DP IB Chemistry: HL Your notes

Electronic Configurations
Contents
The Electromagnetic Spectrum
Emission Spectra
Energy Levels, Sublevels & Orbitals
Writing Electron Configurations
Ionisation Energy from an Emission Spectrum (HL)
Successive Ionisation Energies (HL)

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 The Electromagnetic Spectrum
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The Electromagnetic Spectrum
The electromagnetic spectrum is a range of frequencies that covers all
electromagnetic radiation and their respective wavelengths and energy
It is divided into bands or regions, and is very important in analytical chemistry.
The spectrum shows the relationship between frequency, wavelength and energy
Frequency is how many waves pass per second, and wavelength is the distance
between two consecutive peaks on the wave
Gamma rays, X-rays and UV radiation are all dangerous - you can see from that end of
the spectrum that it is high frequency and high energy, which can be very damaging to
your health

The electromagnetic spectrum diagram

The electromagnetic spectrum spans a broad spectrum from very long radio waves to
very short gamma rays
All light waves travel at the same speed; what distinguishes them is their different
frequencies
The speed of light (symbol ‘c’) is constant and has a value of 3.00 x 108 ms-1

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 As you can see from the spectrum, frequency (symbol ‘f') is inversely proportional to
wavelength (symbol ‘λ')
Your notes
In other words, the higher the frequency, the shorter the wavelength
The equation that links them is c = fλ
Since c is constant you can use the formula to calculate the frequency of radiation given
the wavelength, and vice versa
Continuous versus line spectrum
A continuous spectrum in the visible region contains all the colours of the spectrum
This is what you are seeing in a rainbow, which is formed by the refraction of white light
through a prism or water droplets in rain

Continuous spectrum diagram

A continuous spectrum shows all frequencies of light

However, a line spectrum only shows certain frequencies

Helium spectrum diagram

The line spectrum of helium which shows only certain frequencies of light
This tells us that the emitted light from atoms can only be certain fixed frequencies - it is
quantised (quanta means 'little packet')
Electrons can only possess certain amounts of energy - they cannot have any energy
value

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 Your notes
Examiner Tips and Tricks
The formula that relates frequency and wavelength is printed in Section 1 of the IB
Chemistry Data Booklet so you don’t need to learn it
You will also find the speed of light and other useful constants in Section 2

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 Emission Spectra
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Emission Spectra
Electrons move rapidly around the nucleus in energy shells
If their energy is increased, then they can jump to a higher energy level
The process is reversible, so electrons can return to their original energy levels
When this happens, they emit energy
The frequency of energy is exactly the same, it is just being emitted rather than
absorbed:

Absorption and Emission diagram

The difference between absorption and emission depends on whether electrons are
jumping from lower to higher energy levels or the other way around
The energy they emit is a mixture of different frequencies
This is thought to correspond to the many possibilities of electron jumps between
energy shells
If the emitted energy is in the visible region, it can be analysed by passing it through a
diffraction grating
The result is a line emission spectrum

Line emission spectra

Spectrum of hydrogen diagram

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 The line emission (visible) spectrum of hydrogen
Each line is a specific energy value Your notes
This suggests that electrons can only possess a limited choice of allowed energies
These packets of energy are called 'quanta' (plural quantum)
What you should notice about this spectrum is that the lines get closer together towards
the blue end of the spectrum
This is called convergence and the set of lines is converging towards the higher energy
end, so the electron is reaching a maximum amount of energy
This maximum corresponds to the ionisation energy of the electron
These lines were first observed by the Swiss school teacher Johannes Balmer, and they
are named after him
We now know that these lines correspond to the electron jumping from higher levels
down to the second or n = 2 energy level
A larger version of the hydrogen spectrum from the infrared to the ultraviolet region
looks like this

Full hydrogen spectrum diagram

The full hydrogen spectrum

In the spectrum, we can see sets or families of lines
Balmer could not explain why the lines were formed - an explanation had to wait until the
arrival of Planck's Quantum Theory in 1900
Niels Bohr applied the Quantum Theory to electrons in 1913, and proposed that
electrons could only exist in fixed energy levels
The line emission spectrum of hydrogen provided evidence of these energy levels and it
was deduced that the families of lines corresponded to electrons jumping from higher
levels to lower levels

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Diagram to show the energy transitions for the hydrogen atom
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Electron jumps in the hydrogen spectrum

The findings helped scientists to understand how electrons work and provided the
backbone to our knowledge of energy levels, sublevels and orbitals
The jumps can be summarised as follows:

