Amongst Ni(CO)4, [Ni(CN)4]2- and [NiCl4]2- :

Ni(CO) 4 is diamagnetic and [NiCl 4 ] 2- and [Ni(CN) 4 ] 2- are paramagnetic

Ni(CO) 4 is diamagnetic and [NiCl 4 ] 2- and [Ni(CN) 4 ] 2- are paramagnetic

[NiCl 4 ] 2- and [Ni(CN) 4 ] 2- are diamagnetic and Ni(CO) 4 is paramagnetic

Ni(CO) 4 and [Ni(CN) 4 ] 2- are diamagnetic and [NiCl 4 ] 2- is paramagnetic

Ni(CO) 4 is diamagnetic and [NiCl 4 ] 2- and [Ni(CN) 4 ] 2- are paramagnetic

Ni(CO) 4 and [NiCl 4 ] 2- are diamagnetic and [Ni(CN) 4 ] 2- is paramagnetic

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Amongst Ni(CO)₄, [Ni(CN)₄]^2- and [NiCl₄]^2- :

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[NiCl₄]^2- and [Ni(CN)₄]^2- are diamagnetic and Ni(CO)₄ is paramagnetic

Ni(CO)₄ and [Ni(CN)₄]^2- are diamagnetic and [NiCl₄]^2- is paramagnetic

Ni(CO)₄ is diamagnetic and [NiCl₄]^2- and [Ni(CN)₄]^2- are paramagnetic

Ni(CO)₄and [NiCl₄]^2- are diamagnetic and [Ni(CN)₄]^2- is paramagnetic

answer is C.

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When two atomic orbitals join to generate a hybrid orbital in a molecule, the energy of the individual atoms' orbitals is redistributed to give orbitals of equivalent energy. This method is known as hybridization.
Hybridization of [Ni(CO)₄] is sp³ and [Ni(CN)₄]^2-is dsp³, since all the electrons are paired in both of these compounds they are diamagnetic while in[NiCl₄]^2-, Cl is a weak ligand and hence is unable to pair up all the electrons of 3d.

Hence, the correct option is option (C) Ni(CO)₄ and [Ni(CN)₄]^2- are diamagnetic and [NiCl₄]^2- is paramagnetic.

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