Chemistry

PENERBIT
ILMU
BAKTI SDN. BHD.

Contents

PENERBITChapter 1 Matter
ILMU
BAKTI SDN. BHD.1.1 Atoms and Molecule 1

1.2 Mole Concept 8

1.3 Stoichiometry 19

Summative Practice 1 26

Chapter 2 Atomic Structure 29
38
2.1 Bohr’s Atomic Model 43
2.2 Quantum Mechanics 48
2.3 Electronic Configuration
Summative Practice 2

Chapter 3 Periodic Table 51
57
3.1 Classification of Elements 74
3.2 Periodicity
Summative Practice 3

Chapter 4 Chemical Bonding 77
88
4.1 Lewis Structure 93
4.2 Molecular Shape and Polarity 100
4.3 Orbital Overlaping and Hybridisation 104
4.4 Intermolecular Forces 107
4.5 Metallic Bond
Summative Practice 4

iii

Chapter 5 States of Matter

5.1 Gas 111

5.2 Liquids 120
PENERBIT
ILMU5.3 Solids 123
BAKTI SDN. BHD.
5.4 Phase Diagram 129

Summative Practice 5 134

Chapter 6 Chemical Equilibrium 137
140
6.1 Dynamic Equilibrium 146
6.2 Equilibrium Constants 154
6.3 Le Chatelier’s Principle
Summative Practice 6

Chapter 7 Ionic Equilibria 157
170
7.1 Acids and Bases 175
7.2 Acid–base Titrations 181
7.3 Solubility Equilibria
Summative Practice 7

Semester Examination 183
Answers 189
Periodic Table of Elements 212

iv

PENERBIT 1CHAPTER
ILMU
BAKTI SDN. BHD.Matter

1.1 Atoms and Molecules

Learning Outcomes

• Understand the synthesis of atomic structures.
• Determine the number of neutrons, protons, electrons and nucleon number.
• Write isotopic notations.
• Interpret mass spectrum.
• Calculate the average atomic mass of an element given the relative

abundances of isotopes or mass spectrum.
• Calculate the relative atomic mass based on carbon-12.

1 An atom consists of positively charged protons, negatively charged
electrons and neutral neutrons. These particles are known as the subatomic
particles of an atom.

2 The nucleus of an atom consists of protons and neutrons, which together
they are called the nucleon and are surrounded by electron clouds.

–

Nucleus
– (6 Protons + 6 Neutrons)
+
– + + –
+
Neutron
++

–

+ Proton

– – Electron
Carbon atom

Figure 1.1 Structure of a carbon atom

1

 Matter

3 The properties of subatomic particles of proton, neutron and electron are given
in Table 1.1.

Table 1.1 Properties of subatomic particles

Particle Symbol Relative mass Approximate Relative
relative mass charge
Proton p+ 1.007
Neutron n0 1.009 1.0 +1
Electron e– 0.0005486 or
1.0 0
1
1 834 0.0 –1
BAKTI SDN. BHD.
Chapter 1 4 An isotopic notation provides information that differentiates isotopes.

5 The superscript and subscript are added to the left of the element symbol.

6 In general, the atomic structure of an atom or isotope can be represented by
the isotopic notation of the atomic structure, which can be written as,

A X
Z

where X = symbol of the element
A = nucleon number
Z = proton number

7 Nucleon number, A is the total number of protons and neutrons present in the
nucleus of an atom.

8 Proton number, Z is the number of protons in the nucleus of an atom.
9 Therefore,

Number of neutrons = Nucleon number (A) – Proton number (Z)
Smart Tips! ILMU

PENERBIT Atoms are neutral Example 1
particles. Therefore, the
number of protons in the Magnesium has 12 protons and 12 neutrons. Write the symbol of the
nucleus is equal to the
number of electrons. magnesium atom in the form of A X.
Z
Smart Tips!
Solution
• The nucleon number is
also known as mass Proton number = 12
number. Nucleon number = Number of protons + Number of neutrons
= 12 + 12 = 24
• The proton number is
also known as atomic Therefore, the magnesium atom can be written as 24 Mg.
number. 12

2

Matter 

Example 2

What is the number of protons, neutrons and electrons in 62 Ni?
28

Solution

Number of protons = 28PENERBIT Chapter 1
Number of neutrons = Nucleon number – Proton numberILMU
= 62 – 28 BAKTI SDN. BHD.
= 34
Number of electrons = Number of protons
= 28

Isotopes

1 Isotopes are two or more atoms that have the same proton number but a
different nucleon number.

2 Isotopes of the same element have the same number of protons but a
different number of neutrons.

3 Table 1.2 shows some examples of isotopes and their relative abundance.

Table 1.2 Examples of isotopes

Element Isotope Atomic Number Number Number Relative
mass of of of abundance

protons electrons neutrons (%)

