STPM Scheme of Assessment
Term of Paper Code Type of Test Mark Duration Administration
Study and Name (Weighting)
Written Test 60
(26.67%)
Section A 15
15 compulsory multiple-choice questions
to be answered.
962/1
First Section B 1 Central
Term Chemistry 2 compulsory structured questions to be 1—hours assessment
Paper 1 15 2
answered.
Section C
2 questions to be answered out of 3 essay
questions.
30
All questions are based on topics 1 to 6.
Written Test 60
(26.67%)
Section A 15
15 compulsory multiple-choice questions
to be answered.
962/2
Second Chemistry Section B 1 Central
Term 2 compulsory structured questions to be 1—hours assessment
Paper 2 answered. 15 2
Section C
2 questions to be answered out of 3 essay
questions.
30
All questions are based on topics 7 to 13.
Written Test 60
(26.67%)
Section A 15
15 compulsory multiple-choice questions
to be answered.
962/3 Section B
Central
1
Chemistry 2 compulsory structured questions to be 1—hours assessment
Third Paper 3 answered. 15 2
Term
Section C
2 questions to be answered out of 3 essay
questions.
30
All questions are based on topics 14 to 21.
962/5 Written Practical Test 45
1
Chemistry 3 compulsory structured questions to be (20%) 1—hours Central
Paper 5 answered. 2 assessment
First, School-based Assessment of Practical 225
Second 962/4 13 compulsory experiments and one to be scaled Through-
and Chemistry project to be carried out. to 45 out the School-based
Third Paper 4 (20%) three terms assessment
Terms
vi
STPM Scheme.indd 6 3/26/18 2:28 PM
CONTENTS
Chapter
1 ATOMS, MOLECULES AND STOICHIOMETRY 1
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
1.1 Fundamental Particles of an Atom 2
1.2 Relative Atomic, Isotopic, Molecular and Formula Masses 13
1.3 The Mole and the Avogadro’s Constant 24
Summary 30
STPM Practice 1 30
QQ 35
Chapter
2 ELECTRONIC STRUCTURE OF ATOMS 36
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
2.1 Electronic Energy Levels of Atomic Hydrogen 37
2.2 Atomic Orbitals: s, p and d 48
2.3 Electronic Configurations 51
2.4 Classification of Elements in the Periodic Table 55
Summary 59
STPM Practice 2 60
QQ 63
Chapter
3 CHEMICAL BONDING 64
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
3.1 Ionic Bonding 65
3.2 Covalent Bonding 70
3.3 Metallic Bonding 107
3.4 Intermolecular Forces: van der Waals Forces and Hydrogen Bonding 110
Summary 124
STPM Practice 3 125
QQ 130
Chapter
4 STATES OF MATTERS 131
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
4.1 Gases 132
4.2 Liquids 147
4.3 Solids 152
4.4 Phase Diagrams 161
Summary 174
STPM Practice 4 175
QQ 180
Chapter
5 REACTION KINETICS 181
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
5.1 Rate of Reaction 182
5.2 Rate Law 191
5.3 The Effect of Temperature on Reaction Kinetics 193
5.4 The Role of Catalysts in Reactions 194
5.5 Order of Reactions and Rate Constants 201
vii
Contents.indd 7 3/26/18 2:30 PM
Summary 213
STPM Practice 5 214
QQ 220
Chapter
6 CHEMICAL EQUILIBRIA 221
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
6.1 Reversible Reaction and Dynamic Equilibrium 222
6.2 Law of Mass Action and Equilibrium Constant 224
6.3 Deduce Equilibrium Constant, K and K p 225
c
6.4 Calculation Involving Equilibrium Constant 228
6.5 Nitrogen Dioxide and Stratospheric Ozone 232
6.6 Factors Affecting Equilibrium System and Le Chatelier's Principle 233
6.7 Equilibrium and Temperature 240
6.8 Industrial Processes 243
Summary 245
STPM Practice 6 246
QQ 251
Chapter
7 IONIC EQUILIBRIA AND SOLUBILITY EQUILIBRIA 252
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
7.1 Electrolytes 253
7.2 Acids and Bases 254
7.3 Conjugate Acid-base Pair 257
7.4 Relative Strength of BrØnsted-Lowry Acids and Bases 259
7.5 Acid-base Titration 273
7.6 Buffer Solution 280
7.7 Important Buffers in Human Blood 290
7.8 Solubility Equilibria: Sparingly Soluble Salts 291
7.9 Calculating K from Solubility and Vice Versa 292
sp
7.10 Common Ion Effect 293
7.11 Predicting Precipitation 296
7.12 Solubility Equilibria and Water Softening 298
Summary 300
STPM Practice 7 301
QQ 305
Chapter
8 PHASE EQUILIBRIA 306
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
8.1 Miscible Liquids 307
8.2 Mixture of Two Miscible Liquids 308
8.3 Fractional Distillation 316
8.4 Non-ideal Solutions 319
8.5 Fractional Distillation under Reduced Pressures 330
Summary 332
STPM Practice 8 333
QQ 336
• STPM Model Paper (962/1) 337
• Appendix 343
• Glossary 351
• Answers 357
• Index 382
viii
Contents.indd 8 3/26/18 2:30 PM
CHAPTER ELECTRONIC STRUCTURE
2 OF ATOMS
Concept Map
Electronic Structure of Atoms
Electromagnetic Line Spectrum Bohr’s Model of Modern Atomic Filling of
Radiation of Hydrogen Hydrogen Atom Model Electrons
Electronic Electronic Ionisation Energy Atomic
Configuration Configuration of Hydrogen Orbitals
of Ions • Ionisation energies
and electronic shells
L earning O utcomes
Learning Outcomes
Students should be able to: • define and apply Aufbau principle, Hund’s rule and
Electronic energy levels of atomic hydrogen Pauli exclusion principle.
• explain the formation of the emission line spectrum Classification of elements into s, p, d and f
of atomic hydrogen in the Lyman and Balmer series blocks in the Periodic Table
using Bohr’s Atomic Model. • identify the position of the elements in the Periodic
Atomic orbitals: s, p and d Table as 1
• deduce the number and relative energies of the s, p (a) block s, with valence shell configurations s and
s ,
2
and d orbitals for the principal quantum numbers 1,
2 and 3, including the 4s orbitals; (b) block p, with valence shell configurations from
s p to s p ,
2 1
2 6
• describe the shape of the s and p orbitals. (c) block d, with valence shell configurations from
1 2
10 2
Electronic configuration d s to d s ;
• predict the electronic configuration of atoms and ions • identify the position of elements in block f of the
given the proton number (and charge); Periodic Table.
02 Chapter 2.indd 36 3/26/18 3:14 PM
Chemistry Term 1 STPM
2.1 Electronic Energy Levels of
Atomic Hydrogen
Electromagnetic Radiations
1 Electromagnetic radiations consist of radiations/waves such as CHAPTER
the gamma radiations, X-rays, ultraviolet radiations, infrared 2
radiations, visible light, microwaves and many others.
2 Each type of radiation is characterised by their wavelengths or
frequencies.
3 The wavelength (λ) is defined as the distance between two adjacent
peaks or troughs as shown below. It has the unit of meter (m) or
nanometer (nm).
[1 nm = 1 × 10 m]
–9
Wavelength (λ)
Peak
Trough
4 Frequency (f) is the number of waves that passes through a certain Another symbol for frequency is
point per second. It has a unit of Hertz (Hz) or per second (s ). The ‘v’.
–1
diagram below shows two waves, one of high frequency and one of
low frequency.
High frequency
Low frequency
5 The frequency of a wave is inversely proportional to its wavelength.
1
f ∝
λ
37
02 Chapter 2.indd 37 3/26/18 3:14 PM
Chemistry Term 1 STPM
6 The relationship between the wave length (λ) and the frequency (f)
of the electromagnetic radiation is given by:
λf = C
–1
8
[where C = velocity of light = 3.0 10 m s ]
CHAPTER 7 The diagram below shows the various types of electromagnetic
radiations.
2
Wavelength increasing
10 1 10 –2 10 –4 10 –6 10 –8 10 –10 10 –12 10 –14 Meter
Radio TV Microwaves Infrared Ultraviolet X-rays y-rays Cosmic
waves waves radiation Visible light radiation rays
3 10 8 3 10 10 3 10 12 3 10 14 3 10 16 3 10 18 3 10 20 3 10 22 1
Second
Frequency increasing
700 Red Orange Yellow Green Blue Violet 400 nm
Quantum theory 8 In 1900, Max Planck put forward his quantum theory which
states that energy in the form of electromagnetic radiation can be
emitted or absorbed in discrete amount (or packets) called quanta
(singular: quantum).
Concept of quanta 9 The magnitude of a quantum depends on the frequency (or
wavelength) of the radiation. The energy (in J) associated with a
quantum of energy (or a photon) is given by the Planck’s equation:
E = hf
Where h = Planck’s constant = 6.63 10 J s or
–34
= 3.99 10 kJ mol s
–1
–13
Quick Check 2.1
1 A quantum of a particular electromagnetic radiation has energy of 10 J. Calculate
–18
(a) the energy in kJ mol ,
–1
(b) the frequency, and
(c) the wavelength of the radiation.
2 A gamma radiation has a frequency of 1 10 Hz. Calculate
24
(a) its wavelength,
(b) its energy in J, and
–1
(c) its energy in kJ mol .
38
02 Chapter 2.indd 38 3/26/18 3:14 PM
Chemistry Term 1 STPM
The Line Spectrum of Hydrogen 2013/P1/Q18(a)
2015/P2/Q4
1 When sunlight passes through a prism, the colours of the rainbow
are observed. This type of spectrum is known as a continuous 2016/P1/Q3, Q18(c)
spectrum because it contains all the possible wavelengths (or 2017/P1/Q2
CHAPTER
frequencies) of the visible light. 2018/P1/Q2, Q19(a)
2
~ 700 nm
Red
Violet
ips
T
Sunlight Tips
Exam
Exam
~ 400 nm
The Continuous Spectrum Continuous spectrum
contains all the wavelengths
2 However, when electricity is passed through a discharge tube that make up visible light.
filled with hydrogen gas at low pressure, a pink light is emitted.
ips
T
When this light passes through a prism or is observed through a Exam Tips
Exam
spectroscope a line spectrum is obtained.
The spectrum consists of a
few coloured lines, each
corresponding to a particular
Slits wavelength or frequency.
Prism (Red: 656.2 nm, blue-green:
486.1 nm, blue-violet:
434.0 nm and violet:
410.0 nm)
Blue- Blue-
Violet violet green Red
Gas discharge
tube containing
hydrogen 410.0 nm 434.0 nm 486.1 nm 656.2 nm
Line Spectrum of
Atomic Hydrogen
INFO
400 450 500 550 600 650 700 750
Wavelength
39
02 Chapter 2.indd 39 3/26/18 3:14 PM
Chemistry Term 1 STPM
The hydrogen line spectrum 3 Using more sophisticated instruments, lines in other parts of the
spectrum were also detected. These lines are in the ultraviolet and
the infrared regions. Part of the line spectrum of hydrogen is shown
below:
Info Chem
CHAPTER Characteristics of the series: Wavelength
Frequency
2 • Each series consists of
discrete lines with specific
frequencies.
• The interval between the lines
Ultraviolet Visible Infrared
gets smaller towards the high (Lyman series) (Balmer series) (Paschen series)
frequency end.
4 Each set of lines is known as a series and is named after its
• Each series ends with a line
with maximum frequency.
discoverer.
Region Name of series
Ultraviolet Lyman
Visible Balmer
Infrared Paschen
Infrared Bracket
Infrared Pfund
5 In each series, the intervals between the lines get smaller towards
T
ips
Exam
Exam Tips the high frequency (low wavelength) end of the spectrum and
Converging spectrum.
finally merge to form a continuous spectrum which ends at a
maximum frequency value. Such spectrum is called a converging
spectrum.
6 The line spectra of other elements can also be obtained in the
same way. No two elements have identical line spectra. Hence, it is
possible to identify the element from its line spectrum.
7 Mathematically, the wavelength of the lines in the emission spectrum
of hydrogen can be calculated using the Rydberg’s equation:
1 = R 1 – 1
Tips
Exam T ips λ H n 2 n 2
Exam
1 2
where R = Rydberg’s constant
The Rydbergʼs equation can H
7
be used to calculate the = 1.097 10 m –1
wavelengths of the lines in
the hydrogen spectrum only. n and n = Integers
2
1
n can take any values from (n + 1), (n + 2) .....∞
2 1 1
8 The values of n and n for each series are given in the table below:
1 2
Series Region n n
1 2
Lyman Ultraviolet 1 2, 3, 4 .....∞
Balmer Visible 2 3, 4, 5, 6 .....∞
Paschen Infrared 3 4, 5, 6, 7 .....∞
Bracket Infrared 4 5, 6, 7, 8 .....∞
40
02 Chapter 2.indd 40 3/26/18 3:14 PM
Chemistry Term 1 STPM
9 The first line in the Lyman series corresponds to n = 1 and n = 2,
2
1
while the second line corresponds to n = 1 and n = 3 and so on.
