Statement I: Fe²⁺ is more stable than Fe³⁺ ion

Statement II: Fe³⁺ ion contain half-filled 3d subshell

Statement I is false but statement II is true

The correct answer is option 4. Statement I is false but statement II is true.

Let us break down the explanation in a clear and detailed manner.

Statement I: Fe²⁺ is more stable than Fe³⁺ ion

The statement is incorrect. In aqueous solutions and many chemical environments, Fe³⁺ is generally more stable than Fe²⁺. The reasons for this increased stability are:

Ionic Potential: Fe³⁺, with a +3 charge, has a higher charge density compared to Fe²⁺, which has a +2 charge. This higher charge density means Fe³⁺ has a stronger attraction to surrounding anions or ligands, leading to greater stability.

Lattice Energy: Compounds with Fe³⁺ often have higher lattice energies due to the higher charge, which makes the crystal lattice more stable.

Ligand Field Stabilization: In coordination chemistry, Fe³⁺ often forms more stable complexes than Fe²⁺ because the ligand field stabilization energy is higher for Fe³⁺.

Redox Potential: The standard reduction potential for the Fe³⁺/Fe²⁺ couple is +0.77 V. This positive potential indicates that Fe³⁺ is more thermodynamically stable in aqueous solution than Fe²⁺. Essentially, Fe³⁺ has a stronger tendency to be reduced to Fe²⁺ than Fe²⁺ has to be oxidized to Fe³⁺.

Statement II: Fe³⁺ ion contains a half-filled 3d subshell

The statement is correct. To understand this, consider the electronic configurations:

Neutral Iron (Fe): The electron configuration is [Ar] 3d⁶ 4s².

Fe²⁺ Ion: Two electrons are removed, typically from the 4s orbital, resulting in [Ar] 3d⁶.

Fe³⁺ Ion: One more electron is removed, now from the 3d orbital, resulting in [Ar] 3d⁵.

The 3d⁵ configuration in Fe³⁺ is particularly stable because it is a half-filled subshell. This half-filled state is stable due to:

Symmetry: A half-filled subshell has a symmetrical distribution of electrons.

Exchange Energy: The exchange energy is maximized when electrons are unpaired and distributed among the available orbitals, which minimizes electron-electron repulsions within the same shell. This leads to extra stability.

Therefore, the correct answer is 4. Statement I is false but statement II is true.