Carbon Carbon-12 12 6 6 6 98.90

Carbon-13 13 6 6 7 1.10

Chlorine Chlorine-35 35 17 17 18 75.77

Chlorine-37 37 17 17 20 24.23

Bromine Bromine-79 79 35 35 44 50.70

Bromine-81 81 35 35 46 49.30

Oxygen Oxygen-16 16 8 8 8 99.76

Oxygen-17 17 8 8 9 0.04

Oxygen-18 18 8 8 10 0.20

4 The percentage abundance of an isotope can be determined using the Smart Tips!
following formula.
Abundance can be
Percentage abundance of isotope (%) in terms of relative
= Abundance of an isotope × 100 abundance, peak
height or intensity. Total
Total abundance of isotopes of the element percentage abundance
of isotopes of an element
= 100%.

3

 Atomic Structure

Chapter 2 Electronic Configurations of d-block Elements BAKTI SDN. BHD.

ILMU 1 The d-block elements or the transition elements in the Periodic Table such as
scandium, titanium, vanadium, chromium, manganese, iron, cobalt and nickel
PENERBIT contain partially filled d-orbitals.

2 When filling up electrons in the orbitals, the Aufbau principle must be
followed. The 4s orbital is filled before the higher energy 3d orbital.

3 In the case of transition elements, when the 3d orbital is filled with electrons,
they will repel the 4s electrons to a higher energy level.

4 If a transition metal atom forms an ion, the electrons in the higher energy 4s
orbital are removed first before removing the electrons in the 3d orbitals.

Example 4

Write the spdf notation of (a) Ni atom, (b) Ni2+ ion and (c) Ni3+ ion.

Solution

(a) Nickel atom, Ni has 28 electrons.
Electrons are filled into orbitals according to the Aufbau principle, where

the 4s orbital is filled before the 3d orbital.
Therefore, the spdf notation of Ni: 1s2 2s2 2p6 3s2 3p6 4s2 3d8
Note: Electrons are removed from the 4s orbital followed by the 3d orbital.

Although the 4s orbital is energetically more stable, the electrons are
repelled to a higher energy level as soon as the 3d orbital is occupied with
electrons.
(b) Ni → Ni2+ + 2e–
For Ni2+ ion, 2 electrons are donated to another atom. So, there are 26
electrons remaining.
Therefore, the spdf notation of Ni2+: 1s2 2s2 2p6 3s2 3p6 3d8
(c) Ni → Ni3+ + 3e–
For Ni3+ ion, 3 electrons are donated to another atom. So, there are 25
electrons remaining.
Therefore, the spdf notation of Ni3+: 1s2 2s2 2p6 3s2 3p6 3d7

The Anomalous Electronic Configurations of Copper and
Chromium

1 There is some inconsistency between the electronic configuration predicted by
the Aufbau principle and the actual electronic configuration of chromium, Cr
and copper, Cu.

2 Chromium, Cr has 24 electrons.
3 The electronic configuration of Cr according to the Aufbau principle is

expected to be 1s2 2s2 2p6 3s2 3p6 4s2 3d4.
4 However, the actual electronic configuration of Cr is 1s2 2s2 2p6 3s2 3p6 4s1

3d5.

46

Atomic Structure 

5 The actual electronic configuration of Cr is obtained when an electron is
transferred from 4s orbital to 3d orbital to obtain a more stable half-filled 3d
orbitals.


1s 2s 2p 3s 3p 4s 3d

PENERBIT 6 Copper, Cu has 29 electrons.
ILMU
BAKTI SDN. BHD. 7 The electronic configuration of Cu according to the Aufbau principle is
expected to be 1s2 2s2 2p6 3s2 3p6 4s2 3d9.

8 However, the actual electronic configuration of Cu is 1s2 2s2 2p6 3s2 3p6 4s1
3d10.

9 The actual electronic configuration of Cu is obtained when an electron is Chapter 2
transferred from 4s orbital to 3d orbital to obtain a more stable fully-filled 3d
orbitals.



1s 2s 2p 3s 3p 4s 3d

Quick Check 2.3

1 What is the Aufbau principle?
2 The following is an element.

28 X

(a) Write the electronic configuration (spdf notation and orbital diagram) of
the element.

