From the magnetic behaviour of ${[{NiCl}_{4}]}^{2 -}$ (paramagnetic) and $[Ni (CO)_{4}]$ (diamagnetic), choose the correct geometry and oxidation state.

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${[{NiCl}_{4}]}^{2 -}$ : Ni^II, tetrahedral
$[Ni (CO)_{4}]$ : Ni(0), tetrahedral

${[{NiCl}_{4}]}^{2 -}$ : Ni^II, tetrahedral
$[Ni (CO)_{4}]$ : Ni^II, square planar

${[{NiCl}_{4}]}^{2 -}$ : Ni(0), tetrahedral
$[Ni (CO)_{4}]$ : Ni(0), square planar

${[{NiCl}_{4}]}^{2 -}$ : Ni^II , square planar
$[Ni (CO)_{4}]$ : Ni(0), square planar

answer is B.

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Detailed Solution

${[{NiCl}_{4}]}^{2 -}$
${Ni}^{+ 2} - [Ar] 3 d^{8} 4 s^{0} \to {sp}^{3}$ Tetrahedral
Number of unpaired electron =2 paramagentic
$[Ni (CO)_{4}]$ ,
$Ni (0) \to [Ar] 3 d^{10} 4 s^{0}$ (After rearrangement)
No unpaired electron
sp³, Tetrahedral, Diamagnetic