A 3d transition metal M forms two octahedral complexes: [ M ( H ₂ O )₆]²⁺ and [ M ( CN )₆]⁴⁻ . The difference between the squares of their spin-only magnetic moments (in B.M. ^2 ) is exactly 24 . The transition metal M is:

Step-by-step solution

Let the number of unpaired electrons in the high-spin and low-spin complexes be n₁ and n₂ respectively. The square of the spin-only magnetic moment is given by ^2 = n(n+2) . The given condition is ₁^2 - ₂^2 = n₁(n₁+2) - n₂(n₂+2) = 24 . Both complexes contain the metal M in the +2 oxidation state. H ₂ O acts as a weak field ligand (forming a high-spin complex) and CN ^- acts as a strong field ligand (forming a low-spin complex). Let us evaluate the given options for M ²⁺ : For Mn ²⁺ ( 3d^5 ): High-spin n₁ = 5 ₁^2 =