Ionisation Energies

The first ionisation energy is the enthalpy change when one mole of gaseous atoms loses one electron per atom to form one mole of gaseous ions with a single positive charge.

Equation (general form):

X(g) → X⁺(g) + e⁻

• Always defined in the gaseous state.

• Units: kJ mol⁻¹.

The attraction between the nucleus and an electron can be described using

Coulomb’s Law:

F ∝ 1 / r²

Where:

• F = attractive

• r = distance between nucleus and electron.

1. Nuclear Charge

• More protons = stronger positive nuclear charge.

Worked Example 1:

Compare H (Z = 1) and He (Z = 2).

• Both outer electrons are in the 1s orbital, so similar distance (r) and shielding.

• He has double the number of protons than H.

• So the attraction between the electrons and nucleus in He is stronger than that of H

• Consequence: helium has a much higher ionisation energy than hydrogen.

2. Distance from the Nucleus (r in the denominator)

• Electrons further away from the nucleus experience less attraction.

• Because F ∝ 1/r², even a small increase in r causes a big decrease in attraction.

Worked Example 2:

Compare Li (2,1) and Na (2,8,1).

• Both have outer electrons in an s orbital, but Li’s is 2s, Na’s is 3s (further from nucleus).

• Despite Na having more protons, the increased r and shielding weaken the attraction.

• Consequence: Na has a lower ionisation energy than Li.

3. Shielding (Effective Nuclear Charge)

• Inner shells repel outer electrons, reducing the attraction from the nucleus.

• Coulomb’s Law still applies, but the effective nuclear charge is less than the actual nuclear charge (number of protons).

Worked Example 3:

Compare Li (1s² 2s¹) and K

• In potassium, increase in proton number (nuclear charge) is offset by the increased distance of the outer electron to the nucleus

• And the outer electron is more shielded by the core electrons

• Resulting in a weaker force of attraction between the valence electron and positive nucleus

• So less energy is required to remove it so a lower ionisation energy

4. Electron Repulsion in Orbitals

• Two electrons sharing the same orbital repel each other.

• Repulsion makes it easier to remove one electron.

Worked Example 4:

• N (1s² 2s² 2p³): three p orbitals singly occupied.

• O (1s² 2s² 2p⁴): one orbital now has a pair, causing repulsion.

• Consequence: oxygen has a lower ionisation energy than nitrogen.

Across a Period

• Ionisation energy generally increases.

• More protons

• Outer electrons in the same shell (r roughly constant, shielding unchanged).

• Consequence: stronger nuclear attraction → more energy required to remove electrons.

Down a Group

• Ionisation energy decreases.

• Electrons are further from nucleus (r increases).

• More shielding from inner shells.

• Although nuclear charge increases, distance and shielding dominate.

Key Anomalies

1. Be → B (Mg → Al)

• B’s outer electron enters a new p orbital (higher in energy, further from nucleus).

2. N → O (P → S)

• O’s p orbital now has a paired electron.

• Extra electron–electron repulsion lowers ionisation energy.

Worked Example

Question: Why does Na have a much lower first ionisation energy than Ne?

Step 1 – Compare shells

• Ne: outer electrons in 2p (close to nucleus).

• Na: outer electron in 3s (further away).

Step 2 – Consider shielding

• Na’s outer electron shielded by 2 inner shells.

• Ne’s only shielded by 1 inner shell.

Final Answer: Na’s electron is further away from the positive nucleus and more shielded by the core electrons, so easier to remove.

Explain why the second ionisation energy of calcium is greater than the first.

• The first ionisation energy of calcium involves removing an electron from a neutral atom

  (Ca → Ca⁺ + e⁻).

• The second ionisation energy involves removing an electron from a Ca⁺ ion

  (Ca⁺ → Ca²⁺ + e⁻).

• The Ca⁺ ion has a higher effective nuclear charge and a smaller radius, compared to the neutral atom, making the second electron harder to remove.​

• This means that the second ionisation energy of calcium will be higher as it is harder to remove an electron from an already positive ion than it is to remove an electron from a neutral atom

Successive Ionisation Energies

What does “successive” mean?

• Successive means one after another, in sequence.

• Successive ionisation energies are the series of values you get if you keep removing electrons from the same atom/ion, step by step.

Example:

• First IE (IE₁): X(g) → X⁺(g) + e⁻

• Second IE (IE₂): X⁺(g) → X²⁺(g) + e⁻

• Third IE (IE₃): X²⁺(g) → X³⁺(g) + e⁻

…and so on. Each one is harder, because the ion left behind is more positively charged and holds on more strongly to its remaining electrons.

Why do successive ionisation energies increase?

Nuclear charge stays the same – same number of protons. But each electron is harder to remove, because the ion left behind is more positively charged and holds on more strongly to the remaining electrons. The result is that each successive IE is bigger than the last.

Using Successive Ionisation Energies to Find Group Number

The key idea:

• Small increases between successive values = removing electrons from the same shell.

• One very large jump = you’ve moved into a new, inner shell (closer to nucleus, much stronger attraction).

Worked Example 1 (Magnitude of Jump)

Ionisation energies (kJ mol⁻¹):

IE₁ = 600, IE₂ = 1200, IE₃ = 5000, IE₄ = 7000, IE₅ = 9000

• Between IE₁ and IE₂ → energy roughly doubles (small increase, same shell).

• Between IE₂ and IE₃ → energy suddenly jumps by more than 4×.

• Interpretation → first two electrons were in outer shell, next electron comes from inner shell.

• Conclusion: Element is in Group 2 (two outer electrons).

Worked Example 2 (Orders of Magnitude)

Ionisation energies (kJ mol⁻¹):

• IE₁ = 800, IE₂ = 1600, IE₃ = 2500, IE₄ = 11000, IE₅ = 15000…

• IE₁ → IE₂ → IE₃ → gradual increases (outer shell).

• IE₃ → IE₄ → sudden jump of more than 4× (order of magnitude).

• Big jump after the 3rd IE → 3 outer electrons.

• Conclusion: Element is in Group 13 (3 outer electrons).

How to Spot the “Big Jump” Step by Step

1. Look down the list of IEs.

2. Compare each increase – is it gradual (same shell) or sudden, huge (new shell)?

3. The number of electrons before the jump = group number.