Electronic configuration of d6 when Δo < P

The correct answer is Option (3) → t_2g⁴ e_g²

In an octahedral crystal field the five d orbitals split into the lower t_2g set (three orbitals) and the higher e_g set (two orbitals), separated by Δ_o. After three electrons have entered t_2g, each further electron has a choice: pair up in t_2g (costing the pairing energy P) or go up to e_g (costing Δ_o).

• If Δ_o < P (weak field ligands), it is cheaper to go up: electrons fill all five orbitals singly before pairing — high spin.

• If Δ_o > P (strong field ligands), electrons pair in t_2g first — low spin.

For d⁶ with Δ_o < P:

    t_2g: [↑↓] [↑] [↑]   e_g: [↑] [↑] → t_2g⁴ e_g², 4 unpaired electrons

Why the other options are incorrect

• Option (1) t_2g⁵ e_g¹ — Incorrect. This pairs two electrons in t_2g while e_g still has an empty orbital; it fits neither the weak-field nor the strong-field filling.

• Option (2) t_2g⁶ e_g⁰ — Incorrect. This is the low spin arrangement, found when Δ_o > P (e.g. \( \ce{[Co(NH3)6]^{3+}} \)).

• Option (4) t_2g² e_g⁴ — Incorrect. The higher e_g level is filled before the lower t_2g level, which is never energetically favoured.

NCERT (2026-27 edition):
Crystal Field Theory – “… For d⁴ ions, two possible patterns of electron distribution arise: (i) the fourth electron could either enter the t_2g level and pair with an existing electron, or (ii) it could avoid paying the price of the pairing energy by occupying the e_g level. Which of these possibilities occurs, depends on the relative magnitude of the crystal field splitting, Δ_o and the pairing energy, P (P represents the energy required for electron pairing in a single orbital). The two options are:
(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.”