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The correct combination is: [JEE-Main On line-2018]

1. [NiCl4]2− - square-planar; [Ni(CN)4]2− - paramagnetic
2. [Ni(CN)4]2− - tetrahedral; [Ni(CO)4] - paramagnetic
3. [NiCl4]2− - paramagnetic; [Ni(CO)4] - tetrahedral
4. [NiCl4]2− - diamagnetic; [Ni(CO)4] - square-planar

Correct answer: [NiCl4]2− - paramagnetic; [Ni(CO)4] - tetrahedral

Solution

The correct option states that [NiCl4]2− is paramagnetic due to the presence of unpaired electrons in its d-orbitals, while [Ni(CO)4] is tetrahedral and diamagnetic because CO is a strong field ligand that causes pairing of electrons.

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