Select the correct statement.

• A

  $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is diamagnetic whereas $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ is paramagnetic.
• Correct answer
  B

  $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ and $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ both are diamagnetic.
• C

  $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is $s p^3$ hybridised and square planar in shape whereas $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ is $d s p^2$ hybridised and tetrahedral in shape.
• D

  $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is paramagnetic and $d s p^2$ hybridised whereas $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ is diamagnetic and $s p^3$ hybridised.

Solution

Both $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ and $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ are diamagnetic. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$
Electronic configuration of $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8$. $\mathrm{CN}^{-}$being strong field ligand cause pairing of electrons.
There are no unpaired electrons. Therefore, it is diamagnetic. Shape : square planar $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ Electronic configuration of $\mathrm{Ni}=[\mathrm{Ar}] 3 d^8 4 s^2$ CO being strong field ligand cause pairing of electrons.
No unpaired electrons are there. Therefore, it is diamagnetic in nature. Shape : Tetrahedral.

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