TEACHER COPY – WITH SUGGESTED SOLUTIONS

Anglo−Chinese School (Independent)

Year 5 (2022) IBDP Chemistry HL

TOPIC 2 ATOMIC STRUCTURE

(IBDP syllabus Topic 2)

2.1 The nuclear atom

- Essential Idea: the mass of an atom is concentrated
in its minute positively charged nucleus.

2.2 Electron configuration

- Essential Idea: the electron configuration of an atom
can be deduced from its atomic number.

(IBDP syllabus Topic 12)

12.1 Electrons in atoms

- Essential Idea: the quantized nature of energy
transitions is related to the energy states of electrons
in atoms and molecules

IBDP Chemistry HL/Atomic Structure Page 1

2.1 The nuclear atom
Essential idea: The mass of an atom is concentrated in its minute, positively charged nucleus.

 Nature of Science:
• Evidence and improvements in instrumentation—alpha particles were used in the
development of the nuclear model of the atom that was first proposed by Rutherford. (1.8)
• Paradigm shifts—the subatomic particle theory of matter represents a paradigm shift in
science that occurred in the late 1800s. (2.3)

Understandings:
• Atoms contain a positively charged dense nucleus composed of protons and neutrons
(nucleons).
• Negatively charged electrons occupy the space outside the nucleus.
• The mass spectrometer is used to determine the relative atomic mass of an element from
its isotopic composition.

Applications and skills:

• Use of the nuclear symbol notation ΖΑ X to deduce the number of protons, neutrons and
electrons in atoms and ions.
• Calculations involving non-integer relative atomic masses and abundance of isotopes from
given data, including mass spectra.

2.2 Electronic configuration

Essential idea: The electronic configuration of an atom can be deduced from its atomic
number.

 Nature of Science:
• Developments in scientific research follow improvements in apparatus—the use of
electricity and magnetism in Thomson’s cathode rays.(1.8)
• Theories being superseded—quantum mechanics is among the most current models of
the atom. (1.9)
• Use theories to explain natural phenomena—line spectra explained by the Bohr model of
the atom. (2.2)

Understandings:
• Emission spectra are produced when photons are emitted from atoms as excited electrons
return to a lower energy level.
• The line emission spectrum of hydrogen provides evidence for the existence of electrons
in discrete energy levels, which converge at higher energies.
• The main energy level or shell is given an integer number, n, and can hold a maximum
number of electrons, 2n2.
• A more detailed model of the atom describes the division of the main energy level into s,
p, d and f sub-levels of successively higher energies.
• Sub-levels contain a fixed number of orbitals, regions of space where there is a high
probability of finding an electron.
• Each orbital has a defined energy state for a given electronic configuration and chemical
environment and can hold two electrons of opposite spin.

IBDP Chemistry HL/Atomic Structure Page 2

Applications and skills:
• Description of the relationship between colour, wavelength, frequency and energy across
the electromagnetic spectrum.
• Distinction between a continuous spectrum and a line spectrum.
• Description of the emission spectrum of the hydrogen atom, including the relationships
between the lines and energy transitions to the first, second and third energy levels.
• Recognition of the shape of an s atomic orbital and the px, py and pz atomic orbitals.
• Application of the Aufbau principle, Hund’s rule and the Pauli Exclusion Principle to write
electron configurations for atoms and ions up to Z = 36.

Guidance:
• Details of the electromagnetic spectrum are given in the data booklet in section 3.
• The names of the different series in the hydrogen line emission spectrum are not required.
• Full electron configurations (eg 1s22s22p63s23p4) and condensed electron configurations
(eg [Ne] 3s23p4) should be covered.
• Orbital diagrams should be used to represent the character and relative energy of orbitals.
• The electron configurations of Cr and Cu as exceptions should be covered.

12.1 Electronic configuration

Essential idea: The quantized nature of energy transitions is related to the energy states of
electrons in atoms and molecules.

 Nature of Science:
• Experimental evidence to support theories – emission spectra provide evidence for the
existence of energy levels.

Understandings:
• In an emission spectrum, the limit of the convergence at higher frequency corresponds to
the first ionisation energy
• Trends in first ionisation energy across periods account for the existence of main energy
levels and sub-levels in atoms.
• Successive ionisation energy data for an element give information that shows relations to
electron configuration.

Applications and skills:

• Solving problems using E = hv
• Calculation of the value of the first ionisation energy from spectral data which gives the
wavelength or frequency of the convergence limit.
• Deduction of the group of an element from its successive ionisation energy data.
• Explanation of the trends and discontinuities in the first ionisation energy across a period.

Guidance:
• The value of Planck’s constant (h) and E = hv are given in the data booklet in sections 1
and 2.
• Use of the Rydberg formula is not expected in calculations of ionisation energy.

Important information for the topic

𝑐𝑐 = 𝑣𝑣λ
𝐸𝐸 = ℎ𝑣𝑣
Speed of light = 3.00 × 108 m s−1
Planck’s constant (ℎ) = 6.63 × 10−34 J s

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2.1. The Nuclear Atom

2.1.1 The Subatomic Particles In An Atom

NOS - Background to atomic theory and Dalton’s atomic theory.