Electron Jumps & Energy Table

Jumps Region Energy

n∞→ n3 Infrared Low

n∞ → n 2 Visible ↓

n∞ → n 1 Ultraviolet High

Worked Example
Which electron transition in the hydrogen atom emits visible light?
A. n = 1 to n = 2

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B. n = 2 to n = 3
C. n = 2 to n = 1 Your notes
D. n = 3 to n = 2
Answer
Option D is correct
Emission in the visible region occurs for an electron jumping from any higher
energy level to n = 2

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 Energy Levels, Sublevels & Orbitals
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Energy Levels
What are electron shells?
The arrangement of electrons in an atom is called the electronic configuration
Electrons are arranged around the nucleus in principal energy levels or principal
quantum shells
Principal quantum numbers (n) are used to number the energy levels or quantum shells
The lower the principal quantum number, the closer the shell is to the nucleus
The higher the principal quantum number, the greater the energy of the electron
within that shell
Each principal quantum number has a fixed number of electrons it can hold
n = 1 : up to 2 electrons
n = 2 : up to 8 electrons
n = 3 : up to 18 electrons
n = 4 : up to 32 electrons
There is a pattern here - the mathematical relationship between the number of electrons
and the principal energy level is 2n2
So for example, in the third shell n = 3 and the number of electrons is 2 x (32) = 18

Principle quantum shells

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 Your notes

Electrons are arranged in principal quantum shells, which are numbered by principal
quantum numbers

What are subshells?

The principal quantum shells are split into subshells which are given the letters s, p and d
Elements with more than 57 electrons also have an f subshell
The energy of the electrons in the subshells increases in the order s < p < d
The order of subshells overlap for the higher principal quantum shells as seen in the
diagram below:

Principle Quantum Number and Sub-Shells

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 Your notes

Electrons are arranged in principal quantum shells, which are numbered by principal
quantum numbers

What are orbitals?

The subshells contain one or more atomic orbitals
Orbitals exist at specific energy levels and electrons can only be found at these specific
levels, not in between
Each atomic orbital can be occupied by a maximum of two electrons
The orbitals have specific 3D shapes:

The shape of s and p orbitals

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 Your notes

Representation of orbitals (the dot represents the nucleus of the atom) showing spherical
s orbitals (a), p orbitals containing ‘lobes’ along the x, y and z axis
Note that the shape of the d orbitals is not required for IB Chemistry

Summary of s and p orbitals

An overview of the shells, subshells and orbitals in an atom

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Ground state
The ground state is the most stable electronic configuration of an atom which has the Your notes
lowest amount of energy
This is achieved by filling the subshells of energy with the lowest energy first (1s)
This is called the Aufbau Principle
The order of the subshells in terms of increasing energy does not follow a regular pattern
at n = 3 and higher

The Aufbau Principle

The Aufbau Principle - following the arrows gives you the filling order

Sublevels & Orbitals

The principal quantum shells increase in energy with increasing principal quantum
number
E.g. n = 4 is higher in energy than n = 2
The subshells increase in energy as follows: s ＜ p ＜ d ＜ f
The only exception to these rules is the 3d orbital which has slightly higher energy
than the 4s orbital, so the 4s orbital is filled before the 3d orbital

Energy Levels

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 Your notes

Relative energies of the shells and subshells

Each shell can be divided further into subshells, labelled s, p, d and f
Each subshell can hold a specific number of orbitals:
s subshell : 1 orbital
p subshell : 3 orbitals labelled px, py and pz
d subshell : 5 orbitals
f subshell : 7 orbitals
Each orbital can hold a maximum number of 2 electrons so the maximum number of
electrons in each subshell is as follows:
s : 1 x 2 = total of 2 electrons
p : 3 x 2 = total of 6 electrons
d : 5 x 2 = total of 10 electrons
f : 7 x 2 = total of 14 electrons
In the ground state, orbitals in the same subshell have the same energy and are said to
be degenerate, so the energy of a px orbital is the same as a py orbital

Division of Shells Diagram

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 Your notes

Shells are divided into subshells which are further divided into orbitals
Summary of the Arrangement of Electrons in Atoms Table

rinciple Subshells Orbitals Orbitals Electrons Electrons

quantum possible per per per per shell
number, n (s, p, d, f) subshell principle subshell
(shell) quantum
number

1 s 1 1 2 2

2 s 1 4 2 8

p 3 6

3 s 1 9 2 18

p 3 6

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 d 5 10
Your notes
4 s 1 16 2 32

p 3 6

d 5 10

f 7 14

What is the shape of an s orbital?

