-1

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

oshhh
  • 806
  • 2
  • 9
  • 24
  • https://www.quora.com/If-4s-orbitals-are-higher-in-energy-than-3d-orbitals-then-why-do-electrons-fill-up-in-4s-before-filling-up-in-3d – Kenny Lau May 27 '16 at 14:16
  • In short, 3d>4s only in group I and II. – Kenny Lau May 27 '16 at 14:16
  • Bottom of http://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html – Kenny Lau May 27 '16 at 14:17
  • In AO level diagrams, it is important to note the large differences of gaps: 
    Not: 1s < 2s < 2p < 3s < 3p < 4s <3d < 4p <5s <4d with equal distances

    But correctly, according to both spectroscopy and quantum chemistry:

    1s << 2s < 2p << 3s < 3p << 3d< 4s < 4p << 4d <5s

    Note the d < s order given here, which is the correct one for most cases, with the exception of group 1 and 2 atoms.

     – Kenny Lau May 27 '16 at 14:17
  • Because the energy of 3d is less than the energy of 4s. – Kenny Lau May 27 '16 at 14:19
  • @kenny but while filling of atomic orbitals 3d>4s... why is that so then – oshhh May 27 '16 at 14:23
  • The first website already explained it. "The 4s sublevel is only lower in energy if there are no electrons in the 3d sublevel." – Kenny Lau May 27 '16 at 14:24
  • The second website provided extra information: the energy levels of 3d and 4s are very close. – Kenny Lau May 27 '16 at 14:24
  • 2
    $\ce{PCl5}$ is poorly described by hybridization. – bon May 27 '16 at 15:47
  • 1
    It simply does not work like that - significant contribution of d orbitals in p black compounds was disproved many years ago. – Mithoron May 28 '16 at 22:07
  • related http://chemistry.stackexchange.com/questions/29142/hypervalency-and-the-octet-rule http://chemistry.stackexchange.com/questions/18427/why-does-f-replace-the-axial-bond-in-pcl5 – Mithoron May 28 '16 at 22:10

1 Answers1

2

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:

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

Jan
  • 67,989
  • 12
  • 201
  • 386