1 2
10 The same applies to the other series.
Example 2.1 CHAPTER
Calculate the wavelengths and frequencies of the following lines in 2
the emission spectrum of hydrogen.
(a) First line in the Balmer series.
(b) Last line in the Balmer series.
Solution
(a) Using n = 2 and n = 3
1
2
1 = 1.097 10 1 –
1
7
λ 2 2 3 2
1 = 1.534 10 m –1
6
λ
–7
∴ λ = 6.52 10 m or
= 652 nm
C 3.0 × 10 8
f = = = 4.60 × 10 Hz
14
λ 6.52 × 10 –7
(b) Using n = 2 and n = ∞
1
1
1 = 1.097 × 10 1 – 1
7
λ 2 2 ∞ 2
1 = 2.74 × 10 m –1
6
λ
∴λ = 3.65 10 m or
–7
= 365 nm
C 3.0 × 10 8
f = = –7 = 8.22 10 Hz
14
λ 3.65 × 10
Bohr's Model of the Hydrogen Atom 2009/P1/Q3
1 In order to explain the formation of the line spectrum of hydrogen, 2014/P1/Q18(a)
Niels Bohr, a Danish physicist, put forward his model of the
hydrogen atom in 1913. [He received the Nobel Prize for physics Bohr's Model of the
in 1922 for his theory explaining the emission line spectrum of Hydrogen Atom
hydrogen]. VIDEO
2 He postulated that electrons revolve round the nucleus in fixed Concept of orbit:
circular paths of different radii called orbits, much like the planets A fixed circular path
revolving around the Sun.
3 The energy of the orbits is quantised. Each orbit is represented by
an integer, n, which is called the principle quantum number.
41
02 Chapter 2.indd 41 3/26/18 3:14 PM
Chemistry Term 1 STPM
Quantised energy levels 4 The single electron in the hydrogen atom can be located in any of
can be visualised as the the orbits (depending on its energy) but not in spaces between the
shelves in a book shelf.
The books can be on any orbits.
shelf, but not the space
n = 7
between the shelves. n = 6
CHAPTER n = 5
n = 4
2 Book n = 3
n = 2
Shelf
n = 1
+
5 Each orbit is also known as a principle shell and is represented by
an alphabet.
Orbit, n Principle shell
1 K
2 L
3 M
4 N
5 O
6 P
7 Q
6 When the electron is in the orbit closest to the nucleus (i.e. at orbit
In the ground state, the
electron is stable. n = 1), it is in its ground state and is stable. It will not lose energy as it
revolves round the nucleus. (This is in contradiction to the classical
law of physics which states that an electron revolving around a
positive nucleus must lose
Contradiction to classical
Physics energy continuously and Energy absorbed
will eventually fall into the
nucleus).
Nucleus
7 The electron can be excited
Excitation of the electron
to higher energy levels by
absorbing the required
energy. This is called the
excited state.
Energy released
8 The electron in the excited
Returning to the ground state state will ‘drop’ back to
lower levels by emitting the Nucleus
excess energy in the form of
electromagnetic radiations.
42
02 Chapter 2.indd 42 3/26/18 3:14 PM
Chemistry Term 1 STPM
9 The frequency of the radiation emitted is given by Plank’s
equation: Exam Tips
ips
T
Exam
∆E = hf
where ∆E = energy difference between the two levels concerned
10 All transitions from higher energy levels to level n = 1 emit 2009/P1/Q3
radiations with frequencies in the ultraviolet region, giving rise to CHAPTER
Lyman Series. 2
11 Transitions from higher levels back to level n = 2 emit radiations
with frequencies in the visible region, giving rise to Balmer Series.
This is summarised in the diagram below:
n = 6
n = 5
Tips
Exam
n = 4 Exam T ips
n = 3
Paschen series
The intervals between levels/
T
ips
Tips
orbits gets smaller with
Exam
Exam
n = 2 increasing value of n.
Balmer series
n = 1
Lyman series
T
ips
Frequency Tips
Exam
Exam
Example 2.2
ips
T
Exam
Exam Tips
The difference in energy between the first and second levels in
–1
the hydrogen atom is 985 kJ mol . Calculate the frequency of the Since ∆E is given in terms of
radiation emitted when an electron falls from the second level to kJ mol , h will have the value
–1
–1
the first. of 3.99 × 10 –13 kJ mol s and
not 6.63 × 10 –34 Js.
Solution
ips
T
Exam Tips
Exam
∆E = 985 kJ mol –1
This frequency corresponds
to the ultraviolet region.
Using the equation:
∆E = hf
–13
985 = (3.99 10 ) f
∴ f = 2.47 10 Hz
15
43
02 Chapter 2.indd 43 3/26/18 3:14 PM
Chemistry Term 1 STPM
Quick Check 2.2
1 The difference in energy between two levels in the hydrogen atom is 3.03 10 J.
–19
(a) Calculate the frequency of light emitted when transition takes place between these two levels.
CHAPTER 2 The line spectrum of sodium consists of a line with wavelength of 590 nm.
(b) In what part of the electromagnetic spectrum is the light emitted?
2 (a) Calculate its frequency.
–1
(b) Calculate the difference in energy (in kJ mol ) between the two levels where the
transition occurs.
3 The energies of the orbits in the hydrogen atom are as shown below:
1277 n = 6
1261 n = 5
1231 n = 4
y
1168 n = 3
985 n = 2
x
0 n = 1
Calculate the frequencies of the lines marked x and y.
Ionisation Energy of Hydrogen
2013/P1/Q18(b)
Definition of the ionisation energy 1 Ionisation energy of hydrogen is the minimum energy required to
remove the lone electron in the hydrogen atom per mole of gaseous
of hydrogen
hydrogen atom under standard conditions (298 K and 101 kPa).
+
H(g) → H (g) + e
2 For the hydrogen atom, the ionisation energy corresponds to the
difference in energy between the levels, n = 1 and n = ∞.
n = ∞
∆E = ionisation energy
n = 1
The line with maximum 3 The electronic transition between these two levels will result in a
frequency is known as line with maximum frequency in the Lyman series of the hydrogen
the convergence limit or
convergence frequency. line spectrum.
44
02 Chapter 2.indd 44 3/26/18 3:14 PM
Chemistry Term 1 STPM
n = ∞
f
(maximum)
CHAPTER
n = 1
2
15
3.29 10 Hz
Frequency
4 Using the formula:
∆E = hf
–13
15
= (3.99 10 )(3.29 10 ) kJ mol
–1
= 1312.7 kJ mol –1
Example 2.3
The frequencies of the first four lines in the Lyman Series are given Graphical method to
below: calculate the ionisation
15
2.56; 2.92; 3.08; 3.16 ( 10 Hz) energy of hydrogen
(a) Plot a suitable graph to determine the frequency of the last
line in the Lyman Series.
(b) Use your answer to question (a), calculate the ionisation
energy of hydrogen.
Solution
Using the equation:
The Rydberg’s equation for the Lyman series is:
1 1 1
= R –
λ H 1 2 n 2
1
Or, f = CR 1 –
H 1 2 n 2
Rearranging:
1
f = CR – CR
H H n 2
1
A graph of frequency, f, against will give a straight line where
n 2
the maximum frequency can be obtained by the intercept at the
frequency axis (where n = ∞) and 1 = 0
n 2
15
f (× 10 )/s –1 2.56 2.92 3.08 3.16
n 2 3 4 5
1
0.25 0.11 0.063 0.040
n 2
45
02 Chapter 2.indd 45 3/26/18 3:14 PM
Chemistry Term 1 STPM
15
f(× 10 )/s –1
3.29
CHAPTER 1
2 n 2
The maximum frequency is 3.29 10 Hz.
15
(b) Using the equation:
∆E = hf
15
–3
∴ ∆E = (3.99 10 )(3.29 10 ) 10 = 1312.7 kJ mol –1
–10
Alternative method:
The convergence frequency of the Lyman series occurs when the
difference in frequency of successive lines (∆f) is zero. Thus, if
we plot a graph of frequency (f) against ∆f, the intersection on
the f axis (where ∆f = 0) would give the value of the convergence
frequency.
f(× 10 )/Hz 2.56 2.92 3.08 3.16
15
∆f(× 10 )/Hz – 0.36 0.16 0.08
15
15
ƒ (× 10 ) / Hz
3.29
Δƒ (× 10 ) / Hz
15
Using the equation:
∆E = hf
= (3.99 × 10 )(3.29 × 10 ) × 10 –3
–10
15
= 1312.7 kJ mol –1
Ionisation Energies and Electronic Shells
1 Evidence of energy levels or electronic shells is evident when one
studies the successive ionisation energies of an atom.
2 A graph of ionisation energy against order of electrons removed
provides information on the number of principle shells or energy
levels and the distribution of the electrons in the shells.
Ionisation
energy
3 The table below lists the successive ionisation energies of carbon
atom, C.
6
No. of electrons removed 1 2 3 4 5 6
Ionisation energy/kJ mol –1 1090 2350 4610 6220 37 830 42 280
4 A plot of the successive ionisation energies gives the graph on the
1 2 3 4 5 6
left.
Order of electron removed
46
02 Chapter 2.indd 46 3/26/18 3:14 PM
Chemistry Term 1 STPM
5 The first and second ionisation energies correspond to the process:
–1
+
C(g) → C (g) + e ∆H = 1090 kJ mol
2+
C (g) → C (g) ∆H = 2350 kJ mol –1
+
6 The ionisation energy increases with the number of electrons
removed. This is because successive species have ever increasing CHAPTER
proton ratio. This causes the remaining electrons to be more
electron 2
tightly held by the nucleus.
7 The graph shows that there is a gradual increase in the first four Sucessive ionisation energy
ionisation energies. However, there is a very large increase in energy increases because electrons
in removing the fifth electron. are removed from cations
8 This suggests that the first four electrons and the fifth electron of increasing charge.
occupy different energy levels or shells.
9 The ionisation energies of the last two electrons are very large
compared to the first four, suggesting that the last two electrons are
closest to the nucleus in energy level or orbit n = 1, while the first
four electrons are in the outermost energy level or orbit, n = 2.
10 The distribution of the five electrons in the boron atom is:
Orbit, n Principle shell No. of electrons
1 K 2
2 L 3
11 A similar study shows that silicon, Si, has the following
14
electronic distribution.
Orbit, n Principle shell No. of electrons
1 K 2
2 L 8
3 M 4
Quick Check 2.3
1 Use appropriate equations to represent the following ionisation:
(a) the 5th ionisation energy of sodium.
(b) the 12th ionisation energy of aluminium.
2 The following graph shows the successive ionisation energies of an element X.
Ionisation
energy
Order of electrons removed
47
02 Chapter 2.indd 47 3/26/18 3:14 PM
Chemistry Term 1 STPM
(a) How many electrons are there in one atom of X?
(b) How many energy levels are filled with electrons?
(c) State the distribution of the electrons in X.
3 Element W is in Group 15 of the Periodic Table. Sketch a graph of successive ionisation
CHAPTER energies for the first 8 electrons.
2
2.2 Atomic Orbitals: s, p and d
2015/P1/Q5
Short-comings of Bohr’s model. 1 Although Bohr’s model can satisfactorily explain the formation
of the line spectrum of hydrogen or one-electron species such as
He and Li . It cannot explain the formation of the line spectra of
+
2+
multielectron atoms.
2 Bohr’s model also cannot explain the appearance of extra lines in
the emission spectra of elements when a magnetic field is applied.
3 Modern theories still embrace the concept of quantised energy
levels but totally reject the circular orbit concept.
Energy Subshell
1 A detailed study of the line spectrum of atoms shows that each
principle shell is further made up of subshells with slightly
different energies.
2 The number of subshells in a principle shell is equal to the
n
quantum number of the shell concerned.
3
Principle shell n No. of subshells
2 First, K 1 1
Second, L 2 2
1
Third, M 3 3
Fourth, N 4 4
Fifth, O 5 5
3 In any one subshell, the level with the lowest energy is assigned the
letter s, followed by p, d and f.
For example,
Energy
Principle shell n No. of subshells Symbol
d
First, K 1 1 1s
p Second, L 2 2 2s, 2p
Third, M 3 3 3s, 3p, 3d
s
Fourth, N 4 4 4s, 4p, 4d, 4f
[In all cases, the s orbital has the lowest energy in each shell.]
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02 Chapter 2.indd 48 3/26/18 3:14 PM
Chemistry Term 1 STPM
4 The total number of electrons that can occupy any principle shell
2
(with quantum number n) is 2n .
Principle shell n No. of subshells
First, K 1 2
CHAPTER
Second, L 2 8
Third, M 3 18
2
Fourth, N 4 32
5 Further experiments on the line spectra shows that each subshell is
further made up of orbital where the electrons are placed.