(b) Draw the dxy and dyz orbitals.
(c) Give a possible value of the following quantum numbers for the dyz

orbital of the element.
(i) Principal quantum number, n
(ii) Angular momentum quantum number, l
(iii) Magnetic quantum number, m

3 Write the electronic configuration (spdf notation and orbital diagram) of the
following ions.
(a) K+ ion
(b) Cl– ion
(c) O2– ion
(d) Mg2+ ion

4 Write the electronic configuration of 26X.

47

Summative Practice 2

Objective Questions BAKTI SDN. BHD. 6 Calculate the frequency and the energy of blue
light that has a wavelength of 410 nm.
1 Which atom has an incomplete subshell? [h = 6.63 × 10–34 J s]
A Ge
Chapter 2 B Zn Frequency, υ (Hz) Energy, E (J)
C He A 7.32 × 1014 4.85 × 10–19
D Ar B 7.32 × 1015 4.85 × 10–20
C 7.32 × 10–16 4.85 × 10–22
2 Which of the following line spectrum series of D 7.32 × 1024 4.85 × 10–23
hydrogen atom is found in ultraviolet region?
A Paschen series 7 Calculate the wavelength and the energy of light
B Balmer series that has a frequency of 1.52 × 1015 Hz.
C Lyman series
D Pfund series

3 How many electrons can the d sublevel hold? Wavelength, λ (m) Energy, E (J)
A 3 A 3.00 × 10–15 19.95 × 10–19
B 6 B 3.10 × 10–10 9.95 × 10–20
C 7 C 1.97 × 10–7 1.01 × 10–18
D 10 D 2.01 × 10–8 18.95 × 10–23

4 Calculate the energy and the wavelengthILMU 8 A photon of light has a wavelength of 0.051 cm.
of a photon of light with a frequency of Calculate its energy.
6.166 × 1014 Hz. A 3.90 × 10–20 J
B 3.90 × 10–22 J
Energy, E (J) Wavelength, λ (m) C 3.90 × 10–13 J
D 3.90 × 1022 J
A 5.6 × 10–19 8.0 × 10–7
PENERBIT 9 Electrons occupy orbitals of the lowest energy
B 4.1 × 1019 4.87 × 107 first. Which of the following rules or principles
states this?
C 4.1 × 10–23 4.87 × 10–13 A Hund’s rule
B Balmer rule
D 4.1 × 10–19 4.87 × 10–7 C Aufbau principle
D Pauli exclusion principle
5 What is the ionisation energy of 1 mole of
hydrogen atoms at ground state?
A 2.18 × 10–8 J mol–1
B 6.023 × 1023 mol–1
C 1 313 kJ mol–1
D 1 313 J mol–1

48

10 The electronic configuration of a sulphur atom is Atomic Structure 
1s2 2s2 2p6 3s2 3p4. Which of the following pairs 14 Which of the following statements is correct
of valence shell electrons and unpaired electrons
is correct? about the Hund’s rule?
I Electrons must occupy the degenerate
Valence shell Unpaired
electrons electrons orbital first with the same spin before adding
electrons with opposite spins to the same
A6 4 orbital
B4 2 II Each electron will occupy the lowest energy
C6 2 orbital first
D2 1 III A maximum of two electrons can occupy a
PENERBIT single atomic orbital, only if the electrons
11 The atomic number of phosphorus is 15. WhichILMU have opposite spins Chapter 2
atomic orbitals are occupied with electrons in aBAKTI SDN. BHD.A I only
phosphorus atom? B I and II
A 2s 2p 2d C II and III
B 2p 3s 3d D I, II and III
C 2s 2p 3s
D 1p 2s 2p 15 Which of the following statements is correct
about the Pauli exclusion principle?
12 Electrons are paired based on their spin I A maximum of two electrons can occupy a
directions, clockwise or anticlockwise. Which of single atomic orbital, only if the electrons
the following is related to the above statement? have opposite spins
A Hund’s rule II Each electron will occupy the lowest energy
B Balmer rule orbital first
C Aufbau principle III Electrons must occupy the degenerate
D Pauli exclusion principle orbital first with the same spin before adding
electrons with opposite spins to the same
13 Figure 1 shows the electronic configuration of an orbital
element. A I only
B I and II
1s 2s 2p C II and III
D I, II and III
Figure 1
16 Which pair of ions have the same electronic
Which of the following statements is correct configuration?
about Figure 1? I 13Al3+ and 9F–
A In the 2p orbitals, there should only be one II 12Mg2+ and 13Al3+
III 21Sc3+ and 9F–
electron in the first two orbitals with the A I only
same spinning direction B I and II
B Both arrows in the 2p orbitals should be C II and III
pointing down D I, II and III
C There is nothing incorrect with this diagram
D All the arrows should be pointing up

49

 Atomic Structure

Structured Questions

1 The ion of Y contains 10 inner electrons and 8 outermost electrons. The charge is –2.
(a) Discuss all the rules and principles involved to fill the electrons in the orbitals of atom Y.
(b) Give the electronic configuration of atom and ion Y.

2 Fill up the table by stating the number of orbitals filled with electrons in an isolated atom of
36Kr. The electrons in 36Kr are distributed in s, p and d orbitals.
BAKTI SDN. BHD.
Type of orbitals s p d

Number of orbitals

Chapter 2 3 Figure 2 shows the number and relative energy of electrons. Complete the orbital diagram of
atom 14Si.

