

5.1 Formation of Compounds 5 Which of the following substances 9 Element P and element Q
has particles bonded with very
1 Which of the statements below strong electrostatic forces? are located in Group 2 and
best explains why elements of A Carbohydrate
Group 18 of the Periodic Table B Graphite ’10 Group 17 in the Periodic
such as neon, argon and krypton C Napthalene
are chemically inert? D Magnesium sulphide Table respectively. Element
A Exists as monatoms
B Have 8 valence electrons 6 The proton numbers of three P reacts with element Q to
C Have 8 electrons in every elements represented by the letters
5 electron shell P, Q and R are given as follows: form a compound. What is
D Have weak covalent bonds
between atoms P: 4 Q: 13 R: 16 the chemical formula of the

2 Which of the following proton What will be the charges of the compound?
numbers belong to an element most stable ions of P, Q and R?
that does not form chemical A PQ
bonds?
B P2Q
I 8 C PQ2
II 10 D P2Q3
III 18
IV 20 10 Element P reacts with element Q

A I and II only P ion Q ion R ion to form an ionic compound with
B I and III only
C II and III only A +4 +3 +6 cfoornmfiugularaPt2ioQn3.fIof rthQeies l2e.c8tr.6on, ic
D II and IV only B +2 +3 +6
C +2 +3 –2 what is the possible electron
3 Which of the following D +4 –3 +2
statements best explains the arrangement of P?
formation of chemical bonds
between elements? A 2.8.2 C 2.8.5
A The elements have high
reactivities. B 2.8.3 D 2.8.6
B The atoms of the elements
have less than eight valence 7 The table shows the proton 11 An ionic nitride compound has
electrons. numbers for four elements R, S,
C The atoms of the elements T and U. formula X3N2. What is the possible
have too many electrons. proton number of atom X if nitrogen
D The elements are
either electropositive or Element R S TU is in Group 15 of the Periodic Table?
electronegative.
Proton number 8 11 13 20 A 2 C 13

B 12 D 15

Which of the following ions have 12 Which of the following represents

the same number of electrons the valence electron arrangement
’06 for the compound sodium oxide?
as the sulphide ion?
[Proton number: O, 8; Na, 11]
[Proton number of S = 16] A
A U 2+ C T 3+
B S + D R 2–

8 The proton number of sodium and B
oxygen are 11 and 8 respectively.
C
’07 Which of the following occurs
5.2 Formation of Ionic Bonds when sodium metal is burned in
air to produce sodium oxide?
4 Oxide and fluoride ions have the A One oxygen atom receives one D
electron from one sodium atom.
same number of B One oxygen atom receives
two electrons from one
’03 [Proton number: O, 8; F, 9] sodium atom.
C One oxygen atom receives
A charge C neutrons two electrons, one from each
sodium atom.
B electrons D protons D One sodium atom and one
oxygen atom share two
electrons.

Chemical Bonds 128

13 The following diagram shows the 18 The following diagram shows the X Y
electron arrangement for the J 2+ electron arrangements of atoms A2 14
B4 6
’07 ion. An atom of element J has 20 ’11 X and Y. X and Y are not the C 14 16
neutrons. actual symbols of the elements. D 18 18

What is the nucleon number of X Y 22 Element R reacts with element S

element J ? Which pair of formula and the to produce a covalent compound
with the formula R2S3. Which
A 18 C 38 type of compounds is correct? of the following electron
arrangements of atoms R and S
B 20 D 40 Formula Type of are true?
compound
5
14 G is an element in Group 2 and X is
an element in Group 16. Which of A XY3 Ionic
B XY3 Covalent Electron Electron
the following is the formula of the C X3Y
compound formed by G and X? D X3Y Ionic arrangement arrangement
Covalent
A GX C GGX3X2 of atom R of atom S
B G2 X6 D
A 2.8.2 2.3

19 The electronic configuration B 2.8.2 2.5
of the atoms of four elements
5.3 Formation of Covalent represented by the letters P, Q, C 2.8.3 2.6
Bonds R and S are shown in the table
below. D 2.8.5 2.6
15 The diagram shows the electron
arrangement of an atom of 23 The electronic configuration of
element X.
Element P Q R S atom E is 2.8.6 and atom G has

Electronic ’03 four valence electrons. What is
configuration
1 2.7 2.8 2.8.8.2 the formula of the compound

formed between E and G?

Which of the following pairs of A GG4EE22 C GG22EE4
B D
elements will form a covalent

The atom of element X can compound? 24 The diagram shows the symbols
form a covalent bond with of atoms P and Q.
another atom through the A P and Q C Q and S
A acceptance of two electrons.
B donation of two electrons. B Q and R D P and R
C sharing of two pairs of
20 Which of the following 28 16
electrons. P Q
D sharing of six electrons. compounds have covalent
14 8
double bonds?
Atom P reacts with atom Q to
AIII ONI a22 n d II only IIVII CCOCl24
16 An element forms a covalent B III and IV only form a compound. Calculate the
compound with hydrogen. C I and IV only
Atoms of this element also form D I, II and IV only relative molecular mass for the
covalent molecules themselves.
This element may be compound formed.

