Amongst the following ions which one of the following has the highest magnetic moment value?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → [Fe(H2O)6]2+
The spin-only magnetic moment depends on the number of unpaired electrons (n):
\( \mu = \sqrt{n(n+2)}\ \text{BM} \)
So the ion with the most unpaired electrons has the highest moment. Strong field ligands (CO, \( \ce{CN^{-}} \)) pair the d electrons; weak field \( \ce{H2O} \) does not.
K[Co(CO)4]: \( \ce{K^{+}} \) balances one negative charge, so cobalt is −1. Co (Z = 27) is [Ar]3d74s2; \( \ce{Co^{-}} \) has 10 valence electrons, arranged 3d10 with the strong field CO. n = 0, μ = 0.
K4[Mn(CN)6]: x + 6(−1) = −4 gives \( \ce{Mn^{2+}} \). Mn (Z = 25) is [Ar]3d54s2; \( \ce{Mn^{2+}} \) is 3d5. \( \ce{CN^{-}} \) is strong field: t2g5, n = 1, μ = 1.73 BM.
[Fe(H2O)6]2+: Fe (Z = 26) is [Ar]3d64s2; \( \ce{Fe^{2+}} \) is 3d6. \( \ce{H2O} \) is weak field: t2g4eg2, n = 4, μ = √24 = 4.90 BM.
[Cr(H2O)6]3+: Cr (Z = 24) is [Ar]3d54s1; \( \ce{Cr^{3+}} \) is 3d3: t2g3, n = 3, μ = √15 = 3.87 BM.
\( \ce{[Fe(H2O)6]^{2+}} \) has the most unpaired electrons (4), so it has the highest magnetic moment.
Shortcut: no square roots are needed; μ rises with n, so compare n only: 0, 1, 4, 3.
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., \( \mu = \sqrt{n(n+2)} \) 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).”