Which Element Would You Expect to Have the Lowest Third Ionization Energy? Scandium, Calcium, or Potassium Explained

Potassium has the lowest third ionization energy of the three, at roughly 4,420 kJ/mol, compared with about 4,910 kJ/mol for calcium and 4,250–4,400 kJ/mol for scandium depending on the source table. The surprising part is that potassium, the element with the smallest nuclear charge, does not win by a wide margin—and in some datasets scandium edges it out. The ordering is not the simple monotonic trend most periodic-table summaries imply.

Third ionization energy (IE₃) is the energy required to remove the third electron from a neutral gaseous atom after two electrons are already gone. For the three elements in question, that third removal strips the valence shell entirely, so the value depends less on periodic position than on which orbital is being emptied and how many protons remain to hold it. Potassium loses its third electron from the 2p subshell, calcium from 3p, and scandium from 3d—three different orbital environments with three different effective nuclear charges.

Why orbital identity and effective nuclear charge set the third ionization energy

Ionization energy is governed by the electrostatic attraction between the remaining electrons and the nucleus, filtered through shielding and orbital penetration. The third ionization energy is the energy to remove electron number three from the +2 cation. By the time you reach that third removal, the valence shell configuration is fixed: K²⁺ is [Ar], Ca²⁺ is [Ar], and Sc²⁺ is [Ar]3d¹. So the third electron comes from the argon-like core in K and Ca, but from a 3d orbital in Sc.

Effective nuclear charge (Z_eff) rises across a period because protons are added without new shielding shells. Potassium has Z = 19, calcium Z = 20, scandium Z = 21. Using Slater's rules, the 2p electrons in K²⁺ experience substantial screening from the 1s and 2s electrons but sit closer to the nucleus than 3p or 3d electrons. That proximity makes them comparatively hard to remove despite the low nuclear charge.

Orbital penetration is the second lever. In a multi-electron atom, s orbitals penetrate closest to the nucleus, p orbitals less so, and d orbitals least of all. The 2p electrons in K²⁺ are in a compact shell that sits well inside the argon core radius. The 3p electrons in Ca²⁺ sit farther out, where shielding by the 3s electrons is less complete relative to their distance, raising the energy needed to remove one. The 3d electron in Sc²⁺ is the most diffuse of the three, poorly penetrating and well screened, which lowers the removal energy for scandium even though its Z_eff is highest.

Those two effects pull in opposite directions. Higher Z pushes IE₃ up in the order K < Ca < Sc. Orbital compactness and penetration reshuffle the order. The net result lands Ca highest, K lowest in most tables, and Sc close to K, sometimes above it depending on the reference set.

Quantified context: how the three IE₃ values actually compare

Published IE₃ values cluster tightly, and the ordering is sensitive to which dataset and which revision you consult. The values below are representative ranges from standard thermochemical compilations such as the NIST Atomic Spectra Database and the CRC Handbook; exact figures differ by a few percent between editions.

The reference rows matter because they reveal the mechanism. Magnesium and sodium's third ionization energies are far higher than potassium's because removing the third electron from the [Ne] core (Mg²⁺ is [Ne]3s⁰, so the third electron comes from 2p of the neon-like core) costs more than removing it from the argon-like core. Across the K–Ca–Sc triad, the increases from Mg to Ca and from Na to K show the periodic trend that higher nuclear charge raises IE₃ for a fixed orbital type—but that trend is broken when the orbital type changes.

The scandium number sits in a bracket because the 3d¹ electron is a single unpaired electron in a compact but poorly penetrating orbital. Its removal energy is lower than for the 3p electron of Ca²⁺ because the 3d electron is shielded more effectively by the 3s and 3p electrons than the 3p electron is by the 3s electrons alone. The scandium value is close to potassium's because those two offsetting effects almost cancel.

Edge cases and dataset dependence

Three practical wrinkles make this comparison harder than a single-number answer suggests.

First, IE₃ values are not measured directly for most elements. They are inferred from spectroscopic lines, extrapolated from Rydberg series, or computed with multi-configuration Dirac–Fock methods. Different compilers adopt different extrapolation procedures, and the spread for scandium spans roughly 150 kJ/mol—enough to flip its position relative to potassium in some tables. Any claim that scandium is "lower" or "higher" than potassium without naming the dataset overstates precision.

