JEE Chemistry 2025 Question with Solution

Given: The magnetic behaviour of $[NiCl_4]^{2-}$ is paramagnetic and that of $[Ni(CO)_4]$ is diamagnetic.

Find: The correct oxidation state and geometry of both complexes.

For $[NiCl_4]^{2-}$, nickel is in the $+2$ oxidation state. Chloride, $Cl^-$, is a weak field ligand, so it does not pair the $d$-electrons. Hence the complex remains high spin with unpaired electrons and is paramagnetic. This is consistent with tetrahedral geometry.

For $[Ni(CO)_4]$, nickel is in the $0$ oxidation state. Carbon monoxide, $CO$, is a strong field ligand, leading to electron pairing. The complex becomes diamagnetic. The hybridization is $sp^3$, so the geometry is tetrahedral.

Therefore, $[NiCl_4]^{2-}$ is $Ni^{2+}$, tetrahedral and $[Ni(CO)_4]$ is $Ni(0)$, tetrahedral. The correct option is B.

Given: $[NiCl_4]^{2-}$ is paramagnetic and $[Ni(CO)_4]$ is diamagnetic.

Find: Match each complex with its oxidation state and geometry.

1. Analyze $[NiCl_4]^{2-}$

The oxidation state of nickel is obtained as:

So nickel is $Ni^{2+}$.

Since $Cl^-$ is a weak field ligand, it does not force pairing of electrons. Therefore, the complex has unpaired electrons and is paramagnetic. For four-coordinate $Ni^{2+}$ with weak field ligands, the geometry is tetrahedral.

2. Analyze $[Ni(CO)_4]$

Carbon monoxide is a neutral ligand, so the oxidation state of nickel is:

Thus nickel is $Ni(0)$.

Because $CO$ is a strong field ligand, electrons pair up and the complex becomes diamagnetic. The complex uses $sp^3$ hybridization and therefore has tetrahedral geometry.

Conclusion:

• $[NiCl_4]^{2-}$ : $Ni^{2+}$, tetrahedral
• $[Ni(CO)_4]$ : $Ni(0)$, tetrahedral

Hence, the correct option is B.