JEE Main 2026 — Chemistry Inorganic Chemistry

2026mcqmedium

The correct statement among the following is:

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Options

A
$\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are diamagnetic and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ is paramagnetic.
B
$\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ are diamagnetic and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ is paramagnetic.
C
$\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ are diamagnetic and $\mathrm{Ni}(\mathrm{CO})_{4}$ is paramagnetic.
D
$\mathrm{Ni}(\mathrm{CO})_{4}$ is diamagnetic and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are paramagnetic.

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Solution

In $Ni(CO)_4$ , Nickel is in $0$ oxidation state with configuration $[Ar]3d^8 4s^2$ . $CO$ is a strong field ligand, causing pairing of $4s$ electrons into $3d$ orbitals, resulting in $3d^{10}$ configuration. It is $sp^3$ hybridized and diamagnetic (zero unpaired electrons).

In $[Ni(CN)_4]^{2-}$ , Nickel is in $+2$ oxidation state with configuration $[Ar]3d^8$ . $CN^-$ is a strong field ligand, causing pairing of the two unpaired electrons in $3d$ orbitals. It is $dsp^2$ hybridized (square planar) and diamagnetic.

In $[NiCl_4]^{2-}$ , Nickel is in $+2$ oxidation state with configuration $[Ar]3d^8$ . $Cl^-$ is a weak field ligand and cannot cause pairing of electrons. It has two unpaired electrons in $3d$ orbitals, making it $sp^3$ hybridized (tetrahedral) and paramagnetic.

Therefore, $Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic, while $[NiCl_4]^{2-}$ is paramagnetic.

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Options

$\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are diamagnetic and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ is paramagnetic.

$\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ are diamagnetic and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ is paramagnetic.

$\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ are diamagnetic and $\mathrm{Ni}(\mathrm{CO})_{4}$ is paramagnetic.

$\mathrm{Ni}(\mathrm{CO})_{4}$ is diamagnetic and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are paramagnetic.

Solution

Therefore, $Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic, while $[NiCl_4]^{2-}$ is paramagnetic.

JEE Main 2026 — Chemistry Inorganic Chemistry

Options

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JEE Main 2026 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

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