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VBT for [Fe(CN)₆]³⁻ inner/outer orbital

Short answer

CN- pairs four 3d5 electrons, leaving one unpaired and vacating two 3d orbitals for d2sp3 hybridization. The complex is inner orbital, octahedral, and paramagnetic with one unpaired electron (1.73 BM).

How does valence bond theory explain the formation of [Fe(CN)₆]³⁻? Also explain whether this complex is inner or outer orbital and why?

Fe (Z = 26); in [Fe(CN)₆]³⁻ the oxidation state of Fe is +3, so Fe³⁺ is 3d⁵.

Formation: CN⁻ is a strong-field ligand, so the five 3d electrons are forced to pair up in two orbitals (with one electron left unpaired), vacating two 3d orbitals. Two 3d, one 4s and three 4p orbitals form six d²sp³ hybrid orbitals, which accept six lone pairs from CN⁻.

Fe³⁺ 3d: [↑↓][↑↓][↑ ][  ][  ]  4s [ ]  4p [ ][ ][ ]
                   \_______ d²sp³ _______/
  • Geometry: octahedral.
  • Inner or outer orbital: it is an inner orbital complex, because the inner (n−1)d orbitals take part in hybridisation, and low spin because electrons are paired.
  • Magnetism: one unpaired electron, so paramagnetic, μ = √3 = 1.73 BM.

✓ Answer checked by Claude Opus 5.5 (high effort)

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