4s
3p
3s
2p
2s
1s

Figure 2

4 (a) Give the definition of the term ‘first ionisation energy’.
(b) What are the methods used to calculate the ionisation energy of the hydrogen atom?

5 Figure 3 shows a type of orbital.
ILMU

PENERBIT Figure 3

Name the type of orbital.

6 The highest projecting line in the line spectrum of mercury is found at 253.651 nm.
(a) What is the frequency of the line?
(b) What is the energy of one photon with this wavelength?

50

BAKTI SDN. BHD. 3CHAPTER

Periodic Table

3.1 Classification of Elements

Learning Outcomes

• Describe periods, groups and blocks (s, p, d, f).
• Deduce the position of elements in the Periodic Table from their electronic

configuration.
ILMU
1 Elements in the Periodic Table are arranged based on their proton number (Z). Smart Tips!

2 There are three classifications of elements which are known as period, group Group
and block.
Period
3 The elements in horizontal rows are arranged in order of increasing proton
number from left to right. These rows are known as periods.

4 The vertical columns are known as groups. Elements of the same group have
the same number of valence electrons.

5 Blocks are divided into four major regions according to the outermost orbital
of the elements (s-block, p-block, d-block and f-block).
PENERBIT
Periods in the Periodic Table

1 There are seven periods in the Periodic Table. Period 1 to Period 3 are short
periods, while Period 4 to Period 7 are long periods.

2 The short periods (Period 1 to Period 3)
(a) In Period 1, there are only two elements, hydrogen (1s1) and helium (1s2).
The quantum physics of this period states that electrons can fill up the 1s
orbital. Therefore, a maximum of two electrons will complete the valence
shell. The 1s orbital can only hold two electrons.

51

 Periodic Table

Chapter 3 (b) In Period 2, there are eight elements. The outermost shell that will be BAKTI SDN. BHD.
filled first is the 2s orbital followed by the 2p orbital. The elements and
their valence electrons are lithium (2s1), beryllium (2s2), boron (2s2 2p1),
carbon (2s2 2p2), nitrogen (2s2 2p3), oxygen (2s2 2p4), fluorine (2s2 2p5) and
neon (2s2 2p6).

(c) In Period 3, there are eight elements. The outermost shell that will be
filled first is the 3s orbital followed by the 3p orbital. The elements and
their valence electrons are sodium (3s1), magnesium (3s2), aluminium
(3s2 3p1), silicon (3s2 3p2), phosphorus (3s2 3p3), sulphur (3s2 3p4), chlorine
(3s2 3p5) and argon (3s2 3p6).

3 The long periods (Period 4 to Period 7)
(a) In Period 4, there are 18 elements. The outermost shell that will be filled
first is the 4s orbital then the 3d orbital followed by the 4p orbital. The
elements start from potassium (1s2 2s2 2p6 3s2 3p6 4s1) to krypton (1s2 2s2
2p6 3s2 3p6 3d10 4s2 4p6).
(b) In Period 5, there are 18 elements. The outermost shell that will be filled
first is the 5s orbital then the 4d orbital followed by the 5p orbital. The
elements start from rubidium (1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s1) to xenon
(1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6).
(c) Period 6 elements have an outermost shell that includes the 6s and 6p
orbitals. This period is the first period to include the f-block elements
called the lanthanides (consists of 14 elements). The f-block elements
from Period 6 are removed and placed at the bottom of the Periodic
Table. There are 32 elements in Period 6, including the lanthanides from
caesium to radon.
(d) Period 7 elements have an outermost shell that includes the 7s and
7p orbitals. This period also includes the f-block elements called the
actinides (consists of 14 elements). The f-block elements from
Period 7 are removed and placed below the lanthanides. Presently, there
are 32 elements in Period 7 starting with francium to oganesson, which is
the heaviest element discovered.
ILMU
Table 3.1 Period 1 to Period 7 elements in the Periodic Table

PENERBIT Short periods Long periods

Period number Period 1 Periods 2 and 3 Periods 4 and 5 Period 6 Period 7

Total number of 2 elements 8 elements 18 elements 32 elements 32 elements
elements
Period 2: 2s including the including the
Valence orbitals 1s1 to 1s2 orbital followed
filled with by 2p orbital lanthanide series actinide series
electrons
Period 3: 3s Period 4: 4s 6s orbital, then 7s orbital, then
orbital followed 5f orbital, then
by 3p orbital orbital, then 3d 4f orbital, then 6d orbital,
followed by 6p
orbital, followed 5d orbital, orbital

by 4p orbital followed by 6p

Period 5: 5s orbital

orbital, then 4d

orbital, followed

by 5p orbital

52

4CHAPTERPENERBIT
ILMU
Chemical Bonding BAKTI SDN. BHD.

