Which of the following complexes is expected to be diamagnetic where…

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

A complex is diamagnetic 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 \( \ce{[Co(NH3)6]^{3+}} \): x + 6(0) = +3, so x = +3. Co (Z = 27) is [Ar]3d⁷4s²; losing the two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \): 3d⁶. \( \ce{NH3} \) is a strong field ligand for \( \ce{Co^{3+}} \), so all six electrons pair in t_2g:

    \( \displaystyle \mathrm{Co^{3+}}\ (d^6),\ \text{low spin}:\ t_{2g}^{6}\,e_g^{0}\ \Rightarrow\ n=0 \)

The complex is inner orbital (d²sp³) and diamagnetic.

Why the other options are incorrect

• Option (1) [FeF₆]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. \( \ce{F^{-}} \) is weak field, so t_2g³e_g²: 5 unpaired electrons.

• Option (2) [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 (3) [NiCl₄]²⁻ — Incorrect. x − 4 = −2, so Ni is +2. Ni (Z = 28) is [Ar]3d⁸4s²; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d⁸. \( \ce{Cl^{-}} \) is weak field and the complex is tetrahedral (sp³), so the d⁸ ion keeps 2 unpaired electrons.

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: