Ionization Enthalpy (IE)

The greater the nuclear charge (atomic number Z), the stronger the electrostatic attraction between the nucleus and the valence electrons. Coulomb's law states that the force F = kq₁q₂/r². As Z increases across a period, the force holding each valence electron increases and IE rises. Sodium (Z = 11) has IE₁ = 496 kJ mol⁻¹, while chlorine (Z = 17) has IE₁ = 1251 kJ mol⁻¹.

Coulomb's law also shows that force decreases with the square of distance. Down a group, valence electrons occupy higher principal quantum shells (n = 1, 2, 3...) and are therefore much further from the nucleus. Lithium (n = 2 valence shell, IE₁ = 520 kJ mol⁻¹) versus caesium (n = 6 valence shell, IE₁ = 376 kJ mol⁻¹) illustrates how increasing atomic radius dramatically lowers IE.

Inner-shell (core) electrons repel valence electrons and partially cancel the attractive effect of the nucleus. This is called shielding or screening. The effective nuclear charge felt by a valence electron is Zeff = Z − σ, where σ is the shielding constant. Using Slater's rules, each inner-shell electron contributes approximately 0.85–1.00 to σ, while same-shell electrons contribute about 0.35. As more inner shells accumulate going down a group, shielding increases and Zeff for the valence electrons barely rises, causing IE to fall.

Not all orbitals of the same principal quantum number are equally effective at penetrating toward the nucleus. The radial probability distributions show that s orbitals have a nonzero probability density at the nucleus, while p and d orbitals have nodes at the nucleus. The penetration order is s > p > d > f. A 2s electron penetrates inner shells more than a 2p electron and therefore experiences a higher Zeff and is harder to remove. This explains why boron (2p¹, IE₁ = 801 kJ mol⁻¹) has a lower IE than beryllium (2s², IE₁ = 900 kJ mol⁻¹) — the 2p electron of boron is shielded by the 2s electrons and has lower penetration.

Certain electron configurations confer extra stability, causing anomalous (lower than expected) ionization enthalpies for the next element. Nitrogen [He]2s²2p³ has a half-filled 2p sub-shell with electron exchange energy providing extra stability; its IE₁ = 1402 kJ mol⁻¹ is higher than oxygen's IE₁ = 1314 kJ mol⁻¹, even though oxygen has greater nuclear charge. In oxygen, the fourth 2p electron must pair with an existing 2p electron, introducing electron–electron repulsion that makes it easier to remove. The same reasoning applies to the anomaly between phosphorus and sulfur in Period 3.

Each successive ionization enthalpy is larger because the same nuclear charge now acts on fewer electrons, increasing Zeff per remaining electron. Crucially, enormous jumps occur when the electron being removed comes from a new, inner principal shell. Magnesium data (all kJ mol⁻¹): IE₁ = 738, IE₂ = 1451, IE₃ = 7733. The massive jump between IE₂ and IE₃ reveals that Mg has two valence electrons (Group 2). This pattern is exploited analytically to determine group membership.

• X(g) — Gaseous atom (—): The neutral atom must be in the gaseous phase and in its electronic ground state.
• X^+(g) — Gaseous cation (—): The singly charged positive ion formed after electron removal; also in the gaseous phase.
• e^- — Electron (—): The electron removed; assumed to be at rest (zero kinetic energy) at infinite distance from the ion.
• \Delta H — Ionization enthalpy (kJ mol⁻¹): Always positive (endothermic) because energy must be supplied to overcome nuclear attraction.
• Z_{eff} — Effective nuclear charge (dimensionless): The net nuclear charge felt by the valence electron after accounting for shielding: Z_eff = Z − σ.

Problem. The successive ionization enthalpies (kJ mol⁻¹) of an unknown element X are: IE₁ = 578, IE₂ = 1817, IE₃ = 2745, IE₄ = 11578, IE₅ = 14831. (a) In which group of the periodic table does element X belong? (b) Identify the element if it is in Period 3. (c) Explain why IE₃ to IE₄ shows the largest jump.

Answer. Element X is in Group 13; it is Aluminium (Al). The large jump between IE₃ and IE₄ occurs because the 4th electron is removed from the inner n = 2 shell (core) rather than the n = 3 valence shell.

• Electron Affinity — Electron affinity is the energy change when a gaseous atom gains an electron — the complementary process to ionization enthalpy. Together, they define how strongly atoms attract or repel electrons and underpin electronegativity.
• Effective Nuclear Charge (Z_eff) and Slater's Rules — Z_eff is the core quantitative concept behind all ionization enthalpy trends. Understanding Slater's rules for calculating shielding constants allows numerical prediction of IE across the periodic table.
• Atomic Radius and Periodic Trends — Atomic radius directly affects ionization enthalpy via Coulomb's law (force ∝ 1/r²). The same nuclear charge and shielding factors that control IE also govern how atomic radius changes across periods and down groups.
• Electronegativity and the Pauling Scale — Electronegativity is derived partly from ionization enthalpy and electron affinity. Elements with high IE tend to have high electronegativity and form polar covalent bonds by attracting bonding electrons.
• Orbital Penetration and Shielding — Penetration (s > p > d > f) explains anomalies in IE trends such as why B has lower IE than Be, and why d-block contraction occurs. It is essential for a complete mechanistic understanding of IE values.