Second, condensed-phase effects intrude if you ask the question in a different context. The question specifies gaseous atoms, which is the standard definition. In solution or in a solid, the energetics of removing three electrons are dominated by solvation or lattice energies, and the IE₃ ranking becomes largely irrelevant to observable chemistry. The values only compare cleanly in the gas phase.

Third, relativistic effects shift scandium slightly. The 3d orbital contracts under relativistic influence in heavier elements; for scandium the effect is modest but still measurable in high-precision calculations. Potassium and calcium, with only s, p, and (for Ca) no open d electrons in the third-removal ion, are less sensitive to relativity. This is a small correction, but it belongs in any honest accounting of why scandium sits between high and low.

The counterintuitive detail: scandium, the element with the highest nuclear charge in the set, does not have the highest third ionization energy. The trend that holds across a period for first and second ionization energies breaks at IE₃ because the orbital being emptied changes identity rather than position.

Misconceptions that produce the wrong answer

Several errors reliably push students and even professional-style explainers toward the wrong element.

Claiming the highest nuclear charge always means the highest ionization energy. Z_eff ranks K < Ca < Sc for a fixed orbital, but IE₃ involves three different orbital types, so the ranking does not transfer directly. Assuming the leftmost element always has the lowest IE at every step. First and second ionization energies do follow that trend roughly (K < Ca < Sc for IE₁ is not exact—Ca > K—and IE₂ is Ca < K < Sc depending on configuration), but third ionization energies follow the orbital, not the column. Treating IE₃ as a smooth extension of IE₁ and IE₂. The third removal crosses a shell boundary for K and Ca but not for Sc, so the energy jump between IE₂ and IE₃ is enormous for K and Ca and comparatively modest for Sc. Potassium's IE₂ is about 3,060 kJ/mol and its IE₃ jumps to about 4,420—a step of roughly 1,360. Calcium's IE₂ is about 1,150 and jumps to about 4,910—a step near 3,760. Scandium's IE₂ is about 1,240 and IE₃ lands near 4,400—a step of roughly 3,160. The jump size, not just the absolute value, encodes the shell-crossing physics. Confusing the question with "which element loses three electrons most easily in practice." In chemical compounds, ionization is coupled to lattice or solvation energy. The gas-phase IE₃ ranks do not predict, for example, the stability of +3 oxidation states in solution. Relying on a single reference value without stating the dataset. If two textbooks disagree on whether Sc or K is lower, both may be internally consistent. The honest answer names the range and the orbital reason.

Practical actions for checking the answer yourself

If you need to resolve this comparison for a course, a paper, or an exam, the following steps produce a defensible answer.

Work in the gas phase and specify neutral atoms converting to +3 cations. Condensed-phase numbers are not comparable. Pull IE₃ from at least two authoritative compilations—NIST's Atomic Spectra Database and the CRC Handbook are the standard pair—and report the values you find rather than a single number. Write the +2 electron configuration first. For K²⁺, Ca²⁺, and Sc²⁺ that is [Ar], [Ar], and [Ar]3d¹ respectively. The third removed electron is then unambiguous. Use the resulting orbital identity to predict ordering: 2p (K) versus 3p (Ca) versus 3d (Sc). This prediction matches the observed grouping, with potassium typically lowest. Check the IE₂-to-IE₃ step size. A large step signals a shell crossing; a small step signals removal within an incompletely filled or newly opened subshell. Scandium's smaller step is the tell that its 3d electron is being removed. If two sources flip the K and Sc order, report the bracket (roughly 4,300–4,450 kJ/mol for both) and note the dataset dependence rather than declaring a single winner.

A practical threshold: if the reported IE₃ gap between K and Sc is under about 100 kJ/mol, treat the order as within the uncertainty of thermochemical compilations and say so. That is what the data support.

Limits of the comparison and what to say about it

The clean statement is that potassium has the lowest third ionization energy in most compilations, calcium has the highest, and scandium sits near or just below potassium depending on the dataset. The mechanistic reason is that the third removal empties the argon-like core in K and Ca but takes a 3d electron in Sc, and the orbital compactness of the 2p core makes potassium's third electron relatively easy to remove despite potassium's lower Z_eff.

What the comparison cannot tell you is anything about triplet or quartet state energies in real samples, solvation effects, or the practical ease of forming +3 ions in compounds. Those require thermodynamic cycles, not bare IE₃ values. If a source presents a single exact number for IE₃ without a dataset reference, treat it as a representative approximation rather than a measured fact. That caution keeps the answer accurate without overclaiming precision the physics does not deliver.