The s orbitals are spherical in shape
The size of the s orbitals increases with increasing shell number
E.g. the s orbital of the third quantum shell (n = 3) is bigger than the s orbital of the
first quantum shell (n = 1)

s orbital diagram

The s orbitals become larger with increasing principal quantum number

What is the shape of a p orbital?

The p orbitals are dumbbell-shaped
Every shell has three p orbitals except for the first one (n = 1)
The p orbitals occupy the x, y and z axes and point at right angles to each other, so are
oriented perpendicular to one another
The lobes of the p orbitals become larger and longer with increasing shell number

p orbital diagram

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 Your notes

The p orbitals become larger and longer with increasing principal quantum number

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 Writing Electron Configurations
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Writing Electron Configurations
The electron configuration gives information about the number of electrons in each
shell, subshell and orbital of an atom
The subshells are filled in order of increasing energy

Electron Configuration Key

The electron configuration shows the number of electrons occupying a subshell in a

specific shell
Electrons can be imagined as small spinning charges which rotate around their own axis
in either a clockwise or anticlockwise direction

Spin pair repulsion diagram

Electrons can spin either in a clockwise or anticlockwise direction around their own axis.
The spin creates a tiny magnetic field with N-S pole pointing up or down, although you are
not required to know this for the exam
Electrons with the same spin repel each other which is also called spin-pair repulsion

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 Therefore, electrons will occupy separate orbitals in the same subshell first to
minimise this repulsion and have their spin in the same direction
Your notes
They will then pair up, with a second electron being added to the first p orbital, with
its spin in the opposite direction
This is known as Hund's Rule
E.g. if there are three electrons in a p subshell, one electron will go into each px, py
and pz orbital
Hund's Rule

Electron configuration: three electrons in a p subshell

The principal quantum number indicates the energy level of a particular shell but also
indicates the energy of the electrons in that shell
A 2p electron is in the second shell and therefore has an energy corresponding to n =
2
Even though there is repulsion between negatively charged electrons, they occupy the
same region of space in orbitals
An orbital can only hold two electrons and they must have opposite spin - the is known
as the Pauli Exclusion Principle
This is because the energy required to jump to a higher empty orbital is greater than the
inter-electron repulsion
For this reason, they pair up and occupy the lower energy levels first
The electron configuration can also be represented using the orbital spin diagrams
Each box represents an atomic orbital
The boxes are arranged in order of increasing energy from lower to higher (i.e. starting
from closest to the nucleus)
The electrons are represented by opposite arrows to show the spin of the electrons
E.g. the box notation for titanium is shown below

Electron box notation for titanium diagram

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 The electrons in titanium are arranged in their orbitals as shown. Electrons occupy the
lowest energy levels first before filling those with higher energy
Your notes
Writing out the electronic configuration tells us how the electrons in an atom or ion are
arranged in their shells, subshells and orbitals
This can be done using the full electron configuration or the shorthand version
The full electron configuration describes the arrangement of all electrons from the
1s subshell up
The shorthand electron configuration includes using the symbol of the nearest
preceding noble gas to account for however many electrons are in that noble gas,
followed by the rest of the electron configuration
Ions are formed when atoms lose or gain electrons
Negative ions are formed by adding electrons to the outer subshell
Positive ions are formed by removing electrons from the outer subshell
The transition metals fill the 4s subshell before the 3d subshell, but they also lose
electrons from the 4s first rather than from the 3d subshell
The Periodic Table is split up into four main blocks depending on their electronic
configuration:
s block elements (valence electron(s) in s orbital)
p block elements (valence electron(s) in p orbital)
d block elements (valence electron(s) in d orbital)
f block elements (valence electron(s) in f orbital)

s, p, d and f blocks in the Periodic Table

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 The elements can be divided into four blocks according to their outer shell electron
configuration
Your notes
Exceptions to the Aufbau Principle
Chromium and copper have the following electron configurations:
Cr is [Ar] 3d5 4s1 not [Ar] 3d4 4s2
Cu is [Ar] 3d10 4s1 not [Ar] 3d9 4s2
This is because the [Ar] 3d5 4s1 and [Ar] 3d10 4s1 configurations are energetically
favourable
By promoting an electron from 4s to 3d, these atoms achieve a half full or full d-subshell,
respectively

