JEE Chemistry 2026 Question with Solution

Given: We must identify the correct magnetic behaviour of $\mathrm{Ni(CO)_4}$, $[\mathrm{Ni(CN)_4}]^{2-}$ and $[\mathrm{NiCl_4}]^{2-}$.

Find: Which option gives the correct diamagnetic or paramagnetic nature of these complexes.

Concept: Magnetic behaviour depends on oxidation state of the metal, nature of ligands, and geometry of the complex.

Step 1: Analyze $\mathrm{Ni(CO)_4}$

Oxidation state of $\mathrm{Ni}$ is $0$, so its configuration is $3d^{10}$. All electrons are paired.

Step 2: Analyze $[\mathrm{Ni(CN)_4}]^{2-}$

Oxidation state of $\mathrm{Ni}$ is $+2$, so it is $d^8$. $\mathrm{CN^-}$ is a strong-field ligand and gives square planar geometry with low spin configuration. All electrons are paired.

Step 3: Analyze $[\mathrm{NiCl_4}]^{2-}$

Oxidation state of $\mathrm{Ni}$ is $+2$, so it is $d^8$. $\mathrm{Cl^-}$ is a weak-field ligand and gives tetrahedral geometry with high spin configuration. Two unpaired electrons are present.

Therefore, $\mathrm{Ni(CO)_4}$ and $[\mathrm{Ni(CN)_4}]^{2-}$ are diamagnetic, while $[\mathrm{NiCl_4}]^{2-}$ is paramagnetic. The correct option is D.

Given: Strong-field ligands such as $\mathrm{CN^-}$ and $\mathrm{CO}$ favour electron pairing, while weak-field ligands such as $\mathrm{Cl^-}$ favour unpaired electrons.

Find: Use ligand strength and geometry to identify the correct statement.

For $\mathrm{Ni(CO)_4}$, nickel is in oxidation state $0$ and has configuration $3d^{10}$, so no unpaired electron is present.

For $[\mathrm{Ni(CN)_4}]^{2-}$, nickel is $d^8$ and the strong-field ligand $\mathrm{CN^-}$ produces a square planar low-spin complex, so electrons pair up and the complex is diamagnetic.

For $[\mathrm{NiCl_4}]^{2-}$, nickel is also $d^8$ but $\mathrm{Cl^-}$ is weak field, giving a tetrahedral high-spin complex with unpaired electrons, so it is paramagnetic.

Hence the only correct statement is Option D.