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Recently I had given a test where I was asked to specify the hybridisation of $\ce{OSF4}$ and its structure. I guessed that the structure would be trigonal bipyramidal, but I was unsure since I hadn't dealt with molecules with double bonds.

So my first question is: Should having double bonds in the structure affect my geometry considerations of the structures or will it be the same as in structures only containing single bonds (say $\ce{PCl5}$) ?

Also, if I am supposed to draw its structure then where would I place the double bonded oxygen? Axial or equatorial position? Why? I know that when lone pairs are considered they are placed at the equatorial position (after considering repulsions). Is there anything common in this case?

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    $\begingroup$ The Trigonal Bipyramidal is right. There would be a linear F-S-F through the apex of the bipyramid. In the plane I'd expect the O-S-F bond to be larger than the F-S-F bond. See: en.wikipedia.org/wiki/Thionyl_fluoride $\endgroup$ – MaxW Nov 10 '15 at 7:24
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    $\begingroup$ @Jan. Your improvements on my edits noted. Will take care. $\endgroup$ – Shailesh Nov 10 '15 at 13:34
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So my first question is: Should having double bonds in the structure affect my geometry considerations of the structures or will it be the same as in structures only containing single bonds (say ) ?

Also, if I am supposed to draw its structure then where would I place the double bonded oxygen? Axial or equatorial position? Why? I know that when lone pairs are considered they are placed at the equatorial position (after considering repulsions). Is there anything common in this case?

The answer to your question is NO. A double bond shouldn't make any difference. It would be a Trigonal Bipyramidal where the 2 F's will occupy the axial positions according to Bent's Rule and the remaining two F's along with the Oxygen atom will occupy the equatorial position.

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