Which of the following ligands is most likely to form high-spin compl…

The correct answer is Option (1) → Cl⁻

• In an octahedral field the d orbitals split into a lower t_2g set and a higher e_g set, separated by Δ_o. A weak field ligand gives a small Δ_o, smaller than the pairing energy P (Δ_o < P). The electrons then go into the e_g orbitals rather than pair up, giving a high spin complex.

• So the ligand most likely to give high spin complexes is the one lowest in the spectrochemical series (the order of increasing field strength). Of the four, Cl⁻ comes first:

      \( \displaystyle \ce{Cl^-} \lt \ce{OH^-} \lt \ce{C2O4^2-} \lt \ce{CN^-} \)

Why the other options are incorrect

Option (2) OH⁻ — Incorrect. OH⁻ stands above Cl⁻ in the spectrochemical series, so it splits the d orbitals more and is less likely to give a high spin complex.

Option (3) C₂O₄²⁻ — Incorrect. Oxalate is stronger in field than both Cl⁻ and OH⁻, so Δ_o is larger still.

Option (4) CN⁻ — Incorrect. CN⁻ is near the top of the series, a strong field ligand (Δ_o > P) that forces pairing and forms low spin complexes.

NCERT (2026-27 edition):
5.5.4 Crystal Field Theory – “… In general, ligands can be arranged in a series in the order of increasing field strength as given below: I^– < Br^– < SCN^– < Cl^– < S^2– < F^– < OH^– < C₂O₄^2– < H₂O < NCS^– < edta^4– < NH₃ < en < CN^– < CO Such a series is termed as spectrochemical series. … (i) If Δ_o < P, the fourth electron enters one of the e_g orbitals giving the configuration t_2g³ e_g¹. Ligands for which Δ_o < P are known as weak field ligands and form high spin complexes.”