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Octahedral Crystal Field Splitting Orbital Degeneracy
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Octahedral Crystal Field Splitting Orbital Degeneracy
The crystal field splitting is based on where the ligands (modelled as point charges) are in relation to the orbitals. In an octahedral complex, the ligands are all at 90° from each other and are placed on each of the $x,$$y,$$z$ axes. The orbitals that lie on these axes will experience the most repulsion and will rise in energy, while the orbitals between the axes $(\mathrm{t_{2g}})$ will lower in energy as they experience less repulsion from the ligands, and the average overall energy is maintained.
I can only assume that the degree by which the $\mathrm{e_g}$ orbitals are raised is the same due to the $\mathrm d_{z^2}$ orbital technically being a linear combination of what would have been the $\mathrm d_{z^2-x^2}$ and $\mathrm d_{z^2-y^2}$ orbitals. This means that the $\mathrm{e_g}$ orbitals lie on the axes to the same extent as each other, so experience the same overall repulsion from the ligands.
The crystal field splitting is based on where the ligands (modelled as point charges) are in relation to the orbitals. In an octahedral complex, the ligands are all at 90° from each other and are placed on each of the $x,$$y,$$z$ axes. The orbitals that lie on these axes will experience the most repulsion and will rise in energy, while the orbitals between the axes $(\mathrm{t_{2g}})$ will lower in energy as they experience less repulsion from the ligands, and the average overall energy is maintained.
I can only assume that the degree by which the $\mathrm{e_g}$ orbitals are raised is the same due to the $\mathrm d_{z^2}$ orbital technically being a linear combination of what would have been the $\mathrm d_{z^2-x^2}$ and $\mathrm d_{z^2-y^2}$ orbitals. This means that the $\mathrm{e_g}$ orbitals lie on the axes to the same extent as each other, so experience the same overall repulsion from the ligands.
It is because perfect octahedral symmetry is normally assumed; the two $\mathrm{e_g}$ levels and three $\mathrm{t_{2g}}$ are degenerate. If there was a distortion, say by lengthening both $z$-axis ligand positions then the $\mathrm{e_g}$ degeneracy would be removed as the $\mathrm d_{z^2}$ becomes more stable than $\mathrm d_{x^2-y^2}$. This happens because the $z$-axis ligand has more effect on $\mathrm d_{z^2}$ than on $\mathrm d_{x^2-y^2}$ orbitals simply due to its position. The $\mathrm d_{xy}$ orbital also increases in energy removing the degeneracy of the $\mathrm{t_{2g}}$ and the $\mathrm d_{yz},$$\mathrm d_{zx}$ lowered.
It is because perfect octahedral symmetry is normally assumed; the two $\mathrm{e_g}$ levels and three $\mathrm{t_{2g}}$ are degenerate. If there was a distortion, say by lengthening both $z$-axis ligand positions then the $\mathrm{e_g}$ degeneracy would be removed as the $\mathrm d_{z^2}$ becomes more stable than $\mathrm d_{x^2-y^2}$. This happens because the $z$-axis ligand has more effect on $\mathrm d_{z^2}$ than on $\mathrm d_{x^2-y^2}$ orbitals simply due to its position. The $\mathrm d_{xy}$ orbital also increases in energy removing the degeneracy of the $\mathrm{t_{2g}}$ and the $\mathrm d_{yz},$$\mathrm d_{zx}$ lowered.
The crystal field splitting is based on where the ligands (modelled as point charges) are in relation to the orbitals. In an octahedral complex, the ligands are all at 90° from each other and are placed on each of the $x,$ $y,$ $z$ axes. The orbitals that lie on these axes will experience the most repulsion and will rise in energy, while the orbitals between the axes $(\mathrm{t_{2g}})$ will lower in energy as they experience less repulsion from the ligands, and the average overall energy is maintained.
I can only assume that the degree by which the $\mathrm{e_g}$ orbitals are raised is the same due to the $\mathrm d_{z^2}$ orbital technically being a linear combination of what would have been the $\mathrm d_{z^2-x^2}$ and $\mathrm d_{z^2-y^2}$ orbitals. This means that the $\mathrm{e_g}$ orbitals lie on the axes to the same extent as each other, so experience the same overall repulsion from the ligands.
The crystal field splitting is based on where the ligands (modelled as point charges) are in relation to the orbitals. In an octahedral complex, the ligands are all at 90° from each other and are placed on each of the $x,$ $y,$ $z$ axes. The orbitals that lie on these axes will experience the most repulsion and will rise in energy, while the orbitals between the axes $(\mathrm{t_{2g}})$ will lower in energy as they experience less repulsion from the ligands, and the average overall energy is maintained.
I can only assume that the degree by which the $\mathrm{e_g}$ orbitals are raised is the same due to the $\mathrm d_{z^2}$ orbital technically being a linear combination of what would have been the $\mathrm d_{z^2-x^2}$ and $\mathrm d_{z^2-y^2}$ orbitals. This means that the $\mathrm{e_g}$ orbitals lie on the axes to the same extent as each other, so experience the same overall repulsion from the ligands.
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It is because perfect octahedral symmetry is normally assumed; the two $\mathrm{e_g}$ levels and three $\mathrm{t_{2g}}$ are degenerate. If there was a distortion, say by lengthening both $z$-axis ligand positions then the $\mathrm{e_g}$ degeneracy would be removed as the $\mathrm d_{z^2}$ becomes more stable than $\mathrm d_{x^2-y^2}$. This happens because the $z$-axis ligand has more effect on $\mathrm d_{z^2}$ than on $\mathrm d_{x^2-y^2}$ orbitals simply due to its position. The $\mathrm d_{xy}$ orbital also increases in energy removing the degeneracy of the $\mathrm{t_{2g}}$ and the $\mathrm d_{yz},$ $\mathrm d_{zx}$ lowered.
It is because perfect octahedral symmetry is normally assumed; the two $\mathrm{e_g}$ levels and three $\mathrm{t_{2g}}$ are degenerate. If there was a distortion, say by lengthening both $z$-axis ligand positions then the $\mathrm{e_g}$ degeneracy would be removed as the $\mathrm d_{z^2}$ becomes more stable than $\mathrm d_{x^2-y^2}$. This happens because the $z$-axis ligand has more effect on $\mathrm d_{z^2}$ than on $\mathrm d_{x^2-y^2}$ orbitals simply due to its position. The $\mathrm d_{xy}$ orbital also increases in energy removing the degeneracy of the $\mathrm{t_{2g}}$ and the $\mathrm d_{yz},$ $\mathrm d_{zx}$ lowered.
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