6 The number of orbital depends on the type of subshell as shown
below: Energy
d
Subshell No. of orbitals Symbol
s 1 s p
p 3 p , p , p
x y z s
d 5 , d 2
d , d , d , d 2 – y 2
xy yz xz x z Subshell Orbital
7 The three p orbitals have equivalent energies. In other words, they
degenerate. The same goes for the five d orbitals.
Shapes of the Orbitals
1 The concept of orbitals arises from the fact that an electron has dual
nature. It is a particle as well as a wave. Shapes of the
Orbitals
2 As a wave, it extends in all directions in space. VIDEO
3 However, we can define the region in space within which the ‘wave
density’ of a particular electron is maximum.
4 An orbital is defined as the region (or volume) in space around Definition of orbital
the nucleus where the probability of finding a particular electron
is maximum ( 95%). This is opposed to the idea of orbit which
refers to a fixed circular path where the electron moves.
(a) (b) (c)
49
02 Chapter 2.indd 49 3/26/18 3:14 PM
Chemistry Term 1 STPM
5 The s orbital is spherical and non-directional.
Y
CHAPTER X
2
Z
The shaded area represents the region in which the chance of
finding the s electron is more that 95%. The size of the orbital
increases in the order 1s 2s 3s .....
6 The p orbitals have dumb-bell shapes and are directional.
The diagram below shows the distribution of three p orbitals around
the nucleus.
y
z
x
50
02 Chapter 2.indd 50 3/26/18 3:14 PM
Chemistry Term 1 STPM
7 The shapes of the five d orbitals are:
z z
y y
CHAPTER
2
x x
d xz d yz
z z
y y
x
x
d xy d x 2 y 2
–
z
y
x
d x 2
8 Note that in the set of d orbitals, two of them, d 2 2 and d 2, have
x – y z
their ‘lobes’ along the x, y and z axes while the other three have their
'lobes' in between the axes.
2.3 Electronic Configurations 2008/P2/Q1(a)
2010/P1/Q4
2011/P1/Q3, Q5
1 There are three rules that govern the filling of electrons in an atom:
The Aufbau Principle, the Pauli’s exclusion Principle and Hund’s 2011/P2/Q5(a)
rule. 2013/P1/Q3, Q18(c)
2014/P1/Q3
2 The Aufbau Principle states that in the ground state of an atom, the 2014/P1/Q18(b)
electrons must occupy orbitals in order of increasing energy. The 2015/P1/Q6
diagrams below show the arrangement of the energy levels and the 2015/P1/Q18
order they are filled.
2016/P1/Q4
2018/P1/Q19(b)
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02 Chapter 2.indd 51 3/26/18 3:14 PM
Chemistry Term 1 STPM
6s
5p
4d
1s
CHAPTER 2s 2p 5s 4p
2 3s 3p 3d Energy 3d
4s
4s 4p 4d 4f
3p
3s
5s 5p 5d
2p
6s 6p
2s
1s
7s
Electrons with opposite 3 The Pauli’s Exclusion Principle states that an orbital can
accommodate a maximum of two electrons only. Both the
spins attract one another.
electrons must have opposite spins. (Otherwise they will repel one
Info Chem another).
T
Tips
ips
The spins of the electrons are 4 The total number of electrons in an orbital or a set of degenerate
Exam
Exam ’ and ‘ ’.
represented by ‘
orbital is given below:
Orbital Maximum number of electrons
s 1 2 = 2
p 3 2 = 6
d 5 2 = 10
ips
T
ips
T
Tips
Exam Tips f 7 2 = 14
Exam
Exam
Exam
5 Hund’s Rule states that in a set of degenerate orbitals, electrons
Another statement of Hund's
rule: 'If two or more orbitals must occupy the orbital singly (with parallel spins) first before
with equal energy are pairing.
available, the orbitals are
filled with one electron, with 6 An example of Hund’s rule is illustrated below:
the electrons having parallel If the p orbitals are occupied by only two electrons, the distribution
spins, before a second
electron of the opposite spin is:
is added.'
T
ips
Exam Tips and not:
Exam
Two electrons occupying the
same orbit experience
repulsion due to interaction of 7 In the above diagram, each box represents an orbital. The boxes that
their electron clouds. are joined together represent degenerate orbitals.
52
02 Chapter 2.indd 52 3/26/18 3:14 PM
ips
Tips
T
Exam
Exam
Chemistry Term 1 STPM
8 Following the three rules introduced, the electronic configurations
Tips
of the first 30 elements in the Periodic Table are given below: Exam T ips
Exam
Proton Electronic Condensed
Element The outermost shell cannot
number configuration configuration
have more than 8 electrons.
H 1 1s 1 1
1 CHAPTER
He 2 1s 2 2
2
2
2
Li 3 1s 2s 1 2.1
3
2
Be 4 1s 2s 2 2.2
4
2
2
B 5 1s 2s 2p 1 2.3
5
2
2
C 6 1s 2s 2p 2 2.4
6
2
2
N 7 1s 2s 2p 3 2.5
7
2
2
O 8 1s 2s 2p 4 2.6
8
2
2
F 9 1s 2s 2p 5 2.7
9
2
2
Ne 10 1s 2s 2p 6 2.8
10
2
2
6
Na 11 1s 2s 2p 3s 1 2.8.1
11
6
2
2
Mg 12 1s 2s 2p 3s 2 2.8.2
12
2
2
6
2
Al 13 1s 2s 2p 3s 3p 1 2.8.3
13
2
6
2
2
Si 14 1s 2s 2p 3s 3p 2 2.8.4
14
2
2
2
6
P 15 1s 2s 2p 3s 3p 3 2.8.5
15
2
2
6
2
S 16 1s 2s 2p 3s 3p 4 2.8.6
16
2
2
2
6
Cl 17 1s 2s 2p 3s 3p 5 2.8.7
17
2
2
2
6
Ar 18 1s 2s 2p 3s 3p 6 2.8.8
18
6
2
2
6
2
K 19 1s 2s 2p 3s 3p 4s 1 2.8.8.1
19
2
6
2
2
6
Ca 20 1s 2s 2p 3s 3p 4s 2 2.8.8.2
20
2
1
2
6
6
2
Sc* 21 1s 2s 2p 3s 3p 3d 4s 2 2.8.9.2
21
6
2
6
2
2
2
Ti* 22 1s 2s 2p 3s 3p 3d 4s 2 2.8.10.2
22
6
2
6
2
2
3
V* 23 1s 2s 2p 3s 3p 3d 4s 2 2.8.11.2
23
6
2
5
2
6
2
Cr* 24 1s 2s 2p 3s 3p 3d 4s 1 2.8.13.1
24
6
5
2
6
2
2
Mn* 25 1s 2s 2p 3s 3p 3d 4s 2 2.8.13.2
25
2
2
6
6
6
2
Fe* 26 1s 2s 2p 3s 3p 3d 4s 2 2.8.14.2
26
2
6
7
6
2
2
Co* 27 1s 2s 2p 3s 3p 3d 4s 2 2.8.15.2
27
6
2
6
2
8
2
Ni* 28 1s 2s 2p 3s 3p 3d 4s 2 2.8.16.2
28
6
6
2
2
2
Cu* 29 1s 2s 2p 3s 3p 3d 10 4s 1 2.8.18.1
29
2
2
6
6
2
Zn* 30 1s 2s 2p 3s 3p 3d 10 4s 2 2.8.18.2
30
[* denotes d-block elements]
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02 Chapter 2.indd 53 3/26/18 3:14 PM
Chemistry Term 1 STPM
9 Note that for potassium, the 19th electron, occupies the 4s orbital
and not the 3d because in an empty atom, the 4s is of lower energy
than the 3d.
10 For elements with proton number 21 to 30 (known as the d-block
CHAPTER elements), the energy level of the 4s and 3d are reversed. This is
because once the 3d orbital is/are filled, the 3d electrons repel the 4s
2 electrons to a higher energy level.
Reversal of the 4s and 3d
Energy
orbitals
3d 4s
4s 3d
Empty 3d orbitals Filled 3d orbitals
6
2
10
6
10
5
3
2
s , p , d, p and d are more 11 Orbitals that are fully filled (e.g. s , p or d ) or are half-filled
3
5
stable because of the even (e.g. p or d ) have extra stability due to their symmetrical charge
distribution of charge
distribution. This is evident in the electronic configuration of Cr
24
and Cu.
29
12 The configuration of chromium is [Ar] 3d 4s and not
5
1
4
2
[Ar] 3d 4s .
1
Similarly, the configuration of copper is [Ar] 3d 4s and not
10
9
2
[Ar] 3d 4s .
3d 4s
Cr:
3d 4s
Cu:
Electronic Configuration of Ions
1 In the formation of cations, electrons are lost in the reverse order
from electron filling. That is, last to be in, first to be out.
2 The electronic configuration of aluminium is 2.8.3 or
2
2
6
1
2
1s 2s 2p 3s 3p .
1s 2s 2p 3s 3p
54
02 Chapter 2.indd 54 3/26/18 3:14 PM
Chemistry Term 1 STPM
+
In the formation of Al , the electron removed is from the 3p orbital:
2
2
2
6
The configuration of Al is 2.8.2 or 1s 2s 2p 3s .
+
3 In the formation of anions, electrons are added in the same CHAPTER
manner as filling of electrons. 2
5
4 The electronic configuration of fluorine is 2.7 or 1s 2s 2p .
2
2
1s 2s 2p
The additional electron is added to one of the 2p orbitals which
have only one electron.
1s 2s 2p
2
2
–
6
The electronic configuration of F is 2.8 or 1s 2s 2p .
Quick Check 2.4
1 Write the electronic configurations for the following atoms:
(a) Ge (c) Pb
32 82
(b) Br
35
2 Iron (proton number 26) is a d block element. It can form two stable ions, Fe and Fe .
3+
2+
Write the electronic configuration for:
(a) the iron atom, and
(b) the Fe and Fe ions.
2+
3+
(c) Based on the electronic configuration in (b), state and explain which oxidation state of iron is
more stable?
2.4 Classification of Elements in 2017/P1/Q3
the Periodic Table 2018/P1/Q3
1 In the modern Periodic Table, the elements are arranged in order of
increasing proton numbers.
2 In 1984, the International Union of Pure and Applied Chemistry
(IUPAC) formally recommended the format as shown:
55
02 Chapter 2.indd 55 3/26/18 3:14 PM
Chemistry Term 1 STPM
p-block
1 18
1 2
CHAPTER H 2 4 1 Proton number 13 14 15 16 17 He
8
7
3
5
6
10
9
2 Li Be H Symbol B C N O F Ne
11 12 d-block 13 14 15 16 17 18
Na Mg 3 4 5 6 7 8 9 10 11 12 Al Si P S Cl Ar
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
Cs Ba La Hf Ta W Re Os Ir Pl Au Hg T1 Pb Bi Po At Rn
87 88 89 104 105 [106] [107] [108] [109]
Fr Ra Ac Rf Ha
s-block 58 59 60 61 62 63 64 65 66 67 68 69 70 71
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
f-block
90 91 92 93 94 95 96 97 98 99 100 101 102 103
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lw
Info Chem 3 The Table consists of 7 horizontal rows called Periods and 18
vertical columns called Groups.
7 Periods
18 Groups 4 It is interesting to note that only 11 of all the known elements are
4 Blocks gases under room conditions. They are H , He, Ne, Ar, Kr, Xe, Rn,
2
F , Cl , N and O .
2 2 2 2
5 Only 2 elements are liquids at room conditions: Br and Hg
2
(mercury).
Electronic Configuration and the Periodic Table
1 The elements in the Periodic Table can be divided into four blocks
depending on their valence shell electronic configurations.
2 They are the s block, p-block, d-block and f-block elements.
3 The s and p-blocks are called the representative elements, the
d-blocks are the transition elements, and the f-block is called the
inner transition elements.
4 All elements in the same group have the same number of valence
electrons with similar electronic configurations.
The s-block Elements
1 The s-block elements consist of elements from Group 1 and Group 2.
2 Group 1 elements (also called the alkaline metals) have one valence
1
electron in the outermost orbital with the configuration of s .
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02 Chapter 2.indd 56 3/26/18 3:14 PM
Chemistry Term 1 STPM
Element Valence shell configuration
H 1
1 1s
Li 2s 1
3
Na 3s 1
11
K 4s 1
19 CHAPTER
Rb 1
37 5s
Cs 1 2
55 6s
87 Fr 7s 1
3 Group 2 elements (also called the alkaline earth metals) have two
electrons in the outermost orbital with the configuration of s .
2
Element Valence shell configuration
Be 2s 2
4
Mg 3s 2
12
Ca 2
20 4s
Sr 2
38 5s
Ba 6s 2
56
88 Ra 7s 2
The p-block Elements
1 Elements of Group 13 to Group 18 are known as the p-block elements
because the outermost orbitals that are filled with electrons are the
p orbitals.
