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The coordination number of \(\mathrm{Ni}^{2+}\) is 4 .
\(\mathrm{NiCl}_2+\mathrm{KCN}\) (excess) \(\rightarrow A\) (cyano complex)
\(\mathrm{NiCl}_2+\) conc. \(\mathrm{HCl}\) (excess) \(\rightarrow B\) (chloro complex)
Question:
Predict the magnetic nature of \(A\) and \(B\) are

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
\(\mathrm{Ni}^2+=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^8, 4 s^0\)
In \(\left.\left[\mathrm{Ni}_{\left(\mathrm{Cl}_4\right)}\right)\right]^2\) ion, \(\mathrm{Ni}^{2+}\) is present as follows due to weaker ligand character is \(\mathrm{Cl}^{-}\)ion. \(\left(\mathrm{Cl}^{-}\right.\)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.