Which among the following will form low spin complexes?(A) t32g e1g(B…

The correct answer is Option (2) → (B) and (D) only

In an octahedral complex, the five d orbitals split into a lower set of three (t_2g) and a higher set of two (e_g). The energy gap is the crystal field splitting energy, Δ_o. With a strong field ligand (Δ_o > pairing energy) electrons pair in t_2g before e_g is used, giving low spin.

So in a low spin arrangement the e_g set stays empty until t_2g holds six electrons. Low spin d⁴, d⁵ and d⁶ ions are t_2g⁴e_g⁰, t_2g⁵e_g⁰ and t_2g⁶e_g⁰.

• (A) t³_2g e¹_g — Incorrect. This is d⁴ with the fourth electron in e_g instead of pairing in t_2g, so it is the high spin arrangement.

• (B) t⁴_2g e⁰_g — Correct. This is low spin d⁴: the fourth electron pairs in t_2g, leaving 2 unpaired electrons.

• (C) t⁵_2g e¹_g — Incorrect. This d⁶ arrangement places an electron in e_g while t_2g still has room, so it is not low spin. Low spin d⁶ is t_2g⁶e_g⁰, and high spin d⁶ is t_2g⁴e_g².

• (D) t⁵_2g e⁰_g — Correct. This is low spin d⁵, with only 1 unpaired electron (high spin d⁵ would be t_2g³e_g² with 5).

NCERT (2026-27 edition):
Crystal Field Theory – …