Find the oxidation state and electronic configuration of Ni in each complex

• For $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$:

  • CO is a neutral ligand.
  • So oxidation state of Ni is $0$.
  • Atomic number of Ni = 28
  • Ground-state configuration of Ni: $\mathrm{Ni}: [\mathrm{Ar}]\,3d^8 4s^2$
  • Hence $\mathrm{Ni}^0$ effectively has $3d^{10}$ configuration in the complex.

• For $\left[\mathrm{NiCl}_4\right]^{2-}$:

  • Let oxidation state of Ni be $x$. $x + 4(-1) = -2$ $x = +2$
  • So Ni is $\mathrm{Ni}^{2+}$.
  • Configuration of $\mathrm{Ni}^{2+}$: $[\mathrm{Ar}]\,3d^8$

Determine magnetic nature of $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$

• CO is a strong-field ligand.
• $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ is a tetrahedral-type carbonyl complex with Ni in $d^{10}$ state.
• A $d^{10}$ configuration has all electrons paired.
• Therefore, $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ is diamagnetic.

Determine magnetic nature of $\left[\mathrm{NiCl}_4\right]^{2-}$

• $\mathrm{Cl}^-$ is a weak-field ligand.
• $\left[\mathrm{NiCl}_4\right]^{2-}$ is tetrahedral.
• For tetrahedral $\mathrm{Ni}^{2+}$, we have $d^8$ configuration.
• Tetrahedral splitting is small, so electrons remain unpaired.
• A tetrahedral $d^8$ complex has 2 unpaired electrons.
• Therefore, $\left[\mathrm{NiCl}_4\right]^{2-}$ is paramagnetic.

Match with the options

• $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$: Diamagnetic
• $\left[\mathrm{NiCl}_4\right]^{2-}$: Paramagnetic

Hence, the correct option is D.

Comparison with stored answer

• Stored correct answer: D
• Derived answer: D
• They match.