Worked Example
Write down the full and shorthand electron configuration of the following elements:
1. Potassium
2. Calcium
3. Gallium
4. Ca2+
Answer 1:
Potassium has 19 electrons so the full electronic configuration is:
1s2 2s2 2p6 3s2 3p6 4s1
The 4s orbital is lower in energy than the 3d subshell and is therefore filled first
The nearest preceding noble gas to potassium is argon which accounts for 18
electrons so the shorthand electron configuration is:
[Ar] 4s1
Answer 2:
Calcium has 20 electrons so the full electronic configuration is:
1s2 2s2 2p6 3s2 3p6 4s2
The 4s orbital is lower in energy than the 3d subshell and is therefore filled first
The shorthand version is [Ar] 4s2 since argon is the nearest preceding noble gas
to calcium which accounts for 18 electrons
Answer 3:
Gallium has 31 electrons so the full electronic configuration is:
Full: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1
Shorthand: [Ar] 3d10 4s2 4p1
Answer 4:
If you ionise calcium and remove two of its outer electrons, the electronic
configuration of the Ca2+ ion is identical to that of argon:

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 Ca2+ is 1s2 2s2 2p6 3s2 3p6
Ar is also 1s2 2s2 2p6 3s2 3p6 so the shorthand version is [Ar]
Your notes

Examiner Tips and Tricks

Orbital spin diagrams can be drawn horizontally or vertically, going up or down the
page - there is no hard and fast rule about this
The important thing is that you label the boxes and have the right number of
electrons shown
The arrows you use for electrons can be full or half-headed arrows, but they must
be in opposite directions in the same box.

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 Ionisation Energy from an Emission Spectrum (HL)
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Ionisation Energy from an Emission Spectrum
Emission Spectra
Electrons move rapidly around the nucleus in energy shells
Heat or electricity can be used to excite an electron to a higher main energy level
These range from n = 1 (ground state) to n = ∞
When the electrons 'fall' back down they must lose the energy difference between the
two energy levels. This loss of energy is performed by releasing electromagnetic energy
in the form of infrared, visible light or ultraviolet radiation.
When the electron falls back to n = 1 (ground state) the energy released is in the
ultraviolet region of the spectrum
This corresponds to the Lyman series

Diagram to show the release of a photon when an electron is

promoted

Promotion of an electron from the ground state (n=1) to n=2

Jumps in the hydrogen spectrum diagram

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 Your notes

Electron jumps in the hydrogen spectrum

This gives evidence for Bohr's model which is the idea that electrons exist in discrete
energy levels so an exact amount of energy is required for an electron to 'jump' an
energy level, a little like a ladder
There are however limitations to this model
Assumes positions of electrons are fixed
Assumes energy levels are spherical in nature
Bohr limited calculations to hydrogen only, so does not explain the line spectra of
other elements containing more than one electron

The Limit of Convergence

As the line spectra is produced the lines will become closer together
Where the lines appear to meet is called the limit of convergence
The convergence limit is the frequency at which the spectral lines converge
The energy required for an electron to escape the atom, or reach the upper limit of
convergence, is the ionisation energy
The frequency of the radiation in the emission spectrum at the limit of convergence can
be used to determine the first ionisation energy or IE1
In the Lyman series for the hydrogen atom (UV region), the frequency at the limit of
convergence relates to the energy given out when an electron falls from n = ∞ to n = 1
For hydrogen, the lines converge to a limit with a wavelength of 91.16 nm or 91.16 × 10−9 m

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Limit of Convergence diagram for hydrogen
Your notes

Lyman series (ultra-violet radiation) corresponds to transitions between higher shells and
the ground state (n=1)

Calculating First Ionisation Energy

When dealing with the Lyman series, the largest transitions represent the fall from the
infinite level to n=1
In reverse, it can be considered to be equal to the ionisation energy (note that ionisation
energy is given per mole of atoms)
Therefore, the first ionisation energy (IE1) of an atom can be calculated using the
frequency (or wavelength) of the convergence limit
We can do this by using the following equations
ΔE = h ν

c=νλ
In order to calculate the first ionisation energy, (IE1), we must first calculate the
frequency using the given data and rearranging:
c=νλ
as
ν=c÷λ

Once we know the frequency, we can use this to calculate the ionisation energy

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 E = Energy (J)
h = Planck's constant (6.63 x 10–34 J s) Your notes
v = frequency (s–1)
λ = wavelength (m)

c = speed of light (3.00 x 108 m s–1)

