Energy Comparison of d Orbitals in Tetrahedral Complex

In a tetrahedral crystal field, the approach of the four ligands causes the five degenerate $d$-orbitals of the central metal ion to split into two distinct energy sets due to electrostatic repulsion:

1. $t_2$ set (Triply Degenerate): Comprising $d_{xy}$, $d_{yz}$, and $d_{xz}$ orbitals. These orbitals point more directly toward the direction of the approaching ligands, resulting in greater electrostatic repulsion and thus higher energy.
2. $e$ set (Doubly Degenerate): Comprising $d_{x^2-y^2}$ and $d_{z^2}$ orbitals. These orbitals lie along the coordinate axes between the approaching ligands, resulting in lesser electrostatic repulsion and lower energy.

Therefore, the relative energies of the $d$-orbitals in a tetrahedral field satisfy the following fundamental relationship: $E(d_{xy}) = E(d_{yz}) = E(d_{xz}) > E(d_{x^2-y^2}) = E(d_{z^2})$

Now, let us evaluate the given statements:

• Statement A: $d_{xy} = d_{xz} > d_{x^2-y^2}$
  Since $d_{xy}$ and $d_{xz}$ both belong to the higher-energy $t_2$ set, $E(d_{xy}) = E(d_{xz})$. Also, any $t_2$ orbital has higher energy than any $e$ orbital ($d_{x^2-y^2}$). Thus, Statement A is True.

• Statement B: $d_{xy} = d_{yz} > d_{z^2}$
  Since $d_{xy}$ and $d_{yz}$ both belong to the $t_2$ set, $E(d_{xy}) = E(d_{yz})$. Also, $t_2$ orbitals are higher in energy than the $e$ orbital ($d_{z^2}$). Thus, Statement B is True.

• Statement C: $d_{x^2-y^2} > d_{z^2} > d_{xz}$
  The $e$ set orbitals $d_{x^2-y^2}$ and $d_{z^2}$ are degenerate ($E(d_{x^2-y^2}) = E(d_{z^2})$) and have lower energy than the $t_2$ orbital $d_{xz}$. Thus, Statement C is False.

• Statement D: $d_{x^2-y^2} = d_{z^2} < d_{xz}$
  Both $d_{x^2-y^2}$ and $d_{z^2}$ belong to the $e$ set, making them equal in energy, and both have lower energy than $d_{xz}$ from the $t_2$ set. Thus, Statement D is True.

Hence, statements A, B, and D only are correct.

Correct Option: A