The crystal field stabilization energy for low spin d^6 octahedral co…

The correct answer is Option (2) → \(-2.4 \Delta _O\)

In an octahedral field the five d orbitals split into a lower t_2g set and a higher e_g set. Each electron in t_2g lowers the energy by 0.4Δ_o (2/5 Δ_o) and each electron in e_g raises it by 0.6Δ_o (3/5 Δ_o).

In a low spin complex the ligand field is strong (Δ_o > P), so the electrons pair up in t_2g before entering e_g. For d⁶, low spin gives t_2g⁶ e_g⁰:

    CFSE = 6 × (−0.4Δ_o) + 0 × (0.6Δ_o) = −2.4Δ_o (the pairing energy is not included in the options)

Why the other options are incorrect

• Option (1) −2.0Δ_o — Incorrect. This is t_2g⁵, the low spin d⁵ value; one electron is missing.

• Option (3) −0.4Δ_o — Incorrect. This is the high spin d⁶ value, t_2g⁴ e_g²: −1.6 + 1.2 = −0.4Δ_o.

• Option (4) −1.6Δ_o — Incorrect. This is t_2g⁴ e_g⁰, the low spin d⁴ value.

NCERT (2026-27 edition):
Crystal Field Splitting in Octahedral Coordination Entities – “… This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation is denoted by Δ_o (the subscript o is for octahedral). Thus, the energy of the two e_g orbitals will increase by (3/5) Δ_o and that of the three t_2g will decrease by (2/5)Δ_o.”

Crystal Field Theory – “… (ii) If Δ_o > P, it becomes more energetically favourable for the fourth electron to occupy a t_2g orbital with configuration t_2g⁴ e_g⁰. Ligands which produce this effect are known as strong field ligands and form low spin complexes.”