I sodium III chlorine A 21 C 60
II oxygen IV nitrogen
B 43 D 113
A I and II only
B III and IV only 21 The diagram shows the electron 5.4 Properties of Ionic and
C I, II and IV only arrangement in a compound Covalent Compounds
D II, III and IV only
’05 formed by elements X and Y. 25 Element W and element Y have
proton numbers 17 and 19 as
17 How many pairs of electrons are shown below.

shared by the oxygen atoms in a

molecule of oxygen gas? [Proton Which of the following is the
correct groups of X and Y in the
number: O, 8] C 3 Periodic Table of Elements?
A 1
B 2 D 4

129 Chemical Bonds

Which of the following are true of C they are insoluble in water. D Magnesium oxide has a high
the compound formed between D they can become vapour melting point whereas carbon
element W and element Y? dioxide has a low melting
I Has high melting point easily when heated. point.
II Dissolves in organic solvents
5 III Dissolves in water 30 Covalent compounds have 33 Element X and element Y form a
IV Conducts electricity in the compound that has a low melting
low melting and boiling points point and does not conduct
solid state electricity. Element X and Y
A I and III only because maybe
B II and IV only A they have weak covalent
C III and IV only I hydrogen and sulphur
D I, III and IV only bonds. II chlorine and oxygen
B they have weak III sodium and sulphur
26 Which of the following elements IV potassium and oxygen
will form an oxide that conducts intermolecular forces of
electricity when dissolved in A I and II only
water? attraction. B I and III only
A Potassium C they are very volatile. C II and IV only
B Silicon D they cannot withstand strong D III and IV only
C Hydrogen
D Aluminium heating. 34 Which of the following
compounds have a high melting
27 The electronic configuration of 31 Two elements represented by point and are soluble in water?
element X is 2.5. Which of the the letters X and Y have proton
following statements are true of numbers as given in the table. I Calcium chloride
element X? II Sulphur dichloride
Element XY III Copper(II) chloride
I X forms a hydride with formula IV Tetrachloromethane
XH3. Proton number 16 20
A I and II only
II Molecule X has a triple bond. From the information given in the B III and IV only
III X forms an oxide with high C I and III only
table, it can be deduced that D II and IV only
melting point. I X and Y will form a compound
IV Chlorine and X form a 35 Element X has an electronic
with formula XY2. configuration of 2.8.2. Which of
covalent compound. II Y can form ionic compounds the following are true of element
A I and III only X?
B II and IV only but X can form both ionic and
C III and IV only I It can form negative ions.
D I, II and IV only covalent compounds. II It is a metallic element.
III element X is a non-conductor III It can form a covalent
28 Which of the following
statements best explains why of electricity whereas element compound with chlorine.
bromine liquid is very volatile? Y is a conductor of electricity. IV It can form a basic oxide.
A Bromine consists of diatomic
covalent molecules. IV the compound formed A I and II only
B Intermolecular forces of between element X and B II and IV only
attraction between bromine element Y has a high melting C I and III only
molecules are weak. D III and IV only
C Size of bromine molecules is point.
small. A I and II only 36 Which of the following
D The covalent bond in bromine B III and IV only compounds can be used
molecules is weak. C I, II and III only as solvents for covalent
D II, III and IV only compounds?
29 Covalent compounds do not
conduct electricity because 32 Both carbon and magnesium I Water
A they do not have free moving form compounds with oxygen. II Ethanol
ions. Which of the following is the III Benzene
B they have strong covalent difference between magnesium IV Tetrachloromethane
bonds. oxide and carbon dioxide?
A Magnesium oxide is soluble in A I and II only
water whereas carbon dioxide B II and IV only
is insoluble in water. C III and IV only
B Magnesium oxide solid D II, III and IV only
conducts electricity whereas
carbon dioxide solid does
not.
C Magnesium oxide is basic
whereas carbon dioxide is
neutral.

Chemical Bonds 130

Structured Questions

1 Diagram 1 shows the Periodic Table of elements. The letters P, Q, R, S, T, U, V and W represent elements
and are not the symbols of the actual elements.

12 13 14 15 16 17 18

1

2 PQU

3R ST

4V W

5

6

Diagram 1 5

Answer the following questions with reference only to the letters given above.

(a) (i) State the type of chemical bond found in the compound formed between R and T. [1 mark]

(ii) Draw a diagram showing the electron arrangement of the compound formed in (i). [2 marks]

(iii) State a physical property of the compound formed in (i). [1 mark]

(b) (i) Write the electronic arrangement of an ion formed by element Q. [1 mark]

(ii) Q exists as diatomic molecules. Draw a diagram that shows the electron arrangement of

the diatomic molecule. [2 marks]

(c) Can element U form a compound? Explain your answer. [2 marks]

(d) Write the equation for the reaction between V and water. [1 mark]

2 Diagram 2 shows two elements represented by the letters P and Q. 23 16
(a) Write the electron arrangement of atom P and atom Q.
P Q
[2 marks]

(b) State the groups of elements P and Q in the Periodic Table. 11 8
[2 marks]

(c) Write an equation for the formation of Diagram 2

(i) ion P. (ii) ion Q. [2 marks]

(d) P can combine with Q to form a compound.

