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

Solution & Explanation

Spin-only moment µ = √(n(n+2)) BM rises with the number of unpaired electrons n, so find the species with the most unpaired electrons. First get the d-configuration of each metal ion, then apply the field strength of the ligand (CN⁻ and NH₃ are strong-field → pairing; H₂O is weak/moderate). [Mn(CN)₆]³⁻: Mn³⁺ is 3d⁴; strong-field CN⁻ gives low-spin t₂g⁴ ⇒ 2 unpaired. [Fe(CN)₆]³⁻: Fe³⁺ is 3d⁵; CN⁻ low-spin t₂g⁵ ⇒ 1 unpaired. [Co(NH₃)₆]³⁺: Co³⁺ is 3d⁶; strong-field gives t₂g⁶ ⇒ 0 unpaired. [Ti(H₂O)₆]³⁺: Ti³⁺ is 3d¹ ⇒ 1 unpaired. Maximum unpaired electrons (2) is in [Mn(CN)₆]³⁻, giving the highest µ ≈ 2.83 BM. Answer: option A.