Worked Example
The convergence limit for the sodium atom has a frequency of 1.24 × 1015 s−1. Calculate
the first ionisation energy of sodium in kJ mol−1.
Answer:
Step 1: Write out the equation to calculate the first ionisation energy (IE1)
ΔE = h ν

Step 2: Substitute in numbers from question and data booklet to give energy change
per atom
IE1 = 6.63 × 10−34 × 1.24 × 1015
IE1 = 8.22 × 10−19 J atom−1
Step 3: Calculate the first ionisation energy per mole by multiplying by Avogadro's
constant
IE1 = 8.22 × 10−19 × 6.02 × 1023
IE1 = 494 916 J mol−1
Step 4: Convert J mol−1 to kJ mol−1 by dividing by 1000
IE1 = 495 kJ mol−1
So the first ionisation energy (IE1) of sodium has been calculated as 495 kJ mol−1

Worked Example
The convergence limit for the hydrogen atom has a wavelength of 91.16 nm. Calculate
the ionisation energy for hydrogen in kJ mol−1.
Answer:
Step 1: Calculate the frequency of the convergence limit, converting wavelength
into m (nm to m = × 10−9)
c=νλ

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 ν=c÷λ

ν = 3.00 × 108 ÷ 91.16 × 10−9 Your notes

ν = 3.29 × 1015 s−1

Step 2: Substitute into the equation to calculate IE1 for one atom of hydrogen in J
mol−1
ΔE = h ν

IE1 = 6.63 × 10−34 × 3.29 × 1015

IE1 = 2.18 × 10-18 J atom−1
Step 3: Calculate IE1 for 1 mole of hydrogen atoms
IE1 = 2.18 × 10−18 × 6.02 × 1023
IE1 = 1 313 491 J mol−1
Step 4: Convert J mol−1 to kJ mol−1
IE1 = 1313 kJ mol−1
So the first ionisation energy (IE1) of hydrogen has been calculated as 1313 kJ mol−1

Examiner Tips and Tricks

These equations are found in the data booklet so you don't need to learn them
Also, be careful to calculate the first ionisation energy (IE1) per mole by using
Avogadro's constant (NA) 6.02 × 1023 and converting units to kJ mol−1
Finally, when working through calculations, keep the numbers in your calculator to
avoid rounding up too early.

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 Successive Ionisation Energies (HL)
Your notes
Successive Ionisation Energies
Successive ionisation energies of an element
The successive ionisation energies of an element increase
This is because once you have removed the outer electron from an atom, you have
formed a positive ion
Removing an electron from a positive ion is more difficult than from a neutral atom
As more electrons are removed, the attractive forces increase due to decreasing
shielding and an increase in the proton to electron ratio
The increase in ionisation energy, however, is not constant and is dependent on the
atom’s electronic configuration
Taking calcium as an example:

Ionisation Energies of Calcium Table

Electronic 1s2 2s2 2p6 3s2 1s2 2s2 2p6 3s2 1s2 2s2 2p6 1s2 2s2 2p6
Configuration 3p6 4s2 3p6 4s1 3s2 3p6 3s2 3p5

IE First Second Third Fourth

IE (kJ mol-1) 590 1150 4940 6480

Successive Ionisation Energies of Calcium

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 Graph to show the successive ionisation energies for the element calcium
The first electron removed has a low IE1 as it is easily removed from the atom due to the Your notes
spin-pair repulsion of the electrons in the 4s orbital
The second electron is more difficult to remove than the first electron as there is
no spin-pair repulsion
The third electron is much more difficult to remove than the second one corresponding
to the fact that the third electron is in a principal quantum shell which is closer to the
nucleus (3p)
Removal of the fourth electron is more difficult as the orbital is no longer full, and there is
less spin-pair repulsion
The graph shows there is a large increase in successive ionisation energy as the
electrons are being removed from an increasingly positive ion
The big jumps on the graph show the change of shell and the small jumps are the change
of subshell
Successive ionisation data can be used to:
Predict or confirm the simple electronic configuration of elements
Confirm the number of electrons in the outer shell of an element
Deduce the group an element belongs to in the Periodic Table
By analysing where the large jumps appear and the number of electrons removed when
these large jumps occur, the electron configuration of an atom can be determined
Na, Mg and Al will be used as examples to deduce the electronic configuration and
positions of elements in the Periodic Table using their successive ionisation energies

Successive Ionisation Energies Table

Element Atomic Number Ionisation Energy (kJ mol-1)