(i) Write the formula of the compound formed. [1 mark]

(ii) What type of bond is formed in this compound? [1 mark]

(iii) Draw the electron arrangement of this compound. [1 mark]

(e) Q can combine with itself to form a compound. State a physical property of the compound formed. [1 mark]

3 Table 1 shows a few properties of substances P, Q, R and S.

Substance Melting point Solubility in Solubility in Conductivity
(°C) water organic solvent
P Solid state Liquid state
Q 114 Insoluble Soluble
R 782 Soluble Insoluble Non-conducting Non-conducting
S 840 Insoluble Insoluble Non-conducting
2750 Insoluble Insoluble Conducting
Conducting
Non-conducting Conducting

(Does not exist in
the liquid state)

Table 1 [1 mark]
[1 mark]
Based on the information given in Table 1, answer the questions below.
(a) What type of bond is found in the particles of [2 marks]

(i) substance P?
(ii) substance Q?

(b) Explain why P has a low melting point.

131 Chemical Bonds

(c) Name a substance that has the same physical (d) (i) State two atoms in Table 2 that are elements
in the same group in the Periodic Table.
property as Q. [1 mark] [1 mark]

(d) Which of the above substances is a metal? (ii) Explain your answer in (i). [1 mark]
[1 mark]

(e) Give reasons why Q conducts electricity in the (e) Z can react with heated iron to form a compound.
(i) Write the formula of this compound.
liquid state but does not do so in the solid state.
[1 mark]
[2 marks] (ii) State one physical property of this
compound.
(f) Suggest one difference between the types of [1 mark]

particles in substance R and in substance S.

Give an example each for substance R and 5 Diagram 3 shows the diagram of the valence electron
arrangement of a compound formed by two types of
substance S. [2 marks] atoms represented by the letters P and Q.

5 4 Table 2 shows the number of protons and neutrons
for a few atoms represented by the letters V, W, X, Y
and Z.

Atoms Number of protons Number of neutrons Diagram 3

V 3 4
W 6 6
X 6 7
Y 9 10
Z 17 18

Table 2 (a) What is the formula of this compound? [1 mark]

Use the information in Table 2 to answer the (b) Predict the groups to which the atoms P and Q
questions below.
belong to in the Periodic Table. [2 marks]

(a) What is the nucleon number of element Y? (c) Name the type of bond formed in the compound

[1 mark] shown in Diagram 3. [1 mark]

(b) W can combine with Y to form a compound. (d) State three physical properties of this compound.
(i) Name the type of chemical bond found in [3 marks]
this compound. [1 mark]
(ii) Write the formula of this compound. [1 mark] (e) Q is an element that exists as diatomic
(iii) Predict the relative molecular mass of this
compound. [1 mark] molecules.

(i) Draw a diagram to show the valence electron

arrangement of the molecule. [2 marks]

(c) Name two atoms that are isotopes. Explain your (ii) Explain why element Q has a low boiling
answer. [2 marks]
point. [1 mark]

Essay Questions

1 Diagram 1 shows the chemical symbols for three elements: carbon, sodium and chlorine:

Diagram 1

(a) Construct a table to compare the three elements in terms of the number of protons, the number
of neutrons, the electron arrangement and the number of valence electrons. [4 marks]

(b) Carbon reacts with chlorine to form a covalent compound whereas sodium reacts with [8 marks]
chlorine to form an ionic compound. Explain how these ionic and covalent compounds are formed.

(c) State four differences in physical properties between the two types of compounds formed in (b). [4 marks]

Chemical Bonds 132

2 Element Group Period

W1 3

Y 14 2

Z 17 3

Table 1 [14 marks] 5
[6 marks]
(a) With the help of an electron arrangement diagram, explain how two elements from Table 1 can
combine to form [15 marks]

(i) an ionic compound, and [5 marks]
(ii) a covalent compound.

(b) State the differences between an ionic compound and a covalent compound.

3 (a) Using suitable examples, explain the meaning of single covalent bond, double covalent bond and
triple covalent bond.

(b) Explain why ionic compounds can conduct electricity in the liquid state and in aqueous solution
whereas covalent compounds cannot.

Experiments

1 A group of students carried out an experiment to determine the types of particles in three compounds X, Y
and Z. The results obtained is recorded in Table 1.

Chemical compound Physical state Observation when shaken with

X Solid Water Acetone
Y Liquid
Z Liquid Forms a colourless solution X does not dissolve in acetone
Potassium chloride Forms two immiscible layers Y and acetone mix completely
Z and water mix completely Z and acetone mix completely

Table 1

(a) If the experiment is repeated using potassium chloride, predict what will be observed and complete [3 marks]
Table 1.

(b) Deduce the solubility of X, Y and Z in water and in acetone. [3 marks]

(c) Classify the four chemical compounds used in Table 1 into [3 marks]
(i) ionic compound,
(ii) covalent compound.

(d) Potassium nitrate is used as a nitrogenous fertiliser in agriculture. Which of the compounds X, Y and Z [3 marks]
in the experiment may be potassium nitrate? Explain your answer.

