Classification of Elements and Periodicity in Properties · Lesson 6 of 10
Ionization Enthalpy and Electron Removal
“Understand the energy needed to remove electrons and the exceptions that reveal how orbitals are occupied.”
• Define first and successive ionization enthalpies with the correct gaseous processes. • Explain their signs, units, and changes after successive electron removals. • Relate period and group trends to attraction, distance, shielding, and penetration. • Explain the Be–B and N–O exceptions using configurations. • Use ionization data to identify valence-shell changes and compare elements.
Measuring how strongly an electron is held
Electron loss is central to the chemistry of metals, but it does not happen without an energy cost for an isolated gaseous atom. Ionization enthalpy measures that cost under clearly specified conditions. Specifying the state matters: we want to compare atoms themselves, rather than include energy changes caused by a crystal, a liquid, or chemical bonds.
The enthalpy required to remove one electron from each atom in one mole of isolated gaseous atoms in their ground states, forming gaseous singly positive ions. It is represented by and commonly measured in .
The ground state is the lowest-energy electron arrangement. An excited atom could require less additional energy to lose an electron. “Isolated gaseous” excludes interactions with neighbours, making the comparison consistent from one element to another. The electron is taken away from the atom, rather than merely promoted to a different bound orbital.
A second removal starts with the positive ion formed by the first removal, not with a fresh neutral atom. The remaining electrons experience less repulsion and a stronger attraction relative to their number. Each successive removal from the same species therefore costs more energy, with especially large jumps when removal begins from a compact filled inner shell.
Problem
Why does sodium have a low first ionization enthalpy but a very high second one?
- 1.Neutral Na has configuration . Its first removal takes the outer 3s electron.
- 2.The resulting has the compact closed-shell configuration .
- 3.The second removal must take a 2p core electron rather than another 3s electron.
- 4.That electron is much closer to the nucleus and held strongly. The large jump signals that the single valence electron has already been removed.
The general trends and their causes
Across a main-group period, nuclear charge rises while the outer electrons occupy the same shell. Shielding by the core does not rise enough to cancel the additional attraction. Ionization enthalpy therefore generally increases as radius decreases. Down a group, greater distance and shielding usually outweigh the increase in nuclear charge, making the outer electron easier to remove.
| Group 1 element | Li | Na | K | Rb | Cs |
|---|---|---|---|---|---|
| First ionization enthalpy / kJ mol−1 | 520 | 496 | 419 | 403 | 374 |
The broad first-ionization graph across many elements has recurring minima at alkali metals and maxima at noble gases. A single loosely held outer electron follows each completed noble-gas arrangement. This is another observable pattern generated by recurring configurations. The detailed graph is not a smooth staircase, because orbital energy and electron pairing also matter.
| Period 2 element | Li | Be | B | C | N | O | F | Ne |
|---|---|---|---|---|---|---|---|---|
| First ionization enthalpy / kJ mol−1 | 520 | 899 | 801 | 1086 | 1402 | 1314 | 1681 | 2080 |
Why boron and oxygen depart from the simple trend
Beryllium has an outer configuration. Boron’s extra electron occupies 2p, giving . Within the same shell an s electron penetrates closer to the nucleus than a p electron. The 2p electron is more shielded and less tightly held than the 2s electron removed from Be, so boron has the lower first ionization enthalpy despite its larger nuclear charge.
The extent to which an electron’s orbital brings its probability distribution close to the nucleus. Greater penetration usually means less shielding and stronger nuclear attraction when comparing subshells in the same shell.
Nitrogen has three 2p electrons occupying separate p orbitals. Oxygen has four, so one p orbital contains a pair. Repulsion between the paired electrons makes one of them easier to remove. Consequently oxygen has a smaller first ionization enthalpy than nitrogen, while the broader rise resumes towards fluorine and neon. “A half-filled subshell is stable” is a useful shorthand, but the pairing and repulsion give the actual reasoning here.
Problem
Na, Mg, and Si have first ionization enthalpies 496, 737, and 786 kJ mol−1. Is Al more likely to have a value near 575 or 760?
- 1.The general left-to-right trend alone might suggest a value near or above Mg.
- 2.But Mg loses a 3s electron, whereas Al, with , loses a 3p electron.
- 3.The 3p electron is less penetrating and more shielded than a 3s electron.
- 4.The value should therefore be below Mg’s, making 575 kJ mol−1 the better estimate. This parallels the Be–B exception.
Problem
A hydrogen electron in its ground state has energy . Estimate the energy to ionize one mole of ground-state H atoms.
- 1.The separated electron is assigned zero energy, so the energy required per atom is .
- 2.Multiply by the number of atoms per mole: .
- 3.The result is , which rounds to or .
- 4.This is the elementary energy-based estimate; a precise thermodynamic value also specifies temperature and thermal contributions.
Problem
An element has successive ionization enthalpies 738, 1451, and approximately 7730 kJ mol−1. What does the jump suggest?
- 1.The first and second electrons cost energies of the same broad scale.
- 2.The third removal costs far more, indicating that it starts from a much more tightly bound inner arrangement.
- 3.Two outer electrons were available before that closed core was reached.
- 4.For a main-group metal this is consistent with a Group 2 atom, such as Mg; the jump supports the inference rather than giving a universal group rule for every element.
Across a period, “generally increases” does not mean “always increases”. Identify the orbital from which the electron is removed and whether it is paired. Also compare successive removals of the same element separately from first removals across different elements.
Down Group 13, the values B 801, Al 577, Ga 579, In 558, and Tl 589 kJ mol−1 do not fall smoothly. Added d and f electrons do not shield as effectively as a simple extra-shell picture might suggest. The resulting stronger effective attraction helps account for Ga’s value being slightly above Al’s and Tl’s being above In’s. Radius and shielding must be examined together; a one-direction slogan is insufficient.
Quiz
What is the correct starting species for second ionization enthalpy?
Why must a definition specify a ground-state gaseous atom?
Why is the first ionization enthalpy of B lower than that of Be?
Why is oxygen’s first ionization enthalpy below nitrogen’s?
A huge increase occurs between an atom’s second and third ionization enthalpies. What is the strongest simple inference?
Which factor usually explains decreasing first ionization enthalpy down Group 1?
Practice Problems
- Write the first and second gaseous ionization equations for Mg and explain why the second requires more energy.
- Use electron configurations to compare the first and second ionization enthalpies of Na and Mg.
- Explain both dips in the second-period first-ionization graph.
- Calculate the molar ionization estimate for H from per atom.
- Explain why increasing nuclear charge does not prevent first ionization enthalpy from generally decreasing down Group 1.
- Use the given Group 13 values to explain why a simple monotonic trend is inadequate.
- Would two isotopes of the same element have essentially equal first ionization enthalpies at this level of comparison? Explain using nuclear charge and electronic configuration; tiny isotope shifts are outside this approximation.
Key Takeaways
Ionization enthalpy describes removal from an isolated gaseous ground-state atom or ion and is positive. Successive removals require progressively more energy; a large jump often signals the start of core-electron removal. The general trends reflect nuclear attraction, size, shielding, and penetration. The Be–B and N–O exceptions follow from subshell energy and electron pairing. Use configurations and data when competing effects prevent a smooth trend.