4.1 Lewis Structure

Learning Outcomes

• State the octet rule.
• Describe how atoms obtain the octet configuration of noble gas, pseudo-

noble gas and half-filled orbital.
• Describe the formation of the following bonds using Lewis dot symbol:
(a) Ionic or electrovalent bond
(b) Covalent bond
(c) Dative or coordinate bond
• Draw Lewis structure of molecules and polyatomic ions with single, double

and triple bonds.
• Compare the bond length between single, double and triple bonds.
• Determine the formal charge and the most plausible Lewis structure.
• Explain the exception to the octet rule:
(a) Incomplete octet
(b) Expanded octet
(c) Odd number electron
• Describe the concept of resonance.

Lewis Structure

1 Lewis structure is also known as electron-dot structure or Lewis symbols.
The Lewis structure shows only the valence electrons of atoms and
monoatomic ions.

2 A dot or cross symbol represents the valence electrons of an element.

77

 Chemical Bonding

3 For example, the electronic configuration of magnesium is 1s2 2s2 2p6 3s2.
The valence electronic configuration is 3s2. So, the Lewis dot symbol of
magnesium is •Mg•.

Table 4.1 Lewis symbol of several elements

Element Symbol Group BAKTI SDN. BHD.Electronic Number Lewis
configuration of symbol

valence
electrons

Lithium Li 1 1s2 2s1 1 Li or Li

Chapter 4 Beryllium Be 2 1s2 2s2 2 Be or Be

Boron B 13 1s2 2s2 2p1 3 B or B

Aluminium Al 13 1s2 2s2 2p6 3s2 3p1 3 Al or Al

Silicon Si 14 1s2 2s2 2p6 3s2 3p2 4 Si or Si

Nitrogen N 15 1s2 2s2 2p3 5 N or N

Sulphur S 16 1s2 2s2 2p6 3s2 3p4 6 S or S
Chlorine Cl or Cl
Cl 17 1s2 2s2 2p6 3s2 3p5 7

ILMUNeon Ne 18 1s2 2s2 2p6 8 Ne or Ne

Example 1

Draw the Lewis electron-dot symbol for the following atoms.

(a) Rubidium, Rb (c) Iodine, I
PENERBIT
(b) Silicon, Si (d) Krypton, Kr

Solution

(a) Rb has 1 valence electron. (c) I has 7 valence electrons.

Rb I

(b) Si has 4 valence electrons. (d) Kr has 8 valence electrons.

Si Kr

78

Octet Rule Chemical Bonding 

1 Octet rule refers to the tendency of atoms to achieve eight electrons in the 79
valence shell (except hydrogen) by accepting, donating or sharing electrons.

2 An atom that has fewer than eight electrons, tends to react and form a
more stable compound with an octet electronic configuration. Whereas, the
hydrogen atom achieves a duplet electronic configuration by sharing electrons
to form a molecule or a polyatomic ion.

3 Atoms can achieve octet/duplet electronic configuration in two ways:
(a) Donate electrons to form positive ions or accept electrons to form negative
ions.
(b) Share electrons with other atoms to form covalent bonds.

4 Atoms that are already stable (achieved octet or duplet electronic
configuration) are atoms of elements in Group 18.

5 Table 4.2 shows the stable duplet and octet electronic configurations of Group
18 elements.
PENERBIT Chapter 4
ILMU
Table 4.2 Duplet and octet electron configurations of Group 18 elementsBAKTI SDN. BHD.

Electronic configuration Element
Duplet electronic configuration 2He: 1s2
10Ne: 1s2 2s2 2p6
Octet electronic configuration 18Ar: [Ne] 3s2 3p6
36Kr: [Ar] 3d10 4s2 4p6
54Xe: [Kr] 4d10 5s2 5p6
86Rn: [Xe] 4f 14 5d10 6s2 6p6

Ionic and Covalent Bonds

Ionic (electrovalent) bond

1 Ionic bond is defined as a type of chemical bonding that involves the
electrostatic attraction force between oppositely charged ions. Ionic bond is
also known as an electrovalent bond.

2 Formation of ionic bonds involves the following:
(a) Atoms of metal elements lose one or more electrons to be donated to
non-metal elements to form positively charged ions known as cations.
(b) Atoms of non-metal elements gain one or more electrons from metal
elements to form negatively charged ions known as anions.

3 Ionic bond is formed between a metal element with a non-metal element.

4 For example, lithium fluoride, LiF forms an ionic bonding where Li loses
an electron to form a positive charge ion, lithium ion, Li+ to be donated to
fluorine atom, F.
F gains an electron from Li to form a negatively charged ion, fluorine ion, F–.
There is an electrostatic force of attraction between the cation and anion. The
Lewis symbols of the formation of ions are shown below.

