Consider the nickel(II) complexes [NiCl4]^2-, [Ni(CN)4]^2- and Ni(CO)4 (atomic number of Ni =28).

Ni(II) is $d^8$. With the weak-field ligand $\mathrm{Cl^-}$ the complex $\mathrm{[NiCl_4]^{2-}}$ is $sp^3$ tetrahedral; in the tetrahedral field the $d^8$ configuration $e^4t_2^4$ leaves two unpaired electrons, so it is paramagnetic and (A) is correct. The strong-field ligand $\mathrm{CN^-}$ forces pairing of all eight electrons in four orbitals, freeing one $3d$ orbital for $dsp^2$ hybridisation, so $\mathrm{[Ni(CN)_4]^{2-}}$ is square planar and diamagnetic and (B) is correct. In $\mathrm{Ni(CO)_4}$ the strong-field CO ligands push the two $4s$ electrons into $3d$, giving $3d^{10}$ nickel(0); it is $sp^3$ tetrahedral but has no unpaired electron, so it is diamagnetic and (C) is wrong. Since $\mathrm{[NiCl_4]^{2-}}$ has two unpaired electrons ($\mu=\sqrt{8}\approx2.83$ BM) while $\mathrm{[Ni(CN)_4]^{2-}}$ has none, (D) is wrong.