Ionisation energy

Defining First Ionisation Energy

The first ionisation energy (IE₁) is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous +1 ions.

Every part of this definition earns a separate mark. The species must be gaseous, the electrons must come from atoms (not ions), and the quantities must be expressed per mole. The unit of ionisation energy is kJ mol⁻¹.

Ionisation energies arise because of the electrostatic attraction between the positively charged nucleus and the negatively charged outer electron. Energy must be supplied to overcome this attraction and pull the electron away to infinity. Ionisation is therefore always endothermic, so values are always positive.

The general equation for the first ionisation energy of element X is:

State symbols are essential. Writing X without (g) loses the mark, even if the rest of the equation is correct.

Examiner InsightExaminers check three things in the IE definition: “gaseous”, “one mole”, and “+1 ions”. Missing any one of these costs a mark, even if the meaning is otherwise clear. Exam cue: Always include “gaseous” and “per mole” when defining IE.

Equations for Successive Ionisation Energies

Successive ionisation energies describe the removal of electrons one at a time from the same atom, after the first electron has already been removed.

The second ionisation energy (IE₂) removes an electron from the +1 ion to form a +2 ion. The third ionisation energy (IE₃) removes an electron from the +2 ion to form a +3 ion. Each successive value is larger than the previous one.

For magnesium, the first three ionisation energies are represented as:

Each equation must show only one electron being removed and must include all state symbols. The species on the left is always gaseous, and the product is always one charge unit higher.

MisconceptionSuccessive ionisation energies do not remove electrons from a neutral atom each time. IE₂ removes an electron from a +1 ion, not from the original atom. Writing X(g) → X²⁺(g) + 2e⁻ is wrong for IE₂. Exam cue: One electron per ionisation step, with charge increasing by +1 each time.

Factors Influencing Ionisation Energy

Four factors determine the magnitude of an ionisation energy: nuclear charge, atomic/ionic radius, shielding by inner electrons, and spin-pair repulsion.

Nuclear charge is the total positive charge from the protons in the nucleus. A higher nuclear charge increases the attraction on the outer electron, so more energy is required to remove it. This raises the ionisation energy.

Atomic radius is the distance between the nucleus and the outer electron. A larger radius weakens the electrostatic attraction because attraction decreases sharply with distance. This lowers the ionisation energy.

Shielding is the repulsion of the outer electron by inner-shell electrons, which reduces the effective nuclear charge felt by the outer electron. More inner shells produce more shielding and lower the ionisation energy.

Spin-pair repulsion occurs when two electrons occupy the same orbital. The two paired electrons repel each other, which makes one of them easier to remove. This lowers the ionisation energy compared to an unpaired electron in a similar environment.

Trends Across a Period and Down a Group

The general trend across a period is an increase in first ionisation energy, while the trend down a group is a decrease.

Across Period 2 or Period 3, the nuclear charge increases as protons are added. Electrons enter the same outer shell, so shielding stays roughly constant. The atomic radius decreases because the stronger nuclear pull contracts the shell. Together these effects raise the ionisation energy.

Two dips break this rising trend and must be explained:

Dip from Group 2 to Group 13 (e.g. Mg to Al). The outer electron in Al is in a 3p orbital, which is higher in energy and slightly further from the nucleus than the 3s orbital in Mg. The 3p electron is also shielded by the filled 3s sub-shell. Less energy is therefore required to remove it.

Dip from Group 15 to Group 16 (e.g. P to S). In P, the three 3p electrons each occupy separate orbitals. In S, the fourth 3p electron must pair up in an orbital already containing one electron. The resulting spin-pair repulsion makes this paired electron easier to remove.

Down a group, the first ionisation energy decreases. Each element down the group has an additional electron shell, so the atomic radius increases and shielding by inner shells increases. Both effects weaken the attraction on the outer electron, which outweighs the increase in nuclear charge.

Successive Ionisation Energies and Electron Shells

Successive ionisation energies of a single element show large jumps that reveal the shell structure of the atom.

Each electron removed comes from a higher-energy position than the last, because the remaining ion has a greater positive charge attracting fewer electrons. The increase between consecutive values is gradual within a single shell or sub-shell.

A large jump occurs when an electron must be removed from a shell closer to the nucleus. This new shell has less shielding and a smaller radius, so the attraction is much stronger. The jump indicates that all electrons in the outer shell have already been removed.

For sodium (1s² 2s² 2p⁶ 3s¹), the first ionisation energy is small because the 3s¹ electron is far from the nucleus and well shielded. The second ionisation energy is much larger because the next electron is removed from the 2p sub-shell, which is closer to the nucleus.

Deducing Electronic Configuration and Position

Successive ionisation energy data can be used to deduce both the group number and the electron configuration of an element.

The number of electrons removed before the first large jump equals the number of electrons in the outer shell. This number gives the group number of the element directly. For Group 1 there is one electron before the jump, for Group 2 there are two, and so on up to Group 18.

The pattern of jumps reveals the full shell structure. A jump after one electron, then after a further eight electrons, then after a further eight, indicates a 2,8,1 configuration — characteristic of sodium.

Worked Example: Deducing the Identity of Element X

Scenario

Element X has successive ionisation energies (kJ mol⁻¹): 786, 1580, 3230, 4360, 16100, 19800, 23800. Deduce the group and identify X, given that X is in Period 3.

Step-by-step solution

Step 1: Identify the first large jump. The values rise gradually from 786 to 4360. The jump from 4360 to 16100 is unusually large.

$$\frac{16100}{4360}\approx 3.7$$

Step 2: Count the electrons removed before the jump. Four electrons (IE₁ to IE₄) are removed before the large jump.

Step 3: Deduce the group number. The outer shell of X contains four electrons. X is therefore in Group 14.

Step 4: Identify the element. A Period 3 element in Group 14 is silicon (Si).

Interpretation

The fifth electron requires far more energy because it comes from the inner 2p sub-shell, which is closer to the nucleus and less shielded. This confirms the configuration 1s² 2s² 2p⁶ 3s² 3p².