Question

Which of the following d-orbitals experience more repulsion in the crystal field splitting of a tetrahedral complex?

• A. $d_{x^2-y^2}$, $d_{z^2}$
• B. $d_{x^2-y^2}$, $d_{xy}$
• C. $d_{xy}$, $d_{yz}$, $d_{z^2}$
• D. $d_{xy}$, $d_{yz}$, $d_{xz}$

Hint:

In tetrahedral fields, the $t_2$ set is higher in energy (unlike octahedral where the $e_g$ set is higher). $\Delta_t = \frac{4}{9}\Delta_o$.

Answer 1

The Correct Option is D

Step 1: Geometry of a tetrahedral field.
In a tetrahedral complex, four ligands approach the metal from alternate corners of a cube, NOT along the x, y, z axes.
Step 2: Orientation of $d$-orbital sets.
The $e$ set ($d_{z^2}$, $d_{x^2-y^2}$) has lobes pointing directly along the axes, which are far from tetrahedral ligand positions. The $t_2$ set ($d_{xy}$, $d_{yz}$, $d_{xz}$) has lobes pointing between the axes, which is closer to where the tetrahedral ligands actually are.
Step 3: Which set feels more repulsion?
Because the $t_2$ set orbitals point toward the ligands more directly, electrons in these orbitals experience greater electrostatic repulsion and are raised to higher energy.
Step 4: Conclusion.
The $d_{xy}$, $d_{yz}$, $d_{xz}$ orbitals (the $t_2$ set) experience more repulsion in a tetrahedral crystal field. This is the reverse of the octahedral case.
\[ \boxed{d_{xy},\, d_{yz},\, d_{xz}} \]