[The Group 17 elements are also called the halogens, and the Group
18 elements are called the noble gases.]
2 They have valence shell electronic configurations of s p to s p .
2
1
6
2
Group Representative element Valence shell configuration
2
13 Al 3s 3p 1
13
2
14 Si 3s 3p 2
14
2
15 P 3s 3p 3
15
2
16 S 3s 3p 4
16
2
17 Cl 3s 3p 5
17
2
18 18 Ar 3s 3p 6
The d-block Elements
1 Elements of Group 3 through 12 are known as d-block elements.
2 They have valence shell electronic configurations of d s to d s .
10 2
1 2
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02 Chapter 2.indd 57 3/26/18 3:14 PM
Chemistry Term 1 STPM
Group Element Valence shell configuration
1
3 Sc 3d 4s 2
21
2
4 Ti 3d 4s 2
22
CHAPTER 5 6 23 Cr 3d 4s 2 1
3
V
5
3d 4s
2 7 24 Mn 3d 4s 2
5
25
6
8 Fe 3d 4s 2
26
7
9 Co 3d 4s 2
27
10 Ni 3d 4s 2
8
28
10
11 Cu 3d 4s 1
29
10
12 30 Zn 3d 4s 2
3 They are typical metals with very high melting points and boiling
points.
The f-block Elements
1 Elements with proton numbers 58 – 71 (known as the
lanthanides) and from 90 – 103 (also known as the actinides) are
f-block elements.
1
10 2
2 They have valence shell electronic configuration of f d s to
f d s .
10 2
14
Valence Shell Configuration from the Periodic Table
1 Most chemical recations involve only the valence shell electrons.
Hence, the valence shell electronic configuration of an element is
T
ips
Exam
Exam Tips of utmost importance in predicting the chemical properties of an
element.
2 The valence shell configuration of the representative elements
(Group 1, 2, 13 to 18) can be deduced from its position in the
Periodic Table.
3 The period in which the atom is found corresponds to the Principle
Quantum Number of the valence shell.
4 The group in which the atom is in corresponds to the number of
valence electrons.
5 For example, aluminium (proton number 13) is in Period 3 and
Tips
Exam
Exam T ips Group 13 of the Periodic Table. This means that the outermost
energy level that is filled with electrons is the 3rd shell, and it has
3 (not 13) valence electrons. Hence, its valence shell configuration
The number of valence shell
1
2
electrons of Group 13 to 18 is 3s 3p .
elements is given by: 6 Lead (proton number 82) is in the 6th Period and Group 14. Its
(Group number – 10)
valence shell configuration is 6s 6p .
2
2
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02 Chapter 2.indd 58 3/26/18 3:14 PM
Chemistry Term 1 STPM
Example 2.4
Write the valence shell electronic configuration of the following
elements.
Element Period Group CHAPTER
A 4 17
2
B 6 1
C 2 18
Solution
2
A . . . . . 4s 4p 5
B . . . . . 6s 1
2
C . . . . . 2s 2p 6
SUMMARY
SUMMARY
1 Electromagnetic radiations are characterised 7 The wavelengths of the lines in the
by their wavelengths (λ) and frequencies (f ). hydrogen spectrum can be calculated using
2 The relationship between wavelength and the equation:
frequency is:
c 1 = R 1 – 1
λ = λ H n 1 2 n 2 2
f
3 The energy associated with a electromagnetic 8 Evidence of energy levels or shells is
radiation is given by: supported by the successive ionisation
energy graphs of elements.
E = hf
9 Aufbau’s Principle states that electrons will
4 The line spectrum of hydrogen: fill orbitals of lower energy before those of
(a) shows that the energy levels in the higher energies are filled.
hydrogen atom are quantised, 10 Pauli’s exclusion Principle states that an
(b) shows the electron in the hydrogen atom orbital can accommodate a maximum of two
can have certain fixed values only and electrons with opposite spins only.
not any values, 11 Hund’s Rule states that electrons must first
(c) shows that electronic transitions fill a set of degenerate orbitals singly before
between energy levels are possible, pairing occurs.
(d) can be used to calculate the ionisation
energy of the hydrogen atom. 12 An orbital is the region/volume in space
5 The lines in the Lyman series (ultra-violet around the nucleus where the probability
region) is a result of transitions between of finding an electron is maximum
higher energy levels and level n = 1. ( 95%).
6 The lines in the Balmer series (visible region)
is a result of transitions between higher
energy levels and level n = 2.
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02 Chapter 2.indd 59 3/26/18 3:14 PM
Chemistry Term 1 STPM
STPM PRACTICE 2
Objective Questions
CHAPTER 1 Which of the following statements is true 6 An orbital always contains
2 regarding the line spectrum of hydrogen? A 0, 1 or 2 electrons C 1 or 2 electrons
A The first line in each series has the longest
D 8 electrons
B 2 electrons
wavelength.
B The ionisation energy of hydrogen can be 7 A d subshell contains how many degenerate
calculated using the line with the lowest orbitals?
frequency in the ultraviolet region. A 1 C 5
C The lines are all equally spaced. B 3 D 7
D The lines converge when the wavelength
increases. 8 In the building up of the electronic configuration
of atoms, electrons are added to the 4s subshell
2 What is the maximum number of atomic orbitals before the 3d subshell because
with the principle quantum number 3? A the 3d subshell can hold more electrons than
A 3 C 9 the 4s subshell.
B 6 D 12 B the 3d subshell is closer to the nucleus.
C the 4s subshell is spherical.
3 Which of the following statements is not true of D the 3d subshell is of higher energy.
the line spectrum of hydrogen atom?
A The lines become closer as frequency 9 Which of the following correctly represents the
increases Balmer series in the line spectrum of hydrogen?
B There are no more lines after the line with A
maximum frequency in the Lyman series
C The spectrum consists of only two series,
Lyman series and Balmer series Wave number
D The lines are produced when the electron B
absorbed energy and gets ‘promoted’ to
higher energy levels
Wave number
4 What is the maximum number of electronic C
transitions possible between the first 7 energy
levels in the hydrogen atom?
A 7 Wave number
B 14 D
C 21
D 5040
Wave number
5 Which of the following statements explains why 10 Tungsten ( W) is used as the element in
the emission spectrum of hydrogen is a line incandescent bulbs. Which of the following is not
74
spectrum? true about the element?
A The hydrogen has only one electron. A Tungsten oxide is amphoteric
B The lone electron in the hydrogen atom can B Tungsten exhibits variable oxidation states in
revolve round the nucleus in fixed orbits only.
its compound
C The energy levels in the hydrogen atom are C The maximum oxidation state of tungsten is
quantised.
D Electrons have dual nature. +6
D Tungsten compounds are used as catalyst
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02 Chapter 2.indd 60 3/26/18 3:14 PM
Chemistry Term 1 STPM
11 What is the proton number of an element that has 17 Which of the following elements has 5 unpaired
6 unpaired electrons in its ground state? electrons in its ground state?
A 14 C 24 A Nitrogen
B 22 D 30 B Phosphorous
C Manganese
12 The lines in the emission spectrum of hydrogen D Iron
are the results when CHAPTER
A electrons are removed from the atom. 18 Which of the following valence configuration is 2
B electrons absorb energy. for an atom that can form positive ions of different
oxidation state?
C electrons lose energy. A 3d 4s C 3d 4s 2
10
10
1
D the hydrogen molecules split to form atoms. B 3s 3p D 3d 4s 2
1
1
2
13 Part of the lines in the Lyman series of an emission 19 Which of the following species would have a line
spectrum of atomic hydrogen is shown below. spectrum similar to that of the hydrogen atom?
A He 2+ C Na +
B Li D K +
2+
X 20 Which of the following statements is not true?
Which of the following statements is not true? A 3d orbitals have the same energy as the 4s
orbital.
A The lines have frequencies corresponding to B p orbitals are filled with electrons according
ultra-violet radiation to Hund’s rule.
B The wavelength of line X can be used to C The principle quantum number of p orbitals
calculate the ionisation energy of hydrogen start with n = 2.
C Each line corresponds to electronic transition D An s orbital is spherical.
between higher energy levels and level n = 1
D The hydrogen atom has 6 electrons 21 An atom X has six valence electrons and forms a
3+
stable X ion. In which group of the Periodic
14 In a hydrogen atom, electron in which orbital can Table does X belong to?
absorb energy but cannot emit energy? A 3 C 13
A 1s C 3s B 6 D 16
B 2p D 3d
22 The proton number for species X is 8 and its
15 Which of the following electronic transitions nucleon number is 18. Species X has two electrons
would emit visible light of highest frequency? less than the number of neutron. What is the
A n → n
4 2 charge on X?
B n → n
2 6 A 0 C +1
C n → n B –1 D +2
5 1
D n → n 2
1
23 Which of the following electronic arrangements
16 Which statement is not true about the s, p and d violates Hund’s rule?
orbitals? A
A The five d orbitals have the same shape and
same energy B
B The three p orbitals have the same shape and
same energy C
C In any energy shell, the energy of the orbitals
increases in the order s < p < d D
D d orbitals are directional whereas s orbitals
are non-directional
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02 Chapter 2.indd 61 3/26/18 3:14 PM
Chemistry Term 1 STPM
Structured and Essay Questions
1 In the stratosphere, chlorine molecules absorb electromagnetic radiation and dissociate to produce
chlorine atoms which are responsible for the destruction of ozone.
CHAPTER Cl—Cl(g) → 2Cl(g)
–1
2 Given that the bond energy of chlorine is 242 kJ mol , calculate the wavelength of the light absorbed.
2 Part of the Lyman series in the hydrogen emission spectrum is as shown below:
8.22 10.27 10.62 10.97
6
Wave number ( 10 ) m –1
(a) Draw an energy level diagram to show how the lines are produced.
(b) (i) Define ionisation energy of hydrogen. Illustrate your answer with an appropriate equation.
(ii) Calculate the ionisation energy (in kJ mol ) of hydrogen from the above spectrum.
–1
3 The diagram below shows the lines in the Balmer series in the emission spectrum of hydrogen atom.
(a) Draw a labelled energy diagram to show how the lines marked X and Y are formed.
(b) Name two species that would produce similar emission spectrum as that of the hydrogen atom. Explain
your reasoning.
X Y
Frequency
4 The frequencies of the first five lines in the Lyman series of the hydrogen atom are given below:
2.47, 2.92, 3.08, 3.16, 3.20 ( 10 s )
5 –1
Plot a suitable graph to determine the ionisation energy of the hydrogen atom
–1
(a) in J, and (b) kJ mol .
–1
5 The first seven ionisation energies (kJ mol ) of an element D are as follows:
496, 4563, 6913, 9544, 13 352, 16 611, 20 115
(a) What do you understand by the term ionisation energy?
(b) Explain why the ionisation energy increases with the number of electrons removed.
(c) Plot a suitable graph or otherwise, determine the Group number of D in the Periodic Table. Explain how
you arrived at the answer.
6 (a) Sketch the energy level diagram for the orbitals of an atom with the principle quantum numbers of
n = 1 to n = 3 inclusive of the 4s orbital.
(b) Using arrows to represent electrons, show the electronic configuration of
(i) a carbon atom in its ground state.
(ii) a carbon atom in its excited state of lowest energy.
(iii) a carbon atom in its excited state with the highest energy.
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02 Chapter 2.indd 62 3/26/18 3:14 PM
Chemistry Term 1 STPM
7 Explain the following observations:
(a) An orbital can accommodate a maximum of only two electrons with opposite spins.
5
1
4
2
(b) The electronic configuration of chromium is [Ar]3d 4s instead of [Ar]3d 4s .
8 (a) State Hund’s rule and Pauli’s exclusion principle.
CHAPTER
(b) Draw an energy level diagram showing the electron distribution in an N ion.
2–
9 The number of electrons occupying the different orbitals of atom X is shown in the following table. 2
Orbital s p d
Number of electrons 7 12 10
Write the electronic configuration of X, and explain how each of these orbitals is filled with electrons.
10 (a) The lines in the emission line spectrum of hydrogen are the result of electronic transitions.
(i) Draw an energy level diagram to show the transitions that give rise to the first and last lines in both
the Lyman series and the Balmer series and state in which part of the electromagnetic spectrum that
each of the series belongs to.
(ii) The energy difference between two of the levels in the hydrogen atom is 987.3 kJ mol . Calculate
–1
the frequency and the wavelength of the line produced when electronic transition occurs between
these two levels.
[h = 6.63 × 10 Js; C = 3.00 × 10 ms ; L = 6.02 × 10 mol ]
8
–1
–1
–34
23
(b) An atom X has 24 electrons in its nucleus.
(i) Draw an energy diagram to show how the electrons are arranged in its atom.
(ii) Explain any abnormality in the electronic arrangement.
