Which metal ion is likely to form a square planar complex ion with CN^-?

Ni²⁺

The correct answer is option 3. \(Ni^{2+}\).

The formation of square planar complexes typically involves transition metal ions with a \(d^8\) electron configuration. In the case of \(Ni^{2+}\), it has the electron configuration \([Ar] 3d^8\), where there are 8 d-electrons available for coordination.

Square planar geometry is favored for metal ions with a \(d^8\) electron configuration because it allows for maximum ligand field splitting and achieves greater stability compared to other geometries. In a square planar complex, the metal ion is surrounded by four ligands arranged at the corners of a square in the same plane.

The cyanide ion \((CN^-)\) is a strong-field ligand known for forming stable square planar complexes with transition metal ions, particularly those with a \(d^8\) electron configuration. The strong ligand field produced by \(CN^-\) effectively splits the d orbitals into two sets: a lower energy set \((t_{2g})\) and a higher energy set \((e_g)\). This splitting results in the filling of the \(t_{2g}\) orbitals, leaving the eg orbitals empty, which is characteristic of a d8 configuration and facilitates the formation of square planar geometry.

Therefore, \(Ni^{2+}\) is the metal ion most likely to form a square planar complex ion with \(CN^-\).