(e) Compound Z is used as a solvent in medicine and perfume. Name a substance that may be compound Z. Explain

your answer. [3 marks]

2 Covalent compounds and ionic compounds differ in their ability to conduct electricity in the liquid state.

By referring to the statement above, design a laboratory experiment relating to the difference in electrical conductivity of

ethanol and sodium nitrate. In designing your experiment, the following aspects must be included:

(a) Problem statement

(b) Hypothesis

(c) All the variables

(d) List of substances and apparatus

(e) Experimental procedures

(f) Tabulation of data [17 marks]

133 Chemical Bonds

6CHAPTER FORM 4

THEME: Interaction between Chemicals

Electrochemistry

SPM Topical Analysis

Year 2008 2009 2010 2011

Paper 1 2 31 2 31 2 31 2 3

Section ABC ABC ABC ABC

Number of questions 5 – 1 – – 6 – 1 – 1 4 1 – 1 – 5 1 – – –

ONCEPT MAP

ELECTROCHEMISTRY
types of substances

Electrolytes and non-electrolytes

conversion of electrical energy conversion of chemical
into chemical energy energy into electrical energy

Electrolysis Voltaic cells

Electrolysis of molten Electrolysis of aqueous Various types of voltaic cells: can be used
compounds solution • non-rechargeable cell and to determine
• rechargeable cell
products products Electrochemical
series
• Metals are formed at Selective discharge of ions
the cathode determined by: To construct the Uses of the
• Position of ions in the electrochemical series: electrochemical series:
• Non-metals are formed • Potential difference • Predict the products of
at the anode electrochemical series
• Concentration of ions between two different electrolysis
• Types of electrodes metals • Determine the terminals
• Displacement reaction
Uses of electrolysis: of metals of cells
• Extraction of metals • Prediction of
• Purification of metals
• Electroplating of metals displacement reactions

Introduction 6 Metals which are conductors are not regarded as
electrolytes because they are not dec­omposed
1 Electrochemistry is the study of the inter­ by the passage of an electric current.
conversion of chemical energy and electrical
energy. 7 The electrical conductivity of any substance is
based on the movement of charged particles.
2 The energy change in electrochemistry consists (a) In metals, electricity is conducted by freely
of the moving electrons.
(a) conversion of electrical energy into
chemical energy (in electrolysis), (b) In electrolytes, electricity is conducted
(b) conversion of chemical energy into by freely moving ions: positive ions
electrical energy (in voltaic cells). and negative ions.

6.1 Electrolytes and Non- 8 Most covalent substances are non-electrolytes Experiment 6.1 6
electrolytes SPM because they consist of molecules, which do
not form ions in aqueous solution.
’08/P1
9 Some covalent substances such as hydrogen
1 An electrolyte is a chemical compound which chloride and ammonia are electrolytes
conducts electricity in the molten state or in because they react with water to produce
an aqueous solution and undergoes chemical freely moving ions.

changes. Examples:
2 When an electrolyte conducts electricity, chemical
HCl(g) + H2O(l) → H3O+(aq) + Cl–(aq)
changes occur and the electrolyte decomposes
NH3(g) + H2O(l) → NH4+(aq) + OH–(aq)
into its component elements at the electrodes.
10 Ions are charged (positive or negative)
Electrolysis is said to have taken place. particles. All metal ions and hydrogen ions
3 Examples of electrolytes are acid solutions, are positive ions (also known as cations). All
non-metal ions are negative ions (also known
alkali solutions, molten salts or aqueous salt as anions). All electrolytes will dissociate into
cations and anions in the molten states or
solutions. aqueous solutions.
4 A non-electrolyte is a chemical compound

which does not conduct electricity in any state.
5 Examples of non-electrolytes are metals and

covalent substances such as naphthalene,

latex and sugar solution.

6.1 Apparatus Crucible, spatula, carbon (graphite)
Materials electrodes, batteries, light bulb,
To identify electrolytes and non-electrolytes switch, rheostat, connecting wires,
tripod stand, clay pipe triangle and
Problem statement Bunsen burner.
How to identify electrolytes and non-electrolytes?
Glucose, naphthalene, lead(II) bromide
Hypothesis and potassium iodide.
Substances that, in the molten state or in aqueous
solution, conduct electricity and then undergo chemical Procedure
reactions are electrolytes. Substances that do not
conduct electricity in any state are non-electrolytes. (A) To investigate the electrical conductivity of
substances in the solid state and in the molten
Variables state
(a) Manipulated variable : Types of compounds
(b) Responding variable : Electrical conductivity 1 A crucible is half-filled with lead(II) bromide
( c) Constant variable : Numbers of batteries, solid.

type of light bulb and
amount of substance used

135 Electrochemistry

6 2 The crucible with its contents is placed on a clay (B) To investigate the electrical conductivity of
triangle on a tripod stand. substances in the solid state and in aqueous
solutions
3 Two carbon electrodes are dipped in the lead(II)
bromide solid and are connected to the batteries, 1 Three spatulas of potassium iodide solid are put
rheostat, switch and a light bulb with connecting
wires (Figure 6.1). in a beaker.
2 Two carbon electrodes are dipped in the
4 The switch is turned on and the light bulb is
checked if it lights up. potassium iodide solid and then connected to the

