Which of the following metal complexes possess the same spin only mag…

The correct answer is Option (3) → (A), (B) and (D) only

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

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

So complexes with the same n have the same μ. To find n, work out the d-electron count of the metal ion and fill the octahedral t_2g and e_g levels. A strong-field ligand such as \( \ce{CN^{-}} \) pairs the electrons in t_2g (low spin); a weak-field ligand such as \( \ce{H2O} \) does not (high spin). This choice matters only for d⁴ to d⁷ ions.

• (A) \( \ce{[Ni(H2O)6]^{2+}} \) — Ni (Z = 28) is [Ar]3d⁸4s²; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d⁸, i.e. t_2g⁶e_g² and n = 2.

• (B) \( \ce{[Ti(H2O)6]^{2+}} \) — Ti (Z = 22) is [Ar]3d²4s²; losing the two 4s electrons gives \( \ce{Ti^{2+}} \): 3d²: t_2g² and n = 2.

• (C) \( \ce{[Mn(CN)6]^{4-}} \) — x + 6(−1) = −4 gives x = +2. Mn (Z = 25) is [Ar]3d⁵4s²; losing the two 4s electrons gives \( \ce{Mn^{2+}} \): 3d⁵. \( \ce{CN^{-}} \) is strong field: t_2g⁵ and n = 1.

• (D) \( \ce{[Mn(CN)6]^{3-}} \) — x + 6(−1) = −3 gives x = +3; removing the two 4s electrons and one 3d electron from Mn gives \( \ce{Mn^{3+}} \): 3d⁴. \( \ce{CN^{-}} \) is strong field: t_2g⁴ and n = 2.

So (A), (B) and (D) each have 2 unpaired electrons:

    \( \displaystyle \mu=\sqrt{2(2+2)}=\sqrt{8}=2.83\ \mathrm{BM} \)

(C) has only one unpaired electron (μ = √3 = 1.73 BM).