Would the lone pair be in the equatorial plane or the axial plane for bromine pentafluoride? [duplicate]

I've been learning chemical bonding and I'm confused about the structure of $\ce{BrF5}$. It has a square bipyramidal geometry but a square pyramidal shape - I got that. I don't get why the lone pair gets placed at the top perpendicular to the equatorial plane instead of the equatorial plane, where it's more stable.

And why does the angle change a lot from 90 degrees to 84 degrees? Similarly for $\ce{ICl4-}$- why do the lone pairs go to the axial plane?

marked as duplicate by bon, Mithoron, pentavalentcarbon, Todd Minehardt, TyberiusMar 1 '18 at 19:10

• There are no axial and equatorial positions here. – Ivan Neretin Mar 1 '18 at 4:33
• Oh! It didn’t show that way, now raajsuriya fixed it i suppose. But wait is there a square biphramidal geometry known? Or is it equivalent to octahedral? – MollyCooL Mar 1 '18 at 6:33
• Square bipyramidal geometry means octahedral geometry. All the vertices of an octahedron are equivalent. – Apoorv Potnis Mar 1 '18 at 7:57
• @MollyCooL Yes, a square bipyramidal geometry is called an octahedral geometry because of its eight sides! – dr.drizzy Mar 1 '18 at 7:59
• Okay thank you! Just got a bit confused between the two. – MollyCooL Mar 1 '18 at 8:00

As @IvanNeretin said in the comments, "There are no axial and equatorial positions here". This is a symmetric 3-D structure. Think of this structure as two planes cutting each other perpendicularly and you can rotate it any way you want. This is well reprsented by this image.

So your line "I don't get why the lone pair gets placed at the top perpendicular to the equatorial plane instead of the equatorial plane, where it's more stable." does not make any sense. No matter where you place your lone pair it will still have no effect on the stability of the compound because everytime you think you are shifting your lone pair to a different position, it is forming the same molecule again.

Angle changes from 90 degrees to 84.8 degrees due to the lone-pair present. This creates a lone pair bond-pair repulsion and therefore slightly pushes the 4 Floruines, perpendicular to the lone pair, away from it making an 84.8 angle and not a 90.

Again in the case for $\ce{ICl4-}$ there is no axial or equatorial plane. Here's a diagram that should be helpful:

Keep in mind that (a), (b), (c), (d) can form a plane as well and according to your theory, it will become the equatorial plane. The main key point here is that the lone pairs must be furthest away from each other to be symmetric.