63
02 Chapter 2.indd 63 3/26/18 3:14 PM
Chemistry Term 1 STPM
ANSWERS
Chapter 1 Atoms, Molecules and Quick Check 1.5
Stoichiometry 1 (a) 714.15 g (b) 28.88 g (c) 3369.8 g
2 (a) 23.46 g (b) 8.56 g (c) 1.15 g
Quick Check 1.1 4 25
222 3 (a) 5.56 10 (b) 3.34 10
1 (a) Rn
86 Quick Check 1.6
(b) 234 Pa
91 1 (a) CaCO + 2HCl → CaCl + H O + CO 2
2
2
3
(c) 38 Ca (b) 2.00 g
20 (c) 80.0%
(d) 212 Pb
82 2 2
(e) 208 Tl; 208 Pb 3 1.95 mol dm –3
81 82
2 228 Y STPM Practice 1
90 Objective Questions
Quick Check 1.2 1 D 2 C 3 C 4 B 5 B
1 39.985 6 A 7 C 8 B 9 C 10 C
2 (a) 1.083 times 11 D 12 B 13 D 14 D 15 C
13
(b) 12 C : C = 9.9 : 0.1 16 A 17 B 18 D 19 D 20 B
3 12.0038 21 B 22 B 23 C 24 B 25 D
26 B 27 D 28 D 29 C 30 B
Quick Check 1.3 31 D 32 A
1 110.32 Structured and Essay Questions
2 (a) 3
(b) 206 M; 207 M; 208 M 1 (a) Number of protons in the nucleus of an atom.
(b) Total number of protons and neutrons in the
82
82
82
2+
(c) It is caused by M ion nucleus of an atom.
206
3 (a) (c) Atoms of the same element but with different
nucleon number.
2 Refer to section isotopes.
30.4 69.9
20 21 22
3 (a) Mole ratio N : O = 14 : 16
(b) 20.18
(c) 1.68 times = 2.1 : 4.4
= 1 : 2
Quick Check 1.4 Empirical formula is NO
1 (b) The molecular peak = 92 2
(NO )n = 92
2
46n = 92
n = 2
Molecular formula is N O
158 160 162 2 4
+
(c) m/e (46) = [NO ] or [N O ] 2+
Ratio: 158 : 160 : 162 = 1 : 2 : 1 m/e (94) = [N O O] + 2 4
2
16
18
2 (a) 3 : 2 (b) 12
4 (a) Refer to text
3 [Not to scale]
(b) Refer to text
(c) (i)
29.8
14 18 28 32
1.5
4 1.0
28 29 30 m/e
14 16 18 30 32 46 48 50 (28 29.8) + (29 1.5) + (30 1.0)
(ii) A =
r (29.8 + 1.5 + 1.0)
14 16 18 30 32 46 48 50
= 28.11
16
16
16
18
16
18
18
N + 16 O + 16 O + N O + N O + N O O + N O O + N O O + 28.11
2 (iii) = 2.34 times
12
357
12 Answers.indd 357 3/26/18 4:06 PM
Chemistry Term 1 STPM
5 (a) Negative. [Positive particles are attracted towards the P Cl Cl + 103
35
37
negative plate]
37
37
+
(b) H . It is lighter and hence deflected more. P Cl Cl + 105
m P Cl Cl Cl + 136
35
35
35
+
2
(c) (i) of H = 2
e +
35
35
37
1 P Cl Cl Cl 138
∴ Angle = 4° = 2°
2 P Cl Cl Cl + 140
37
37
35
(ii) Zero. Neutron is neutral and is not deflected. P Cl Cl Cl + 142
37
37
37
79
79
35
6 C H Br Br Cl : 220 ≡ (1 1 3) = 3 (c) 66 : 68 = 3 : 1
2 3
35
81
79
C H Br Br Cl : 222 ≡ 2(1 1 3) = 6 103 : 105 : 107 = (3 3) : 2(3 1) : (1 1)
2 3
81
35
81
C H Br Br Cl : 224 ≡ (1 1 3) = 3 = 9 : 6 : 1
2 3
79
79
37
C H Br Br Cl : 222 ≡ (1 1 1) = 1 138 : 140 : 142 : 144
2 3 = (3 3 3) : 3(3 3 1) : 3(3 1 1) : (1 1 1)
79
37
81
C H Br Br Cl : 224 ≡ 2(1 1 1) = 2 = 27 : 27 : 9 : 1
2
3
81
81
37
C H Br Br Cl : 226 ≡ (1 1 1) = 1 12 (a) Due to the presence of isotopes.
2
3
Relative abundance: 220 : 222 : 224 : 226 (b) Let the % abundance of X = a%
35
= (3 + 1) : (6 + 1) : (3 + 2) : 1 % abundance of X = 100 – a%
37
= 4 : 7 : 5 : 1 35.5 = 35a + 37(100 – a)
7 (a) Isotope with unstable nucleus and undergoes a = 75% 100
spontaneous disintegration to form nucleus of (c)
smaller isotopes. 75%
(b) (i) 227 W and 227 X
88 89 25%
4 0
(ii) A ≡ 4 He + e (4 α-particles and 1 β-particle)
2 –1
4 0 35 37
B ≡ He + 2 e
–1
2
(iii) They are isotopes.
13 (a) Relative molecular mass of Y = 12 × 9.5
8 (a) The amount of substance that contains the same = 114
number of particles as the number of atoms in 12 g Let the molecular formula of Y be C H 2n+2 .
n
of C. 12n + 2n + 2 = 114
12
(b) (i) 180 g n = 8
4.8 × 10 –3 Alkane Y is C H .
23
(ii) 180 (6.02 10 ) 25 8 18
19
= 1.61 10 molecules (b) C H + 2 O → 8CO + 9H O
2
2
18
2
8
89.5 25
(iii) Mass = 2.5 g = 2.24 g Volume of O required = × 0.15 dm 3
100 2 2
9 (a) Atoms having the same number of protons and the = 1.88 dm 3
same number of electrons, but different number of 100
neutrons. ∴ Volume of air needed = 20 × 1.88
(b) (i) 35 36 37 38 = 9.40 dm 3
1
35 CI + 35 CI H + 37 CI + 37 CI H +
1
Chapter 2 Electronic Structure of
(ii) 35 Cl H 2+ Atom
1
The energy required to form ion of +2 charge is
much higher. Quick Check 2.1
–1
10 (a) 18 34 60 1 (a) 602 kJ mol
15
(b) 1.51 10 Hz
16
H O CH OH CH COOH
18
2 3 3 –7
(c) 1.99 10 m
(b) CH —C— O— CH 2 (a) 3.0 10 m
18
–16
3 3
|| (b) 6.63 10 J
–10
O (c) 3.99 10 kJ mol –1
11
11 (a), (b) Species m/e value Quick Check 2.2
14
35
P Cl + 66 1 (a) 4.57 10 Hz (b) Visible
–1
14
2 (a) 5.08 10 Hz (b) 202.9 kJ mol
37
P Cl + 68
14
15
3 x = 3.20 10 Hz. y = 2.33 10 Hz
P Cl Cl + 101
35
35
358
12 Answers.indd 358 3/26/18 4:06 PM
Chemistry Term 1 STPM
Quick Check 2.3 (ii) E = hf
4+
5+
1 (a) Na (g) → Na (g) + e = h λ c
(b) Al 11+ (g) → Al 12+ (g) + e = (3.99 10 )(3.0 10 )(10.97 10 )
–13
8
6
2 (a) 13 (b) 3 (c) 2.8.3 = 1313.1 kJ mol
–1
3
Ionisation 3 (a) n = ∞
energy
n = 4
n = 3
n = 2
No. of electrons removed
n = 1
Quick Check 2.4
1 (a) 2.8.18.4
(b) 2.8.18.7
(c) 2.8.18.32.18.4 X Y
6
2 (a) [Ar] 3d 4s 2 Frequency
+
2+
2+
(b) Fe : [Ar] 3d 6 (b) Li and He . Both the ions are one electron species.
Fe : [Ar] 3d 5 4 f (× 10 )/s –1
3+
5
3+
(c) Fe . All the five 3d orbitals are singly occupied. 3.24
3.2
STPM Practice 2
Objective Questions 3.0
1 A 2 C 3 D 4 C 5 C 2.8
6 A 7 C 8 D 9 B 10 A 2.6
11 C 12 C 13 B 14 A 15 A
16 A 17 C 18 A 19 B 20 A 2.4 0 10
21 B 22 A 23 C 5 1 –2 15 20 25
n 2 (× 10 )
Structured and Essay Questions (a) Using E = hf
–34
1 E = hf = (6.63 10 )(3.24 10 )
15
–18
242 = (3.99 10 ) f = 2.15 10 J
–13
23
–18
f = 6.07 10 Hz (b) E = (2.15 10 )(6.02 10 ) 10 –3
14
–1
c 3.0 × 10 8 = 1294.3 kJ mol
–7
Wavelength = = 6.07 × 10 14 = 4.94 10 m or 5 (a) The minimum energy required to remove one electron
f
from every atom in one mole of gaseous atoms.
= 494 nm (b) This is because successive species have ever increasing
2 (a) n = ∞
proton ratio. This causes the remaining electrons
electron
n = 4 to be more tightly held by the nucleus.
(c)
n = 3
I.E. 496 4563 6913 9544 13 352 16 611 20 115
n = 2
log I.E. 2.7 3.7 3.8 4.0 4.1 4.2 4.3
n = 1 Log(ionisation energy)
4
8.22 10.27 10.62 10.97
Wave number ( 10 6 ) m –1
(b) (i) Ionisation energy of hydrogen is the minimum 3
energy required to remove the lone electron in
the hydrogen atom per mole of gaseous
hydrogen atom under standard conditions
(298 K and 101 kPa).
2
H(g) → H (g) + e 1 2 3 4 5 6 7
+
Order of electron removed
359
12 Answers.indd 359 3/26/18 4:06 PM
Chemistry Term 1 STPM
D is in Group 1 of the Periodic Table. There is a very The lines in the Lyman series are in the ultra-
nd
st
large increase between the 1 and 2 I.E. violet region, while the lines in the Balmer series
are in the visible region.
6 (a) 3d (ii) Using the equation:
Energy 4s ∆E = hf
23
–34
3
3p 987.3 × 10 = (6.63 × 10 )(6.02 × 10 )(f)
15 –1
3s f = 2.47 × 10 s
Or,
Using the relationship:
2p
c
λ = —
2s
f
3.00 × 10
= —–———— 8
1s 2.47 × 10 15
–7
(b) (i) 1s 2s 2p = 1.21 × 10 m
= 121 nm
(b) (i) The electronic configuration is:
1s 2s 2p 3s 3p 3d 4s 1
5
6
6
2
2
2
(ii)
4s
3d
(iii)
3p Energy
7 (a) The third electron will have the same spin as one
of the other two electrons and will experience 3s
repulsion. 2p
(b) A set of 3d orbitals that are fully half-filled has extra
stability compared to a partially filled 3d orbitals. 2s
8 (a) Hund’s rule states that in a set of degenerate orbitals 1s
electrons will occupy the orbitals singly first before
pairing occurs. (ii) The electronic arrangement violates the
Pauli’s exclusion principle states that an orbital can Aufbau’s principle which states that electrons
accommodate a maximum of two electrons only with will fill available orbitals with lower energy first
opposite spins. before orbitals with higher energies are filled. In
(b) The electronic configuration of N is 1s 2s 3p . an empty atom, the 4s orbital is of lower energy
2
5
2
2–
than the 3d. Thus, by right, two electrons must
2p be filled in 4s before the 3d orbitals are filled to
2
2
2
4
6
6
2s give it a configuration of 1s 2s 2p 3s 3p 3d
6
4s instead of 1s 2s 2p 3s 3p 3d 4s . This
2
2
2
1
6
5
2
1s abnormality arises because a half-filled 3d sub-
5
9 Each orbital can accept two electrons only. shell (3d ) is energetically more stable than a
4
2
The seven s electrons will have configuration of 1s , 2s , 3s 2 partially filled 3d sub-shell (3d ).
2
and 4s . 1
6
The twelve p electrons will have configuration of 2p and Chapter 3 Chemical Bonding
3p .
6
The ten d electrons will have configuration of 3d . Quick Check 3.1
10
2
2
2
6
10
Electronic configuration of X is 1s 2s 2p 3s 3p 3d 4s . 1
6
1 (a) (f)
10 (a) (i) n = ∞
O Cl –
x x • x x
• • •
– x x – x x
O • S • O Cl • I • Cl
1st Last • • •
x x x • •
n = 2 O Cl
Balmer series
1st Last
n = 1
Lyman series
360
12 Answers.indd 360 3/26/18 4:06 PM
Chemistry Term 1 STPM
(b) (g) 4 (a) (b)
O O
x x x •• xx x x • x x O • x •
• • x • x •
O • N • x O •
x • • x • N N
– x x – • x
O • S • O O x • O • O x •
xx x x x • x • • •
O O
•• x x xx
Quick Check 3.2
(c) (h) 1 Pyramidal 8 Linear
••
xx xx 2 Linear 9 Square planar
3 Tetrahedral 10 Bent
x x x • x
x I I I – x • C x • O x
x • x x • xx 4 Trigon bipyramidal 11 Square-based pyramidal
5 Octahedral 12 Bent
xx •• •• xx 6 Tetrahedral 13 Trigon planar
7 Pyramidal 14 Bent
(d) (i)
–
O Quick Check 3.3
x • xx xx
1 Nitrogen is more electronegative than phosphorous. The
x x x bonding electrons in NH are closer to one another and
– x x – Cl S S Cl 3
O • P • O • • • experience greater repulsion.
x • • x xx xx 2 H S. Sulphur is more electronegative than Se.
