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This question already has an answer here:

Consider a molecule such as $\ce{ClF_3}$.
Shouldn't the electron clouds (which are more negative than the Chlorine atom) want to repel each other and so the $\ce{ClF_3}$molecule should arrange into a trigonal planar shape where the electron clouds are as far away as possible?

I think at a deeper level than VSEPR, Chlorine is either partially ionized to a hybrid of $\ce{F^- + Cl^+F_2}$ configurations or has some electrons promoted up to d shaped orbitals but I don't know how to calculate the resulting shape. I think the main thing throwing me off is that the $\ce{F^- + Cl^+F_2}$ shapes should have the $\ce{F^-}$ ion repulsed from the lone pair electrons.

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marked as duplicate by bon, M.A.R., ron, Todd Minehardt, Jon Custer Jul 17 '16 at 15:49

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  • $\begingroup$ There is a 3-center 4-electron bond between the chlorine and two fluorine atoms. This bond is linear, so the fluorines are necessarily on opposite sides of the chlorine. $\endgroup$ – f'' Jul 16 '16 at 20:23
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    $\begingroup$ A more general question is $$\mathrm{Why\ does\ VSEPR\ predict\ a\ T-shaped\ geometry\ for AX_3 E_2 \ and\ not\ trigonal\ planar?}$$ We also have to remember that VSEPR was developed to rationalize what we were learning about electrons, bonding, and quantum mechanics withe known geometries. $\endgroup$ – Ben Norris Jul 16 '16 at 21:08
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    $\begingroup$ see chemistry.stackexchange.com/questions/34610/… $\endgroup$ – ron Jul 16 '16 at 21:52