Ionisation energy is a quantitative measure of how strongly an atom holds onto its electrons. Studying how these energies change across elements provides solid evidence for the existence of principal shells and sub-shells.
First Ionisation Energy
The energy required to remove one mole of the most loosely held electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.
All ionisation processes are endothermic (\( \Delta H \) is positive) because energy must be supplied to overcome the electrostatic attraction between the negatively charged electron and the positively charged nucleus.
Always include state symbols in ionisation energy equations. Both the starting reactant and the resulting ion must be in the gaseous state, i.e. (g). Leaving out state symbols will cost you the mark in the exam.
Factors Affecting Ionisation Energy
The magnitude of the attraction between the nucleus and the outermost electron is determined by three main factors:
- Nuclear Charge: The number of protons in the nucleus. More protons mean a larger positive charge, which exerts a stronger pull on the outer electrons, increasing ionisation energy.
- Electron Shielding: Inner shells of electrons repel the outer shell electrons. This shielding reduces the net electrostatic attraction felt by outer electrons, decreasing ionisation energy.
- Atomic Radius (Distance): The distance of the outer electron from the positive nucleus. Electrostatic attraction drops off rapidly with distance. A larger radius means weaker attraction, decreasing ionisation energy.
Trends in First Ionisation Energy
1. Trend Down a Group
First ionisation energy decreases down any group. Although the nuclear charge increases (more protons), this is completely outweighed by two factors:
- The outer electron is in a **new shell**, so it is **further away** from the nucleus.
- There is **more shielding** from inner electron shells.
As a result, the electrostatic attraction is weaker, and the electron is easier to remove.
2. Trend Across a Period
First ionisation energy generally increases across a period. This is because:
- The nuclear charge increases (more protons).
- The shielding remains relatively constant (electrons are added to the same principal shell).
- Therefore, the atomic radius decreases as the outer shell is pulled closer, increasing the attraction on the outer electron.
Exceptions to the Period Trend (Crucial for Exams)
While the overall trend across Period 3 is an increase, there are two distinct dips that provide key evidence for sub-shell structure:
A. The Dip at Aluminium (Group 13)
Magnesium's configuration is [Ne] 3s². Aluminium's configuration is [Ne] 3s² 3p¹.
Al's outer electron resides in a 3p sub-shell, which is higher in energy than the 3s sub-shell. It is also shielded by the 3s² electrons. This shielding and higher initial energy make it easier to remove, causing the dip.
B. The Dip at Sulfur (Group 16)
Phosphorus's configuration is [Ne] 3s² 3p³. Sulfur's configuration is [Ne] 3s² 3p⁴.
In P, the 3p orbitals are singly occupied (Hund's rule). In S, one of the 3p orbitals contains paired electrons. The mutual repulsion between the two paired electrons in the same orbital makes it easier to remove one of them.
Successive Ionisation Energies
Successive ionisation energies are the energies required to remove subsequent electrons from the same atom one after the other:
\[ \text{1st: } \text{X(g)} \rightarrow \text{X}^+\text{(g)} + \text{e}^- \]
\[ \text{2nd: } \text{X}^+\text{(g)} \rightarrow \text{X}^{2+}\text{(g)} + \text{e}^- \]
\[ \text{3rd: } \text{X}^{2+}\text{(g)} \rightarrow \text{X}^{3+}\text{(g)} + \text{e}^- \]
Successive ionisation energies always increase because as electrons are removed, the remaining electrons experience a stronger net pull from the positive nucleus (reduced shielding and electron repulsion).
Identifying Shell Structure from Jumps
When plotting successive ionisation energies on a logarithmic scale, we observe **sharp jumps** in energy. A large jump indicates that an electron is being removed from a shell that is **closer to the nucleus** (a new principal energy level) with significantly less shielding.
Step 1: Look for the largest ratio jump between successive values:
- Ratio 2nd / 1st: \( \frac{1817}{578} = 3.14 \)
- Ratio 3rd / 2nd: \( \frac{2745}{1817} = 1.51 \)
- Ratio 4th / 3rd: \( \frac{11578}{2745} = 4.22 \) (A very large jump!)
- Ratio 5th / 4th: \( \frac{14831}{11578} = 1.28 \)
Step 2: Interpret the jump:
The large jump between the 3rd and 4th ionisation energies indicates that the 4th electron is being removed from a inner shell, which is closer to the nucleus and less shielded. This means the atom has **3 electrons in its outer shell**.
Step 3: Deduce element:
An element with 3 outer electrons belongs to Group 13. Since it is in Period 3, the element is Aluminium (Al).
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