JEE Main 2025 Jan 22 Shift 1, Chemistry Q19: Colour, Magnetism And Stability Of Co-ordination Compounds

JEE Main2025Jan 22, Shift 1Chemistry

From the magnetic behaviour of [NiCl $_{4}$ ] $^{2 -}$ (paramagnetic) and [Ni(CO) $_{4}$ ] (diamagnetic), choose the correct geometry and oxidation state.

A
[NiCl $_{4}$ ] $^{2 -}$ : Ni $^{I I}$ , square planar
[Ni(CO) $_{4}$ ] : Ni(0), square planar
B
[NiCl $_{4}$ ] $^{2 -}$ : Ni $^{I I}$ , tetrahedral
[Ni(CO) $_{4}$ ] : Ni(0), tetrahedral
C
[NiCl $_{4}$ ] $^{2 -}$ : Ni $^{I I}$ , tetrahedral
[Ni(CO) $_{4}$ ] : Ni $^{I I}$ , square planar
D
[NiCl $_{4}$ ] $^{2 -}$ : Ni(0), tetrahedral
[Ni(CO) $_{4}$ ] : Ni(0), square planar

Solution

[NiCl $_{4}$ ] $^{2 -}$ : Cl $^{-}$ is a weak-field ligand. Ni is in the +2 state with $d^{8}$ configuration; with weak-field Cl $^{-}$ the geometry is tetrahedral (sp $^{3}$ ). Two unpaired electrons remain $\Rightarrow$ paramagnetic. $✓$

[Ni(CO) $_{4}$ ]: CO is neutral, so Ni is in the 0 oxidation state with $d^{10}$ configuration. Even though CO is a strong-field ligand, a $d^{10}$ system has no unpaired electrons, so it is diamagnetic. Hybridisation is sp $^{3}$ and geometry is tetrahedral.

$∴$ Both complexes are tetrahedral; Ni is +II in [NiCl $_{4}$ ] $^{2 -}$ and 0 in [Ni(CO) $_{4}$ ].

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JEE Main 2025 Jan 22 Shift 1, Chemistry Q19: Colour, Magnetism And Stability Of Co-ordination Compounds

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