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In case of hybridization of PCl5 why does the electron move to 3d orbital though 4s orbital has a lower energy?

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  • $\begingroup$ quora.com/… $\endgroup$ – Kenny Lau May 27 '16 at 14:16
  • $\begingroup$ In short, 3d>4s only in group I and II. $\endgroup$ – Kenny Lau May 27 '16 at 14:16
  • $\begingroup$ Bottom of meta-synthesis.com/webbook/39_diatomics/diatomics.html $\endgroup$ – Kenny Lau May 27 '16 at 14:17
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    $\begingroup$ $\ce{PCl5}$ is poorly described by hybridization. $\endgroup$ – bon May 27 '16 at 15:47
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    $\begingroup$ It simply does not work like that - significant contribution of d orbitals in p black compounds was disproved many years ago. $\endgroup$ – Mithoron May 28 '16 at 22:07
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It doesn’t. It simply doesn’t. The $\mathrm{3d}$ orbital has a similar energy to the $\mathrm{4s}$ one — nobody considers $\mathrm{4s}$ to take part in bonding but historically, invoking $\mathrm{d}$-orbitals was very popular.

The better description is to have an $\mathrm{sp^2}$ hybridised phosphorus atom which has a free p-orbital. The $\mathrm{sp^2}$ orbitals bond with three chlorides in traditional two-electron-two-centre bonds while the remaining $\mathrm{p}$-orbital — populated by two electrons — participates in a four-electron-three-centre bond with the two remaining chlorines. You can think of this as two resonance structures:

$$\ce{Cl-{P+}Cl3\bond{...}Cl- <-> Cl- \bond{...}{P+}Cl3-Cl}$$

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