From the magnetic behaviour of $\left[\mathrm{NiCl}_4\right]^{2-}$ (paramagnetic) and…
- $\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{\mathrm{II}}$, tetrahedral
$\left[\mathrm{Ni}(\mathrm{CO})_4\right]: \mathrm{Ni}^{\mathrm{II}}$, 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 - $\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{\mathrm{II}}$, tetrahedral
$\left[\mathrm{Ni}(\mathrm{CO})_4\right]: \mathrm{Ni}(0)$, tetrahedral - $\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}(0)$, tetrahedral
$\left[\mathrm{Ni}(\mathrm{CO})_4\right]: \mathrm{Ni}(0)$, square planar
Solution
$\mathrm{Ni}^{+2}-[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 \rightarrow \mathrm{sp}^3$, 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
Asked in: JEE Main 2025 (22 Jan Shift 1)