Which one of the following complex ions is diamagnetic in nature?
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To determine if a complex is para- or diamagnetic, follow these steps:
1. Find the oxidation state of the central metal.
2. Determine the number of d-electrons in the metal ion.
3. Classify the ligand as strong-field (causes pairing, low-spin) or weak-field (no pairing, high-spin).
4. Fill the d-orbitals (\(t_{2g}\) and \(e_g\)) accordingly.
5. If there are any unpaired electrons, it's paramagnetic. If all electrons are paired, it's diamagnetic.
Co\(^{3+}\) and \(d^6\) ions with strong-field ligands are a common source of diamagnetic complexes.
A complex is diamagnetic if it has no unpaired electrons. We need to analyze the electron configuration of the central metal ion in each complex.
(A) [CoF\(_6\)]\(^{3-}\): The oxidation state of Co is +3. Co([Ar]\(3d^7 4s^2\)) \(\rightarrow\) Co\(^{3+}\)([Ar]\(3d^6\)). Fluoride (F\(^-\)) is a weak-field ligand, so this is a high-spin complex. The six \(d\) electrons will occupy the orbitals as \(t_{2g}^4 e_g^2\), resulting in 4 unpaired electrons. It is paramagnetic.
(B) [Co(ox)\(_3\)]\(^{3-}\): The oxidation state of Co is +3, so it is a \(d^6\) ion. Oxalate (\(ox^{2-} = C_2O_4^{2-}\)) is a bidentate ligand. While it is not as strong as CN\(^-\), it is generally considered a strong enough field ligand to cause pairing for Co\(^{3+}\), which has a large crystal field splitting energy. This forms a low-spin complex. The six \(d\) electrons will pair up in the lower energy orbitals, giving the configuration \(t_{2g}^6 e_g^0\). There are 0 unpaired electrons. This complex is diamagnetic.
(C) [Mn(CN)\(_6\)]\(^{3-}\): The oxidation state of Mn is +3. Mn([Ar]\(3d^5 4s^2\)) \(\rightarrow\) Mn\(^{3+}\)([Ar]\(3d^4\)). Cyanide (CN\(^-\)) is a strong-field ligand, so this is a low-spin complex. The four \(d\) electrons will occupy the \(t_{2g}\) orbitals as \(t_{2g}^4 e_g^0\), resulting in 2 unpaired electrons (using Hund's rule within the \(t_{2g}\) level). It is paramagnetic.
(D) [Fe(CN)\(_6\)]\(^{3-}\): The oxidation state of Fe is +3. Fe([Ar]\(3d^6 4s^2\)) \(\rightarrow\) Fe\(^{3+}\)([Ar]\(3d^5\)). Cyanide (CN\(^-\)) is a strong-field ligand, forming a low-spin complex. The five \(d\) electrons will occupy the orbitals as \(t_{2g}^5 e_g^0\), resulting in 1 unpaired electron. It is paramagnetic.
Therefore, the only diamagnetic complex is [Co(ox)\(_3\)]\(^{3-}\).
