Paragraph: 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

• A

  both are diamagnetic
• B

  $A$ is diamagnetic and $B$ is paramagnetic with one unpaired electron
• Correct answer
  C

  $A$ is diamagnetic and $B$ is paramagnetic with two unpaired electrons
• D

  both are paramagnetic

Solution

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.

Hence, the correct answer is option: 'C'.