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I learned in general chemistry that hybridization of n orbitals produces n hybrid orbitals with the same energy (degenerate).

However, in Housecroft and Sharpe inorganic chemistry, I read the following:

The mixing of $\mathrm{s}$, $\mathrm{p}_x$, $\mathrm{p}_y$, $\mathrm{p}_z$, and $\mathrm{d}_{z^2}$ atomic orbitals gives a set of five $\mathrm{sp^3d}$ hybrid orbitals, the mutual orientations of which correspond to a trigonal bipyramidal arrangement. The five $\mathrm{sp^3d}$ hybrid orbitals are not equivalent and divide into sets of two axial and three equatorial orbitals; the axial orbital lobes lie along the $z$ axis.

The above excerpt seems to imply that the 5 resulting $\mathrm{sp^3d}$ hybrid orbitals are not degenerate and have different energies. However, this is contradictory to what I learned in general chemistry. Why is this so?

andselisk
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    They never say a word about energy. The orbitals are geometrically different, that's the point. – Ivan Neretin Mar 22 '19 at 15:27
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    Also notice here that the energy of the orbitals is for the unbonded atom in free space. When you bond the five orbitals then they are not necessary equivalent anymore because of the geometric arrangement of the substituents. – MaxW Mar 22 '19 at 16:55

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