Arrange the following ions in the ascending order of their paramagnet…

The correct answer is Option (1) → (B), (D), (A), (C)

Paramagnetism comes from unpaired electrons, so the more unpaired electrons, the more paramagnetic the ion. For each octahedral complex, find the oxidation state of the metal, its d electron count, and whether the ligand is strong field (electrons pair in t_2g) or weak field (electrons spread over t_2g and e_g).

• (A) \( \ce{[Cr(H2O)6]^{3+}} \): Cr (Z = 24) is [Ar]3d⁵4s¹; \( \ce{H2O} \) is neutral, so the metal is +3, and removing the 4s electron and two 3d electrons gives \( \ce{Cr^{3+}} \), 3d³, t_2g³ whatever the ligand: 3 unpaired electrons

• (B) \( \ce{[Co(en)3]^{3+}} \): Co (Z = 27) is [Ar]3d⁷4s²; en is neutral, so the metal is +3, and removing the two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \), 3d⁶, en is a strong field ligand, t_2g⁶ e_g⁰: 0 unpaired electrons (diamagnetic)

• (C) \( \ce{[CoF6]^{3-}} \): x + 6(−1) = −3 gives x = +3, so the metal is again \( \ce{Co^{3+}} \), 3d⁶, \( \ce{F^{-}} \) is a weak field ligand, t_2g⁴ e_g²: 4 unpaired electrons

• (D) \( \ce{[Ni(H2O)6]^{2+}} \): Ni (Z = 28) is [Ar]3d⁸4s²; the metal is +2, and removing the two 4s electrons gives \( \ce{Ni^{2+}} \), 3d⁸, t_2g⁶ e_g² whatever the ligand: 2 unpaired electrons

    \( \displaystyle \text{Unpaired electrons: (B) }0 < \text{(D) }2 < \text{(A) }3 < \text{(C) }4 \)

Shortcut: \( \ce{[Co(en)3]^{3+}} \) (B) is diamagnetic, so it must come first. Only Option (1) starts with (B).