5 The lead(II) bromide solid in the crucible is batteries, rheostat, switch and a light bulb with
heated up until it melts. The switch is turned on
again to check if the light bulb lights up. connecting wires (Figure 6.2).
3 The switch is turned on and the light bulb is
6 Steps 1 to 5 of the experiment are repeated using
naphthalene in place of lead(II) bromide. checked if it lights up.
4 Distilled water is added to the beaker and the

mixture is stirred until all the potassium iodide

has dissolved.
5 The switch is turned on again and the light bulb

is checked if it lights up.
6 Steps 1 to 5 of the experiment is repeated using

glucose in place of potassium iodide.

Figure 6.1 To investigate the conductivity of Figure 6.2 To investigate the conductivity of
substances in the solid state and in substances in the solid state and in
the molten state aqueous solution

Results Physical state Does the light Observation: Inference
Chemical bulb light up? Does reaction occur?
substance Solid Non-electrolyte
Liquid (molten) No No noticeable change Electrolyte
Lead(II) bromide Solid Yes Non-electrolyte
Liquid (molten) No Brown gas is evolved Non-electrolyte
Naphthalene Solid No Non-electrolyte
Aqueous solution No No noticeable change Electrolyte
Potassium iodide Yes
No noticeable change Non-electrolyte
Glucose Solid No Non-electrolyte
Aqueous solution No No noticeable change

Solution turns to a brown
colour

No noticeable change

No noticeable change

Disscussion the ionic bond. The ions are free to move in
the molten state. Hence ionic compounds are
1 Electrolytes can conduct electricity because they electrolytes in the molten state.
have freely moving charged ions. 4 When ionic compounds are dissolved in water,
the water molecules separate the ions into freely
2 In the solid state, an ionic compound is not an moving ions. Hence aqueous solutions of ionic
electrolyte because the ions are held together by compounds are electrolytes.
strong ionic bonds and are not free to move.

3 When an ionic compound is heated to its melting
point, the heat energy supplied overcomes

Electrochemistry 136

5 Covalent compounds such as glucose and Conclusion

naphthalene do not conduct electricity because 1 Lead(II) bromide is an electrolyte in the liquid
they consist of molecules which are uncharged state but not in the solid state.

particles. 2 Potassium iodide is an electrolyte in aqueous
6 When electricity passes through molten lead(II) solution but not in the solid state.

bromide, the brown gas evolved is bromine gas. 3 Lead(II) bromide and potassium iodide are ionic
7 When electricity passes through aqueous potassium compounds. Ionic compounds are electrolytes
in the molten state or aqueous solution but are
iodide solution, the brown iodine solution is formed. non-electrolytes in the solid state.
8 The presence of freely moving ions enable a
4 Naphthalene and glucose are covalent compounds
molten compound or aqueous solution to conduct and are non-electrolytes in any state.

electricity.

6

All liquid or aqueous solutions of ionic compounds are 6.2 Electrolysis of Molten
electrolytes. Compounds
All covalent compounds (except those that can dissociate
to form ions when dissolved in water) are non-electrolytes. Meaning of Electrolysis and Electrolytic
An electrolyte can conduct electricity because it has Cell
freely moving ions.
• Electrolysis is a process of decomposition of an
6.1 electrolyte by an electric current.

1 (a) (i) What is meant by the term electrolyte? • An electrolytic cell consists of batteries, a
Give two examples of electrolytes. cathode, an anode and an electrolyte consisting
of cations and anions.
(ii) What is meant by the term non-electrolyte?
Give two examples of non-electrolytes. • During electrolysis, anions (negatively-charged
ions) move towards the anode and cations
(b) State the difference between a conductor and (positively-charged ions) move towards the
an electrolyte. cathode.

2 Classify the ten substances below into electrolytes • Graphite or platinum is usually used as
and non-electrolytes. electrodes because they are inert; they do not
react with the electrolyte or the products of
Molten sulphur, molten zinc oxide, zinc oxide solid, electrolysis.
aqueous zinc chloride solution, zinc metal, molten
zinc, acetone, aqueous glucose solution, aqueous Electrons flow from the anode to the
ethanoic acid solution, molten sodium chloride. cathode through the connecting wire
in the external circuit.
3 Explain why magnesium chloride solid cannot
conduct electricity but becomes an electrolyte
when it is in the molten state.

A rheostat can be used to control the quantity An ammeter or a bulb can be used to
of electric current that flows through the indicate the flow of electric current.
electrolyte. • Cathode is the electrode that
• Anode is the electrode connected to the is connected to the negative
terminal of the batteries.
positive terminal of the batteries. • At the cathode, cations discharge
• At the anode, anions discharge by by accepting electrons,
Example: Na+ + e– → Na
donating electrons.
Example: 2Cl– → Cl2 + 2e–
An electrolyte conducts electricity in aqueous solution or in
the molten state as a result of the presence of freely moving
cations (positive ions) and anions (negative ions).

