Which metal ion is most likely to form a square planar complex ion wi…

The correct answer is Option (3) → Ni²⁺

A square planar complex uses dsp² hybridisation, which needs one vacant inner 3d orbital. For \( \ce{Ni^2+} \) with \( \ce{CN^-} \):

• Configuration: Ni (Z = 28) is [Ar]3d⁸4s²; losing the two 4s electrons gives \( \ce{Ni^2+} \): 3d⁸.

• Effect of CN⁻: \( \ce{CN^-} \) is a strong field ligand, so the eight 3d electrons pair up in four 3d orbitals and one 3d orbital is left empty.

• Hybridisation: that 3d orbital, the 4s and two 4p orbitals mix to give four dsp² orbitals for the four \( \ce{CN^-} \) electron pairs.

• Result: \( \ce{[Ni(CN)4]^2-} \), square planar.

Why the other options are incorrect

Option (1) Cu⁺ — Incorrect. Cu (Z = 29) is [Ar]3d¹⁰4s¹, so \( \ce{Cu^+} \) is 3d¹⁰: the 3d orbitals are full and no 3d orbital is free for dsp².

Option (2) Ag⁺ — Incorrect. Ag (Z = 47) is [Kr]4d¹⁰5s¹, so \( \ce{Ag^+} \) is 4d¹⁰, with no vacant inner d orbital; with cyanide it forms the two-coordinate \( \ce{[Ag(CN)2]^-} \).

Option (4) Zn²⁺ — Incorrect. Zn (Z = 30) is [Ar]3d¹⁰4s², so \( \ce{Zn^2+} \) is 3d¹⁰: no 3d orbital is free for dsp².

NCERT (2026-27 edition):
5.5.1 Valence Bond Theory – “In the square planar complexes, the hybridisation involved is dsp². An example is [Ni(CN)₄]^2–. Here nickel is in +2 oxidation state and has the electronic configuration 3d⁸.”