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VBT for [Fe(CN)₆]⁴⁻: geometry, magnetism

Short answer

CN- as a strong-field ligand pairs the 3d6 electrons of Fe2+, vacating two 3d orbitals. This yields d2sp3 hybridization, an octahedral inner orbital low-spin complex, and diamagnetic behavior.

Explain the formation of [Fe(CN)₆]⁴⁻ complex on the basis of VBT. Also predict its geometry and magnetic behavior.

Fe (Z = 26); in [Fe(CN)₆]⁴⁻ the oxidation state of Fe is +2 (6 × (−1) + x = −4). Fe²⁺ has the configuration 3d⁶.

CN⁻ is a strong-field ligand, so the six 3d electrons pair up in three d orbitals, leaving two 3d orbitals empty. These two 3d, the 4s and three 4p orbitals mix to give d²sp³ hybrid orbitals, which accept six lone pairs from CN⁻.

Fe²⁺ 3d: [↑↓][↑↓][↑↓][  ][  ]  4s [ ]  4p [ ][ ][ ]
                      \_______ d²sp³ _______/
                      filled by 6 CN⁻ lone pairs
  • Geometry: octahedral (inner orbital, low spin complex).
  • Magnetism: no unpaired electron, so diamagnetic.

✓ Answer checked by Claude Opus 5.5 (high effort)

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