Question

Which of the following complex ion is diamagnetic?

• A. [MnCl\(_6\)]\(^{3-}\)
• B. [Fe(CN)\(_6\)]\(^{3-}\)
• C. [Co(C\(_2\)O\(_4\))\(_3\)]\(^{3-}\)
• D. [FeF\(_6\)]\(^{3-}\)
• E. [CoF\(_6\)]\(^{3-}\)

Hint:

Cobalt(III) complexes with strong or intermediate field ligands (like NH\(_3\), en, oxalate) are usually low-spin and diamagnetic.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Diamagnetic complexes have no unpaired electrons. We need to check the electronic configuration of the central metal ion and the ligand field strength to determine the pairing of electrons.

Step 2: Detailed Explanation:
Let's analyze each option:
• (A) [MnCl\(_6\)]\(^{3-}\): Mn is in +3 state (d\(^4\)). Cl\(^-\) is a weak field ligand, so it's high spin. d\(^4\) in weak field = 4 unpaired electrons. Paramagnetic.
• (B) [Fe(CN)\(_6\)]\(^{3-}\): Fe is in +3 state (d\(^5\)). CN\(^-\) is a strong field ligand, so it's low spin. d\(^5\) in strong field = 1 unpaired electron. Paramagnetic.
• (C) [Co(C\(_2\)O\(_4\))\(_3\)]\(^{3-}\): Co is in +3 state (d\(^6\)). C\(_2\)O\(_4^{2-}\) (oxalate) is a moderately strong field ligand. For d\(^6\), it usually causes pairing, resulting in 0 unpaired electrons. Diamagnetic.
• (D) [FeF\(_6\)]\(^{3-}\): Fe is in +3 state (d\(^5\)). F\(^-\) is a weak field ligand, so high spin. d\(^5\) in weak field = 5 unpaired electrons. Paramagnetic.
• (E) [CoF\(_6\)]\(^{3-}\): Co is in +3 state (d\(^6\)). F\(^-\) is a weak field ligand, so high spin. d\(^6\) in weak field = 4 unpaired electrons. Paramagnetic.

Step 3: Final Answer:
[Co(C\(_2\)O\(_4\))\(_3\)]\(^{3-}\) is diamagnetic.