NOS – Thomson’s “plum-pudding” model of the atom.
NOS – Rutherford’s gold foil experiment

An atom is the smallest component of an element. Atoms contain a positively charged dense
nucleus composed of protons and neutrons (nucleons). Negatively charged electrons occupy
the space outside the nucleus.
Electrons carry a negative
charge of –1e where
e = 1.6 × 10–19 C

Protons carry a positive

charge of +1e Scan me to
view a video
on the
Neutrons carry no charge evolution of
models of the
Relative masses and charges of the three subatomic particles: atom

Subatomic particle Proton (p) Neutron (n) Electron (e)

Mass (kg) 1.672 ×10−27 1.674 ×10-27 9.109 ×10−31
Charge (× 10−19 C) 1.602 ×10−19 0 -1.602 ×10−19
Relative charge +1 0 –1

An atom can be represented by the nuclear symbol notation ΖΑ X , where X is the symbol of the
element.
• A = total number of protons and neutrons (mass number or nucleon number)
• Z = number of protons (atomic number or proton number)
• (A − Z) = number of neutrons.
• Atoms are electrically neutral as the number of protons is equal to the number of electrons.

• Isoelectronic: Atoms / ions that have the same number of electrons.

• Isotonic: Atoms / ions that have the same number of neutrons.

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2.1.2 Isotopic Masses

Isotopes are atoms of the same element with a different number of neutrons. They have the
same atomic number but different mass numbers.
• Isotopes have the same chemical properties but different physical properties.
• Isotopes of elements have different nucleon numbers, relative isotopic masses and
numbers of neutrons.
• The isotope with more neutrons is referred to as the heavier isotope.
• Isotopes of some elements are radioactive as the nuclei of these atoms are unstable and
breakdown spontaneously. When they break down, they emit radiation which may be
alpha particles (α), beta particles (β) and gamma (γ) rays.
• Isotopes are used as biochemical tracers in nuclear medicine for diagnostics, treatment
and research. They are also used as “chemical clocks” in geological and archaeological
dating. In nuclear medicine, PET (positron emission tomography) scanners give three–
dimensional images of tracer concentration in the body and are used to detect cancers.

The relative atomic mass (Ar) of an element is the weighted average atomic mass of its
naturally occurring isotopes. Modern mass spectrometer is used to determine the relative
atomic mass of an element from its isotopic composition. Each peak in the mass spectrum
indicates an isotope of the element and m/z values can be used to identify the isotopes. The
height of the peak in the mass spectrum indicates the relative abundance of the isotopes.

Mass spectrometry is an analytical technique which allows the determination of relative

isotopic masses, relative molecular masses and structural features of organic compound as
well as identification of unknown compounds, e.g. in forensic science.

Instrumentation and Operation of a Mass Spectrometer (Not required by the syllabus)

The essential functions of a mass spectrometer and the associated components are given
below:

B C D

Scan me to
view a video
on the
operation of
a mass
spectrometer

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A: Vaporisation
The sample, in minute quantity, must be vaporised either before it is injected into the
spectrometer or immediately after it has been injected.

e.g. Mg (s) → Mg (g)

H2O (l) → H2O (g)

B: Ionisation
The sample of a compound is ionized in the ionisation chamber, usually to unipositive
cations by loss of an electron. This is achieved by bombarding the vapour with a
stream of high-energy electrons. The positive ions can either be:
• molecular ions, or
e.g. CH3CH3 (g) + e- → [CH3CH3]+ (g) + 2 e-
• fragment ions
e.g. CH3CH3 (g) + e- → 2 [CH3]+ (g) + 3 e-

C: Acceleration
The positive ions produced are attracted and accelerated by using an electric field with
a high negative potential applied across parallel plates. The ions are focused to
produce a narrow beam of ions with high constant speed when they pass through slits
in these plates.

D: Deflection
The accelerated ions passed through a magnetic field where they are deflected. The
amount of deflection depends both on the mass (m) of the ion and its charge (z).

1
Angle of deflection ∝
mass / charge (m/z)

1
Radius of curvature ∝
deflection
mass m
∝ or
charge z

Heavier and less highly charged ions will be deflected less than the lighter and more
highly charged ions. Ions with a particular mass-to-charge ratio are recorded on a
detector. In other words, for particles with the same charge, the extent of deflection
depends on mass—the larger the mass, the less is the angle of deflection.

E: Detector
Ions impart a positive charge on collector plate. The charge is amplified and recorded as
a peak on a moving chart. The greater the abundance of the ions, the greater the
charge imparted on the plate, the greater the current produced, thus the higher the peak
on the recorder.

By varying the strength of the electric or magnetic field, the various ion beams
(charged particles of various masses) can be brought (or selected) to strike the collector
plate and the results are displayed on a chart showing a series of peaks known as a
mass spectrum.

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Reading a mass spectrum:

Relative
abundance

Vertical or y–axis:
Signal intensity or peak height
This axis gives the relative
abundance of the ion which gives
rise to that peak.

m/z

Horizontal or x–axis:
Mass/charge ratio (m/z)
This axis measures the mass/charge (m/z)
ratio and not that of the mass (m). However,
since ions which have a single charge are
much more abundant than ions with multiple
charges, the ratio m/z is numerically equal to
m, the mass of the ion.

Fragmentation of molecules usually occur in the ionisation chamber and the resulting mass
spectrum tends to be complicated especially if isotopes and species with multiple charges are
also present. Each peak indicates a molecular ion or fragment ion. Fragmentation usually begins
with the molecular ion which breaks down to give a series of daughter ions or fragment ions.
The process of fragmentation gives rise to characteristic patterns in the mass spectrum.
These patterns can be used to elucidate the structure of a molecule.

The diagram below shows the mass spectra of pentan–2–one and pentan–3–one (isomers of
pentanone).

(Reference: http://www.chemguide.co.uk/analysis/masspec/fragment.html#top)

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Example 1:
79 81
Bromine contains 54.90% of Br and 45.10% of Br, calculate the relative atomic mass of
bromine.

