Concept 02 / 05Classification of Elements and Periodicity
Ionisation Enthalpy: Trends, Exceptions and Successive Ionisation Energies
Definition of first and successive ionisation enthalpies, factors (nuclear charge, size, shielding, penetration, stable configurations), the general trends, the exceptions Be > B, N > O, Mg > Al, the jump in successive ionisation energies that reveals valence electrons, and applications.
The first ionisation enthalpy is the energy needed to remove the most loosely bound electron from an isolated gaseous atom in its ground state: . It is always positive (endothermic). Successive ionisation enthalpies (, , ...) remove further electrons from the increasingly positive ion, and each is larger than the previous one, because the same nuclear charge holds fewer electrons more tightly.
Factors. Ionisation enthalpy is higher when the nuclear charge is larger, the atom is smaller, shielding by inner electrons is less, the electron penetrates closer to the nucleus (s > p > d > f), and the configuration being disturbed is especially stable (half-filled or fully filled subshells).
General trends (see figure, first elements).
Across a period it generally increases, because increases and radius decreases. Each period starts with an alkali metal (a minimum) and ends with a noble gas (a maximum).
Down a group it generally decreases, because the outer electron is farther away and better shielded. (In groups and this trend is irregular because of poor d and f shielding: Ga > Al, Pb > Sn.)
Exceptions across a period, both clearly visible in the figure.
Be () > B (): boron's electron is removed from a orbital, which is higher in energy and less penetrating than beryllium's . The same explains Mg > Al.
N () > O (): nitrogen has a stable half-filled set; in oxygen () one orbital holds a pair of electrons, whose mutual repulsion makes one easier to remove. The same explains P > S.
Successive ionisation energies reveal the number of valence electrons. There is a very large jump when an electron must be removed from a noble-gas core. For magnesium (): , , then kJ/mol. The jump after the second shows two valence electrons, which is why magnesium forms and never . For sodium the jump comes after the first (, then ).
Applications. Low ionisation enthalpy means strong metallic and reducing character (the alkali metals, especially Cs and K, are the strongest reducing metals in the gas phase and eject electrons in the photoelectric effect with visible light). High ionisation enthalpy is associated with non-metals. The energy to form cations is a key term in lattice energy calculations (Born-Haber cycle).
Key Takeaways:
- always.
- Increases across a period, decreases down a group (with exceptions in groups 13 and 14).
- Be > B and Mg > Al (s vs p electron); N > O and P > S (half-filled stability, pair repulsion).
- A large jump in successive values marks the end of the valence electrons.
- Noble gases have the highest, alkali metals the lowest in each period.
- Pitfall: predicting O > N by the general trend.
Brain-Teaser 1: The successive ionisation enthalpies of an element (kJ/mol) are , , , , . Identify its group and write its likely ion.
Solution: The big jump comes after the third value (), so the element has three valence electrons: group . It forms . These are the values for aluminium (). Reading the jump is more reliable than looking at individual magnitudes.
Brain-Teaser 2: Why is the second ionisation enthalpy of sodium much higher than that of magnesium, even though the first is lower?
Solution: is , so its second ionisation removes an electron from a stable noble-gas core (, close to the nucleus): about kJ/mol. is , and its second ionisation still removes a valence electron: kJ/mol. Which shell the electron comes from dominates everything else.
Brain-Teaser 3: Arrange in increasing first ionisation enthalpy: B, C, N, O, Be.
Solution: B () < Be () < C () < O () < N (). Two exceptions are at work: B below Be (p versus s electron) and O below N (paired electron versus half-filled set). This order is a standard examination question because the naive trend gets two places wrong.
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