What is the hybridisation in BrF5 [duplicate]








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[Why does F replace the axial bond in PCl5?](/questions/18427/why-does-f-replace-the-axial-bond-in-pcl5)

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Closed 6 years ago.



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 $\ce{PH3}$.




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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