Figure 6.3 Electrolytic cell

137 Electrochemistry

Electrolysis Process 2 When the solid is heated until it melts (in
the molten form), the heat energy supplied
1 In the solid state, the cations and anions is used to overcome the electrostatic force
of an electrolyte are unable to move freely
because they are held together by strong of attraction between the ions. Hence, the
ionic bonds in fixed positions in a lattice. cations and the anions are free to move.

3 Generally, a molten compound AnBm
produces Am+ cations and Bn– anions.

6 SPM AnBm(s) → nAm+(l) + mBn–(l)
cation anion
’09/P1



Example: 4 Two steps occur during electrolysis.

PbBr2(s) → Pb2+(l) + 2Br–(l) (a) Movement of ions to the electrodes
lead(II) ion bromide ion Cations (positive ions) move towards
the cathode (negative electrode) whereas
• Examples of cations are hydrogen ions, anions (negative ions) move towards
H+, ammonium ions, NH4+ and metal the anode (positive electrode).
ions such as K+, Mg2+, Pb2+ and Al3+.
(b) Discharge of ions at the electrodes
• Examples of anions are Cl–, Br–, I–, OH–, Cations are discharged by accepting
SO42– and NO3–. (gaining) electrons. Generally:

An+ + ne– → A
Anions are discharged by donating
(losing) electrons. Generally:

Bn– → B + ne–

5 The electrons donated by anions at the anode are accepted by the cations at the cathode.
The discharge of ions at the anode and cathode results in the
(a) conduction of electricity by the electrolyte.
(b) decomposition of the electrolyte into its component elements.

Writing Half-equations for the Discharge of Ions at the Anode and the Cathode

1 The reactions that take place at the anodes 4 Addition of the two half-equations at the
or the cathodes involve ions and electrons. anode and cathode gives the overall chemical
The equations representing these reactions are equation for the reaction.
known as half-equations.
Half-equation at the anode:
2 The half-equation of the anode shows the
discharge of the anions by the loss of electrons Bn– → B + ne– ………… equation 1
to produce neutral atoms. Half-equation at the cathode:

Half-equation at the anode: Bn– → B + ne– An+ + ne– → A ………… equation 2
Overall chemical reaction equation:
3 The half-equation of the cathode shows
the discharge of the cations by the gain of
electrons to produce neutral atoms. An+ + Bn– → A + B …… equation 1 + equation 2

Half-equation at the cathode: An+ + ne– → A

Electrochemistry 138

1 Step 2: Balance the number of atoms on both sides
of the equation.
Write the half-equation for the discharge of oxide
ions, O2– at the anode. 2O2– → O2

Solution Step 3: Balance the charge by adding in the right
Step 1: Write the formulae of the reactant and the number of electrons.

product of the reaction. 2O2– → O2 + 4e–

O2– → O2

2 At the cathode: Cations receive electrons Activity 6.1 6

When molten aluminium oxide, Al2O3 is electrolysed, Al3+ + 3e– → Al ……… (2)
aluminium metal and oxygen gas are produced.
(a) Write half-equations for the reactions at the (Number of electrons donated at the anode
anode and cathode. (b) Write the overall chemical = number of electrons accepted at the cathode)
equation of electrolysis.
Equation (1)  3: 6O2– → 3O2 + 12e–
Solution Equation (2)  4: 4Al3+ + 12e– → 4Al
At the anode: Anions lose electrons Overall chemical equation:
4Al3+(l) + 6O2–(l) → 4Al(l) + 3O2(g)
2O2– → O2 + 4e– ……… (1)

To investigate the electrolysis of molten lead(II) bromide SPM
’06/P2

Apparatus
Crucible, spatula, graphite electrodes, batteries, light
bulb, ammeter, switch, rheostat, connecting wires,
tripod stand, clay pipe triangle and Bunsen burner.

Materials
Lead(II) bromide.

Procedure Figure 6.4 To investigate the electrolysis of lead(II)
bromide
1 A crucible is half-filled with lead(II) bromide
solid.

2 The solid lead(II) bromide is heated strongly
until it melts to a molten state.

3 Two carbon electrodes are dipped in the molten
lead(­I­I) bromide and are then connected to
batteries, rheos­tat, switch and light bulb by the
connecting wires (Figure 6.4).

4 Electric current is allowed to flow through for 15
minutes and the changes that occur at the light
bulb, ammeter, cathode and anode are recorded.

