Which of the following will be having maximum magnetic moment? 1. [Fe…

The correct answer is Option (3) → \(Na_3[FeCl_6]\)

The spin-only magnetic moment is μ = √[n(n + 2)] BM, where n is the number of unpaired electrons, so the complex with the most unpaired electrons has the highest moment. Strong field ligands (\( \ce{CN-} \), and \( \ce{NH3} \) with \( \ce{Co^3+} \)) pair the d electrons; weak field ligands (\( \ce{H2O} \), \( \ce{Cl-} \)) do not.

• (1) \( \ce{[Fe(H2O)6]Cl2} \): x + 0 + 2(−1) = 0, so Fe is +2. Fe (Z = 26) is [Ar]3d⁶4s²; losing the two 4s electrons gives \( \ce{Fe^2+} \): 3d⁶. With weak field \( \ce{H2O} \): t_2g⁴e_g², 4 unpaired, μ = √24 = 4.90 BM.

• (2) \( \ce{K4[Mn(CN)6]} \): 4(+1) + x + 6(−1) = 0, so Mn is +2. Mn (Z = 25) is [Ar]3d⁵4s²; \( \ce{Mn^2+} \) is 3d⁵. With strong field \( \ce{CN-} \): t_2g⁵, 1 unpaired, μ = √3 = 1.73 BM.

• (3) \( \ce{Na3[FeCl6]} \): 3(+1) + x + 6(−1) = 0, so Fe is +3. Fe (Z = 26) is [Ar]3d⁶4s²; losing the two 4s electrons and one 3d electron gives \( \ce{Fe^3+} \): 3d⁵. With weak field \( \ce{Cl-} \): t_2g³e_g², 5 unpaired, μ = √35 = 5.92 BM.

• (4) \( \ce{[Co(NH3)6]Cl3} \): x + 0 + 3(−1) = 0, so Co is +3. Co (Z = 27) is [Ar]3d⁷4s²; \( \ce{Co^3+} \) is 3d⁶. \( \ce{NH3} \) pairs these electrons: t_2g⁶, 0 unpaired, diamagnetic.

So the maximum magnetic moment is shown by Na₃[FeCl₆] (5.92 BM).

Shortcut: a high spin d⁵ ion has the largest possible number of unpaired electrons (five); \( \ce{Fe^3+} \) with the weak field \( \ce{Cl-} \) is the only high spin d⁵ case here.

NCERT (2026-27 edition):
Magnetic Properties – “… Paramagnetism arises from the presence of unpaired electrons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e., μ = √n(n+2) where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of 1.73 Bohr magnetons (BM). The magnetic moment increases with the increasing number of unpaired electrons.”