Chemistry · Coordination Compounds

JEE Main 2025 — 22 January, Morning Shift — Question 42

From the magnetic behaviour of [NiCl4]2−\left[\mathrm{NiCl}_{4}\right]^{2-} (paramagnetic) and [Ni(CO)4]\left[\mathrm{Ni}(\mathrm{CO})_{4}\right] (diamagnetic), choose the correct geometry and oxidation state.

  1. Option A:

    [NiCl4]2−\left[\mathrm{NiCl}_{4}\right]^{2-} : NiII \mathrm{Ni}^{\text {II }}, square planar[Ni(CO)4]\left[\mathrm{Ni}(\mathrm{CO})_{4}\right] : Ni(0)\mathrm{Ni}(0), square planar

  2. Option B:

    [NiCl4]2−:NiII \left[\mathrm{NiCl}_{4}\right]^{2-}: \mathrm{Ni}^{\text {II }}, tetrahedral[Ni(CO)4]:Ni(0)\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]: \mathrm{Ni}(0), tetrahedral

    Correct
  3. Option C:

    [NiCl4]2−\left[\mathrm{NiCl}_{4}\right]^{2-} : NiII\mathrm{Ni}^{\mathrm{II}}, tetrahedral[Ni(CO)4]:NiII\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]: \mathrm{Ni}^{\mathrm{II}}, square planar

  4. Option D:

    [NiCl4]2−:Ni(0)\left[\mathrm{NiCl}_{4}\right]^{2-}: \mathrm{Ni}(0), tetrahedral[Ni(CO)4]:Ni(0)\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]: \mathrm{Ni}(0), square planar

Answer: B

Step-by-step solution

[NiCl4]2−\left[\mathrm{NiCl}_{4}\right]^{2-}

Ni+2−[Ar]3 d84 s0→sp3\mathrm{Ni}^{+2}-[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0} \rightarrow \mathrm{sp}^{3}, Tetrahedral

Number of unpaired electron =2=2 paramagentic [Ni(CO)4]\left[\mathrm{Ni}(\mathrm{CO})_{4}\right],

Ni(0)→[Ar]3 d104 s0\mathrm{Ni}(0) \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^{0} (After rearrangement)

No unpaired electron

sp3\mathrm{sp}^{3}, Tetrahedral, Diamagnetic

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Chemistry
Chapter
Coordination Compounds
Topic
Properties and importance of Coordination Complexes
From the magnetic behaviour of [ NiCl 4 ] 2- (paramagnetic) and [ Ni… | JEE Main 2025 PYQ with Solution · DhiX AI