1. Find the oxidation state and electronic configuration of Ni in each complex

• For $[Ni(Cl)_4]^{2-}$:

  Let oxidation state of Ni be $x$. $x+4(-1)=-2 \Rightarrow x=+2$ So, $Ni^{2+}$ has configuration: $Ni: [Ar]3d^8 4s^2 \, ,\, Ni^{2+}:[Ar]3d^8$

• For $[Ni(CN)_4]^{2-}$:

  Let oxidation state of Ni be $x$. $x+4(-1)=-2 \Rightarrow x=+2$ Again, $Ni^{2+}$ is $3d^8$.

• For $Ni(CO)_4$:

  CO is a neutral ligand, so oxidation state of Ni is $0$. $Ni^0:[Ar]3d^8 4s^2$

2. Determine geometry and magnetic nature

(i) $[Ni(Cl)_4]^{2-}$

• $Cl^-$ is a weak field ligand.
• For $Ni^{2+}(d^8)$ with weak ligands, the complex is usually tetrahedral.
• In tetrahedral field, electrons remain unpaired.
• Hence it is paramagnetic.

So: $[Ni(Cl)_4]^{2-} : \text{tetrahedral, paramagnetic}$

(ii) $[Ni(CN)_4]^{2-}$

• $CN^-$ is a strong field ligand.
• For $Ni^{2+}(d^8)$ with strong ligands, pairing occurs and the complex becomes square planar ($dsp^2$).
• All electrons are paired.
• Hence it is diamagnetic.

So: $[Ni(CN)_4]^{2-} : \text{square planar, diamagnetic}$

(iii) $Ni(CO)_4$

• CO is a strong field neutral ligand.
• $Ni(0)$ forms $Ni(CO)_4$ using $sp^3$ hybridization.
• Geometry is tetrahedral.
• All electrons are paired, so it is diamagnetic.

So: $Ni(CO)_4 : \text{tetrahedral, diamagnetic}$

3. Check each option

Option A

$[Ni(CN)_4]^{2-}$ tetrahedral ; $[Ni(CO)_4]$ paramagnetic

• First part wrong: $[Ni(CN)_4]^{2-}$ is square planar, not tetrahedral.
• Second part wrong: $Ni(CO)_4$ is diamagnetic, not paramagnetic.

A is incorrect.

Option B

$[Ni(Cl)_4]^{2-}$ paramagnetic ; $[Ni(CO)_4]$ tetrahedral

• $[Ni(Cl)_4]^{2-}$ is indeed paramagnetic.
• $Ni(CO)_4$ is indeed tetrahedral.

B is correct.

Option C

$[Ni(Cl)_4]^{2-}$ square-planar ; $[Ni(CN)_4]^{2-}$ paramagnetic

• First part wrong: $[Ni(Cl)_4]^{2-}$ is tetrahedral, not square planar.
• Second part wrong: $[Ni(CN)_4]^{2-}$ is diamagnetic, not paramagnetic.

C is incorrect.

Option D

$[Ni(Cl)_4]^{2-}$ diamagnetic ; $[Ni(CO)_4]$ square-planar

• First part wrong: $[Ni(Cl)_4]^{2-}$ is paramagnetic.
• Second part wrong: $Ni(CO)_4$ is tetrahedral.

D is incorrect.

4. Final Answer

The correct combination is: $\boxed{B}$

5. Comparison with stored correct answer

Stored correct answer = $B$.

My derived answer also = $B$. So they agree.