Among $\left[\mathrm{Ni}(\mathrm{CO}{)}_4\right]·{\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}\text{ and }{\left[{\mathrm{NiCl}}_4\right]}^{2-}$

[Ni(CO)₄]
Ni has the oxidation number O.
The atomic number is 28.
Ni = [Ar]3d⁸ 4s² 
hybridization sp ³ (tetrahedral)
The complex is diamagnetic because there are no unpaired electrons.
Spin magnetic moment = 0 

[Ni(CN)₄]²⁻
Ni has an oxidation number of +2.
The atomic number is 28.
Ni = [Ar]3d⁸ 4s²
Ni 2+= [Ar]3d⁸
Hybridization of dsp² (square planar)
The complex is diamagnetic because there are no unpaired electrons.
Magnetic moment of spin = 0
 

[NiCl₄​]²⁻
$$\begin{gathered} \mathrm{\mu } = \sqrt{\mathrm{n}(\mathrm{n}+2}\mathrm{BM} \\ = \sqrt{2(2+2)}\mathrm{BM} = \sqrt{8}\mathrm{BM} \end{gathered}$$

Hence, the correct option (C). 

($\left[\mathrm{Ni}(\mathrm{CO}{)}_4\right]\text{ and }{\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}\text{ are diamagnetic and }{\left[{\mathrm{NiCl}}_4\right]}^{2-}\text{ is paramagnetic}$)