1. Find the oxidation state of nickel in $[NiX_4]^{2-}$.

Since each ligand $X^-$ has charge $-1$,

So the metal ion is $\text{Ni}^{2+}$.

2. Write the electronic configuration of $\text{Ni}^{2+}$.

Nickel $(Z=28)$ has ground-state configuration

Removing two electrons gives

So it is a $d^8$ system.

3. Use the given information: the complex is paramagnetic.

The complex is $[NiX_4]^{2-}$ with a uninegative monodentate ligand. Such 4-coordinate nickel(II) complexes with weak-field ligands like halides commonly form tetrahedral complexes.

For a tetrahedral field, crystal field splitting is small, so electrons remain unpaired as much as possible.

For a $d^8$ tetrahedral configuration, the arrangement gives 2 unpaired electrons.

4. Check square planar possibility.

A square planar $d^8$ complex of $\text{Ni}^{2+}$ is usually diamagnetic because electrons pair up completely in the lower-energy orbitals.

But the question says the complex is paramagnetic, so square planar is ruled out.

5. Conclusion.

Thus, the complex $[NiX_4]^{2-}$ has:

• 2 unpaired electrons
• tetrahedral geometry

So the correct option is B.