Spin free complex among the following is:

The correct answer is Option (2) → [CoF₆]³⁻

In an octahedral complex the five d orbitals split into a lower t_2g set and a higher e_g set, separated by the crystal field splitting energy Δ_o. For d⁴ to d⁷ ions, the extra electrons have two choices:

• Δ_o < P (pairing energy): electrons enter e_g unpaired. The complex is high spin, also called spin-free. This happens with weak field ligands such as \( \ce{F^{-}} \), \( \ce{Cl^{-}} \) and \( \ce{H2O} \).

• Δ_o > P: electrons pair up in t_2g. The complex is low spin, or spin-paired. This happens with strong field ligands such as \( \ce{CN^{-}} \), and with \( \ce{NH3} \) or oxalate on a highly charged ion like \( \ce{Co^{3+}} \).

In \( \ce{[CoF6]^{3-}} \) cobalt is +3 (x + 6(−1) = −3). Co (Z = 27) is [Ar]3d⁷4s²; removing the two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \): 3d⁶. \( \ce{F^{-}} \) is a weak field ligand. The configuration is t_2g⁴e_g² with four unpaired electrons, and the complex uses outer sp³d² orbitals. So it is spin-free.

Why the other options are incorrect

• Option (1) [Co(NH₃)₆]³⁺ — Incorrect. \( \ce{Co^{3+}} \) is again 3d⁶, and for this highly charged ion \( \ce{NH3} \) acts as a strong field ligand, so the configuration is t_2g⁶ (inner d²sp³). It is diamagnetic and spin-paired.

• Option (3) [Co(C₂O₄)₃]³⁻ — Incorrect. Oxalate carries −2, so x + 3(−2) = −3 and Co is +3 (3d⁶). The six electrons are paired in t_2g (t_2g⁶), so the complex is diamagnetic and spin-paired.

• Option (4) [Mn(CN)₆]³⁻ — Incorrect. Mn (Z = 25) is [Ar]3d⁵4s²; removing the two 4s electrons and one 3d electron gives \( \ce{Mn^{3+}} \): 3d⁴. With \( \ce{CN^{-}} \) all four electrons stay in t_2g, leaving two unpaired electrons instead of four. It is spin-paired.

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

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