Among the species given below, the spin-only magnetic moment is highest for (Given : Atomic number of Ti =22 , Mn =25 , Fe =26 and Co =27 )

Step-by-step solution

The spin-only magnetic moment is given by = n(n+2) BM, where n is the number of unpaired electrons. We determine n for each complex: For [ Ti(H ₂ O )₆]³⁺ : Ti is in +3 oxidation state. The electronic configuration of Ti ³⁺ is [ Ar ] 3d^1 . Number of unpaired electrons, n = 1 . = 1(1+2) = 3 BM. For [ Mn(CN) ₆]³⁻ : Mn is in +3 oxidation state. The electronic configuration of Mn ³⁺ is [ Ar ] 3d^4 . Since CN ^- is a strong field ligand, pairing of electrons takes place. The configuration in the octahedral field is t_ 2