NEET2026ChemistryChapterActual
In the lanthanoid series, the +3 oxidation state is the most characteristic and stable state in aqueous solutions, despite the large amount of energy required to remove three electrons. Which of the following thermodynamic factors primarily compensates for the high sum of the first three ionization enthalpies?
Options
- ALow enthalpy of atomization of the solid metals
- BLarge negative hydration enthalpy of Ln³⁺ ions
- CHigh exchange energy of the 4f electrons
- DLarge crystal field stabilization energy in aqueous medium
Correct answer
B. Large negative hydration enthalpy of Ln³⁺ ions
Step-by-step solution
The stability of an oxidation state in aqueous solution is determined by the overall enthalpy change of the process Ln(s) Ln³⁺(aq) + 3e⁻ . This process involves three steps: 1. Sublimation/Atomization: Ln(s) Ln(g) (requires _ atom H ) 2. Ionization: Ln(g) Ln³⁺(g) + 3e⁻ (requires IE₁ + IE₂ + IE₃ ) 3. Hydration: Ln³⁺(g) Ln³⁺(aq) (releases _ hyd H ) The sum of the first three ionization enthalpies is very high and endothermic. However, the Ln³⁺ ions have a high charge (+3) and a relatively small ionic radius, which re