Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the…

🧠 Removing Axial Ligands ⟶ Square Planar

Starting with octahedral splitting (5 d-orbitals split into $t_{2g}^3 + e_g^2$):

• $t_{2g}$: $d_{xy}, d_{xz}, d_{yz}$ (lower).
• $e_g$: $d_{x^2-y^2}, d_{z^2}$ (upper).

Now remove the two ligands along the z-axis (axial). The orbitals with z-component of electron density are now under less ligand repulsion → they drop in energy:

• $d_{z^2}$ (mostly along z) → drops sharply.
• $d_{xz}, d_{yz}$ (have z-components) → drop somewhat.

Orbitals without z-component see no change:

• $d_{xy}$ (purely in xy plane) → stays put relative to before.
• $d_{x^2-y^2}$ (purely in xy plane, points at the remaining 4 ligands) → stays at the top.

🗺️ Final Sq. Planar Order (Low → High)

$d_{xz}, d_{yz} < d_{z^2} < d_{xy} < d_{x^2-y^2}$

Key observations:

• $d_{x^2-y^2}$ is highest — it points directly at the four remaining ligands.
• $d_{z^2}$ has dropped below $d_{xy}$ — this is the critical change from oct splitting.

⚡ Why d⁸ Sq Planar Is Diamagnetic

The huge gap between $d_{x^2-y^2}$ (top, empty in d⁸) and the rest (filled, 8 electrons paired) makes square planar d⁸ complexes (Ni(II), Pd(II), Pt(II), Au(III)) overwhelmingly diamagnetic.

⚠️ Don't Memorise the Order Backwards

A common error: putting $d_{z^2}$ above $d_{xy}$ (carrying over the oct ordering). Once axial ligands are removed, $d_{z^2}$ has no axial repulsion → it drops to the middle of the diagram, below $d_{xy}$.

$\boxed{\text{Answer: (2) } d_{x^2-y^2}\text{ highest, } d_{z^2}\text{ drops below } d_{xy}}$