Which one of the following is a low spin complex?

The correct answer is Option (1) → \([Co(NH_3)_6]^{3+}\)

In an octahedral complex the ligands split the d orbitals into a lower t_2g set and a higher e_g set. A strong field ligand gives a large splitting, so the electrons pair up in t_2g: a low spin (inner orbital, d²sp³) complex. A weak field ligand gives a small splitting, the electrons spread over all five d orbitals, and the complex is high spin (outer orbital, sp³d²).

In \( \ce{[Co(NH3)6]^{3+}} \) the charge balance gives \( \ce{Co^{3+}} \): x + 6(0) = +3. Co (Z = 27) is [Ar]3d⁷4s²; losing three electrons gives \( \ce{Co^{3+}} \): 3d⁶. \( \ce{NH3} \) is a strong field ligand towards \( \ce{Co^{3+}} \), so the six electrons pair up in t_2g (t_2g⁶e_g⁰): d²sp³, diamagnetic, low spin.

Why the other options are incorrect

• Option (2) \([Fe(H_2O)_6]^{2+}\) — Incorrect. Fe (Z = 26) [Ar]3d⁶4s² gives \( \ce{Fe^{2+}} \) 3d⁶; \( \ce{H2O} \) is a weak field ligand, so the complex is high spin with four unpaired electrons (t_2g⁴e_g²).

• Option (3) \([CoF_6]^{3-}\) — Incorrect. Again \( \ce{Co^{3+}} \) 3d⁶ (x + 6(−1) = −3), but \( \ce{F^{-}} \) is a weak field ligand: sp³d², outer orbital, high spin, four unpaired electrons.

• Option (4) \([Mn(H_2O)_6]^{2+}\) — Incorrect. Mn (Z = 25) [Ar]3d⁵4s² gives \( \ce{Mn^{2+}} \) 3d⁵; with weak field \( \ce{H2O} \) all five electrons stay unpaired (t_2g³e_g²), high spin.

NCERT (2026-27 edition):
Valence Bond Theory – “… In the diamagnetic octahedral complex, [Co(NH₃)₆]³⁺, the cobalt ion is in +3 oxidation state and has the electronic configuration 3d⁶. The hybridisation scheme is as shown in diagram. Six pairs of electrons, one from each NH₃ molecule, occupy the six hybrid orbitals. Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron. In the formation of this complex, since the inner d orbital (3d) is used in hybridisation, the complex, [Co(NH₃)₆]³⁺ is called an inner orbital or low spin or spin paired complex.”