Anomalies and successive IECourse level: AP Chemistry

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

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.

Write the last occupied subshell before explaining a dip. Be is 2s² and B is 2s² 2p¹; the electron removed from B is a less penetrating, higher-energy 2p electron. Mg versus Al repeats the 3s²/3p¹ pattern. This mechanism is different from the N/O dip: there the electron is in the same p subshell but its pairing changes. Naming “extra repulsion” for every dip misses the distinction.

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: state the process and sign convention

Consider X(g) + e⁻ → X⁻(g). When electron affinity is reported as the enthalpy change for this process, exothermic attachment has a negative value; more exothermic means more negative. Some tables instead report the positive energy released, so always read the convention before comparing numbers. A second electron added to an already negative ion is a different process and is generally endothermic for an isolated gaseous ion.

First electron-affinity trends are irregular because shell, subshell and electron-repulsion effects compete. Chlorine's electron attachment is more exothermic than fluorine's: the extra electron enters a roomier 3p cloud rather than compact 2p, reducing the repulsion penalty. Filled s subshells, half-filled p subshells and closed shells can make attachment less favorable than a simple across-period arrow suggests. These exceptions are reasons to inspect the electron configuration and supplied data, not to invent an exact affinity from position alone.

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.

Successive removals involve different starting ions: IE₁ is X → X⁺, IE₂ is X⁺ → X²⁺, and IE₃ is X²⁺ → X³⁺, each in the gas phase with an electron produced. Later electrons are removed from increasingly positive species, so the values rise even within the same shell. A very large jump indicates a qualitative change to a tightly bound core electron. For a main-group atom with a jump between IE₂ and IE₃, two valence electrons were removed before the core was reached, so group 2 is the usual inference. This simple group assignment is not a general shortcut for every transition metal.

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.

Locate the jump before assigning a group

A large jump between IEₙ and IEₙ₊₁ means n electrons were removed before a core electron had to be taken.

The jump position counts valence electrons, but it does not by itself name the element: a group 1 element in different periods can show the same pattern.

To identify an element, combine the jump with another clue such as its period or the magnitude of the first IE. Compare ratios between consecutive values, not merely which IE is largest; later removals always cost more.

Check that the proposed neutral configuration has n outer electrons.

Two different orbital arguments explain the N/O pairing exception and Mg’s jump after removal of two valence electrons.

Diagnose a dip before invoking the trend

A first-ionization-energy dip can come from two different orbital effects. Starting a higher-energy p subshell makes the first p electron easier to remove than an s electron in the preceding element.

Pairing two p electrons adds repulsion, making one paired electron easier to remove than a single p electron in the preceding half-filled arrangement. State which orbital change occurs in the pair you compare.

Group 2→13: new p electron. Group 15→16: paired p electron.

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 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.

    0 of 3 steps revealed.

    Worked example 2

    Why does oxygen have a lower first ionization energy than nitrogen, despite having one more proton?

      0 of 3 steps revealed.

      Worked example 3

      An unknown period-3 element has IE₁ = 496, IE₂ = 4560 and IE₃ = 6910 kJ/mol. Identify the element.

      Try it first: Look for the first large jump and count removals before it.

        0 of 3 steps revealed.

        Worked example 4

        Why is the first ionization energy of Al lower than Mg even though Al has another proton?

        Try it first: Write just the outer configurations.

          0 of 3 steps revealed.

          Common mistakes

          Common mistakes and better checks
          MistakeWhy it's wrongCheck instead
          Explaining the Be to B dip by saying boron has more shielding electrons in its core.Both have the same 1s² core. The difference is that boron ionizes a 2p electron rather than a 2s electron.Name the subshell of the removed electron in your explanation.
          Calling the N to O dip a violation of Hund''s rule.Oxygen obeys Hund''s rule. The pairing in one orbital is required once the fourth p electron is added.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.Later ionizations are all large, so absolute gaps grow naturally without marking the core.Compare successive values as ratios and look for the outlier.
          Assuming fluorine has the most exothermic electron affinity.Fluorine is so compact that the added electron suffers extra repulsion, so chlorine releases more energy.Quote chlorine as the most exothermic first electron affinity.

          Scroll horizontally if needed to read all columns.

          After the lesson · Question bank

          Independent practice

          Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.

          40 topic questions across 3 difficulty levels.

          Loading practice questions…

          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

          Your next step

          Lesson: Not marked completed

          Marking a lesson completed records that you finished reading it. Practice establishes mastery separately. Saved on this device.

          Try a short set of different questions to check what you can do independently.

          Watch and Learn

          An optional video to support this lesson.

          Loading video resources…

          Supplementary learning resources

          More resources for this topic →

          External resources belong to their respective creators. Chem Help is not affiliated with or endorsed by these providers. Links open the original provider; their terms apply.

          Sources and further reading

          • Chemistry 2e, Section 6.5: Periodic Variations in Element Properties
            Flowers, Theopold, Langley, Robinson · OpenStax, Rice University · Chapter 6.5
            View source

            Lesson-specific rights and provenance need review.

            The current online Chemistry 2e edition lists CC BY-NC-SA 4.0. The source version used for this lesson and its applicable rights remain unverified.

            Recorded use: reference. This describes the stored citation, not verified authorship or reuse permission.

            Recorded attribution / license claim (applicability unverified): Access for free at openstax.org

            Source rights and review status
          • NIST Atomic Spectra Database: ionization energies
            National Institute of Standards and Technology
            View source

            Lesson-specific rights and provenance need review.

            NIST rights depend on the dataset and work category. A generic government-work or public-data label does not establish rights for this lesson's source material.

            Recorded use: further reading. This describes the stored citation, not verified authorship or reuse permission.

            Recorded attribution / license claim (applicability unverified): Data from the NIST Atomic Spectra Database

            Source rights and review status