(54.90 × 79) + (45.10 × 81)

Ar of Bromine = = 79.90
100

Example 2:

Using the mass spectrum and relative abundance data below, calculate the relative atomic
mass of neon.

m/z Species present Relative abundance Ar

20 Ar
20 10Ne
10
(10 × 20) + (1 × 22)
=
22 22
1 11
10Ne = 20.18
Exercise 1

1. Fill in information in the table below:

Symbol Mass No. (A) Atomic No. (Z) Protons Neutrons Electrons

Zn 65 30 30 35 30

Sr2+ 88 38 38 50 36

35
17Cl
35 17 17 18 17

37
17Cl
37 17 17 20 17

37 −
17Cl
37 17 17 20 18

2 +
2 1 1 1 0
1H

2 −
1H
2 1 1 1 2

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2. Suggest why the isotopes of an element have the same chemical properties, though
they have different relative isotopic masses?

All the isotopes have same number of protons and electron configurations. As the
chemical properties are dependent on the electron configurations, hence they will
have the same chemical properties.

104
3. The species 44Ru contains

A. 44 neutrons and 44 electrons

B. 44 protons and 44 electrons
C. 44 protons and 104 neutrons
D. 60 neutrons and 60 electrons

Answer: B

Mass number = 104

Atomic number = 44
Number of protons = atomic number = 44
Number of electrons = number of protons = 44 (since the element is not charged)
Number of neutrons = 104 – 44 = 60

4. Chlorine occurs as two natural isotopes, chlorine–35 and chlorine–37. Which of the
following statements about chlorine atoms is false?

A. All chlorine atoms have the same nuclear charge.

B. The nuclei have diameters which are approximately 1/10000 of the diameter of
chlorine atoms.
C. Some naturally occurring chlorine nuclei contain 18 protons.
D. Some naturally occurring chlorine nuclei contain 20 neutrons.

Answer: C

Atoms of all chlorine isotopes contain 17 protons.

Atoms that contain 18 protons will be argon.
Atoms of chlorine–37 contain (37 – 17) = 20 neutrons.

5. Iridium has a relative atomic mass of 192.2. Its isotopes are 191Ir and 193Ir. Calculate
the percentage abundances of the two isotopes.

Let the percentage of 191Ir = x and percentage of 193Ir = 100 – x.

(x × 191) + [(100−x) × 193]
= 192.2
100
191x + 19300 – 193x = 192.2 × 100
2x = 19300 – 19220
x = 40
Iridium is 40% of 191Ir and 60% 193Ir.

6. A sample of the element rubidium (symbol: Rb) contains two isotopes, 85Rb and 87Rb,
in the ratio of 18 : 7. Calculate the relative atomic mass of rubidium.

(18 × 85) + (7 × 87)

Relative atomic mass of rubidium = = 85.56
25

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2.2 Electron Configuration

2.2.1 Electromagnetic Spectrum

The different electromagnetic radiations have different wavelengths. The visible light region
contains radiation having wavelengths between 400 nm – 700 nm and the different colours
correspond to radiations of different wavelengths. In addition to visible light there are many
other electromagnetic radiations, such as X–rays, ultraviolet rays (uv), infra–red rays (IR),
microwaves and radio waves.

The arrangement of all the electromagnetic radiations in the increasing order of their
wavelengths or decreasing order of their frequencies is termed the electromagnetic
spectrum.

wavelength / m

The frequency of the radiation can be determined using the equation:

c = λν

where c is the speed of light = 3.00 × 108 m s−1

A photon is characterised by either wavelength, denoted by λ or energy, by E. The inverse

relationship between the energy of a photon (E) and the wavelength of the light (λ) is shown
in the equation:

hc
E=
λ

where ℎ is the Planck's constant = 6.63 × 10−34 J s

The inverse relationship means that light consisting of high energy photons has a short
wavelength. Light consisting of low energy photons has a long wavelength.

hc
As 𝑐𝑐 = νλ, the equation E = can also be written as:
λ
E = ℎν

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Example 3:

A certain energy source emits radiation of wavelength 500.0 nm. What is the energy of one
photon of this radiation?

500.0 nm = 500.0 x 10–9 m = 5.000 x 10–7 m

c = λν
(5.000 x 10–7 m) (ν) = 3.00 x 108 m s–1
ν = 6.00 x 1014 s–1
E = ℎν
E = (6.63 x 10–34 J s) (6.00 x 1014 s–1)
E = 3.978 x 10–19 J

2.2.2 Energy Level and Bohr Model

In atomic physics, the Bohr model, introduced by Niels Bohr in 1913, depicts the atom as a
small, positively charged nucleus surrounded by electrons that travel in circular orbits around
the nucleus—similar in structure to the solar system, but with attraction provided by
electrostatic forces rather than gravity. Bohr modified the Rutherford model by requiring that
the electrons move in orbits of fixed size and energy. According to Bohr’s model, the energy
of the electron in an atom is quantized. Quantization means that a quantity cannot vary
continuously to have any arbitrary value but can change only discontinuously to have specific
or discrete values.

Bohr found that the electron had the least energy when n = 1 which corresponds to the first
Bohr energy level. This lowest energy state is called the ground state and electrons in this
energy level are closest to the nucleus.

NOS – Models of the atom and electron arrangements.

NOS – The quantum mechanical model of the atom.

2.2.3 Hydrogen Emission Spectrum

When a sample of gaseous hydrogen atoms at low pressure is subjected to a high amount of
energy, such as from an electric discharge or heat, the atoms will be able to emit
electromagnetic radiation. The electron in the hydrogen atom may absorb energy and undergo
an electronic transition (‘jump’) to higher energy levels. The electron at the higher energy level
is said to be in an excited state. Then when the electron returns to a lower energy level,
energy is emitted in the form of light/ photon. For example, an electron may return from n = 5
to n = 4 then to n = 2 and finally to n = 1 or it may be from n = 3 to n = 1. The electron cannot
change its energy in a continuous way, in the same way you cannot stand in between the
steps; it can only change in discrete amounts.