2
O
Quick Check 3.4
(e) (j) 1 Polar 4 Non-polar
2 Polar 5 Non-polar
Cl
3 Non-polar 6 Non-polar
x •
Quick Check 3.5
x + x
Cl • P • Cl
x x • Chromium can make use of the electrons from the 3d as well
H • C x • N • as 4s sub-shell to form metallic bonds. Calcium can only make
x • x • •
use of electrons from the 4s sub-shell to form metallic bonds.
Cl
Quick Check 3.6
2 (a) (b) 1 H O CH Cl 2
2
2
4
Number of electrons in the molecules increases.
Cl
Cl Cl Cl 2 Size of the molecules as well as the total number of
x • • x • x
x • • • electrons in the molecules increases from CH to SnH . As
4
4
a result, the van der Waals forces get stronger.
Cl x • Al x • Cl Al Al
x • x •
• • • x 3 (a) Relative Total Boiling
Cl Cl Cl Compound molecular number of point/°C
mass electrons
Silane 32.1 18 –112
3 (a)
Hydrogen 33.1 18 –61
Cl x Be x Cl sulphide
• •
(b) H S is polar but SiH is non-polar.
2 4
(b) Presence of empty orbitals in beryllium and lone-pair 4 CH CN. CH CN is polar but C H is non-polar.
3
3
3
8
electrons in the ammonia molecule. Quick Check 3.7
(c) (a) and (d)
H Cl H
x • • x • Quick Check 3.8
x x x x 1 (a), (b), (d), (e) and (f)
H • N x Be x N • H
2 Ethanol: Strong hydrogen bonding
x • • x • Dimethyl ether: Weak van der Waals forces
H Cl H
3 HCOOH can form more intermolecular hydrogen
bonding compared to ethanol.
361
12 Answers.indd 361 3/26/18 4:06 PM
Chemistry Term 1 STPM
4 1-propanol. It is bigger and has more electrons than (b) NH and H O have intermolecular hydrogen
3
2
ethanol. bonding while weak van der Waals forces exist
Quick Check 3.9 between CH molecules. This accounts for the
4
higher boiling point of NH and H O.
2
3
1 CH COOH. The large non-polar C H group in H O has two lone pair of electrons compared to
5
3
6
C H COOH decreases the solubility of benzoic acid. 2
6 5 only one for NH . Thus, H O can form more
2
3
2 1,2,3-propantriol has three —OH group. As a result, it intermolecular hydrogen bonding with one another.
can form more hydrogen bonds with water molecule On top of that, oxygen being more electronegative
compared to ethanol. than nitrogen forms stronger hydrogen bonds.
3 C H OH. It can form hydrogen bonds with water (c)
2
5
molecules, but C H SH cannot. HC—(CHOH) —CH —O—H
4
2
2 5 ||
4 (C H ) N cannot form hydrogen bonds with water O
5 3
2
because there are no hydrogen atoms attached directly to H H
the nitrogen atom.
5 NH can form hydrogen bonds with water molecules, but O
3
PH cannot.
3
3 (a) (i) HCl is polar but CO is non-polar.
2
STPM Practice 3 (ii) Intermolecular forces between NH molecules
3
Objective Questions are the hydrogen bonds.
1 C 2 C 3 B 4 B 5 C Weak intermolecular van der Waals forces exist
6 A 7 D 8 D 9 B 10 D between HCl molecules
11 B 12 B 13 D 14 A 15 C (b) NH can form hydrogen bonds with water
3
16 B 17 B 18 C 19 D 20 D molecules.
21 C 22 A 23 B 24 C 25 D HCl reacts with water to form water-soluble ions:
+
26 C 27 D 28 B 29 C 30 C HCl(g) + H O(l) → H O (aq) + Cl (aq)
–
3
2
31 B 32 C 33 D 34 D 35 B CO is non-polar.
2
36 B 37 C 38 D 39 A 40 C 4 (a)
41 C 42 C
Structured and Essay Questions Cl Cl x x
x • x •
1 Ionic bond is the electrostatic attraction between a pair
of opposite charged ions formed by the complete transfer Al Be H • x N • x H
of electrons from one atom to the other. An example is Cl • x • x Cl
sodium chloride. Sodium transfers its lone valence x • x •
electron to the chlorine atom. This results in the Cl H
–
+
formation of Na and Cl ions with octet configurations Trigonal planar Linear Pyramidal
Na + Cl → Na + Cl –
+
2.8.1 2.8.7 2.8 2.8.8 (b) The aluminium atom in AlCl and the beryllium
3
atom in BeCl do not have eight electrons in their
–
+
Attraction between Na and Cl results in the formation valence shells. 2
of the ionic bond.
Covalent bond is the electrostatic attraction between (c) (i) AlCl •NH 3
3
adjacent nuclei and the electrons that are shared between
(ii)
them. An example is Cl . Each chlorine atom shares one Cl H
2
of its unpaired valence electron with another chlorine x • •
atom so that both achieve octet configurations. x
Cl + Cl → Cl — Cl Cl • Al N • H
2.8.7 2.8.7 2.8.8 2.8.8 x • •
The attraction between the two nuclei of chlorine atom Cl H
and the shared pair of electrons hold the chlorine atoms
together in Cl molecule. (d) (i) BeCl •2NH
2 2 3
2 (a) In the HF molecule, the hydrogen atom is bonded
to a very small and highly electronegative fluorine (ii) H Cl H
atom. The covalent bond in H—F is greatly x • • x •
polarised. x
H • N Be N H
δ ++ H—F δ–– x x •
The dipole-dipole attractions between HF molecules x • • x •
are stronger than the ‘ordinary’ van der Waals force. H Cl H
This dipole-dipole attraction is called hydrogen
bonding.
362
12 Answers.indd 362 3/26/18 4:06 PM
Chemistry Term 1 STPM
(iii) The beryllium atom in BeCl is 4 electrons short (b) H + H + H – + – –
2
of octet. Hence, it will combine with
H
N
H
N
N
two NH molecules to achieve octet. H N H H H H H H N H H H N N H H N H N
3
5 (a) (b) H H H H H H H H H
x x x x
NH : Trigonal pyramidal (3 bond-pairs + 1 lone-pair)
sp 3 sp 3 3
+
O N NH : Tetrahedral (4 bond-pairs)
4
NH : Bent (2 bond-pairs + 2 lone-pairs)
–
x x x x x 2
• • •
H H H (c)
x • x • N N P P
H H
CI CI CI CI CI CI CI CI
(c) x π π •
sp • x CI CI CI CI
x x NCl : sp hybridisation
3
3
H • C x • N x PCl : sp hybridisation
sp 3 3
NCl has a larger bond angle. N is more electronegative
3
than P, the bonding electrons will be closer to N in
6 (a) sp 2 sp 2 NCl compared to P in PCl . Thurs, the bond-pair
3
3
repulsion in NCl is stronger and pushed the bonds
3
O further apart.
+ 9 (a) (i) BeCl + 2NH → BeCl (NH )
CH N 2 3 2 3 2
3 (ii)
H H
sp 3 O CI N
–
sp 3 H
(b) Trigonal planar Be
(c) – H
C N
N CI
H H
(d) The carbon-nitrogen bond in CN is a triple bond
–
while that in the CH NO is a single bond.
3 2 (iii) BeCl : sp hybridisation
2
7 (a) Mixing of two or more non-equivalent atomic BeCl (NH ) : sp hybridisation
3
3 2
2
orbitals to form a set of equivalent hybrid orbitals.
3
(b) (i) Nitrogen undergoes sp hybridisation. (b) (i) P
x x
H
H
H
N
(ii) Ammonia. Nitrogen is more electronegative than
x x •
H • H phosphorus. As a result, the bonding electrons
x • in the N—H bonds are nearer to the nitrogen
H
atom than the phosphorus atom in PH . As a
3
3
(ii) Carbon undergoes sp hybridisation. result, the three N—H bonds experience greater
mutual repulsion which pushed them further
H
x • apart than the P—H bonds in PH .
3
10 (a) H (g) + Cl (g) → 2HCl(g)
C 2 2
x • • x (b) Boiling point
x •
Cl Cl
HF
Cl
HI
8 (a) In the reaction, NH accepts a proton from another HBr
3
+
NH to form NH . Thus, one of the NH molecule HCI
3
4
3
acts as a Bronsted base (proton acceptor) and the M of HX
other a Bronsted-Lowry acid (proton donor). r
363
12 Answers.indd 363 3/26/18 4:06 PM
Chemistry Term 1 STPM
(c) The intermolecular forces between the HCl (ii) Hydrogen chloride is a covalent compound that
molecules, the HBr molecules and the HI molecules exists as the simple HCl molecules with strong
are the weak van der Waals forces. The strength of covalent bond holding the atoms together in the
the van der Waals force increases with the increase in molecule. There are no ions present in liquid
the size of the hydrogen halide molecules as well as hydrogen chloride. As a result it does not
the number of electrons in the molecules from HCl conduct electricity.
to HI. This is reflected in the increase in the boiling However, when hydrogen chloride dissolves in
point from HCl to HI. HF, despite its smallest size has water it undergoes complete dissociation to
the highest boiling point. This is due to the presence produce free aqueous ions.
of hydrogen bonding between the HF molecules. HCl(g) + H O(l) → H O (aq) + Cl (aq)
–
+
2
3
11 (a) Lead(II) chloride is an ionic compound with ionic The production of the mobile aqueous ions is
2+
–
bonds binding the Pb and Cl ions in the giant responsible for its electrical conductivity.
2+
ionic lattice. Due to the large size of the Pb ion, the (b) (i) Ion ClO – ClO – ClO –
ionic bond is rather weak. On the other hand, silicon 2 3 4
dioxide exists in the form of giant covalent structure Oxidation state +3 +5 +7
with strong covalent bonds holding the silicon and (ii)
oxygen atoms together. A lot of energy is required
to break the covalent bonds. This accounts for the O CI O – (Bent)
higher melting point of silicon dioxide compared to
lead(II) chloride.
(b) (i)
O O
O
O - S O - O CI O – (Trigonal pyramidal)
S
O O
O
O
(ii) O
O –
O CI O (Tetrahedral)
S
S
O O O O - O - O
Square planar (120°) Tetrahedron (109.5°)
12 (a) (i) The melting point of ice decreases with Chapter 4 States of Matter
increasing pressure shows that ice is less dense
than water. The melting of ice is accompanied Quick Check 4.1
by a decrease in volume. 1 0.179 dm
3
H O(s) H O(l) Volume decreases
2 2 2 143.2 kPa
Increasing pressure would shift the equilibrium
to the right-hand side and more ice will melt. 3 27.8 kPa
This is because, in ice the water molecules are 4
held rigidly in place by hydrogen bonds. This Volume
creates a lot of empty spaces in the ice structure.
400 K
H O H
Hydrogen bond 300 K
H H H
O O O Pressure
H H H
H H H
O O
H Empty space The higher the temperature, the higher is the pressure.
H
H O H O H 5
H H H
O O O
H H H Pressure
O
H H Condensation
When ice melts, the rigid structure collapses
and the mobile water molecules flow to fill up 1
those empty spaces causing its volume to V
decrease.
364
12 Answers.indd 364 3/26/18 4:06 PM
Chemistry Term 1 STPM
At high enough pressure, carbon dioxide undergoes 2
condensation causing a very large decrease in the volume
1
(or a large increase in ). After that, pressure has little Pressure/atm
V
effect on the volume because liquid carbon dioxide is Solid Liquid
difficult to compress. Gas
Quick Check 4.2
3
1 3.15 dm 3 329 °C Temperature
2 323 °C 4 18.93 dm 3
Quick Check 4.3 3 (a)
1 (a) 8.75 dm (b) 50.0 dm 3 Pressure/atm
3
2 (a) 227 (b) 1481.8 dm 3
3 C H 4 C H
3 6 3 8
Quick Check 4.4
3
1 5.00 10 m 2 27.3
–3
3 0.0173 m –3 4 31.08 dm 3 Temperature
Quick Check 4.5 (b) Melting of iron is accompanied by a decrease in
1 (a) 9.01 kPa (b) 5.01 kPa volume. As a result, the melting point decreases with
increasing pressure.
