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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 $\ce{PH3}$.

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  • $\begingroup$ I am wondering is this kind of question morally justified? Since the d orbital contribution is already deputable in hypervalent molecule such as, SF6 $\endgroup$ – Rodriguez Dec 19 '16 at 14:10
  • $\begingroup$ What dou you mean by "Morally justified" ? $\endgroup$ – InquisitiveMind Dec 19 '16 at 14:13
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    $\begingroup$ There is, in-general, no one-to-one mapping between molecular geometry and hybridization. $\endgroup$ – Rodriguez Dec 19 '16 at 14:21
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    $\begingroup$ Just a side quotation, 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" $\endgroup$ – Rodriguez Dec 19 '16 at 14:43
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    $\begingroup$ @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. $\endgroup$ – Jan Dec 20 '16 at 0:53
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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.

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