2.1.3Physical Chemistry
Trends of Period 3 Elements: First Ionisation Energy — AQA AS Chemistry Revision Notes & Topical Questions
5 min read 4 parts Lesson 54 of 102
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 4The General Increase
First ionisation energy is the energy needed to remove one electron from each atom in one mole of gaseous atoms: X(g) → X⁺(g) + e⁻.
Nuclear charge
- Nuclear charge Moving across the period, the number of protons increases, so the attraction to outer electrons becomes stronger.
Constant shell
- Constant shell Outer electrons stay in the same shell, so shielding hardly changes and the atomic radius falls.
Greater energy needed
- Greater energy needed A stronger pull from a closer nucleus means more energy is needed to remove the outermost electron.
Overall trend
- Overall trend First ionisation energy rises from sodium to argon, apart from two small decreases.
Part 2 of 4Two Dips in the Pattern
Two Dips in the Pattern
Mg to Al dip
- Mg to Al dip Aluminium loses a 3p electron,
- which is higher in energy and slightly more shielded by the 3s electrons than the 3s electron in magnesium.
P to S dip
- P to S dip Sulfur has a pair of electrons in one 3p orbital, and the repulsion between them makes one easier to remove than in phosphorus.
Rest of the pattern
- Rest of the pattern Between the dips, the increasing nuclear charge dominates, so ionisation energy keeps rising to argon.
| Element | First IE (kJ mol⁻¹) | Comment |
|---|---|---|
| Na | 496 | Lowest in the period |
| Mg | 738 | Rises: 3s² more tightly held |
| Al | 578 | Falls: 3p electron removed |
| Si | 786 | Rises again |
| P | 1012 | Half-filled 3p³ |
| S | 1000 | Falls: paired 3p electron |
| Cl | 1251 | Rises |
| Ar | 1521 | Highest in the period |
Part 3 of 4Writing a Full Explanation
1
State the trend first, for example that first ionisation energy generally increases across the period.
2
Link it to the increase in nuclear charge, with outer electrons in the same shell and similar shielding.
3
Identify the exceptions at Mg → Al and P → S by naming the sub-shell involved in each case.
4
Explain the Mg → Al fall using the higher energy and extra shielding of the 3p sub-shell compared with 3s.
5
Explain the P → S fall using the repulsion between two electrons paired in the same 3p orbital.
Part 4 of 4Exam Checklist & Tips
2.1.3 – Key Points for the Exam
- First ionisation energy The energy needed to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.
- Sub-shell A set of orbitals of the same type within a shell, such as 3s or 3p.
- Electron pair repulsion The repulsion between two electrons sharing one orbital, which makes one easier to remove.
- Gaseous state Ionisation energy is defined for isolated gaseous atoms, so no interactions between particles are involved.
Common Mistakes
- Leaving out state symbols or the word "gaseous" in the definition of ionisation energy.
- Explaining the Mg → Al dip using electron repulsion instead of the higher energy of the 3p sub-shell.
- Saying the 3p electron in sulfur is further from the nucleus; the key point is the paired-electron repulsion.
- Forgetting that the electron being removed is the outermost one, not a random electron.
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