JEE Main 2025 — Coordination Compounds Question with Solution

The magnetic behavior of complexes can provide insights into their geometry and oxidation states. Given two complexes, $[{\mathrm{NiCl}}_4{]}^{2-}$ and $[\mathrm{Ni}(\mathrm{CO}{)}_4]$, we need to analyze the magnetic properties to deduce these characteristics.

$[{\mathrm{NiCl}}_4{]}^{2-}$

Oxidation State: In $[{\mathrm{NiCl}}_4{]}^{2-}$, nickel is in the +2 oxidation state (${\mathrm{Ni}}^{2+}$). The electron configuration for ${\mathrm{Ni}}^{2+}$ is $[\mathrm{Ar}]3d^{8}4s^0$.

Geometry and Hybridization: The complex shows paramagnetic properties, indicating the presence of unpaired electrons. The hybridization in this case is ${\mathrm{sp}}^3$, which corresponds to a tetrahedral geometry. Thus, there are two unpaired electrons leading to its paramagnetism.

$[\mathrm{Ni}(\mathrm{CO}{)}_4]$

Oxidation State: In this complex, nickel is in the zero oxidation state ($\mathrm{Ni}(0)$). The electron configuration is $[\mathrm{Ar}]3d^{10}4s^0$ after rearrangement.

Geometry and Hybridization: The compound is diamagnetic, indicating all electrons are paired. Its hybridization is also ${\mathrm{sp}}^3$, meaning the geometry is tetrahedral, resulting in no unpaired electrons and confirming its diamagnetic nature.

In summary, $[{\mathrm{NiCl}}_4{]}^{2-}$ is a tetrahedral complex with nickel in a +2 oxidation state and is paramagnetic. In contrast, $[\mathrm{Ni}(\mathrm{CO}{)}_4]$ is a tetrahedral complex with nickel in a 0 oxidation state and is diamagnetic.