1. Find oxidation state of Ni in each complex

   (i) In $[\mathrm{NiCl}_4]^{2-}$

   Let oxidation state of Ni be $x$.

   Each $\mathrm{Cl}^-$ ligand has charge $-1$.

   $x + 4(-1) = -2$ $x - 4 = -2$ $x = +2$

   So, Ni is in $+2$ oxidation state.

   (ii) In $[\mathrm{Ni}(\mathrm{CO})_4]$

   CO is a neutral ligand, so

   $x + 4(0) = 0$ $x = 0$

   So, Ni is in $0$ oxidation state.

2. Use magnetic behavior to infer geometry

   (i) $[\mathrm{NiCl}_4]^{2-}$ is paramagnetic

   Ni$^{2+}$ has electronic configuration:

   $\mathrm{Ni}: [\mathrm{Ar}]3d^8 4s^2$ $\mathrm{Ni}^{2+}: [\mathrm{Ar}]3d^8$

   With weak-field ligand $\mathrm{Cl}^-$, a 4-coordinate Ni$^{2+}$ complex is typically tetrahedral.

   Tetrahedral $d^8$ complexes have unpaired electrons, hence are paramagnetic.

   Therefore, $[\mathrm{NiCl}_4]^{2-}$ is tetrahedral.

   (ii) $[\mathrm{Ni}(\mathrm{CO})_4]$ is diamagnetic

   Ni$(0)$ has configuration:

   $\mathrm{Ni}(0): [\mathrm{Ar}]3d^8 4s^2$

   In this complex, due to strong-field CO ligand and metal-ligand bonding, electrons become paired, and the complex is known to be tetrahedral and diamagnetic.

3. Match with options

   • For $[\mathrm{NiCl}_4]^{2-}$: Ni$^{2+}$, tetrahedral
   • For $[\mathrm{Ni}(\mathrm{CO})_4]$: Ni$(0)$, tetrahedral

   This matches Option B.

4. Check other options briefly

   • A: wrong oxidation state for $[\mathrm{NiCl}_4]^{2-}$ and wrong geometry for $[\mathrm{Ni}(\mathrm{CO})_4]$
   • C: wrong oxidation state for $[\mathrm{Ni}(\mathrm{CO})_4]$
   • D: wrong geometry for both complexes

Hence, the correct answer is B.