# 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?

• 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

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