Based on CFT, in octahedral crystal field, \( d \) orbitals split into orbitals of \( t_{2g} \) and \( e_g \) sets. Which d orbitals forms the \( t_{2g} \) set?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(d_{xy}, d_{yz}, d_{xz}\)
In crystal field theory the six ligands of an octahedral complex are treated as point charges placed along the x, y and z axes. They repel the electrons in the metal d orbitals, and the repulsion is greater for an orbital that points towards the ligands than for one that points between them.
The three orbitals whose lobes lie between the axes, \( d_{xy} \), \( d_{yz} \) and \( d_{xz} \), are repelled less and form the lower \( t_{2g} \) set.
The energy gap between the two sets is the crystal field splitting, \( \Delta_o \). So the \( t_{2g} \) set is \( d_{xy}, d_{yz}, d_{xz} \).
Why the other options are incorrect
Option (2) \(d_{xy}, d_{yz}, d_{z^2}\) — Incorrect. \( d_{z^2} \) points along the z axis, towards two ligands, so it belongs to the \( e_g \) set.
Option (3) \(d_{x^2-y^2}, d_{z^2}\) — Incorrect. These two orbitals form the \( e_g \) set, the higher-energy pair.
Option (4) \(d_{xy}, d_{yz}, d_{x^2-y^2}\) — Incorrect. \( d_{x^2-y^2} \) points along the x and y axes, so it belongs to the \( e_g \) set.
NCERT (2026-27 edition):
Crystal Field Splitting in Octahedral Coordination Entities