Li + F → [ Li ]+ [ F ]–

1s22s1 1s22s22p5 1s2 1s22s22p6

States of Matter 

Example 7PENERBIT Chapter 5
ILMU
Explain why gases behave almost ideally at high temperatures and lowBAKTI SDN. BHD.
pressures.

Solution

At high temperatures, the kinetic energy of gas particles increases as the
particles are moving very fast. Because of this high energy, gas particles are
able to overcome the intermolecular attractive forces between them. Therefore,
the attractive and repulsive forces between the gas particles are negligible.

At low pressures, the gas particles have very low force of attraction between
each other because they are very far apart from each other. The volume
occupied by the gas molecules is negligible because the gas particles occupy
only a small fraction compared to the volume of container. Therefore, the
volume of gas can be assumed to be negligible compared to the volume of
container.

Quick Check 5.1

1 Explain the Boyle’s, Charles's and Avogadro’s laws.
2 State the assumptions of the kinetic molecular theory of gases.
3 State the conditions of real gas that deviate from the behaviour of ideal gas.

Explain your answer.
4 State two conditions of real gases that behave close to ideal gas.
5 Propane gas is burnt completely in oxygen gas in a furnace to produce carbon

dioxide and water.
(a) Write a balanced equation of the combustion of propane.
(b) A sample of 5.01 g of propane gas is burnt completely at 1 980 °C at

1 bar. Calculate the volume of propane gas by assuming the gas behaves
ideally.
[Given R = 0.08206 L atm mol–1 K–1]
(c) Calculate the final pressure of gas, if the temperature drops to 200 °C at a
constant volume.
(d) Calculate the partial pressure of carbon dioxide, CO2 and water H2O at
the final pressure and constant volume.

119

 States of Matter 5.2 Liquids

120 Learning Outcomes

• Explain the properties of liquid.
• Explain vaporisation and condensation processes based on kinetic

molecular theory and intermolecular forces.
• Define vapour pressure and boiling point.
• Explain boiling process.
• Explain the relationship between intermolecular forces and vapour pressure.
• Explain the relationship between vapour pressure and temperature.
BAKTI SDN. BHD.
Chapter 5 Properties of Liquid

Intermolecular forces

Surface of a liquid
Interior of a liquid

Liquid particleILMU

PENERBIT Figure 5.3 Particles in liquid

1 The properties of liquid can be explained as follows:
(a) Arrangement: Liquid particles are closely packed together but not at
fixed position.
(b) Shape and volume: Liquid has a fixed volume but no fixed shape. Liquid
takes the shape of its container because particles of liquid move around
each other.
(c) Compressibility: Liquid is slightly compressible because its particles are
closely packed with small empty spaces between them.
(d) Movement: The particles of liquid are constantly moving randomly. The
particles vibrate, rotate and move around each other.
(e) Intermolecular forces: The intermolecular attractive forces between the
particles of liquid are strong enough to hold them together. However, the
particles are still allowed to move around the neighbouring particles. As
the temperature increases, the average kinetic energy of the particles also
increases.
(f) Diffusion: Liquid has a lower diffusion rate compared to gas because
the empty spaces between the liquid particles are smaller than in gas.
The particles will move from a high concentration or pressure to a lower
concentration or pressure.
(g) Viscosity: Liquid has a significant resistance to flow, depending on the
temperature, size of particles and intermolecular attractive forces. The
higher the temperature, the lower the viscosity of the liquid. The larger the
size of particles or the stronger the intermolecular forces, the higher the
viscosity of the liquid and the more difficult for the liquid to flow.

(h) Surface tension: Liquid requires a significant amount of energy to States of Matter 
stretch its surface by a unit area. The particles on the surface are pulled
downwards and sideways from the neighbouring particles. However, the Recap!
interior particles experience attractive forces in all directions from the
neighbouring particles. The stronger the intermolecular attractive forces, Liquid
the higher the surface tension of the liquid. Surface tension in liquid

Vaporisation and Condensation Processes

1 The vaporisation process is the process where the molecules change from a
liquid phase to a gas phase, in which the liquid particles gain sufficient energy
to escape from the surface and transform into gas.

2 Vaporisation occurs when liquid particles at the surface of a liquid gain
enough kinetic energy to overcome the attractive forces that bind them
together.

3 The particles will escape from the liquid surface and vapourise into gas.
PENERBIT Chapter 5
ILMU
BAKTI SDN. BHD.
Gas particle

Liquid particles with a higher
kinetic energy will escape
from the liquid surface and
move away as gas

Liquid particle

Figure 5.4 Vaporisation

4 The condensation process is the process where particles change from vapour
to liquid, in which gas particles lose kinetic energy and transform from a gas
phase to a liquid phase.

5 Condensation occurs when temperature reduces and the kinetic energy of
particles decreases.

6 The gas particles that have lower kinetic energy become closer to each other
and turn into liquid if they cannot overcome the intermolecular attractive
forces between them.

