Question

Draw a diagram to show the splitting of d-orbitals in an octahedral crystal field.

Hint:

Five degenerate d-orbitals split into lower \(t_{2g}\) (3 orbitals) and higher \(e_g\) (2 orbitals), separated by \(\Delta_o\).

Answer 1

Step 1: Starting point. In a free (gaseous) metal ion the five d-orbitals \((d_{xy}, d_{yz}, d_{zx}, d_{x^2-y^2}, d_{z^2})\) are degenerate, i.e. all have the same energy.
Step 2: Effect of six ligands. In an octahedral complex, six ligands approach the metal along the \(x, y\) and \(z\) axes. The two orbitals that point directly along the axes, \(d_{x^2-y^2}\) and \(d_{z^2}\) (the \(e_g\) set), face the ligands head-on and are repelled more, so their energy rises. The three orbitals lying between the axes, \(d_{xy}, d_{yz}, d_{zx}\) (the \(t_{2g}\) set), are repelled less, so their energy falls.
Step 3: The splitting. The degenerate level splits into a lower \(t_{2g}\) triplet and an upper \(e_g\) doublet, separated by the crystal field splitting energy \(\Delta_o\). Relative to the average (barycentre), \(t_{2g}\) is lowered by \(0.4\,\Delta_o\) each and \(e_g\) is raised by \(0.6\,\Delta_o\) each.
Step 4: Diagram (energy increases upward):
\[ \begin{array}{c} e_g\ (d_{x^2-y^2},\ d_{z^2}) \quad \text{(higher, } +0.6\,\Delta_o)\\ \text{------------ average energy (barycentre) ------------}\\ t_{2g}\ (d_{xy},\ d_{yz},\ d_{zx}) \quad \text{(lower, } -0.4\,\Delta_o)\\ \end{array} \]
The gap between the two sets is \(\Delta_o\), the octahedral crystal field splitting energy.
\[\boxed{t_{2g}\ (\text{3 lower}) \ \text{and}\ e_g\ (\text{2 higher}),\ \text{gap} = \Delta_o}\]