What is the correct order of spin-only magnetic moment (in B.M.) of M…

The correct answer is Option (3) → \(Mn^{2+} > Cr^{2+} > V^{2+}\)

The spin-only magnetic moment of an ion comes from its unpaired electrons. For ions of the first transition series it is given by:

    \( \displaystyle \mu=\sqrt{n(n+2)}\,\mathrm{BM} \)

where n is the number of unpaired electrons. The more unpaired electrons, the larger the magnetic moment.

• \( \ce{Mn^2+} \) — first (highest).

  • Configuration: Mn (Z = 25) is [Ar]3d⁵4s².

  • Electrons removed: removing the two 4s electrons gives \( \ce{Mn^2+} \): 3d⁵.

  • Unpaired electrons: the five d electrons occupy the five d orbitals singly, so n = 5:

      \( \displaystyle \mu=\sqrt{5(5+2)}=\sqrt{35}=5.92\,\mathrm{BM} \)

• \( \ce{Cr^2+} \) — second.

  • Configuration: Cr (Z = 24) is [Ar]3d⁵4s¹ (half-filled d subshell).

  • Electrons removed: removing the 4s electron and one 3d electron gives \( \ce{Cr^2+} \): 3d⁴.

  • Unpaired electrons: four unpaired electrons, so n = 4:

      \( \displaystyle \mu=\sqrt{4(4+2)}=\sqrt{24}=4.90\,\mathrm{BM} \)

• \( \ce{V^2+} \) — third (lowest).

  • Configuration: V (Z = 23) is [Ar]3d³4s².

  • Electrons removed: removing the two 4s electrons gives \( \ce{V^2+} \): 3d³.

  • Unpaired electrons: three unpaired electrons, so n = 3:

      \( \displaystyle \mu=\sqrt{3(3+2)}=\sqrt{15}=3.87\,\mathrm{BM} \)

NCERT (2026-27 edition):
Magnetic Properties – “The magnetic moment increases with the increasing number of unpaired electrons. … The magnetic moments calculated from the ‘spin-only’ formula and those derived experimentally for some ions of the first row transition elements are given in Table 4.7.”

Table 4.7: Calculated and Observed Magnetic Moments (BM)