139 Electrochemistry

Results

Apparatus Observation Inference
Light bulb Light bulb lights up Molten lead(II) bromide
Ammeter conducts electricity
Anode Ammeter needle is deflected
Bromine gas is evolved
Cathode Pungent brown gas that changes damp blue litmus paper to red is
evolved Lead metal is formed

Shiny grey metal is deposited

6 Discussion 5 At the cathode, a lead(II) ion discharges by
accepting 2 electrons to form a lead metal atom.
1 Solid lead(II) bromide consists of lead(II) ions,
Pb2+ and bromide ions, Br– that are held by Pb2+ + 2e– → Pb
strong ionic bonds.
6 A larger quantity of the products would be
2 When heated until it melts to form the molten formed at the anode and the cathode if the
state, lead(II) ions and bromide ions are free to electroysis is carried out
move. (a) for a longer time period,
(b) with a larger current.
PbBr2(s) → Pb2+(l) + 2Br –(l)
Conclusion
3 During electrolysis, bromide ions are attracted 1 The lighting up of the bulb and the deflection of
to the anode while lead(II) ions are attracted to
the cathode. the ammeter needle shows that molten lead(II)
bro­mide is an electrolyte and can conduct
4 At the anode, a bromide ion discharges by electricity.
releasing one electron to form a bromine atom. 2 Electrolysis of molten lead(II) bromide produces
Two bromine atoms combine to form a bromine bromine gas at the anode and lead metal at the
molecule that exists as a brown gas. cathode.

2Br – → Br2 + 2e–

Prediction of the Products of Electrolysis formed at the cathode while the non-metal
of Molten Electrolytes component is formed at the anode.
2 Products of electrolysis of molten electrolytes
1 When molten compound is electrolysed,
the metal component of the compound is can be predicted by the following steps.

Step 1 Step 2 Step 3
Identifying the cations and
anions that are present in the Identify the cathode and Write the half-equations for
molten electrolyte. Generally, anode and the movement the discharge of the ions at the
a salt with formula AxBy will of ions in the electrolyte electrode.
produce ions as below. to the electrodes. • At the cathode: Cations
• Cations move to
AxBy → xAy+ + yBx– discharge (losing the positive
cation anion cathode (electrode charge) by accepting electrons.
connected to the
negative terminal of Ay+ + ye– → A
the battery).
• Anions move to • At the anode: Anions discharge
anode (electrode (losing the negative charge) by
connected to the donating electrons.
positive terminal of
the battery). Bx– → B + xe–

Electrochemistry 140

Electrolysis of molten sodium Electrolysis of molten lead(II) oxide, PbO 6
chloride, NaCl 1 Cations and anions produced from the

1 The cations and anions produced from molten lead(II) oxide are lead(II) ions, Pb2+
molten sodium chloride are sodium ions, and oxide ions, O2–.
Na+ and chloride ions, Cl–.
PbO → Pb2+ + O2–
NaCl → Na+ + Cl–
2 Lead(II) ions are attracted to the cathode
2 The sodium ions are attracted to the cathode while the oxide ions are attracted to the anode.
while the chloride ions are attracted to the
anode. 3 At the cathode, a lead(II) ion accepts 2 electrons
to form a lead metal atom (grey metal).
3 At the cathode, a sodium ion accepts an Pb2+ + 2e– → Pb
electron to form a sodium metal atom
(grey metal). 4 At the anode, an oxide ion donates 2 electrons
to form an oxygen atom. Two oxygen atoms
Na+ + e– → Na combine to form an oxygen gas molecule.

4 At the anode, a chloride ion donates an 2O2– → O2 + 4e–
electron to form a chlorine atom. Two
chlorine atoms combine to form a chlorine The term electrolysis was introduced by Michael
molecule (greenish-yellow gas). Faraday. ‘Lysis’ means loosening in Greek. Electrolysis
means loosening by electric current.
2Cl– → Cl2 + 2e–

6.2

1 Write the formulae of the ions produced in the molten chemical compounds in the table below. Identify the ions that
move to the anode and to the cathode respectively during electrolysis by completing the table below.

Name of compound Ions produced Ions that move to the

(a) Zinc chloride anode cathode
(b) Magnesium oxide

2 For every compound below, write the half-equations that occur at the anode and the cathode, as well as the products
formed during electrolysis.

Name of compound Half-equation at the Products formed at the

(a) Calcium oxide anode cathode anode cathode
(b) Aluminium iodide

3 The figure shows the arrangement of apparatus used in an experiment. The switch is turned

on for 20 minutes.
(a) Predict the observations at (i) electrode P, (ii) electrode Q.
(b) What are the products formed at (i) electrode P, (ii) electrode Q?
(c) Write the half-equations for the reactions that take place at (i) electrode P, (ii) electrode Q.

141 Electrochemistry

6.3 Electrolysis of Aqueous (A) Positions of ions in the electrochemical
Solutions series

Identifying Cations and Anions in 1 The tendency of ions to be selectively discharged
Aqueous Solutions at electrodes depends on their positions in a
series known as the electrochemical series.
1 A molten salt consists of one type of
cation and one type of anion only. During 2 The tendency of the ions to be discharged is
shown below in ascending order.
electrolysis, the products formed are the
6 component elements of the compound. Cation tendency Anion
2 An aqueous solution is produced when a K+ to F–
solute is dissolved in water. An aqueous Na+ SO42–
solution of a salt consists of two types of Mg2+ discharge NO3–
cations (cations of the salt and hydrogen Al3+ increases Cl–
ions, H+) and two types of anions (anions of Zn2+ Br–
the salt and hydroxide ions, OH–). Fe2+ I–
3 H+ ions and OH– ions are always present Sn2+ OH–
Pb2+
together with the ions produced from the H+
Cu2+
dissociation of salts in aqueous solutions. Hg+
Ag+
This is because water dissociates partially to
3 The lower the position of the ion in the
form hydrogen ion and hydroxide ion. electrochemical series, the easier the ion
will be discharged.
H2O H+ + OH–
4 For example, in the electrolysis of aqueous
4 For example, in an aqueous sodium chloride
solution, 4 types of ions are present in the sodium sulphate (Na2SO4) solution, with
carbon electrodes,
solution from the dissociation of NaCl and H2O.

