Which of the following possess the calculated magnetic moment of 3.87…

The correct answer is Option (3) → [V(H₂O)₆]²⁺

The spin-only magnetic moment depends only on the number of unpaired electrons (n):

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

n = 1, 2, 3, 4, 5 gives μ = 1.73, 2.83, 3.87, 4.90, 5.92 BM. Working backwards from the given value:

    \( \displaystyle \sqrt{n(n+2)}=3.87 \)

    \( \displaystyle n(n+2)=15 \)

    \( \displaystyle n=3 \)

All four are M²⁺ aqua ions. Water is a weak field ligand, so the electrons stay unpaired as far as possible (high spin). The number of unpaired electrons then follows from the 3dⁿ configuration of each M²⁺ ion. V (Z = 23) is [Ar] 3d³4s²; losing the two 4s electrons gives \( \ce{V^{2+}} \): 3d³, with 3 unpaired electrons:

    \( \displaystyle \mu=\sqrt{3\times5}=\sqrt{15}\approx3.87\ \mathrm{BM} \)

So the complex with a magnetic moment of 3.87 BM is [V(H₂O)₆]²⁺.

Why the other options are incorrect

• Option (1) [Ti(H₂O)₆]²⁺ — Incorrect. Ti (Z = 22) is [Ar] 3d²4s²; losing the two 4s electrons gives \( \ce{Ti^{2+}} \): 3d², with 2 unpaired electrons, μ = √8 ≈ 2.83 BM.

• Option (2) [Fe(H₂O)₆]²⁺ — Incorrect. Fe (Z = 26) is [Ar] 3d⁶4s²; losing the two 4s electrons gives \( \ce{Fe^{2+}} \): 3d⁶; high spin t_2g⁴e_g² has 4 unpaired electrons, μ = √24 ≈ 4.90 BM.

• Option (4) [Mn(H₂O)₆]²⁺ — Incorrect. Mn (Z = 25) is [Ar] 3d⁵4s²; losing the two 4s electrons gives \( \ce{Mn^{2+}} \): 3d⁵, with 5 unpaired electrons, μ = √35 ≈ 5.92 BM.

NCERT (2026-27 edition):
Magnetic Properties – …