The figure shows the splitting of the d orbitals of a metal ion in a…

The correct answer is Option (2) → The e set, which is lowered by (3/5)Δ_t below the average energy

In the ground state an electron occupies the lowest energy level available. In a tetrahedral field the splitting is inverted compared with the octahedral field, so the lower level is the set of two orbitals.

• Reading the figure — the upper set of three orbitals (d_xy, d_yz, d_xz) is t₂, raised by (2/5)Δ_t; the lower set of two orbitals (d_x²−y², d_z²) is e, lowered by (3/5)Δ_t.

• The d¹ electron — goes into the e set, the lower one, (3/5)Δ_t below the average energy.

Why the other options are incorrect

Option (1) The t₂ set, which is raised by (2/5)Δ_t above the average energy — Incorrect. The t₂ set is the higher level. A single electron in its ground state takes the lower e set.

Option (3) The e set, which is lowered by (2/5)Δ_t below the average energy — Incorrect. The set is right, but the figure marks the fall of the e set as (3/5)Δ_t; (2/5)Δ_t is the rise of the t₂ set.

Option (4) The t₂ set, which is lowered by (3/5)Δ_t below the average energy — Incorrect. In a tetrahedral field the t₂ set is raised, not lowered. It is in the octahedral field that the set of three orbitals (t_2g) lies lower.

NCERT (2026-27 edition):
5.5.4 Crystal Field Theory – “In tetrahedral coordination entity formation, the d orbital splitting (Fig. 5.9) is inverted and is smaller as compared to the octahedral field splitting.”

5.5.5 Colour in Coordination Compounds – “This is an octahedral complex where the single electron (Ti³⁺ is a 3d¹ system) in the metal d orbital is in the t_2g level in the ground state of the complex. The next higher state available for the electron is the empty e_g level.”

Fig. 5.9: d orbital splitting in a tetrahedral crystal field.