From chemguide.co.uk:

If the phosphorus is going to form $\ce{PCl5}$ it has first to generate 5 unpaired electrons. It does this by promoting one of the electrons in the $\mathrm{3s}$ orbital to the next available higher energy orbital. Which higher energy orbital? It uses one of the $\mathrm{3d}$ orbitals. You might have expected it to use the $\mathrm{4s}$ orbital because this is the orbital that fills before the $\mathrm{3s}$ when atoms are being built from scratch. Not so! Apart from when you are building the atoms in the first place, the $\mathrm{3s}$ always counts as the lower energy orbital."

Could you please explain why this is so?

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    $\begingroup$ This is because the commonly taught method of building electronic configurations, by assuming that all atoms have the energetic diagram for their atomic orbitals, is not a valid hypothesis any more when you read the d block. So you can learn rule of thumbs, but the reality is that energy levels of polyelectronic atoms are much harder to understand (graduate-level quantum chemistry). Here's one link that may help $\endgroup$ – F'x Sep 4 '13 at 7:34
  • $\begingroup$ @F'x That's already halfway to an answer! Would you care to turn it into a full answer on its own? $\endgroup$ – tschoppi May 3 '14 at 19:54
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    $\begingroup$ It is also from the misconception that d-orbitals are actually necessary to build up the hyper-coordinated species. $\endgroup$ – Martin - マーチン May 12 '14 at 12:34

In P element, the energy of 3d is lower than 4s. Look at the link provided below. Only in K and Ca, 4s has lower energy than 3d.


Another reason could be because the symmetry of hybridized orbital with d orbital will be more preferred in bonding with that of s orbital, and the molecule obtain more energy compensation with sp3d over sp3s.

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    $\begingroup$ Ian your link is now broken. Could you somehow fix it please? $\endgroup$ – M.A.R. Feb 24 '16 at 19:35

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