The correct statement among the following is:

The correct statement among the following is:
  1. $\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are diamagnetic and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ is paramagnetic.
  2. $\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ are diamagnetic and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ is paramagnetic.
  3. $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ are diamagnetic and $\mathrm{Ni}(\mathrm{CO})_{4}$ is paramagnetic.
  4. $\mathrm{Ni}(\mathrm{CO})_{4}$ is diamagnetic and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are paramagnetic.

Solution

In $Ni(CO)_4$, Nickel is in $0$ oxidation state with configuration $[Ar]3d^8 4s^2$. $CO$ is a strong field ligand, causing pairing of $4s$ electrons into $3d$ orbitals, resulting in $3d^{10}$ configuration. It is $sp^3$ hybridized and diamagnetic (zero unpaired electrons).
In $[Ni(CN)_4]^{2-}$, Nickel is in $+2$ oxidation state with configuration $[Ar]3d^8$. $CN^-$ is a strong field ligand, causing pairing of the two unpaired electrons in $3d$ orbitals. It is $dsp^2$ hybridized (square planar) and diamagnetic.
In $[NiCl_4]^{2-}$, Nickel is in $+2$ oxidation state with configuration $[Ar]3d^8$. $Cl^-$ is a weak field ligand and cannot cause pairing of electrons. It has two unpaired electrons in $3d$ orbitals, making it $sp^3$ hybridized (tetrahedral) and paramagnetic.
Therefore, $Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic, while $[NiCl_4]^{2-}$ is paramagnetic.

Asked in: JEE Main 2026 (28 Jan Shift 1)

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