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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 lig

Options

  1. AStatement I is true but Statement II is false
  2. BBoth Statement I and Statement II are false
  3. CBoth Statement I and Statement II are true
  4. DStatement I is false but Statement II is true

Correct answer

C. Both Statement I and Statement II are true

Step-by-step solution

In Statement I, for K₃[Co(CO₃)₃] , Cobalt is in +3 oxidation state ( Co³⁺: 3d^6 ). Carbonate ( CO₃²⁻ ) 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⁻ , Ni²⁺ is 3d^8 . CN^- is a strong field ligand, causing pairing. Hybridisation is dsp^2 , geometry is square planar, and = 0 BM. For [MnBr₄]²⁻ , Mn²⁺ is 3d^5 . Br^- is a weak field ligand. Hybridisation is

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