Scan me to
view a
simulation on
IBDP Chemistry HL/Atomic Structure
the hydrogen
Page 11
emission
spectrum
 Source: http://www.cas.miamioh.edu/~yarrisjm/F06101incl.html

The emitted photon from the atom on passing through a very thin slit (diffraction grating) and
then through a prism. The emission lines correspond to photons of energies that are emitted
when the excited electrons transit from higher energy levels (e.g. n = 4, 5, 6) to a lower energy
level (e.g. n = 2 or n = 1).

The hydrogen emission spectrum is a line spectrum and the lines in the spectrum
correspond to the photons of a particular wavelength (frequency). For each element, the
emission spectrum is characteristic and unique to the element and it can be used to identify
the element. (Similar to a flame test of cations). The emission spectrum provides crucial
evidence for the existence of electrons in discrete energy levels which converge at higher
energies (frequencies). The energy level is said to be quantised. If energy of an atom is not
quantised, the emission spectrum would be continuous.

The spectrum formed from white light contains all colours or frequencies and is known as a
continuous spectrum. Continuous spectra are produced by all incandescent solids and
liquids and by gases under high pressure.

Source: http://astro.unl.edu/naap/hr/hr background1.html

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The hydrogen emission spectrum consists of four coloured lines separated by dark bands.
The wavelengths of the four lines are 656.3 nm (red), 486.1 nm (blue-green), 434.0 nm (indigo)
and 410.1 nm (violet). This set of coloured lines is found in the visible region, are associated
with electron transitions to n = 2 energy level.

There are other sets of lines (series) in the ultraviolet and infrared red region, they are
associated with electronic transitions to n = 1 and n = 3 energy level respectively.

Source: http://www.ibchem.com/IB/ibnotes/full/ato_htm/2.3.htm

Each series in the hydrogen emission spectrum consists of lines of fixed frequencies. This
shows that the energy of the electron is quantised, i.e. it has a fixed amount of energy as the
electron is in a discrete energy level. These energy levels are characterized by a whole
number called the principal quantum number, n.

Red-hot H2 Colored
discharge tube Lines
(Balmer)

The hydrogen emission spectrum above shows that with an increase in frequency, the spacing
between adjacent lines in the emission spectrum decreases. The lines are closer to each
other and each series converges to a limit, called the convergence or series limit. The lines
are so close together in that region that the spectrum looks continuous.

The gap between the lines in the spectrum indicates that the energy levels of the atom. There
is a convergence of the energy levels at higher frequencies and wider energy gaps when the
energy level is closer to the nucleus.

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Further Reading: Fireworks
The colours of fireworks are the result of emissions of light (photons) by energetically excited
metal atoms. The gun powder used in fireworks heats and excites the electrons in metal. A
fraction of a second later, these excited electrons return to the ground state and energy is
released as light of various colours.

Exercise 2

1. Which of the following statements is/are consistent with the Bohr model?

I. The lowest energy level is closest to the nucleus.

II. An electron can transit from n = 2 energy level to n = 3 energy level by emitting a
photon of definite energy.
III. An electron can remain in a particular energy level as long it emits a radiation of a
constant frequency.

A. I only
B. II only
C. III only
D. I and II only

Answer: A

I is true.
II is false as energy is absorbed for an electron to move from a lower to a higher
energy level.
III is false as an electron has a fixed amount of energy (quantisation).

2. The emission spectrum of an atom appears as a

A. continuous spectrum.
B. series of dark lines in an otherwise continuous spectrum.
C. series of evenly-spaced bright lines.
D. series of bright lines that converge at high frequency.

Answer: D

3. What is the energy of a photon with a wavelength of 586 nm?

A. 3.39 × 10–28
B. 3.39 × 10–26
C. 3.39 × 10–19
D. 3.39 × 10–17

Answer: C

3 × 108 = ν × 586 × 10–9

ν = 5.12 × 1014 Hz
E = 6.626 × 10–34 × 5.12 × 1014 = 3.392 × 10–19 J

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4. The spectral line that corresponds to the electronic transition from n = 3 to n = 2 in the
hydrogen atom is red in colour. What type of electromagnetic radiation will be emitted
during an electronic transition from n = 2 to n = 1?

A. Ultraviolet
B. Red light
C. Infrared
D. Radio waves

Answer: A

Transition to n = 1 is in the ultraviolet range.

5. If it takes 3.36 x 10–19 J of energy to eject an electron from the surface of a certain metal,
calculate the longest possible wavelength, in nanometers, of light that can ionise the
metal.

3.36 x 10–19 = (6.626 x 10–34) (ν)

ν = 5.071 x 1014 s–1
λν = c
(λ) (5.071 x 1014) = 3.00 x 108
λ = 592 nm

2.2.4 Energy Levels, Sub–Levels, Orbitals and Electron Spin

The Bohr model is the idea of a main energy level, described by n, which is called the principal
quantum number. n determines the energy and the average distance of an electron from the
nucleus. The main energy level or shell is given an integer number and can hold a maximum
of 2n2 electrons.

The main energy level or shell is divided into s, p, d and f sub–levels. Sub–levels contain a
fixed number of orbitals which are regions of space with a high probability of finding an electron.
Each orbital can hold a maximum of 2 electrons. According to Pauli Exclusion Principle, the
two electrons occupying the same orbital must have different spin states.

