Energies of d orbitals relative to the barycentre 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?
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

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

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