In the above diagram, where do the 3/5 and 2/5 come from? Also, how are the $\ce{e_g}$ orbitals degenerate with each other?
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$\begingroup$ This isn't a homework question. After the semester ended (I don't go to MIT), I ended up on MIT open courseware to watch some videos about areas of chemistry I haven't covered yet or haven't covered well. I am asking the question here because I have no other avenue in which to ask questions to other people. I'm just trying to use my time during the pandemic to build my knowledge of chemistry $\endgroup$– Bob SmithCommented May 18, 2020 at 22:52
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The total energy of the d orbitals before and after the application of the octahedral (cubic) field has to be the same. As three levels fall and two rise, the ratio is as shown. The $\Delta_0$ arises from the solution of the equations. (In older literature the total gap is $10Dq$ with the levels $6Dq$ and $-4Dq$.)
If $V_0$ is the interaction energy $3\Delta_0/5=\int\psi(e_g)^*V_0\psi(e_g)d\tau$ and $-2\Delta_0/5=\int\psi(t_{2g})^*V_0\psi(t_{2g})d\tau$. In practice $\Delta_0$ is found empirically.
The order of levels is reversed in a tetrahedral field and the size of $\Delta_0$ is also changed to $-9/4\Delta_0$.
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$\begingroup$ "The total energy of the d orbitals before and after the application of the octahedral field has to be the same". Why? $\endgroup$– Ian BushCommented May 15, 2020 at 7:43
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$\begingroup$ You are not creating nor destroying energy, you are just moving electron density into metal centered orbitals that do not lie in the path of the approaching ligands. To do this those orbitals that are in the way have to be higher in energy so the other ones are filled first. But if you lift one you have to lower another one to conserve energy, that's the lever principle. $\endgroup$ Commented May 15, 2020 at 10:06