Which of the following metal ion will be forming most stable coordinate compound (with same particular ligand)

\(Co^{2+}\)

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

To determine which metal ion forms the most stable coordination compound with a particular ligand, we should examine several factors related to coordination chemistry, particularly focusing on the nature of the metal ion and its ability to stabilize the ligand-metal bond. Here’s a detailed analysis of the metal ions listed:

Electronic Configuration and Oxidation State

1. \(Fe^{2+}\) (Iron(II))

Electronic Configuration: \([Ar] 3d^6\)

In an octahedral field, the \(3d^6\) configuration can either be high-spin or low-spin depending on the ligand. For weak field ligands (like chloride), it tends to be high-spin, where all electrons are unpaired, leading to relatively lower stability due to less ligand field stabilization.

2. \(Co^{2+}\) (Cobalt(II))

Electronic Configuration: \([Ar] 3d^7\)

In an octahedral field, \(3d^7\) configuration generally leads to a high-spin complex with weak field ligands, but cobalt(II) often forms stable complexes even with weak field ligands. Cobalt can achieve stabilization through additional factors like Jahn-Teller distortion, which can enhance stability in certain coordination environments.

3. \(Mn^{2+}\) (Manganese(II))

Electronic Configuration: \([Ar] 3d^5\)

The \(3d^5\) configuration is particularly stable due to its half-filled nature, which provides extra stability (exchange energy). However, \(Mn^{2+}\) typically forms high-spin complexes which may not be as stabilized by the ligand field effects as compared to metal ions with lower spin configurations.

4. \(Cr^{2+}\) (Chromium(II))

Electronic Configuration: \([Ar] 3d^4\)

In an octahedral field, the \(3d^4\) configuration can lead to high-spin arrangements. Chromium(II) complexes are less common due to the relative instability of \(3d^4\) in high-spin configurations, making them less stable compared to other metal ions.

Ligand Field Stabilization Energy (LFSE)

The ligand field stabilization energy is a key factor in determining the stability of a coordination compound. This energy is influenced by:

Ligand Field Strength: Strong field ligands (like CN⁻ or NH₃) cause larger splitting of the d-orbitals, leading to low-spin complexes and generally higher stabilization for metal ions with d^6 or d^8 configurations.

Electronic Configuration: The stability of the coordination compound is enhanced if the metal ion can achieve a more stable electronic configuration through interaction with the ligand field.

Comparison of Metal Ions

Iron(II) (\(Fe^{2+}\)): With \(3d^6\), it can form both high-spin and low-spin complexes depending on the ligand. In low-spin configurations, it can achieve significant stabilization.

Cobalt(II) (\(Co^{2+}\)): With \(3d^7\), cobalt(II) often forms stable complexes even with weaker ligands. Cobalt’s ability to undergo Jahn-Teller distortion also adds to its stability in certain complexes.

Manganese(II) (\(Mn^{2+}\)): With \(3d^5\), while stable due to half-filled configuration, the high-spin nature in octahedral complexes might not be as stable as low-spin configurations of other metal ions.

Chromium(II) (\(Cr^{2+}\)): With \(3d^4\), its high-spin nature in octahedral complexes typically results in less stabilization compared to other metal ions.

Conclusion

Cobalt(II) (\(Co^{2+}\)) is generally the most stable metal ion in coordination complexes with a particular ligand due to its ability to form stable complexes even with weak field ligands, and because of additional stabilization mechanisms like Jahn-Teller distortion.

So, the metal ion that will form the most stable coordination compound with the same ligand is 2. \(Co^{2+}\).