Which among the following will show square planar geometry?

The correct answer is Option (4) → [PtCl₄]²⁻

The geometry of a four-coordinate complex depends on the hybridisation of the metal ion:

• sp³ hybridisation gives a tetrahedral complex.

• dsp² hybridisation gives a square planar complex. It needs one vacant inner d orbital, which a d⁸ ion gets when its electrons pair up.

For 3d⁸ \( \ce{Ni^{2+}} \), pairing occurs only with strong field ligands such as \( \ce{CN^{-}} \).

Pt (Z = 78) is [Xe] 4f¹⁴5d⁹6s¹; losing the 6s electron and one 5d electron gives \( \ce{Pt^{2+}} \): 5d⁸, and \( \ce{[PtCl4]^{2-}} \) is square planar (dsp², diamagnetic), as the book states.

So the complex with square planar geometry is [PtCl₄]²⁻.

Why the other options are incorrect

• Option (1) [Ni(Cl)₄]²⁻ — Incorrect. Ni (Z = 28) is [Ar] 3d⁸4s²; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d⁸, and \( \ce{Cl^{-}} \) is a weak field ligand, so the electrons do not pair; the complex is sp³, tetrahedral and paramagnetic with 2 unpaired electrons.

• Option (2) [Ni(CO)₄] — Incorrect. Nickel is in the zero oxidation state, Ni (Z = 28): [Ar] 3d⁸4s²; the strong field of CO pushes the two 4s electrons into 3d, giving 3d¹⁰, so it uses sp³ orbitals and is tetrahedral.

• Option (3) [Cr(CO)₆] — Incorrect. Cr (Z = 24) is [Ar] 3d⁵4s¹; in Cr(0) the six electrons pair up in three 3d orbitals under the strong field of CO, leaving two 3d orbitals for d²sp³ hybridisation with six ligands, so it is octahedral.

NCERT (2026-27 edition):
Definitions of Some Important Terms Pertaining to Coordination Compounds – …