Question:
Arrange the following complex ions in increasing order of their crystal field splitting energy (\( \Delta_0 \)):
\(\text{[Co(NH}_3\text{)}_6]^{3+}, \, \text{[CoF}_6]^{3-}, \, \text{[Co(CN)}_6]^{3-}\)
Arrange the following complex ions in increasing order of their crystal field splitting energy (\( \Delta_0 \)):
\(\text{[Co(NH}_3\text{)}_6]^{3+}, \, \text{[CoF}_6]^{3-}, \, \text{[Co(CN)}_6]^{3-}\)
\(\text{[Co(NH}_3\text{)}_6]^{3+}, \, \text{[CoF}_6]^{3-}, \, \text{[Co(CN)}_6]^{3-}\)
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Ligand field strength influences the crystal field splitting energy (\( \Delta_0 \)). Cyanide (\( \text{CN}^- \)) is a strong field ligand, followed by ammonia (\( \text{NH}_3 \)), and fluoride (\( \text{F}^- \)) is a weak field ligand.
Updated On: Feb 17, 2025
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Solution and Explanation
Crystal Field Splitting Energy (∆0) and Ligand Field Strength
The crystal field splitting energy (∆0) depends on the ligand field strength. The order of field strength for ligands is:
CN− > NH3 > F−
Order of Increasing ∆0:
[CoF6]3− < [Co(NH3)6]3+ < [Co(CN)6]3−
Explanation:
- Cyanide (CN−) is a strong field ligand, causing a large splitting of the d-orbitals, leading to a high ∆0.
- Ammonia (NH3) is a moderate field ligand, causing moderate splitting.
- Fluoride (F−) is a weak field ligand, leading to the smallest splitting of d-orbitals.
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