Sub–level Number of orbitals in sub–level Maximum number of electrons

s 1 2
p 3 6
d 5 10
f 7 14

Each orbital has a defined energy state for a given configuration and chemical environment.
It may be noted that all the orbitals of a particular energy level will have the same energy. For
example, all the 2p orbitals have the same energy. Orbitals having equal energy are termed
degenerate.

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The energies of the orbitals depend not only upon the nuclear charge but also upon the other
electrons present in the atom. We have to consider the attraction between the electrons and
the nucleus as well as the repulsive interactions between the electrons.

An electron in an atom is stable when the total attractive interactions are greater than the total
repulsive interactions.

2.2.5 Types of Orbitals

The s orbitals have spherical shape and they are non–directional. For every quantum shell, there
is only one s orbital and the size of the s orbital increases with the principal quantum number.

y
y
z z

x x

1s orbital 2s orbital

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The p orbitals have dumb–bell shape and they are directional – the three p orbitals have different
directions in space. The size of the p orbitals also increases with the principal quantum number.

px orbital lies along py orbital lies along pz orbital lies along

the x axis the y axis the z axis

A p subshell consisting of
three p orbitals.
x
z

Enrichment:
The d orbitals are directional and their sizes increase with the principal quantum number.

IBDP Chemistry HL/Atomic Structure Page 17

 Energy
4f

4d

4p
3d
4s
3p

3s

2p

2s

1s

Main shell 1 2 3 4

Note: 4s orbital (when empty) has lower energy than the 3d orbitals despite its
higher quantum number.

For a given main energy level, the energy of d-orbitals > p-orbitals > s-orbitals.

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2.2.6 Aufbau Principle, Hund’s Rule and Pauli Exclusion Principle

Orbital diagrams refer to arrow–in–box diagrams and they are used to represent the character
and relative energy of orbitals. Electrons are arranged in orbitals according to a set of rules.

If the above rules are followed, electrons occupy the lowest possible energy and the resultant
arrangement of electrons is known as ground state electronic configuration (i.e. lowest
energy state). When one or more electrons absorb energy and are promoted to higher energy
level(s), the atom is said to be in an excited state.

Rule 1: Aufbau Principle

Electrons are added progressively to the orbitals starting with the lowest energy.

IMPORTANT NOTE
 4s has lower energy level than 3d when empty.
 When adding electrons, fill up 4s before 3d.
 4s has higher energy level than 3d when filled.
 When removing electrons, remove from 4s before 3d.

Order of orbitals in terms of increasing energy:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, etc.

Rule 2: Pauli Exclusion Principle

Each orbital can hold a maximum of two electrons. Paired electrons can only be stable when
they spin in opposite directions so that the magnetic attraction which results from their opposite
spins counterbalances the electrical repulsion.

NOT

Rule 3: Hund’s Rule

When filling a sub–level, each orbital must be occupied singly before they are occupied in
pairs i.e. electrons only ‘pair–up’ when each sub–level is already half–filled.

NOT NOT

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The electronic configuration of an atom or ion refers to the arrangement of electrons in its main
energy levels, sub–levels and orbitals.

Notation for writing electronic configuration:

main energy (superscript)

level 2p1 number of electrons in
sub–level

sub–level
Ground State Electronic
Element Atomic Number
Configuration
Ne 10 1s2 2s2 2p6
Cl 17 1s2 2s2 2p6 3s2 3p5
Ca 20 1s2 2s2 2p6 3s2 3p6 4s2
V 23 1s2 2s2 2p6 3s2 3p6 3d3 4s2

Exceptions to the Aufbau principle

At higher principal quantum numbers, the energy difference between certain sub–levels is
smaller than that between the 3d and 4s sub–levels. As a result, there are some exceptions to
the Aufbau principle among the heavier transition elements.

Element Symbol Atomic no. Electron configuration

Scandium Sc 21 [Ar] 3d 1 4s2
Titanium Ti 22 [Ar] 3d 2 4s2
Vanadium V 23 [Ar] 3d 3 4s2
Chromium Cr 24 [Ar] 3d 5 4s1
Manganese Mn 25 [Ar] 3d 5 4s2
Iron Fe 26 [Ar] 3d 6 4s2
Cobalt Co 27 [Ar] 3d 7 4s2
Nickel Ni 28 [Ar] 3d 8 4s2
Copper Cu 29 [Ar] 3d 10 4s1
Zinc Zn 30 [Ar] 3d 10 4s2

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From the electron configuration of the transition metals, it may be noted that chromium and
copper atoms have 5 and 10 electrons in the 3d orbitals instead of 4 and 9 electrons respectively.
The reason is because fully–filled orbitals and half–filled orbitals have extra stability. Thus, p3,
p6, d 5, d 10, f 7 and f 14 configurations which are either fully–filled or half–filled are more stable.
Therefore, to acquire greater stability, one of the 4s electron enters the 3d orbitals so that the
set of 3d orbitals become half–filled or fully–filled as seen in the configurations of chromium and
copper atoms, respectively.

The extra stability of half–filled and fully–filled electronic configurations can be explained in
terms of symmetry and exchange energy (see enrichment for explanation on exchange
energy).

Symmetrical arrangement
The electronic configurations in which all the orbitals of same sub–shell are either fully–filled
or half–filled will have a relatively more symmetrical distribution of electrons and therefore
greater stability.

e.g. Chromium, 24Cr (Complete the electron-in box diagram)

Expected configuration: Less symmetrical
1s2 2s2 2p6 3s2 3p6 3d 4 4s2 3d 4s
Actual configuration: More symmetrical
2 2 6 2 6 5 1
1s 2s 2p 3s 3p 3d 4s 3d 4s

e.g. Copper, 29Cu

Expected configuration: Less symmetrical
1s2 2s2 2p6 3s2 3p6 3d 9 4s2 3d 4s
Actual configuration: More symmetrical
2 2 6 2 6 10 1
1s 2s 2p 3s 3p 3d 4s 3d 4s

Enrichment: Exchange energy

The half–filled or fully–filled degenerate orbitals have a greater number of exchanges and
consequently, they have larger exchange energy of stabilization. The “exchange” refers
to shifting electrons from one orbital to another within the same sub–level. Let us compare
the number of exchanges in the 3d 4 4s2 and in the 3d 5 4s1 configuration in chromium.

