If the lone pairs of elections are put on axial or equatorial position such as to minimize the intera-electronic repulsion, and for the same reason in $\ce{SF4}$, the election pair is kept on the equatorial position(at 90 degrees to only two domains), why then in $\ce{BrF5}$ the lone pair is kept on the axial position, when it is at 90 degrees with 4 domains?

  • $\begingroup$ If you see the structure of $BrF_5$ which is square pyramidal, and with the lone pair it is octahedral, you will find that there is no other option. All the bonds are equiangular from each other. So, it is equivalent to put the lone pair in axial or equatorial position. $\endgroup$
    – Soumik Das
    Apr 5, 2018 at 13:45

1 Answer 1


There is no axial or equatorial position in an octahedral structure. All places are equivalent and you can put the lone pair on position.

  • 1
    $\begingroup$ Not necessarily equivalent - not if there's Jahn-Teller effect. $\endgroup$
    – Mithoron
    Apr 5, 2018 at 21:50

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