3.1.8Additional lessons
Ionisation Energy Trends Across a Period (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 55 of 172
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.
Section 1 of 5General Trend and Dips
General rise
- General rise Across period 3, first ionisation energy rises because nuclear charge increases while shielding stays almost constant.
Mg to Al dip
- Mg to Al dip Aluminium's outer electron is in a 3p sub-level, which is higher in energy and slightly shielded by the 3s electrons.
P to S dip
- P to S dip Sulfur has a paired electron in one 3p orbital, so repulsion between the pair makes it easier to remove.
Noble gas peak
- Noble gas peak Argon has the highest value in the period because its full shell has the greatest effective nuclear charge.
Evidence
- Evidence These dips are experimental evidence for the existence of sub-levels and for electron pairing in orbitals.
| Pair of elements | Expected from nuclear charge | Observed | Reason |
|---|---|---|---|
| Mg to Al | Increase | Decrease | 3p higher than 3s |
| P to S | Increase | Decrease | Pair repulsion in 3p |
Section 2 of 5Successive Ionisation Energies
1
Removing each further electron from the same atom needs more energy because the remaining electrons feel the same nuclear charge with less repulsion.
2
A very large jump between two successive values shows that the next electron is coming from a shell closer to the nucleus.
3
The number of values before the first large jump equals the number of valence electrons, which gives the group number.
4
Plotting log₁₀ of the ionisation energy against electron number shows these jumps as steps, which reveal the shell structure.
5
Smaller jumps within a shell can show the change from one sub-level to another.
Section 3 of 5Applying the Trend
Applying the Trend
Comparing atoms
- Comparing atoms Use nuclear charge, shielding and sub-level to compare ionisation energy of neighbouring elements.
Reactivity
- Reactivity Elements with low first ionisation energy form positive ions readily, which links to metallic character.
Sketching graphs
- Sketching graphs Draw a rising trend with a drop at Group 13 and another at Group 16, and a final peak at the noble gas.
Units
- Units Ionisation energies are quoted in kJ mol⁻¹ for one mole of gaseous atoms.
Section 4 of 5Worked Example
Question
- The first four ionisation energies (kJ mol⁻¹) of an element are 590, 1145, 4912 and 6491.
- Deduce its group.
- Compare successive values590 to 1145 is about 1.9 times.
- 1145 to 4912 is a jump of about 4.3 times, much larger than the others.
- The third electron must come from an inner shell, so two valence electrons were removed first.
- An element with two valence electrons is in Group 2.
Answer Group 2 (calcium is consistent)
Section 5 of 5Exam Checklist & Tips
3.1.8 – Key Points for the Exam
- Successive ionisation energies Energies needed to remove the first, second, third electron and so on from one atom.
- Sub-level A set of orbitals of the same type in one energy level, such as 3p.
- Spin-pair repulsion Repulsion between two electrons sharing one orbital.
- Valence shell The highest occupied energy level.
Common Mistakes
- Explaining the Mg to Al dip as due to a smaller nuclear charge, when the nuclear charge actually increases.
- Describing the P to S dip as caused by a half-filled sub-level, instead of pair repulsion.
- Forgetting that ionisation energy is for gaseous atoms.
- Reading a big jump as the end of the previous shell rather than the removal of the next electron from an inner shell.
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