Worked example 1
A period 3 element has successive ionization energies of 738, 1451, 7733 and 10540 kJ/mol. Identify it.
Try it first: Look for the biggest ratio between consecutive values, not the biggest difference.
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Periodic TrendsAnomalies and successive IEContent level: Challenge 25 min
Trend Anomalies and Successive Ionization Energies
What you'll be able to do: Explain the two classic ionization energy dips and use successive ionization data to identify an element's group.
Best after: Atomic Radius, Ionization Energy and Electronegativity, Orbital Diagrams, Hund's Rule and Magnetism
Introduction
The general trends have two famous exceptions in every period, and exam questions target them deliberately. The same subshell reasoning also lets you read a table of successive ionization energies and name the group an unknown element belongs to.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Explain why B has a lower IE₁ than Be
- Explain why O has a lower IE₁ than N
- Interpret electron affinity comparisons
- Use a jump in successive ionization energies to find the valence electron count
Lesson
Dip one: group 2 to group 13
Beryllium has IE₁ = 900 kJ/mol but boron only 801 kJ/mol, against the general increase. Beryllium removes an electron from a filled 2s subshell, while boron removes a 2p electron that is higher in energy and partly shielded by the 2s pair. The subshell change beats the extra proton.
Same reasoning repeats for Mg versus Al in period 3.
Dip two: group 15 to group 16
Nitrogen has IE₁ = 1402 kJ/mol but oxygen only 1314 kJ/mol. Nitrogen is 2p³, with one electron in each p orbital. Oxygen is 2p⁴, so one orbital holds a pair, and the repulsion between those two same-orbital electrons makes one of them easier to remove.
Electron affinity is not a clean trend
Electron affinity becomes more exothermic across a period and less so down a group, but with exceptions. Chlorine, not fluorine, has the most exothermic first electron affinity, because fluorine is so small that adding an electron adds severe repulsion. Group 2 and group 18 elements have positive (endothermic) affinities, since the added electron must enter a new higher subshell or shell.
More negative electron affinity means energy is released, so the process is more favourable.
Successive ionization energies
Each removal costs more than the one before, because the remaining electrons feel a higher charge-to-electron ratio. The informative feature is a sudden large jump: it appears when the valence shell is exhausted and the next electron must come from the core. Counting the removals before the jump gives the valence electron count, and therefore the group.
IE₁ < IE₂ < IE₃ < ...
Reading a data table
For a period 3 element with IE values 578, 1817, 2745 and then 11577 kJ/mol, three modest values are followed by a jump of about four times. Three valence electrons means group 13, and in period 3 that is aluminium.
Key ideas
Key concept
Subshell dip
Group 13 elements ionize a p electron that lies above the filled s subshell, lowering IE₁ below the group 2 neighbour.
Key concept
Pairing dip
Group 16 elements have one paired p orbital, and that repulsion lowers IE₁ below the group 15 neighbour.
Rule
Ionization jump
A large jump between IEn and IEn+1 means the valence shell held exactly n electrons.
Definition
Electron affinity
Energy change when a gaseous atom gains an electron, negative when energy is released.
Worked examples
Worked example 2
Why does oxygen have a lower first ionization energy than nitrogen, despite having one more proton?
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Common mistakes
Explaining the Be to B dip by saying boron has more shielding electrons in its core.
Why it's wrong: Both have the same 1s² core. The difference is that boron ionizes a 2p electron rather than a 2s electron.
Check instead: Name the subshell of the removed electron in your explanation.
Calling the N to O dip a violation of Hund''s rule.
Why it's wrong: Oxygen obeys Hund''s rule. The pairing in one orbital is required once the fourth p electron is added.
Check instead: Attribute the dip to repulsion between the two electrons sharing one 2p orbital.
Choosing the largest absolute difference instead of the largest ratio when finding the IE jump.
Why it's wrong: Later ionizations are all large, so absolute gaps grow naturally without marking the core.
Check instead: Compare successive values as ratios and look for the outlier.
Assuming fluorine has the most exothermic electron affinity.
Why it's wrong: Fluorine is so compact that the added electron suffers extra repulsion, so chlorine releases more energy.
Check instead: Quote chlorine as the most exothermic first electron affinity.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
What you should now know
Ionization energy dips at the group 13 and group 16 elements because of subshell energy and p-orbital pairing repulsion. A large jump between successive ionization energies marks the point where the valence shell is empty and a core electron must be removed.
- Group 13 dips because a 2p electron is removed instead of a 2s electron
- Group 16 dips because of pairing repulsion in one p orbital
- A large ratio jump in successive IE marks the core
- Chlorine, not fluorine, has the most exothermic electron affinity
Sources and further reading
- Chemistry 2e, Section 6.5: Periodic Variations in Element PropertiesFlowers, Theopold, Langley, Robinson · OpenStax, Rice University · Chapter 6.5View source
Access for free at openstax.org License
- NIST Atomic Spectra Database: ionization energiesNational Institute of Standards and TechnologyView source
Data from the NIST Atomic Spectra Database