Which one of the following will exhibit maximum paramagnetic character?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \([Mn(H_2O)_6]^{2+}\)
Paramagnetism comes from unpaired electrons; the more unpaired electrons, the larger the magnetic moment, \( \mu = \sqrt{n(n+2)} \) BM. Water is a weak field ligand, so in all four hexaaqua ions the d electrons first occupy the five 3d orbitals singly (Hund’s rule) and pair only when every orbital has one electron (high spin).
Each metal is +2 (six neutral \( \ce{H2O} \) ligands, charge 2+). The 4s electrons are removed first:
[Fe(H2O)6]2+: Fe (Z = 26) is [Ar] 3d6 4s2; \( \ce{Fe^{2+}} \) is [Ar] 3d6: [↑↓] [↑] [↑] [↑] [↑] → 4 unpaired, μ = √24 = 4.90 BM.
[Cr(H2O)6]2+: Cr (Z = 24) is [Ar] 3d5 4s1; \( \ce{Cr^{2+}} \) is [Ar] 3d4: [↑] [↑] [↑] [↑] [ ] → 4 unpaired, μ = 4.90 BM.
[Co(H2O)6]2+: Co (Z = 27) is [Ar] 3d7 4s2; \( \ce{Co^{2+}} \) is [Ar] 3d7: [↑↓] [↑↓] [↑] [↑] [↑] → 3 unpaired, μ = √15 = 3.87 BM.
[Mn(H2O)6]2+: Mn (Z = 25) is [Ar] 3d5 4s2; \( \ce{Mn^{2+}} \) is [Ar] 3d5: [↑] [↑] [↑] [↑] [↑] → 5 unpaired, μ = √35 = 5.92 BM.
A half-filled d5 ion has the largest number of unpaired electrons possible, so [Mn(H2O)6]2+ is the most paramagnetic.
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).”