On the basis of valence bond theory,discuss the geometrical shapes and magnetic properties of the following $4$-coordinated complexes:
$(i)$ $[NiCl_{4}]^{2-}$
$(ii)$ $[Ni(CN)_{4}]^{2-}$
$(iii)$ $[Ni(CO)_{4}]$
$(iv)$ $MnO_{4}^{-}$

(N/A) $(i)$ $[NiCl_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state. $Ni^{2+}$ has $3d^{8}$ configuration. $Cl^-$ is a weak field ligand,so no pairing occurs. It undergoes $sp^3$ hybridization. Geometry: Tetrahedral. Magnetic property: Paramagnetic ($n=2$,$\mu = 2.82 \ B.M.$).
$(ii)$ $[Ni(CN)_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state. $Ni^{2+}$ has $3d^{8}$ configuration. $CN^-$ is a strong field ligand,causing pairing of $3d$ electrons. It undergoes $dsp^2$ hybridization. Geometry: Square planar. Magnetic property: Diamagnetic ($n=0$,$\mu = 0 \ B.M.$).
$(iii)$ $[Ni(CO)_{4}]$: $Ni$ is in $0$ oxidation state. $Ni$ has $3d^{8} 4s^{2}$ configuration. $CO$ is a strong field ligand,causing pairing of $4s$ electrons into $3d$ orbitals. It undergoes $sp^3$ hybridization. Geometry: Tetrahedral. Magnetic property: Diamagnetic ($n=0$,$\mu = 0 \ B.M.$).
$(iv)$ $MnO_{4}^{-}$: $Mn$ is in $+7$ oxidation state. $Mn^{7+}$ has $3d^{0}$ configuration. It undergoes $sd^3$ (or $d^3s$) hybridization. Geometry: Tetrahedral. Magnetic property: Diamagnetic $(n=0)$.