In an octahedral complex, if Δo < P (P = energy required for electron…

The correct answer is Option (1) → High spin complex

• In an octahedral field the d orbitals split into a lower t_2g set and a higher e_g set, separated by Δ_o.

• From the fourth d electron on, an electron either pairs in t_2g (costing the pairing energy P) or goes up to e_g (costing Δ_o).

• When Δ_o < P, entering e_g costs less, so electrons stay unpaired as long as possible. The ligand is a weak field ligand and the complex is high spin.

Why the other options are incorrect

Option (2) Low spin complex — Incorrect. Low spin complexes form when Δ_o > P (strong field ligands), so pairing in t_2g is cheaper.

Option (3) Inner orbital complex — Incorrect. This valence bond term describes complexes using inner (n−1)d orbitals, which are low spin (for example \( \ce{[Co(NH3)6]^3+} \)), not the high spin case.

Option (4) Complex with maximum pairing of the d electrons — Incorrect. Maximum pairing is the low spin arrangement found when Δ_o > P; with Δ_o < P pairing is avoided.

NCERT (2026-27 edition):
5.5.4 Crystal Field Theory – “(i) If Δ_o < P, the fourth electron enters one of the e_g orbitals giving the configuration t_2g³e_g¹. Ligands for which Δ_o < P are known as weak field ligands and form high spin complexes.”