JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Coordination Compounds

Pulling both ligands off the $z$-axis of an octahedron leaves a square-planar field with ligands only in the $xy$-plane, so orbitals that point into that plane are pushed highest. The $d_{x^2-y^2}$ orbital, aimed straight at the four ligands, rises to the top; the $d_{xy}$ orbital, in the plane but between the ligands, comes next; the $d_{z^2}$ orbital drops because the $z$-axis is now empty; and the $d_{xz},d_{yz}$ pair lies lowest. The resulting order from highest to lowest energy is $d_{x^2-y^2}>d_{xy}>d_{z^2}>(d_{xz},d_{yz})$.