Assuming \([NiCl_4]^{2-}\) as Complex-A and \([Ni(CO)_4]\) as Complex-B, choose the correct statement. (At.No. of \(Ni\) is 28.)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Complex-A is paramagnetic and complex-B is diamagnetic
By valence bond theory, a four-coordinate complex is tetrahedral (sp3) when the inner d orbitals are not made free, and square planar (dsp2) when the ligand is strong enough to pair the d electrons and free one 3d orbital.
Complex-A, [NiCl4]2−: x + 4(−1) = −2 gives \( \ce{Ni^{2+}} \). Ni (Z = 28) is [Ar]3d84s2; \( \ce{Ni^{2+}} \) is 3d8. \( \ce{Cl^{-}} \) is a weak ligand and does not pair the electrons, so two 3d electrons stay unpaired and the 4s and 4p orbitals give sp3 (tetrahedral). With 2 unpaired electrons the complex is paramagnetic.
Complex-B, [Ni(CO)4]: CO is neutral, so nickel is 0. Ni(0) has 10 valence electrons; with the strong ligand CO they are all paired as 3d10, and the 4s and 4p orbitals give sp3 (tetrahedral). No unpaired electron: the complex is diamagnetic.
Why the other options are incorrect
Option (1) State of hybridization of Ni in complex-A is dsp2 and in complex-B is sp3 — Incorrect. \( \ce{[NiCl4]^{2-}} \) is sp3 (tetrahedral); dsp2 is found in \( \ce{[Ni(CN)4]^{2-}} \), where the strong \( \ce{CN^{-}} \) pairs the electrons.
Option (3) Oxidation state of Ni in both complexes is +2 — Incorrect. Nickel is +2 in complex-A but 0 in \( \ce{[Ni(CO)4]} \).
Option (4) CO is a weaker ligand than Cl− — Incorrect. The reverse is true. CO is a strong field ligand (it also forms back bonds with the metal); \( \ce{Cl^{-}} \) is weak.
NCERT (2026-27 edition):
Valence Bond Theory – …
