Coordination Compounds JEE Main 2025 Solution

JEE MAIN 2025

22-01-2025 SHIFT-1

Question

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.

Select the correct option:

$\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}(0),$ tetrahedral $\quad\left[\mathrm{Ni}(\mathrm{CO})_4\right]: \mathrm{Ni}(0),$ square planar

$\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{\text {II }}$ , tetrahedral $\quad\left[\mathrm{Ni}(\mathrm{CO})_4\right]: \mathrm{Ni}(0),$ tetrahedral

$\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{\mathrm{II}}$, square planar $\quad\left[\mathrm{Ni}(\mathrm{CO})_4\right]: \mathrm{Ni}(0)$, square planar

$\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{\mathbb{I I}},$ tetrahedral $\quad\left[\mathrm{Ni}(\mathrm{CO})_4\right]: \mathrm{Ni}^{\mathbb{I I}},$ square planar

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\left[\mathrm{NiCl}_4\right]^{2-}$
$\mathrm{Ni}^{+2}-[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0 \rightarrow \mathrm{sp}^3$, Tetrahedral
Number of unpaired electron $=2$ paramagnetic
$\left[\mathrm{Ni}(\mathrm{CO})_4\right]$,
$\mathrm{Ni}(0) \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^0$ (After rearrangement) No unpaired electron
$\mathrm{sp}^3$, Tetrahedral, Diamagnetic

Question Tags

JEE Main

Chemistry

Medium

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