JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Coordination Compounds

The correct statement among the following is:

(A) $[Ni (CN) X_{4}]^{2 -}$ and $[NiCl X_{4}]^{2 -}$ are diamagnetic and $Ni (CO) X_{4}$ is paramagnetic.
(B) $Ni (CO) X_{4}$ and $[NiCl X_{4}]^{2 -}$ are diamagnetic and $[Ni (CN) X_{4}]^{2 -}$ is paramagnetic.
(C) $Ni (CO) X_{4}$ and $[Ni (CN) X_{4}]^{2 -}$ are diamagnetic and $[NiCl X_{4}]^{2 -}$ is paramagnetic.✓ Correct
(D) $Ni (CO) X_{4}$ is diamagnetic and $[NiCl X_{4}]^{2 -}$ and $[Ni (CN) X_{4}]^{2 -}$ are paramagnetic.

Answer: (C)

Solution

Ni (CO) X_{4}

nickel is in the zero oxidation state,

3 d^{10} 4 s^{0}

; CO pairs the electrons into

s p^{3}

hybrids with none unpaired, so it is diamagnetic. In

[Ni (CN) X_{4}]^{2 -}

nickel is

Ni X^{2 +}

3 d^{8}

; the strong-field

CN X^{-}

forces pairing and gives

d s p^{2}

square-planar geometry with no unpaired electrons, so it too is diamagnetic. In

[NiCl X_{4}]^{2 -}

nickel is again

3 d^{8}

, but the weak-field

Cl X^{-}

causes no pairing, giving

s p^{3}

tetrahedral geometry with two unpaired electrons, so it is paramagnetic. Thus

Ni (CO) X_{4}

and

[Ni (CN) X_{4}]^{2 -}

are diamagnetic while

[NiCl X_{4}]^{2 -}

is paramagnetic.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option