Nickel $$(Z = 28)$$ combines with a uninegative monodentate ligand $$X^-$$ to form a paramagnetic complex $$[NiX_4]^{2-}$$. The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively

The complex is $$[NiX_4]^{2-}$$.

Step 1 : Oxidation state and d-electron count
Each $$X^-$$ is uninegative and the overall charge is $$-2$$, so for nickel
$$x + 4(-1) = -2 \;\Rightarrow\; x = +2.$$
Atomic configuration of Ni (Z = 28) is $$[Ar]\,3d^{8}\,4s^{2}$$.
Removing two electrons (from 4s) gives
$$Ni^{2+} : [Ar]\,3d^{8}.$$
Thus the metal centre is $$d^{8}$$.

Step 2 : Nature of the ligand and preferred geometry
Halide-type ligands $$X^-$$ lie at the weak-field end of the spectrochemical series.
For a four-coordinate $$d^{8}$$ ion, a weak-field environment favours tetrahedral geometry (sp3 hybridisation) because the crystal-field splitting energy $$\Delta_t$$ is smaller than the electron-pairing energy. A square-planar (dsp2) arrangement is adopted only with strong-field ligands such as $$CN^-$$ or $$NH_3$$.

Step 3 : Crystal-field splitting in a tetrahedral field
In a tetrahedral field the five d orbitals split into a lower $$e$$ set (2 orbitals) and a higher $$t_2$$ set (3 orbitals). For a high-spin $$d^{8}$$ ion the electron filling is

$$e : \uparrow\downarrow \; \uparrow\downarrow \; (4\;e^-)$$
$$t_2 : \uparrow\downarrow \; \uparrow \; \uparrow \; (4\;e^-)$$

The $$t_2$$ set contains two singly occupied orbitals, so

Number of unpaired electrons = 2.

Step 4 : Paramagnetism and conclusion
Two unpaired electrons make the complex paramagnetic, and the geometry is tetrahedral.

Hence, the correct combination is:
Option B which is “two, tetrahedral”.