Question

Which of the following are true about the energy of the given d-orbitals of a tetrahedral complex?

A. $d_{xy} = d_{yz}>d_{x^2-y^2}$
B. $d_{xy} = d_{yz}>d_{z^2}$
C. $d_{x^2-y^2}>d_{z^2}>d_{xz}$
D. $d_{x^2-y^2} = d_{z^2}<d_{xz}$

• A. A, B and D only
• B. A and B only
• C. B and D only
• D. B, C and D only

Hint:

In tetrahedral splitting, the t2 set (xy, yz, xz) is higher in energy than the e set (z^2, x^2-y^2).

Answer 1

The Correct Option is A

The crystal field splitting in tetrahedral geometry ($T_d$) splits d-orbitals into a lower energy e-set ($d_{z^2}, d_{x^2-y^2}$) and a higher energy $t_2$-set ($d_{xy}, d_{yz}, d_{xz}$).

Let the energy of the e orbitals be $E_e$ and the energy of the $t_2$ orbitals be $E_{t2}$. We know that $E_{t2}>E_e$.

From orbital degeneracies:
$E(d_{xy}) = E(d_{yz}) = E(d_{xz}) = E_{t2}$
$E(d_{z^2}) = E(d_{x^2-y^2}) = E_e$

Checking the conditions:
A. $E_{t2} = E_{t2}>E_e$ (Correct)
B. $E_{t2} = E_{t2}>E_e$ (Correct)
C. $E_e>E_e>E_{t2}$ (Incorrect, as $E_e$ is lower than $E_{t2}$ and $d_{x^2-y^2}$ energy equals $d_{z^2}$)
D. $E_e = E_e<E_{t2}$ (Correct)

Since statements A, B, and D satisfy the energy splitting rules of a tetrahedral complex, option 1 is the correct choice.