Question

The crystal field stabilization energy (CFSE) of [Fe(H₂O)₆]Cl₂ and K₂ [NiCl₄], respectively, are:
1. - 0.6Δ_o and - 0.8Δ_t
2. - 0.4Δ_o and - 0.8Δ_t
3. - 2.4Δ_o and - 1.2Δ_t
4. - 0.4Δ_o and - 1.2Δ_t

Answer 1

Correct Answer - Option 2 : - 0.4Δ_o and - 0.8Δ_t

Concept:

The Crystal Field Stabilization Energy (CFSE) is defined as the energy of the electron configuration in the ligand field minus the energy of the electronic configuration in the isotropic field.

CFSE is determined by the given formula

CFSE = ΔE = E_{ligand field} – E_{isotropic field}

Calculation:

The CFSE is found by the following formula

CFSE = ΔE = E_{ligand field} – E_{isotropic field}

For [Fe(H₂O)₆]Cl₂:

Fe²⁺ → [Ar]3d⁶ → (t_2g)⁴ (e_g)²

H₂O is a weak field ligand so pairing does not take place.

C.F.S.E. = 4 × (-0.4Δ₀) + 2 × 0.6Δ₀

= - 1.6Δ₀ + 1.2Δ₀

∴ C.F.S.E. = - 0.4Δ₀

For K₂ [NiCl₄]:

K₂ [NiCl₄], Ni²⁺ → [Ar]3d⁸ → (e_g)⁴ (t_2g)⁴

Cl^- → Weak field ligand, Δ₀ paring does not take place and have tetrahedral geometry.

C.F.S.E. = 4 × (-0.6Δ_t) + 4 × (0.4Δ_t)

= -2.4Δ₀ + 1.6Δ₀

∴ C.F.S.E. = - 0.8Δ_t

Thus, the required CFSE are - 0.4Δ_o and - 0.8Δ_t