Identify the homoleptic complex(es) that is/are low spin.
[(A)] \([\text{Fe(CN)}_5\text{NO}]^{2-}\)
[(B)] \([\text{CoF}_6]^{3-}\)
[(C)] \([\text{Fe(CN)}_6]^{4-}\)
[(D)] \([\text{Co(NH}_3)_6]^{3+}\)
[(E)] \([\text{Cr(H}_2\text{O)}_6]^{2+}\)
Choose the correct answer from the options given below:
[(A)] \([\text{Fe(CN)}_5\text{NO}]^{2-}\)
[(B)] \([\text{CoF}_6]^{3-}\)
[(C)] \([\text{Fe(CN)}_6]^{4-}\)
[(D)] \([\text{Co(NH}_3)_6]^{3+}\)
[(E)] \([\text{Cr(H}_2\text{O)}_6]^{2+}\)
Choose the correct answer from the options given below:
Show Hint
- C and D only
- B and E only
- A and C only
- C only
The Correct Option is C
Solution and Explanation
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 \( \text{Fe(CN)}_6^{4-} \).
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