Successive Ionisation Energies

Learning Objectives (OCR Spec 3.1.1)

By the end of this lesson you should be able to:

• Define successive ionisation energy (second, third, n-th)
• Use patterns in successive ionisation energies to deduce the electron shell structure of an element
• Use successive ionisation energies to determine the group of an unknown element
• Explain why successive ionisation energies always increase
• Explain the large jumps that occur when moving from one shell to another

Defining Successive Ionisation Energies

The n-th ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous n-1 positive ions to form one mole of gaseous n positive ions.

First: X(g) → X+(g) + e- ΔH = IE1 Second: X+(g) → X2+(g) + e- ΔH = IE2 Third: X2+(g) → X3+(g) + e- ΔH = IE3 ... n-th: X^(n-1)+(g) → X^n+(g) + e- ΔH = IEn

Each successive ionisation energy is larger than the previous one because:

1. After each electron is removed, the remaining electrons experience a greater effective nuclear charge per electron (same protons, fewer electrons)
2. The ion becomes smaller, so remaining electrons are closer to the nucleus and held more tightly
3. Shielding from the departing electron is lost

Evidence for Shell Structure

While successive IEs always rise, they do not rise smoothly. There are large jumps when the next electron must be removed from an inner shell, because that electron is much closer to the nucleus and less shielded.

Example: Sodium (Z = 11)

Na has configuration 1s2 2s2 2p6 3s1 - that is, 1 electron in shell 3, 8 in shell 2, 2 in shell 1.

Ionisation	IE / kJ mol-1	Electron removed from
1st	496	3s (shell 3)
2nd	4563	2p (shell 2)
3rd	6913	2p
4th	9544	2p
5th	13352	2p
6th	16611	2s
7th	20115	2s
8th	25491	2p? (no, 2s completed)
9th	28934	1s
10th	141367	1s
11th	159076	1s (very last e-)

The really dramatic features are:

• A huge jump between IE1 (496) and IE2 (4563) - roughly 9x increase - because IE1 removes the lone 3s electron, while IE2 must come from the much closer, less shielded 2p shell
• Another large jump between IE9 and IE10 - because IE10 comes from the 1s (innermost) shell

Counting the Shells

If you plot log(IE) against ionisation number, you see flat(-ish) sections separated by sudden rises. The number of electrons in each flat section tells you the number of electrons in each shell:

• Shell 3: 1 electron (IE1 only) → matches 3s1
• Shell 2: 8 electrons (IE2 - IE9) → matches 2s2 2p6
• Shell 1: 2 electrons (IE10 - IE11) → matches 1s2

Total: 1 + 8 + 2 = 11 electrons → Z = 11, sodium.

Using Successive IEs to Identify an Unknown Group

If you are not told the identity of the element, you can still work out its group by looking at where the first big jump occurs: