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Chemistry · Coordination Compounds

JEE Main 2026 — 24 January, Morning Shift — Question 48

Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II :

Statement-I: Hybridisation, shape and spin only magnetic moment of K3[Co(CO3)3]\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{CO}_{3}\right)_{3}\right] is sp3 d2\mathrm{sp}^{3} \mathrm{~d}^{2}, octahedral and 4.9 BM respectively.

Statement-II: Geometry, hybridisation and spin only magnetic moment values (BM) of the ions [Ni(CN)4]2−,[MnBr4]2−\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-},\left[\mathrm{MnBr}_{4}\right]^{2-} and [CoF6]3−\left[\mathrm{CoF}_{6}\right]^{3-} respectively are square planar, tetrahedral, octahedral : dsp2,sp3\mathrm{dsp}^{2}, \mathrm{sp}^{3}, sp3 d2\mathrm{sp}^{3} \mathrm{~d}^{2} and 0, 5.9, 4.9.

In the light of the above statements, choose the correct answer from the options given below

  1. Option A:

    Both statement-I and statement-II are false

  2. Option B:

    Statement I is false but statement-II is true

  3. Option C:

    Both statement-I and statement-II are true

    Correct
  4. Option D:

    Statement-I is true but statement-II is false

Answer: C

Step-by-step solution

In K3[Co(CO3)3]⇒sp3 d2\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{CO}_{3}\right)_{3}\right] \Rightarrow \mathrm{sp}^{3} \mathrm{~d}^{2} hybridized, octahedral

⇒4 unpaired electron ⇒4.9 B.M. [Ni(CN)4]2−⇒dsp2 hybridized , square plai ⇒0 unpaired electron ⇒0 B.M. [MnBr4]2−⇒sp3 hybridized , tetrahedral ⇒5 unpaired electron ⇒5.9 B.M. [CoF6]3−⇒sp3 d2 hybridized, octahedral ⇒4 unpaired electron ⇒4.9 B.M. \begin{aligned} & \Rightarrow 4 \text { unpaired electron } & \Rightarrow 4.9 \text { B.M. } {\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-} } & \Rightarrow \mathrm{dsp}^{2} \text { hybridized , square plai } & \Rightarrow 0 \text { unpaired electron } & \Rightarrow 0 \text { B.M. } {\left[\mathrm{MnBr}_{4}\right]^{2-} } & \Rightarrow \mathrm{sp}^{3} \text { hybridized , tetrahedral } & \Rightarrow 5 \text { unpaired electron } & \Rightarrow 5.9 \text { B.M. } {\left[\mathrm{CoF}_{6}\right]^{3-} } & \Rightarrow \mathrm{sp}^{3} \mathrm{~d}^{2} \text { hybridized, octahedral } & \Rightarrow 4 \text { unpaired electron } & \Rightarrow 4.9 \text { B.M. } \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Morning Shift
Subject
Chemistry
Chapter
Coordination Compounds
Topic
Theories of Bonding in Coordination Compounds

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