In the 3d 4 arrangement, there are 6 electron exchanges which imply that there are six
possible arrangements with parallel spin in 3d4 configurations. Whereas in the 3d5
arrangement, there are 10 electron exchanges which imply that there are ten possible
arrangements with parallel spin in the 3d5 configurations. It is quite evident from the above
description that the total number of electron exchanges in the 3d 5 arrangement is larger;
which lends it relatively greater stability.

In a similar way, it can be established that the number of exchanges in 3d 10 configurations

is larger than in 3d 9 configurations which makes the 3d 10 configurations relatively more
stable.

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2.2.7 Electron Configurations of Ions

Ions are formed when the atom loses or gains electron(s). A positive ion (cation) is formed
when an atom loses electron(s) whereas a negative ion (anion) is formed when an atom gains
electron(s).

The electron configuration of ions is written using the same rules as the atoms. However, it
must be noted that when forming cations, the electrons are lost from the orbital with the highest
energy. The anions are formed by adding electrons to vacant orbital of highest energy.

Example 4:

Write the electron configuration of Fe2+ and Fe3+ ions.

26Fe : 1s2 2s2 2p6 3s2 3p6 3d6 4s2

2+
26Fe : 1s2 2s2 2p6 3s2 3p6 3d6
3+
26Fe : 1s2 2s2 2p6 3s2 3p6 3d5

Exercise 3

1. Which principles or rules are violated in the following electron configurations of atoms?
Write the correct electron configuration for each of the elements.
(a) boron: 1s2 2s3

Pauli’s exclusion principle: each orbital can hold 2 electrons (spin pair)
Boron atom: 1s2 2s2 2p1

(b) nitrogen: 1s2 2s2 2px2 2py1 2pz0

Hund’s rule: no of unpaired electrons is maximized within a subshell/orbital

Nitrogen atom: 1s2 2s2 2p3 (2 px1, 2py1 and 2pz1)

(c) beryllium: 1s2 2px2

Aufbau principle: fill the orbitals in order of increasing potential energy

Beryllium atom: 1s2 2s2

2. Write the full electron configurations for the following ions and draw their orbital diagrams
(electron–in–box diagram).
(a) 24Cr3+

1s2 2s2 2p6 3s2 3p6 3d3

+
(b) 29Cu

1s2 2s2 2p6 3s2 3p6 3d10

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3. For each of the following, decide whether the electron configurations shown represent a
neutral atom, a positive ion or a negative ion of the element shown,
(a) H : 1s2

H– (hydride ion) (ground state)

(b) S : 1s2 2s2 2p6 3s2 3p4

Atom (ground state)

(c) N : 1s2 2s1 2p3

N+ (unipositive ion) (excited state)

(d) F : 1s2 2s2 2p5 3s1

F– (fluoride ion) (excited state)

Enrichment: Quantum Numbers

Four quantum numbers, n, l, ml, and ms, define the electron’s position in the atom.

The principal quantum number, n, represents the main energy levels, or shells, the electrons
can occupy in an atom, and the whole-number values 1, 2, 3….

The secondary quantum number, l, represents sub-levels and it has value ranging from 0 to n
– 1, and the letters s, p, d, f. l describes the shape of the orbital

The magnetic quantum number, ml, describes the amount of energy levels in a sub–levels and
has values – l to + l. It determines the number of orbitals and their orientation within a sub-
level.

The spin quantum number, ms, represents electron spin and has values + ½ or - ½. The Pauli
Exclusion Principle states that no two electrons in an atom can have the same set of four
quantum numbers (n, l, ml, and ms).

TOK – Historical development in the discovery of the structure of atom.

NOS – Determining the wavelengths of lines in spectra.

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12.1 Electrons in an atom (AHL – Topic 12)

12.1.1 Ionisation Energy

When enough energy is supplied to move the electron in hydrogen atom from the ground state
to n = ∞, the ionisation of hydrogen atom takes place. In an emission spectrum, the limit of
convergence at higher frequency corresponds to the first ionisation energy. The 1st ionisation
energy per electron is therefore a measure of the distance between the n = 1 level and n = ∞
level.

The 1st ionisation energy is defined as the minimum energy required in removing one
mole of valence electrons from one mole of gaseous atoms to form 1 mole of singly
positively charge gaseous ion.

The second ionisation energy (2nd IE) is defined as the energy required in removing one mole
of electrons from one mole of its gaseous ions M+(g) in its ground state. Ionisation energies
always have positive values (endothermic) since energy is required to remove an electron.

In general, X (g) → X+ (g) + e− ∆H1 = 1st IE

X+ (g) → X2+ (g) + e− ∆H2 = 2nd IE
X2+ (g) → X3+ (g) + e− ∆H3 = 3rd IE
X (g) → X3+ (g) + 3e− ∆H = summation of the 1st, 2nd & 3rd IE

Example 5:

Calculate the 1st IE of hydrogen using the list of the frequencies of the seven most widely spaced
lines in the Lyman series.
ν (Hz)
2.467 x 1015
2.924 x 1015
3.084 x 1015
3.158 x 1015
3.198 x 1015
3.223 x 1015
3.238 x 1015

E = ℎν
= 6.63 x 10-34 x 3.238 x 1015
= 2.147 x 10-18 J

st
2.147 × 10−18 × 6.02 × 1023
1 IE =
1000
= 1292 kJ mol–1

IBDP Chemistry HL/Atomic Structure Page 24

12.1.2 Factors Affecting Ionisation Energy

(a) Nuclear charge

• Nuclear charge is the total charge of the protons in the nucleus. This is also the
same as the number of protons.
• Given atoms or ions with the same number of filled quantum shells, the greater
the nuclear charge, the greater the ionisation energy.