7 More vapour particles will come in contact with the liquid surface and the
particles are attracted to the liquid surface. Thus, gas is condensed into liquid.

Gas particle loses kinetic
energy and moves to the
surface of liquid

Vapour turns
into liquid

Liquid particle

Figure 5.5 Condensation

121

 States of Matter 5.2 Liquids

120 Learning Outcomes

• Explain the properties of liquid.
• Explain vaporisation and condensation processes based on kinetic

molecular theory and intermolecular forces.
• Define vapour pressure and boiling point.
• Explain boiling process.
• Explain the relationship between intermolecular forces and vapour pressure.
• Explain the relationship between vapour pressure and temperature.
BAKTI SDN. BHD.
Chapter 5 Properties of Liquid

Intermolecular forces

Surface of a liquid
Interior of a liquid

Liquid particleILMU

PENERBIT Figure 5.3 Particles in liquid

1 The properties of liquid can be explained as follows:
(a) Arrangement: Liquid particles are closely packed together but not at
fixed position.
(b) Shape and volume: Liquid has a fixed volume but no fixed shape. Liquid
takes the shape of its container because particles of liquid move around
each other.
(c) Compressibility: Liquid is slightly compressible because its particles are
closely packed with small empty spaces between them.
(d) Movement: The particles of liquid are constantly moving randomly. The
particles vibrate, rotate and move around each other.
(e) Intermolecular forces: The intermolecular attractive forces between the
particles of liquid are strong enough to hold them together. However, the
particles are still allowed to move around the neighbouring particles. As
the temperature increases, the average kinetic energy of the particles also
increases.
(f) Diffusion: Liquid has a lower diffusion rate compared to gas because
the empty spaces between the liquid particles are smaller than in gas.
The particles will move from a high concentration or pressure to a lower
concentration or pressure.
(g) Viscosity: Liquid has a significant resistance to flow, depending on the
temperature, size of particles and intermolecular attractive forces. The
higher the temperature, the lower the viscosity of the liquid. The larger the
size of particles or the stronger the intermolecular forces, the higher the
viscosity of the liquid and the more difficult for the liquid to flow.

Answers

PENERBIT1 Matter (c) 1.8 × 1027 = 2 990 mol KCl
ILMU 6.02 × 1023
BAKTI SDN. BHD.Quick Check 1.1

1 Isotopes are atoms of the same element that have  3 Element H O
Mass (g)
the same number of protons but different number of Number of 0.44 6.92
moles
neutrons. 0.44 g 6.92 g
Ratio of 1 g mol–1 16 g mol–1
2 (a) Given 35Cl = 3.127, Assume 35Cl = 3.127 and 37Cl = 1 moles
37Cl
= 0.440 mol = 0.433 mol
Percentage abundance of 35Cl
0.440 0.433
= 3.127 × 100 = 75.77% 0.433 = 1 0.433 = 1
4.127

Percentage abundance of 37Cl The empirical formula is HO.

= 1 × 100 = 24.23% 34.0 g mol–1
4.127 n = 17 g mol–1 = 2

(b) Average atomic mass of Cu (HO)2 = H2O2

= ∑Qimi = (34.9689 × 75.77) + (36.9659 × 24.23) Therefore, the molecular formula is H2O2.
∑Qi 75.77 + 24.23
0.4 L–1
= 35.45 a.m.u. 4 (a) Molarity = 0.125 L = 3.2 mol

3 (a) Given 63Cu = 2.235, Assume 63Cu = 2.235 and (b) Molar mass of C6H12O6 = 180 g mol–1
65Cu
65Cu = 1 Number of moles of C6H12O6
1 mol
Percentage abundance of 63Cu = 275 g × 180 g = 1.528 mol

= 2.235 × 100 = 69.09% Molarity = 1.528 mol = 1.946 mol L–1
3.235 0.785 L

Percentage abundance of 65Cu 5 (a) Assume the mass of the solution = 100 g

= 1 × 100 = 30.91% Mass of HCl = 39 g
3.235
Mass of water = 61 g
(b) Average atomic mass of Cl
MSolution 100 g
= ∑Qimi = (62.930 × 69.09) + (64.928 × 30.91) Use ρ = VSolution =V = 2.1977 g ml–1 = 45.50 ml
∑Qi 69.09 + 30.91
MV
= 63.55 a.m.u. Number of moles = 1 000

4 (a) Number of protons = 36 39 g

(b) Number of electrons = 36 = 36.5 g mol–1 = 1.07 mol

(c) Number of neutrons = 84 – 36 = 48 Molarity = 1.07 × 1 000 = 23.52 M
45.50
5 (a) 29444Pu
(b) 2693Cu (b) Molality = 1.07 = 17.54 mol L–1
(c) 8305Br (100 – 39) ÷ 1 000

