Which of the following complex will be repelled by the applied magnet…

The correct answer is Option (4) → [Co(NH₃)₆]³⁺

A complex is repelled by an applied magnetic field when it is diamagnetic, i.e. when it has no unpaired electrons. To decide, find the oxidation state of the metal, its d-electron count, and then whether the ligand pairs the electrons:

• Strong field ligands (\( \ce{CN^{-}} \), \( \ce{NH3} \), en) give a large splitting Δ_o, so electrons pair in the lower t_2g set (low spin).

• Weak field ligands (\( \ce{F^{-}} \), \( \ce{Cl^{-}} \), \( \ce{H2O} \)) give a small splitting, so electrons stay unpaired (high spin).

In \( \ce{[Co(NH3)6]^{3+}} \), Co is +3 (3d⁶) and \( \ce{NH3} \) pairs all six electrons in t_2g, so n = 0 and the complex is diamagnetic.

Why the other options are incorrect

• Option (1) [CoF₆]^3- — Incorrect. x − 6 = −3, so Co is +3 and, as above, \( \ce{Co^{3+}} \) is 3d⁶; but \( \ce{F^{-}} \) is weak field, so t_2g⁴e_g²: 4 unpaired electrons (high spin, paramagnetic).

• Option (2) [Cr(NH₃)₆]³⁺ — Incorrect. x + 6(0) = +3, so Cr is +3. Cr (Z = 24) is [Ar]3d⁵4s¹; losing the 4s electron and two 3d electrons gives \( \ce{Cr^{3+}} \): 3d³. The three electrons occupy the t_2g set singly whatever the ligand, so t_2g³: 3 unpaired electrons.

• Option (3) [Fe(CN)₆]^3- — Incorrect. x − 6 = −3, so Fe is +3. Fe (Z = 26) is [Ar]3d⁶4s²; losing the two 4s electrons and one 3d electron gives \( \ce{Fe^{3+}} \): 3d⁵. \( \ce{CN^{-}} \) is strong field, so the electrons pair as far as possible, t_2g⁵: still 1 unpaired electron.

NCERT (2026-27 edition):
Valence Bond Theory – …

Note: The same/similar question has been asked in different years by NTA as per the details given below: