Chemistry - Coordination Chemistry Question with Solution | TestHub

In it unpaired orbital is not present, so it is diamagnetic in character (square planar shape) In chloro complex ${\mathrm{K}}_2\left[\mathrm{Ni}(\mathrm{Cl}{)}_4\right]$, complex ion is ${\left[{\mathrm{Ni}}^2\left({\mathrm{Cl}}_4\right)\right]}^{2-}$ and $\mathrm{Ni}$ is present ${\mathrm{Ni}}^{2+}$ or $\mathrm{Ni}(\mathrm{II})$, so $${\mathrm{Ni}}^2+=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^8,4s^0$$ In ${\left[{\mathrm{Ni}}_{\left({\mathrm{Cl}}_4\right)}\right)]}^2$ ion, ${\mathrm{Ni}}^{2+}$ is present as follows due to weaker ligand character is ${\mathrm{Cl}}^{-}$ion. $({\mathrm{Cl}}^{-}$is weak field ligand). So, it is unable to pair up the electron and ${\mathrm{Ni}}^{2+}$ needs four empty orbitals to accommodate four ${\mathrm{Cl}}^{-}$ligand. Thus, ${\left({\mathrm{NiCl}}_4\right)}^{2-}$ shows $s{p}^3$-hybridisation (Tetrahedral shapes)
Hence, due to presence of unpaired orbitals, it is paramagnetic in character.