Which one of the following ions has the highest magnetic moment? (Atomic numbers: Ti = 22, Mn = 25, Ni = 28, Cu = 29)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Mn2+
For ions of the first transition series the magnetic moment comes from the unpaired electrons, given by the spin-only formula \( \mu=\sqrt{n(n+2)} \) BM. The ion with the most unpaired electrons has the highest moment.
\( \ce{Cu^2+} \): Cu (Z = 29) is [Ar]3d104s1; losing the 4s electron and one 3d electron gives 3d9, so n = 1 and \( \mu=\sqrt{1\times3}=1.73 \) BM.
\( \ce{Ti^3+} \): Ti (Z = 22) is [Ar]3d24s2; losing the two 4s electrons and one 3d electron gives 3d1, so n = 1 and \( \mu=1.73 \) BM.
\( \ce{Ni^2+} \): Ni (Z = 28) is [Ar]3d84s2; losing the two 4s electrons gives 3d8, so n = 2 and \( \mu=\sqrt{2\times4}=2.83 \) BM.
\( \ce{Mn^2+} \): Mn (Z = 25) is [Ar]3d54s2; losing the two 4s electrons gives 3d5, one electron in each d orbital, so n = 5 and \( \mu=\sqrt{5\times7}=5.92 \) BM.
\( \ce{Mn^2+} \), with 5 unpaired electrons, has the highest magnetic moment.
NCERT (2026-27 edition):
4.3.9 Magnetic Properties – “… 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 = \( \sqrt{n(n+2)} \) where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM).”
“The magnetic moment increases with the increasing number of unpaired electrons.”