# What is the hybridisation in BrF5 [duplicate]

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 $\ce{PH3}$.

## marked as duplicate by Jan, Klaus-Dieter Warzecha, Jon Custer, ron, Todd MinehardtDec 22 '16 at 23:42

• @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 0:53
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.