Which of the following metal ion will be forming most stable coordina…

The correct answer is Option (2) → \(Co^{2+}\)

With the same ligand, the stability of a complex depends on the metal ion. Two factors decide it for ions of the same charge: a smaller ionic radius (higher charge density, so a stronger metal–ligand bond) and a larger crystal field stabilisation energy (CFSE).

All four ions are M²⁺ ions of the first transition series. Their configurations, with the high-spin octahedral CFSE (weak ligand, Δ_o < P), are:

• Cr (Z = 24) [Ar]3d⁵4s¹ → \( \ce{Cr^{2+}} \) 3d⁴: t_2g³e_g¹, CFSE = −0.6 Δ_o; radius about 80 pm.

• Mn (Z = 25) [Ar]3d⁵4s² → \( \ce{Mn^{2+}} \) 3d⁵: t_2g³e_g², CFSE = 0; radius about 83 pm.

• Fe (Z = 26) [Ar]3d⁶4s² → \( \ce{Fe^{2+}} \) 3d⁶: t_2g⁴e_g², CFSE = −0.4 Δ_o; radius about 78 pm.

• Co (Z = 27) [Ar]3d⁷4s² → \( \ce{Co^{2+}} \) 3d⁷: t_2g⁵e_g², CFSE = −0.8 Δ_o; radius about 75 pm.

\( \ce{Co^{2+}} \) is the smallest of the four and has the largest CFSE, so it forms the most stable complex. This fits the observed stability order of bivalent ions, \( \ce{Mn^{2+}} \) < \( \ce{Fe^{2+}} \) < \( \ce{Co^{2+}} \) < \( \ce{Ni^{2+}} \) < \( \ce{Cu^{2+}} \) > \( \ce{Zn^{2+}} \) (Irving–Williams series).

Why the other options are incorrect

• Option (1) Fe²⁺ — Incorrect. Larger than \( \ce{Co^{2+}} \) and with a smaller CFSE (−0.4 Δ_o).

• Option (3) Mn²⁺ — Incorrect. The largest of the four, and its half-filled d⁵ ion has zero CFSE; its complexes are the least stable.

• Option (4) Cr²⁺ — Incorrect. Larger than \( \ce{Co^{2+}} \), with a smaller CFSE (−0.6 Δ_o).

Note: The question is from beyond the NCERT textbook.