Successive Ionisation Energies (HL) (DP IB Chemistry): Revision Note
Successive ionisation energies
Successive ionisation energies of an element increase
This is because once the outer electron is removed, the atom becomes a positive ion, making further electron removal more difficult
As more electrons are removed:
Shielding decreases
The proton-to-electron ratio increases
Attraction between nucleus and remaining electrons increases
The increase is not constant, it depends on the electronic configuration
Examiner Tips and Tricks
Databases such as NIST or the IB Chemistry Data Booklet (Section 9) are useful for compiling successive ionisation energy values and plotting graphs to identify group trends.
Taking calcium as an example:
Ionisation energies of calcium table
Electronic Configuration | 1s2 2s2 2p6 3s2 3p6 4s2 | 1s2 2s2 2p6 3s2 3p6 4s1 | 1s2 2s2 2p6 3s2 3p6 | 1s2 2s2 2p6 3s2 3p5 |
---|---|---|---|---|
IE | First | Second | Third | Fourth |
IE (kJ mol-1) | 590 | 1150 | 4940 | 6480 |
Successive ionisation energies of calcium

The first electron is easy to remove due to spin-pair repulsion in the 4s orbital, resulting in a low first ionisation energy (IE₁)
The second electron is harder to remove as there’s no spin-pair repulsion
The third electron is much harder to remove as it comes from the 3p subshell, which is closer to the nucleus
The fourth electron is also difficult to remove as there is less spin-pair repulsion in the 3p orbital
Successive ionisation energies increase as electrons are removed from an increasingly positive ion
Large jumps in IE show a change in shell, while smaller jumps show changes within a subshell
Uses of successive ionisation energy data
Predict or confirm the electronic configuration
Identify the number of outer electrons
Determine the group number in the Periodic Table
Large jumps help identify the shell where electrons are being removed
Used for deducing configurations of Na, Mg, Al and placing them in the Periodic Table
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
A large increase between the first and second ionisation energies shows it is much easier to remove the first electron
This means the first electron removed is from the valence shell, so Na is in Group 1
The jump corresponds to removing an electron from the 3s to the full 2p subshell
Na: 1s2 2s2 2p6 3s1
Magnesium
A large increase between the second and third ionisation energies suggests the first two electrons are easier to remove
Therefore, Mg has two valence electrons, placing it in Group 2
The jump corresponds to removing an electron from the 3s to the full 2p subshell:
Mg: 1s2 2s2 2p6 3s2
Aluminium
A large increase between the third and fourth ionisation energies shows the first three electrons are easier to remove
These are the 3p and 3s electrons, which are farther from the nucleus and experience less nuclear charge than 2p electrons
This suggests Al has three valence electrons, so it belongs to Group 13 (Group III)
The jump corresponds to removing an electron 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.
Successive ionisation energies of an unknown element
Ionisation number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
---|---|---|---|---|---|---|---|---|
Ionisation energy (kJ mol-1) | 1314 | 3388 | 5301 | 7469 | 1089 | 13327 | 71337 | 84080 |
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 energy shows periodicity; a repeating trend across a period in the Periodic Table
Group 1 metals have low first ionisation energies; noble gases have very high values due to stable configurations
Four key factors affect first ionisation energy:
Nuclear charge (more protons = stronger attraction)
Distance of outer electrons from the nucleus
Shielding by inner electrons
Spin–pair repulsion between paired electrons
Across a period:
first ionisation energy increases due to higher nuclear charge and same shielding
Down a group:
first ionisation energy decreases due to greater atomic radius and more shielding
Ionisation energies of hydrogen to sodium

Ionisation energy across a period
Across a period, ionisation energy increases because:
Nuclear charge increases, pulling electrons closer
Atomic radius decreases, reducing the distance to outer electrons
Shielding stays roughly constant, since electrons are added to the same shell
Outer electrons are held more tightly, so more energy is needed to remove them
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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