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

In general, low-spin complexes are formed by transition metals with higher oxidation states and/or ligands that create strong crystal field splitting (such as \( \text{CN}^- \)). Based on this: - \( [\text{Fe(CN)}_5\text{NO}]^{2-} \) and \( [\text{Fe(CN)}_6]^{4-} \) are low-spin because \( \text{CN}^- \) is a strong field ligand. - \( [\text{CoF}_6]^{3-} \) and \( [\text{Cr(H}_2\text{O})_6]^{2+} \) are high-spin due to weaker ligands or lower oxidation states. Thus, the correct complexes that are low-spin are \( \text{Fe(CN)}_5\text{NO}^{2-} \) and \( \text{Fe(CN)}_6^{4-} \).