Two complexes of nickel have same geometry but different magnetic beh…

The correct answer is Option (3) → (B) and (C) only

By valence bond theory, the geometry follows from the hybridisation (dsp² square planar, sp³ tetrahedral, sp³d² octahedral), and the magnetic behaviour from the number of unpaired electrons left on the metal. A strong-field ligand (\( \ce{CN^{-}} \), CO) forces the d electrons to pair; a weak-field ligand (\( \ce{Cl^{-}} \)) does not.

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

• (A) \( \ce{[Ni(CN)4]^{2-}} \): \( \ce{Ni^{2+}} \), 3d⁸. Strong-field \( \ce{CN^{-}} \) pairs the two odd electrons and frees one 3d orbital:

      3d: [↑↓] [↑↓] [↑↓] [↑↓] [  ] → dsp², square planar, 0 unpaired, diamagnetic

• (B) \( \ce{[Ni(CO)4]} \): Ni(0), 3d⁸4s². Strong-field CO pushes the 4s electrons into 3d, which becomes full:

      3d: [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] → sp³, tetrahedral, 0 unpaired, diamagnetic

• (C) \( \ce{[NiCl4]^{2-}} \): \( \ce{Ni^{2+}} \), 3d⁸. Weak-field \( \ce{Cl^{-}} \) causes no pairing; by Hund’s rule two orbitals stay singly filled:

      3d: [↑↓] [↑↓] [↑↓] [↑ ] [↑ ] → sp³, tetrahedral, 2 unpaired, paramagnetic

• (D) \( \ce{[Ni(NH3)6]^{2+}} \): \( \ce{Ni^{2+}} \), 3d⁸. Six ligands need six orbitals; with 3d⁸ two inner d orbitals cannot be emptied, so outer 4d orbitals are used:

      3d: [↑↓] [↑↓] [↑↓] [↑ ] [↑ ] → sp³d², octahedral, 2 unpaired, paramagnetic

NCERT (2026-27 edition):
Valence Bond Theory – …

Note: The same/similar question has been asked in different years by NTA as per the details given below: