Which of the following divalent metal ion form the most stable complexes?

Cu²⁺

The correct answer is option 4. \(Cu^{2+}\).

The stability of complexes formed by divalent metal ions can be understood using the Irving-Williams series, which ranks the relative stability of complexes formed by the first-row transition metals in the +2 oxidation state. The Irving-Williams series is as follows:

\(\text{Mn}^{2+} < \text{Fe}^{2+} < \text{Co}^{2+} < \text{Ni}^{2+} < \text{Cu}^{2+} > \text{Zn}^{2+}\)

This series indicates that as we move from left to right across the first-row transition metals, the stability of their \(+2\) complexes generally increases up to copper \((Cu)\) and then decreases slightly for zinc \((Zn)\). Here’s a detailed explanation of why this trend occurs:

Crystal Field Stabilization Energy (CFSE): The stability of a metal complex is influenced by the crystal field stabilization energy, which arises from the splitting of d-orbitals in the presence of a ligand field. Transition metals such as \(Ni^{2+}\) and \(Cu^{2+}\) have higher CFSE compared to \(Mn^{2+}\) and \(Fe^{2+}\), contributing to the increased stability of their complexes.

Ligand Field Effects: \(Cu^{2+}\) has a strong preference for forming stable complexes due to its high ligand field stabilization. This is because \(Cu^{2+}\) has a \(d^9\) electronic configuration, which leads to a significant stabilization when forming complexes, as it effectively utilizes the ligand field to lower its energy.

Jahn-Teller Distortion: \(Cu^{2+}\) often experiences Jahn-Teller distortion, which further stabilizes its complexes. This distortion occurs because the \(d^9\) configuration leads to an uneven distribution of electrons in the \(e_g\) orbitals, causing the complex to distort to lower its overall energy.

Charge Density and Ionic Radius: As we move from \(Mn^{2+}\) to \(Cu^{2+}\), the charge density (charge per unit volume) of the metal ion increases due to the decreasing ionic radius. This increased charge density allows for stronger electrostatic attraction between the metal ion and the ligands, enhancing the stability of the complex.

Given these factors, \(Cu^{2+}\) forms the most stable complexes among the divalent metal ions listed \((Mn^{2+},\, \ Fe^{2+},\, \ Ni^{2+},\, \  Cu^{2+})\). Thus, the divalent metal ion that forms the most stable complexes is: \({\text{Cu}^{2+}}\)