JEE MAIN - Chemistry (2025 - 22nd January Morning Shift - No. 5)
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.
$\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}(0)$, tetrahedral
$\left[\mathrm{Ni}(\mathrm{CO})_4\right]: \mathrm{Ni}(0)$, square planar
$\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{\mathrm{II}}$, tetrahedral $\left[\mathrm{Ni}(\mathrm{CO})_4\right]: \mathrm{Ni}(0)$, tetrahedral
$\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{\mathrm{II}}$, tetrahedral $\left[\mathrm{Ni}(\mathrm{CO})_4\right]: \mathrm{Ni}^{I I}$, square planar
$\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{\mathrm{II}}$, square planar $\left[\mathrm{Ni}(\mathrm{CO})_4\right]: \mathrm{Ni}(0)$, square planar
Explanation
$$ \left[\mathrm{NiCl}_4\right]^{2-} $$
$\mathrm{Ni}^{+2}-[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 \rightarrow \mathrm{sp}^3$, Tetrahedral
Number of unpaired electron $=2$ paramagentic
$$ \left[\mathrm{Ni}(\mathrm{CO})_4\right] $$
$\mathrm{Ni}(0) \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^0$ (After rearrangement)
No unpaired electron
$\mathrm{sp}^3$, Tetrahedral, Diamagnetic
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