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Rather than referring to the energy of an isolated metal atom, the weighted mean of these two sets of perturbed orbitals $(\mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$ and $\mathrm d_{x^2 - y^2}$, $\mathrm d_{z^2} )$ is taken as zero. This sometimes is called the barycentre. The difference in energy between the two $\mathrm d$ levels is given either of symbols $\Delta_\mathrm O$ or $10~\mathrm {Dq}$. It follows that $\mathrm{e_g}$ orbitals have $+0.6 \Delta_\mathrm O$ above average level and $\mathrm{t_{2g}}$ orbitals are $-0.4\Delta_\mathrm O$ below the average.

It might be basic but why is orbitals split in $-0.4$ and $0.6$? If the barycentre is at the average then the orbitals should be split in $-0.5$ and $0.5$ below and above average, no?

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    $\begingroup$ No, there are 3 low and 2 high energy orbitals. You should put the barycenter at the average location. $\endgroup$ Jul 31, 2017 at 1:45
  • $\begingroup$ Can you please add the source from which you quoted this passage. $\endgroup$ Jul 31, 2017 at 7:23
  • $\begingroup$ @orthocresol Sure. $\endgroup$ Jul 31, 2017 at 7:27
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    $\begingroup$ @PrittBalagopal Do you mean something like $B = \dfrac{3L + 2H}{5}$ ? $\endgroup$ Jul 31, 2017 at 7:32
  • $\begingroup$ @123 Precisely. $\endgroup$ Jul 31, 2017 at 11:56

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The passage you quoted is perfectly correct and contains one word which you may have overlooked:

the weighted mean of these two sets of perturbed orbitals is taken as zero [...]

In this case the weight corresponds to the degeneracy of the set. The lower-energy $\mathrm{t_{2g}}$ set contains three orbitals and the higher-energy $\mathrm{e_g}$ set contains two. If we place the $\mathrm{t_{2g}}$ set at $x \Delta_\mathrm{O}$ and the $\mathrm{e_g}$ set at $y\Delta_\mathrm{O}$, then

$$\begin{align} y - x &= 1 & &\text{(difference must be equal to }\Delta_\mathrm{O}) \\ \frac{3x + 2y}{3+2} &= 0 & &(\textit{weighted}\text{ mean is equal to zero)} \end{align}$$

and $(x,y) = (-0.4, 0.6)$.

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