0

What is the hybridisation of $\ce{BrF5}$ ? I find different sources giving different answers.

When I approach this problem , I don't find any exceptional case like $\ce{SH6}$ (in which hybridisation doesn't take place.) So it follows normal rules, but I find the opposite in many different books ?

I think it square pyramidal $\ce{sp^3d}$. I also face the same problem for other compounds like $\ce{H2S}$ and $\ce{PH3}$.

Melanie Shebel
  • 6,704
  • 10
  • 45
  • 86
InquisitiveMind
  • 98
  • 1
  • 2
  • 12
  • I am wondering is this kind of question morally justified? Since the d orbital contribution is already deputable in hypervalent molecule such as, SF6 – Rodriguez Dec 19 '16 at 14:10
  • What dou you mean by "Morally justified" ? – InquisitiveMind Dec 19 '16 at 14:13
  • I mean, this question looks like an innocent homework assignment. But it is a complicated problem at the frontier of computational chemistry. – Rodriguez Dec 19 '16 at 14:14
  • No , not a homework problem . I myself was practicing by writing some compounds , my intuition says the above one is square pyramidal . When I searched a library book a different story was written there . – InquisitiveMind Dec 19 '16 at 14:18
  • 1
    There is, in-general, no one-to-one mapping between molecular geometry and hybridization. – Rodriguez Dec 19 '16 at 14:21
  • 1
    Just a side quotation, http://depa.fquim.unam.mx/amyd/archivero/VBHIBRIDOSd_26401.pdf, "Figure 4. From a VB standpoint, the bonding in SF6 can be described using two, 2-electron bonds from sulfur sp hybrids pointing 180 away from each other and two, 2-electron bonds from sulfur p orbitals with the remaining four electrons located on two fluorine atoms" – Rodriguez Dec 19 '16 at 14:43
  • Thats a good source , but what about BrF5 ? – InquisitiveMind Dec 19 '16 at 14:46
  • To be honest, I never run an ab initio valence bond calculation as described in that reference. I suggest someone interested in this topic, run an ab initio valence bond calculation. – Rodriguez Dec 19 '16 at 14:47
  • 2
    @InquisitiveMind $\ce{BrF5}$ should be composed of a standard 2e2c $\ce{Br-F}$ bond and two elongated 4e3c $\ce{F\bond{...}Br\bond{...}F}$ bonds (four-electron-three-centre bonds), giving it a square pyramidal structure with the central atom at the base of the pyramid. However, it probably fluctuates quickly between that and and the pentagonal bipyramid. In any case, remove d orbitals from your argumentation; they do not take part. – Jan Dec 20 '16 at 00:53
  • @Jan Hey my query on the hybridization of H2S and PH3 – InquisitiveMind Dec 23 '16 at 14:55

1 Answers1

0

Hybridisation of $\ce{BrF5}$ is $\ce{sp^3d^2}$ (involving one 4s, three 4p and two 4d orbitals) giving rise to octahedral geometry. But one hybrid orbital is occupied by lone pairs. So the effective shape of molecule is square pyramidal.

aventurin
  • 7,200
  • 3
  • 27
  • 38
Srishti
  • 204
  • 1
  • 4