Q23.95P - Chemistry: Molecular Nature Of Matter And Change

Question

${\mathit{d}}_{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}$The two ${\mathit{e}}_{\mathbf{g}}$ orbitals are identical in energy in an octahedral complex but have different energies in a square planar complex, with the ${\mathit{d}}_{{\mathbf{z}}^{\mathbf{2}}}$ orbital being much lower in energy than the ${\mathit{d}}_{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}$. Explain with orbital sketches.

Step-by-Step Solution

Answer

$d_{x^2-y^2}$ orbital has higher energy than the $d_{xy}$ orbital because the former has lobes that are along the x-axis and y-axis.

Step 1: Square planar

Square planar coordination can be imagined to result when two ligands on the z-axis of an octahedron are removed from the complex, leaving only the ligands in the x-y plane. As the z-ligands move away, the ligands in the square plane move a little closer to the metal.

Step 2: Orbital sketches

The orbital splitting diagram for square planar coordination can thus be derived from the octahedral diagram. As ligands move away along the z-axis, d-orbitals with a z-component will fall in energy. The ${\mathrm{d}}_{z^2}$ orbital falls the most, as its electrons are concentrated in lobes along the z-axis. The $d_{xz}$ and $d_{yz}$ orbitals also drop in energy, but not as much. Conversely, ${\mathrm{d}}_{x^2-y^2}$ and $d_{xy}$ orbitals increase in energy. The splitting diagram for square planar complexes is more complex than for octahedral and tetrahedral complexes and is shown below with the relative energies of each orbital.