Successive Ionisation Energies (HL) (DP IB Chemistry): Revision Note

Philippa Platt

Written by: Philippa Platt

Reviewed by: Richard Boole

Updated on

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

Graph showing ionisation energy versus electrons removed. Labels indicate shell changes with big jumps and subshell changes with small jumps.
Graph to show the successive ionisation energies for the element 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 

Graph of first ionisation energy versus atomic number ranging from 0 to 11, showing peaks at helium and neon with a red line connecting data points.
A graph showing the ionisation energies of the elements 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 2porbital, 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

You've read 0 of your 5 free revision notes this week

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Philippa Platt

Author: Philippa Platt

Expertise: Chemistry Content Creator

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

Richard Boole

Reviewer: Richard Boole

Expertise: Chemistry Content Creator

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.