Nickel ($Z=28$) combines with a uninegative monodentate ligand ${\text{X}}^{-}$ to form a paramagnetic complex $[{\text{NiX}}_4{]}^{2-}$. The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively:

1. Oxidation state: $\text{Ni}+4(-1)=-2⟹{\text{Ni}}^{2+}$.2. Electronic Configuration: ${\text{Ni}}^{2+}$ is $d^8$.3. Paramagnetism: For $d^8$ to be paramagnetic, there must be unpaired electrons.If it were square planar, the $d^8$ configuration would pair up electrons in the $d_{x^2-y^2}$ vacancy (diamagnetic).In a tetrahedral geometry, $d^8$ fills as $e^4{t}_2^4$, leaving $2$ unpaired electrons in the $t_2$ orbitals.Thus, the complex is tetrahedral with two unpaired electrons.