First Second Third Fourth

Na 11 494 4560 6940 9540

Mg 12 736 1450 7740 10500

Al 13 577 1820 2740 11600

Sodium
For sodium, there is a huge jump from the first to the second ionisation energy,
indicating that it is much easier to remove the first electron than the second

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 Therefore, the first electron to be removed must be the last electron in
the valence shell thus Na belongs to group I
Your notes
The large jump corresponds to moving from the 3s to the full 2p subshell
Na 1s2 2s2 2p6 3s1
Magnesium
There is a huge increase from the second to the third ionisation energy, indicating that it
is far easier to remove the first two electrons than the third
Therefore the valence shell must contain only two electrons indicating that magnesium
belongs to group II
The large jump corresponds to moving from the 3s to the full 2p subshell
Mg 1s2 2s2 2p6 3s2

Aluminium
There is a huge increase from the third to the fourth ionisation energy, indicating that it is
far easier to remove the first three electrons than the fourth
The 3p electron and 3s electrons are relatively easy to remove compared with the 2p
electrons which are located closer to the nucleus and experience
greater nuclear charge
The large jump corresponds to moving from the third shell to the second shell
Al 1s2 2s2 2p6 3s2 3p1

Worked Example
Values for the successive IEs of an unknown element are:
IE1 = 899 kJ mol-1,
IE2 = 1757 kJ mol-1,
IE3 = 14850 kJ mol-1,
IE4 = 21005 kJ mol-1
Deduce which group of the periodic table of elements you would expect to find the
unknown element.
Answer:
The largest jump is between IE2 and IE3 which will correspond to a change in energy
level. Therefore, the unknown element must be in group 2.

Worked Example
The table shows successive ionisation energies for element X in period 2.

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 Successive Ionisation Energies of Unknown Element
Ionisation number 1 2 3 4 5 6 7 8 Your notes

Ionisation energy (kJ 1314 3388 5301 7469 1089 13327 71337 84080
mol-1)

Identify element X.
Answer:
The largest jump in ionisation energy is between IE6 and IE7 meaning that the 7th
electron is being removed from an energy level closer to the nucleus
Therefore, element X must be group 16 (6)
If element X is in group 16 (6) and in period 2, it must be oxygen

Ionisation energies show periodicity – a trend across a period of the Periodic Table
As could be expected from their electron configuration, the Group 1 metals have a
relatively low ionisation energy, whereas the noble gases have very high ionisation
energies
The size of the first ionisation energy is affected by four factors:
Size of the nuclear charge
Distance of outer electrons from the nucleus
Shielding effect of inner electrons
Spin-pair repulsion
First ionisation energy increases across a period and decreases down a group

Ionisation Energies of Hydrogen to Sodium

A graph showing the ionisation energies of the elements hydrogen to sodium

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Ionisation energy across a period
The ionisation energy across a period generally increases due to the following factors: Your notes
Across a period the nuclear charge increases
This causes the atomic radius of the atoms to decrease, as the outer shell is pulled
closer to the nucleus, so the distance between the nucleus and the outer
electrons decreases
The shielding by inner shell electrons remain reasonably constant as electrons are
being added to the same shell
It becomes harder to remove an electron as you move across a period; more
energy is needed
So, the ionisation energy increases

Dips in the trend

There is a slight decrease in IE1 between beryllium and boron as the fifth electron in
boron is in the 2p subshell, which is further away from the nucleus than the 2s subshell of
beryllium
Beryllium has a first ionisation energy of 900 kJ mol-1 as its electron configuration
is 1s2 2s2
Boron has a first ionisation energy of 800 kJ mol-1 as its electron configuration
is 1s2 2s2 2px1
There is a slight decrease in IE1 between nitrogen and oxygen due to spin-pair
repulsion in the 2px orbital of oxygen
Nitrogen has a first ionisation energy of 1400 kJ mol-1 as its electron configuration is
1s2 2s2 2px1 2py1 2pz1
Oxygen has a first ionisation energy of 1310 kJ mol-1 as its electron configuration is
1s2 2s2 2px2 2py1 2pz1
In oxygen, there are 2 electrons in the 2px orbital, so the repulsion between those
electrons makes it slightly easier for one of those electrons to be removed

From one period to the next

There is a large decrease in ionisation energy between the last element in one period,
and the first element in the next period
This is because:
There is increased distance between the nucleus and the outer electrons as you
have added a new shell
There is increased shielding by inner electrons because of the added shell
These two factors outweigh the increased nuclear charge

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