Answer the question on the basis of passage given below:
Transition metals form a large number of complexes or coordination compounds in which the metal atoms are bound to a number of anions or neutral molecules. The valence bond theory explains the formation, magnetic behaviour and geometrical shapes while the crystal field theory explains the effect of different crystal fields on the degeneracy of d-orbital energies of the central metal atom/ion. This provides for the quantitative estimation of orbital separation energies, magnetic moments and spectral and stability parameters.
Two complexes of nickel have same geometry but different magnetic behaviour are:
(A) \([Ni(CN)_4]^{2-}\)
(B) \([Ni(CO)_4]\)
(C) \([NiCl_4]^{2-}\)
(D) \([Ni(NH_3)_6]^{2+}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
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:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 10-06 S3 22-05 S3 | – |