Question: Which of the following d-orbitals experience more repulsion in the crystal field splitting of octahedral complex?
Step 1 (Governing Principle): [0.5 Mark]
In an octahedral complex, the central metal ion is surrounded by six ligands approaching along the axes ($x, y,$ and $z$). According to Crystal Field Theory (CFT), the d-orbitals are split into two sets based on their orientation relative to the incoming ligands:
- $e_g$ set: Orbitals directed along the axes ($d_{x^2-y^2}$ and $d_{z^2}$).
- $t_{2g}$ set: Orbitals directed between the axes ($d_{xy}, d_{yz},$ and $d_{xz}$).
Step 2 (Step-by-Step Analysis): [0.25 Mark]
Since the ligands approach the metal ion directly along the $x, y,$ and $z$ axes, the electron density of the orbitals lying along the axes experiences greater electrostatic repulsion from the lone pairs of the ligands compared to those lying between the axes.
Mathematically, the energy of the $e_g$ orbitals increases relative to the barycenter:
$$E(e_g) > E(t_{2g})$$The orbitals $d_{x^2-y^2}$ and $d_{z^2}$ point directly towards the ligands, resulting in higher energy due to increased repulsion.
Step 3 (Final Concluded Answer): [0.25 Mark]
The correct option is (B) $d_{x^2-y^2}, d_{z^2}$.
💡 EL10 Concept Insight: Imagine the ligands are like people walking straight down the hallways (the axes); the d-orbitals that are also positioned right in those hallways ($d_{x^2-y^2}$ and $d_{z^2}$) get "bumped" or repelled the most, while the ones tucked into the corners ($t_{2g}$) stay safer and have less energy.