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NEETChemistryCoordination Compounds

An octahedral complex [M(H₂O)₆]²⁺ has a spin-only magnetic moment of 3.87 BM . Which of the following correctly identifies the possible metal ion M and its d -orbital electronic configuration?

Options

  1. AV²⁺ with configuration t_ 2g ^3 e_g^0
  2. BCo²⁺ with configuration t_ 2g ^5 e_g^2
  3. CNi²⁺ with configuration t_ 2g ^6 e_g^2
  4. DBoth (1) and (2)

Correct answer

D. Both (1) and (2)

Step-by-step solution

The spin-only magnetic moment is given by = n(n+2) BM . Given = 3.87 BM , we have n(n+2) = 3.87 , which gives n = 3 unpaired electrons. The complex is [M(H₂O)₆]²⁺ , so the metal ion is in the +2 oxidation state and the ligand ( H₂O ) is a weak field ligand. For V²⁺ ( 3d^3 ): The configuration in an octahedral field is t_ 2g ^3 e_g^0 . Number of unpaired electrons n = 3 . For Co²⁺ ( 3d^7 ): In the presence of a weak field ligand, the configuration is t_ 2g ^5 e_g^2 . Number of unpaired electrons n = 3 . For Ni²⁺ ( 3

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