Ionisation Energy

First ionisation energy (IE) — the energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.

$\text{X}(g) \rightarrow \text{X}^+(g) + e^-$

Four words earn the marks: one mole, gaseous (g), 1+ ion, and per electron removed.

Successive ionisation energies remove electrons one at a time from the increasingly positive ion:

• 1st IE: $\text{Mg}(g) \rightarrow \text{Mg}^+(g) + e^-$
• 2nd IE: $\text{Mg}^+(g) \rightarrow \text{Mg}^{2+}(g) + e^-$
• 3rd IE: $\text{Mg}^{2+}(g) \rightarrow \text{Mg}^{3+}(g) + e^-$

Successive IEs always increase, because each electron is pulled away from an ion with one more net positive charge and a smaller radius, so the attraction is stronger.

Ionisation energy reflects the attraction between the nucleus and the outer electron being removed. Four factors decide how strong that attraction is:

Factor	Effect on IE
Nuclear charge (number of protons)	Greater charge → stronger attraction → higher IE
Atomic radius / distance of outer electron	Greater distance → weaker attraction → lower IE
Shielding by inner shells	More inner shells/electrons → weaker net attraction → lower IE
Spin-pair repulsion	Paired electrons in the same orbital repel → easier to remove → lower IE

Every trend explanation in this topic is just a competition between these factors — say which factor wins and why.

General trend across Period 3 (Na → Ar): IE increases.

• Nuclear charge increases (more protons).
• Outer electrons enter the same shell, so shielding is roughly constant and the atomic radius decreases.
• The outer electron is held more tightly → IE rises.

But there are two dips you must be able to explain:

Dip 1 — Mg → Al (Group 2 → 13). Mg loses a $3s$ electron; Al loses a $3p$ electron. The 3p sub-shell is higher in energy than 3s (further out and slightly shielded by the 3s electrons), so Al's outer electron is easier to remove → IE drops.

Dip 2 — P → S (Group 15 → 16). P is $3p^3$ (three singly-occupied 3p orbitals — no pairing). S is $3p^4$, so one 3p orbital now holds a pair of electrons. Spin-pair repulsion makes one of that pair easier to remove → IE drops.

Down a group, first ionisation energy decreases. Compare the factors:

• The outer electron is in a new, higher shell each step down → greater distance from the nucleus.
• There are more inner shells → more shielding of the outer electron from the nuclear charge.
• Although nuclear charge increases, the increases in distance and shielding dominate, so the outer electron is less strongly held → IE falls.

Example: $\text{IE}(\text{Li}) > \text{IE}(\text{Na}) > \text{IE}(\text{K})$.

Plotting (or listing) the successive IEs of one element shows a steady rise with big jumps at certain points. A large jump occurs when the next electron must come from a shell closer to the nucleus (smaller radius, less shielding, so much harder to remove).

How to deduce the Group / configuration:

• Count how many electrons are removed before the first big jump — that is the number of outer-shell (valence) electrons, which gives the Group.

Worked. An element has successive IEs (kJ mol⁻¹): 590, 1150, 4940, 6480, 8120, …

• The huge jump comes after the 2nd electron (1150 → 4940).
• So there are 2 outer electrons → Group 2 → outer configuration $\dots ns^2$ (this pattern fits calcium).

Reading the shell structure. The pattern of jumps maps the shells. For sodium ($2,8,1$) the successive IEs jump after the 1st electron (1 in the outer shell), then after the next 8 (the second shell), revealing $2, 8, 1$.