molecular and electron geometry for PF5 SF6 SF4
XeF2 BrF5 ClF3 XeF4 ?

Question

molecular and electron geometry for PF5 SF6 SF4
XeF2 BrF5 ClF3 XeF4 ?

abb0t · Answer

$PF_5$ is one of those hypervalent atoms (basically it violates the octet rule by allowing more atoms). Since you've used up the 4 orbitals, $P_x$, and $P_y$, you're only left with $P_z$. Try to visualize this on a 3-dimensional plane in which you have x, y, and z. 
x points left to right, y points up and down, and z points right at you and away from you.
This makes $PF_5$ a trigonal pyrdamidal species. Does that make sense? It's about the same concept with the rest of the molecules you have there.

abb0t · Answer

but in essence, the middle open circle is P and the closed circles are the fluorine

anonymous · Answer

those are sp3d hybridization examples

abb0t · Answer

let me rephrase that. i think it's trigonal bipyramidal?

abb0t · Answer

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