Quick Check 1.2 Quick Check 1.3

1 (a) 0.23 × (6.02 × 1023) = 1.3846 × 1023 Al atoms 1 Given MnO4– + S2– → Mn2+ + S

(b) 1.2 × (6.02 × 1023) = 7.224 × 1023 Cl2 molecules Step 1
(c) 0.5 × (6.02 × 1023) = 3.01 × 1023 OH– ions
[O]: S2– → S
2 (a) 4× 1024 = 6.64 mol C
6.02 × 1023 [R]: MnO – → Mn2+
4

(b) 2.4 × 1027 = 3 986 mol H2O Step 2 (For reduction equation, add 4H2O on the right
6.02 × 1023 side and 8H+ ion on the left side)

189

 Answers

[O]: S2– → S Structured Questions

[R]: 8H+ + MnO4– → Mn2+ + 4H2O  1 Element/ Number of Number of Numbers of
Step 3 ( Add electrons on both sides to balance the Ion electrons protons neutrons

charges) 58 Fe3+ 23 26 32
26
[O]: S2– → S + 2e–

[R]: 5e– + 8H+ + MnO – → Mn2+ + 4H2O 244 Pu 94 94 150
4 94
AnswersPENERBIT
ILMUStep 4 ( Oxidation equation is multiplied by 5, 32 S2– 18 16 16
BAKTI SDN. BHD. 16
reduction equation is multiplied by 2)
2 0
[O]: 5S2– → 5S + 10e– 3 NH3 is the limiting reactant because the number of

[R]: 10e– + 16H+ + 2MnO–4 → 2Mn2+ + 8H2O moles of NH3 given is less than the number of moles of
Step 5 (Add the [O] and [R] equations, then cancel out NH3 needed.
4 (a) Assume the mass of oxalic acid = 100 g
10e–)

Balanced chemical equation:

5S2– + 16H+ + 2MnO4– → 5S + 2Mn2+ + 8H2O Element H C O
2 Limiting reactant is the reactant that reacts completely
Mass (g) 2.22 26.66 71.12
in a reaction.
Number 2.22 g 26.66 g 71.12 g
3 (a) Aluminium 0.5 g
of moles 1 g mol–1 12 g mol–1 16 g mol–1
Number of moles of Al = 27 g mol–1 = 0.0185 mol
= 2.22 mol = 2.22 mol = 4.45 mol

Number of moles of Cu(NO3)2 Ratio of 2.22 = 1.00 2.22 = 1.00 4.45 = 2.00
moles 2.22 2.22 2.22
= 0.35 × 100 = 0.035 mol
1 000 The empirical formula of oxalic acid is CHO2.
Number of moles of Al 90 g mol–1
Actual ratio = Number of moles of Cu(NO3)2
(b) n = [12 + 1 + 2(16)]g mol–1 = 2
= 0.0185 mol = 0.529
0.035 mol (CHO2)2 = C2H2O4

The limiting reactant from the stoichiometric ratio. Therefore, the molecular formula of oxalic acid is

Stoichiometric ratio = 2 mol = 0.66 C2H2O4.
3 mol 10 g

2 mol Al produce 3 mol Cu. (c) Number of moles = 90 g mol–1 = 0.111 mol
0.0185 mol Al produces 0.02775 mol Cu.
Since the number of moles of Cu(NO3)2 given Molarity = 0.111 mol = 0.222 M
0.5 L
is more than the number of moles of Cu(NO3)2
needed, Cu(NO3)2 is the excess reactant, while Al is 5 (a) Relative abundance
the limiting reactant.
28Si
63.6 g
mol Cu 29.8

(b) Mass of Cu = 0.02775 mol Cu × 1 = 1.765 g

(c) 2 mol Al produce 3 mol Cu(NO3)2 1.5 29Si
0.0185 mol Al produces 0.02775 mol Cu(NO3)2 1.0 30Si
Mass of used Cu(NO3)2
28 29 30 m/e
187.6 g
= 0.02775 mol × 1 mol = 5.206 g (b) Average atomic mass of Si

Mass of Cu(NO3)2 in the solution = 28(29.8) + 29(1.5) + 30(1.0) = 28.1 a.m.u.
(29.8 + 1.5 + 1.0)
187.6 g
= 0.035 mol × 1 mol = 6.566 g

Mass of excess reactant = 6.566g – 5.206 g = 1.36 g 2 Atomic Structure

Summative Practice 1 Quick Check 2.1

Objective Questions 1 2 1 λ1 = RH1 – 1
1 B 2 A n12 n22
6 A 7 D 3 A 4 D 5 C 1 2
11 C 12 A 8 A 9 B 10 B = 1.097 × 107 m–1 × 1 – 1
13 D 14 B 15 D 12 42

λ = 9.7235 × 10–8 m = 97.23 nm

190

PENERBIT ILMU
BAKTI SDN. BHD.