cation anion Na2SO4 → 2Na+ + SO42–

NaCl ⎯→ Na+ + Cl– H2O H+ + OH–

H2O H+ + OH– The cations present are Na+ ions and H+
ions. The anions present are SO42– ions and
The four types of ions present are Na+, Cl–, H+ OH– ions.
and OH–. 5 Both the Na+ ions and the H+ ions are
attracted to the cathode. However, only the
5 During electrolysis of an aqueous salt solution,
H+ ions are discharged at the cathode because
2 types of cations will move to the cathode the H+ ion is lower in position than the Na+
ion in the electrochemical series.
while 2 types of anions will move to the anode.
Half-equation at the cathode:
For example, during electrolysis of aqueous
2H+ + 2e– → H2
sodium chloride solution, the cations that move
to the cathode are Na+ ions and H+ ions. The Hence, hydrogen gas is produced at the
anions that move to the anode are Cl– ions and
OH– ions. However, only one type of cation and cathode.

anion will be discharged at each electrode. 6 Both the SO42– ions and the OH– ions are
attracted to the anode. However, only the
Factors that Determine the Selective
Discharge of Ions at the Electrodes OH– ions are discharged at the anode because

The factors that determine the types of ions to be the OH– ion is lower in position than the
discharged at the electrodes are:
(a) Positions of ions in the electrochemical series SO42– ion in the electrochemical series.
(b) Concentration of ions in the solution Half-equation at the anode:
(c) Types of electrodes used
4OH– → 2H2O + O2 + 4e–
Hence, oxygen gas is produced at the anode.

Electrochemistry 142

(C) Effect of types of electrodes used SPM
’04/05
(B) Effect of concentration of ions in the P2
solution
1 The common materials used as electrodes are
1 When the concentration of a particular type
of ion is high, that ion will more likely to carbon and platinum because they are inert.
be discharged in electrolysis irrespective of
its position in the electrochemical series. Both of these materials do not react with the

2 Usually, in the electrolysis of concentrated electrolytes or the products of electrolysis.
halide (Cl–/Br–/I– ions) solutions, the
concentration of the halide ion is always 2 The types of electrodes used can determine
higher than the hydroxide ion, OH–. Hence,
halide ions will be selectively discharged at the types of ions discharged in electrolysis.
the anode.
3 For example, in the electrolysis of aqueous
3 For example, in the electrolysis of aqueous
copper(II) chloride solution, CuCl2 with copper(II) sulphate (CuSO4) solution,
carbon electrodes,
CuSO4 → Cu2+ + SO42–

H2O H+ + OH–

6

CuCl2 → Cu2+ + 2Cl– both the anions, sulphate ions, SO42– and
hydroxide ions, OH– are attracted to the
H2O H+ + OH– anode.
(a) If carbon is used as the electrodes,

OH– ions are discharged at the anode
because of the position of OH– ion in
the elect­rochemical series.

Half-equation at the anode:

Both types of anions, chloride ion, Cl– and 4OH– → 2H2O + O2 + 4e–
hydroxide ions, OH– are attracted to the
anode. Hence, oxygen gas is produced at the
(a) If a concentrated solution is electrolysed, anode.
(b) If copper is used as the anode, both SO42–
chloride ions, Cl–, will be selectively ions and OH– ions are not discharged.
discharged at the anode because the Instead the copper anode dissolves by
concentration of the chloride ions is releasing electrons to form copper(II)
higher than that of the hydroxide ions. ions, Cu2+.
Half-equation at the anode: Half-equation at the anode:

2Cl– → Cl2 + 2e– Cu → Cu2+ + 2e–

Hence chlorine gas is evolved at the Hence, the mass of anode decreases.
anode. Copper acts as an active electrode here
(b) If a dilute solution is electrolysed, because it takes part in the chemical
hydroxide ions, OH–, will be discharged reaction during electrolysis.
at the anode because the concentration
of the chloride ions is low and OH– ion Generally, in the electrolysis of a halide solution using
is more easily discharged because of its carbon electrodes:
position in the electrochemical series. • A concentration of more than 0.5 mol dm–3 halide
Half-equation at the anode:
solution is concentrated solution whereby the halide
4OH– → 2H2O + O2 + 4e– ions will be selectively discharged at the anode.
• A concentration of less than 0.005 mol dm–3 halide
Hence, oxygen gas is evolved at the solution is considered a dilute solution whereby the
anode. hydroxide ions will be selectively discharged at the
anode.
• Electrolysis of a solution with a concentration of
between 0.005 mol dm–3 and 0.5 mol dm–3 may
result in two types of products: halogen and oxygen
to be produced at the anode.

143 Electrochemistry