UPSC Civil Services (Mains), Chemistry optional · 2016 · Paper I

Q5(a)(i). Splitting of d orbitals in a square-planar field

Question

Explain the pattern of splitting of d-orbitals in a square-planar ligand field. Label each energy level properly.

d_(x²-y²) (b_(1g)) ≫ d_(xy) (b_(2g)) > d_(z²) (a_(1g)) > d_(xz),d_(yz) (e_g): four levels, the highest pointing straight at the four ligands.

A square-planar complex is an octahedron from which the two ligands on the z axis have been removed; the four ligands lie on ± x and ± y.
Orbitals with a z component lose repulsion and fall: d_(z²) strongly; d_(xz), d_(yz) (a degenerate pair, e_g) a little.
The in-plane orbitals rise (centre of gravity kept): d_(x²-y²), whose lobes point at the ligands, becomes the highest (b_(1g)); d_(xy), lying between the ligands in the plane, comes next (b_(2g)).
The gap d_(x²-y²)–d_(xy) equals Δₒ, so the top orbital lies far above the rest. A d⁸ ion with strong-field ligands fills the four lower levels and leaves d_(x²-y²) empty: [Ni(CN)₄]²-, [PtCl₄]²-, [PdCl₄]²- are square planar and…

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

A square-planar complex is an octahedron from which the two ligands on the z axis have been removed; the four ligands lie on ± x and ± y.
Orbitals with a z component lose repulsion and fall: d_(z²) strongly; d_(xz), d_(yz) (a degenerate pair, e_g) a little.
The in-plane orbitals rise (centre of gravity kept): d_(x²-y²), whose lobes point at the ligands, becomes the highest (b_(1g)); d_(xy), lying between the ligands in the plane, comes next (b_(2g)).
The gap d_(x²-y²)–d_(xy) equals Δₒ, so the top orbital lies far above the rest. A d⁸ ion with strong-field ligands fills the four lower levels and leaves d_(x²-y²) empty: [Ni(CN)₄]²-, [PtCl₄]²-, [PdCl₄]²- are square planar and diamagnetic.

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