From the magnetic behaviour of $\left[\mathrm{NiCl}_4\right]^{2-}$ (paramagnetic) and $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ (diamagnetic), choose the correct geometry and oxidation state.

<p>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.</p> <h3>$[\mathrm{NiCl}_4]^{2-}$</h3> <p><p><strong>Oxidation State:</strong> In $[\mathrm{NiCl}_4]^{2-}$, nickel is in the +2 oxidation state ($\mathrm{Ni}^{2+}$). The electron configuration for $\mathrm{Ni}^{2+}$ is $[\mathrm{Ar}]\, 3d^8\, 4s^0$.</p></p> <p><p><strong>Geometry and Hybridization:</strong> 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.</p></p> <h3>$[\mathrm{Ni}(\mathrm{CO})_4]$</h3> <p><p><strong>Oxidation State:</strong> 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.</p></p> <p><p><strong>Geometry and Hybridization:</strong> 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.</p></p> <p>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.</p>