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
A complex ion is diamagnetic if it has no unpaired electrons in its d-orbitals. To determine this, we need to find the oxidation state of the central metal, its d-electron count, and then consider the effect of the ligands (weak field or strong field) on the electron distribution (high spin or low spin) based on Crystal Field Theory.
Step 2: Detailed Explanation
(A) [MnCl₄]²⁻: Oxidation state of Mn: \(x + 4(-1) = -2 \implies x = +2\). Mn(II) is a \(d^5\) ion. Cl⁻ is a weak field ligand. The geometry is tetrahedral. The complex will be high spin, with 5 unpaired electrons. It is strongly paramagnetic.
(B) [Fe(CN)₆]³⁻: Oxidation state of Fe: \(x + 6(-1) = -3 \implies x = +3\). Fe(III) is a \(d^5\) ion. CN⁻ is a strong field ligand. The complex is octahedral and will be low spin. The electron configuration is \(t_{2g}^5 e_g^0\). There is 1 unpaired electron. It is paramagnetic.
(C) [Co(C₂O₄)₃]³⁻: Oxidation state of Co: \(x + 3(-2) = -3 \implies x = +3\). Co(III) is a \(d^6\) ion. C₂O₄²⁻ (oxalate) is considered a relatively strong field ligand, especially with Co(III). The complex is octahedral and will be low spin. The electron configuration is \(t_{2g}^6 e_g^0\). All 6 electrons are paired up in the \(t_{2g}\) orbitals. There are 0 unpaired electrons. The complex is diamagnetic.
(D) [FeF₆]³⁻: Oxidation state of Fe: \(x + 6(-1) = -3 \implies x = +3\). Fe(III) is a \(d^5\) ion. F⁻ is a weak field ligand. The complex is octahedral and will be high spin. The electron configuration is \(t_{2g}^3 e_g^2\). There are 5 unpaired electrons. It is strongly paramagnetic.
(E) [CoF₆]³⁻: Oxidation state of Co: \(x + 6(-1) = -3 \implies x = +3\). Co(III) is a \(d^6\) ion. F⁻ is a weak field ligand. The complex is octahedral and will be high spin. The electron configuration is \(t_{2g}^4 e_g^2\). There are 4 unpaired electrons. It is paramagnetic.
Step 3: Final Answer
The only complex ion with zero unpaired electrons is \([\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}\), which is therefore diamagnetic.