The correct combination is :
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- \([\mathrm{Ni(CN)_4}]^{2-}\) - tetrahedral ; \([\mathrm{Ni(CO)_4}]\) - paramagnetic
- \([\mathrm{NiCl_4}]^{2-}\) - paramagnetic ; \([\mathrm{Ni(CO)_4}]\) - tetrahedral
- \([\mathrm{NiCl_4}]^{2-}\) - square-planar ; \([\mathrm{Ni(CN)_4}]^{2-}\) - paramagnetic
- \([\mathrm{NiCl_4}]^{2-}\) - diamagnetic ; \([\mathrm{Ni(CO)_4}]\) - square-planar
The Correct Option is B
Solution and Explanation
\([\mathrm{Ni(CO)_4}]\): Ni(0) is \(3d^{8}4s^{2}\); CO is a strong-field ligand, so the electrons rearrange to \(3d^{10}\), hybridisation \(sp^3\), tetrahedral, no unpaired electrons, diamagnetic.
\([\mathrm{Ni(CN)_4}]^{2-}\): Ni(II) \(d^8\) with strong-field CN-, \(dsp^2\), square planar, diamagnetic.
Checking options: (1) wrong (the cyanide complex is square planar, the carbonyl is diamagnetic); (2) correct; (3) wrong; (4) wrong.
Answer: (2) \([\mathrm{NiCl_4}]^{2-}\) - paramagnetic ; \([\mathrm{Ni(CO)_4}]\) - tetrahedral
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