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VBT for [Fe(CN)₆]³⁻ and [FeF₆]³⁻ magnetism

Short answer

[Fe(CN)6]3-: Fe3+ (3d5) with strong CN- pair electrons, leaving one unpaired, d2sp3 (inner orbital) octahedral, paramagnetic (μ=1.73 BM). [FeF6]3-: weak F- leaves five unpaired, sp3d2 (outer orbital) octahedral, paramagnetic (μ=5.92 BM).

Explain the formation of [Fe(CN)₆]³⁻ and [FeF₆]³⁻ on the basis of Valence Bond Theory and also predict their magnetic property.

[Fe(CN)₆]³⁻

Fe³⁺ is 3d⁵. CN⁻ is a strong-field ligand, so the electrons pair up, leaving one unpaired electron and two empty 3d orbitals. d²sp³ hybridisation (inner orbital) gives an octahedral complex.

3d: [↑↓][↑↓][↑ ][  ][  ]  4s 4p  → d²sp³

One unpaired electron: paramagnetic, μ = √3 = 1.73 BM (low spin).

[FeF₆]³⁻

F⁻ is a weak-field ligand, so the 3d⁵ electrons remain unpaired. The outer 4s, 4p and two 4d orbitals form sp³d² hybrids (outer orbital), also octahedral.

3d: [↑ ][↑ ][↑ ][↑ ][↑ ]  4s 4p 4d  → sp³d²

Five unpaired electrons: paramagnetic, μ = √35 = 5.92 BM (high spin).

Both are paramagnetic, but [FeF₆]³⁻ is far more strongly paramagnetic.

✓ Answer checked by Claude Opus 5.5 (high effort)

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