(b) Shielding effect by inner electrons

• Electrons in the inner shells repel the valence electrons, giving rise to the
shielding effect. Shielding effect is usually given by the number of electrons in
the inner quantum shells.
• The greater the shielding effect, the lower will be the ionisation energy.

The nuclear charge of an atom serves to attract the electrons towards the nucleus but the
shielding effect of the inner core electrons serves to shield the electrostatic forces of attraction
of the nucleus on the valence electrons. This balance can be roughly measured by the
effective nuclear charge. Effective nuclear charge (Zeff) is difference between nuclear charge
and shielding effect. This gives a measure of the actual nuclear charge experienced by
electrons in the valence shell, hence how tightly the electrons are attracted to the nucleus.

12.1.3 Successive Ionisation Energy

The successive ionisation energies of an atom increase with the removal of each electron, 1st
IE < 2nd IE < 3rd IE.

Reason:

The general increase in successive ionisation energy is due to increasing amount of energy
required to remove successive electrons from an increasingly positive ion due to an increasing
electrostatic force of attraction between the nucleus and valence electrons.

The electron configurations of atoms can be investigated by experimentally measuring the

successive ionisation energies. A large increase in ionisation energy indicates that the electron
removed was more strongly held by the nucleus than the earlier ones; i.e. it could have come
from an inner quantum shell.

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Example 6:
Deduce the electron configuration of potassium using its successive ionisation energies.

The potassium atom has a total of 19 electrons which fall into four groups:

1. The first electron (valence electron) is relatively easy to remove.

2. The next eight electrons are more difficult to remove (closer to the nucleus) and require
higher energy.
3. The next subsequent group of eight electrons are even more difficult to remove (even
closer to the nucleus) and require even higher energy.
4. The last two electrons are most difficult to remove as they are closest to the nucleus.

This suggests that the potassium atom has two electrons closest to the nucleus (in quantum
shell, n = 1), eight electrons further out (in quantum shell, n = 2). It also has another eight
electrons even further out (in quantum shell, n = 3) and one valence (in quantum shell, n = 4).
Hence, the electron arrangement in potassium is 1s2 2s2 2p6 3s2 3p6 4s1

Example 7:
Consider the successive lg IE values of the third quantum shell of a period 4 main group
element:

There is a steady rise in the lg IE values for the successive removal of the first six electrons,
followed by a larger increase in the lg IE value. This indicates that the last two electrons are
more strongly attracted to the nucleus. This larger increase is a result of the 7th and 8th
electrons being removed from a sub–level of lower energy. This therefore shows that the third
quantum shell is further divided into sub–levels (3s sub–level and 3p sub–level).

IBDP Chemistry HL/Atomic Structure Page 26

Example 8:
The first seven ionisation energies (in kJ mol−1) of an element are as follows:

1012 1903 2912 4956 6273 22233 25397

Deduce the group number of the element and explain your answer.

• Biggest increase between 5th and 6th ionisation energy.

• 6th electron is removed from an inner quantum shell which is closer to the nucleus.
• Hence, there is much stronger electrostatic attraction between the 6th electron and the
nucleus hence a lot more energy is thus required to remove it.
• 1st 5 electrons are removed from the valence shell.
• There are 5 valence electrons in the valence shell.
• Thus, the element is in Group 15.

12.1.4 Trends in the 1st Ionisation Energy

The diagram below shows the graph of the 1st IE of some elements.

Across a period (from left to right),

• There is an increase in effective nuclear charge, Zeff. This is because nuclear charge
increases but shielding effect remains relatively constant since the inner quantum shells of
electrons remain the same.
• The valence electrons are drawn closer to the nucleus, so the electrostatic attractions
between the valence electrons and the nuclei increase.
• More energy is required to remove a valence electron from the gaseous atom.

Down a group,
• The number of filled quantum shells increases and the size of the atoms increase.
• The valence electrons occupy energy levels that are increasingly further from the nucleus.
• This increased distance reduces the electrostatic attractions between the protons in the
nuclei and the valence electrons.
• Hence the electrostatic forces of attraction of the nucleus for the valence electrons
decreases and less energy is required to remove them.

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The diagram shows two deviations from the 1st IE general trend:

1. The first ionisation energy of aluminium is lower than that of magnesium.

Mg : 1s2 2s2 2p6 3s2

Al : 1s2 2s2 2p6 3s2 3p1

It is because less energy is required to remove a 3p electron in the aluminium atom

than a 3s electron in the magnesium atom as the 3p sub–level is of higher energy than
the 3s sub–level and the 3p electron is further from the nucleus. The 3p electron also
experiences shielding by the 3s electrons. As a result, the 3p electron experiences a
weaker electrostatic force of attraction from the nucleus.

2. The first ionisation energy of the sulfur atom is lower than that of a phosphorus atom.

P : 1s2 2s2 2p6 3s2 3p3

S : 1s2 2s2 2p6 3s2 3p4

It is because in a sulfur atom, there are two electrons occupying the same 3p orbital
and this gives rise to inter–electronic repulsion. Thus, less energy is required to
remove a paired 3p electron from a sulfur atom compared to the energy required to
remove an unpaired 3p electron from a phosphorus atom.