2 8.61
4 (a) & (b)
3 (a)
Gas CO N 2 SO 2
Pressure/atm Critical point
Pressure/kPa 120 72 48
(b) (i) P (CO) = 120 kPa 1 X
P (N ) = 72 kPa Liquid
2
(ii) Total pressure = 192 kPa Solid
Gas
4 (a) 498.6 kPa
(b) 2493 kPa
(c) 997.2 kPa Temperature/°C
M.p. B.p 132
The water vapour pressure is negligible.
(c) Yes. The critical temperature is higher than room
Quick Check 4.6 temperature.
23
–1
1 6.0 10 mol (d) On the graph
2 8 atoms (e) (i) Nothing happens
3 MS (ii) Solid → Liquid → Gas
(f) Density of solid is higher than the liquid
Quick Check 4.7
5 CO , NH , C H OH, H O
1 (a) Conversion of a substance from solid to gas/vapour 2 3 2 5 2
without passing through the liquid phase Quick Check 4.9
(b) Reducing pressure to below 0.006 atm 1 77.44 g
2 (a) Vapour 2 0.838
(b) A Condensation 3 2.95 kPa 3x
Temperature Freezing 4 (a) 2 (c) 7.5x
x
2
(b) 2x
(d)
2
Quick Check 4.10
1 100.17 °C
B
Time 2 –0.084 °C
(c) Increasing the pressure. Vapour → Liquid 3 CaCl
2
Quick Check 4.8 4 (a) Ethanoic acid undergoes association in benzene.
2CH COOH → (CH COOH)
1 The molecules are more closely packed in solid CO 3 3 2
2 (b) Aluminium chloride undergoes association.
compared to liquid CO .
2 2AlCl → Al Cl
3 2 6
365
12 Answers.indd 365 3/26/18 4:06 PM
Chemistry Term 1 STPM
5 (a) 60 (b) (i) Volume of one argon atom
(b) 119.9 4 3
(c) 88.3% = 3 πr
4
STPM Practice 4 = π (0.192 10 ) m 3
–9 3
Objective Questions 3 –29 3
1 B 2 C 3 C 4 D 5 B = 2.96 10 m
6 D 7 D 8 B 9 D 10 B Volume of one mole of argon atom
23
–29
11 B 12 D 13 D 14 C 15 C = (2.96 10 )(6.02 10 )
–5
16 C 17 B 18 D 19 C 20 A = 1.78 10 m 3
21 A 22 A 23 B 24 D 25 B (ii) PV = nRT
26 B 27 B (101 10 )V = 1 8.31 273
3
–2
Structured and Essay Questions V = 2.25 10 m 3
1 (a) A gas that obeys the gas law: PV = nRT under all (iii) % = 1.78 × 10 –5
–2 100% = 0.079%
2.25 × 10
conditions
(b) (i) When temperature increases, the kinetic energy (iv) The volume of the gas molecules is indeed
of the particles increases. This leads to an negligible when compared to the volume
increase in the rate of collision with the walls of occupied by the gas.
the container. At the same time, the collisions 5 (a) Increasing pressure pushes the gas particles closer
are more energetic. to one another. The attractive force is then strong
(ii) Increasing pressure pushes the molecules closer enough to hold the particles together causing it to
to one another. This reduces the amount of condense.
empty space between the particles causing the (b) Presence of intermolecular hydrogen bonding pulls
volume to decrease. the molecules closer.
(iii) When temperature increases, the pressure (c) There are no intermolecular forces between gas
exerted by the gas will increase. To maintain the particles. They are free to move around and thus will
original pressure, the volume occupied by the distribute themselves throughout the container.
gas must increase so as to reduce the rate of (d) Going down the group, the size of the molecules gets
collision. bigger and the strength of the intermolecular forces
2 (a) PV = nRT between the molecules gets stronger causing the
(b) Presence of intermolecular forces. deviation to be more profound.
The gas particles have finite volume. (e) At high pressures, intermolecular repulsion exists
(c) Helium between the molecules. This makes the gas more
(d) At low pressures, intermolecular attraction operates difficult to compress.
between the molecules (In this case, it is the hydrogen At moderate pressures, intermolecular attraction
bonding). Thus, the volume actually occupied by the exists between the molecules, thus making them
gas is less than expected. This makes PV nRT. easier to compress.
6 (a) Refer to text
PV
nRT (b) P (× 10 )/Pa 4 8 15 20
5
(b)
2
PV (× 10 )/Pa m 3 23.2 22.9 22.4 22.0
2
PV (× 10 )/Pa m 3
1 (a)
23.5
P
(e) High temperature and low pressure
23
3 Miscible: There are no intermolecular forces between the
gas particles.
Pressure: Collision of the gas particles with the walls of
the containers.
Compressible: No intermolecular forces and presence of
a lot of empty space.
22
4 (a) No intermolecular forces between gas particles. 0 2 4 6 8 10 12 14 16 18 20
5
P(× 10 )/Pa
The volume of the gas particles is negligible when
compared to the volume of the container in which
the gas is placed.
366
12 Answers.indd 366 3/26/18 4:06 PM
Chemistry Term 1 STPM
2
PV at P = 0, is 23.5 10 Pa m 3 (ii) Total pressure
5
5
3
Using PV = nRT = (1.30 10 ) + (3.75 10 ) + (5.60 10 ) Pa
2
23.5 10 = n 8.31 298 = 5.11 10 Pa
5
∴ n = 0.949 mol (iii) P(Ar) = 1.30 10 Pa
5
32.5
5
∴ M = 0.949 = 34.2 P(Ethane) = 3.75 10 Pa
r
5
Total pressure = (1.30 + 3.75) 10 Pa
7 (a) PV = nRT = 5.05 10 Pa
5
m
PV = —–RT 9 Solid:
M r • Strong attractive forces between particles.
1
m
P = —(—–) RT • Particles have no translational motion.
V M r
m • Particles vibrate and rotate in their mean position.
But — = d (density)
V Liquid:
d • Intermediate strength attractive forces between
∴ P = —–RT
M r particles.
d M r • Particles can move freely in the body of the liquid but
or — = —–
P RT not out of it.
(b) (i) When pressure increases, the volume occupied Gas:
by the gas decreases. This causes the density • No intermolecular forces between particles.
Volume to increase. • Particles can move freely throughout the container
Mass
(ii) where it is placed.
10 (a) The pressure exerted by water vapour which is in
2
Pressure (× 10 )/Pa 33.7 67.6 100.0
equilibrium with liquid water at a fixed temperature.
The pressure is due to the collisions of the water
Density 2.85 10 –5 2.86 10 –5 2.93 10 –5
–3
Pressure /kg m Pa –1 vapour molecules with the walls of the container.
(b)
d /kg m Pa –1
–3
P Pressure
3.0 × 10 –5
2.9 × 10 –5
Temperature
Increasing temperature increases the kinetic energy
2.8 × 10 –5 of the water molecules. More water will undergo
evaporation. At the same time, the collisions between
2.7 × 10 –5 the vapour molecules and the walls of the container is
0 10 20 30 40 50 60 70 80 90 100 more energetic.
2
P (× 10 )/Pa (c) Decreases. Fraction of the liquid surface for water to
evaporate is decreased.
d –5
P at P = 0 is 2.80 10 11 (a) Melting point: The temperature at which a solid is in
M
Using d = r equilibrium with its liquid at 101 kPa.
P RT Boiling point: The temperature at which the vapour
–5
∴ M = (2.80 10 ) (8.31 301.5)
r pressure of a liquid is equal to the external pressure
= 0.0702 kg mol –1 (101 kPa).
∴ Relative molecular mass = 70.2 (b)
8 (a) Refer to text
(b) (i) For argon: Pressure/atm Sea
5
(2.0 10 ) 2.60 = P 4.0 1.0
P(Ar) = 1.30 10 Pa Liquid
5
For ethane: Solid
–5
(1.2 10 ) 12.5 = P 4.0
5
P(Ethane) = 3.75 10 Pa Vapour
For CO :
2 Temperature/°C
4
(2.8 10 ) 0.80 = P 4.0 0 100
3
P(CO ) = 5.60 10 Pa
2
367
12 Answers.indd 367 3/26/18 4:06 PM
Chemistry Term 1 STPM
(c) Presence of dissolved salts makes the orderly (b) Increasing the pressure on the carbon dioxide gas
arrangement of the ice crystal structure more pushes the molecules close together. The attractive
difficult to form. force ‘binds’ the molecules in clusters causing the gas
The volume of water in the sea is much larger. to liquefied.
12 (a) The temperature and pressure where the three (c) When the liquefied carbon dioxide is released, it
phases of a substance can exist in equilibrium. absorbs heat from the surroundings and vaporised.
(b) Refer to text This causes the water vapour in the surrounding air
(c) Decreases. Presence of another substance makes the to condense causing a white mist to form.
solid crystal structure of CO more difficult to form.
2
(d) Because it’s triple point pressure is higher than 1 atm. Chapter 5 Reaction Kinetics
13 (a) (i) Melting point
(ii) Boiling point Quick Check 5.1
(b) 1 57.0 s
Temperature d 2 (a) 0.0125 mol dm min
–3
–1
–3
(b) 0.25 mol dm
3 [A]/mol dm –3
c
1
b
0.9
a Rate = 0.021
0.8
Time
0.7
(c) Water. Because the solid/liquid line has a negative 0.6
slope. 0.5
14 (a) Refer to text 0.4 Rate = 0.0085
(b) The triple point. It is the only conditions where solid, 0.3 Rate = 0.0040
liquid and vapour can exist in equilibrium. 0.2
(c) Freezing point refers to an external pressure of 0.1
1 atm. The triple point refers to when water is under 0 10 20 30 40 50 60 70 80
its own vapour pressure (0.006 atm). Time/min
(d) It is the temperature above which water vapour –3 –1
cannot be condensed by increasing pressure alone. (a) 0.021 mol dm min
–3
15 (a) Helium. It has the smallest size and the least number (b) 0.0085 mol dm min –1
–1
–3
of electrons amongst all gases. As a result, the volume (c) 0.0040 mol dm min
of the molecules and the intermolecular forces can
be ignored. Quick Check 5.2
(b) –273 °C 1 (a) 4.5 10 mol dm s
–3
–3 –1
(c) and (d) (b) 3.0 10 mol dm s
–3 –1
–3
Volume –
2+
p 1 ∆[Cu ] 1 ∆[CN ] 1 ∆[Cu(CN) ]
–
2
2 (a) – 2 ∆t = – 6 ∆t = ∆t
2
(d) p’ ∆[(CN) ]
–
= ∆t 2
–3 –1
(b) (i) 0.168 mol dm s
(ii) 0.028 mol dm s
–3 –1
–3
–3 –1
X 0 Temperature / °C 3 5.0 10 mol dm s
16 (a) 73 C Quick Check 5.3
Pressure/atm B 1 (a) Rate = k[A][B]
2
–2
6 –1
(b) mol dm s
(c) (i) 0.5x
Solid
Liquid
(ii) 0.25x
(iii) 18x
T
5
Gas
A
–57 31 Temperature/°C
368
12 Answers.indd 368 3/26/18 4:06 PM
Chemistry Term 1 STPM
Quick Check 5.4 (b) 9.97 10 mol dm s
–1
–4
3 –1
1 In k 3 [A]/mol dm –3
–2
1.0
–4
9 0.8
Gradient =
–6 0.42 × 10 –3
0.6
–8
0.4
–10
0.2
–12
–14
0 5 10 15 20 25 30 35 40 Time/min
1 (× 10 )/K –1
–3
1.2 1.3 1.4 1.5 1.6 1.7 1.8 T Zero order
–3
–1
178.1 kJ mol 4 (a) Rate/mol dm s –1
4
Quick Check 5.5 2.6
Gradient =
2
1 Rate = k[NO] [O ] 2.35
2 = 1.11 s –1
2 Rate = k[A][B] 2 3
3 (a) (i) +4
(ii) +5 2
(iii) +3
(b) (i) 2
(ii) 1
–2
6 –1
(c) 17.39 mol dm s 1 0 1 2 3
(d) 0.34 mol dm s [N O ]/mol dm –3
–3 –1
5
2
First order
Quick Check 5.6 (b) 1.11 s –1
1 (a) [X]/mol dm –3 5 (a) (c)
Rate [H 2 O 2 ]
1.0
0.8
0.6
0.4 [H O ]
2 2 Time
(b)
0.2
Rate
25 25
Time/min
0 10 20 30 40 50
First order
–2
(b) 2.77 10 min –1
2 (a)
–2
[Sucrose] (× 10 )/mol dm –3 2
[H O ]
2 2
40
7 (a) Fraction of molecules having energy equal to or
35
greater than the activation energy.
30
(b) Ratio between 35 °C to 25 °C = 2.34 : 1
25 (c) The fraction of molecules having energy equal to or
20 greater than the activation energy increases about
2 times for 10° rise in temperature. As a result, the
15
rate will increase by about 2 times.
