$\ce{SF6}$ is an octahedral shape which makes perfect sense. $\ce{SF5^+}$ arranges 5 pairs of electrons in a trigonal bipyramidal structure. But what if you add a lone pair to this? Will the electron pairs arrange into an octahedral shape, but due to extra repulsion of lone pair, or will it have some distortion?

Will it be similar to $\ce{BrF5}$ structure?


Yes, $\ce{SF5^-}$ does infact have an octahedral shape. It's structure would be

enter image description here

Yes, it is similar to the structure of $\ce{BrF5}$

  • 2
    $\begingroup$ That's not octahedral, it's derived from octahedral but the geometry is square pyramidal. I realise the question asked about the disposition of electron pairs, but it's not good to reinforce this slightly sloppy phrasing. $\endgroup$
    – orthocresol
    May 7 '17 at 15:37

More imformation fore above question

It is not Octahedral because octahedral formed in 0 lone elctrones and 6 bond electron pairs

It should be squre pyramidal according to this reason,

Valence electrones by sulfur atom-6

Electrons by 5 florin atoms-5 (1*5)

(- )charge electron-1

Therefore total electrons-6+5+1=12

Therefore repultion items-12÷2=6

Bond pairs between 5 Florine atoms and sulfur atom-5 (because the bond is SF5-)

Therefore lone electron pairs-6-5=1

So molecular shape: squre pyramidal(5,1)

therefore it is same to the molecular shape of BrF5.

  • 1
    $\begingroup$ I have deleted the other answer since they were duplicates by different user accounts. You may want to contact the team to merge your accounts. $\endgroup$ May 8 '17 at 10:52

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