Exercise 4

1. Deduce the group number of the elements A to D using the ionisation energies.

Ionisation Energy / kJ mol−1

Element 1st 2nd 3rd 4th
A 500 4600 6900 9500
B 740 1500 7700 10500
C 900 1800 14800 21000
D 580 1800 2700 11600

A Group 1 (large ‘jump’ after 1st IE)

B Group 2 (large ‘jump’ after 2nd IE)
C Group 2 (large ‘jump’ after 2nd IE)
D Group 13 (large ‘jump’ after 3rd IE)

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Summary

• All atoms of the same element have the same atomic number (Z); that is, they have equal
numbers of protons in their nuclei.
• The mass number (A) of an atom is the total number of protons and neutrons. Thus the
number of neutrons = A – Z.
• The isotopes of an element are atoms with the same atomic number but different mass
numbers. They have the same number of protons and electrons, but different numbers of
neutrons.
• Mass spectra of elements enable isotopic abundances and relative atomic masses to be
determined.
• If an atom is given sufficient energy (electrical or thermal), electrons can be excited from
a lower energy level to a higher energy level. When the excited electron falls from a higher
to a lower energy level, a quantum (photon) of electromagnetic radiation is emitted. This
can be seen as a line in the emission spectrum of the element.
• Electrons can occupy a discrete energy level and gain or lose quanta (‘packets’) or photons
of energy. When they return from higher energy levels to a lower energy level, a photon of
energy is emitted.
• The main energy levels or shells are given principal quantum numbers n = 1, 2, 3, 4, etc.
The first shell (n=1) is closest to the nucleus.
• The quantum shell is divided into sub–levels known as s, p, d or f and each sub–levels
consists of atomic orbitals (solutions to the Schrodinger equation).
• Sub-level s, p, d and f have 1, 3, 5 and 7 orbitals, respectively. s orbitals are spherical and
p orbitals are dumb–bell shaped.
• Each orbital holds a maximum of two electrons and the two electrons in an orbital are spin
paired.
• The first ionisation energy of an element is the minimum energy required to remove one
mole of electrons from a mole of gaseous atoms to form one mole of singly charged
gaseous cations. Ionisation energies data gives crucial evidence to the existence of main
energy level and sub–levels.

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Suggested Answers

Exercise 1
1.

Symbol Mass No. (A) Atomic No. (Z) Protons Neutrons Electrons

Zn (35 + 30) = 65 30 30 35 30

Sr2+ (38 + 50) = 88 38 38 50 36

35 (35 – 17)
17Cl
35 17 17 17
=18
37 (37 – 17)
17Cl
37 17 17 17
= 20
37 − (37 – 17)
17Cl
37 17 17 18
= 20
2 +
2 1 1 1 0
1H

2 −
1H
2 1 1 1 2

2. All the isotopes have same number of protons and electron configurations. As the
chemical properties are dependent on the electron configurations, hence they will have
the same chemical properties.

3. B
104
The species 44Ru contains 44 protons, 44 electrons and (104 – 44) = 60 neutrons.

4. C
Atoms of all chlorine isotopes contain 17 protons. Atoms that contain 18 protons will be
argon. Atoms of chlorine–37 contain (37 – 17) = 20 neutrons.

5. Let the percentage of 191Ir = x and percentage of 193Ir = 100 – x.

(x × 191) + [(100−x) × 193]
= 192.2
100
191x + 19300 – 193x = 192.2 × 100
2x = 19300 – 19220
x = 40
Iridium is 40% of 191Ir and 60% 193Ir.

(18 × 85) + (7 × 87)

6. Relative atomic mass of rubidium = = 85.56
25

IBDP Chemistry HL/Atomic Structure Page 30

Exercise 2
1. A
I is true; II is false – energy is absorbed and III is false – they have a fixed amount of
energy (quantisation).

2. D

3. C
3 × 108 = ν × 586 × 10–9
ν = 5.12 × 1014 Hz
E = 6.626 × 10–34 × 5.12 × 1014 = 3.392 × 10–19 J

4. A
Transition to n = 1 is in the ultraviolet range.

5. 3.36 x 10–19 = (6.626 x 10–34) (ν)

ν = 5.071 x 1014 s–1
λν = c
(λ) (5.071 x 1014) = 3.00 x 108
λ = 592 nm

Exercise 3
1. (a) Pauli’s exclusion principle: each orbital can hold 2 electrons (spin pair)
Boron atom: 1s2 2s2 2p1
(b) Hund’s rule: no of unpaired electrons is maximized within a subshell/orbital
Nitrogen atom: 1s2 2s2 2p3 (2 px1, 2py1 and 2pz1)
(c) Aufbau principle: fill the orbitals in order of increasing potential energy
Beryllium atom: 1s2 2s2

2. (a) [Ar] 3d3

1s2 2s2 2p6 3s2 3p6 3d3
(b) [Ar] 3d10
1s2 2s2 2p6 3s2 3p6 3d10

3. (a) H atom is 1s1 H– (hydride ion) (ground state)

(b) S atom is 1s2 2s2 2p6 3s2 3p4 atom (ground state)
(c) N atom is 1s2 2s2 2p3 N+ (unipositive ion) (excited state)
(d) F atom is 1s2 2s2 2p5 F– (fluoride ion) (excited state)

Exercise 4
1. A Group 1 (large ‘jump’ after 1st IE)
B Group 2 (large ‘jump’ after 2nd IE)
C Group 2 (large ‘jump’ after 2nd IE)
D Group 3 (large ‘jump’ after 3rd IE)

IBDP Chemistry HL/Atomic Structure Page 31