10
5
84
44
0 20 40 60 80 100120140 160 Time/min
Second order
369
12 Answers.indd 369 3/26/18 4:06 PM
Chemistry Term 1 STPM
STPM Practice 5 (b) (i)
Objective Questions In k
1 C 2 D 3 C 4 C 5 B 1
6 C 7 C 8 C 9 C 10 B
11 B 12 D 13 C 14 C 15 B
16 C 17 B 18 C 19 A 20 A Gradient = 1.8
21 C 22 B 23 B 24 C 25 C 0 0.22 × 10 –3
26 D 27 A 28 A 29 B 30 B = –8.18 × 10 –3
31 A 32 D
Structured and Essay Questions –1
1 (a) Catalyst
(b) So that its concentration remains almost constant
and will not affect the rate of reaction. –2
(c) 3.1 3.2 3.3 1 (× 10 )/K –1
–3
–1
[Ester] (× 10 )/mol dm –3 T
–E
10 3 a
–8.18 10 =
R
8
∴ E = 67 975.8 J
a
= 67.98 kJ
6
(ii) Adding a suitable catalyst
–1
4 (iii) First order. The rate constant has a unit of s .
t 2 + –
2 4 (a) H O + 2H + 2I → 2H O + I 2
2
2
2
(b) As one of the reagent
t 1
+
–
0 10 20 30 40 50 60 70 80 (c) Rate = k[IO ][H ]
–
Time/min However, the rate of formation of IO is given by:
Rate = k’[H O ][I ]
–
(i) 20 minutes 2 2
(ii) 19.5 minutes Substitute into the first rate equation:
+
–
(d) First order Rate = kk’[H O ][I ][H ]
2
2
+
–
2 (a) Rate = k[NO] [O ] 5 (a) H O + 2H + 2Cl → 2H O + Cl
2
2 2 2 2 2
Slow (b) Energy
(b) 2NO + O 9: N O
2 2 4
Slow Uncatalysed
N O 9: 2NO reaction
2 4 2
(c) (i) 9.6 × 10 = k(0.80) (0.26) Catalysed
–3
2
k = 5.7 × 10 mol dm s –1 reaction
–2
6
–2
(ii) Let original rate = x
x = k[NO] [O ]
2
2
1
The new rate = k[—NO] [2O ] Reaction pathway
2
4 2 (c) Increasing the temperature
1
2
= —k[NO] [O ] 6 (a) Let the rate equation be:
8 2 Rate = k[BrO ] [Br ] [H ]
– x
+ z
– y
3
1
= —x Comparing experiment II and I:
8 x –3
(c) 0.60 = 3.75 × 10 –3
[H ] + 0.40 2.50 × 10
∴ x = 1
Comparing experiment III and II:
y
0.56 = 7.50 × 10 –3
0.28 3.75 × 10 –3
∴ y = 1
Comparing experiment IV and I:
Time z –2
0.062 = 1.0 × 10 –3
3 (a) k = Rate constant 0.031 2.50 × 10
A = Constant z = 2
E = Activation energy Rate equation is:
a
–
–
+ 2
R = Gas constant Rate = k[BrO ][Br ][H ]
3
370
12 Answers.indd 370 3/26/18 4:06 PM
Chemistry Term 1 STPM
(b) Using experiment I: 2 (a) Q = 6.12 ≠ K
c
2.50 × 10 = k[0.40][0.28][0.031] 2 (b) Net reaction: From right to left
–3
k = 23.2 mol dm s –1
–3
9
7 (a) Quick Check 6.3
Concentration of PCl /mol dm 1.0 0.6
-3
1 (b) and (e) only
5
Half-life/min 190* 190**
2 (a) and (d) only
[NOTE: * The time taken for the concentration to
decrease from 1.0 to 0.5 mol dm . ** The time taken 3 (a) Increases
-3
for the concentration to decrease from 0.6 to (b) Increases
(c) No change
-3
0.3 mol dm ]
(b) Rate = k[PCl ] 4 [H ] K
5
[Since the half-life is a constant it is first order with 2 c
respect to PCl ] (a) Increases No change
5
(c) The half-life is given by: (b) Decreases No change
1
In 2
T— = ——– (c) No change No change
2 k
When temperature increases, the rate constant, k, (d) Decreases Decreases
also increases. This will cause the half-life to decrease.
[Alternatively: When temperature increases, the rate STPM Practice 6
of reaction will increase. Hence, it takes shorter time Objective Questions
for the original concentration to decrease to half its
original value.] 1 C 2 C 3 D 4 A 5 B
6 B 7 A 8 B 9 A 10 B
11 A 12 C 13 C 14 D 15 A
Chapter 6 Chemical Equilibrium 16 D 17 A 18 D 19 D 20 B
21 C 22 C 23 D 24 C 25 C
26 B
Quick Check 6.1
1 (a) (i) H O(g) + C(s) CO(g) + H (g) Structured and Essay Questions
2 2
P(H )P(CO) 1 (a) Let the degree of dissociation = a.
(ii) K = 2 Pa N O 2NO
p P(H O) 2 4 2
2
(b) (i) 180 kPa Original/mol 1 0
(ii) 281.7 kPa Equilibrium 1 – a 2a
2 (a) Catalyst Total particle present at equilibrium = (1 – a) + 2a
(b) P(SO ) = 0.10 atm = 1 + a
2
P(O ) = 0.050 atm ∴ 92 = 1 + a
2
(c) 60 1
P/atm
a = 0.53 (or 53.0%)
2a
(b) Partial pressure of NO = P
2 2 1 + a
1.9 SO
3 (1 – a)
Partial pressure of N O = P
2 4 (1 + a)
P (NO )
2
1 2
K = P(N O )
p
0.1 SO 2 2 4 2
0.05 O 2a
2 (1 + a) P
=
(d) 7220 atm –1 (1 – a) P
(1 + a)
6
3 (a) 7.79 10 Pa 4a P
2
–2
(b) 8.33 10 mol dm –3 = (1 + a)(1 – b)
(c) 2.73 10 Pa
6
By substitution:
Quick Check 6.2 4(0.53) 2
–1
1 (a) Q = 4.78 atm K p K = (1 + 0.53)(1 – 0.53) × 100
p
System not in equilibrium = 153 kPa
(b) From the right to the left
371
12 Answers.indd 371 3/26/18 4:06 PM
Chemistry Term 1 STPM
2
(c) Let the degree of dissociation = b P (SO )
3
(e) (i) K =
2
4b 2 p P (SO )P(O )
K = × 1 000 2 2
p (1 + b)(1 – b) = 0.80 2
2
4b 2 (0.10) (0.70)
–1
156 = × 1 000 = 91.4 atm
(1 + b)(1 – b) (ii) 2SO + O 2SO
∴ b = 0.21 (or 21.0%) Final/mol 0.10 2 0.70 2 0.80 3
According to Le Chatelier’s Principle, increasing
pressure would shift the equilibrium to the left-hand Initial pressure of SO = 0.10 + 0.80 = 0.90 atm
2
side as the number of moles of gaseous particles is Initial pressure of O = 0.70 + 0.40 = 1.10 atm
2
less. This is would decrease the degree of dissociation 4 (a) (i), (ii) & (iii)
of N O as shown by the calculation.
2 4
2 (a) When a reversible reaction has achieved dynamic
equilibrium, the ratio of the molar concentrations
of the products to the molar concentration of the E a
reactants, both raised to their respective stoichiometric E ’ a
coefficient, is a constant at constant temperature.
(b) Concentration, pressure and temperature. –288 kJ
Only temperature will affect the equilibrium
constant.
P (NH ) E = Activation energy for the forward reaction
2
3
(c) K = a
3
p P(N )P (H )
2 2 E ' = Activation energy of the reverse reaction
a
(b) Higher. Due to the smaller size of the chlorine atom,
N + 3H 2NH the Cl—Cl bond is stronger than the I—I bond.
2 2 3
Initial/mol 1 3 0 [PCl ][Cl ]
Final/mol 0.4 1.2 1.2 5 (a) K = 3 2
c [PCl ]
1.2 5
% NH = 100% = 42.9% P[PCl ]P[Cl ]
2
3
3 2.8 K =
p
1.2 P[PCl ]
5
% H = 100% = 42.9%
2 2.8 (b) At 200 °C:
% N = 100 – 2(42.9) = 14.2%
2 PCl PCl + Cl
(0.429 × P) 2 5 3 2
K = Original/mol 1 0 0
3
p (0.429 × P) (0.142P) Final/mol 0.5 0.5 0.5
= 16.42(P) –2
4 –2
= 16.42 (3.0 10 ) P(PCl ) = P(Cl ) = P(PCl ) = 0.5 200 = 66.67 kPa
–8
= 1.82 10 kPa 3 2 5 1.5
K = 66.67 kPa
3 (a) Low temperature and high pressure p
(b) It is not too low that the rate becomes too slow. It is At 350 °C:
not too high that the yield becomes too low.
(c) (i) Vanadium(V) oxide PCl PCl + Cl 2
3
5
(ii) No effect. It just shortens the time taken for Original/mol 1 0 0
equilibrium to established. Final/mol 0.15 0.85 0.85
(d) 0.85
Energy Uncatalysed P(PCl ) = P(Cl ) = 1.85 200 = 91.89 kPa
2
3
P(PCl ) = 200 – 2(91.89) = 16.22 kPa
5
Catalysed (91.89) 2
K = = 520.6 kPa
p 16.22
(c) Endothermic. K increases with increasing
c
temperature.
(d) (i) Decreases
(ii) Decreases
(iii) No change
372
12 Answers.indd 372 3/26/18 4:06 PM
Chemistry Term 1 STPM
6 1 (× 10 )/K –1 2.86 1.97 1.82 9 (a) (i) pV = nRT pV
–3
T ∴ For an ideal gas: —––— = 1
ln K 1.22 7.60 8.70 nRT
c pV
When a graph of —––— is plotted against p, a
In K c nRT
straight line parallel to the p axis will be obtained
and it cuts the y-axis at 1.0
9
(ii) Positive deviation occurs when carbon dioxide
8 is under high pressure. The molecules are
pushed so close to one another that the
7 4.6
Gradient = molecules repel one another and make the gas
0.64 × 10 –3
6 less compressible.
= –7187.5
5 [NOTE: In this case, the volume decrease less
than expected and the product of pV becomes
4
pV
3 greater than nRT causing —––— > 1]
nRT
2
(b) (i)
1 d
1.8 2.0 2.2 2.4 2.6 2.8 p
1 (× 10 )/K –1
–3
T
Gradient = – 7187.5 K
–∆H
– 7187.5 =
R
∆H = +59 728 J or
= +59.73 kJ p
M
7 (a) 2SO + O 2 2SO 3 Given: d = p —–––
2
Initial/mol: 2 1 0 RT
Final/mol: 0.2 0.1 1.8 d M
p
∴ — = —–––
1.8 RT
Partial pressure of SO = × 5 = 4.29 atm
3 2.1 M
0.2 At constant temperature, —––– = constant.
RT
Partial pressure of SO = × 5 = 0.48 atm
2 2.1 d
Hence, a plot of — against p would give a
p
Partial pressure of O = 5 – 4.29 – 0.48 = 0.23
2
4.29 2 straight line parallel to the p axis.
K = = 347.3 atm –1 M
p (0.48) (0.23) (ii) Using the expression: d = p —–––
2
(b) (i) Increasing temperature favours the reverse RT
M
reaction which is endothermic. The yield of SO 0.57 = (101 × 10 ) ————–––––
–3
3
decreases. –2 8.31 × 308
–1
(ii) Increasing temperature increases both the Or, M = 1.44 × 10 kg mol
–1
= 14.4 g mol
forward and reverse reactions. As a result, the The relative molecular mass of the gas = 14.4
rate of attainment of equilibrium increases.
10 (a) (i) The magnitude of the equilibrium constant
8 (a) The pressure of a gas in a mixture of gases is the decreases when the temperature increases. This
pressure that will be exerted by that gas if it alone shows that, the equilibrium shifts to the left-
occupies the same volume at the same temperature.
hand side when heat is supplied. Therefore,
p(Z)
(b) K = the reverse reaction is endothermic and the
2
p p(X)p (Y) forward reaction is exothermic.
1.2 (ii) Using the expression:
(c) K = K
∆H
1
1
p (0.4)(0.9) 2 ln —–– = —––– —– – —–
2
= 3.7 K 1 R T 1 T 2
1
1.6
1
∆H
(d) The volume occupied by the mixture will increase in ln —–– = —––– ——––– – ——–––
R
order to maintain the original pressure. This amounts 2.4 ∆H 1 000 1 300
to decreasing the pressure on the original equilibrium –0.41 = —–––(2.31 × 10 )
–4
8.31
mixture. As a result, the equilibrium will shift to the 4 –1
left-hand side to increase the number of moles of gas. ∆H = 1.48 × 10 J mol
Or, = –14.8 kJ mol
–1
373
12 Answers.indd 373 3/26/18 4:06 PM
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