The magnetic moment of the complex species- [Fe(H2O)6]2+ and K2[MnCl4] are 5.3 and 5.9 respectively. The number of unpaired electrons respectively are:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4, 5
For first-row transition metal ions the magnetic moment depends mainly on the number of unpaired electrons n, by the spin-only formula:
\( \displaystyle \mu = \sqrt{n(n+2)}\ \mathrm{BM} \)
n = 3: μ = √15 = 3.87 BM; n = 4: μ = √24 = 4.90 BM; n = 5: μ = √35 = 5.92 BM
[Fe(H2O)6]2+: Fe (Z = 26) is [Ar] 3d64s2; \( \ce{Fe^{2+}} \) is 3d6. \( \ce{H2O} \) is a weak-field ligand, so no pairing beyond Hund’s rule:
3d: [↑↓] [↑] [↑] [↑] [↑] → 4 unpaired (spin-only 4.90 BM; the observed 5.3 BM includes a small orbital contribution)
K2[MnCl4]: charge balance 2(+1) + x + 4(−1) = 0 gives \( \ce{Mn^{2+}} \). Mn (Z = 25) is [Ar] 3d54s2; \( \ce{Mn^{2+}} \) is 3d5:
3d: [↑] [↑] [↑] [↑] [↑] → 5 unpaired (5.92 BM, matching 5.9)
So the numbers of unpaired electrons are 4 and 5.
Shortcut: 5.9 BM is the value for 5 unpaired electrons; only Options (1) and (3) end in 5, and 5.3 BM is well below 5.9, so the first complex has 4.
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).”