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JEE Main Chemistry Coordination Compounds 2026 JEE Main 2026 (24 January Shift 1)

JEE Main Chemistry Question (2026) — Solution

Question

Given below are two statements: Statement I : Hybridisation, shape and spin only magnetic moment of K _ 3 [ Co ( CO _ 3 )_ 3 ] is sp ^ 3 ~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- , [ MnBr _ 4 ]^ 2- and [ CoF _ 6 ]^ 3- respectively are square planar, tetrahedral, octahedral; dsp ^ 2 , sp ^ 3 , sp ^ 3 ~d ^ 2 and 0,5.9,4.9. In the light of the above statements, choose the correct answer from the options given below

Options

  1. A. Statement I is true but Statement II is false
  2. B. Both Statement I and Statement II are false
  3. C. Both Statement I and Statement II are true
  4. D. Statement I is false but Statement II is true

Answer

C. Both Statement I and Statement II are true

Step-by-step solution

In Statement I, for K_3[Co(CO_3)_3], Cobalt is in +3 oxidation state (Co^ 3+ : 3d^6). Carbonate (CO_3^ 2- ) is a weak field ligand. Thus, no pairing occurs, resulting in 4 unpaired electrons. Hybridisation is sp^3d^2, shape is octahedral, and = 4(4+2) = 24 4.9 BM. Statement I is true. In Statement II: For [Ni(CN)_4]^ 2- , Ni^ 2+ is 3d^8. CN^- is a strong field ligand, causing pairing. Hybridisation is dsp^2, geometry is square planar, and = 0 BM. For [MnBr_4]^ 2- , Mn^ 2+ is 3d^5. Br^- is a weak field ligand. Hybridisation is sp^3, geometry is tetrahedral, and = 5(5+2) = 35 5.9 BM. For [CoF_6]^ 3- , Co^ 3+ is 3d^6. F^- is a weak field ligand. Hybridisation is sp^3d^2, geometry is octahedral, and = 4(4+2) = 24 4.9 BM. Statement II is true. Both statements are true.

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