Question

Identify the homoleptic complex(es) that is/are low spin.

• A. \([Fe(CN)_5NO]^{2−}\)
• B. \([CoF_6]^{3−}\)
• C. \([Fe(CN)_6 ]^{4−}\)
• D. \( [Co(NH_3 )_6]^{3+}\)

Hint:

For complexes with transition metals, the crystal field splitting is influenced by both the oxidation state and the nature of the ligand. Strong field ligands (like \( \text{CN}^- \)) cause low-spin complexes, while weak field ligands (like \( \text{F}^- \) and \( \text{H}_2\text{O} \)) tend to lead to high-spin complexes.

Answer 1

The Correct Option is C

The complex \([Fe(CN)_6]^{4-}\) exhibits a low spin configuration.

Rationale

• The cyanide ion (\(CN^-\)) is classified as a strong field ligand. This characteristic leads to a significant splitting of the d-orbitals (\(\Delta_{o}\)).
• For the iron atom in this complex, the substantial d-orbital splitting promotes the pairing of electrons within the lower energy d-orbitals, resulting in a low-spin electron configuration.

Comparative Analysis

• In \([CoF_6]^{3-}\), fluoride (F⁻) acts as a weak field ligand, typically resulting in a high-spin configuration.
• Ammonia (NH₃) is a stronger field ligand than fluoride. While it can induce low-spin configurations for certain metal ions and oxidation states (for instance, Co(III) with a d⁶ configuration frequently results in low spin), \([Fe(CN)_6]^{4-}\) serves as the most definitive example of a low-spin complex within this set, owing to the potent strong-field nature of the cyanide ligand.