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Since the molecule has 5 bonding pairs and zero lone pairs, the shape should be trigonal bipyramidal. But one thing i am confused about is which atom(s) will occupy the axial and the planar positions, respectively, and why?

Thanks

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    $\begingroup$ Related (duplicate?): Why does F replace the axial bond in PCl5? (permeakra's comment on that question sums it up quite nicely.) So, I'd surmise the first two fluorines preferentially go axial. The third fluorine goes equatorial because there are no more axial positions left. $\endgroup$ – orthocresol Jun 26 '17 at 18:44
  • $\begingroup$ But why axial ligands have more significant negative charge on them ? Thanks $\endgroup$ – user8167818 Jun 27 '17 at 1:49
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The axial positions see 3 substituents at 90 degrees, while the equatorial positions only see 2.

To first order, it is generally favorable to place the smaller substituents in the axial positions. Unfortunately, when you have fluorine, higher order effects (like minimizing dipole moment) may overcome steric considerations, but I would still prefer to place the fluorines axially as my first guess.

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I'm not entirely sure about the background of this question, but $\ce{AsF3Cl2}$ before it decomposes above $-75 ^{\circ}$C in liquid phase (1, p. 100) has the following trigonal bipyramidal structure with two axial F atoms:

enter image description here

Crystal structure of solid compound though relates to the mixed halide of $\ce{AsCl4[AsF6]}$ formula (2) with discrete $\ce{AsCl4}$ and $\ce{AsF6}$ units:

enter image description here

(1) Chemistry of arsenic, antimony, and bismuth, 1st ed.; Norman, N. C., Ed.; Blackie Academic & Professional: London ; New York, 1998.
(2) Preiss, H. Z. anorg. allg. Chem. 1971, 380 (1), 45–50. DOI 10.1002/zaac.19713800109

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